Category: Class 11

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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 Physics Model Question Paper 1 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 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: 3 Hours
Maximum Marks: 70

PART – I

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

Question 1.
A cyclist moving on a circular track of radius 40 m completes half a revolution in 40 sec average velocity is ………………..
(a) 0
(b) 2 m/s
(c) 4 m/s
(d) 2π m/s
Hint:
Displacement of the cyclist in half revolution is
d = diameter of the circular track
i.e., d= 80 m
Time taken, t = 40 s
Average velocity, V =  = \(\frac{80}{40}\)
V = 2 m/s
Answer:
(b) 2 m/s

Question 2.
A wheel has angular acceleration of 3.0 rad/s² and an initial angular speed of 2.00 rad/s. In a time of 2 seconds it has rotated through an angle of (in radian) ………………..
(a) 10
(b) 12
(c) 4
(d) 6
Answer:
(a) 10

Question 3.
If the origin of co-ordinate system lies at the centre of mass. The sum of the moments of the masses of the system about the centre of mass is …………………
(a) May be greater than zero
(b) May be less than zero
(c) May be equal to zero
(d) Always zero
Answer:
(d) Always zero

Question 4.
Dimensional formula for co-efficient of viscousity
(a) ML^-2 T^-2
(b) ML^-2 T^-1
(c) ML^-1 T^-1
(d) M^-1 L^-1 T^-1
Answer:
(c) ML^-1 T^-1

Question 5.
Action and reaction …………………
(a) Acts on same object
(b) Acts on two different objects
(c) Have resultant not zero
(d) Acts on the same direction
Answer:
(b) Acts on two different objects

Question 6.
A spring is stretched by applying load to its free end. The strain produced in the spring is …………………
(a) Volumetric
(b) Shear
(c) Longitudinal
(d) Longitudinal and shear
Answer:
(d) Longitudinal and shear

Question 7.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?
(a) 0.25 rad s^-2
(b) 25 rad s^-2
(c) 5 ms^-2
(d) 25 ms^-2

Hint:
τ = F × r
Iα = F × r
MR² × α = 30 × \(\frac{40}{100}\); \(\frac { 3\times 40\times 40\times \alpha }{ 100\times 100 } \) = 12
\(\frac { 3\times 16\times \alpha }{ 100 } \) = 12; α = 25 rad/s²
Answer:
(b) 25 rad s^-2

Question 8.
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (E is total energy) ………………
(a) \(\frac{2}{3}\) E
(b) \(\frac{1}{3}\) E
(c) \(\frac{1}{4}\) E
(d) \(\frac{1}{2}\) E
Hint:
PE = \(\frac{1}{2}\) kx²
⇒ \(P E_{V_{2}}=\frac{1}{2} K\left(\frac{A}{2}\right)^{2}=\frac{1}{4}\left(\frac{1}{2} K A^{2}\right)\)
Answer:
(c) \(\frac{1}{4}\) E

Question 9.
A particle executes simple harmonic motion with an angular velocity and maximum acceleration of 3.5 rad/s and 7.5 m/s² respectively. Amplitude of the oscillation is ………………
(a) 0.36
(b) 0.28
(c) 0.61
(d) 0.53
Hint:
x = A sin ωt
∴ a = \(\frac { d^{ 2 }x }{ dt^{ 2 } } \) = -Aω² sinωt
∴Maximum acceleration |a_max| = Aω²
Now Aω² = 7.5
A = \(\frac { 7.5 }{ \omega ^{ 2 } } \) = \(\frac { 7.5 }{ (3.5)^{ 2 } } \) = 0.61
Answer:
(c) 0.61

Question 10.
If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved, then its fundamental frequency will become ………………….
(a) \(\frac{n}{4}\)
(b) \(\sqrt{2n}\)
(c) n
(d) \(\frac { n }{ \sqrt { 2 } } \)
Hint:

Answer:
(c) n

Question 11.
The theory of refrigerator is based on …………….
(a) Joule-Thomson effect
(b) Newton’s particle theory
(c) Joule’s effect
(d) None of the above
Answer:
(d) None of the above

Question 12.
Work done by 0.1 mole of a gas at 27°C to double its volume at constant pressure is ………………..
(a) 54 cal
(b) 60 cal
(c) 546 cal
(d) 600 cal
Hint:
Workdone (W) = – p.dv = nRT
= 0.1 × (0.2 cal) × (273 + 27) = 0.1 × 2 × 300
W = 60 cal
Answer:
(b) 60 cal

Question 13.
When a lift is moving upwards with acceleration a, then time period of simple pendulum in it will ………………..
(a) 2π\(\sqrt { \frac { 1 }{ g+a } } \)
(b) 2π\(\sqrt { \frac { g+a }{ l } } \)
(c) \(\frac{1}{2π}\)\(\sqrt { \frac { 1 }{ g+a } } \)
(d) \(\frac{1}{2π}\)\(\sqrt { \frac { g+a }{ l } } \)
Answer:
(a) 2π\(\sqrt { \frac { 1 }{ g+a } } \)

Question 14.
A disc is rotating with angular speed ω. If a child sits on it, what is conserved?
(a) Linear momentum
(b) Angular momentum
(c) Kinetic energy
(d) Potential energy
Answer:
(b) Angular momentum

Question 15.
The vectors \(\vec { A } \) and \(\vec { B } \) are such that |\(\vec { A } \) + \(\vec { B } \)| = |\(\vec { A } \) – \(\vec { B } \)|. The angle between the two vector is ………………..
(a) 45°
(b) 60°
(c) 75°
(d) 90°
Hint:
The angle between two vector is always 90°.
Answer:
(d) 90°

PART – II

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

Question 16.
Velocity – time graph of a moving object is shown below. What is the acceleration of the object? Also draw displacement – time graph for the motion of the object?
Answer:
The given graph shows that the velocity of the object is constant. That is, the velocity of the object is not changing, so the acceleration of the object is zero. Since the acceleration of an object is given by


Displacement – time graph for the motion of the object is shown in the figure above.

Question 17.
Can a body subjected to a uniform acceleration always move in a straight line?
Answer:
It will be a straight line in one dimensional motion but not applicable for two dimensional motion because the projectile has a parabolic path but it has a uniform acceleration.

Question 18.
Calculate the viscous force on a ball of radius 1mm moving through a liquid of viscosity 0.2 Nsm^-2 at a speed of 0.07 ms^–
Answer:
Radius of the ball (a) = 1mm = 1 × 10^-3m
Co-effecient of viscosity of liquid (η) = 0.2 Nsm^-2
Speed of the ball (v) = 0.07 ms^-1
According to Stoke’s law
Viscous force F = 6 π η av
= 6 × 3.14 × 1 × 10^-3 × 0.2 × 0.07
= 0.26376 × 10^-3 = 2.64 × 10^-4N

Question 19.
Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10ms^-2)
Answer:
Given data: F = 30 N, load (m) 2 kg; height = 10m, g = 10 ms^-2
Gravitational forcc F = mg = 30 N
The distance moved h = 10 m
Work done on the object W = Fh = 30 × 10 = 300 J

Question 20.
Why are shockers used in automobiles like car?
Answer:
In the event of jump or jerk, the lime of action of force increases. Since the product of force aid time is constant in a given situation. therefore the force decreases.

Question 21.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighted 250 N on the surface?
Answer:
As g_d = g (1 – \(\frac{d}{R}\)) ⇒ mg_d = mg(1 – \(\frac{d}{R}\))
Here d = \(\frac{R}{2}\)
∴mg_d = (250) × (1 – \(\frac { R/2 }{ R } \)) = 250 × \(\frac{1}{2}\) = 125N

Question 22.
How do you deduce that two vectors are perpendicular?
Answer:
If two vectors \(\vec { A } \) and \(\vec { B } \) are perpendicular to each other than their scalar product \(\vec { A } \).\(\vec { B } \) = O because cos 90° = 0. Then the vectors \(\vec { A } \) and \(\vec { B } \) are said to be mutually orthogonal.

Question 23.
An air bubble of radius r in water is at a depth of h below the water surface at some instant. If P is atmospheric pressure and d and T are the density and surface tension of ater respectively. Calculate the pressure P inside the bubble?
Answer:
Excess ot pressure inside the air bubble in water = \(\frac{2T}{r}\)
∴ Total pressure inside the air bubble
= atmospheric pressure + pressure due to liquid column + Excess pressure due to surface tension
= P + hρg + \(\frac{2T}{r}\)

Question 24.
Define beats?
Answer:
Formation of beats: When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second.
n = |f₁ – f₂| per second.

PART – III

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

Question 25.
Define centripetal acceleration and give any two examples?
Answer:
The acceleration that is directed towards the centre of the circle along the radius and perpendicular to the velocity of the particle is known as centripetal or radial or normal acceleration.
Example:-

1. In the case of planets revolving round the Sun or the moon revolving round the earth, the centripetal force is provided by the gravitational force of attraction between them.
2. For an electron revolving round the nucleus in a circular path, the electrostatic force of attraction between the electron and the nucleus provides the necessary centripetal force.

Question 26.
Write any six 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 \(\vec { A } \) and \(\vec { B } \) 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 θ = 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 θ = 0, i.e., θ = 0° or θ = 180°
(\(\vec { A } \) × \(\vec { B } \))_min = 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.

(VI) The self-vector products of unit vectors are thus zero.
\(\hat { i } \) × \(\hat { i } \) = \(\hat { j } \) × \(\hat { j } \) = \(\hat { k } \) × \(\hat { k } \) = 0

Question 27.
Show that the pressure of the gas is equal to two third of mean kinetic energy per unit volume?
Answer:
The internal energy of the gas is given by
U = \(\frac{3}{2}\) NkT
The above equation can also be written as
U = \(\frac{3}{2}\) PV
Since PV = NkT
P = \(\frac{2}{3}\) \(\frac{U}{V}\) = \(\frac{2}{3}\) u
From the equation (1), we can state that the pressure of the gas is equal to two thirds of internal energy per unit volume or internal energy density (u = \(\frac{U}{V}\))
Writing pressure in terms of mean kinetic energy density using equation.
P = \(\frac{1}{3}\) nm\(\overline { V^{ 2 } } \) = \(\frac{1}{3}\) ρ\(\overline { V^{ 2 } } \)
where ρ = nm = mass density (Note n is number density)
Multiply and divide R.H.S of equation (2) by 2, we get
P = \(\frac{2}{3}\)(\(\frac{ρ}{2}\) \(\overline { V^{ 2 } } \))
P = \(\frac{2}{3}\) \(\overline { KE } \)
From the equation (3), pressure is equal to \(\frac{2}{3}\) of mean kinetic energy per unit volume.

Question 28.
Derive an expression for gravitational potential energy?
Answer:
The gravitational tbrce is a conservative force and hence we can define a gravitational potential energy associated with this conservative force field.
Two masses m₁ and m₂, are initially separated by a distance r’. Assuming m₁ to be fixed in its position. work must be done on m₂ to move the distance from r’ to r.


To move the mass m₂, through an infinitesimal displacement d\(\vec { r } \) from \(\vec { r } \) to \(\vec { r } \) + d\(\vec { r } \) , work has to be done externally. This infinitesimal work is given by
dW = \(\vec { F } \)_ext . d\(\vec { r } \) ……………….. (1)
The work is done against the gravitational force, therefore,
\(\vec { F } \)_ext = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \) ………………. (2)
Substituting equation (2) in (1), we get
dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \).d\(\vec { r } \) ………………… (3)
d\(\vec { r } \) = dr \(\hat { r } \) ⇒ dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \).(dr \(\hat { r } \))
\(\hat { r } \).\(\hat { r } \) = 1 (Since both are unit vectors)
∴ dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \) dr …………….. (4)
Thus the total work done for displacing the particle from r’ to r is

This work done W gives the gravitational potential energy difference of the system of masses and m₁ and m₂ when the separation between them are r and r’ respectively.

Question 29.
A satellite orbiting the Earth in a circular orbit of radius 1600 km above the surface of the Earth. What is the acceleration experienced by satellite due to Earth’s gravitational force?
Answer:
g’ = g(1 – \(\frac { 2h }{ R_{ e } } \))
= g(\(\frac { 1-2\times 1600\times 10^{ 3 } }{ 6400\times 10^{ 3 } } \)) = g(1 – \(\frac{2}{4}\))
g’ = g (1 – \(\frac{1}{2}\)) = \(\frac{g}{2}\)
g’ = g (1- \(\frac{1}{2}\)) = \(\frac{g}{2}\)
g’ = \(\frac{8}{2}\) (or) g’ = \(\frac{9.8}{2}\) = 4.9ms^-2

Question 30.
Explain the v ariation of a g with latitude?
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 object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.
This centrifugal force is given by mωR’.
\(\mathrm{OP}_{z}, \cos \lambda=\frac{\mathrm{PZ}}{\mathrm{OP}}=\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\)
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 – ω²R 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.

Question 31.
A bullet of mass 50g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g = 10 ms^-2
Answer:
m₁ = 50 g = 0.05 kg; m₂ = 450 g = 0.45 kg
The speed of the bullet is u₁ The second body is at rest (u₂ = 0). Let the common velocity of the bullet and the object after the bullet is embedded into the object is v.
v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\)
v = \(\frac{0.05 u_{1}+(0.45 \times 0)}{(0.05+0.45)}\) = \(\frac{0.05}{0.50}\)u₁
The combined velocity is the initial velocity for the vertical upward motion of the combined bullet and the object. From second equation of motion,
v = \(\sqrt{2gh}\)
v = \(\sqrt{2 \times 10 \times 1.8}\) = \(\sqrt{36}\)
v = 6 ms^-1
Substituting this in the above equation, the value of u₁ is
6 = \(\frac{0.05}{0.50}\)u₁ or u₁ = \(\frac{0.05}{0.50}\) × 6 = 10 × 6
u₁ = 60ms^-1

Question 32.
If the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? If not, why?
Answer:
When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Question 33.
Calculate how many times more intense is 90 dB sound compared to 40 dB sound?
Answer:
Given Data:
L = log \(\frac { I }{ I_{ 0 } } \) = log I – log I₀
We get 90 dB = 9 B = log I₁ – log I₀ ……………….. (1)
40 dB = 4 B = logI₂ – logI₀ ………………….. (2)
Subtract (2) from (1)
50 dB = 5B = log I₁ – logI₂
5 = log₁₀ (\(\frac{I_{1}}{I_{2}}\))
\(\frac{I_{1}}{I_{2}}\) = 10⁵

PART – IV

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

Question 34 (a)
Obtain an expression for the time period T of a simple pendulum. The time period T depends on

1. Mass of the bob(m)
2. Length of the pendulum (l)
3. Acceleration due to gravity (g) at the place where the pendulum is suspended, (constant k = 2π)

Answer:
Example:
An expression for the time period T of a simple pendulum can be obtained by using this method as follows.
Let true period T depend upon

1. Mass m of the bob
2. Length l of the pendulum and
3. Acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is k = 2π.

Solution:

Here k is the dimensionless constant. Rewriting the above equation with dimensions.

Comparing the powers of M, L and T on both sides, a = 0, b + c = 0, -2c = 1
Solving for a, b and c ⇒ a = 0, b = 1/2, and c = -1/2
From the above equation
T = \(\mathrm{k} \mathrm{m}^{0} l^{1}=g^{-1}-2\)
T = \(k\left(\frac{1}{g}\right)^{1}\) = \(k \sqrt{1 / g}\)
Experimentally k = 2π, hence
T = \(2 \pi \sqrt{1 / g}\)

[OR]

(b) Obtain an expression for the escape speed in detail?
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v_i the initial total energy of the object is
\(\mathrm{E}_{i}=\frac{1}{2} \mathrm{M} v_{i}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) ………………. (1)
where, M_E is the mass of the Earth and R_E the radius of the Earth.
The term \(-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) is the potential energy of the mass M.
When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
E_F = 0
E_i = E_f ……………….. (2)
Substituting (1) in (2) we get,

Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v_i with v_e i.e..

From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms^-2) and R_e = 6400 km, the escape speed of the Earth is v_e = 11.2 kms^-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity.

Question 35 (a).
Derive an expression for loss of kinetic energy in perfect inelastic collision?
Answer:
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,
\(\mathrm{KE}_{i}=\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}\) …………….. (1)
Total kinetic energy after collision
KE_f = \(\frac{1}{2}\) (m₁ + m₂)v² …………….. (2)
Then the loss of kinetic energy is
Loss of KE, ∆Q = KE_f – KE_i = \(\frac{1}{2}\) (m₁ + m₂)v² – \(\frac{1}{2}\) m₁ u₁² – \(\frac{1}{2}\) m₂ u₂² ………………. (3)
Substituting equation v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\) in equation (3), and on simplying (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

(b) A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. Calculate the initial velocity of the shell?
Answer:
Given Data :
m = 200 gm = 0.2 kg; M = 4 kg.
Energy generated = 1.05 KJ = 1.05 × 10³ J
According to law of Conservation of linear momentum
mv = Mv’
∴v’ = (\(\frac{m}{M}\)) v
Total K.E of the gun and bullet

[OR]

(c) State parallel axis theorem?
Answer:
Parallel axis theorem: Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.

If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is given by the relation,
I = I_C + Md²

(d) Calculate the moment of inertia of uniform circular disc of mass 500 G radius 10 cm about

1. The diameter of the disc
2. The axis, tangent to the disc and parallel to its diameter
3. The axis through the centre of the disc and perpendicular to its plane

Answer:
1. Given Data: M = 500 g = 0.5 kg. R = 10 cm = 10 × 10^-2 m
Moment of inertia of disc about diameter = Id = \(\frac{1}{4}\) MR²
Id = \(\frac{1}{4}\) × 0.5 × 0.1 kg m² = 0.0125 kg m²

2. Apply a parallel axes theorem, moment of inertia of the disc about a tangent to the disc and parallel to the diameter of the disc
= \(\frac{1}{4}\) MR² + MR² = \(\frac{5}{4}\) MR² = \(\frac{5}{4}\) × 0.5 × 1
= 0.0625 kgm²

3. Moment of inertia of the disc about an axis passing through the centre of disc and perpendicular to the plane of the disc
= \(\frac{1}{2}\) MR² = \(\frac{1}{2}\) × 0.5 × 0.1 = 0.025 kgm²

Question 36 (a).
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it?
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum.

When two particles interact with each other, they exert equal and opposite forces on each other. The particle 1 exerts force \(\vec { F } \)₁₂ on particle 2 and particle 2 exerts an exactly equal and opposite force \(\vec { F } \)₁₂ on particle 1 according to Newton’s third law.
\(\vec { F } \)₂₁ = –\(\vec { F } \)₁₂ …………… (1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as
\(\vec { F } \)₁₂ = \(\frac{d \bar{p}_{1}}{d t}\) and \(\vec { F } \)₂₁ = \(\frac{d \vec{p}_{2}}{d t}\) ……………… (2)

Here \(\vec { P } \)₁ is the momentum of particle 1 which changes due to the force \(\vec { F } \)₁₂ exerted by particle 2. Further \(\vec { P } \)₂ is the momentum of particle 2. This changes due to \(\vec { F } \)₂₁ exerted by particle 1.
Substitute equation (2) in equation (1)

It implies that \(\vec { P } \)₁ + \(\vec { P } \)₂ = (constant vector always)

\(\vec { P } \)₁ + \(\vec { P } \)₂ is the total linear momentum of the two particles ( \(\vec { P } \)_tot = \(\vec { P } \)₁ + \(\vec { P } \)₂). It is also called as total linear momentum of the system. Flere, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system ( \(\vec { P } \)_tot) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec { P } \)₁ and \(\vec { P } \)₂ can vary,
in such a way that \(\vec { P } \)₁ + \(\vec { P } \)₂ is a constant vector.

The forces \(\vec { F } \)₁₂ and \(\vec { F } \)₁₂ are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

To find the recoil velocity of a gun when a bullet is fired from it:
Consider the firing of a gun. Here the system is Gun + bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec { P } \)₁ be the momentum of the bullet and \(\vec { P } \)₂ the momentum of the gun before firing. Since initially both are at rest.

\(\vec { P } \)₁ = 0, \(\vec { P } \)₂ = 0.

Total momentum before firing the gun is zero, \(\vec { P } \)₁ + \(\vec { P } \)₂ = 0

According to the law of conservation of linear momentum, total linear momemtum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec { P } \)₁ + \(\vec { P } \)₂. To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec { P } \)₂ to \(\vec { P } \)₂. Due to the conservation of linear momentum, \(\vec { P } \)₁+ \(\vec { P } \)₂‘= 0. It implies that \(\vec { P } \)₁‘ = –\(\vec { P } \)₂ the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum \(\vec { P } \)₂. It is called ‘recoil momentum’. This is an example of conservation of total linear momentum.

[OR]

(b) Derive an expression for escape speed?
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v;, the initial total energy of the object is
\(E_{i}=\frac{1}{2} M v_{i}^{2}-\frac{G M M_{E}}{R_{E}}\) ………………. (1)
where, M_E is the mass of the Earth and RE the radius of the Earth. The term \(-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) is the potential energy of the mass M.

When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
E_f = 0

According to the law of energy conservation,
E_i – E_f = 0 …………….. (2)

Substituting (1) in (2) we get,
\(\frac{1}{2} \mathrm{M} v_{i}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{F}}}=0\)
\(\frac{1}{2} \mathrm{M} v_{i}^{2}=\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) …………….. (3)

Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v_i with v_e, i.e.,

From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms^-2) and R_e = 6400 km, the escape speed of the Earth is v_e = 11.2 kms^-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity.

Question 37 (a).
Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{dQ}{dt}\) ∝ (T – T_s) …………….. (1)

The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,

T = Temperature of the object
T_s = Temperature of the surrounding

From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.

Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount of T in time dt, then the amount of heat lost is,
dQ = msdT ………………. (2)
Dividing both sides of equation (2) by dt
\(\frac{dQ}{dt}\) = \(\frac{msdT}{dt}\) ……………….. (3)
From Newton’s of cooling
\(\frac{dQ}{dt}\) ∝ (T – T_s)
\(\frac{dQ}{dt}\) = -a(T – T_s) ……………….. (4)

Where a is some positive constant.
From equation (3) and (4)

Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides, we get
\(\mathrm{T}=\mathrm{T}_{\mathrm{s}}+b_{2} e^{\frac{-a}{m s}}\) ……………. (6)
Here b₂ = _e^b1 = Constant

[OR]

(b) Derive an expression for pressure exerted by the gas on the wall 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 in 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 sanie 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 bêfore 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 = 2mv_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 &. 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 ii is number of molecules per
unit volume \(\frac{N}{V}\) 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.

Te no.of molecules that hit the right side wall in a time interval ∆t

= \(\frac{n}{2} \mathrm{A} v_{x} \Delta t\) ……………….. (1)
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 = \(\frac{\Delta p}{\Delta t}=n m \mathrm{A} v_{x}^{2}\) ……………….. (3)
Pressure P = force divided by the area of the wall

P = \(\frac{F}{A}\) = nmv²_x …………………. (4)
Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term v²_x by the average \(\bar{v}_{x}^{2}\) in equation (4)

P = nm\(\bar{v}_{x}^{2}\) ………………… (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 = \(\frac{1}{3}\)nm \(\bar{v}^{2}\) or P = \(\frac{1}{3}\) \(\frac{N}{V}\) m\(\bar{v}^{2}\) as [n = \(\frac{N}{V}\)] ……………… (6)

Question 38 (a).
Explain with graphs the difference between work done by a constant force and by a variable force. Arrive at an expression for power and velocity. Give some examples for the same?
Answer:
Work done by a constant force: When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
dW = (F cos θ) dr ………………. (1)
The total Work done in producing a displacement from initial position r_i to final position r_f is,

The graphical representation of the work done by a constant force is shown in figure given below. The area under the graph shows the work done by the constant force.

Work done by a variable force:
When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation.

dW = F cos θ dr [F cos θ is the component of the variable force F]
where, F and θ are variables. The total work done for a displacement from initial position r_i to final position r_f is given by the relation,
W = \(\int_{r_{i}}^{r_{f}} d \mathrm{W}=\int_{r_{i}}^{r_{f}} \mathrm{F} \cos \theta d r\) ………………. (4)

A graphical representation of the work done by a variable force is shown in figure given below. The area under the graph is the work done by the variable force.

Expression for power and velocity

The work done by a force \(\vec { F } \) for a displacement \(\bar { dr } \) is
W = ∫\(\vec { F } \).\(\vec { dr } \) ……………. (1)
Left hand side of the equation (1) can be written as
W = ∫dW = ∫\(\frac{dW}{dt}\) (multiplied and divided by dt) ………………… (2)
Since, velocity is \(\vec { v } \) = \(\frac{d \vec{r}}{d t}\); \(\vec { dr } \) = \(\vec { v } \) dt. Right hand side of the equation (I) can be written as

Substituting equation (2) and equation (3) in equation (1), we get

This relation is true for any arbitrary value of di. This implies that the term within the bracket must be equal to zero, i.e.,
\(\frac{dW}{dt}\) – \(\vec { F } \).\(\vec { v } \) = 0 Or \(\frac{dW}{dt}\) = \(\vec { F } \).\(\vec { v } \)
Hence power P = \(\vec { F } \).\(\vec { v } \)

[OR]

(b) Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{dQ}{dt}\) ∝ (T – T_s)
The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T_s = Temperature of the surrounding
From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.
Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in
time dt, then the amount of heat lost is,
dQ = msdT ………………. (2)
Dividing both sides of equation (2) by dt

\(\frac{dQ}{dt}\) = \(\frac{msdT}{dt}\) ……………….. (3)
From Newton’s law of cooling
\(\frac{dQ}{dt}\) ∝ (T – T_s)
\(\frac{dQ}{dt}\) -a(T – T_s) …………………. (4)
Where a is a positive constant.
From equation (3) and (4)
– a(T – T_s) = ms \(\frac{dT}{dt}\)
\(\frac{d T}{T-T_{s}}\) = -a\(\frac{a}{ms}\) dt ………………. (5)
Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides we get,
\(\mathrm{T}=\mathrm{T}_{s}+b_{2} e^{\frac{-a}{m s}}\) ………………… (6)
Here b₂ = e^b1 = constant

Students can Download Tamil Nadu 11th Maths Model Question Paper 5 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 5 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 – I

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

Question 1.
If n((A × B) ∩ (A × C)) = 8 and n(B ∩ C) = 2 then n(A) = …………………
(a) 6
(b) 4
(c) 8
(d) 16
Answer:
(b) 4

Question 2.
The value of log₃ \(\frac{1}{81}\) is ……………….
(a) -2
(b) -8
(c) -4
(d) -9
Answer:
(c) -4

Question 3.
The value of log₃ 11 log₁₁ 13 log₁₃ 15 log₁₅ 27 log₂₇ 81 is ……………………
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 4.
The value of sin(45° + θ) – cos (45° – θ) is …………………..
(a) 2 cos θ
(b) 1
(c) 0
(d) 2 sin θ
Answer:
(c) 0

Question 5.
If tanα and tan β are the roots of x² + ax + b = 0 then \(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}\) is equal to ……………………..
(a) \(\frac{b}{a}\)
(b) \(\frac{a}{b}\)
(c) –\(\frac{a}{b}\)
(d) –\(\frac{b}{a}\)
Answer:
(c) –\(\frac{a}{b}\)

Question 6.
If a² – aC₂ = a² – aC₄ then the value of a is …………………….
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 7.
If ⁿP_r = 840, ⁿC_r= 35 then n = …………………..
(a) 1
(b) 6
(c) 5
(d) 4
Answer:
(a) 1

Question 8.
If 2x² + 3xy – cy² = 0 represents a pair of perpendicular lines then c = …………………….
(a) -2
(b) \(\frac{1}{2}\)
(c) – \(\frac{1}{2}\)
(d) 2
Answer:
(d) 2

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.
The point on the line 2x – 3y = 5 is equidistance from (1, 2) and (3, 4) is …………………..
(a) (7, 3)
(b) (4, 1)
(c) (1,-1)
(d) (3, 4)
Answer:
(b) (4, 1)

Question 11.
Let A and B be two symmetric matrices of same order. Then which one of the following statement is not true?
(a) A + B is a symmetric matrix
(b) AB is a symmetric matrix
(c) (AB) = (BA)^T
(d) A^TB = AB^T
Answer:
(d) A^TB = AB^T

Question 12.
If [3 -1 2] B = [5, 6] then the order of B is ………………….
(a) 3 × 2
(b) 2 × 3
(c) 3 × 1
(d) 1 × 1
Answer:
(a) 3 × 2

Question 13.
1f \(\underset { x\rightarrow 0 }{ lim } \) \(\frac{sin px}{tan 3x}\) = 4 then the value of p is …………………….
(a) 6
(b) 9
(c) 12
(d) 4
Answer:
(c) 12

Question 14.
For \(\vec { a } \) = \(\vec { i } \) + \(\vec { j } \) – 2\(\vec { k } \), \(\vec { b } \) = \(\vec { i } \) + 2\(\vec { j } \) + \(\vec { k } \) and \(\vec { c } \) = \(\vec { i } \) – 2\(\vec { j } \) + 2\(\vec { k } \) the unit vector parallal to is \(\vec { a } \)
+ \(\vec { b } \) + \(\vec { c } \) is ……………………….
(a) \(\frac{\vec{i}+\vec{j}-\vec{k}}{\sqrt{3}}\)
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)
(c) \(\frac{\vec{i}+\vec{j}+\vec{k}}{3}\)
(d) \(\frac{\vec{i}-\vec{j}+\vec{k}}{\sqrt{6}}\)
Answer:
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)

Question 15.
The differential co-efficient of log₁₀ x with respect to log_x 10 is …………………….
(a) 1
(b) -(log₁₀x)²
(c) (log_x10)²
(d) \(\frac { x^{ 2 } }{ 100 } \)
Answer:
(b) -(log₁₀x)²

Question 16.
\(\frac{d}{dx}\)(e^x+5logx) is …………………..
(a) e^xx⁴(x + 5)
(b) e^xx(x + 5)
(c) e^x + \(\frac{5}{x}\)
(d) e^x – \(\frac{5}{x}\)
Answer:
(a) e^xx⁴(x + 5)

Question 17.
If f(x) = x tan^-1x thenf'(1) = ……………………..
(a) 1 + \(\frac { \pi }{ 4 } \)
(b) \(\frac{1}{2}\) + \(\frac { \pi }{ 4 } \)
(c) \(\frac{1}{2}\) – \(\frac { \pi }{ 4 } \)
(d) 2
Answer:
(b) \(\frac{1}{2}\) + \(\frac { \pi }{ 4 } \)

Question 18.
∫cosec xdx = …………………….
(a) log tan \(\frac{x}{2}\) + c
(b) -log (cosec x + cot x) + c
(c) log (cosecx – cot x) + c
(d) all of them
Answer:
(d) all of them

Question 19.
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) \(\frac{7}{128}\)

Question 20.
Two items are chosen from a lot containing twelve items of which four are defective. Then the probability that atleast one of the item is defective is …………………..
(a) \(\frac{19}{33}\)
(b) \(\frac{17}{33}\)
(c) \(\frac{23}{33}\)
(d) \(\frac{13}{34}\)
Answer:
(a) \(\frac{19}{33}\)

PART – II

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

Question 21.
For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A?
Answer:
A × A = 16 elements = 4 × 4
⇒ A has 4 elements
∴ A = {0, 1, 2, 3}

Question 22.
Prove that \(\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}\) = \(\frac{1+sinθ}{cosθ}\)
Answer:

Question 23.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Answer:
No. of non-collincar points = 15
To draw a Triangle we need 3 points
∴ Selecting 3 from 15 points can be done in ¹⁵C₃ ways.
∴ No. of Tnangle formed ¹⁵C₃
= \(\frac{15 \times 14 \times 13}{3 \times 2 \times 1}\) = 455

Question 24.
Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1?
Answer:
Givc sum of the intercepts = 1
⇒ when x intercept a then y intercept = 1 – a
Equation of the line is \(\frac{x}{a}\) + \(\frac{y}{1-a}\) = 1
The line passes through (8, 3) ⇒ \(\frac{8}{a}\) + \(\frac{3}{1-a}\) = 1
(i.e) 8 (1 – a) + 3a = a(1 – a)
8 – 8a + 3a = a – a²
a² – 6a + 8 =0
(a – 2)(a – 4) = 0 ⇒a = 2 or 4
1. When a = 2 equation of the line is \(\frac{x}{2}\) + \(\frac{y}{-2}\) = 1 (i.e) \(\frac{x}{2}\) – y = 1 ⇒ x – 2y = 2
2.When a = 4 equation of the line is \(\frac{x}{4}\) + \(\frac{y}{1-4}\) = 1 (i.e) \(\frac{x}{4}\) – \(\frac{y}{3}\) = 1 ⇒ 3x – 4y = 12

Question 25.
Find the values of p, q, r, & s if \(\left[\begin{array}{ccc}
p^{2}-1 & 0 & -31-q^{3} \\
7 & r+1 & 9 \\
-2 & 8 & s-1
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & -4 \\
7 & \frac{3}{2} & 9 \\
-2 & 8 & -\pi
\end{array}\right]\)
Answer:
When two matrices (of some order) are equal then their correspondings entries are equal.

Question 26.
Find |\(\vec { a } \) × \(\vec { b } \)| where \(\vec { a } \) = 3\(\vec { i } \) + 4\(\vec { j } \) and \(\vec { b } \) = \(\vec { i } \) + \(\vec { j } \) + \(\vec { k } \)
Answer:

Question 27.
At the given point x₀ discover whether the given function is continous or discontinous citing the reasons for your answer?
Answer:

Question 28.
Evaluate y = xe^-x2
Answer:

Question 29.
Evaluate ∫\(\frac{1}{x logx}\) dx?
Answer:

Question 30.
Evaluate [((256)^-1/2)^-1/4]³
Answer:

PART – III

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

Question 31.
Find the largest possible domain for the real valued functionsf defined by f(x) = \(\sqrt { x^{ 2 }-5x+6 } \)

Question 32.
Show that tan 75° + cot 75° = 4?

Question 33.
There are 10 bulbs in a room. Each one of them can be operated independently. Find the number of ways in which the room can be illuminated?

Question 34.
Find the \(\sqrt [ 3 ]{ 126 } \) approximately to two decimal places?

Question 35.
Find the equation of the line passing through the point of intersection 2x + y = 5 and x + 3y + 8 = O and parallel to the line 3x + 4y = 7?

Question 36.
If \(\left|\begin{array}{ccc}
a & b & a \alpha+b \\
b & c & b \alpha+c \\
a \alpha+b & b \alpha+c & 0
\end{array}\right|\) = 0 prove that a, b, c are in G.P. or a is a root of ax² + 2bx + c =0

Question 37.
Evaluate \(\underset { x\rightarrow 3 }{ lim } \)\(\frac { x^{ 2 }-9 }{ x-3 } \) ¡f it exists by finding f(3^–) and f(3⁺)

Question 38.
Find the derivative of tan^-1 (1 + x²) with respect x² + x + 1?

Question 39.
Evaluate: ∫x⁵e^{x2 dx}

Question 40.
Prove that the line segment joining the mid points of the adjacent sides of a quadrilateral from parllelogram?

PART – IV

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

Question 41.
(a) Graph the functions f(x) = x³ and g(x) = \(\sqrt [ 3 ]{ x } \) on the same coordinate plane. Find fog and graph it on the plane as well. Explain your results.

[OR]

(b) 1f x = -2 is one root of x³ – x² – 17x = 22 then find the other roots of the equation?

Question 42.
(a) lf A + B + C= it prove that cosA + cos B + cosC = 1 + 4 sin (\(\frac{A}{2}\)) sin (\(\frac{B}{2}\) sin (\(\frac{C}{2}\))

[OR]

(b) If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) show that A² – 4A – 5I = O

Question 43.
(a) If ⁿ⁺¹C₈ : ^(n-3)P₄ = 57 : 16, find the value of n?

[OR]

(b) If the letters of the word IITJEE arc permuted in all possible ways and the strings thus formed are arranged in the lexicographic order, find the rank of the word IITJEE?

Question 44.
(a) The line \(\frac{x}{a}\) + \(\frac{y}{b}\)= 1 moves in such a way that \(\frac { 1 }{ a^{ 2 } } \) + \(\frac { 1 }{ b^{ 2 } } \) = \(\frac { 1 }{ c^{ 2 } } \) where c is a constant. Find the locus of the foot of the perpendicular from the origin on the given line?

[OR]

(b) Show that the equation 4x² + 4xy + y² – 6x – 3y – 4 = 0 represents a pair of parallel lines. Find the distance between them?

Question 45 (a).
Prove that \(\left|\begin{array}{lll}
1 & x^{2} & x^{3} \\
1 & y^{2} & y^{3} \\
1 & z^{2} & z^{3}
\end{array}\right|\) = (x – y) (y – z) (z – x) (xy + yz + zx)

(b) Evaluate \(\underset { x\rightarrow \infty }{ lim } \) \(\frac{3}{x-2}\) – \(\frac{2 x+11}{x^{2}+x-6}\)

Question 46.
(a) Evaluate \(\frac{1}{6 x-7-x^{2}}\)

(b) Evaluate ∫e^tan-1x (\(\frac { 1+x+x^{ 2 } }{ 1+x^{ 2 } } \)) dx

Question 47 (a).
Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?

[OR]

(b) Firm manufactures PVC pipes in three plants viz. X, Y and Z. The daily production volumes from the three firms X, Y and Z are respectively 2000 units, 3000 units and 5000 units. It is known from the past experience that 3% of the output from plant X, 4% from plant Y and 2% from plant Z are defective. A pipe is selected at random from a day’s total production,

1. find the probability that the selected pipe is a defective one.
2. if the selected pipe ¡s a defective, then what is the probability that it was produced by plant Y?

Students can Download Tamil Nadu 11th Maths Model Question Paper 4 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 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 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 – I

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

Question 1.
If n(A) = 2 and n(B∪C) = 3 then n[(A × B) ∪ (A × C)] is ………………..
(a) 2³
(b) 3²
(c) 6
(d) 5
Answer:
(c) 6

Question 2.
For any two sets A and B, A∩(A∪B) = …………………….
(a) B
(b) ∅
(c) A
(d) none of these
Answer:
(c) A

Question 3.
cos 1° + cos 2° + cos 3° + cos 4° + cos 179° = …………………
(a) 0
(b) 1
(c) -1
(d) 89
Answer:
(a) 0

Question 4.
The value of log₉ 27 is ……………………
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{3}{4}\)
(d) \(\frac{4}{3}\)
Answer:
(b) \(\frac{3}{2}\)

Question 5.
The value of \(\frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta}\) = ………………..
(a) tan3θ
(b) tan6θ
(c) cot3θ
(d) cot6θ
Answer:
(b) tan6θ

Question 6.
In 3 fingers the number of ways 4 rings can be worn in ……………………. ways.
(a) 43 – 1
(b) 34
(c) 68
(d) 64
Answer:
(d) 64

Question 7.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is ……………….
(a) 11
(b) 12
(c) 10
(d) 6
Answer:
(b) 12

Question 8.
The H.M of two positive number whose AM and G.M. are 16, 8 respectively is ………………..
(a) 10
(b) 6
(c) 5
(d) 4
Answer:
(d) 4

Question 9.
The co-efficient of the term independent of x in the expansion of (2x+\(\frac{1}{3x}\))⁶ is …………………
(a) \(\frac{160}{27}\)
(b) \(\frac{160}{27}\)
(c) \(\frac{80}{3}\)
(d) \(\frac{80}{9}\)
Answer:
(a) \(\frac{160}{27}\)

Question 10.
The value of \(\left|\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right|^{2}\) is ………………..
(a) abc
(b) -abc
(c) 0
(d) a²b²c²
Answer:
(d) a²b²c²

Question 11.
The value of x for which the matrix A = \(\left[\begin{array}{cc}
e^{x-2} & e^{7+x} \\
e^{2+x} & e^{2 x+3}
\end{array}\right]\) is singular is …………………..
(a) 9
(b) 8
(c) 7
(d) 6
Answer:
(b) 8

Question 12.
If |\(\vec { a } \) + \(\vec { b } \)| = 60, |\(\vec { a } \) – \(\vec { b } \)| = 40 and |\(\vec { b } \)| = 46 then |\(\vec { a } \)| is …………………
(a) 42
(b) 12
(c) 22
(d) 32
Answer:
(c) 22

Question 13.
Given \(\vec { a } \) = 2\(\vec { i } \) + \(\vec { j } \) – 8\(\vec { k } \) and \(\vec { b } \) = \(\vec { i } \) + 3\(\vec { j } \) – 4\(\vec { k } \) then |\(\vec { a } \) + \(\vec { b } \)| = ………………….
(a) 13
(b) \(\frac{13}{3}\)
(c) \(\frac{4}{13}\)
(d) \(\frac{3}{13}\)
Answer:
(a) 13

Question 14.
If f(x) = \(\left\{\begin{array}{ccc}
k x & \text { for } & x \leq 2 \\
3 & \text { for } & 2
\end{array}\right.\) is continous at x = 2 then the value of k is ……………………
(a) \(\frac{3}{4}\)
(b) 0
(c) 1
(d) \(\frac{4}{3}\)
Answer:
(c) 1

Question 15.
If f: R→R is defined by f(x) = |x – 3| + |x – 4| for x∈R then \(\lim _{x \rightarrow 3^{-}}\) f(x) is equal to ………………..
(a) -2
(b) -1
(c) 0
(d) 1
Answer:
(c) 0

Question 16.
\(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+5 x+3}{x^{2}+x+3}\right)^{x}\) is ………………..
(a) e⁴
(b) e²
(c) e³
(d) 1
Answer:
(a) e⁴

Question 17.
\(\int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1}\) dx is ………………..
(a) e tan^-1(x + 1)
(b) tan^-1(e^x) + c
(c) e^x \(\frac{\left(\tan ^{-1} x\right)^{2}}{2}\) + c
(d) e^xtan^-1x + c
Answer:
(d) e^xtan^-1x + c

Question 18.
∫ \(\frac { secx }{ \sqrt { cos2x } } \) dx = …………………..
(a) tan^-1(sin x) + c
(b) 2 sin^-1(tan x) + c
(c) tan^-1(cos x) + c
(d) sin^-1 (tan x) + c
Answer:
(d) sin^-1 (tan x) + c

Question 19.
\(\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx = ………………..
(a) x + c
(b) \(\frac { x^{ 3 } }{ 3 } \) + c
(c) \(\frac { 3 }{ x^{ 3 } } \) + c
(d) \(\frac { 1 }{ x^{ 2 } } \) + c
Answer:
(b) \(\frac { x^{ 3 } }{ 3 } \) + c

Question 20.
It is given that the events A and B are such that P(A) = \(\frac{1}{4}\), P(A/B) = \(\frac{1}{2}\), P(B/A) = \(\frac{2}{3}\) then P(B) = …………………
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(b) \(\frac{1}{3}\)

PART – II

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

Question 21.
In the set Z of integers, define mRn if m – n is a multiple of 12. Prove that R is an equivalence relation?
Answer:
As m – m = 0 and 0 = 0 × 12, we have 0 is a multiple of 12; hence mRm proving that R is reflexive.
Let mRn. Then m – n = 12k for some integer k; thus n – m = 12(-k) and hence nRm.
This shows that R is symmetric.
Let mRn and nRp: then m – n = 12k and n – p = 12l for some integers k and l.
So m – p = 12(k + l) and hence mRp. This shows that R is transitive.
Thus R is an equivalence relation.

Question 22.
Simplify:
\(\frac { 1 }{ 2+\sqrt { 3 } } \) + \(\frac { 3 }{ 4-\sqrt { 5 } } \) + \(\frac { 6 }{ 7-\sqrt { 8 } } \)
Answer:

Question 23.
Find the value of sin 22 \(\frac{1}{2}\)°?
Answer:
We know that cos θ = 1 – 2 sin² \(\frac{θ}{2}\) ⇒ sin \(\frac{θ}{2}\) = ±\(\sqrt { \frac { 1-cos2\theta }{ 2 } } \)
Take θ = 45°, we get sin \(\frac{45°}{2}\) = ±\(\sqrt { \frac { 1-cos45°}{ 2 } } \), (taking positive sign only, since 22\(\frac{1}{2}\)° lies in the first quadrant)
Thus, sin 22\(\frac{1}{2}\)° = \(\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}\).

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 2nd hour, 4th
hour and wth hour?
Answer:
No. of bacteria at the beginning = 30
No. of bacteria after 1 hour = 30 × 2 = 60
No. of bactena 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.
If the sum of the distance of a moving point in a plane from the axis is 1, then find the locus of the point?
Answer:
Let coordinates of a moving point P be (x, y).
Given that the sum of the distances from the axis to the point is always 1.
∴|x| + |y| = 1 ⇒ x + y = 1
⇒- x- y = 1 ⇒ x + y = 1
⇒ x – y = 1
Hence, these equations give us the locus of the point P which is a square.

Question 26.
If (a, a + b, a + b + c) is one set of direction ratios of the line joining (1, 0, 0) and (0, 1, 0) then find a set of values of a, b, c?
Answer:
Let A be the point (1, 0, 0) and B be the point (0, 1, 0) (i.e.,) \(\overline { OA } \) = \(\hat { i } \) and \(\overline { OB } \) = \(\hat { j } \).
Then \(\overline { AB } \) = \(\overline { OB } \) – \(\overline { OA } \) = \(\hat { j } \) – \(\hat { i } \) = –\(\hat { i } \) + \(\hat { j } \)
= (-1, 1, 0)
= (a, a + b, a + b + c)
⇒ a = -1, a + b = 1 and a + b + c = 0
⇒ a = -1, ⇒ -1 + b = 1; a + b + c = 0
⇒ b = 2; -1 + 2 + c = 0 ⇒ c + 1 = 0
⇒ c = -1
∴ a = -1, b = 2, c = -1.
Note: if we taken \(\overline { BA } \) then we get a = 1, b = -2 and c = 1.

Question 27.
Find \(\frac{dy}{dx}\) for y = (x² + 4x + 6)⁵
Answer:
Let u = x² + 4x + 6
⇒ \(\frac{du}{dx}\) = 2x + 4
Now y = u⁵ = \(\frac{dy}{du}\) = 5u⁴
∴ \(\frac{dy}{dx}\) = \(\frac{dy}{du}\) × \(\frac{du}{dx}\) = 5u⁴ (2x + 4)
= 5(x² + 4x + 6)⁴ (2x + 4)
= 5(2x + 4) (x² + 4x + 6)⁴

Question 28.
Evaluate ∫\(\sqrt { 25x^{ 2 }-9 } \) dx?
Answer:

Question 29.
A bag contains 5 white and 7 black balls. 3 balls are drawn at random. Find the probability that

1. all are white
2. one white and 2 black.

Answer:
Number of white balls = 5
Number of black balls = 7
Total number of balls = 12
Selecting 3 from 12 balls can be done in
¹²C₃ = \(\frac{12 \times 11 \times 10}{3 \times 2 \times 1}\) = 220 ways
∴n(S) = 220

1. Let A be the selecting 3 white balls.
∴n(A) = ⁵C₃ = ⁵C₂ = \(\frac{5×3}{2×1}\) = 10
∴P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{10}{220}\) = \(\frac{1}{22}\)

2. Let B be the event of selecting one white and 2 black balls.
∴n(B) = ⁵C₁ × ⁷C₂ = (5) (\(\frac{7×6}{2×1}\)) = 5(21) = 105
∴P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{105}{220}\) = \(\frac{21}{44}\).

Question 30.
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k?
Answer:
Area of ∆ with vertices (k, 2) (2, 4) and (3, 2) = \(\frac{1}{2}\) \(\left|\begin{array}{lll}
k & 2 & 1 \\
2 & 4 & 1 \\
3 & 2 & 1
\end{array}\right|\) = 4 given
⇒ \(\left|\begin{array}{lll}
k & 2 & 1 \\
2 & 4 & 1 \\
3 & 2 & 1
\end{array}\right|\) = 2(4) = 8
(i.e.,) k(4 – 2) – 2(2 – 3) + 1(4 – 12) ± 8
(i.e.,) 2k – 2(-1) + 1(-8) = ± 8
(i.e.,) 2k + 2 – 8 = 8
(i.e.,) 2k = 8 + 8 – 2 = 14
k = 14/2 = 7
∴k = 7
So k = 7 (or) k = -1.

2k + 2 – 8 = -8
⇒2k = – 8 + 8 – 2
2k = – 2
k = -1

PART – III

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

Question 31.
In the set Z of integers define mRn if m – n is a multiple of 12. Prove that R is an equivalence relation?

Question 32.
Prove that \(\frac{sin4x+sin2x}{cos4x+cos2x}\) = tan 3x?

Question 33.
A polygon has 90 diagonals. Find the number of its sides?

Question 34.
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)ⁿ are equal?

Question 35.
Find the equation of the line passing through the point (1, 5) and also divides the co-ordinate axes in the ratio 3 : 10?

Question 36.
Prove that \(\left|\begin{array}{ccc}
1 & x & x \\
x & 1 & x \\
x & x & 1
\end{array}\right|^{2}=\left|\begin{array}{ccc}
1-2 x^{2} & -x^{2} & -x^{2} \\
-x^{2} & -1 & x^{2}-2 x \\
-x^{2} & x^{2}-2 x & -1
\end{array}\right|\)

Question 37.
If G is the centroid of a traiangle ABC prove that \(\overline { GA } \) + \(\overline { GB } \) + \(\overline { GC } \) = 0

Question 38.
Find \(\frac{dy}{dx}\) for y = \(\sqrt { 1+tan2x } \)?

Question 39.
Evaluate \(\frac { \sqrt { x } }{ 1+\sqrt { x } } \) dx?

Question 40.
Find the relation between a and b if \(\underset { x\rightarrow 3 }{ lim } \) f(x) exists where f(x) = \(\left\{\begin{array}{cc}
a x+b & \text { if } x>3 \\
3 a x-4 b+1 \text { if } x<3
\end{array}\right.\)

PART – IV

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

Question 41 (a).
From the curve y = |x|, draw(i) y = |x-1| + 1

1. y = |x + 1| + 1
2. y = |x + 2| – 3

[OR]

(b) Resolve into partial fraction \(\frac { x+4 }{ (x^{ 2 }-4)(x+1) } \)

Question 42 (a).
Find the number of positive integers greater than 6000 and less than 7000 which arc divisible by 5, provided that no digit is to be repeated?

[OR]

(b) If ⁿP_r = ⁿP_r+1 and ⁿC_r = ⁿC_r-1, find the values of n and r?

Question 43 (a).
In a ∆ABC, prove that b² sin 2C + c² sin 2B = 2bc sin A?

[OR]

(b) Differentiate the following s(t) = \(\sqrt[4]{\frac{t^{3}+1}{t^{3}-1}}\)?

Question 44 (a).
Find the equation of the lines make an angle 60° with the positive x axis and at a distance 5\(\sqrt{2}\) units measured from the point (4, 7) along the line x – y + 3 = 0

[OR]

(b) If y = A cos4x + B sin 4x, A and B are constants then Show that y₂ + 16y = 0

Question 45 (a).
Find the sum up to the 17^th term of the series \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}\) + …………..

[OR]

(b) A shopkeeper in a Nuts and Spices shop makes gifi packs of cashew nuts, raisins and almonds?

1. Pack I contains 100 gm of cashew nuts, 100 gm of raisins and 50 gm of almonds.
2. Pack-II contains 200 gm of cashew nuts, 100 gm of raisins and 100 gm of almonds.
3. Pack-III contains 250 gm of cashew nuts, 250 gm of raisins and 150 gm of almonds.
4. The cost of 50 gm of cashew nuts is ₹50, 50 gm of raisins is ₹10. and 50gm of almonds is ₹60. What is the cost of each gift pack?

Question 46 (a).
Find matrix C if A = \(\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\), B = \(\begin{bmatrix} -3 & 2 \\ 4 & -1 \end{bmatrix}\) and 5C + 28= A?

[OR]

(b) The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that

1. it will get at least one of the two awards
2. it will get only one of the awards.

Question 47 (a).
\(\underset { \alpha \rightarrow 0 }{ lim } \) \(\frac { sin(\alpha ^{ n }) }{ (sin\alpha )^{ m } } \)

[OR]

(b) Evaluate I = sin^-1 (\(\frac { 2x }{ (1+x)^{ 2 } } \)) dx?

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 Maths Model Question Paper 2 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 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 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 set of all real numbers. Consider the following subsets of the plane R × R:
S= {(x, y) : y = x + 1 and 0 < x < 2} and T = {(x, y): x – y is an integer}
Then which of the following is true?
(a) T is an equivalence relation but S is not an equivalence relation.
(b) Neither S nor T is an equivalence relation
(c) Both S and T are equivalence relation
(d) S is an equivalence relation but T is not an equivalence relation.
Answer:
(a) T is an equivalence relation but S is not an equivalence relation.

Question 2.
If the set A has m elements the set B has n elements and the number of elements in A × B is …………………
(a) m + n
(b) mn
(c) \(\frac{m}{n}\)
(d) m²
Answer:
(b) mn

Question 3.
If \(\frac{ax}{(x+2)(2x-3)}\) = \(\frac{2}{x+2}\) + \(\frac{3}{2x-3}\) then a = ……………….
(a) 8
(b) 7
(c) 5
(d) 4
Answer:
(b) 7

Question 4.
The number of solutions of x² + |x – 1| = 1 is ………………….
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
(c) 2

Question 5.
If a, 8, b are in A.P. a, 4, b are in G.P. and a, x, b are in H.P then x = ………………..
(a) 2
(b) 1
(c) 4
(d) 16
Answer:
(a) 2

Question 6.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of points of intersection are ……………….
(a) 45
(b) 40
(c) 10!
(d) 2¹⁰
Answer:
(a) 45

Question 7.
The value of e^2logx = …………………..
(a) 2x
(b) x²
(c) \(\sqrt{x}\)
(d) \(\frac{x}{2}\)
Answer:
(b) x²

Question 8.
The n^th term of the sequence 1, 2, 4, 7, 11 …. is …………………
(a) n³ + 3n² + 2n
(b) n³ – 3n² + 3n
(c) n\(\frac{(n+1)(n+2)}{3}\)
(d) \(\frac { n^{ 2 }-n+2 }{ 2 } \)
Answer:
(d) \(\frac { n^{ 2 }-n+2 }{ 2 } \)

Question 9.
The last term in the expansion (2+\(\sqrt{3}\))⁸ is ………………
(a) 81
(b) 27
(c) 9
(d) 3
Answer:
(a) 81

Question 10.
A line perpendicular to the line 5x -y = 0 forms a triangle with the coordinate axes. If the area of the triangle is 5sq.units, then its equation is …………………..
(a) x + 5y ± 5\(\sqrt{2}\) = 0
(b) x – 5y ± 5\(\sqrt{2}\) = 0
(c) 5x + y ± 5\(\sqrt{2}\) = 0
(d) 5x – y ± 5\(\sqrt{2}\) = 0
Answer:
(a) x + 5y ± 5\(\sqrt{2}\) = 0

Question 11.
A factor of the determinant \(\left|\begin{array}{ccc}
x & -6 & -1 \\
2 & -3 x & x-3 \\
-3 & 2 x & x+2
\end{array}\right|\) is ……………….
(a) x + 3
(b) 2x – 1
(c) x – 2
(d) x – 3
Answer:
(a) x + 3

Question 12.
If λ\(\vec { a } \) + 2λ\(\vec { j } \) + 2λ\(\vec { k } \) is a unit vector then the value of λ is ………………
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{2}\)
Answer:
(a) \(\frac{1}{3}\)

Question 13.
One of the diagonals of parallelogram ABCD with \(\vec { a } \) and \(\vec { b } \) are adjacent sides is \(\vec { a } \) + \(\vec { b } \). The other diagonal BD is ………………….
(a) \(\vec { a } \) – \(\vec { b } \)
(b) \(\vec { a } \) – \(\vec { b } \)
(c) \(\vec { a } \) + \(\vec { b } \)
(d) \(\frac{\vec{a}+\vec{b}}{2}\)
Answer:
(b) \(\vec { a } \) – \(\vec { b } \)

Question 14.
If (1, 2, 4) and (2, -3λ, -3) are the initial and terminal points of the vector \(\vec { i } \) + 5\(\vec { j } \) – 7\(\vec { k } \) then the value of λ …………………..
(a) \(\frac{7}{3}\)
(b) –\(\frac{7}{3}\)
(c) \(\frac{5}{3}\)
(d) \(\frac{-5}{3}\)
Answer:
(b) –\(\frac{7}{3}\)

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.
The derivative of (x + \(\frac{1}{x}\))² w.r.to. x is ………………..
(a) 2x – \(\frac { 2 }{ x^{ 3 } } \)
(b) 2x + \(\frac { 2 }{ x^{ 3 } } \)
(c) 2(x + \(\frac{1}{x}\))
(d) 0
Answer:
(a) 2x – \(\frac { 2 }{ x^{ 3 } } \)

Question 17.
If f(x) is \(\left\{\begin{array}{cc}
a x^{2}-b, & -1<x<1 \\
\frac{1}{|x|}, & \text { elsewhere }
\end{array}\right.\) is differentiable at x = 1, then …………………
(a) a = \(\frac{1}{2}\), b = \(\frac{-3}{2}\)
(b) a = \(\frac{-1}{2}\), b = \(\frac{3}{2}\)
(c) a = \(\frac{-1}{2}\), b = \(\frac{-3}{2}\)
(d) a = \(\frac{1}{2}\), b = \(\frac{3}{2}\)
Answer:
(c) a = \(\frac{-1}{2}\), b = \(\frac{-3}{2}\)

Question 18.
∫\(\frac { \sqrt { tanx } }{ sin2x } \) dx is ………………
(a) \(\sqrt{tanx}\) + c
(b) 2\(\sqrt{tanx}\) + c
(c) \(\frac{1}{2}\) \(\sqrt{tanx}\) + c
(d) \(\frac{1}{4}\) \(\sqrt{tanx}\) + c
Answer:
(a) \(\sqrt{tanx}\) + 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.
It is given that the events A and B are such that P(A) = \(\frac{1}{4}\), P(A/B) = \(\frac{1}{2}\), and P(B/A) = \(\frac{2}{3}\) then
P(B) = ………………….
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(b) \(\frac{1}{3}\)

PART – II

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

Question 21.
If n(P(A)) = 1024, n(A∪B) = 15 and n(P(B)) = 32 then find n(A∩B)
Answer:
n(P(A)) = 1024 = 2¹⁰ ⇒ n(A) = 10
n(A∪B) = 15
n(P(B)) = 32 = 2⁵ ⇒ n(B) = 5
We know n(A∪B) = n(A) + n(B) – n(A∩B)
(i.e) 15 = 10 + 5 – n(A∩B)
⇒ n(A∩B) = 15 – 15 = 0

Question 22.
Simplify (343)^2/3
Answer:
(343)^2/3 = (7³)^2/3 = 7^3×2/3 = 7² = 49

Question 23.
Show that cos36° cos 72° cos 108° cos 144° = \(\frac{1}{16}\)
Answer:
LHS = cos36° cos(90° – 18°) cos(90° – 18°) cos(90° + 18°) cos(180° – 36°)
= sin² 18° cos² 36°
= (\(\frac { \sqrt { 5-1 } }{ 4 } \))² (\(\frac { \sqrt { 5+1 } }{ 4 } \))² = \(\frac{1}{16}\) = RHS

Question 24.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour?
Answer:
There are 6 red balls, 5 white balls and 5 blue balls.
We have to select 3 balls of each colour.
∴Number of ways of selection = ⁶C₃ × ⁵C₃ × ⁵C₃
= \(\frac { 6! }{ 3!3! } \) × \(\frac { 5! }{ 3!2! } \) × \(\frac { 5! }{ 3!2! } \)
= 20 × 10 × 10 = 2000

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.
For any vector prove that \(\vec { r } \) = [\(\vec { r } \).\(\vec { i } \)) + (\(\vec { r } \).\(\vec { j } \))j + [\(\vec { r } \).\(\vec { k } \)}k
Answer:
Let \(\vec { r } \) = x\(\hat { i } \) + y\(\hat { j } \) + z\(\hat { k } \)

Question 27.
Calculate \(\lim _{x \rightarrow-2}\) (x³ – 3x + 6) (-x² + 15)
Answer:

Question 28.
Evaluate y = e^x sin x
Answer:

Question 29.
Integrate the following with respect to x
\(\frac{4}{3+4x}\) + (10x + 3)⁹ – 3 cosec(2x + 3) cot (2x + 3)
Answer:

Question 30.
P(A) = 0.6, P (B) = 0.5 and P(A∩B) = 0.2 find P(A/B)
Answer:
Given that P(A) = 0.6, P(B) = 0.5, and P(A∩B) = 0.2
P(A/B) = \(\frac { p(A∩B) }{ p(B) } \) = \(\frac{0.2}{0.5}\) = \(\frac{2}{5}\)

PART – III

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

Question 31.
A quadratic polynomial has one of its zeros 1 + \(\sqrt{5}\) and it satisfies p(1) = 2. Find the quadratic polynomial?

Question 32.
Prove that

1. tan^-1 (\(\frac{1}{7}\)) + tan^-1(\(\frac{1}{13}\)) = tan^-1(\(\frac{2}{9}\))
2. cos^-1\(\frac{4}{5}\) + tan^-1\(\frac{3}{5}\) = tan^-1\(\frac{27}{11}\)

Question 33.
The product of three increasing numbers in GP is 5832. If we add 6 to the second number and 9 to the third number, then resulting numbers form an AP. Find the numbers in GP?

Question 34.
Find the equation of the line passing through the point (5, 2) and perpendiular to the line joining the points (2, 3) and (3, -1)?

Question 35.
Find the area of the triangle whose vertices are (0,0), (1,2) and (4,3)?

Question 36.
If \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) are three vectors such that \(\vec { a } \) + 2\(\vec { b } \) + \(\vec { c } \) = 0 and |\(\vec { a } \)| = 3, |\(\vec { b } \)| = 4, |\(\vec { c } \)| = 7 fimd the angle between \(\vec { a } \) and \(\vec { b } \)

Question 37.
Evaluate: \({ \underset { x\rightarrow 0 }{ lim } }\) \(\frac { 3^{ x }-1 }{ \sqrt { 1+x-1 } } \)

Question 38.
Find \(\frac{dy}{dx}\) for y = tan^-1 \((\frac { cosx+sinx }{ cosx-sinx } )\)

Question 39.
Evaluate: ∫x⁵ e^x2

Question 40.
How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?

PART – IV

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

Question 41.
(a) If f : R – {-1, 1} → R is defined by f(x) = \(\frac { x }{ x^{ 2 }-1 } \), verify whether f is one-to-one or not?

[OR]

(b) Solve: log₂ x + log₄ x + log₈ x = 11

Question 42.
(a) Prove that \(\frac{\sin x+\sin 3 x+\sin 5 x+\sin 7 x}{\cos x+\cos 3 x+\cos 5 x+\cos 7 x}\) = tan 4x

[OR]

(b) If x + y + z = xyz, then prove that \(\frac { 2x }{ 1-x^{ 2 } } \) + \(\frac { 2y }{ 1-y^{ 2 } } \) + \(\frac { 2z }{ 1-z^{ 2 } } \) = \(\frac { 2x }{ 1-x^{ 2 } } \) \(\frac { 2y }{ 1-y^{ 2 } } \) \(\frac { 2z }{ 1-z^{ 2 } } \)

Question 43.
(a) If the letters of the word GARDEN are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, then and the ranks of the words

1. GARDEN
2. DANGER

[OR]

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

Question 44.
(a) 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?

[OR]

(b) Evalute \(\sqrt { x^{ 2 }+y^{ 2 } } \) = tan^-1(\(\frac{y}{x}\))

Question 45.
Let \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) be three vectors such that |\(\vec { a } \)| = 3, |\(\vec { b } \)| = 4, |\(\vec { c } \)| = 5 and each one of them being perpendicular to the sum of the other two, find |\(\vec { a } \) + \(\vec { b } \) + \(\vec { c } \)|.

[OR]

(b) Evaluate ∫sec³ 2xdx

Question 46.
(a) Find all the equations of the straight lines in the family of the lines y = mx – 3, for which m and the x-coordinate of the point of intersection of the lines with x – y = 6 are integers?

[OR]

(b) There are two identical boxes containing respectively 5 white and 3 red balls, 4 white and 6 red balls. A box is chosen at random and a ball is drawn from it

1. Find the probability that the ball is white
2. If the ball is white, what is the probability that it from the first box?

Question 47 (a).
If A_i B_i, C_i are the cofactors of a_i, b_i, c_i, respectively, i = 1 to 3 in

[OR]

(b) Express the matrix \(\left(\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right)\) as the sum of symmetric martix and a skew-symmetric martix?

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?

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Let us look at these Government of Tamil Nadu State Board 11th Commerce Model Question Papers Tamil Medium with Answers 2020-21 Pdf. Students can view or download the Class 11th Commerce New Model Question Papers 2021 Tamil Nadu English Medium Pdf for their upcoming Tamil Nadu HSC Board Exams. Students can also read Tamilnadu Samcheer Kalvi 11th Commerce Guide.

TN State Board 11th Commerce Model Question Papers 2020 2021 English Tamil Medium

Tamil Nadu 11th Commerce Model Question Papers English Medium 2020-2021

• Tamil Nadu 11th Commerce Model Question Paper 1 English Medium
• Tamil Nadu 11th Commerce Model Question Paper 2 English Medium
• Tamil Nadu 11th Commerce Model Question Paper 3 English Medium
• Tamil Nadu 11th Commerce Model Question Paper 4 English Medium
• Tamil Nadu 11th Commerce Model Question Paper 5 English Medium

Tamil Nadu 11th Commerce Model Question Papers Tamil Medium 2020-2021

• Tamil Nadu 11th Commerce Model Question Paper 1 Tamil Medium
• Tamil Nadu 11th Commerce Model Question Paper 2 Tamil Medium
• Tamil Nadu 11th Commerce Model Question Paper 3 Tamil Medium
• Tamil Nadu 11th Commerce Model Question Paper 4 Tamil Medium
• Tamil Nadu 11th Commerce Model Question Paper 5 Tamil Medium

11th Commerce Model Question Paper Design 2020-2021 Tamil Nadu

Types of Questions	Marks	No. of Questions to be Answered	Total Marks
Part-I Objective Type	1	20	20
Part-II Very Short Answers (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily)	2	7	14
Part-Ill Short Answers (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily)	3	7	21
Part-IV Essay Type	5	7	35
Total Marks			90

Tamil Nadu 11th Commerce Model Question Paper Weightage of Marks

Purpose	Weightage
1. Knowledge	30%
2. Understanding	40%
3. Application	20%
4. Skill/Creativity	10%

It is necessary that students will understand the new pattern and style of Model Question Papers of 11th Standard Commerce Tamilnadu State Board Syllabus according to the latest exam pattern. These Tamil Nadu Plus One 11th Commerce Model Question Papers State Board Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN HSLC Board Exams and Score More marks.

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