Category: Class 11

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TN State Board 11th Physics Model Question Paper 4 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – 1

Answer all the questions: [15 × 1 = 15]

Question 1.
In some region, the gravitational field is zero. The gravitational potential in this region is …………………….
(a) A variable
(b) A constant
(c) Zero
(d) Can’t be zero
Answer:
(b) A constant

Question 2.
The angle between two vectors 2\(\hat { i } \) + 3\(\hat { j } \) + \(\hat { k } \) and -3\(\hat { j } \) + 6k is …………………..
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Hint:
The angle between the two vector is 90°.
[cos θ = \(\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\mathrm{AB}}\)]
Answer:
(d) 90°

Question 3.
A stretched rubber has
(a) Increased kinetic energy
(b) Increased potential energy
(c) Decreased kinetic energy
(d) The axis of rotation
Answer:
(b) Increased potential energy

Question 4.
The direction of the angular velocity vector is along ……………………..
(a) The tangent to the circular path
(b) The inward radius
(c) The outward radius
(d) The axis of rotation
Answer:
(d) The axis of rotation

Question 5.
For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is …………………….
(a) \(\frac{1}{2}\)
(b) \(\frac { 1 }{ \sqrt { 2 } } \)
(c) 2
(d) \(\sqrt{2}\)
Hint:

Answer:
(a) \(\frac{1}{2}\)

Question 6.
If the linear momentum of the object is increased by 0.1 %, then the kinetic energy is increased by ………………………..
(a) 0.1%
(b) 0.2%
(c) 0.4%
(d) 0.01%
Hint:
The relation b/w linear momentum and kinetic energy
P = \(\sqrt{2 \mathrm{mE}_{\mathrm{K}}} \Rightarrow \mathrm{E}_{\mathrm{K}}=\frac{\mathrm{P}^{2}}{2 \mathrm{m}}\)
If L is increased by 0.1% 1′ = P + \(\frac{0.1}{100}\)P

Answer:
(c) 0.4%

Question 7.
A block of mass 4 kg is suspended through two light spring balances A and B. Then A and B will read respectively ………………………
(a) 4 kg and 0 kg
(b) 0 kg and 4 kg
(c) 4 kg and 4 kg
(d) 2kg and 2 kg
Hint:
Tension is uniformly transmitted if the springs are massless.
Answer:
(c) 4 kg and 4 kg

Question 8.
At the same temperature, the mean kinetic energies of molecules of hydrogen and oxygen are in the ratio of ………………………
(a) 1 : 1
(b) 1 : 16
(c) 8 : 1
(d) 16 : 1
Hint:
Average kinetic energy of a molecule is proportional to the absolute temperature.
Answer:
(a) 1 : 1

Question 9.
A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. Which of the following plots shows the correct variation of speed v with height h from the lower end?
(a)
(b)
(c)
(d)
Answer:

(d)

Question 10.
In a simple hormonic oscillation, the acceleration against displacement for one complete oscillation will be ………………………
(a) An ellipse
(b) A circle
(c) A parabola
(d) A straight line
Answer:
(d) A straight line

Question 11.
A shell is fired from a canon with velocity v m/s at an angle 0 with the horizontal direction. At the highest point in the path it explodes into two pieces of equal mass. One of the pieces retraces it path of the cannon and the speed in m/s of the other piece immediately after the. explosion is ………………………..
(a) 3v cos θ
(b) 2v cos θ
(c) \(\frac{3}{2}\)v cos θ
(d) \(\frac { \sqrt { 3 } }{ 2 } \) cos θ
Hint:
Velocity at the highest point = horizontal component of velocity = V cos θ
Momentum of shell before explosion = mv cos θ
Momentum of two pieces after explosion = \(\frac{m}{2}\) (- V cos θ) + \(\frac{m}{2}\) v
Law of conservation of momentum mv cos θ = –\(\frac{mv}{2}\)cos θ + \(\frac{m}{2}\) v
∴ V = 3V cos θ
Answer:
(a) 3v cos θ

Question 12.
In which process, the p – v indicator diagram is a straight line parallel to volume axis?
(a) Isothermal
(b) Adiabatic
(c) Isobaric
(d) Irreversible
Answer:
(c) Isobaric

Question 13.
For a liquid to rise in capillary tube, the angle of contact should be ………………………….
(a) Acute
(b) Obtuse
(c) Right
(d) None of these
Answer:
(a) Acute

Question 14.
The increase in internal energy of a system is equal to the work done on the system which process does the system undergo?
(a) Isochoric
(b) Adiabatic
(c) Isobaric
(d) Isothermal
Answer:
(b) Adiabatic

Question 15.
The change in frequency due to Doppler effect does not depend on …………………………
(a) The speed of the source
(b) The speed of the observer
(c) The frequency of the source
(d) Separation between the source and the observer
Answer:
(d) Separation between the source and the observer

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
What are the advantages of SI system?
Answer:

1. This system makes use of only one unit for one physical quantity, which means a rational system of units
2. In this system, all the derived units can be easily obtained from basic and supplementary units, which means it is a coherent system of units.
3. It is a metric system which means that multiples and submultiples can be expressed as powers of 10.

Question 17.
Using the principle of homogunity of dimensions, check dimensionally the given equations are correct,
(a) \(\mathbf{T}^{2}=\frac{4 \pi^{2} r^{3}}{\mathbf{G}}\)
(b) \(T^{2}=\frac{4 \pi^{2} r^{3}}{G M}\)?
Answer:
Here G-gravitational constant
r – radius of orbit M – mass
Dimensional formula for T = T
Dimensional formula for r = L
Dimensional formula for G = M^-1L³T^-2
Dimensional formula for M = M
(a) T² M° L° = [L³] [M L^-3T^-2]
T² M° L° = L° MT² – Not correct

(b) T² = [L³][ML^-3T²][M^-1]
T² = T² – Dimensionally correct

Question 18.
Find out the workdone required to extract water from the well of depth 20 m. Weight of water and backet is 2.8 kg wt?
Answer:
Workdone, W = mgh
W = (Weight) × depth
= 2.8 × 20
W = 56 J

Question 19.
Is a single isolated force possible in nature?
Answer:
No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

Question 20.
State the factors on which the moment of inertia of a body depends?
Answer:

1. Mass of body
2. Size and shape of body
3. Mass distribution w.r.t. axis of rotation
4. Position and orientation of rotational axis

Question 21.
If a drop of water falls on a very hot iron, it takes long time to evaporate. Explain why?
Answer:
When a drop of water falls on a very hot iron it gets insulated from the hot iron due to a thin layer of water vapour which is a bad conductor of heat. It takes quite long to evaporate as heat is conducted from hot iron to the drop through the insulating layer of water vapour very slowly.

Question 22.
What is meant by gravitational field. Give its unit?
Answer:
The gravitational field intensity \(\vec { E } \)₁, at a point is defined as the gravitational force experienced by unit mass at that point. It’s unit N kg^-1.

Question 23.
Why dimensional methods are applicable only up to three quantities?
Answer:
Understanding dimensions is of utmost importance as it helps us in studying the nature of physical quantities mathematically. The basic concept of dimensions is that we can add or subtract only those quantities which have same dimensions. Also, two physical quantities are equal if they have same dimensions, these basic ideas help us in deriving the new relation between physical quantities, it is just like units.

Question 24.
Why does a porter bend forward while carrying a sack of rice on his back?
Answer:
When a porter carries a sack of rice, the line of action of his centre of gravity will go away from the body. It affects the balance, to avoid this he bends. By which centre of gravity will realign within the body again. So balance is maintained.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Explain variation of ‘g’ with latitude?
Answer:
When an object is on the surface for the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.
OP_z, cos λ = \(\frac{PZ}{OP}\) = \(\frac{R’}{R}\)
R’ = R cos λ

where λ is is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
a_PQ = ω²R cos λ = ω²R cos² λ
Since R’ = R cos λ
Therefore, g’ = g – ω’²R cos²λ
From the above expression, we can infer that at equator, λ = 0, g’ = g – ω²R. The acceleration due to gravity is minimum. At poles λ = 90; g1 = g. it is maximum. At the equator, g’ is minimum.

Question 26.
Cap you associate a vector with
(a) the length of a wire bent into a loop
(b) a plane area
(c) a sphere.
Answer:
(a) We cannot associate a vector with the length of a wire bent into a loop, this is cause the length of the loop does not have a definite direction.

(b) We can associate a vector with a plane area. Such a vector is called area vector and its direction is represented by a normal drawn outward to the area.

(c) The area of a sphere does not point in any difinite direction. However, we can associate a null vector with the area of the sphere. We cannot associate a vector with the volume of a sphere.

Question 27.
What is mean by inertia? Explain its types with example?
Answer:
The inability of objects to move on its own or change its state of motion is called inertia.
Inertia means resistance to change its state. There are three types of inertia:

1. Inertia of rest: The inability of an object to change its state of rest is called inertia of rest.
Example:

• When a stationary bus starts to move, the passengers experience a sudden backward push.
• A book lying on the table will remain at rest until it is moved by some external agencies.

2. Inertia of motion: The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:

• When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
• An athlete running in a race will continue to run even after reaching the finishing point.

3. Inertia of direction: The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:

• When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
• When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

Question 28.
A light body and body with greater mass both are having equal kinetic energy. Among these two which one will have greater linear momentum?
Answer:
Given Data:
E₁ – E₂

then P₂ > P₁
i.e; a heavier body has greater linear momemtum.

Question 29.
Three particles of masses m₁ = 1 kg, m₂ = 2 kg and m₃ = 3 kg are placed at the corners of an equilateral triangle of side lm as shown in Figure. Find the position of center of mass?
Answer:
The center of mass of an equilateral triangle lies at its geometrical center G.
The positions of the mass m₁, m₂ and m₃ are at positions A, B and C as shown in the Figure. From the given position of the masses, the coordinates of the masses m₁ and m₂ are easily marked as (0,0) and (1,0) respectively.

To find the position of m3 the Pythagoras theorem is applied. As the ∆DBC is a right angle triangle,
BC² = CD² + DB²
CD² = BC² – DB²
CD² = 1² – (\(\frac{1}{2}\))² = 1 – (\(\frac{1}{4}\)) = \(\frac{3}{4}\)
CD = \(\frac { \sqrt { 3 } }{ 2 } \)

The position of mass m₃ is or (0.5, 0.5\(\sqrt{3}\))
X coordìnate of center of mass,
y_CM = \(\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}}\)
y_CM = \(\frac { \sqrt { 3 } }{ 4 } \) m
∴ The coordinates of center of mass G (x_CM, y_CM) is (\(\frac{7}{12}\), \(\frac { \sqrt { 3 } }{ 4 } \))

Question 30.
Define precision and accuracy. Explain with one example?
Answer:
The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision of a measurement is a closeness of two or more measured values to each other.

The true value of a certain length is near 5.678 cm. In one experiment, using a measuring instrument of resolution 0.1 cm, the measured value is found to be 5.5 cm. In another experiment using a measuring instrument of greater resolution, say 0.01 cm, the length is found to be 5.38 cm. We find that the first measurement is more accurate as it is closer to the true value, but it has lesser precision. On the contrary, the second measurement is less accurate, but it is more precise.

Question 31.
Derive an expression for total acceleration in the non uniform circular motion?
Answer:
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time. Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration as shown in the figure.

The resultant acceleration is obtained by vector sum of centripetal and tangential acceleration. Since centripetal acceleration is \(\frac { v^{ 2 } }{ r } \), the magnitude of this resultant acceleration is given by
a_R = \(\sqrt{a_{t}^{2}+\left(\frac{v^{2}}{r}\right)^{2}}\)

This resultant acceleration makes an angle θ with the radius vector as shown in figure.
This angle is given by tan θ = \(\frac{a_{t}}{\left(v^{2} / r\right)}\)

Question 32.
Calculate the value of adiabatic exponent for monoatomic molecule?
Answer:
Monoatomic molecule:
Average kinetic energy of a molecule = [\(\frac{3}{2}\)kT]
Total energy of a mole of gas = \(\frac{3}{2}\) kT × N_A = \(\frac{3}{2}\)RT
For one mole, the molar specific heat at constant volume

Question 33.
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s surface, (mass of oxygen molecule : 2.76 × 10^-26kg Boltzmann’s constant (k_B) = 1.38 × 10^-23 J mol^-1 k^-1
Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
State kegler’s laws of planetary motion?
Answer:
1. Law of Orbits:
Each planet moves around the Sun in an elliptical orbit with the Sun at one of the foci.

2. Law of area:
The radial vector (line joining the Sun to a planet) sweeps equal areas in equal intervals of time.

3. Law of period:
The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the semi-major axis of the ellipse.
T² ∝ a³
\(\frac { T^{ 2 } }{ a^{ 3 } } \) = Constant

(b) The distance of planet Jupiter from the sun is 5.2 times that of the earth. Find the period of resolution of Jupiter around the sun?
Answer:
Here r₁ = 5.2 r_e; T_J = ?; T_e = 1 year

= 11.86 years

[OR]

(c) Explain the propagation of errors in multiplication?
Answer:
Error in the product of two quantities:
Let ∆A and ∆B be the absolute errors in the two quantities A, and B, respectively. Consider the product Z = AB,
The error AZ in Z is given by Z ± AZ = (A ± ∆A) (B ± ∆B)
= (AB) ± (A ∆ B) ± (B ∆ A) ± (∆A • ∆B)
Dividing L.H.S by Z and R.H.S by AB, we get,
1 ± \(\frac{∆Z}{Z}\) = 1 ± \(\frac{∆B}{B}\) ± \(\frac{∆A}{A}\) ± \(\frac{∆A}{A}\). \(\frac{∆B}{B}\)
As ∆A/A, ∆B/B are both small quantities, their product term \(\frac{∆A}{A}\).\(\frac{∆B}{B}\) can be neglected.
The maximum fractional error in Z is
\(\frac{∆Z}{Z}\) = ± (\(\frac{∆A}{A}\) + \(\frac{∆B}{B}\))

(d) The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42s, 271s and 2.80s respectively. Calculate the average absolute error?
Answer:
Mean absolute error = \(\frac{\Sigma\left|\Delta \mathrm{T}_{i}\right|}{n}\)
∆T_m = \(\frac{0.01+0.06+0.20+0.09+0.18}{5}\)
∆T_in = \(\frac{0.54}{5}\) = 0.108s = 0.1 1s (Rounded of 2^nd decimal place).

Question 35 (a).
Explain in detail the triangle law of vector addition?
Answer:
Let us consider two vectors \(\vec { A } \) and \(\vec { B } \) as shown in figure.

To find the resultant of the two vectors we apply the triangular law of addition as follows:

Represent the vectors \(\vec { A } \) and \(\vec { B } \) by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector \(\vec { A } \) is connected to the tail of the second vector \(\vec { B } \). Let θ be the angle between \(\vec { A } \) and \(\vec { B } \). Then \(\vec { R } \) is the resultant vector connecting the tail of the first vector A to the head of the second vector B.

The magnitude of R (resultant) is given geometrically by the length of \(\vec { R } \) (OQ) and the direction of the resultant vector is the angle between \(\vec { R } \) and \(\vec { A } \). Thus we write \(\vec { R } \) = \(\vec { A } \) + \(\vec { B } \).

1. Magnitude of resultant vector:
The magnitude and angle of the resultant vector are determined by using triangle law of vectors as follows. From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\vec { A } \) and \(\vec { B } \).
cos θ = \(\frac{AN}{B}\) ∴AN = B cos θ ans sin θ = \(\frac{BN}{B}\) ∴ BN = B sin θ
For ∆OBN, we have OB² = ON² + BN²

2. Direction of resultant vectors: If θ is the angle between \(\vec { A } \) and \(\vec { B } \), then

(b) State and prove Bernoulli’s theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid?
Answer:
Bernoulli’s theorem:
According to Bernoulli’s theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant. Mathematically,
\(\frac { P }{ \rho } \) + \(\frac{1}{2}\)v² + gh = Constant
This is known as Bernoulli’s equation.

Proof: Let us consider a flow of liquid through a pipe AB. Let V be the volume of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time. Let a_A, v_A and P_A be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.

Let the force exerted by the liquid at A is
F_A = P_Aa_A

Distance travelled by the liquid in time t is d = v_at

Therefore, the work done is W = F_Ad = P_Aa_Av_At
But a_Av_At = a_Ad = V, volume of the liquid entering at A.

Thus, the work done is the pressure energy (at A), W = P_Ad = P_AV
Pressure energy per unit volume at
A = \($\frac{\text { Pressure energy }}{\text { Volume }}$\) = \(\frac{P_{A} V}{V}\) = P_A

Pressure energy per unit mass
A = \($\frac{\text { Pressure energy }}{\text { Mass }}$\) = \(\frac{P_{A} V}{m}\) = \(\frac{P_{A}}{\frac{m}{V}}=\frac{P_{A}}{\rho}\)

Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
EP_A = P_AV = P_AV × (\(\frac{m}{m}\)) = m\(\frac { P_{ A } }{ \rho } \)

Potential energy of the liquid at A,
PE_A = mg h_A

Due to the flow of liquid, the kinetic energy of the liquid at A,
KE_A = \(\frac{1}{2} m v_{\mathrm{A}^{2}}\)

Therefore, the total energy due to the flow of liquid at A,
E_A = EP_A + KE_A + PE_A
E_A = m \(m \frac{P_{A}}{\rho}+\frac{1}{2} m v_{A}^{2}+m g h_{A}\)

Similarly, let a_B, v_B and P_B be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at B. Calculating the total energy at F_B, we get
E_B = m \(m \frac{P_{B}}{\rho}+\frac{1}{2} m v_{B}^{2}+m g h_{B}\)
From the law of conservation of energy,
E_A = E_B

Thus, the above equation can be written as
\(\frac{P}{\rho g}+\frac{1}{2} \frac{v^{2}}{g}\) + h = Constant

The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction.

This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli’s relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then
h = 0 ⇒ \(\frac{P}{\rho g}+\frac{1}{2} \frac{v^{2}}{g}\) = Constant

Question 36 (a).
What are the limitations of dimensional analysis?
Answer:
Limitations of Dimensional analysis

1. This method gives no information about the dimensionless constants in the formula like 1, 2, …………….. π, e, etc.
2. This method cannot decide whether the given quantity is a vector or a scalar.
3. This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.
4. It cannot be applied to an equation involving more than three physical quantities.
5. It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation.

For example, using dimensional analysis, s = ut + \(\frac{1}{3}\)at² is dimensionally correct whereas the correct relation is s = ut + \(\frac{1}{2}\)at²

(b) The escape velocity v of a body depends on

1. The acceleration due to gravity ‘g’ of the planet
2. The radius R of the planet. Establish dimensionally the relation for the escape velocity?

Answer:
\(v \propto g^{a} \mathrm{R}^{b} \Rightarrow v=k g^{a} \mathrm{R}^{b}\), K → dimensionally proportionality constant.
[v] = [g]^a [R]^b
[M⁰L¹T^-1] = [M⁰L¹T^-2]^a [M⁰L¹T¹⁰]^b
equating powers
1 = a + b
-1 = -2a ⇒ a = \(\frac{1}{2}\)
b = 1 – a = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∴ v = k\(\sqrt{gR}\)

[OR]

(c) Discuss the law of transverse vibrations In stretched string?
Answer:
Laws of transverse vibrations in stretched strings:
There are three laws of transverse vibrations of stretched strings which are given as follows:

(I) The law of length:
For a given wire with tension T (which is fixed) and mass per unit length p (fixed) the frequency varies inversely with the vibrating length. Therefore,
f ∝\(\frac{1}{l}\) ⇒ f = \(\frac{C}{2}\)
⇒1 × f = C, where C is constant

(II) The law of tension:
For a given vibrating length I (fixed) and mass per unit length p (fixed) the frequency varies directly with the square root of the tension T,
f ∝\(\sqrt{T}\)
⇒f = A\(\sqrt{T}\), where A is constant

(III) The law of mass:
For a given vibrating length l (fixed) and tension T (fixed) the frequency varies inverely with the square root of the mass per unit length µ,
f = \(\frac{1}{\sqrt{\mu}}\)
⇒f = \(\frac{B}{\sqrt{\mu}}\), where B is constant

(d) Explain how to determine the frequency of tuning for k using sonometer?
Answer:
Working: A transverse stationary or standing wave is produced and hence, at the knife edges P and Q, nodes are formed. In between the knife edges, anti-nodes are formed. If the length of the vibrating element is l then
l = \(\frac{\lambda}{2}\) ⇒ λ = 2l
Let f be the frequency of the vibrating element, T the tension of in the string and µ the mass per unit length of the string. Then using equation ,we get
f = \(\frac{v}{\lambda}=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}\) in Hertz …………………. (1)

Let ρ be the density of the material of the string and d be the diameter of the string. Then the mass per unit length µ,

Question 37.
(a) To move an object, which one is easier, push or pull? Explain?
Answer:
When a body is pushed at an arbitrary angle θ (o to \(\frac{π}{2}\)), the applied force F can be resolved into two components as F sin e parallel to the surface and F cos θ perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is
no acceleration along the vertical direction the normal force N is equal to
N_push = mg + F cos θ ………………………. (1)
As a result the maximal static friction also increases and is equal to
\(f_{s}^{\max }=\mu_{r} N_{p u s h}=\mu_{s}(m g+F \cos \theta)\) ………………… (2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.
When an object is pulled at an angle θ, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is
N_pull = mg – F cos θ ……………….. (3)

Equation (3) shows that the normal force needs is less than N_push. From equations (1) and (3), it is easier to pull an object than to push to make it move.

[OR]

(b) Derive an expression for the velocities of two objects colliding elastically in one dimension?
Answer:
Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.

In order to have collision, we assume that the mass m₁ moves, faster than mass m₂ i.e., u₁ > u₂. For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same.

From the law of conservation of linear momentum,
Total momentum before collision (ρ_i) = Total momentum after collision (ρ_f)
m₁u₁ + m₂u₂ = m₁v₁ + m₂ v₁ …………………. (1)
or m₁ (u₁ – v₁) = m₂(v₂ – u₂) …………………… (2)
Furthur,

For elastic collision,
Total kinetic energy before collision KE_i = Total kinetic energy after collision KE_f
\(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}\) ………………….. (3)
\(m_{1}\left(u_{1}^{2}-v_{1}^{2}\right)=m_{2}\left(v_{2}^{2}-u_{2}^{2}\right)\)
After simplifying and rearranging the terms,
Using the formula a² – b² = (a + b) (a – b), we can rewrite the above equation as
\(m_{1}\left(u_{1}+v_{1}\right)\left(u_{1}-v_{1}\right)=m_{2}\left(v_{2}+u_{2}\right)\left(v_{2}-u_{2}\right)\) ……………….. (4)
Dividing the equation (4) by (2) we get

Equation (5) can be written as
u₁ – u₂ = -(v₁ – v₂)
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for v₁ and v₂,
v₁ = v₂ + u₂ – u₁ …………………….. (6)
or v₂ = u₁ + v₁ – u₁ …………………….. (7)
To find the final velocities V₁ and v₂:
Substituting equation (5) in equation (2) gives the velocity of m₁ as
m₁ (u₁ – v₁) = m₂ (u₁ + v₁ – u₂ – u₂)
m (u₁ – v₁) = m₂ (u₁ + v₁ – 2u₂)
m₁u₁ – m₁v₁ = m₂u₁ + m₂v₁ + 2m₂u₂
m₁u₁ – m₂u₁ + 2m₂u₂ = m₁v₁ + m₂v₁
(m₁ – m₂)u₁ + 2m₂u₂ = (m₁ + m₂) v₁
or v₁ = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\) u₁  + \(\left(\frac{2 m_{2}}{m_{1}+m_{2}}\right)\) u₂
Similarly, by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final m₂ as
v₂ = \(\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right)\) u₁ + \(\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right)\) u₂ …………………. (9)

Question 38 (a).
Prove that at points near the surface of the Earth the gravitational potential energy of the object is v = mgh?
Answer:
When an object of mass m is raised to a height h, the potential energy stored in the object is mgh. This can be derived using the general expression for gravitational potential energy. Consider the Earth and mass system, with r, the distance between the mass m and the Earth’s centre. Then the gravitational potential energy.
U = –\(\frac{\mathrm{GM}_{e} m}{r}\) …………………… (1)

Here r = R_e + h, where Re is the radius of the Earth, h is the height above the Earth’s surface

By using Binomial expansion and neglecting the higher order

Replace this value and we get,
U = \(-\frac{\mathrm{GM}_{e} m}{\mathrm{R}_{e}}\left(1-\frac{h}{\mathrm{R}_{e}}\right)\) ………………… (4)

We know that, for a mass m on the Earth’s surface,
\(\mathrm{G} \frac{\mathrm{M}_{e} m}{\mathrm{R}_{e}}=m g \mathrm{R}_{e}\) …………………… (5)

Substituting equation (4) in (5) we get,
U = -mgR_e + mgh ………………….. (6)

It is clear that the first term in the above expression is independent of the height h. For example, if the object is taken from height h₁ to h₂, then the potential energy at h₁ is
U(h₁) = -mgR_e + mgh₁ …………………… (7)
and the potential energy at h₂ is
U(h₂) = -mgR_e + mgh₂ …………………… (8)

The potential energy differenqe between h₁ and h₂ is
U(h₂) – U(h₁) = mg(h₁-h₂) …………………. (9)

The term mgR_e in equation (7) and (8) plays no role in the result. Hence in the equation (6) the first term can be omitted or taken to zero. Thus it can be stated that the gravitational potential energy stored in the particle of mass m at a height h from the surface of the Earth is U = mgh. On the surface of the Earth, U = 0, since h is zero.

[OR]

(b) Derive an expression for Radius of gyration?
Answer:
For bulk objects of regular shape with uniform mass distribution, the expression for moment of inertia about an axis involves their total mass and geometrical features like radius, length, breadth, which take care of the shape and the size of the objects.

But, we need an expression for the moment of inertia which could take care of not only the mass, shape and size of objects, but also its orientation to the axis of rotation. Such an expression should be general so that it is applicable even for objects of irregular shape and non-uniform distribution of mass. The general expression for moment of inertia is given as,
I = MK²

where, M is the total mass of the object and K is called the radius of gyration. The radius of gyration of an object is the perpendicular distance from the axis of rotation to an equivalent point mass, which would have the same mass as well as the same moment of inertia of the object.

As the radius of gyration is distance, its unit is m. Its dimension is L. A rotating rigid body with respect to any axis, is considered to be made up of point masses m₁, m₂, m₃, . . . mn at perpendicular distances (or positions) r₁, r₂, r₃ . . . r_n respectively as shown in figure. The moment of inertia of that object can be written as,

I = \(\sum m_{1} r_{1}^{2}=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+m_{3} r_{3}^{2}+\ldots .+m_{n} r_{n}^{2}\)
If we take all the n number of individual masses to be equal
m = m₁ = m, = m₂ = m₃ = ……………… = m_n
then

I = MK²

where, nm is the total mass M of the body and K is the radius of gyration.

The expression for radius of gyration indicates that it is the root mean square (rms) distance of the particles of the body from the axis of rotation. In fact, the moment of inertia of any object could be expressed in the form, I = MK²

For example, let us take the moment of inertia of a uniform rod of mass M and length l. Its moment of inertia with respect to a perpendicular axis passing through the center of mass is,
I = \(\frac{1}{12}\)Ml²
In terms of radius of gyration, I = MK²
Hence, MK² = \(\frac{1}{12}\)Ml²
K² = \(\frac{1}{12}\)l²
K = \(\frac{1}{12}\) or K = \(\frac{1}{2 \sqrt{3}} l\) or K = (0.289)l

Students can Download Tamil Nadu 11th Physics Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
A new unit of length is chosen such that the speed of light in vacuum is unity. The distance between the Sun and the Earth in terms of the new unit, if light takes 8 minute and 20 sec to cover the distance is …………………..
(a) 100 new unit
(b) 300 new unit
(c) 500 new unit
(d) 700 new unit
Hint:
Speed is unity = 1 unit/sec
Time = 8 min and 20 sec = 500 sec
Distance b/w sun and earth = Speed × Time
= 1 × 500 = 500 unit
Answer:
(c) 500 new unit

Question 2.
For a satellite moving in an orbit around the Earth, the ratio of kinetic energy to potential energy is ………………….
(a) 2
(b) 1 : 2
(c) 1 : \(\sqrt{2}\)
(d) \(\sqrt{2}\)
Hint:
\(\frac { GMm }{ R^{ 2 } } \) = mω²R
K.E = \(\frac{1}{2}\)Iω² = \(\frac{1}{2}\)mR²ω² = \(\frac{GMm}{2R}\)
P.E = – \(\frac{GMm}{R}\) ⇒ So \(\frac{K.E}{|P.E|}\) = \(\frac{1}{2}\)
Answer:
(b) 1 : 2

Question 3.
In the equilibrium position a body has ……………….
(a) Maximum potential energy
(b) Minimum potential energy
(c) Minimum kinetic energy
(d) Neither maximum nor minimum potential energy
Answer:
(c) Minimum kinetic energy

Question 4.
The centrifugal force appears to exist ………………….
(a) Only in inertial frames
(b) Only in rotating frames
(c) In any accelerated frame
(d) Both in inertial and non-inertial frames
Answer:
(b) Only in rotating frames

Question 5.
A particle is moving with a constant velocity along a line parallel to positive x-axis. The magnitude of its angular momentum with respect to the origin is …………………..
(a) Zero
(b) Increasing with x
(c) Decreasing with x
(d) Remaining constant
Answer:
(d) Remaining constant

Question 6.
When 8 droplets of water of radius 0.5 mm combine to form a single droplet. The radius of it is …………………
(a) 4 mm
(b) 2 mm
(c) 1 mm
(d) 8 mm
Hint:
Volume of 8 droplets of water = 8 × \(\frac{4}{3}\) π(0.5)³
When each droplet combine to form one volume remains conserved
R³ = 8 × (0.5)³
R³ = (8 × (0.5)³)
R³ = 2 × 0.5 = 1 mm
Answer:
(c) 1 mm

Question 7.
Pressure head in Bernoulli’s equation is …………………
(a) \(\frac { P_{ \rho } }{ g } \)
(b) \(\frac { P }{ \rho g } \)
(c) ρg
(d) Pρg
Answer:
(b) \(\frac { P }{ \rho g } \)

Question 8.
The angle between particle velocity and wave velocity in a transverse wave is ……………………
(a) Zero
(b) π/4
(c) π/2
(d) π
Answer:
(c) π/2

Question 9.
If the masses of the Earth and Sun suddenly double, the gravitational force between them will …………………….
(a) Remains the same
(b) Increase two times
(c) Increase four times
(d) Decrease two times
Answer:
(c) Increase four times

Question 10.
A mobile phone tower transmits a wave signal of frequency 900 MHz, the length of the transmitted from the mobile phone tower ……………………
(a) 0.33 m
(b) 300 m
(c) 2700 × 10⁸m
(d) 1200 m
Hint:
f = 900 MHz = 900 × 10⁶ Hz
Speed of wave (c) = 3 × 10⁶ ms^-1
λ = \(\frac{v}{f}\) = \(\frac { 3\times 10^{ 8 } }{ 900\times 10^{ 6 } } \) = \(\frac{1}{3}\) = 0.33m
Answer:
(a) 0.33 m

Question 11.
The displacement y of a wave travelling in the x direction is given by
y = (2 × 10^-3) sin (300 t – 2x + \(\frac { \pi }{ 4 } \)), where x and y are measured in metres and t in second. The speed of the wave is …………………
(a) 150 ms^-1
(b) 300 ms^-1
(c) 450 ms^-1
(d) 600 ms^-1
Hint:
From standard equation of wave, Y = a sin (ωt – kx + ϕ)
ω = 300 ; k = 2
Speed of wave, V = \(\frac{ω}{k}\) = \(\frac{300}{2}\) = 150ms^-1
Answer:
(a) 150 ms^-1

Question 12.
The increase in internal energy of a system is equal to the workdone on the system. The process does the system undergoes is ……………………
(a) Isochoric
(b) Adiabatic
(c) Isobaric
(d) Isothermal
Answer:
(d) Isothermal

Question 13.
The minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop is ……………………..
(a) \(\sqrt{2gR}\)
(b) \(\sqrt{3gR}\)
(c) \(\sqrt{5gR}\)
(d) \(\sqrt{gR}\)
Answer:
(c) \(\sqrt{5gR}\)

Question 14.
If the rms velocity of the molecules of a gas in a container be doubled then the pressure of the gas will.
(a) Becomes 4 times of the previous value
(b) Becomes 2 times of its previous value
(c) Remains same
(d) Becomes \(\frac{1}{4}\) of its previous value
Hint:

Answer:
(a) Becomes 4 times of the previous value

Question 15.
Gravitational mass is proportional to gravitational ………………….
(a) Intensity
(b) Force
(c) Field
(d) All of these
Answer:
(b) Force

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
Write any four postulates of Kinetic theory of gases?
Answer:

1. A gas consists of a very large number of molecules. Each one is a perfectly identical elastic sphere.
2. The molecules of a gas are in a state of continuous and random motion. They move in all directions with all possible velocities.
3. The size of each molecule is very small as compared to the distance between them. Hence, the volume occupied by the molecule is negligible in comparison to the volume of the gas:
4. There is no force of attraction or repulsion between the molecules and the walls of the container.

Question 17.
Draw the free body diagram of the book at rest on the table?
Answer:

Question 18.
Three block are connected as shown in fig on a horizontal frictionless table. If m₁ = 1 kg, m₂ = 8 kg, m₃ = 27 kg and T₃ = 36 N then calculate tension T₂?
Answer:

Acceleration acquired by all blocks a = \(\frac { T_{ 3 } }{ m_{ 1 }+m_{ 2 }+m_{ 3 } } \) = \(\frac{36}{36}\) = 1ms^-2
∴ Tension T₂ = (m₁ + m₂) a
= (1 + 8) × 1 = 9N

Question 19.
What is power? Give its dimensional formula?
Answer:
The rate of work done is called power. Dimensional formula of power is ML² T^-3

Question 20.
What are geostationary and polar satellites?
Answer:
Geostationary Satellite: It is the satellite which appears at a fixed position and at a definite height to an observer on Earth.
Polar Satellite: It is the satellite which revolves in polar orbit around the Earth.

Question 21.
An iceberg of density 900 kg m^-3 is floating in water of density 1000 kg m^-3. What is the percentage of volume of iceberg outside the water?
Answer:
Fraction of volume inside water = Relative density of the body
\(\frac { V’ }{ V } \) = \(\frac { \rho }{ \rho ‘ } \) = \(\frac{900}{1000}\) = 0.9
Fraction of volume outside water = 1 – 0.9 = 0.1
Percentage of volume outside water = 0.1 × 100 = 10%

Question 22.
State stoke’s law and define terminal velocity?
Answer:
Stoke’s law:
When a body falls through a highly viscous liquid, it drags the layer of the liquid immediately in contact with it. This results in a relative motion between the different layers of the liquid. As a result of this, the falling body experiences a viscous force F.

Stoke performed many experiments on the motion of small spherical bodies in different fluids and concluded that the viscous force F acting on the spherical body depends on

1. Coefficient of viscosity q of the liquid
2. Radius a of the sphere and
3. Velocity v of the spherical body

Dimensionally it can be proved that ∴ F = k ηav
Experimentally Stoke found that k = 6π
This is Stoke’s law

Terminal velocity:
Terminal velocity of a body is defined as the constant velocity acquired by a body while falling through a viscous liquid.

Question 23.
The Earth without its atmosphere would be hospitably cold. Explain why?
Answer:
The lower layers of Earth’s atmosphere reflect infrared radiations from Earth back to the surface of Earth. Thus the heat radiation received by the earth from the Sun during the day are kept trapped by the atmosphere. If atmosphere of Earth were not there, its surface would become too cold to live.

Question 24.
A body A is projected upwards with velocity v₁ Another body B of same mass is proj eeted at an angle of 45°. Both reach the same height. Calculate the ratio of their initial kinetic energies?
Answer:
As A and B attain the same height therefore vertical component of initial velocity of B is equal to initial velocity of A
v₂ cos 45° = V₁ (or) \(\frac { v_{ 2 } }{ \sqrt { 2 } } \) = v₁

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Explain the rules for counting significant figures with examples?
Rules for counting significant figures:
Answer:

Question 26.
Elastic headon collision, consider two particles one is moving and another one is stationary with their respective masses m and \(\frac { M }{ m } \). A moving particle meets collides elastically on stationary particle in the opposite direction. Find the kinetic energy of the stationary particle after a collision?
Answer:
mass of the moving particle m₁ = m (say)
mass of the stationary particle m₁ = \(\frac { 1 }{ m } \) M
Velocity of the moving particle before collision = v_1i (say)
Velocity of the stationary particle before collision = v_2i = 0
Velocity of the stationary particle after collision = v_2f (say)

Kinetic energy of the stationary particle after a collision

Question 27.
Calculate the angle for which a cyclist bends when he turns a circular path of length 34.3 m in \(\sqrt{22}\) s?
Answer:
Given Data:
l = 34.3 m, t = \(\sqrt{22}\) , g = 9.8 ms^-2, θ = ?
If r is radius of circular path, then length of path = 2πr = 34.3 m
r = \(\frac { 33.4 }{ 2\pi } \) and time taken t = \(\sqrt{22}\)s

As tan θ = \(\frac { v^{ 2 } }{ rg } \)
∴ tan θ = (\(\frac { 34.3 }{ \sqrt { 22 } } \))² × \(\frac { 2\pi }{ 34.3\times 9.8 } \)
tan θ = \(\frac { 34.3\times 34.3 }{ 22 } \) × \(\frac{2×22}{7×343×9.8}\) = \(\frac{34.3×2}{68.6}\) = 1 [∴θ = 45°]

Question 28.
Explain how density, moisture affect the velocity of sound in gases?
Answer:
Effect of density:
Let us consider two gases with different densities having same temperature and pressure. Then the speed of sound in the two gases are

Taking ratio of equation (1) and equation (2) we get

For gases having same value of γ,
\(\frac { v_{ 1 } }{ v_{ 2 } } \) = \(\sqrt { \frac { \rho _{ 2 } }{ \rho _{ 1 } } } \)
Thus the velocity of sound in a gas is inversely proportional to the square root of the density of the gas.

Effect of moisture (humidity):
We know that density of moist air is 0.625 of that of dry air, which means the presence of moisture in air (increase in humidity) decreases its density. Therefore, speed of sound increases with rise in humidity. From equation:
v = \(\sqrt { \frac { \gamma \rho }{ \rho } } \)

Let ρ₁, v₁ = and ρ₂, v₂ be the density and speeds of sound in dry air and moist air, respectively.

Since P is the atmospheric pressure, it can be shown that
\(\frac { \rho _{ 2 } }{ \rho _{ 1 } } \) = \(\frac { P }{ p_{ 1 }+0.625p_{ 2 } } \)
where p₁ and p₂ are the partial pressures of dry air and water vapour respectively. Then

Question 29.
Explain
(a) Why there are no lunar eclipse and solar eclipse every month?
(b) Why do we have seasons on earth?
Answer:
(a) If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so during new Moon we can observe solar eclipse.

But Moon’s orbit is tilted 5° with respect to Earth’s orbit. Due to this 5° tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment.

(b) The common misconception is that ‘Earth revolves around the Sun, so when the Earth is very far away, it is winter and when the Earth is nearer, it is summer’.

Actually, the seasons in the Earth arise due to the rotation of Earth around the Sun with 23.5° tilt. Due to this 23.5° tilt, when the northern part of Earth is farther to the Sun, the southern part is nearer to the Sun. So when it is summer in the northern hemisphere, the southern hemisphere experience winter.

Question 30.
When a person breathes, his lungs can hold up to 5.5 1 of air at body temperature 37°C and atmospheric pressure (1 atm = 101 kPa). This air contains 21% oxygen, calculate the number of oxygen molecules in the lungs?
Answer:
We can treat the air inside the lungs as an ideal gas. To find the number of molecules, we can use the ideal gas law.
PV = NkT
Here volume is given in the Litre. 1 Litre is volume occupied by a cube of side 10 cm
1 Litre = 10 cm × 10 cm × 10 cm = 10^-3m^-3
N = \(\frac{PV}{kT}\) = \(\frac { 1.01\times 10^{ 5 }\times 5.5\times 10^{ -3 } }{ 1.38\times 10^{ -23 }\times 310 } \)
= 1.29 × 10²³ × \(\frac{21}{100}\)
Number of oxygen molecules = 2.7 × 10²² molecules

Question 31.
Give any five properties of vector product of two vectors?
Answer:
(I) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors \(\vec { A } \) and \(\vec { B } \), even though the vectors AandB may or may not be mutually orthogonal.

(II) The vector product of two vectors is not commutative, i.e., \(\vec { A } \) × \(\vec { B } \) ≠ \(\vec { B } \) × \(\vec { A } \) . But, \(\vec { A } \) × \(\vec { B } \) = –\(\vec { B } \) × \(\vec { A } \).

Here it is worthwhile to note that |\(\vec { A } \) × \(\vec { B } \)| = |\(\vec { B } \) × \(\vec { A } \)| = AB sin θ i.e., in the case of the product vectors \(\vec { A } \) × \(\vec { B } \) and \(\vec { B } \) × \(\vec { A } \), the magnitudes are equal but directions are opposite to each other.

(III) The vector product of two vectors will have maximum magnitude when sin 0 = 1, i.e., θ = 90° i.e., when the vectors \(\vec { A } \) and \(\vec { B } \) are orthogonal to each other.
(\(\vec { A } \) × \(\vec { B } \))_max = AB\(\hat { n } \)

(IV) The vector product of two non-zero vectors will be minimum when sin 0 = 0, i.e., 0 = 0° or 180°
(\(\vec { A } \) × \(\vec { B } \))_max = 0
i.e., the vector product of two non-zero vectors vanishes, if the vectors are either parallel or antiparallel.

(V) The self-cross product, i.e., product of a vector with itself is the null vector
\(\vec { A } \) × \(\vec { A } \) = AA sin 0°\(\hat { n } \) = \(\vec { 0 } \)
In physics the null vector \(\vec { 0 } \) is simply denoted as zero.

Question 32.
Why does a porter bend forward w hile carrying a sack of rice on his back?
Answer:
When a porter carries a sack of rice, the line of action of his centre of gravity will go away from the body. It affects the balance, to avoid this he bends. By which centre of gravity will realign within the body again. So balance is maintained.

Question 33.
A piece of wood of mass m is floating erect in a liquid whose density is p. If it is slightly pressed down and released, then executes simple harmonic motion. Show that its time period of oscillation is T = 2π\(\sqrt{m/Agρ}\)
Answer:
Spring factor of liquid(k) = Aρg
Inertia factor of wood = m
Time period T = 2π = image 12
T = 2π\(\sqrt{m/Aρg}\)

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
Describe the vertical oscillations of a spring?
Answer:
Vertical oscillations of a spring: Let us consider a massless spring with stiffness constant or force constant k attached to a ceiling as shown in figure. Let the length of the spring before loading mass m be L.

If the block of mass m is attached to the other end of spring, then the spring elongates by a length. Let F, be the restoring force due to stretching of spring. Due to mass m, the gravitational force acts vertically downward. We can draw free-body diagram for this system as shown in figure. When the system is under equilibrium,

F₁ + mg = 0 ……………… (1)
But the spring elongates by small displacement 1, therefore,
F₁ ∝ l ⇒ F₁ = -kl ……………….. (2)
Substituting equation (2) in equation (1) we get
-kl + mg = 0
mg = kl or
\(\frac{m}{k}\) = \(\frac{l}{g}\) ………………….. (3)

Suppose we apply a very small external force on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is y + 1) is
F₂ ∝ (y + l)
F₂ = -k(y + l) = -ky – kl …………………… (4)

Since, the mass moves up and down with acceleration \(\frac{d^{2} y}{d t^{2}}\), by drawing the free body diagram for this case we get
-ky – kl + mg = m \(\frac{d^{2} y}{d t^{2}}\) ……………………. (5)
The net force acting on the mass due to this stretching is
F = F₂ + mg
F = -ky – kl + mg …………………… (6)
The gravitational force opposes the restoring force. Substituting equation (3) in equation (6), we get
F = – ky- kl + kl = -ky
Applying Newton’s law we get
m \(\frac{d^{2} y}{d t^{2}}\) = -ky
\(\frac{d^{2} y}{d t^{2}}\) = –\(\frac{k}{m}\)y ………………… (7)
The above equation is in the form of simple harmonic differential equation. Therefore, we get the time period as
T = 2π\(\sqrt{m/k}\) second ……………………. (8)
The time period can be rewritten using equation (3)
T = 2π\(\sqrt{m/k}\) = 2πl\(\frac{1}{g}\) second ……………………. (9)
The accleration due to gravity g can be computed by the formula
g = 4π²\((\frac { 1 }{ T } )^{ 2 }\)ms^-2 …………………….. (10)

[OR]

(b) Derive poiseuille’s formula for the volume of a liquid flowing per second through a pipe under streamlined flow?
Answer:
Consider a liquid flowing steadily through a horizontal capillary tube. Let v = (\(\frac{1}{g}\)) be the volume of the liquid flowing out per second through a capillary tube. It depends on (1) coefficient of viscosity (η) of the liquid, (2) radius of the tube (r), and (3) the pressure gradient (\(\frac{P}{l}\)) . Then,
v ∝η^ar^b(\(\frac{P}{l}\))^c
v = kη^ar^b(\(\frac{P}{l}\))^c …………………….. (1)

where, k is a dimensionless constant.
Therefore, [v] = \(\frac { Volume }{ Time } \) = [L³T^-1]; [ \(\frac{dP}{dX}\) ] = \(\frac { Pressure }{ Distance } \) = [ML^-2T^-2]
[η] = [Ml^-1T^-1] and [r] = [L]

Substituting in equation (1)
[L³T^-1] = [ML^-1T^-1]^a[L]^b [ML^-2T^-2]^c
M⁰L³T^-1 = M^a+bL^-a+b-2cT^-a-2c = -1

So, equating the powers of M, L and T on both sides, we get
a + c = 0, – a + b – 2c = 3, and – a – 2c = – 1

We have three unknowns a, b and c. We have three equations, on solving, we get
a = – 1, b = 4 and c = 1

Therefore, equation (1) becomes,
v = kη^-1r⁴(\(\frac{P}{l}\))¹

Experimentally, the value of k is shown to be , we have \(\frac{π}{8}\), we have
v = \(\frac{\pi r^{4} \mathrm{P}}{8 \eta /}\)

The above equation is known as Poiseuille’s equation for the flow of liquid through a narrow tube or a capillary tube. This relation holds good for the fluids whose velocities are lesser than the critical velocity (v_c).

Question 35 (a).
Describe briefly simple harmonic oscillation as a projection of uniform circular motion?
Answer:
Consider a particle of mass m moving with unifonn speed v along the circumference of a circle whose radius is r in anti-clockwise direction (as shown in figure). Let us assume that the origin of the coordinate system coincides with the center O of the circle.

If ω is the angular velocity of the particle and θ the angular displacement of the particle at any instant of time t, then θ = ωt. By projecting the uniform circular motion on its diameter gives a simple harmonic motion.

This means that we can associate a map (or a relationship) between uniform circular (or revolution) motion to vibratory motion. Conversely, any vibratory motion or revolution can be mapped to unifonn circular motion. In other words, these two motions are similar in nature.

Let us first project the position of a particle moving on a circle, on to its vertical diameter or on to a line parallel to vertical diameter as shown in figure. Similarly, we can do it for horizontal axis or a line parallel to horizontal axis.

The projection of uniform circular motion on a diameter of SHM:
As a specific example, consider a spring mass system (or oscillation of pendulum). When the spring moves up and down (or pendulum moves to and fro), the motion of the mass or bob is mapped to points on the circular motion.

Thus, if a particle undergoes uniform circular motion then the projection of the particle on the diameter of the circle (or on a line parallel to the diameter) traces straight line motion which is simple harmonic in nature. The circle is known as reference circle of the simple harmonic motion. The simple harmonic motion can also be defined as the motion of the projection of a particle on any diameter of a circle of reference.

[OR]

(b) State and prove Bernoulli’s theorem for a flow of incompressible non viscous and stream lined flow of fluid?
Answer:
Bernoulli’s theorem:
According to Bernoulli’s theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant. Mathematically,
\(\frac{P}{ρ}\) + \(\frac{1}{2}\)v² + gh – constant
This is known as Bernoulli’s equation.
Proof:
Let us consider a flow of liquid through a pipe AB. Let V be the volume of the liquid when it enters A in a time t. Which is equal to the volume of the liquid leaving B in the same time. Let a_A, v_A and P_A be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.

Let the force exerted by the liquid at A is
F_A = P_Aa_A

Distance travelled by the liquid in time t is d = v_At
Therefore, the work done is W = F_Ad = P_Aa_Av_At
But a_Av_At = a_Ad = V, volume of the liquid entering at A.

Thus, the work done is the pressure energy (at A), W = F_Ad = P_AV

Pressure energy per unit volume at
A = \($\frac{\text { Pressure energy }}{\text { Volume }}$\) = \(\frac { P_{ A }V }{ V } \) = P_A

Pressure energy per unit mass at
A = \($\frac{\text { Pressure energy }}{\text { Mass }}$\) = \(\frac { P_{ A }V }{ m } \) = \(\frac { P_{ A } }{ \frac { m }{ V } } \) = \(\frac { P_{ A } }{ \rho } \)

Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
EP_A = P_AV = P_AV × (\(\frac{m}{m}\)) = m\(\frac { P_{ A } }{ \rho } \)

Potential energy of the liquid at A,
PE_A = mgh_A

Due to the flow of liquid, the kinetic energy of the liquid at A,
KE_A = \(\frac{1}{2}\)mv^2_A

Therefore, the total energy due to the flow of liquid at A,
E_A = EP_A + KE_A + PE_A
E_A = \(m \frac{P_{A}}{\rho}+\frac{1}{2} m v_{A}^{2}+m g h_{A}\)

Similarly, let a_B, V_B and P_B be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at B. Calculating the total energy at F_B, we get .
\(\mathrm{E}_{\mathrm{B}}=m \frac{\mathrm{P}_{\mathrm{B}}}{\rho}+\frac{1}{2} m v_{\mathrm{B}}^{2}+m g h_{\mathrm{B}}\)
From the law of conservation of energy.
E_A = E_B

Thus, the above equation can be written as
\(\frac { P }{ \rho g } \) + \(\frac{1}{2}\) \(\frac { v^{ 2 } }{ g } \) + h = Constant

The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction. This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli’s relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then
h = 0 ⇒ \(\frac { P }{ \rho g } \) + \(\frac{1}{2}\) \(\frac { v^{ 2 } }{ g } \) = Constant

Question 36 (a).
Explain perfect inelastic collision and derive an expression for loss of kinetic energy in perfect inelastic collision?
Answer:
In a perfectly inelastic or completely inelastic collision, the objects stick together permanently after collision such that they move with common velocity. Let the two bodies with masses m₁ and m₂ move with initial velocities u₁ and u₂ respectively before collision. Aft er perfect inelastic collision both the objects move together with a common velocity v as shown in figure.
Since, the linear momentum is conserved during collisions,

m₁u₁ + m₂u₂ = (m₁ + m₂) v

The common velocity can be computed by
v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\) ………………….. (1)

Loss of kinetic energy in perfect inelastic collision;
In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat, light etc. Let KE_i be the total kinetic energy before collision and KE_f be the total kinetic energy after collision.
Total kinetic energy before collision,
KE_e = \(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}\) …………………… (2)
Total kinetic energy after collision,
KE_f = \(\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}\) …………………….. (3)
Then the loss of kinetic energy is Loss of KE, ∆Q = KE_f – KE_i
= \(\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}-\frac{1}{2} m_{1} u_{1}^{2}-\frac{1}{2} m_{2} u_{2}^{2}\) ………………….. (4)
Substituting equation (1) in equation (4), and on simplifying (expand v by using the algebra (a + b)² = a² + b² + 2ab), we get
Loss of KE, ∆Q = \(\frac{1}{2}\) \(\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)\) (u₁ – u₂)²

[OR]

(b) Derive an expression for maximum height attained, time of flight, horizontal range for a projectile in oblique projection?
Answer:

Maximum height (h_max):
The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion
\(v_{y}^{2}=u_{y}^{2}+2 a_{y} s\)
Here, u_y= u sin θ, a = -g, s = h_max, and at the maximum height v = 0

Time of flight (T_f):
The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight. This time of flight is the time taken by the projectile to go from point O to B via point A as shown in figure.
We know that s_y = u_yt + \(\frac{1}{2}\)a_yt²
Here, s_y = y = 0 (net displacement in y-direction is zero), u_y = u sin θ, a_y = -g, t = T_f, Then
0 = u sin θ T_f – \(\frac{1}{2} g \mathrm{T}_{f}^{2}\)
T_f = 2u \(\frac{sin θ}{g}\) …………………….. (2)

Horizontal range (R):
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write.

Range R = Horizontal component of velocity % time of flight = u cos θ × T_f = \(\frac{u^{2} \sin 2 \theta}{g}\)
The horizontal range directly depends on the initial speed (u) and the sine of angle of projection (θ). It inversely depends on acceleration due to gravity ‘g’.

For a given initial speed u, the maximum possible range is reached when sin 2θ is
maximum, sin 2θ = 1. This implies 2θ = π/2 or θ = π/4
This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range, given by
R_max = \(\frac { u^{ 2 } }{ g } \).

Question 37 (a).
Explain the work-energy theorem in detail and also give three examples?
Answer:

1. If the work done by the force on the body is positive then its kinetic energy increases.
2. If the work done by the force on the body is negative then its kinetic energy decreases.
3. If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.
4. When a particle moves with constant speed in a circle, there is no change in the kinetic energy of the particle. So according to work energy principle, the work done by centripetal force is zero.

[OR]

(b) (i) Define molar specific heat capacity?
Answer:
Molar specific heat capacity is defined as heat energy required to increase the temperature of one mole of substance by IK or 1°C

(ii) Derive Mayer’s relation for an ideal gas?

Mayer’s relation: Consider p mole of an ideal gas in a container with volume V, pressure P and temperature T.

When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.

If C_V is the molar specific heat capacity at constant volume, from equation.
C_V = \(\frac { 1 }{ \mu } \) \(\frac{dU}{dT}\) …………………… (1)
dU = µC_V dT ………………… (2)

Suppose the gas is heated at constant pressure so that the temperature increases by dT. If ‘Q’ is the heat supplied in this process and ‘dV’ the change in volume of the gas.
Q = pCpdT ……………. (3)

If W is the workdone by the gas in this process, then
W = P dV ………………….. (4)

But from the first law of thermodynamics,
Q = dU + W ………………… (5)

Substituting equations (2), (3) and (4) in (5), we get,
For mole of ideal gas, the equation of state is given by
\(\mu \mathrm{C}_{\mathrm{p}} d \mathrm{T}=\mu \mathrm{C}_{\mathrm{v}} d \mathrm{T}+\mathrm{P} d \mathrm{V}\)

Since the pressure is constant, dP = 0
C_pdT = C_VdT + PdV
∴ C_p = C_V + R (or) C_p – C_V = R …………………… (6)
This relation is called Mayer’s relation It implies that the molar specific heat capacity of an ideal gas at constant pressure is greater than molar specific heat capacity at constant volume.
The relation shows that specific heat at constant pressure (s_p) is always greater than specific heat at constant volume (s_v).

Question 38 (a).
Derive an expression of pressure exerted by the gas on the walls of the container?
Answer:
Expression for pressure exerted by a gas : Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.
The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.

The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision, the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.

A molecule of mass m moving with a velocity \(\vec { v } \) having components (v_x, v_y, v_z) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. This is shown in the figure. The components of velocity of the molecule after collision are (-v_x, v_y, v_z).

The x-component of momentum of the molecule before collision = mv_x
The x-component of momentum of the molecule after collision = – mv_x
The change in momentum of the molecule in x direction
= Final momentum – initial momentum = – mv_x – mv_x = – 2mv_x
According to law of conservation of linear momentum, the change in momentum of the wall = 2 mv_x
The number of molecules hitting the right side wall in a small interval of time ∆t.

The molecules within the distance of v_x∆t from the right side wall and moving towards the right will hit the wall in the time interval ∆t. The number of molecules that will hit the right side wall in a time interval ∆t is equal to the product of volume (Av_x∆t) and number density of the molecules (n).

Here A is area of the wall and n is number of molecules per \(\frac{N}{V}\) unit volume. We have assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.

The number of molecules that hit the right side wall in a time interval ∆t
= \(\frac{n}{2}\) Av_x∆t
In the same interval of time ∆t, the total momentum transferred by the molecules
\(\Delta \mathrm{P}=\frac{n}{2} \mathrm{A} v_{x} \Delta t \times 2 m v_{x}=\mathrm{A} v_{x}^{2} m n \Delta t\) ………………….. (2)
From Newton’s second law, the change in momentum in a small interval of time gives rise to force.

The force exerted by the molecules on the wall (in magnitude)
F = \(\frac{∆p}{∆t}\) = nmAv^2_x ……………………. (3)

Pressure, P = force divided by the area of the wall
P = \(\frac{F}{A}\) = nmAv^2_x ……………………….. (4)
p = \(nm\bar{v}_{x}^{2}\)

Since all the molecules are moving completely in random manner, they do not have same . speed. So we can replace the term vnmAv^2_x by the average \(\bar { v } \)^2_x in equation (4).
P = nm\(\bar { v } \)^2_x ……………………. (5)

Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, \(\bar{v}_{x}^{2}\) = \(\bar{v}_{y}^{2}\) = \(\bar{v}_{z}^{2}\). The mean square speed is written as
\(\bar{v}^{2}\) = \(\bar{v}_{x}^{2}\) + \(\bar{v}_{y}^{2}\) + \(\bar{v}_{z}^{2}\) = 3\(\bar{v}_{x}^{2}\)
\(\bar{v}_{x}^{2}\) = \(\frac{1}{3}\) \(\bar{v}^{2}\)
Using this in equation (5), we get
P = \(\frac{1}{3} n m \bar{v}^{2} \quad \text { or } P=\frac{1}{3} \frac{N}{V} m \bar{v}^{2}\) ………………….. (6)

[OR]

(b) Discuss the simple pendulum in detail?
Answer:
Simple pendulum

A pendulum is a mechanical system which exhibits periodic motion. It has a bob with mass m suspended by a long string (assumed to be massless and inextensible string) and the other end is fixed on a stand. At equilibrium, the pendulum does not oscillate and hangs vertically downward.

Such a position is known as mean position or equilibrium position. When a pendulum is displaced through a small displacement from its equilibrium position and released, the bob of the pendulum executes to and fro motion. Let l be the length of the pendulum which is taken as the distance between the point of suspension and the centre of gravity of the bob. Two forces act on the bob of the pendulum at any displaced position.

• The gravitational force acting on the body (\(\vec { F} \) = m\(\vec { g } \)) which acts vertically downwards.
• The tension in the string T which acts along the string to the point of suspension.

Resolving the gravitational force into its components:

1. Normal component: The component along the string but in opposition to the direction of tension, F_as = mg cos θ.
2. Tangential component: The component perpendicular to the string i.e., along tangential direction of arc of swing, F_ps = mg sin θ.

Therefore, The normal component of the force is, along the string,
\(\mathrm{T}-\mathrm{W}_{a s}=m \frac{v^{2}}{l}\)

Here v is speed of bob
T -mg cos θ = m \(\frac{v^{2}}{l}\)

From the figure, we can observe that the tangential component W_ps of the gravitational force always points towards the equilibrium position i.e., the direction in which it always points opposite to the direction of displacement of the bob from the mean position. Hence, in this case, the tangential force is nothing but the restoring force. Applying Newton’s second law along tangential direction, we have
\(m \frac{d^{2} s}{d t^{2}}+\mathrm{F}_{p s}=0 \Rightarrow m \frac{d^{2} s}{d t^{2}}=-\mathrm{F}_{p s}\)
\(m \frac{d^{2} s}{d t^{2}}=-m g \sin \theta\) …………………. (1)

where, s is the position of bob which is measured along the arc. Expressing arc length in terms of angular displacement i.e.,
s = lθ ………………… (2)
then its acceleration, \(\frac{d^{2} s}{d t^{2}}=l \frac{d^{2} \theta}{d t^{2}}\) …………………. (3)

Substituting equation (3) in equation (1), we get
\(\begin{aligned}
l \frac{d^{2} \theta}{d t^{2}} &=-g \sin \theta \\
\frac{d^{2} \theta}{d t^{2}} &=-\frac{g}{l} \sin \theta
\end{aligned}\) ………………….. (4)

Because of the presence of sin θ in the above differential equation, it is a non-linear differential equation (Here, homogeneous second order). Assume “the small oscillation approximation”, sin θ ~ 0, the above differential equation becomes linear differential equation.
\(\frac{d^{2} \theta}{d t^{2}}=-\frac{g}{l} \theta\) …………………… (5)

This is the well known oscillatory differential equation. Therefore, the angular frequency of this oscillator (natural frequency of this system) is
ω² = \(\frac{g}{l}\) …………………… (6)
∴ ω = \(\sqrt{g/l}\) in rad s^-1 ……………….. (7)

The frequency of oscillation is
f = \(f=\frac{1}{2 \pi} \sqrt{\frac{g}{l}} \text { in } \mathrm{Hz}\) ………………… (8)
and time period of sscillations is
T = 2π\(\sqrt{l/g}\) in second. ……………….. (9)

Students can Download Tamil Nadu 11th Maths Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Maths Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 90

PART – 1

I. Choose the correct answer. Answer all the questions: [20 × 1 = 20]

Question 1.
Let R be the universal relation on a set X with more than one element then R is ………………
(a) Not reflexive
(b) Not symmetric
(c) Transitive
(d) None of the above
Answer:
(c) Transitive

Question 2.
The value of log_ab log_bc log_ca is …………………..
(a) 2
(b) 1
(c) 3
(d) 4
Answer:
(b) 1

Question 3.
If log \(\log _{\sqrt{ }}\) 0.25 = 4 then the value of x is ………………..
(a) 0.5
(b) 2.5
(c) 1.5
(d) 1.25
Answer:
(a) 0.5

Question 4.
The product of r consecutive positive integers is divisible by ………………..
(a) r!
(b) (r-1)!
(c) (r+l)!
(d) r^r
Answer:
(a) r!

Question 5.
The value of tan75° – cot 75° is ……………….
(a) 1
(b) 2 + \(\sqrt{3}\)
(c) 2 – \(\sqrt{3}\)
(d) 2\(\sqrt{3}\)
Answer:
(d) 2\(\sqrt{3}\)

Question 6.
If (1 +x²)²(1 + x)² = a₀ + a₁ x + a₂x² …. + xⁿ⁺⁴ and if a₀, a₁, a₂, are in AP, then n is …………………..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 7.
If _nC₁₂ = _nC₅ then _nC₂ = …………………
(a) 72
(b) 306
(c) 152
(d) 153
Answer:
(d) 153

Question 8.
The line (p + 2q)x + (p- 3q)y =p – q for different values of p and q passes through the point …………………
(a) (\(\frac{3}{5}\), \(\frac{2}{5}\))
(b) (\(\frac{2}{5}\), \(\frac{2}{5}\))
(c) (\(\frac{3}{5}\), \(\frac{3}{5}\))
(d) (\(\frac{2}{5}\), \(\frac{3}{5}\))
Answer:
(d) (\(\frac{2}{5}\), \(\frac{3}{5}\))

Question 9.
The number of terms in the expansion of [(a + b)²]¹⁸ = ………………..
(a) 19
(b) 18
(c) 36
(d) 37
Answer:
(d) 37

Question 10.
A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 5 then its y intercept is …………………
Answer:
(a) \(\frac{3}{4}\)
(b) \(\frac{4}{3}\)
(c) 5
(d) \(\frac{1}{3}\)

Question 11.
If a and b are the roots of the equation x² – kx + 16 = 0 satisfy a² + b² = 32, then the value of k is ………………..
(a) 10
(b) -8
(c) -8, 8
(d) 6
Answer:
(c) -8, 8

Question 12.
If A is a square matrix of order 3 then |kA| = ………………….
(a) k |A|
(b) k²|A|
(c) k³|A|
(d) k|A³|
Answer:
(c) k³|A|

Question 13.
If ABCD is a parallelogram then \(\bar { AB } \) + \(\bar { AD } \) + \(\bar { CD } \) + \(\bar { CD } \) = ………………..
(a) 2(\(\bar { AB } \) + \(\bar { AD } \))
(b) 4\(\bar { AC } \)
(c) 4\(\bar { BD } \)
(d) \(\bar { o } \)
Answer:
(d) \(\bar { o } \)

Question 14.
\(\lim _{x \rightarrow 0}\) x cot x = ………………….
(a) 0
(b) 1
(c) -1
(d) ∞
Answer:
(b) 1

Question 15.
If x = \(\frac { 1-t^{ 2 } }{ 1+t^{ 2 } } \) and y = \(\frac { 2t }{ 1+t^{ 2 } } \) then \(\frac{dy}{dx}\) = ………………..
(a) \(\frac{y}{x}\)
(b) \(\frac{-y}{x}\)
(c) –\(\frac{x}{y}\)
(d) \(\frac{x}{y}\)
Answer:
(c) –\(\frac{x}{y}\)

Question 16.
If y = \(\frac { (1-x)^{ 2 } }{ x^{ 2 } } \) then \(\frac{dy}{dx}\) is …………………..
(a) \(\frac { 2 }{ x^{ 2 } } \) + \(\frac { 2 }{ x^{ 3 } } \)
(b) –\(\frac { 2 }{ x^{ 2 } } \) + \(\frac { 2 }{ x^{ 3 } } \)
(c) –\(\frac { 2 }{ x^{ 2 } } \) – \(\frac { 2 }{ x^{ 3 } } \)
(d) –\(\frac { 2 }{ x^{ 3 } } \) + \(\frac { 2 }{ x^{ 2 } } \)
Answer:
(d) –\(\frac { 2 }{ x^{ 3 } } \) + \(\frac { 2 }{ x^{ 2 } } \)

Question 17.
If y = \(\frac{sinx+cosx}{sinx-cosx}\) then \(\frac{dy}{dx}\) at x = \(\frac { \pi }{ 2 } \) is ………………….
(a) 1
(b) 0
(c) -2
(d) 2
Answer:
(c) -2

Question 18.
\(\int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}\) dx is ……………………
(a) \(\frac{1}{2}\) sin2x + c
(b) –\(\frac{1}{2}\) sin2x + c
(c) \(\frac{1}{2}\) cos 2x + c
(d) – \(\frac{1}{2}\) cos 2x + c
Answer:
(b) –\(\frac{1}{2}\) sin2x + c

Question 19.
An urn contains 5 red and 5 black balls. A balls is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. The probability that the second ball drawn is red will be ………………
(a) \(\frac{5}{12}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{7}{12}\)
(d) \(\frac{1}{4}\)
Answer:
(b) \(\frac{1}{2}\)

Question 20.
Let A and B be two events such that P(\(\bar { AUB } \)) = \(\frac{1}{6}\) , Then the events A and B are P(A∩B) = 1/4 and P(\(\bar { A } \)) = 1/4 is ………………
(a) Equally likely but not independent
(b) Independent but not equally likely
(c) Independent and equally likely
(d) Mutually inclusive and dependent
Answer:
(b) Independent but not equally likely

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
Let A and B are two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2) and (z, 1) are in A × B, find A and B where x, y, z are distinct elements?
Answer:
n(A) = 3 ⇒ set A contains 3 elements
n(B) = 2 ⇒ set B contains 2 elements
we are given (x, 1), (y, 2), (z, 1) are elements in A × B
⇒ A = {x, y, z} and B = {1, 2}

Question 22.
Solve |5x — 12| ← 2
Answer:
5x – 12 > -2 (or) 5x – 12 < 2 ⇒ 5x > -2 + 12 (= 10)
⇒ x > \(\frac{10}{5}\) = 2
x > 2
(or)
5x < 2 + 12 (= 14)
⇒ x < \(\frac{14}{5}\)
so 2 < x < \(\frac{14}{5}\)

Question 23.
If ¹⁰P_r-1 = 2 × 6 P_r, find r?
Answer:
¹⁰P_r-1 = 2 × 6P_r

⇒ (11 – r) (10 – r) (9 – r) (8 – r) (7 – r) = 10 × 9 × 4 × 7
= 5 × 2 × 3 × 3 × 2 × 2 × 7
= 7 × 6 × 5 × 4 × 3
⇒ 11 – r = 7
11 – 7 = r
r = 4

Question 24.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2^nd hour, 4^th hour and n^th hour?
Answer:
No. of bacteria at the beginning = 30
No. of bacteria after 1 hour = 30 × 2 = 60
No. of bacteria after 2 hours = 30 × 2² = 30 × 4 = 120
No. of bacteria after 4 hours = 30 × 2⁴ = 30 × 16 = 480
No. of bacteria after n^th hour = 30 × 2ⁿ

Question 25.
Find |A| if A = \(\left[\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right]\)
Answer:
\(\left[\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right]\)
= 0M₁₁ – sin αM₁₂ + cos αM₁₃
= 0 – sin α(0 – cos α sin β) + cos α(- sin α sin β – 0) = 0

Question 26.
Find the value of λ for which the vectors \(\vec { a } \) = 3\(\hat { i } \) + 2\(\hat { j } \) + 9\(\hat { k } \) and \(\vec { a } \) = \(\hat { i } \) + λ\(\hat { j } \) +3\(\hat { k } \) are parallel?
Answer:
Given \(\vec { a } \) and \(\vec { b } \) are parallel ⇒\(\vec { a } \) = t\(\vec { b } \) (where t is a scalar)
(i.e.,) 3\(\hat { i } \) + 2\(\hat { j } \) + 9\(\hat { k } \) = t\(\hat { i } \) + λ\(\hat { j } \) + 3\(\hat { k } \))
equating \(\hat { i } \) components we get 3 = t
equating \(\hat { j } \) components
(i.e); 2 = tλ
2 = 3λ ⇒λ = 2/3

Question 27.
Evaluate \(\underset { x\rightarrow \pi }{ lim } \) \(\frac{sin 3x}{sin 2x}\)
Answer:

Question 28.
Find the derivative of sinx² with respect to x²?
Answer:

Question 29.
Let the matrix M = \(\begin{bmatrix} x & y \\ z & 1 \end{bmatrix}\) if x, y and z are chosen at random from the set {1, 2, 3}, and repetition is allowed (i.e., x = y = z), what is the probability that the given matrix M is a singular matrix?
Answer:
If the given matnx M is singular, then = \(\begin{vmatrix} x & y \\ z & 1 \end{vmatrix}\) = 0
That is, x – yz = 0
Hence the possible ways of selecting (x, y, z) are
{(1, 1, 1), (2, 1, 2), (2, 2, 1), (3, 1, 3), (3, 3, 1)} = A (say)
The number of favourable cases n(A) = 5
The total number of cases are n(S) = 3³ = 27
The probability of the given matrix is a singular matrix is
P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{5}{27}\)

Question 30.
Evaluate \(\frac { x^{ 2 } }{ 1+x^{ 6 } } \)
Answer:

PART – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
If f, g, h are real valued functions defined on R, then prove that (f + g) o h = f o h + g o h. What can you say about fo(g + h)? Justify your answer?

Question 32.
Solve \(\frac{4}{x+1}\) ≤ 3 ≤ \(\frac{6}{x+1}\), x > 0?

Question 33.
Prove that cos^-1 \(\frac{4}{5}\) + tan^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{27}{11}\)?

Question 34.
There are 15 candidates for an examination. 7 candidates are appearing for mathematics examination while the remaining 8 are appearing for different subjects. In how many ways
can they be seated in a row so that no two mathematics candidates are together?

Question 35.
Prove that if a, b, c are in H.P. if and only if \(\frac{a}{c}\) = \(\frac{a-b}{b-c}\)?

Question 36.
If (-4, 7) is one vertex of a rhombus and if the equation of one diagonal is 5x – y + 7 = 0, then find the equation of another diagonal?

Question 37.
Verify the existence of \(\underset { x\rightarrow 1 }{ lim } \) f(x), where f(x) = \(\left\{\begin{aligned}
\frac{|x-1|}{x-1}, & \text { for } x \neq 1 \\
0, & \text { for } x=1
\end{aligned}\right.\)

Question 38.
If y = sin^-1 x then find y?

Question 39.
Evaluate cot² x + tan² x?

Question 40.
Show that
\(\left|\begin{array}{ccc}
2 b c-a^{2} & c^{2} & b^{2} \\
c^{2} & 2 c a-b^{2} & a^{2} \\
b^{2} & a^{2} & 2 a b-c^{2}
\end{array}\right|=\left|\begin{array}{ccc}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|^{2}\)

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41 (a).
The total cost of airfare on a given route is comprised of the base cost C and the fuel surcharge S in rupee. Both C and S are functions of the mileage m; C(m) = 0.4m + 50 and S(m) = 0.03m. Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying ¡600 miles?

[OR]

(b) Evaluate \(\sqrt { x^{ 2 }+x+1 } \)?

Question 42 (a).
Determine the region in the plane determined by the inequalities y ≥ 2x and -2x + 3y ≤ 6?

[OR]

(b) If y(cos^-1 x)², prove that (1-x²) \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) – x \(\frac{dy}{dx}\) – 2 = 0. Hence find y₂ when x = 0?

Question 43(a).
Prove that ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r

[OR]

(b) If the binomial coefficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42, then find n?

Question 44 (a).
(a) Prove that

1. sin A + sin( 120° + A) + sin (240° + A) = O
2. cos A+ cos (120° + A) + cos (120° – A) = O

[OR]

(b) A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

Question 45 (a).
Show that \(\left|\begin{array}{ccc}
a^{2}+x^{2} & a b & a c \\
a b & b^{2}+x^{2} & b c \\
a c & b c & c^{2}+x^{2}
\end{array}\right|\) is divisible by x⁴?

[OR]

(b) \(\left[\begin{array}{ccc}
0 & p & 3 \\
2 & q^{2} & -1 \\
r & 1 & 0
\end{array}\right]\) is skew-symmetric, find the values of p, q and r?

Question 46 (a).
In a shopping mall there is a hail of cuboid shape with dimension 800 × 800 × 720 units, which needs to be added the facility of an escalator in the path as shown by the dotted line in the figure. Find

1. The minimum total length of the escalator
2. The heights at which the escalator changes its direction
3. The slopes of the escalator at the turning points.

[OR]

(b) Evaluate \(\lim _{x \rightarrow a} \frac{\sqrt{x-b}-\sqrt{a-b}}{x^{2}-a^{2}}(a>b)\)

Question 47 (a).
Evaluate ∫\(\frac { 3x+5 }{ x^{ 2 }+4x+7 } \) dx

[OR]

(b) A factory has two Machines – I and II. Machine-I produces 60% of items and Machine-II produces 40% of the items of the total output. Further 2% of the items produced by Machine-I are defective whereas 4% produced by Machine-II are defective. If an itci is drawn at random what is the probability that it is defective?

Students can Download Tamil Nadu 11th Physics Model Question Paper 2 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 2 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be ………………..
(a) 8%
(b) 2%
(c) 4%
(d) 6%
Hint:
Volume of the sphere, V = \(\frac{4}{3}\) πr³
\(\frac{∆V}{V}\) × 100 = 3 × (\(\frac{∆r}{r}\) × 100) = 3 × 2%
\(\frac{∆V}{V}\) × 100 = 6%
Answer:
(d) 6%

Question 2.
A ball is dropped from a building. It takes 4s to reach the ground. The height of the building is (use g= 10 m/s²)
(a) 20 m
(b) 40 m
(c) 80 m
(d) 75 m
Hint:
s = ut + \(\frac{1}{2}\) at² s = h, g = a, u = 0
h = \(\frac{1}{2}\) gt²
h = \(\frac{1}{2}\) × 10 × (4)²; h = 80m
Answer:
(c) 80 m

Question 3.
For inelastic collision between two spherical rigid bodies ……………………
(a) the total kinetic energy is conserved
(b) the total mechanical energy is not conserved
(c) the linear momentum is not conserved
(d) the linear momentum is conserved
Answer:
(d) the linear momentum is conserved

Question 4.
Two rods OA and OB of equal length and mass are lying on xy plane as shown in figure. Let I_x, I_y and I_z be the moments of inertia of the the rods about x, y and z axis respectively, then …………………….
(a) I_x = I_y > I_z
(b) I_x > I_y > I_z
(c) I_x = I_y < I_z
(d) I_z > I_y > I_x

Hint:
I_x = I_y = 2\(\left[\frac{\mathrm{M} l^{2}}{3} \sin ^{2} 45^{\circ}\right]\) = \(\frac { ml^{ 2 } }{ 3 }\)
I_z = \(\left[\frac{m l^{2}}{3}\right]\) = \(\frac { 2ml^{ 2 } }{ 3 }\)
Answer:
(c) I_x = I_y < I_z

Question 5.
The motion of a rocket is based on the principle of conservation of ………………….
(a) Linear momentum
(b) Mass
(c) Angular momentum
(d) Kinetic energy
Answer:
(a) Linear momentum

Question 6.
The work done by the Sun’s gravitational force on the Earth is ………………….
(a) Always zero
(b) Always positive
(c) Can be positive or negative
(d) Always negative
Answer:
(c) Can be positive or negative

Question 7.
A given glass tube having uniform cross section is filled with water and is mounted on a rotatable shaft as shown. If the tube is rotated with a constant angular velocity ω, then …………………
(a) Water levels in both sections A and B go up
(b) Water level in section A goes up and that in B comes down
(c) Water level in section B goes up and that in A comes down
(d) Water level remain same in both
Answer:
(a) Water levels in both sections A and B go up

Question 8.
The efficiency of a heat engine working between the freezing point and boiling point of water is …………………..
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
Hint:
Freezing point of water T_L = 0°C = 273K
Boiling point of water T_H – 100°C = 373K
∴ Efficiency, η = 1 – \(\frac { T_{ L } }{ T_{ H } } \) = 1- \(\frac{273}{373}\) = 0.2861; η = 26.8%.
Answer:
(c) 26.8%

Question 9.
Two waves represented by the following equation are travelling in the same medium
y₁ = 5 sin 2π (75 t – 0.25 x), y₂ = 10 sin 2π (150 – 0.25 x)
The intensity ratio of the two waves is ………………….
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16
Hint:
I ∝ A² ⇒ \(\frac { I_{ 1 } }{ I_{ 2 } }\) = (\(\frac { A_{ 1 } }{ A_{ 2 } }\))² = (\(\frac{5}{2}\))² = \(\frac{1}{4}\)
Answer:
(b) 1 : 4

Question 10.
A man pushes a wall and fails to displace it. He does …………………..
(a) Negative work
(b) Positive but not maximum work
(c) No work at all
(d) Maximum work
Answer:
(c) No work at all

Question 11.
A car moving on a horizontal road may be thrown out of the road in taking a turn …………………
(а) By the gravitational force
(b) Due to lack of sufficient centripetal force
(c) Due to rolling frictional force between tyre and road
(d) Due to the reaction of the ground
Answer:
(b) Due to lack of sufficient centripetal force

Question 12.
The volume of a gas expands by 0.25 m³ at a constant pressure of 10³ N/m, the workdone is equal to ……………………..
(a) 250 W
(b) 2.5 W
(c) 250 N
(d) 250 J
Hint:
Workdone = P. ∆V = 10³ × 0.25 = 250 J
Answer:
(d) 250 J

Question 13.
When three springs of spring constants k₁, k₂, k₃ connected in parallel, then the resultant spring constant is ……………………
(a) K = k₁ + k₂ + k₃
(b) \(\frac{1}{K}\) = \(\frac { 1 }{ k_{ 1 } } +\frac { 1 }{ k_{ 2 } } +\frac { 1 }{ k_{ 3 } } \)
(c) K = \(\frac { 1 }{ k_{ 1 } } +\frac { 1 }{ k_{ 2 } } +\frac { 1 }{ k_{ 3 } } \)
(d) K = k₁ – k₂ – k₃
Answer:
(a) K = k₁ + k₂ + k₃

Question 14.
The distance of the two planets from Sun are 10¹³ and 10¹² m respectively. The ratio of time period of the planets is …………………..
(a) 100
(b) \(\frac { 1 }{ \sqrt { 10 } } \)
(c) \(\sqrt{10}\)
(d) 10\(\sqrt{10}\)
Hint:
According to Kepler’s third law of planetary motion.
\(\frac { T_{ 1 } }{ T_{ 2 } } \) = \(\sqrt{\frac{R_{1}^{3}}{R_{2}^{3}}}\) = \(\sqrt{\frac{\left(10^{13}\right)^{3}}{\left(10^{12}\right)^{3}}}\) = \(\sqrt{\frac{10^{39}}{10^{36}}}\) = \(\sqrt{10^{3}}\) = 10\(\sqrt{10}\)
Answer:
(d) 10\(\sqrt{10}\)

Question 15.
The dimensional formula of planck’s constant is ………………..
(a) [M L² T^-1]
(b) [M L² T^-3]
(c) [M L T^-1]
(d) [M L³ T^-3]
Answer:
(a) [M L² T^-1]

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
A particle is moving along a circular track of radius lm with uniform speed. What is the ratio of the distance covered and the displacement in half revolution?
Answer:
Distance covered by a particle = π × 1 = πm
Displacement covered by a particle = 2 × 1 = 2m
Ratio between distance and displacement

Question 17.
Give one argument in favour of the fact that frictional force is a non-conservative force?
Answer:
The direction of the frictional force is opposite to the direction of motion. When a body is moved, say from A to B and then back to A, work is required to be done both during forward and backward motion. So, the net workidone in a round trip is not zero. Hence, the frictional force is a non-conservative force.

Question Question 18.
Why does a gas not have a unique value of specific heat?
Answer:
This is because a gas can be heated under different conditions of pressure and volume. The amount of heat required to raise the temperature of unit mass through unit degree is different under different conditions of heating.

Question 19.
A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest path in 15 min. Calculate the velocity of river water in km/h?
Answer:
Resultant velocity = \(\frac{1km}{(15/60)h}\) = 4 km/h
If v is velocity of river, then v² + 4² = 5² ⇒ v = \(\sqrt{2-16}\) =3 km/h

Question 20.
In a dark room would you be able to tell whether a given note had been produced by a Piano or a Violin?
Answer:
Yes, in a dark room we can easily identify a sound produced by a Piano or a Violin by using the knowledge of timber or quality of sound. The two sources even though having the same intensity and fundamental frequency will be associated with different number of overtones of different relative intensities. These overtones combine and produce different sounds which enables us to identify them.

Question 21.
What is mean by P – V diagram?
Answer:
PV diagram is a graph between pressure P and volume V of the system. The P-V diagram is used to calculate the amount of work done by the gas during expansion or on the gas during compression.

Question 22.
Why does a parachute descend slowly?
Answer:
The surface area of a parachute is much larger as compared to the surface area of stone. So, the air resistance in the case of a parachute is much larger than in the case of a stone. This explains as to why parachute descends slowly.

Question 23.
What is Brownian motion?
Answer:
The motion of the particles in a random and zig-zag mannar in a fluid is called Brownian motion.

Question 24.
Write a note on reverberation?
Answer:
The persistence of audible sound after the source has ceased to emit sound is called reverberation.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Write the rules for determining significant figure?
Answer:
Rules for counting significant figures:

Question 26.
Explain Joule’s experiment of the mechanical equivalent of heat?
Answer:
Joule’s mechanical equivalent of heat:
The temperature of an object can be increased by heating it or by doing some work on it. In the eighteenth century, James Prescott Joule showed that mechanical energy can be converted into internal energy and vice versa. In his experiment, two masses were attached with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity, both the masses lose potential energy equal to 2 mgh.

When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel. This causes a rise in temperature of the water. This implies that gravitational potential energy is converted to internal energy of water.

The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat. He found that to raise 1 g of an object by 1°C , 4.186 J of energy is required. In earlier days the heat was measured in calorie. 1 cal = 4.186 J This is called Joule’s mechanical equivalent of heat.

Question 27.
How do you classify the physical quantities on the basis of dimension?
Answer:
(I) Dimensional variables:
Physical quantities, which possess dimensions and have variable values are called dimensional variables. Examples are length, velocity, and acceleration etc.

(II) Dimensionless variables:
Physical quantities which have no dimensions, but have variable values are called dimensionless variables. Examples are specific gravity, strain, refractive index etc.

(III) Dimensional Constant:
Physical quantities which possess dimensions and have constant values are called dimensional constants. Examples are Gravitational constant, Planck’s constant etc.

(IV) Dimensionless Constant:
Quantities which have constant values and also have no dimensions are called dimensionless constants. Examples are n, e, numbers etc.

Question 28.
State laws of simple pendulum?
Answer:
Law of length:
For a given value of acceleration due to gravity, the time period of a simple pendulum is directly proportional to the square root of length of the pendulum.
T ∝ \(\sqrt{l}\)

Law of acceleration:
For a fixed length, the time period of a simple pendulum is inversely proportional to square root of acceleration due to gravity.
T ∝ \(\frac { 1 }{ \sqrt { g } } \)

Question 29.
Explain super position principle for gravitational field?
Answer:
The total gravitational field at a point due to all the masses is given by the vector sum of the gravitational field due to the individual masses. This principle is known as superposition of
\(\overrightarrow{\mathrm{E}}_{\text {total }}=\overrightarrow{\mathrm{E}}_{1}+\overrightarrow{\mathrm{E}}_{2}+\ldots \overrightarrow{\mathrm{E}}_{n}=-\frac{\mathrm{G} m_{1}}{r_{1}^{2}} \hat{r}_{1}-\frac{\mathrm{G} m_{2}}{r_{2}^{2}} \hat{r}_{2}-\ldots \cdot \frac{\mathrm{G} m_{n}}{r_{n}^{2}} \hat{r}_{n}\)
\(\overrightarrow{\mathrm{E}}_{\mathrm{total}}=-\sum_{i=1}^{n} \frac{\mathrm{G} m_{n}}{r_{n}^{2}} \hat{r}_{n}\)

Question 30.
Write a note on static friction?
Answer:
Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force f_s lies between
0 ≤ f_s ≤ µ_sN
where, µ_s – coefficient of static friction
N – Normal force then, equation shows that f_s can take any value from 0 & µ_sN. If object is at rest, when no external force acts on it then f_s = 0. If object is at rest, also external force acts on it, then f_s = F_ext
But still the static friction f_s is less than µ_sN when object begins to slide, the static friction (f_s) acting on the object attains maximum.

Question 31.
A small metal ball falls in liquid with a terminal velocity of V. If a ball of radius twice of first ball but same mass falls through a same medium, calculate the terminal velocity with which it falls?
Answer:
Given v = \(\frac{2 r^{2} \rho g}{9 \eta}\)
mass = \(\frac{4}{3}\) πr³ ρ = \(\frac{4}{3}\)π(2r)³ ρ¹ or ρ¹ = \(\frac{ρ}{8}\)
Terminal velocity of second ball is
v¹ = \(\frac{2(2 r)^{2}\left(\frac{\rho}{8}\right) g}{9 \eta}\) = \(\frac{v}{2}\)
v¹ = \(\frac{v}{2}\)

Question 32.
Derive an expression for co-efficient of performance of refrigerator?
Answer:
Coefficient of performance (COP) (β):
COP is a measure of the efficiency of a refrigerator. It is defined as the ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.
COP = β = \(\frac { Q_{ L } }{ W } \) …………………… (1)
From the equation Q_L + W = Q_H
β = \(\frac{Q_{L}}{Q_{H}-Q_{L}}\)
β = \(\frac{1}{\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{Q}_{\mathrm{L}}}-1}\) ………………. (2)
But we know that \(\frac { Q_{ H } }{ Q_{ L } } \) = \(\frac { T_{ H } }{ T_{ L } } \)
Substituting this equation (1) we get β = \(\frac{1}{\frac{T_{H}}{T_{L}}-1}\) = \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}\)

Question 33.
Derive an expression for Laplace’s correction?
Answer:
Laplace’s correction:
In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat.

Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson’s law (not Boyle’s law as Newton assumed), which is
PV^γ = Constant …………… (4)

where, γ = \(\frac { C_{ p } }{ C_{ v } } \) which is the ratio between specific heat at constant pressure and specific heat at constant volume.

Differentiating equation (4) on both the sides, we get
\(\mathrm{V}^{\gamma} d \mathrm{P}+\mathrm{P}\left(\gamma \mathrm{V}^{\gamma-1} d \mathrm{V}\right)=0\)
or
\(\gamma \mathrm{P}=-\mathrm{V} \frac{d p}{d \mathrm{V}}=\mathrm{B}_{\mathrm{A}}\) ………………… (5)

where, B_A is the adiabatic bulk modulus of air. Now, substituting equation (5) in equation
V = \(\sqrt{\frac{B}{\rho}}\), the speed of sound in air is
v_a = \(\sqrt{\frac{B_{A}}{\rho}}=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\gamma v_{T}}\)

Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take γ = 1.47. Hence, speed of sound in air is v_a = (\(\sqrt{1.4}\)) (280m s^-1) = 331.30 m s^-1, which is very much closer to experimental data.

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
Explain different types of error?
Answer:
The uncertainty in a measurement is called an error. Random error, systematic error and gross error are the three possible errors.

(I) Systematic errors:
Systematic errors are reproducible inaccuracies that are consistently in the same direction. These occur often due to a problem that persists throughout the experiment. Systematic errors can be classified as follows

(i) Instrumental errors:
When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.

(ii) Imperfections in experimental technique or procedure:
These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied

(iii) Personal errors:
These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.

(iv) Errors due to external causes:
The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity, or pressure during measurements may affect the result of the measurement.

(v) Least count error:
Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument’s resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.

(II) Random errors:
Random errors may arise due to random and unpredictable variations in experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer. Who performs the experiment. Random errors are sometimes called “chance error”. When different readings are obtained by a person every time he repeats the experiment, personal error occurs.

For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.

If n number of trial readings are taken in an experiment, and the readings are
a₁, a₂, a₃, ………………. a_n. The arthematic mean is
\(a_{m}=\frac{a_{1}+a_{2}+a_{3}+\ldots \ldots \ldots a_{n}}{n}\) (or) \(a_{m}=\frac{1}{n} \sum_{i=1}^{i=n} a_{i}\)

[OR]

(b) By using equations of motion, derive an expression for range and maximum height reached by the object thrown at an oblique angle θ with respect to the horizontal direction?
Answer:

This projectile motion takes place when the initial velocity is not horizontal, but at some angle with the vertical, as shown in Figure.

(Oblique projectile):
Examples:

1. Water ejected out of a hose pipe held obliquely.
2. Cannot fired in a battle ground.

Consider an object thrown with initial velocity at an angle 0 with the horizontal.
Then,
\(\vec { u } \) = u_x\(\hat { i } \) + u_y\(\hat { j } \)
where u_x = u cos θ is the horizontal component and u_y = u sin θ the vertical component of velocity.

Since the acceleration due to gravity is in the direction opposite to the direction of vertical component u_y, this component will gradually reduce to zero at the maximum height of the projectile.

At this maximum height, the same gravitational force will push the projectile to move downward and fall to the ground. There is no acceleration along the x direction throughout the motion. So, the horizontal component of the velocity (u_x = u cos θ) remains the same till the object reaches the ground.

Hence after the time f, the velocity along horizontal motion v_x = u_x + a_xt = u_x = u cos θ.
The horizontal distance travelled by projectile in time t is s_x = u_xt + \(\frac{1}{2}\)a_xt²
Here, s_x = x, u_x = u cos θ, a_x = 0.

Thus, x = u cos θ.t or t = \(\frac{x}{u cosθ}\) …………….. (1)
Next, for the vertical motion v_y= u_y + a_yt
Here u_y = u sin θ, ay = – g (acceleration due to gravity acts opposite to the motion).
Thus, v_y = u sin θ – gt
The vertical distance travelled by the projectile in the same time t is
Here, s_y = y, u_y = u sin θ, a_x = – g Then,
y = u sin θ t – \(\frac{1}{2}\)gt² ……………….. (2)
Subsitute the value of t from equation (i) and equation (ii), we have
y = \(u \sin \theta \frac{x}{u \cos \theta}-\frac{1}{2} g \frac{x^{2}}{u^{2} \cos ^{2} \theta}\)
y = \(x \tan \theta-\frac{1}{2} g \frac{x^{2}}{u^{2} \cos ^{2} \theta}\) ………………… (3)
Thus the path followed by the projectile is an inverted parabola.

Maximum height (h_max):
The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion
\(v_{y}^{2}=u_{y}^{2}+2 a_{y} s\)
Here, u_y = u sin θ, a = -g, s = h_max, and at the maximum height v_y = 0
Hence, (0)² = u² sin² θ = 2gh_max
or
h_max = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\) …………….. (4)

Time of flight (T_f):
The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight.
This time of flight is the time taken by the projectile to go from point O to B via point A as shown in figure.
We know that s_y = u_yt + \(\frac{1}{2}\) a_yt²
Here, s_y = y = 0 (net displacement in y-direction is zero), u_y = u sin θ, a_y = -g, t = T_f.
Then 0 = u sinθ T_f – \(\frac{1}{2}\)gT^2_f
T_f = 2u \(\frac{sin θ}{g}\) …………………. (5)

Horizontal range (R):
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write

Range R = Horizontal component of velocity x time of flight = u cos θ × T_f = \(\vec{r}_{1} \times \vec{r}_{2}\)

The horizontal range directly depends on the initial speed (u) and the sine of angle of projection (θ). It inversely depends on acceleration due to gravity ‘g’.

For a given initial speed u, the maximum possible range is reached when sin 2θ is maximum,
sin 2θ = 1. This implies 2θ = π/2
or θ = \(\frac{π}{4}\)
This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range is given by.
R_max = \(\frac { u^{ 2 } }{ g }\) ……………….. (6)

Question 35 (a).
Explain in detail the triangle law of vector addition?
Answer:
Let us consider two vectors \(\vec { A } \) and \(\vec { B } \) as shown in figure.

To find the resultant of the two vectors we apply the triangular law of addition as follows: Represent the vectors \(\vec { A } \) and \(\vec { B } \) by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector \(\vec { A } \) is connected to the tail of the second vector \(\vec { B } \). Let 0 be the angle between A and B. Then R is the resultant vector connecting the tail of the first vector \(\vec { A } \) to the head of the second vector \(\vec { B } \).

The magnitude of R (resultant) is given geometrically by the length of \(\vec { R } \)(OQ) and the direction of the resultant vector is the angle between R and A. Thus we write \(\vec { R } \) = \(\vec { A } \) + \(\vec { B } \).
\(\overline { OQ } \) = \(\overline { OP } \) + \(\overline { PQ } \)

(I) Magnitude of resultant vector:
The magnitude and angle of the resultant vector are determined by using triangle law of vectors as follows.
From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\vec { A } \) and \(\vec { B } \).
cos θ = \(\frac{AN}{B}\) ∴ AN = B cos θ and sin θ = \(\frac{BN}{B}\) ∴BN = B sin θ
For ∆OBN, we have OB² = ON² + BN²
⇒ R² = (A + B cos θ)² + (B sin θ)²
⇒ R² = A² + B² cos² θ + 2AB cos θ + B² sin² θ
⇒ R² = A² + B ² (cos² θ + sin² θ) + 2AB cos θ
⇒ R² = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

(II) Direction of resultant vectors:
If θ is the angle between \(\vec { A } \) and \(\vec { B } \), then
|\(\vec { A } \) + \(\vec { B } \)| = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)
IF \(\vec { R } \) makes an angle α with \(\vec { A } \), then in ∆OBN,
tan α = \(\frac{BN}{ON}\) = \(\frac{BN}{OA + AN}\)
tan α = \(\frac{B \sin \theta}{A+B \cos \theta}\) ⇒ α = tan^-1(\(\frac{B \sin \theta}{A+B \cos \theta}\))

[OR]

(b) Arrive at an expression for velocity of objects in one dimensional elastic collision?
Answer:
Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.

In order to have collision, we assume that the mass m₁ moves faster than mass m₂ i.e., u₁ > u₂. For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same.

From the law of conservation of linear momentum,
Total momentum before collision (ρ_i) = Total momentum after collision (ρ_f)
m₁u₁ + m₂u₂ = m₁u₁ + m₂v₁ ………………. (1)
or m₁ (u₁ – v₁) = m₂(v₂ – u₂) ………………. (2)
Further,

Total kinetic energy before collision KE_i = Total kinetic energy after collision KE_f
\(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}\) ……………….. (3)
After simplifying and rearranging terms,
\(m_{1}\left(u_{1}^{2}-v_{1}^{2}\right)=m_{2}\left(v_{2}^{2}-u_{2}^{2}\right)\)
Using the formula a² – b² = (a + b) (a – b), we can rewrite the above equation as
m₁(u₁ + v₁) = m₂(v₂ + u₂) (v₂ – u₂) …………………. (4)
Dividing equation (4) by (2) gives,
\(\frac{m_{1}\left(u_{1}+v_{1}\right)\left(u_{1}-v_{1}\right)}{m_{1}\left(u_{1}-v_{1}\right)}=\frac{m_{2}\left(v_{2}+u_{2}\right)\left(v_{2}-u_{2}\right)}{m_{2}\left(v_{2}-u_{2}\right)}\)
u₁ + v₁ = v₂ + u₁
Rearranging, u₁ – u₂ = v₂ – v₁ ………………… (5)
Equation (5) can be written as
u₁ – u₂ = -(v₁ – v₂)
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for v₁ and v₂,
v₁ = v₂ + u₂ – u₁ …………………. (6)
or v₂ = u₁ + v₁ – u₁ ………………. (7)

To find the final velocities v₁ and v₂:
Substituting equation (7) in equation (2) gives the velocity of m₁ as
m₁ (u₁ – v₁) = m₂ (u₁ + v₁ – u₂ – u₂)
m₁ (u₁ – v₁) = m₂ (u₁ + v₁ – 2a₂)
m₁u₁ – m₁v₁ = m₂u₁ + m₂v₁ + 2m₂u₂
m₁u₁ = m₂u₁ + 2m₂u₂ = m₁v₁ + m₂v₁
(m₁ – m₂) u₁ + 2m₂u₂ = (m₁ + m₂)v₁
or v₁ = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\) u₁ + \(\left(\frac{2 m_{2}}{m_{1}+m_{2}}\right)\) u₂ ……………. (8)
Similarly, by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final velocity of m₂ as
v₂ = \(\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right)\) u₁ + \(\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right)\) u₂ ………………. (9)

Question 36 (a).
Discuss the variation of g with change in altitude and depth?
Answer:
When an object is on the surface fo the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the objecf would have been mg. Elowever, the object experiences an additional centrifugal force due to spinning of the Earth.

This centrifugal force is given by mω²R’.
\(\mathrm{OP}_{z}, \cos \lambda=\frac{\mathrm{PZ}}{\mathrm{OP}}=\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\)
R’ = R cos λ
where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
\(a_{\mathrm{PQ}}=\omega^{2} \mathrm{R} \cos \lambda=\omega^{2} \mathrm{R} \cos ^{2} \lambda\)
Since R’ = R cos λ

Therefore, g’ = g – ω²2 R cos² λ
From the above expression, we can infer that at equator, λ = 0, g’ = g – ω²R. The acceleration due to gravity is minimum. At poles λ = 90; g’ = g, it is maximum. At the equator, g’ is minimum.

Variation of g with depth:
Consider a particle of mass m which is in a deep mine on the Earth. (Example: coal mines in Neyveli). Assume the depth of the mine as d. To calculate g’ at a depth d, consider the following points. The part of the Earth which is above the radius (R_e – d) do not contribute to the acceleration. The result is proved earlier and is given as

g’ = \(\frac{\mathrm{GM}^{\prime}}{\left(\mathrm{R}_{\mathrm{e}}-d\right)^{2}}\) ………………. (1)
Here M’ is the mass of the Earth of radius (R_e – d) Assuming the density of the earth ρ be constant,
ρ = \(\frac{M}{V}\) …………………. (2)

where M is the mass of the Earth and V its volume, Thus,

Here also g’ < g. As depth increases, g’ decreases. It is very interesting to know that acceleration due to gravity is maximum on the surface of the Earth but decreases when we go either upward or downward.

[OR]

(b) Explain in detail the maxwell Boltzmann distribution function?
Answer:
Maxwell-Boltzmann: In speed distribution function:-
Consider an atmosphere, the air molecules-are moving in random directions. The speed of each molecule is not the same even though macroscopic parameters like temperature and pressure are fixed. Each molecule collides with every other molecule and they exchange their speed.

In the previously we calculated the rms speed of each molecule and not the speed of each molecule which is rather difficult. In this scenario we can find the number of gas molecules that move with the speed of 5 ms^-1 to 10 ms^-1 or 10 ms^-1
to 15 ms^-1 etc. In general our interest is to find how many gas molecules have the range of speed from v to v + dv. This is given by Maxwell’s speed distribution function.

The above expression is graphically shown as follows:

From the figure (1), it is clear that, for a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The rms speed, average speed and most probable speed are indicated in the figure (1). It can be seen that the rms speed is greatest among the three.

1. The area under the graph will give the total number of gas molecules in the system.
2. Figure 2 shows the speed distribution graph for two different temperatures.

As temperature increases, the peak of the curve is shifted to the right. It implies that the average speed of each molecule will increase. But the area under each graph is same since it represents the total number of gas molecules.

Question 37 (a).
What is meant by simple harmonic motion?
Answer:
A particle is said to execute simple harmonic motion if it moves to and fro about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is always directed towards the mean position.

(b) The acceleration due to gravity on the surface of the moon is 1.7 ms^-2. What is the time period of simple pendulum on the moon if its time period on the Earth is 3.5 s? Given g on Earth = 9.8 ms^-2?
Answer:

(c) A man with wrist watch on his hand falls from the top of the tower. Does the watch , give correct time during the free fall?
Answer:
Yes. Because the working of a wrist watch does pot depend on gravity at that place but depends on spring action.

[OR]

(d) State Wien’s law?
Answer:
When the animals feed cold, they curl their bodies into the ball so as to decrease the surface area of their bodies. As total energy radiated by a body varies directly as the surface area of the body, the loss of heat due to radiation would be reduced.

(e) Normal human body of the temperature is 98.6°F. During high fever if the temperature increases to 104°F. What is the change in peak wavelength that emitted by our body (Assume human body is a black body)?
Answer:
Normal human body temperature (T) = 98.6°F.
Convert Fahrenheit into Kelvin, \(\frac{F-32}{180}\) = \(\frac{K-273}{100}\)
So, T = 98.6°F = 310K
From Wien’s displacement law
Maximum wavelength λ_max = \(\frac{b}{T}\) = \(\frac { 2.898\times 10^{ -3 } }{ 313 } \)
λ_max = 9348 × 10^-9 m
λ_max = 9348 nm (at 98.6°F)
During high fever, human body temperature
T = 104°F = 313K
Peak wavelength λ_max = \(\frac{b}{T}\) = \(\frac { 2.898\times 10^{ -3 } }{ 313 } \)
λ_max = 9259 × 10^-9 m
λ_max = 9259 (at 104°F)

(f) Animals curl into a ball, when they feel very cold. Why?
Answer:
When animals feel cold, they curl their bodies into the ball so as to decrease the surface area of their bodies. As total energy radiated by a body varies directly as the surface area of the body, the loss of heat due to radiation would be reduced.

Question 38 (a).
Explain the horizontal oscillations of spring?
Answer:
Horizontal oscillations of a spring-mass system: Consider a system containing a block of mass m attached to a massless spring with stiffness constant or force constant or spring constant k placed on a smooth horizontal surface (frictionless surface) as shown in figure.

Let x₀ be the equilibrium position or mean position of mass m when it is left undisturbed. Suppose the mass is displaced through a small displacement x towards right from its equilibrium position and then released, it will oscillate back and forth about its mean position x₀. Let F be the restoring force (due to stretching of the spring) which is proportional to the amount of displacement of block. For one dimensional motion, . mathematically, we have.
F ∝ x
F = -kx ……………… (1)

where negative sign implies that the restoring force will always act opposite to the direction of the displacement. This equation is called Hooke’s law. Notice that, the restoring force is linear with the displacement (i.e., the exponent of force and displacement are unity).

This is not always true; in case if we apply a very large stretching force, then the amplitude of oscillations becomes very large (which means, force is proportional to displacement containing higher powers of x) and therefore, the oscillation of the system is not linear and hence, it is called non-linear oscillation.

We restrict ourselves only to linear oscillations throughout our discussions, which means Hooke’s law is valid (force and displacement have a linear relationship). From Newton’s second law, we can write the equation for the particle executing simple harmonic motion.

\(m \frac{d^{2} x}{d t^{2}}=-k x\) ……………….. (1)
\(\frac{d^{2} x}{d t^{2}}=-\frac{k}{m} x\) ……………….. (2)
Comparing the equation with simple harmonic motion equation, we get
ω² = \(\frac{k}{m}\) ………………….. (3)

which means the angular frequency or natural frequency of the oscillator is
ω = \(\sqrt{\frac{k}{m}} \mathrm{rad} s^{-1}\) ……………….. (4)

The frequency of the oscillation is
f = \(\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \mathrm{Hertz}\) …………………… (5)
and the time period of the oscillation is
T = \(\frac{1}{f}\) = 2π \(\sqrt{m/k}\) seconds …………………. (6)

[OR]

(b) What is capillarity? Obtain an expression for the surface tension of a liquid by capillary rise method?
Answer:
In a liquid whose angle of contact with solid is less than 90° suffers capillar rise. On the other hand, in a liquid whose angle of contact is greater than 90°, suffers capillary fall. The rise or fall of a liquid in a narrow tube is called capillarity or capillary action.

Practical application of capillarity:

1. Due to capillary action, oil rises in the cotton within an earthen lamp. Likewise, sap raises from the roots of a plant to its leaves and branches.
2. Absorption of ink by a blotting paper.
3. Capillary action is also essential for the tear fluid from the eye to drain constantly.
4. Cotton dresses are preferred in summer because cotton dresses have fine pores which act as capillaries for sweat.

Surface Tension by capillary rise method:
The pressure difference across a curved liquid air interface is the basic factor behind the rising up of water in a narrow tube (influence of gravity is ignored). The capillary rise is more dominant in the case of very fine tubes.

But this phenomenon is the outcome of the force of surface tension. In order to arrive a relation between the capillary rise (h) and surface tension (T), consider a capillary tube which is held vertically in a beaker containing water, the water rises in the capillary tube to a height h due to surface tension.

The surface tension force F_T acts along the tangent at the point of contact downwards and its reaction force upwards. Surface tension T, it resolved into two components

1. Horizontal component T sin θ and
2. Vertical component T cos θ acting upwards, all along the whole circumference of the meniscus.

Total upward force = (T cos θ) (2nr) = 2nrT cos θ)

where θ is the angle of contact, r is the radius of the tube. Let ρ be the density of water and h be the height to which the liquid rises inside the tube. Then,

The upward force supports the weight of the liquid column above the free surface, therefore,

If the capillary is a very fine tube of radius (i.e., radius is very small) then \(\frac{r}{3}\) can be neglected when it is compared to the height h. Therefore,
T = \(\frac{r \rho g h}{2 \cos \theta}\)
Liquid rises through a height h
h = \(\frac{2 \mathrm{T} \cos \theta}{r \rho g} \Rightarrow h \alpha \frac{1}{r}\)
This implies that the capillary rise (h) is inversely proportional to the radius (r) of the tube, i.e., the smaller the radius of thd tube greater will be the capillarity.

Students can Download Tamil Nadu 11th Biology Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Biology Model Question Paper 4 English Medium

General Instructions:

1. 1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
   2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
   3. All questions of Part I, II, III and IV are to be attempted separately.
   4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
   5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
   6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
   7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 3:00 Hours
Maximum Marks: 70

Bio – Botany [Maximum Marks: 35]

PART – I

Answer all the questions. Choose the correct answer. [8 × 1 = 8]

Question 1.
Number of domains of life are there according to Carl Woese
(a) 3
(b) 2
(c) 4
(d) 5
Answer:
(a) 3

Question 2.
Which of the following shows coiled RNA strand and capsomeres?
(a) Polio virus
(b) Tobacco mosaic virus
(c) Measles virus
(d) Retrovirus
Answer:
(b) Tobacco mosaic virus

Question 3.
Talipot palm, Bamboo and Agave are examples of …………………….
(a) Polycarpic geophytes
(b) Therophytes
(c) Monocarpic geophytes
(d) Biennial
Answer:
(c) Monocarpic geophytes

Question 4.
The type of inflorescence seen in Caesalpinia is ………………………
(a) Corymb
(b) Compound corymb
(c) Capitulum
(d) Umbel
Answer:
(a) Corymb

Question 5.
Which one of the following organism is correctly matched with its three characteristics?
(a) Pea: C₃ pathway, Endospermic seed,Vexillary aestivation
(b) Tomato: Twisted aestivation, Axile placentation, Berry
(c) Onion: Bulb, Imbricate aestivation, Axile placentation
(d) Maize: C₃ pathway, Closed vascular bundles, scutellum
Answer:
(c) Onion: Bulb, Imbricate aestivation, Axile placentation

Question 6.
Protoplasm is made of ……………………. water.
(a) 80 – 90%
(b) 85 – 90%
(c) 60 – 80%
(d) 75 – 85%
Answer:
(c) 60 – 80%

Question 7.
The term soilless culture refers to …………………….
(a) Aeroponics
(b) Aqua culture
(c) Hydroponics
(d) Drip irrigation
Answer:
(c) Hydroponics

Question 8.
Which of the following statement is NOT true regarding cyclic photophosphorylation?
(a) The primary electron acceptor is FRS
(b) It produces only ATP molecules
(c) It produces only NADPH + H⁺ molecules
(d) Electrons ejected from PSI again cycled back to PSI
Answer:
(c) It produces only NADPH + H⁺ molecules

PART – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Name few plant disease caused by mycoplasma?
Answer:
Little leaf of brinjal, witches broom of legumes, phyllody of cloves and sandal spike are some plant diseases caused by mycoplasma.

Question 10.
Which leaf part acts a bridge between leaf & stem? Define?
Answer:
Petiole is the bridge between leaf and stem. Petiole or leaf stalk is a cylindrical or subcylindrical or flattened structure of a leaf which joins the lamina with the stem.

Question 11.
Mention any two morphological differences between Dicot & Monocots?
Answer:

S.No	Dicot	Monocots
1.	Leaves show reticulate venation	Leaves show parallel venation
2.	Flowers are tetramerous or pentamerous	Flowers are trimerous

Question 12.
What is the cell wall composition of the following organism?

1. Fungi
2. Bacteria
3. Algae

Answer:

1. Fungi – Chitin and fungal cellulose
2. Bacteria – Peptidoglycan
3. Algae – Cellulose, mannan and galactan

Question 13.
How imbibition is important for plants?
Significance of imbibition:
Answer:

1. During germination of seeds, imbibition increases the volume of seed enormously and leads to bursting of the seed coat.
2. It helps in the absorption of water by roots at the initial level.

Question 14.
How does nitrogen fixation occurs non – biologically?
Non – Biological nitrogen fixation:
Answer:

1. Nitrogen fixation by chemical process in industry.
2. Natural electrical discharge during lightening fixes atmospheric nitrogen.

PART – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Name few fungal diseases in Humans?
Answer:

Question 16.
List out any 3 significances of seed?
Answer:

1. The seed encloses and protects the embryo for next generation.
2. Seeds of various plants are used as food, both for animals and human.
3. Seeds are the products of sexual reproduction so they provide genetic variations and recombination in a plant.

Question 17.
Write a note on endosymbiont theory?
Answer:
Two eukaryotic organelles believed to be the descendants of the endosymbiotic prokaryotes. The ancestors of the eukaryotic cell engulfed a bacterium and the bacteria continued to function inside the host cell.

Question 18.
What are sieve tubes? Explain?
Answer:
Sieve Elements:
Sieve elements are the conducting elements of the phloem. They are of two types, namely sieve cells and sieve tubes.

Sieve Cells:
These are primitive type of conducting elements found in Pteridophytes and Gymnosperms. Sieve cdlls have sieve areas on their lateral walls only. They are not associated with companion cells.

Sieve Tubes:
Sieve tubes are long tube like conducting elements in the phloem. These are formed from a series of cells called sieve tube elements. The sieve tube elements are arranged one above the other and form vertical sieve tube.

The end wall contains a number of pores and it looks like a sieve. So it is called as sieve plate. The sieve elements show nacreous thickenings on their lateral walls.

They may possess simple or compound sieve plates. The function of sieve tubes are believed to be controlled by campanion cells. In mature sieve tube, Nucleus is absent. It contains a lining layer of cytoplasm.

A special protein (P. Protein = Phloem Protein) called slime body is seen in it. In mature sieve tubes, the pores in the sieve plate arc blocked by a substance called callosc (callose plug).

The conduction of food material takes place Different types of phloem elements through cytoplasmic strands. Sieve tubes occur only in Angiosperms.

Question 19.
Draw and label the structure of lentical?
Answer:

PART – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
List out the salient features of cyanobacteria?
Answer:

• The members of this group are prokaryotes and lack motile reproductive structures.
• The thallus is unicellular in Chroococcus, colonial in Gloeocapsa and filamentous trichome in Nostoc.
• Gliding movement is noticed in some species (Oscillatorid).
• The protoplasm is differentiated into central region called centroplasm and peripheral region bearing chromatophore called chromoplasm.
• The photosynthetic pigments include c – phydcyanin and c – phycoerythrin along with myxoxanthin and myxoxanthophyll.
• The reserve food material is cyanophycean starch.
• In some forms a large colourless cell is found in the terminal or intercalary position called heterocysts. They are involved in nitrogen fixation.
• They reproduce only through vegetative methods and produce akinetes (thick wall dormant cell formed from vegetative cell), hormogonia (a portion of filament get detached and reproduce by cell division), fission and endospores.
• The presence of mucilage around the thallus is characteristic feature of this group. Therefore, this group is also called Myxophyceae.
• Sexual reproduction is absent.
• Microcystis aeruginosa and Anabaena flos – aquae cause water blooms and release toxins and affect the aquatic organism. Most of them fix atmospheric nitrogen and are used as biofertilizers (Example: Nostoc and Anabaena). Spirulina is rich in protein hence it is used as single cell protein.

[OR]

Differentiate between Mitosis and Meiosis?
Answer:

Mitosis	Meiosis
1. One division	1. Two divisions
2. Number of chromosomes remains the same	2. Number of chromosomes is halved
3. Homologous chromosomes line up separately on the metaphase plate	3. Homologous chromosomes line up in pairs at the metaphase plate
4. Homologous chromosome do not pair up	4. Homologous chromosome pairup to form bivalent
5. Chiasmata do not form and crossing over never occurs	5. Chiasmata form and crossingover occurs
6. Daughter cells are genetically identical	6. Daughter cells are genetically different from the parent cells
7. Two daughter cells are formed	7. Four daughter cells are formed

Question 21.
Write in detail about Passive Absorption of minerals salts?
Answer:
Ion-Exchange:
Ions of external soil solution were exchanged with same charged (anion for anion or cation for cation) ions of the root cells. There are two theories explaining this process of ion exchange namely:

1. Contact exchange and
2. Carbonic acid exchange.

1. Contact Exchange Theory:
According to this theory, the ions adsorbed on the surface of root cells and clay particles (or clay micelles) are not held tightly but oscillate within a small volume of space called oscillation volume. Due to small space, both ions overlap each other’s oscillation volume and exchange takes place.

2. Carbonic Acid Exchange Theory:
According to this theory, soil solution plays an important role by acting as a medium for ion exchange. The CO₂ released during respiration of root cells combines with water to form carbonic acid (H₂CO₃).

Carbonic acid dissociates into H⁺ and HCO_32^– in the soil solution. These H⁺ ions exchange with cations adsorbed on clay particles and the catipns from micelles get released into soil solution and gets adsorbed on root cells.

[OR]

Give an detail account on different types of senescence?
Answer:
1. Overall senescence:
This kind of senescence occurs in annual plants when entire plant gets affected and dies.
Example: Wheat and Soybean. It also occurs in few perennials also. Example: Agave and Bamboo.

2. Top senescence:
It occurs in aerial parts of plants. It is common in perennials, underground and root system remains viable.
Example: Banana and Gladiolus.

3. Deciduous senescence:
It is common in deciduous plants and occurs only in leaves of plants, bulk of the stem and root system remains alive.
Example: Elm and Maple.

4. Progressive senescence:
This kind of senescence is gradual. First it occurs in old leaves followed by new leaves then stem and finally root system. It is common in annuals.

Bio – Zoology [Maximum Marks: 35]

PART – I

Answer all the questions. Choose the correct answer. [8 × 1 = 8]

Question 1.
True species are …………………….
(a) Inter breeding
(b) Sharing the same niche
(c) Reproductively isolated
(d) Feeding on the same food
Answer:
(c) Reproductively isolated

Question 2.
Collar cells are found in ……………………
(a) Aschelminthes
(b) Sponges
(c) Cnidaria
(d) Arthropoda
Answer:
(b) Sponges

Question 3.
Match the List – I and List – II.

List –  I	List – II
1. Adenohypopysis	(i) Epinephrine
2. Adrenal medulla	(ii) Somatotropin
3. Parathyroid gland	(iii) Thymosin
4. Thymus gland	(iv) Calcitonin

Answer:
(a) 1 – (i), 2 – (ii), 3 – (iii), 4 – (iv)
(b) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)
(c) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)
(d) 1 – (iv), 2 – (iii), 3 – (ii), 4 – (i)

Question 4.
BMR stands for ……………………….
(a) Body matabolic rate
(b) Basal matabolic rate
(c) Body mass rate
(d) Basal mass rate
Answer:
(b) Basal matabolic rate

Question 5.
Which of the following statement is correct ………………………….
(a) Testosterone is produced by leyding cells under the influence Luteinizing hormone.
(b) Progesterorie is secreted by corpus luteum.
(c) Oestrogen is secreted by Sertolic cells.
(d) Progesterone produced by Corpus Luteum is biologically different from the one produced by placenta.
Answer:
(a) Testosterone is produced by leyding cells under the influence Luteinizing hormone.

Question 6.
Which of the following pair statement is incorrect?
(a) Humans – ureotelic
(b) Birds – uricotelic
(c) Lizards – uricotelic
(d) Whale – Ammonotelic
Answer:
(d) Whale – Ammonotelic

Question 7.
Inflammation of joints due to accumulation of uric acid crystals is called as ………………………..
(a) Myasthenia gravis
(b) Gout
(c) Osteoporosis
(d) Osteomalacia
Answer:
(b) Gout

Question 8.
The olfactory umpulses are transmitted to the ………………………. lobe of brain.
(a) Parietal
(b) Temporal
(c) Occipital
(d) Frontal
Answer:
(d) Frontal

PART – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
What is Regeneration?
Answer:
The ability to regrow the lost parts is called regeneration eg. planaria.

Question 10.
What are the functions of Tapetum lucidum?
Answer:
A reflective layer of tissue called tapetum lucidum, enhances night time vision in most of the animals like cat.

Question 11.
Give an example for longest species of earthworm?
Answer:
Microchaetus rappi is an African giant earthworm, can reach a length of 6.7 meter (22 feet). Drawida nilamburansis is a south Indian (Kerala) species of earthworm and can reach a maximum length up to f meter (3 feet).

Question 12.
Differences between frog and toad?
Answer:

Question 13.
Brief notes on spinal nerves?
Answer:
Spinal nerves:
31 pairs of spinal nerves emerge out from the spinal cord through spaces called the intervertebral foramina found between the adjacent vertebrae. The spinal nerves are named according to the region of vertebral column from which they originate

1. Cervical nerves (8 pairs)
2. Thoracic nerves (12 pairs)
3. Lumbar nerves (5 pairs)
4. Sacral nerves (5 pairs)
5. Coccygeal nerves (1 pair)

Each spinal nerve is a mixed nerve containing both afferent (sensory) and efferent (motor) fibres. It originates as two roots:

1. A posterior dorsal root with a ganglion outside the spinal cord and
2. An anterior ventral root with no external ganglion.

Question 14.
Draw the structure of Nephron?
Answer:

PART – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
What is phylogenetic classification or cladistics?
Answer:
The classification of organisms based on evolutionary and genetic relationship among them is known as phylogenetic classification.

Question 16.
Compare Schizocoelom with Enterocoelom?
Answer:

Schizocoelom	Enterocoelom
1. The coelom which is formed by splitting of mesoderm is called Schizocoelom.	1. The coelom which is formed from the mesodermal pouches of archenteron is called entercoelom.
2. It is found in lower vertebrates like annelids, arthopods, molluscs.	2. It is found in echinodermis, hemichordates and chordates.

Question 17.
Head of cockroach is called hypognathous. why?
Answer:
The mouth parts of cockroach are directed downwards. The head is small, triangular lies at right angle to the longitudinal body axis. Hence it is called hypognatus.

Question 18.
Draw the diagram and label the parts of micro villi?
Answer:

Question 19.
What are the advantages of artificial insemination?
Answer:

1. It increases the rate of conception
2. It avoids genital diseases
3. Semen can be collected from injured bulls which have desirable traits
4. Superior animals located apart can be bred successfully

PART – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
Write brief notes on Diploblastic and Triploblastic organisation with example?
Answer:
During embryonic development, the tissues and organs of animals originate from two or three embryonic germ layers. On the basis of the origin and development, animals are classified into two categories: Diploblastic and Triploblastic.

Animals in which the cells are arranged in two embryonic layers, the external ectoderm, and internal endoderm are called diploblastic animals. In these animals the ectoderm gives rise to the epidermis (the outer layer of the body wall) and endoderm gives rise to gastrodermis (tissue lining the gut cavity). An undifferentiated layer present between the ectoderm and endoderm is the mesoglea. (Corals, Jellyfish, Sea anemone)

Animals in which the developing embryo has three germinal layers are called triploblastic animals and consists of outer ectoderm (skin, hair, neuron, nail, teeth, etc), inner endoderm (gut, lung, liver) and middle mesoderm (muscle, bone, heart). Most of the triploblastic animals show organ system level of organisation (Flat worms to Chordates).

[OR]

Explain the reproductive system of frog?
Answer:
Reproductive system:
The male frog has a pair of testes which are attached to the kidney and the dorsal body wall by folds of peritonium called mesorchium. Vasa efferentia arise from each testis. They enter the kidneys on both side and open into the bladder canal. Finally, it communicates with the urinogenital duct that comes out of kidneys and opens into the cloaca.

Female reproductive system consists of paired ovaries, attached to the kidneys, and dorsal body wall by folds of peritoneum called mesovarium. There is a pair of coiled oviducts lying on the sides of the kidney. Each oviduct opens into the body-cavity at the anterior end by a funnel like opening called ostia.

Unlike the male frog, the female frog has separate genital ducts distinct from ureters. Posteriorly the oviducts dilated to form ovisacs before they open into cloaca. Ovisacs store the eggs temporarily before they are sent out through the cloaca. Fertilization is external.

Within few days of fertilization, the eggs hatch into tadpoles. A newly hatched tadpole lives off the yolk stored in its body. It gradually grows larger and develops three pairs of gills.

The tadpole grows and metamorphosis into an air – breathing carnivorous adult frog. Legs grow from the body, and the tail and gills disappear. The mouth broadens, developing teeth and jaws, and the lungs become functional.

Question 21.
What are the disorders related to the excretory system?
Answer:
Urinary tract infection:
Female’s urethra is very short and its external opening is close to the anal opening, hence improper toilet habits can easily carry faecal bacteria into the urethra. The urethral mucosa is continuous with the urinary tract and the inflammation of the urethra (urethritis) can ascend the tract to cause bladder inflammation (cystitis) or even renal inflammation (pyelitis or pyelonephritis).

Symptoms include dysuria (painful urination), urinary urgency, fever and sometimes cloudy or blood tinged urine. When the kidneys are inflammed, back pain and severe headache often occur. Most urinary tract infections can be treated by antibiotics.

Renal Failure (Kidney Failure):
Failure of the kidneys to excrete wastes may lead to accumulation of urea with marked reduction in the urine output. Renal failure are of two types, Acute and chronic renal failure.

In acute renal failure the kidney stops its function abruptly, but there are chances for recovery of kidney functions. In chronic renal failure there is a progressive loss of function of the nephrons which gradually decreases the function of kidneys.

Uremia:
Uremia is characterized by increase in urea and other non – protein nitrogenous substances like uric acid and creatinine in blood. Normal urea level in human blood is about 17 – 30mg/100mL of blood. The urea concentration rises as 10 times of normal levels during chronic renal failure.

Renal calculi:
Renal calculi, also called renal stone or kidney stone or nephrolithiasis, is the formation of hard stone like masses in the renal tubules of renal pelvis.

It is mainly due to the accumulation of soluble crystals of salts of sodium oxalates and certain phosphates. This result in severe pain called “renal colic pain” and can cause scars in the kidneys. Renal stones can be removed by techniques like pyleothotomy or lithotripsy.

Glomerulonephritis:
It is also called Bright’s disease and is characterized by inflammation of the glomeruli of both kidneys and is usually due to post-streptococcal infection that occurs in children. Symptoms are haematuria, proteinuria, salt and water retention, oligouria, hypertension and pulmonary oedema.

[OR]

Describe the life cycle of Bombyx mori?
Answer:
Life cycle of Bombyx mori:
The adult of Bombyx mori is about 2.5 cm in length and pale creamy white in colour. Due to heavy body and feeble wings, flight is not possible by the female moth.

This moth is unisexual in nature and does not feed during its very short life period of 2 – 3 days. Just after emergence, male moth copulates with female for about 2 – 3 hours and if not separated, they may die after few hours of copulating with female.

Just after copulation, female starts egg laying which is completed in 1-24 hours. A single female moth lays 400 to 500 eggs depending upon the climatic conditions.

Two types of eggs are generally found namely diapause type and non-diapause type. The diapause type is laid by silkworms inhabiting the temperate regions, whereas silkworms belonging to subtropical regions like India lay non – diapause type of eggs.

The eggs after ten days of incubation hatch into larva called as caterpillar. The newly hatched caterpillar is about 3 mm in length and is pale, yellowish-white in colour. The caterpillars are provided with well developed mandibulate type of mouth-parts adapted to feed easily on the mulberry leaves.

After 1^st, 2^2nd, 3^rd and 4^th moultings caterpillars get transformed into 2^nd, 3^rd, 4^th and 5^th instars respectively. It takes about 21 to 25 days after hatching. The fully grown caterpillar is 7.5 cm in length. It develops salivary glands, stops feeding and undergoes pupation.

The caterpillars stop feeding and move towards the comer among the leaves and secretes a sticky fluid through their silk gland. The secreted fluid comes out through spinneret (a narrow pore situated on the hypopharynx) and takes the form of long fine thread of silk which hardens on exposure to air and is wrapped around the body of caterpillar in the forms of a covering called as cocoon.

It is the white coloured bed of the pupa whose outer threads are irregular while the inner threads are regular. The length of continuous thread secreted by a caterpillar for the formation of cocoon is about 1000 – 1200 metres which requires 3 days to complete. The pupal period lasts for 10 to 12 days and the pupae cut through the cocoon and emerge into adult moth.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3

Solve the following problems by using Factor Theorem:

Question 1.
Show that \(\left| \begin{matrix} x & a & a \\ a & x & a \\ a & a & x \end{matrix} \right| \) = (x – a)² (x + 2a)
Answer:

By putting x = a , we have three rows of |A| are identical. Therefore (x – a)² is a factor of |A|
Put x = – 2a in |A|

∴ x + 2a is a factor of |A|. The degree of the product of the factors (x – a)² (x + 2a) is 3.
The degree of tfie product of the leading diagonal elements x . x . x is 3.
∴ The other factor is the contant factor k.

a³ [ – 1 (1 – 1) – 1 ( – 1 – 1) + 1 (1 + 1)] = k . 4a³
a³ [o + 2 + 2 ] = 4 ka³
4a³ = 4 ka³
k = 1

 

Free handy Remainder Theorem Calculator tool displays the remainder of a difficult polynomial expression in no time.

Question 2.
Show that \(\left| \begin{matrix} b\quad +\quad c & a\quad -\quad c & a\quad -\quad b \\ b\quad -\quad c & c\quad +\quad a & b\quad -\quad a \\ c\quad -\quad b & c\quad -\quad a & a\quad +\quad b \end{matrix} \right| \) = 8 abc
Answer:


since two columns identical
= bc × 0 = 0
∴ a – 0 is a factor. That is, a is a factor.
Put b = 0 in |A|

since two columns identical
= ca × 0 = 0
∴ b – 0 is a factor. That is, a is a factor.
Put c = 0 in |A|

since two columns identical
= ab × 0 = 0
∴ c – 0 is a factor. That is, c is a factor.
The degree of the product of the factors abc is 3.
The degree of the product of leading diagonal elements (b + c) (c + a) (a + b) is 3.
∴ The other factor is the constant factor k.

Question 3.
Solve that \(\left| \begin{matrix} x\quad +\quad a & b & c \\ a & x\quad +\quad b & c \\ a & b & x\quad +\quad c \end{matrix} \right| \) = 0
Answer:

Put x = 0

x = 0 satisfies the given equation. x = 0 is a root of the given equation, since three rows are identical. x = 0 is a root of multiplicity 2. Since the degree of the product of the leading diagonal elements (x + a) (x + b) (x + c) is 3. There is one more root for the given equation.
Put x = – (a + b + c)

∴ x = – (a + b + c) satisfies the given equation.
Hence, the required roots of the given equation are x = 0, 0 , – (a + b + c)

Question 4.
Show that \(\left| \begin{matrix} b\quad +\quad c & a & { a }^{ 2 } \\ c\quad +\quad a & b & { b }^{ 2 } \\ a\quad +\quad b & c & { c }^{ 2 } \end{matrix} \right| \) = (a + b + c) (a – b) (b – c) (c – a)
Answer:

Since two rows are idenctical
|A| = 0
since two rows are idenctical
|A| = 0
∴ a – b is a factor of | A |. The given determinant is in cyclic symmetric form in a , b and c. Therefore, b – c and c – a are also factors. The degree of the product of the factors (a – b) (b – c) (c – a) is 3 and the degree of the product of the leading diagonal elements (b + c) . b . c² is 4.
Therefore, the other factor is k (a + b + c).

5(18 – 12) – 1(36 – 12) + 1(12 – 6) = 12k
5 × 6 – 24 + 6 = 12k
30 – 24 + 6 = 12k
12 = 12 ⇒ k = 1

Question 5.
Solve \(\left| \begin{matrix} 4\quad -\quad x & 4\quad +\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad -\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad +\quad x & 4\quad -\quad x \end{matrix} \right| \) = 0
Answer:

Put x = 0

∴ x = 0 satisfies the given equation. Hence x = 0 is a root of the given equation. since three rows are identical, x = 0 is a root of multiplicity 2.

Since the degree of the product of the leading diagonal elements (4 – x) (4 – x) (4 – x) is 3. There is one more root for the given equation.


∴ x = – 12 is a root of the given equation.
Hence, the required roots are x = 0 , 0 , – 12

Question 6.
Show that \(\left| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \end{matrix} \right| \) = (x – y) (y – z) (z – x)
Answer:

|A| = 0 since two columns identical
∴ x – y is a factor of A. The given determinant is in the cyclic symmetric form in x, y, and z. Therefore, y – z and z – x are also factors of |A|.

The degree of the product of the factors (x – y) (y – z) (z – x) is 3 and the degree of the product of the leading diagonal elements 1, y, z² is 3. Therefore, the other factor is the constant factor k.

Put x = 0, y = 1, z = -1 we get

Expanding along the first column
1 (1 + 1) = 2k
2 = 2k ⇒ k = 1

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Having trouble working out with the Binomial theorem? You’ve come to the right place, our Binomial Expansion Calculator is here to save the day for you.

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
(i) \(\frac{1}{5+x}\)
Answer:

(ii) \(\frac{2}{(3+4 x)^{2}}\)
Answer:

(iii) (5 + x²)^2/3
Answer:

(iv) \((x+2)^{-\frac{2}{3}}\)
Answer:

Expanding binomials calculator is a free online tool that displays the expansion of the given binomial term.

Question 2.
Find \(\sqrt[3]{1001}\) approximately. (two decimal places)
Answer:

Question 3.
Prove that \(\sqrt[3]{x^{3}+6}-\sqrt[3]{x^{3}+3}\) is approximately equal to \(\frac{1}{x^{2}}\) when x is sufficiently large.
Answer:

Question 4.
Prove that \(\sqrt{\frac{1-x}{1+x}}\) is approximately equal to 1 – x + \(\frac{x^{2}}{2}\) when x is very small.
Answer:

Question 5.
Write the first 6 terms of the exponential series
(i) e^5x
Answer:

(ii) e^-2x
Answer:

(iii) e^x/2
Answer:

Question 6.
Write the first 4 terms of the logarithmic series.
(i) log (1 + 4x)
(ii) log (1 – 2x)
(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)
(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)
Find the intervals on which the expansions are valid.
Answer:
(i) log ( 1 + 4x )

(ii) log (1 – 2x)

(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)

(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)

Question 7.

Answer:

Question 8.

Answer:

Question 9.
Find the coefficient of x⁴ in the expansion of \(\frac{3-4 x+x^{2}}{e^{2 x}}\)
Answer:

Question 10.
Find the value of
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 3 Books of Prime Entry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 3 Books of Prime Entry

11th Accountancy Guide Books of Prime Entry Text Book Back Questions and Answers

prime factorization of 38 are numbers that, when multiplied in pairs give the product as 38.

I. Choose the correct answer.

Question 1.
Accounting equation signifies …………….
(a) Capital of a business is equal to assets
(b) Liabilities of a business are equal to assets
(c) Capital of a business is equal to liabilities
(d) Assets of a business are equal to the total of capital and liabilities
Answer:
(d) Assets of a business are equal to the total of capital and liabilities

Question 2.
‘Cash withdrawn by the proprietor from the business for his personal use’ causes …………….
(a) Decrease in assets and decrease in owner’s capital
(b) Increase in one asset and decrease in another asset
(c) Increase in one asset and increase in liabilities
(d) Increase in asset and decrease in capital
Answer:
(a) Decrease in assets and decrease in owner’s capital

Question 3.
A firm has assets of Rs. 1,00,000 and the external liabilities of Rs. 60,000. Its capital would be ……………….
(a) Rs. 1,60,000
(b) Rs. 60,000
(c) Rs. 1,00,000
(d) Rs. 40,000
Answer:
(d) Rs. 40,000

Question 4.
The incorrect accounting equation is ………………….
(a) Assets = Liabilities + Capital
(b) Assets = Capital + Liabilities
(c) Liabilities = Assets + Capital
(d) Capital = Assets – Liabilities
Answer:
(c) Liabilities = Assets + Capital

Question 5.
Accounting equation is formed based on the accounting principle of ………………
(a) Dual aspect
(b) Consistency
(c) Going concern
(d) Accrual
Answer:
(a) Dual aspect

Question 6.
Real account deals with ……………..
(a) Individual persons
(b) Expenses and losses
(c) Assets
(d) Incomes and gains
Answer:
(c) Assets

Question 7.
Which one of the following is representative personal account?
(a) Building A/c
(b) Outstanding salary A/c
(c) Mahesh A/c
(d) Balan & Co
Answer:
(b) Outstanding salary A/c

Question 8.
Prepaid rent is a ……………….
(a) Nominal A/c
(b) Personal A/c
(c) Real A/c
(d) Representative personal A/c
Answer:
(d) Representative personal A/c

Question 9.
Withdrawal of cash from business by the proprietor should be credited to
(a) Drawings A/c
(b) Cash A/c
(c) Capital A/c
(d) Purchases A/c
Answer:
(b) Cash A/c

Question 10.
In double entry system of book keeping, every business transaction affects
(a) Minimum of two accounts
(b) Same account on two different dates
(c) Two sides of the same account
(d) Minimum three accounts
Answer:
(a) Minimum of two accounts

II. Very Short Answer Type Questions

Question 1.
What are source documents?
Answer:
“Source documents are the authentic evidence of financial transactions. These documents show the nature of the transaction, the date, the amount, and the parties involved. Source documents include cash receipt, invoice, debit note, credit note, pay-in-slip, salary bills, wage bills, cheque record slips, etc.

Question 2.
What Is the accounting equation?
Answer:
The relationship of assets with that of liabilities to outsiders and to owners in the equation form is known as the accounting equation.
The accounting equation is a mathematical expression which shows that the total of assets is equal to the total of liabilities and capital.
This is based on the dual aspect concept of accounting.
This means that the total claims of outsiders and the proprietor against a business enterprise will always be equal to the total assets of the business enterprise.
Capital + Liabilities = Assets

Question 3.
Write any one transaction which

1. Decreases the assets and decreases the liabilities
2. Increases one asset and decreases another asset

Answer:

1. Paid creditors
2. Gash sales

Question 4.
What is meant by journalizing?
Answer:
The process of analyzing the business transactions under the heads of debit and credit and recording them in the journal is called journalizing.

Question 5.
What is the real account?
Answer:
All accounts relating to tangible and intangible properties and possessions are called real accounts.

Question 6.
How are personal accounts classified?
Answer:

Question 7.
State the accounting rule for a nominal account.
Answer:

Question 8.
Give the golden rules of double-entry accounting system.
Answer:

III. Short Answer Questions

Question 1.
Write a brief note on the accounting equation approach of recording transactions.
Answer:
The relationship of assets with that of liabilities to outsiders and to owners in the equation form is known as the accounting equation.
The accounting equation is a mathematical expression which shows that the total of assets is equal to the total of liabilities and capital.
This is based on the dual aspect concept of accounting.

This means that the total claims of outsiders and the proprietor against a business enterprise will always be equal to the total assets of the business enterprise. The equation is given under:
Capital + Liabilities = Assets
Capital can also be called owner’s equity and liabilities as outsider’s equity.

As the revenues and expenses will affect capital, the expanded equation may be given as under:
Assets = Liabilities + Capital + Revenues – Expenses
Accounts are classified into five categories: (i) Asset account, (ii) Liability account, (iii) Capital account, (iv) Revenue account and (v) Expense account as follows:

Question 2.
What is an Account? Classify the accounts with suitable examples.
Answer:
Every transaction has two aspects and each aspect affects a minimum of one account.
An account is the basic unit of identification in accounting.
A ledger account is a summary of relevant transactions at one place relating to a particular head.
The account is the systematic presentation of all material information regarding a particular person or item at one place, under one head.

Classification of Accounts:
Personal Account;
Accounting relating to persons is called a personal account. The personal account may be a natural, artificial, or representative personal account.
E.g: Malini account, BHEL account, Prepaid Rent account.

Impersonal Account:
All accounts which do not affect persons are called impersonal accounts. These are further classified into Real and Nominal Accounts.
E.g; Building account, Goodwill, Salaries account.

Question 3.
What are the three different types of personal accounts?
Answer:
Under the double-entry system of book-keeping, for the purpose of recording the various financial transactions, the accounts are classified as personal accounts and impersonal accounts.

1. Natural person’s account: Natural person means human beings. Example: Vinoth account, Malini account.
2. Artificial person’s account: Artificial person refers to persons other than human beings recognized by law as persons. They include business concerns, charitable institutions, etc. Example: BHEL account, Bank account.
3. Representative personal accounts: These are the accounts which represent persons natural or artificial or a group of persons. Example: Outstanding salaries account, Prepaid rent account. When expenses are outstanding, it is payable to a person. Hence, it represents a person.

Question 4.
What is the accounting treatment for insurance premium paid on the of the proprietor?
Answer:
The proprietor has to pay an agreed premium on a monthly or annual basis. The amount should be treated as prepayment which is considered a current asset.
The following entry should be made:
Insurance A/c Dr –
To Cash/Bank A/c –
(Being insurance premium paid.)

Question 5.
State the principles of the double-entry system of book-keeping.
Answer:
Following are the principles of the double-entry system:

1. In every business transaction, there are two aspects.
2. The two aspects involved are the benefit or value receiving aspect and the benefit or value giving aspect.
3. These two aspects involve a minimum of two accounts; at least one debit and at least one credit.
4. For every debit, there is a corresponding and equivalent credit.
5. If one account is debited the other account must be credited.

Question 6.
Briefly explain about steps in journalizing.
Answer:
The process of analyzing the business transactions under the heads of debit and credit and recording them in the journal is called journalizing. An entry made in the journal is called a journal entry. The following steps are followed in journalizing:

Analyze the transactions and identify the accounts (based on aspects) which are involved in the transaction.

Classify the above accounts under Personal account, Real account or Nominal account, Apply the rules of debit and credit for the above two accounts.

Find which account is to be debited and which account is to be credited by the application of rules of the double-entry system. Record the date of the transaction in the date column.

Enter the name of the account to be debited in the particulars column very close to the left hand side of the particulars column followed by the abbreviation ‘Dr/ at the end in the same line. Against this, the amount to be debited is entered in the debit amount column in the same line.

Write the name of the account to be credited in the second line starting with the word ‘To’ prefixed a few spaces away from the margin in the particulars column. Against this, the amount to be credited is entered in the credit amount column in the same line.

Write the narration within brackets in the next line in the particulars column.

Question 7.
What is the double-entry system? State its advantages.
Answer:
A double-entry system of book keeping is a scientific and complete system of recording the financial transactions of an organization. According to this system, every transaction has a two-fold effect. That is, there are two aspects involved, namely, receiving aspect and the giving aspect. It is denoted by debit (Dr.) and credit (Cr.). The basic principle of the double-entry system is that for every debit there must be an equivalent and corresponding credit. Debit denotes an increase in assets or expenses or a decrease in liabilities, income, or capital. Credit denotes an increase in liabilities, income, or capital or a decrease in assets or expenses.

IV. Exercises

Question 1.
Complete the accounting equation.
(a) Assets = Capital + Liabilities
1,00,000 = 80,000 + ?
Answer:
(a) Liabilities = Assets – Capital
Liabilities = 1,00,000 – 80,000
Liabilities = 20,000

(b) Assets = Capital + Liabilities
2,00,000 = ? + 40,000
Answer:
Capital = Assets – Liabilities
Capital = 2,00,000 – 40,000
Capital = 1,60,000

(c) Assets = Capital + Liabilities
? = 1,60,000 + 80,000
Answer:
Assets = Capital + Liabilities
Assets = 1,60,000 + 80,000
Assets = 2,40,000

Question 2.
For the following transactions, show the effect on the accounting equation.
(a) Raj Started business with cash – 40,000
(b) Opened bank account with a deposit of – 30,000
(c) Bought goods from Hari on credit for – 12,000
(d) Raj withdrew cash for personal use – 1,000
(e) Bought furniture by using a debit card for – 10,000
(f) Sold goods to Murugan and cash received – 6,000
(g) Money is withdrawn from bank for office use – 1,000
Answer:
(a) Raj started the business with cash Rs 40,000
Increasing in capital and increasing in Asset Effects:
Cash comes in → Increase in asset
Capital provided by the owner → Increase in capital of owner
Capital = Asset
Capital = Cash
(+)Rs. 40,000 = (+)Rs. 40,000

(b) Opened bank account with a deposit of Rs 30,000
Decrease in one asset and increase in another asset Effects:
Bank is the receiver → Increases in Assets
Cash goes out → Decrease in Asset
Capital = Asset
Capital = Bank + cash
Capital = (+) Rs. 30,000 + (-) Rs. 30,000

(c) Bought goods from Hari on credit for Rs. 12,000
Increasing Asset and Increase in Liabilities Effects:
Stock comes in → Increase in Asset
Creditors arises → Increase in Liabilities
Asset = Liabilities
Stock = Creditors
(+) Rs. 12,000 = (+) Rs. 12,000

(d) Raj withdrew cash for personal use Rs 1,000
Decrease in Asset and decrease in capital Effects:
Cash goes out → Decrease in cash
Drawings proprietor is the receiver → Decrease in capital
Asset = Capital
Cash = Capital
(-) Rs. 1,000 = (-) Rs. 1,000

(e) Bought furniture by using debit card for Rs 10,000
Decrease in one asset and Increase in another asset Effects:
Furniture comes in → Increase in Furniture
Bank is giver → Decrease in a Bank
Asset = Liabilities
Furniture + Bank = Liabilities
(+) Rs. 10,000 + (-) Rs. 10,000 = Liabilities

(f) Sold goods to Murugan and cash received Rs 6,000
Decrease in one asset and increase in another asset Effects:
Cash comes in → Increase in cash
<$> Stock goes out → Decrease in stock
Asset = Liabilities
Cash + Stock = Liabilities
(+) Rs. 6,000 + (-) Rs. 6,000 = Liabilities

(g) Money is withdrawn from bank for office use Rs 1,000
Decrease in one asset and increase in another asset Effects:
Cash comes in → Increase in Cash
The bank is a giver → Decrease in Bank
Asset = Liabilities
Cash + Bank = Liabilities
(+) Rs. 1,000 + (-) Rs. 1,000 = Liabilities

Accounting Equation:

Question 3.
Prepare accounting equation for the following transactions,
(a) Murugan commenced business with cash Rs 80,000
(b) Purchased goods for cash Rs 30,000
(c) Paid salaries by cash Rs 5,000
(d) Bought goods from Kumar for Rs 5,000 and deposited the money in CDM.
(e) Introduced additional capital of Rs 10,000
Answer:
Accounting Equation:

Question 4.
What will be the effect of the following on the accounting equation?
(a) Sunil started the business with Rs 1,40,000 cash and goods worth Rs 60,000
(b) Purchased furniture worth Rs 20,000 in cash
(c) Depredation on furniture Rs 800
(d) Deposited into bank Rs 40,000
(e) Paid electricity charges through net banking Rs 500
(f) Sold goods to Ravi costing Rs 10,000 for Rs 15,000
(g) Goods returned by Ravi Rs 5,000
Answer:
(a) Sunil started the business with Rs 1,40,000 cash and goods worth Rs 60,000
Transactions affecting more than two accounts Effects:
Cash comes in → Increase in asset
Stock comes in → an Increase in asset
Capital provide by the owner → Increase in capital
Capital = Asset
Capital = Cash + Stock
(+) 2,00,000 = (+) 1,40,000 + (+) 60,000

(b) Purchased furniture worth Rs 20,000 in cash
Decrease in one asset and Increase in Another asset Effects:
Furniture comes in → Increase in asset
Cash goes out → Decrease in asset
Liabilities = Assets
Liabilities = Cash + Furniture
Liabilities = (-) 20,000 + (-) 20,000

(c) Depreciation on furniture Rs 800
Decrease in asset and decrease in capital effects:
Furniture goes out → Decrease in Asset
Capital → Decrease in Capital
Capital = Asset
Capital = Furniture
(-) 800 = (-) 800

(d) Deposited into bank Rs 40,000
Increase in one Asset and decrease in another asset Effects:
Cash goes out → Decrease in asset
The bank is a receiver → Increase in asset
Liability = Asset
Liability = Cash + bank
= (-) 40,000 + (+) 40,000

(e) Paid electricity charges through net banking Rs 500
decrease in Asset and Decrease in capital Effects:
The bank is a given → Decrease in Asset
Capital (Expenses) → Decrease in capital
Capital = Asset
Capital = Bank
(-) 500 = (-) 500

(f) Sold goods to Ravi costing Rs 10,000 for Rs 15,000
Increase in asset and decreases in another asset and Increase in Capital Effects:
Creditors → Increases in Liabilities
Stock goes out → Decreases in asset
Capital (Incom(e) → Increase in capital
Liabilities + Capital = Asset
Creditors + Capital = Stock
(+) 15,000 + (+) 5,000 = (-) 5,000

(g) Goods returned by Ravi Rs 5,000
Increase in asset and decrease in Liabilities Effects:
Stock comes in → an Increase in asset
Reducing in creditors → Decreases in Liabilities
Liability = Asset
Creditors = Stock
(-) 5,000 = (+) 5,000
Accounting Equation:

Question 5.
Create an accounting equation on the basis of the following transactions.
(i) Rakesh started the business with a capital of Rs 1,50,000
(ii) Deposited money with the bank Rs 80,000
(iii) Purchased goods from Mahesh and paid through credit card Rs 25,000
(iv) Sold goods (costing Rs 10,000) to Mohan for Rs 14,000 who pays through debit card
(v) Commission received by cheque and deposited the same in the bank Rs 2,000
(vi) Paid office rent through ECS Rs 6,000
(vii) Sold goods to Raman for Rs 15,000 of which Rs 5,000 was received at once
Answer:
Accounting Equation:

Question 6.
Create an accounting equation on the basis of the following transactions.
(i) Started business with cash 80,000 and goods Rs 75,000
(ii) Sold goods to Shanmugam on credit for Rs 50,000
(iii) Received cash from Shanmugam in full settlement Rs 49,000
(iv) Salary outstanding Rs 3,000
(v) Goods costing Rs 1,000 given as charity
(vi) Insurance premium paid Rs 3,000
(vii) Out of insurance premium paid, prepaid is Rs 500
Answer:
Accounting Equation:

Question 7.
Create an accounting equation on the basis of the following transactions:
(i) Opening balance on 1st January, 2018 cash Rs 20,000; stock Rs 50,000 and bank Rs 80,000
(ii) Bought goods from Suresh Rs 10,000 on credit
(iii) Bank charges Rs 500
(iv) Paid Suresh Rs 9,700 through credit card in full settlement.
(v) Goods purchased on credit from Philip for Rs 15,000
(vi) Goods returned to Philip amounting to Rs 4,000
Answer:

Question 8.
Enter the following transactions in the journal of Manohar who is dealing in textiles. 2018 March
Manohar started the business with cash – 60,000
Purchased furniture for cash – 10,000
Bought goods for cash – 25,000
Bought goods from Kamaiesh on credit – 15,000
Sold goods for cash – 28,000
Sold goods to Hari on credit – 10,000
Paid Kamaiesh – 12,000
Paid rent – 500
Received from Hari – 8,000
Withdrew cash for personal use – 4,000
Answer:
journal of Mr. Manohar

Question 9.
Pass journal entries in the books of Sasi Kumar who is dealing in automobiles. 2017 Oct.
1. Commenced business with goods – 40,000
3. Cash introduced in the business – 60,000
4. Purchased goods from Arul on credit – 70,000
6. Returned goods to Arul – 10,000
10. Paid cash to Arul on account – 60,000
15. Sold goods to Chandar on credit – 30,000
18. Chandar returned goods worth – 6,000
20. Received cash from Chandar in full settlement – 23,000
25. Paid salaries through ECS – 2,000
30. Sahil took for personal use goods worth – 10,000
Answer:
journal of Mr. Sasi Kumar:

Question 10.
Pass Journal entries in the books of Hart who is deafer in sports items. 2017 Jan.
1. Commenced business with cash 50,000
2. Purchased goods from Subash on credit of 20,000
4. Sold goods to Ramu on credit of 15,000
8. Ramu paid the amount through cheque
10. Cheque received from Ramu is deposited with the bank
15. Sports items purchased from Gopal on credit 10,000
18. Paid rent for the proprietor’s residence 1,500
20. Paid Gopal in a full settlement after deducting a 5% discount
25. Paid Subash Rs 4,750 and discount received 250
28. Paid by cash: wages Rs 500; electricity charges Rs 3,000 and trade expenses 1,000
Answer:
journal of Hari:

Question 11.
Karthick opened a provisions store on 1st April 2017, Journalise the following transactions in this books:
2017 April?
1. Paid into the bank for opening a current account – 2,00,000
3. Goods purchased by cheque – 40,000
5. Investments made in securities – 40,000
6. Goods sold to Radha for Rs 20,000 and cheque received and deposited into bank
7. Amount withdrawn from bank for office use – 15,000
10. Purchased goods from Kamala and cash deposited in CDM – 10,000
12. Sold goods to Vanitha who paid through the debit card – 10,000
15. Interest on securities directly received by the bank – 1,000
20. Insurance paid by the bank as per standing instructions – 2,000
25. Sales made to Kunal who made payment through CDM – 6,000
Answer:
journal of Karthik:

Question 12.
Journalise the following transactions in the books of Ramesh who is dealing in computers:
2018, May 1. Ramesh started business with cash Rs 3,00,000, Goods Rs 80,000 and Furniture Rs 27,000.
2. Money deposited into bank Rs 2,00,000
3. Bought furniture from M/s Jayalakshmi Furniture for Rs 28,000 on credit.
4. Purchased goods from Asohan for Rs 5,000 by paying through debit card.
5. Purchased goods from Guna and paid through net banking for cash Rs 10,000
6. Purchased goods from Kannan and paid through credit card Rs 20,000
7. Purchased goods from Shyam on credit for Rs 50,000
8. Bill drawn by Shyam was accepted for Rs 50,000
9. Paid half the amount owed to M/s Jayalakshmi Furniture by cheque
10. Shyam’s bill was paid
Answer:
journal of Mr. Ramesh

Question 13.
Journalise the following transactions in the books of Sundar who is a bookseller.
Dec 2017.
1. Commenced business with cash – 2,00,000
2. Bought goods from X and Co. on credit – 80,000
4. Opened a bank account with – 50,000
5. Sold goods to Naresh who paid the amount through net banking – 5,000
6. Sold goods to Devi who paid through credit card – 7,000
7. Sold goods to Ashish on credit – 700
8. Money withdrawn from the bank through ATM for office use – 1,000
9. Purchased furniture and paid through the debit card – 2,000
10. Salaries paid by cash – 6,000
11. Furniture purchased from Y for Rs 25,000 and advance given – 5,000
Answer:
Journal of Mr. Sundar

Question 14.
Raja has a hotel. The following transactions took place in his business. Journalize them.
Jan. 1. Started business with cash – 3,00,000
2. Purchased goods from Rajiv on credit – 1,00,000
3. Cash deposited with the bank – 2,00,000
20. Borrowed loan from the bank – 1,00,000
22. Withdrew from the bank for personal us – 800
23. Amount paid to Rajiv in full settlement through NEFT – 99,000
25. Paid club bill of the proprietor by cheque – 200
26. Paid electricity bill of the proprietor’s house through the debit card – 2,000
31. Lunch provided at free of cost to a charity – 1,000
31. Bank levied charges for locker rent – 1,000
Answer:
journal of Mr. Raja

Question 15.
From the following transactions of Shyam, a stationery dealer, pass journal entries for the month of August 2017,
Aug 1. Commenced business with cash Rs 4,00,000, Goods Rs 5,00,000
2. Sold goods to A and money received through RTGS 12,50,000
3. Goods sold to Z on credit for Rs 20,000
5. Bill drawn on Z and accepted by him Rs 20,000
8. Bill received from Z is discounted with the bank for 119,000
10. Goods sold to M on credit Rs 12,000
12. Goods distributed as free samples for Rs 2,000
16. Goods taken for office use Rs 5,000
17. M became insolvent and only 0.80 per rupee is received in the final settlement
20. Bill of Z discounted with the bank is dishonored
Answer:
Journal of Mr. Shyam

Question 16.
Mary is a cement dealer having a business for more than 5 years. Pass journal entries in her books for the period of March 2018,
1. Cement bags bought on credit from Sibi – 20,000
2. Electricity charges paid through net banking – 500
3. Returned goods bought from Sibi – 5,000
4. Cement bags are taken for personal use – 1,000
5. Advertisement expenses paid – 2,000
6. Goods sold to Mano – 20,000
7. Goods returned by Mano – 5,000
8. Payment received from Mano through NEFT
Answer:
Journal of Mr. Mary

11th Accountancy Guide Conceptual Framework of Accounting Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
A debit note is also called…………….
(a) Credit note
(b) Debit memo
(c) Vouchers
(d) Cash memo
Answer:
(b) Debit memo

Question 2.
Journal is a book of ……….
(a) Original Entry
(b) Final Entry
(c) Assets
(d) None of the above
Answer:
(a) Original Entry

Question 3.
When cash or cheque is deposited in a bank, a form is to be filled by a customer is called…………….
(a) Pay – in – slip
(b) Voucher
(c) Cash memo
(d) Invoice
Answer:
(a) Pay – in – slip

Question 4.
Accounts of persons with whom the business deals is known as ………….
(a) Personal A/c
(b) Real A/c
(c) Nominal A/c
(d) Profit & Loss A/c
Answer:
(a) Personal A/c

Question 5.
Outsider’s equity is otherwise called as …………….
(a) Capital
(b) liabilities
(c) debtors
(d) Assets
Answer:
(b) liabilities

Question 6.
Which one of the following equations is correct?
(a) Owner’s Equity = Liability + Asset
(b) Owner’s Equity = Asset – Liability
(c) Liability = Owner’s Equity + Asset
(d) Asset = Owner’s Equity – Liability
Answer:
(b) Owner’s Equity = Asset – Liability

Question 7.
Goodwill is an example of accounts …………….
(a) Real
(b) Nominal
(c) Tangible real
(d) Intangible real
Answer:
(d) Intangible real

Question 8.
Atul purchased a car for Rs. 5,00,000, by making a down payment of Rs. 1,00,000 and signing an Rs. 4,00,000 bill payable due in 60 days. As a result of this transaction
(a) Total assets increase by Rs. 5,00,000
(b) Total liabilities increase by Rs.4,00,000
(c) Total assets increase by Rs. 4,00,000
(d) Total assets increased by Rs. 4,00,000 with a corresponding increase in liabilities by Rs. 4,00,000
Answer:
(d) Total assets increased by Rs. 4,00,000 with a corresponding increase in liabilities by Rs. 4,00,000

Question 9.
Journal means …………….
(a) daily
(b) monthly
(c) yearly
(d) weekly
Answer:
(a) daily

Question 10.
Amount paid to a supplier is ……….
(a) An event
(b) A Transaction
(c) Both (a) and (b)
(d) Neither (a) Nor (b)
Answer:
(c) Both (a) and (b)

Question 11.
Goods worth Rs. 2,000 were distributed as free samples in the market. The journal entry will be
(a) Drawings A/c Dr. 2,000
To Purchases A/c 2,000
(b) Sales A/c Dr. 2,000
To Cash A/c 2,000
(c) Advertisement A/c Dr. 2,000
To Purchases A/c 2,000
(d) No entry
Answer:
(c) Advertisement A/c Dr. 2,000
To Purchases A/c 2,000

Question 12.
If two or more transactions of the same nature are journalized together, it is known as;
(a) Compound journal entry
(b) Separate journal entry
(c) Posting
(d) None of the above
Answer:
(a) Compound journal entry

Question 13.
An amount of Rs.200 received from A credited to B would affect ………….
(a) Accounts of A and B
(b) A’s Account
(c) Cash Account
(d) B’s Account
Answer:
(a) Accounts of A and B

Question 14.
Contra entries are passed only when …………..
(a) Double Column cash book is prepared
(b) Three-Column Cashbook is prepared
(c) Single Cashbook is prepared
(d) None of the above
Answer:
(b) Three-Column Cashbook is prepared

Question 15.
Which of the following statements is correct?
(a) Accounting profit is the difference between cash receipts and cash paid in a period.
(b) Accounting profit is the total cash sales in the year is the expenses for the period.
(c) Accounting profit is the difference between revenue income and expenses for the period.
(d) Accounting profit is the difference between revenue income and cash payment for the period.
Answer:
(c) Accounting profit is the difference between revenue income and expenses for the period.

Question 16.
Direct Expenses are those which can be identified with the particular products. Which of the following items is a direct cost?
(a) The rent of the factory in which products are made
(b) The wages of the production supervisor
(c) A royalty paid to the holder of a patent after each item is produced
(d) The cost of the glue used to attach labels to the products.
Answer:
(c) A royalty paid to the holder of a patent after each item is produced

Question 17.
Rent payable to the landlord Rs. 5,00,000 is credited to
(a) Cash Account
(b) Landlord Account
(c) Outstanding Rent Account
(d) None of the above
Answer:
(c) Outstanding Rent Account

Question 18.
Bad debts entry is passed in …………
(a) Sales Book
(b) Cash Book
(c) journal Book
(d) None of the above
Answer:
(c) journal Book

Question 19.
Return of cash sales is recorded in …………..
(a) Sales Return book
(b) cash book
(c) Journal proper
(d) None of the above
Answer:
(b) cash book

Question 20.
John purchased goods ‘ Rs, 5900 for cash at 20% trade discount and 5% cash discount. Purchase account Is to fee debited by Rs ………….
(a) Rs, 3,800
(b) Rs. 5,000
(c) Rs. 3,750
(d) Rs. 4,000
Answer:
(d) Rs. 4,000

Question 21.
Out of the following Identify the wrong statement.
(a) Real & Personal accounts are transferred to Balance Sheet
(b) Nominal accounts are transferred to profit and loss account
(c) Cash account is not opened separately in the ledger
(d) Rent account is a personal account and outstand rent account is a Nominal Account
Answer:
(a) Real & Personal accounts are transferred to Balance Sheet

Question 22.
The financial position of a business concern is ascertained on the basis of ………….
(a) Records Prepared under book-keeping process
(b) Trial Balance
(c) Accounting reports
(d) None of the above
Answer:
(c) Accounting reports

Question 23.
Salary “payable to an employee Rs.50,000. Which account is to be credited?
(a) Cash Account
(b) Salaries Account
(c) Outstanding Salaries
(d) None
Answer:
(b) Salaries Account

Question 24.
Loss on sale of machinery Is credited to …………..
(a) Machinery Account
(b) Purchase Account
(c) Profit and Loss Account
(d) None of the above
Answer:
(a) Machinery Account

Question 25.
The Capital of a sole trader Is affected by ……………..
(a) Purchase of raw material
(b) Commission received
(c) Cash received from trade receivables
(d) Purchase of an asset for cash
Answer:
(b) Commission received

II. Very Short Answer Type Questions

Question 1.
From the following information below journalise its
i. Salary paid Rs. 5,000
ii. Rent paid to house owner Rs. 1,000
iii. Cash sales Rs. 10,000
Answer:
Journal Entries

Question 2.
Journalise the following transactions:
2017, Jan 11 Purchased goods for Rs. 1,500
13. Bought furniture for cash Rs. 2,000
19. Drew for private use Rs. 500
Answer:
Journal Entries

III. Short Answer Questions

Question 1.
Journalise the following transitions in the books of G.
i. Started business with cash Rs. 9,000
ii. Deposited into Canara Bank Rs, 3,000
iii. Paid Salary Rs, 5,000
iv. Received commission Rs. 200
v. Cash received from Rajan Rs. 400
Answer:
Journal Entries

IV. Additional Sums

Question 1.
Prepare accounting equation on the basis of the following.
1) Ramya started the business with cash Rs. 1,20,000
2) She Paid Rent Rs. 4,000
3) She Purchases Furniture Rs. 20,000
4) She Purchased goods on credit from Mr. Parthiban Rs. 60,000
5) She Sold goods (Cost Price Rs. 40,000) for Cash Rs. 50,000
Answer:
Accounting Equation

Question 2.
Show the Accounting equation the basis of the following transaction.
1. Pailavan Started business with cash Rs. 50,000
2. Purchased goods from Monisha Rs. 40,000
3. Sold goods to Aruna (Costing Rs. 36,000) for Rs. 50,000
4. wan withdraw from business Rs. 10,000
Answer:
Accounting Equation

Question 3.
Journalise the following transaction of Mrs. Padmini
2018 May 1. Received Cash from Siva – 75,000
8. Paid Cash to Sayed – 45,000
11. Bought Goods for cash – 27,000
12. Bought Furniture – 48,000
15. Sold Goods for cash – 70,000
18. Sold Furniture – 50,000
Answer:
Journal of Mrs. Padmini

Question 4.
Journalise the following transactions of Mr. Dharani.
2018 Jan
1. Mr. Dharani Started Business with Cash – 1,00,000
4. Sold Goods to Mohan – 70,000
10. Purchased goods from Bashyam – 50,000
20. Sold goods for cash from Natarajan – 70,000
25. Paid to Bashyam – 50,000
28. Received from Mohan – 70,000
Answer:
Journal of Mr. Dharani

Question 5.
Journalise the following transaction of Tmt. Rani
2018 Feb
1. Tmt Rani Started Business With cash – 300000
5. Opening a current account with Indian Overseas Bank – 50000
10. Bought goods from Sumathi – 90000
18. Paid to Sumathi – 90000
20. Sold Goods to Chitra – 126000
28. Chitra Settled her account
Answer:
Journal of Tmt. Rani

Question 6.
Journalise the following transactions of Mr.Jeevan.
2018 Mar 5. Sold goods to Arun on Credit – 17,500
10. Bought Goods for cash from Ravi – 22,500
12. Met Travelling Expenses – 2,500
18. Received Rs.80000 from Sivakumar as a loan
21. Paid wages to workers – 3,000
Answer:
Journal of Mr. Jeevan

Question 7.
Give Transaction with imaginary figure involving the following
(i) Increase in assets and capital
(ii) Increase and decrease in assets
(iii) Increase in assets and a liabilities
(iv) Decrease of an asset and owners capital
Answer:
(i) Mr. Raja Commenced Business with Rs. 2,00,000
(ii) Purchased Computer Rs. 15,000
(iii) Purchased goods from Suresh Rs. 80,000
(iv) Drawings Rs. 15,000

Question 8.
Journalise the following transaction of Mrs. Rama
2018 April 1. Mrs. Rama Commenced Business with cash – 30,000
2. Paid into Bank – 21,000
3. Purchases goods and paid through Debit Card – 15,000
7. Drew Cash from Bank for Personal Use – 3,000
15. Purchased goods from Siva – 15,000
20. Cash Sales – 30,000
23. Money Withdrawn from Bank thought ATM for office use – 1,000
25. Paid to Siva – 14,750
Discount Received – 250
31 Paid Rent – 500
Paid Salaries – 2,000
Answer:
Journal of Mrs. Rama

Question 9.
Record the transactions in the books of Shri. Ganesh & Co.
2012 Jan
1. Ganesh started a business with cash – 90,000
1. Paid into bank – 20,000
2. Purchased goods for cash – 50,000
4. Purchased furniture by issuing cheque – 24,000
5. Sold goods to Siva – 17,000
6. Purchased goods from Anand – 30,000
9. Returned goods to Anand – 3,000
10. Cash received from Siva – 16,000
11. Withdraw cash for personal use – 1,500
18. Purchase of stationary – 500
Answer:
Journal Entries in the Books of Shri. Ganesh & Co.

Question 10.
Record the transactions in the books of XYZ & Co:
2017
May 1. Furniture purchased for Rs. 10,000
2. Old Computer worth Rs. 13,700 sold to Sankar on credit
3. Rent paid Rs. 5,000
6. Depreciation on Machinery Rs. 6,400 paid by cheque
10. Paid Rs. 250 for the installation charges of Machinery
13. Cash taken for personal use Rs. 7,000
Answer:
Journal Entries in the books XYZ & Co.

Question 11.
Record the transaction in the books of Y.
2017, April 1. Started Business with cash Rs, 9,000
2. Purchased goods for cash Rs. 21,000
3. Sold goods for cash Rs. 8,000
4. Deposited into Canara Bank Rs. 3,000
5. Cash received from Rajan Rs. 4,000
7. Paid salary Rs. 3,000
8. Paid Rent Rs. 4,000
9. Received Commission Rs. 500
10. Withdrew from Canara Bank Rs. 2,700
Answer:
Journal Entries in the books of Y

Question 12.
Journalize the following transactions:
2016, Dec 1. Purchased goods for cash Rs. 14,000
8. Purchased goods from Anand Stores Rs. 600
12. Sold goods for Rs. 12,500
16. Sold goods to Sridhar Rs. 6,450
18. Bought Machinery for Cash Rs. 1,20,000
20. Goods returned to Anand Stores Rs. 300
22. Sridhar returned goods worth Rs. 450
25. Drew for personal use Rs. 500
25. Electric charges amounted Rs. 1,000
30. Got a loan from Joice Rs. 750
Answer:

Question 13.
Journalise the following transactions in the books of Z.
2016, july 1. Sold goods for Cash Rs. 2,800
2. Purchased goods for cash Rs, 5,900
5. Purchased goods from K Rs. 3,500
8. Sold goods to P Rs. 6,500
5. Received cash from P Rs. 6,200
9. Paid to K Rs. 3,200
9. Paid Telephone Charges Rs: 1,200
10. Paid Salary Rs. 2,600
10. Received Commission Rs. 3,000
10. Computer purchased from S for Rs. 20,000 and an advance of Rs. 8,000 is given as cash
Answer:

Question 14.
Varna is a women entrepreneur, dealing in textiles. From the following transactions, pass journal entries for the month of March 2017.
2017,
March, 1. Commenced business with cash Rs. 50,000 & with goods Rs. 45,000
2. Purchased 20 sarees from V & Co on credit Rs. 60,000
3. Cash deposited into bank Rs. 48,000
4. Paid to V & Co through net banking
5. Sold 10 sarees @ Rs. 3,000 each on credit to X & Co.
6. Paid Travelling expenses Rs. 450
8. Bank Charges levied Rs. 150
9. X becomes insolvent and only Rs. 2,000 per saree received by cash in final settlement.
10. Electricity charges amount to Rs. 1,500
10. Received Commission Rs. 2,000
Answer:

Question 15.
Journalise the following transactions:
2018, March 1. Salary paid Rs. 8,000
2. Rent paid to the house owner Rs. 7,000
3. Credit purchases from Mr. A Rs. 5,800
8. Cash sales Rs. 10,000
9. Deposited cash into bank Rs. 2,500
10. Purchased furniture Rs. 2,400
Answer:

Question 16.
Complete the accounting equation.
(a) Assets = Capital + Liabilities
Rs.2,00,000 = Rs. 1,60,000 + ?
Answer:
Rs. 40,000

(b) Assets = Capital + Liabilities
Rs. 1,00,000 = ? + Rs. 20,000
Answer:
Rs. 80,000

(c) Assets = Capital + Liabilities
? = Rs. 1,60,000 + Rs. 80,000
Answer:
Rs. 2,40,000

Question 17.
Journalise the following transactions in the books of S.
2017, March
1. M commenced business with Rs. 20,000
2. Remitted into bank account Rs. 18,000
3. Paid to K by cheque Rs. 5,000 and discount allowed Rs. 100
10. Cash sales Rs. 4,500
15. Issued a cheque to L for furniture Rs. 2,000
18. Withdrew from bank Rs. 500
25. Cash purchase Rs. 800
31. Paid salaries by cheque Rs. 6,000
Answer:

Question 18.
Write down the transactions for the following.

Answer:
1. Sale of building for Rs. 2,50,000
2. Paid Telephone Charges by Cheque Rs. 2,000
3. Withdrew Cash from Bank Rs. 5,000
4. Withdrew cash for personal use Rs. 3,000
5. Purchased goods for Rs. 6,400 and paid by cheque.

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is …………………….
(a) 0%
(b) 4.4%
(c) 16%
(d) 8.4%
MgCO₃ → MgO + CO₂↑
MgCO₃: (1 × 24) + (1 × 12) + (3 × 16) = 84g
CO₂: (1 × 12) + (2 × 16) = 44g
100% pure 84 g MgCO₃ on heating gives 44 g CO₂
Given that 1 g of MgCO₃ on heating gives 0.44 g CO₂
Therefore, 84 g MgCO₃ sample on heating gives 36.96 g CO₂
Percentage of purity of the sample = \(\frac { 100% }{ 44gCO_{ 2 } } \) × 36.96 g CO₂ = 84%
Percentage of impurity = 16%
Answer:
(c) 16%

Question 2.
The electronic configuration of Eu (atomic no. 63) Gd (atomic no. 64) and Tb (atomic no. 65) are …………………….. [NEET – Phase II]
(a) [Xe] 4f⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(b) [Xe] 4f⁷ 6s² [Xe] 4f⁸ 6s² and [Xe] 4f⁹ 6s²
(c) [Xe] 4f⁷ 6s² [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
(d) [Xe] 4f ⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
Solution:
Eu : [Xe] 4f⁷, 5d⁰, 6s²
Gd : [Xe] 4f⁷, 5d¹, 6s²
Tb : [Xe] 4f⁹, 5d⁰, 6s²
Answer:
(b) [Xe] 4f⁷ 6s² [Xe] 4f⁸ 6s² and [Xe] 4f⁹ 6s²

Question 3.
Match the following:

Answer:

Question 4.
Various successive ionization enthalpies (in kJ mol^-1) of an element are given below.

The element is ……………………….
(a) Phosphorus
(b) Sodium
(c) Aluminium
(d) Silicon
Answer:
(c) Aluminium

Question 5.
Non-stoichiometric hydrides are formed by ……………………..
(a) Palladium, vanadium
(b) Carbon, nickel
(c) Manganese, lithium
(d) Nitrogen, chlorine
Answer:
(a) Palladium, vanadium

Question 6.
Which is the function of sodium – potassium pump?
(a) Maintenance of ion balance
(b) Used in nerve impulse conduction
(c) Transmitting nerve signals
(d) Regulates the blood level
Answer:
(c) Transmitting nerve signals

Question 7.
∆S is expected to be maximum for the reaction ………………………
(a) Ca_(s) + 1/2O_2(g) → CaO_(s)
(b) C_(s) + O_2(g) → CO_2(g)
(c) N_2(g) + O_2(g) → 2NO_(g)
(d) CaCO_3(s) → CaO_(s) + CO_2(g)
Solution:
In CaCO_3(s) → CaO_(s) + CO_2(g) entropy change is positive. In (a) and (b) entropy change is negative; in (c) entropy change is zero.
Answer:
(d) CaCO_3(s) → CaO_(s) + CO_2(g)

Question 8.
Which one of the following is a reversible reaction?
(a) Ripening of a banana
(b) Rusting of iron
(c) Tarnishing of silver
(d) Transport of oxygen by Hemoglobin in our body
Solution:
All the other three reactions are irreversible reactions. But the hemoglobin combines with O₂ in lungs to form oxyhemoglobin. The oxyhemoglobin has a tendency to form hemoglobin by releasing O₂. So it is a reversible reaction.
Answer:
(d) Transport of oxygen by Hemoglobin in our body

Question 9.
P₁ and P₂ are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x₁ represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ………………………….
Solution:
(a) P₁ + x₁ (P₂ – P₁)
(b) P₂ – x₁ (P₂ + P₁)
(c) P₁ – x₂ (P₁ – P₂)
(d) P₁ + x₂ (P₁ – P₂)
P_total = P₁ + P₂
= P₁x₁ + P₂ x₂
= P₁(1 – x₂) + P₂x₂ [∵x₁ + x₂ = 1, x₁ = 1 – x₂]
= P₁ – P₁x₂ + P₂x₂ = P₁ – x₂ = P₁ – x₂ (P₁ – P₂)
Answer:
(c) P₁ – x₂ (P₁ – P₂)

Question 10.
Which of the following molecule contain no n bond?
(a) SO₂

(b) NO₂

(c) CO₂

(d) H₂O

Solution:
Water (H₂O) contains only σ bonds and no π bonds.
Answer:
(d) H₂O

Question 11.
IUPAC name of  is …………………….
(a) Trimethylheptane
(b) 2 -Ethyl -3, 3- dimethyl heptane
(c) 3, 4, 4 – Trimethyloctane
(d) 2 – Butyl -2 -methyl – 3 – ethyl-butane.
Answer:
(c) 3, 4, 4 – Trimethyloctane

Question 12.
Which one of the following is an example for free radical initiators?
(a) Benzoyl peroxide
(b) Benzyl alcohol
(c) Benzyl acetate
(d) Benzaldehyde.
Answer:
(a) Benzoyl peroxide

Question 13.
Some meta-directing substituents in aromatic substitution are given. Which one is most – deactivating?
(a) – COOH
(b) – NO₂
(C) – C ≡ N
(d) – SO₃H
Answer:
(b) – NO₂

Question 14.
Consider the following statements.
(I) S_N² reaction is a bimolecular nucleophilic first order reaction.
(II) S_N² reaction take place in one step.
(III) S_N² reaction involves the formation of a carbocation. Which of the above statements is/are not correct?
(a) (II)
(b) (I) only
(c) (I) & (III)
(d) (III)
Answer:
(c) (I) & (III)

Question 15.
The questions given below consists of an assertion and the reason. Choose the correct option out of the choices given below each question.
Assertion (A): If BOD level of water in a reservoir is more than 5 ppm it is highly polluted.
Reason(R): High biological oxygen demand means high activity of bacteria in water.
(a) Both (A) and R are correct and (R) is the correct explanation of (A)
(b) Both (A) and R are correct and (R) is not the correct explanation of (A)
(c) Both (A) and R are not correct
(d) (A) is correct but( R) is not correct
Answer:
(d) (A) is correct but( R) is not correct

PART – II

Answer any six questions in which question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
What is the actual configuration of copper (Z = 29)? Explain about its stability?
Answer:
Copper (Z = 29)
Expected configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²
Actual configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
The reason is that fully filled orbitals have been found to have extra stability.

Copper has the electronic configuration [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s² due the symmetrical distribution and exchange energies of d electrons.

Symmetry leads to stability. The full filled configuration have symmetrical distribution of electrons and hence they are more stable than unsymmetrical configuration.

Question 17.
What is screening effect?
Answer:
Screening effect:
The repulsive force between inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus the inner shell elections act as a shield between the nucleus and the valence electrons. This effect is called shielding effect (or) screening effect.

Question 18.
Ice is less dense than water at 0°C. Justify this statement?
Answer:
In ice, each oxygen atom is surrounded tetrahedrally by through hydrogen bonds to four water molecules. That is, the presence of two hydrogen atoms and two lone electron pairs (on oxygen) in each water molecule results in a three-dimensional structure. The arrangements creates an open structure, which in turn accounts for the fact that ice is less dense than water at 0°C.

Question 19.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about 120°F may lead to explosions.

So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire, or leave it in the direct sunlight, even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.

1. The gas pressure increases.
2. More of the liquefied propellant turns into a gas.

Question 20.
One mole of PCl₅ is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant?
Answer:
Given that [PCl₅]_initial = \(\frac { 1mol }{ dm^{ 3 } } \)
[Cl₂]_eq = 0.6 mol dm^-3
PCl₅⇄ PCl₃ + Cl₂
[PCl₅]_eq = 0.6 mole dm^-3
[PCl5]_eq = 0.4 mole dm^-3
∴ K_C = \(\frac{0.6×0.6}{0.4}\)
K_C = 0.9

Question 21.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Mostly dissolution of gases in liquid is an exothermic process. It is because the fact that this process involves decrease of entropy.

Thus, increase of temperature tends to push the equilibrium towards backward direction as a result of which solubility of the gas decrease with rise in temperature.
(Gas + Solvent ⇄ Solution + Heat)

Question 22.
Give the general formula for the following classes of organic compounds?
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines.
Answer:
(a) Aliphatic monohydric alcohol

(b) Aliphatic ketones

(c) Aliphatic amines

Question 23.
Identify which of the following shows +1 and -I effect?

(I) -NO₂
(II) -SO₃H
(III) -I
(IV) -OH
(V) CH₃O^–
(VI) CH_3-

Answer:

Question 24.
Write the products A & B for the following reaction?

Answer:

PART – III

Answer any six questions in which question No. 33 is compulsory. [6 × 3 = 18]

Question 25.
Balance by oxidation number method: Mg + HNO₃ → Mg(N0₃)₂ + NO₂ + H₂O.
Answer:
Step 1:

Step 2:
Mg + 2HNO₃ → Mg(NO₃)₂ + NG₂ + H₂O

Step 3:
To balance the number of oxygen atoms and hydrogen atoms 2HNO₃ is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + H₂O

Step 4:
To balance the number of hydrogen atoms, the H₂O molecule is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + 2H₂O

Question 26.
For each of the following, give the sub level designation, the allowable m values and the number of orbitals?
Answer:

1. n = 4, l = 2
2. n = 5, l = 3
3. n = 7, l = 0

1. n = 4, l = 2
If l = 2, ‘m’values are -2,-1, 0, +1, +2. So, 5 orbitals such as d_xy, d_yz ,d_xz, \(d_{ x^{ 2 }-y^{ 2 } }\) and \(d_{ z^{ 2 } }\)

2. n = 5, l = 3
If l = 3, ‘m’ values are -3, -2, -1, 0, +1, +2, +3
So, 7 orbitals such as \(\mathrm{f}_{\mathrm{z}}^{3}, \mathrm{f}_{\mathrm{xz}}^{2}, \mathrm{f}_{\mathrm{yz}}^{2}, \mathrm{f}_{\mathrm{xyz}}, \mathrm{f}_{\mathrm{z}\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{x}\left(\mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{y}\left(3 \mathrm{x}^{2}-\mathrm{z}^{2}\right)}\)

3. n = 7, l = 0
If l = 0, ‘m’ values are 0. Only one value. So, 1 orbital such as 7s orbital.

Question 27.
Prove that ionization energy is a periodic property?
Answer:
Variation in a period:
On moving across a period from left to right, the ionization enthalpy value increases.
This is due to the following reasons:

1. Increase of nuclear charge in a period
2. Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus, thus ionization enthalpy increases. Hence, ionization energy is a periodic property.

Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons:

1. A gradual increase in atomic size
2. Increase of screening effect on the outermost electrons due. to the increase of number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 28.
Explain how heat absorbed at constant pressure is measured using coffee cup calorimeter with neat diagram?
Answer:

1. Measurement of heat change at constant pressure can be done in a coffee cup calorimeter.
2. We know that ∆H = qp (at constant P) and therefore, heat absorbed or evolved, q_p at constant pressure is also called the heat of reaction or enthalpy of reaction, ∆H_r
3. In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and ∆H_r will Reaction also be negative. Mixture
4. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and ∆H_r will also be positive.

Question 29.
Consider the following reactions,
(a) H₂(g) + I₂(g) ⇄ 2 HI(g)
(b) CaCO₂(s) ⇄ CaO(s) + CO₂(g)
(c) S(s) + 3F₂(g) ⇄ SF₆(g)
In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product?
Answer:
(a) H₂(g) + I₂2(g) ⇄ 2HI(g)
In the above equilibrium reaction, volume of gaseous mdlecules is equal on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no change in the yield of product.

(b)
Volume is greater in product side. By decreasing the pressure, volume will increase thus, to get more of product CO₂, the pressure should be decreased or volume should be increased.

(c)
Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.

Question 30.
You are provided with a solid ‘A’ and three solutions of A dissolved in water one saturated, one unsaturated, and one super saturated. How would you determine each solution?
Answer:
(I) Saturated solution:
When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution.

(II) Unsaturated solution:
When minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution.

(III) Super saturated solution:
It is a solution that holds more solute than it normally could in its saturated form.

Example:

1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 g of NaCl in 1 litre of water at 25°C.
2. An unsaturated solution has the capacity to dissolve more of the compound. 36 g of NaCl in 1 litre of water at 25°C.
3. A super saturated solution is the solution in which crystals can start growing. 500 g of NaCl in 1 litre of water at 25°C.

Question 31.
Explain about the salient features of molecular orbital theory?
Answer:

• When atoms combine to fonn molecules, their individual atomic orbitals lose their identity and form new orbitals called molecular orbitals.
• The shape of molecular orbitals depend upon the shapes of combining atomic orbitals.
• The number of molecular orbitals formed is the same as the number of combining atomic orbitals. Half the number of molecular orbitals formed will have lower energy and are called bonding orbitals, while the remaining half molecular orbitals will have higher energy and are called anti-bonding molecular orbitals.
• The bonding molecular orbitals are represented as σ (sigma), π (pi), δ (delta) and the corresponding anti-bonding orbitals are called C*, 7t* and 5*.
• The electrons in the molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follow Aufbau’s Principle, Pauli’s exclusion principle and Hund’s rule as in the case of filling of electrons in the atomic orbitals.
• Bond order gives the number of covalent bonds between the two combining atoms.

The bond order of a molecule can be calculated using the following equation:
Bond order = \(\frac { N_{ b }-N_{ a } }{ 2 } \)
N_b = Number of electrons in bonding molecular orbitals.
N_a = Number of electrons in anti-bonding molecular orbitals.
(vii) A bond order of zero value indicates that the molecule does not exist.

Question 32.
Explain the types of addition reactions?
Answer:
Addition reactions are classified into three types they are,

1. Electrophilic addition reaction
2. Nucleophilic addition reaction
3. Free radical addition reaction

1. Electrophilic addition reaction:
An electrophilic addition reaction can be described as an addition reaction in whifth a reactant with multiple bonds as in a double or triple bond undergoes has its n bond broken and two new a bond are formed.

(ethane) Br Bf

2. Nucleophilic addition reaction:
A nucleophilic addition reaction is an addition reaction where a chemical compound with an electron deficient or electrophilic double or triple bond, a n bond, reacts with a nucleophilic which is an electron rich reactant with the disappearance of the double bond and creation of two new single or a bonds.

3. Free radical addition reaction:
It is an addition reaction in organic chemistry involving free radicals. The addition may occur between a radical and a non radical or between two radicals.

Question 33.
Complete the following reaction and identify A, B and C?

Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
(I) How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular nodes?
(II) How many unpaired electrons are present in the ground state of

(a) Cr³⁺ (Z = 24)
(b) Ne (Z = 10)

[OR]

(b) (I) The electronic configuration of an atom is one of the important factor which affects the value of ionization potential and electron gain enthalpy. Explain?
(II) Explain why cation are smaller and anions are larger in radii than their parent atoms?
Answer:
(a) (I) Formula for total number of nodes = n – 1
For 2s orbital: Number of radial nodes = 1.
For 4p orbital: Number of radial nodes = n – l – 1.
= 4 – 1 – 1 = 2
Number of angular nodes = l
∴Number of angular nodes = l.
So, 4p orbital has 2 radial nodes and 1 angular node.

For 5d orbital:
Total number of nodes = n – 1
= 5 – 1 =4 nodes
Number of radial nodes = n – l – 1
= 5 – 2 – 1 = 2 radial nodes.
Number of angular nodes = l = 2
∴5d orbital have 2 radial nodes and 2 angular nodes.

For 4f orbital:
Total number of nodes = n – 1
= 4 – 1 = 3 nodes
Number of radial nodes = n – l – 1
= 4 – 3 – 1 = 0 node.
Number of angular nodes = l
= 3 nodes
∴ 5d orbital have 0 radial node and 3 angular nodes.

(II) (a) Cr (Z = 24) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Cr³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴.
It contains 4 unpaired electrons.
(b) Ne(Z=10) 1s² 2s² 2p⁶. No unpaired electrons in it.

[OR]

(b) (I)

• Electronic configuration of an atom affects the value of ionization potential and electron gain enthalpy.
• Half filled valence shell electronic configuration and completely filled valence shell electronic configuration are more stable than partially filled electronic configuration.
• For e.g. Beryllium (Z = 4) 1s² 2s² (completely filled electronic configuration)
  Nitrogen (Z = 7) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹ (half filled electronic configuration)
  Both beryllium and nitrogen have high ionization energy due to more stable nature.
• In the case of beryllium (1s² 2s²), nitrogen (1s² 2s² 2p³) the addition of extra electron will disturb their stable electronic configuration and they have almost zero electron affinity.
• Noble gases have stable ns² np⁶ configuration and the addition of further electron is unfavorable and they have zero electron affinity.

(II) A cation is smaller than the parent atom because it has fewer electrons while its nuclear charge remains the same. The size of anion will be larger than that of parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 35 (a).
(I) Arrange NH₃, H₂O and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement?
(II) Can we use concentrated sulphuric acid and pure zinc in the preparation of dihydrogen?

[OR]

(b) (I) Beryllium halides are covalent whereas magnesium halides are ionic. Why?
(II) What happens when
Sodium metal is dropped in water?

1. Sodium metal is heated in free supply of air?
2. Sodium peroxide dissolves in water?

Answer:
(a) 1. Increasing magnitude of hydrogen bonding among NH₃, H₂O and HF is
HF > H₂O > NH₃

2. The extent of hydrogen bonding depends upon electronegativity and the number of hydrogne atoms available for bonding.

3. Among N, F and O the increasing order of their electronegativities are
N < O < F 4. Hence the expected order of the extent of hydrogen bonding is HF > H₂O > NH₃

(II) Conc. H₂SO₄ cannot be used because it acts as an oxidizing agent also and gets reduced to SO₂.
Zn + 2H₂SO₄ (Conc) → ZnSO₄ + 2H₂O + SO₂ Pure Zn is not used because it is non-porous and reaction will be slow. The impurities in Zn help in constitute of electrochemical couple and speed up reaction.

[OR]

(b)
(I) Beryllium ion (Be²⁺) is smaller in size and it is involved in equal sharing of electrons with halogens to form covalent bond, whereas magnesium ion (Mg²⁺) is bigger and it is involved in transfer of electrons to form ionic bond.

(II)

1. 2Na + 2H₂O → 2NaOH + H₂
2. 2Na + O₂ → Na₂O₂
3. Na₂O₂ + 2H₂O → 2NaOH + H₂O₂

Question 36 (a).
(I) Define Gibb’s free energy?
(II) You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below?
Answer:

[OR]

(b) (I) Define mole fraction.
(II) Differentiate between ideal solution and non-ideal solution.
Answer:
(a) (I) Gibbs free energy is defined as the part of total energy of a system that can be converted (or) available for conversion into work.
G = H – TS, where G = Gibb’s free energy
H – enthalpy;
T = temperature;
S = entropy

(II) For ethanol:
Given: T_b = 78.4°C = (78.4 + 273) = 351.4 K

[OR]

(b) (I) Mole fraction of a component is the ratio of number of moles of the component to the total number of moles of all components present in the solution.

(II)
Ideal Solution:

1. An ideal solution is a solution in which each component obeys the Raoult’s law over entire range of concentration.
2. For an ideal solution,
   ΔH_missing = 0, ΔV_mixing = 0
3. Example: Benzene and toulene n – Hexane and n – Heptane.

Non – Ideal Solution:

1. The solution which do not obey Raults’s law over entire range of concentrations are called non-ideal solution.
2. For an ideal solution,
   ΔH_mixing ≠ 0, ΔV_mixing ≠ 0
3. Example: Ethyl alcohol and Cyclo hexane, Benzene and aceton.

Question 37 (a).
(I) What is dipole moment?
(II) Describe Fajan’s rule?

[OR]

(b) (I) How does Huckel rule help to decide the aromatic character of a compound?
(II) Draw cis-trans isomers for the following compounds
(a) 2- chloro-2-butene
(b) CH₃-CCl=CH-CH₂CH₃
Answer:
(a)
(I)

1. The polarity of a covalent bond can be measured in terms of dipole moment which is defined as: m = q × 2d, where m is the dipole moment, q is the charge, 2d is the distance between the two charges.
2. The dipole moment is a vector quantity and the direction of the dipole moment points from the negative charge to positive charge.
3. The unit of dipole moment is Coulomb metre (C m). It is usually expressed in Debye unit (D).
4. 1 Debye = 3.336 × 10^-3o C m

(II)

1. The ability of a cation to polarise an anion is called its polarising ability and the tendency of ah anion to get polarised is called its polarisibility. The extent of polarisation in an ionic compound is given by the Fajans rule.

2. To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the cation greater will be the attraction on the electron cloud of the anion. Similarly higher the magnitude of negative charge on anion, greater is its polarisability. For example, Na⁺ < Mg²⁺ < Al³⁺, the covalent character also follows the order:
NaCl < MgCl₂ < AlCl₃.

3. The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation, e.g., LiCl is more covalent than NaCl.

4. Cation having ns² np⁶ nd⁰ configuration shows greater polarising power than the cations with ns² np⁶ configuration, e.g., CuCl is more covalent than NaCl.

[OR]

(b) (I) A compound is said to be aromatic, if it obeys the following rules:

1. The molecule must be cyclic.
2. The molecule must be co-planar.
3. Complete delocalisation of rc-electrons in the ring.
4. Presence of (4n + 2)π electrons in the ring where n is an integer (n = 0, 1, 2 …)

This is known as Huckel’s rule.
Example –  – Benzene

1. It is cyclic one.
2. It is a co-planar molecule.
3. It has six delocalised n electrons.
4. 4n + 2 = 6

4n = 6 – 2
4n = 4
⇒ n = 1
It obey Huckel’s rule, with n = 1, hence benzene is aromatic in nature.

(II) (a) 2-Chloro-2-butene:

(b)

Question 38 (a).
(I) Reagents and the conditions used in the reactions are given below. Complete the table by writing down the product and the name of the reaction?
Answer:

(II) What is the IUPAC name of the insecticide DDT? Why is their use banned in most of the countries?

[OR]

(b) (I) Explain about green chemistry in day-to-day life?
(II) How acetaldehyde is commercially prepared by green chemistry?
Answer:
(a)
(I)

(II)

•  The IUPAC name of the insecticide DDT is p, p’-dichloro-diphenyl trichloroethane.
• Even DDT is an effective insecticide. Now-adays it is banned because of its long term toxic effects.
• DDT is very persistent in the environment and it has a high affinity for fatty tissues. As a result, DDT gets accumulated in animal tissue fat, in particular that of birds of prey with subsequent thinning of their eggs shells and impacting their rate of reproduction. That is why DDT is banned in most of the countries.

[OR]

(b)
(I)
1. Dry cleaning of clothes:
Solvents like tetrachloroethylene used in dry cleaning of clothes, pollute the ground water and are carcinogenic. In place of tetrachloro ethylene, liquefied CO₂ with suitable detergent is an alternate solvent used. Liquefied CO₂ is not harmful to the ground water. Nowadays H₂O₂ is used for bleaching clothes in laundry, gives better result and utilises less water.

2. Bleaching of paper:
Conventional method of bleaching was done with chlorine. Nowadays H₂O₂ can be used for bleaching paper in the presence of catalyst.

1. Instead of petrol, methaftol is used as a fuel in automobiles.
2. Neem based pesticides have been synthesised, which are more safer than the chlorinated hydrocarbons.

(II) Acetaldehyde is commericially prepared by one step oxidation of ethene in the presence of ionic catalyst in aqueous medium with 90% yield.

Students can Download Tamil Nadu 11th Maths Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Maths Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 90

PART – 1

I. Choose the correct answer. Answer all the questions: [20 × 1 = 20]

Question 1.
For non empty sets A and B if A ⊂ B then (A × B) n (B × A) is equal to ………………….
(a) A ∩ B
(b) A × A
(c) B × B
(d) none of these
Answer:
(b) A × A

Question 2.
The solution set of the inequality |x – 1| ≥ |x – 3| is ………………..
(a) [0, 2]
(b) [2, ∞)
(c) (0, 2)
(d) (-∞, 2)
Answer:
(b) [2, ∞)

Question 3.
The numer of solutions of x² + |x – 1| = 1 is …………………
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
(c) 2

Question 4.
Which of the following is not true?
(a) sin θ = – \(\frac{3}{4}\)
(b) cos θ = -1
(c) tan θ = 25
(d) sec θ = \(\frac{1}{4}\)
Answer:
(d) sec θ = \(\frac{1}{4}\)

Question 5.
Let f_k(x) = \(\frac{1}{k}\) [sin^k x + cos² x] where x ∈ R and k ≥ 1. Then f₄(x) – f₆(x) = …………………
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{12}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{3}\)
Answer:
(b) \(\frac{1}{12}\)

Question 6.
If A and B are coefficients of xⁿ in the expansion of (1 + x)²ⁿ and (1 + x)^2n-1 respectively then \(\frac{A}{B}\) =
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{n}\)
(c) 1
(d) 2
Answer:
(d) 2

Question 7.
The value of 15C₈ + 15C₉ – 15C₆ – 15C₇ is …………………
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(a) 0

Question 8.
The slope of the line which makes an angle 45° with the line 3x -y = – 5 are …………………
(a) 1, -1
(b) \(\frac{1}{2}\), -2
(c) 1, \(\frac{1}{2}\)
(d) 2, \(\frac{-1}{2}\)
Answer:
(b) \(\frac{1}{2}\), -2

Question 9.
The sum of the binomial co-efficients is ………………….
(a) 2n
(b) 2ⁿ
(c) n²
(d) 1
Answer:
(b) 2ⁿ

Question 10.
If the square of the matrix \(\begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix}\) satisfy the relation
(a) 1 + α² + βγ = 0
(b) 1 – α² – βγ = 0
(c) 1 – α² + βγ = 0
(d) 1 + α² – βγ = 0
Answer:
(b) 1 – α² – βγ = 0

Question 11.
The value of x for which the matrix A = \(\begin{bmatrix} e^{ x-2 } & e^{ 7+x } \\ e^{ 2+x } & e^{ 2x+3 } \end{bmatrix}\) is singular is ………………….
(a) 9
(b) 8
(c) 7
(d) 6
Answer:
(b) 8

Question 12.
If A = \(\begin{pmatrix} \lambda & 1 \\ -1 & -\lambda \end{pmatrix}\) then for what value of λ, A² = 0?
(a) 0
(b) ±1
(c) -1
(d) 1
Answer:
(b) ±1

Question 13.
lim_x→1 \(\frac { xe^{ x }-sinx }{ x } \) is ……………….
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(d) 0

Question 14.
If the points whose position vectors 10\({ \vec { i } }\) + 3\({ \vec { j } }\) + 12\({ \vec { i } }\) – 5\({ \vec { j } }\) and a\({ \vec { i } }\) + 11\({ \vec { j } }\) are collinear then the value of a is ……………….
(a) 3
(b) 5
(c) 6
(d) 8
Answer:
(d) 8

Question 15.
If y = mx + c and f(0) = f'(0) = 1 then f(2) = ………………..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
lim_x→1 \(\frac { e^{ x }-e }{ x-1 } \) = ……………….
(a) 1
(b) e
(c) ∞
(d) 0
Answer:
(b) e

Question 17.
∫2^3x+5 dx is ………………..
(a) \(\frac { 3(2^{ 3x+5 }) }{ log2 } \) + c
(b) \(\frac { 2^{ 3x+5 } }{ 2log(3x+5) } \) + c
(c) \(\frac { 2^{ 3x+5 } }{ 2log3 } \) + c
(d) \(\frac { 2^{ 3x+5 } }{ 3log2 } \) + c
Answer:
(d) \(\frac { 2^{ 3x+5 } }{ 3log2 } \) + c

Question 18.
∫x² cos xdx is ………………..
(a) x²sinx + 2xcosx – 2 sinx + c
(b) x² – 2xcosx – 2sinx + c
(c) -x² + 2xcosx + 2sinx + c
(d) -x² – 2xcosx + 2 sinx + c
Answer:
(a) x²sinx + 2xcosx – 2 sinx + c

Question 19.
∫\(\frac { x }{ 1+x^{ 2 } } \) dx = ………………
(a) tan^-1 x + c
(b) log (1 + x²) + c
(c) log x + c
(d) \(\frac{1}{2}\) log (1 + x²) + c
Answer:
(d) \(\frac{1}{2}\) log (1 + x²) + c

Question 20.
Ten coins are tossed. The probability of getting atleast 8 heads is …………….
(a) \(\frac{7}{64}\)
(b) \(\frac{7}{32}\)
(c) \(\frac{7}{128}\)
(d) \(\frac{7}{16}\)
Answer:
(c) \(\frac{7}{128}\)

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
To secure an A grade one must obtain an average of 90 marks or more in 5 subjects each of maximum 100 marks. If A scored 84, 87, 95, 91 in first four subjects, what is the minimum mark be scored in the fifth subject to get an A grade in the course?
Answer:
Required marks = 5 × 90 = 450
Total marks obtained in 4 subjects = 84 + 87 + 95 + 91 = 357
So required marks in the fifth subject = 450 – 357 = 93.

Question 22.
Prove that (1 + tan1°) (1 + tan2°) (1 + tan 3°) …. (1 + tan 44°) is a multiple of 4?
Answer:
45°= 1 + 44 (or) 2 + 43 (or) 3 + 42 (or) 22 + 23
So we have 22 possible pairs
sa the product is (2) (22) = 44
which is ÷ by 4

Question 23.
Find the distinct permutations of the letters of the word MISSISSIPPI?
Answer:
MISSISSIPPI
Number of letters = 11
Here M-1 time, I-4timcs, S-4times, P-2timcs
So total number of arrangement is of this word = \(\frac { 11! }{ 4!4!2! } \)
\(\frac { 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4! }{ 4\times 3\times 2\times 1\times 2\times 1\times 4! } \) = 34650

Question 24.
Compare |A| using Sarrus rule if A = \(\left[\begin{array}{ccc}
3 & 4 & 1 \\
0 & -1 & 2 \\
5 & -2 & 6
\end{array}\right]\)
Answer:

|A| = [3(-1) (6) + 4(2)(5) + 1(0)(-2)] -[5(-1)(1) + (-2)(2)3 + 6(0)(4)] = [-18 + 40 + 0] – [-5 – 12 + 0] = 22 + 17 = 39.

Question 25.
If |\({ \vec { a } }\)|, |\({ \vec { b } }\)| = 6, |\({ \vec { c } }\)| =7 and \({ \vec { a } }\) + \({ \vec { b } }\) + \({ \vec { c } }\) = 0 find \({ \vec { a } }\).\({ \vec { b } }\) + \({ \vec { b } }.\) \({ \vec { c } }\) + \({ \vec { c } }\).\({ \vec { a } }\)
Answer:
Given \(\bar { a } \)+\(\bar { b } \)+\(\bar { c } \) = 0
⇒ (\(\bar { a } \)+\(\bar { b } \)+\(\bar { c } \))² = 0
(i.e;) |\(\bar { a } \)|² + |\(\bar { b } \)|² + |\(\bar { c } \)|² + 2 [\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \)) = 0
⇒ 5²+6²+7²+2(\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \)) = 0
⇒2(\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \)) = – 25 – 36 – 49 = -110
⇒\(\bar { a } \).\(\bar { b } \)+\(\bar { b } \).\(\bar { c } \)+\(\bar { c } \).\(\bar { a } \) = \(\frac{-110}{2}\) = -55

Question 26.
Find \(\lim _{ t\rightarrow 0 }{ \frac { \sqrt { t^{ 2 }+9-3 } }{ t^{ 2 } } } \)
Answer:
We can’t apply the quotient Iieorem it.:iiediatcly. Use the algebra technique of rationalising the numerator.

Question 27.
Evaluate y = e^-x.log x?
Answer:
y = e^-x log x = uv(say)
Here u = e^-x and v = log x
⇒ u’ = -e^-x and v’ = uv’ + vu’
Now y = uv ⇒y’ = uv’ + vu’
(i.e). \(\frac{dy}{dx}\) = e^-x(\(\frac{1}{x}\)) + log x(- e^-x)
= e^-x(\(\frac{1}{x}\) – log x)

Question 28.
Evaluate ∫\(\frac { 1 }{ \sqrt { 1+(4x)^{ 2 } } } \) dx
Answer:
Let I = ∫\(\frac { 1 }{ \sqrt { 1+4x^{ 2 } } } \)dx = ∫\(\frac { 1 }{ \sqrt { 1+(2x^{ 2 }) } } \) dx
Putting 2x = t ⇒ 2 dx = dt ⇒ dx = \(\frac{1}{2}\) dt
Thus, I = \(\frac{1}{2}\)∫\({ \frac { 1 }{ \sqrt { 1^{ 2 }+t^{ 2 } } } }\) dt
I = \(\frac{1}{2}\) log |t+\(\sqrt { t^{ 2 }+1 } \) + c = \(\frac{1}{2}\) log |2x + \(\sqrt { (2x)^{ 2 }+1 } \)| + c
I = \(\frac{1}{2}\) log |2x+\(\sqrt { 4^{ 2 }+1 } \)| + c.

Question 29.
Nine coins are tossed once. Find the probablity to get atleast 2 heads?
Answer:
Let S be the sample and A be the event of getting at least two heads.
Therefore, the event \(\bar { A } \) denotes, getting at most one head.
n(S) = 2⁹ = 521, n(\(\bar { A } \)) = 9C₀ + 9C₁ = 1 + 9 = 10
P(\(\bar { A } \)) = \(\frac{10}{512}\) = \(\frac{5}{256}\)
P(A) = 1- P(\(\bar { A } \)) = 1 – \(\frac{5}{256}\) = \(\frac{251}{256}\)

Question 30.
If P(A) denotes the power set of A, then find n(P(P(ϕ))))?
Answer:
Since P(∅) contains 1 element, P(P(∅)) contains 2¹ elements and hence P(P(P(∅))) contains 2² elements. That is, 4 elements.

PART – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
A simple cipher takes a number and codes it, using the function f(x) = 3x – 4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x (by drawing the lines)?

Question 32.
Show that 4 sin A sin (60° + A). sin(60° – A) = sin 3A?

Question 33.
How many strings of length 6 can be formed using letters of the word FLOWER if (i) either starts with F or ends with R?

Question 34.
Find the distance of the line 4x – y = 0 from the point P(4, 1) measured along the line making an angle 135° with the positive x- axis?

Question 35.
Using factor theorem prove that \(\left|\begin{array}{lll}
x+1 & 3 & 5 \\
2 & x+2 & 5 \\
2 & 3 & x+4
\end{array}\right|\) = (x – 1)² + ( x+ 9)

Question 36.
Find the unit vectors perpendicular to each of the vectors \({ \vec { a } }\) + \({ \vec { b } }\) and \({ \vec { a } }\) – \({ \vec { b } }\). Where \({ \vec { a } }\) = \({ \vec { i } }\) + \({ \vec { j } }\) + \({ \vec { k } }\) and b = \({ \vec { i } }\) + 2\({ \vec { j } }\) + 3\({ \vec { k } }\)?

Question 37.
If y = tan^-1 \(\left(\frac{1+x}{1-x}\right)\), find y’?

Question 38.
Evaluate ∫\(\frac { x }{ \sqrt { 1+x^{ 2 } } } \) dx?

Question 39.
If A and B are two events such that P(A∪B) = 0.7, P(A∩B) = 0.2 and P(B) = 0.5 then show that A and B are independent?

Question 40.
Resolve into partial fractions \(\frac { 2x^{ 2 }+5x-11 }{ x^{ 2 }+2x-3 } \)

PART – IV

IV. Answer all the questIons [7 × 5 = 35]

Question 41.
(a) A photocopy store charges Rs. 1.50 per copy for the first 10 copies and Rs. 1.00 per copy after the 10th copy. Let x be the number of copies, and let y be the total cost of photocopying.
(a) Draw graph of the cost as x goes from 0 to 50 copies?
(b) Find the cost of making 40 copies

[OR]

(b) If the difference of the roots of the equation 2x² – (a + 1)x + a – 1 = 0 is equal to their product then prove that a =2?

Question 42.
(a) Prove that tan 70° – tan 20° – 2 tan 40° = 4 tan 10°?

[OR]

(b) In a ∆ABC, if a = 2\(\sqrt{3}\), b = 2\(\sqrt{2}\) and C = 75° find the other side and the angles?

Question 43.
(a) Use induction to prove that n³ – 7n + 3, is divisible by 3, for all natural numbers n?

[OR]

(b) Evaluate y = (x² + 1) \(\sqrt [ 3 ]{ x^{ 2 }+2 } \)?

Question 44.
(a) Show that f(x) f(y) = f(x + y), where f(x) = \(\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\)

[OR]

(b) The chances of X, Y and Z becoming managers of a certain company are 4 : 2 : 3. The probabilities that bonus scheme will be introduced if X, Y and Z become managers are 0.3, 0.5 and 0.4 respectively. 1f the bonus scheme has been introduced, what is the probability that Z was appointed as the manager?

Question 45.
(a) If the equation λx² – 10xy + 12y² + 5x – 16y – 3 = O represents a pair of straight lines, find

1. The value of λ and the separate equations of the lines
2. Angle between the lines

[OR]

(b) Show that

1. \(\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{3 n^{2}+7 n+2}=\frac{1}{6}\)
2. \(\lim _{n \rightarrow \infty} \frac{1^{2}+2^{2}+\ldots+(3 n)^{2}}{(1+2+\ldots+5 n)(2 n+3)}=\frac{9}{25}\)
3. \(\lim _{n \rightarrow \infty} \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots .+\frac{1}{n(n+1)}=1\)

Question 46.
(a) A man repays an amount of ₹3250 by paying ₹20 in the first month and then increases the payment by ₹15 per month. How long will it take him to clear the amount?

[OR]

(b) Find the area of the triangle whose vertices are A (3, -1, 2), B (1, -1, -3) and C (4, -3, 1)?

Question 47.
(a) Evaluate (x + 1) \(\sqrt { 2x+3 } \)?
[OR]

(b) Evaluate ∫cosec²x dx?