Category: Class 12

Students can Download Tamil Nadu 12th Maths Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th 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 12th Maths Model Question Paper 3 English Medium

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. questions of Part I, II. III and IV are to be attempted separately
4. Question numbers 1 to 20 in Part I are objective type 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-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part – I

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

Question 1.
If A is a 3 × 3 non-singular matrix such that AA^T = A^T A and B = A^-1 A^T, then BB^T = ________.
(a) A
(b) B
(c) I₃
(d) B^T
Answer:
(c) I₃

Question 2.
The rank of the matrix \(\left[\begin{array}{cc}
7 & -1 \\
2 & 1
\end{array}\right]\) is ________.
(a) 9
(b) 2
(c) 1
(d) 5
Answer:
(b) 2

Question 3.
The value of \(\sum_{i=1}^{13}\left(i^{n}+i^{n-1}\right)\) is ________.
(a) 1 + i
(b) i
(c) 1
(d) 0
Answer:
(a) 1 + i

Question 4.
Which of the following is incorrect?
(d) Re(z) ≤ |z|
(b) Im (z) ≤ |z|
(c) \(z \bar{z}=|z|^{2}\)
(d) Re(z) ≥ |z|
Answer:
(d) Re(z) ≥ |z|

Question 5.
According to the rational root theorem, which number is not possible rational zero of 4x⁷ + 2x⁴ – 10x³ – 5?
(a) -1
(b) \(\frac{5}{4}\)
(c) \(\frac{4}{5}\)
(d)5
Answer:
(c) \(\frac{4}{5}\)

Question 6.
If \(\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})=u\), then cos 2u is equal to ______.
(a) tan² α
(b) o
(c) -1
(d) tan 2α
Answer:
(c) -1

Question 7.
The domain of the function defined by f(x) = sin^-1 \(\sqrt{x-1}\) is _______.
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d) [-1, 0]
Answer:
(a) [1, 2]

Question 8.
The area of quadrilateral formed with foci of the hyperbolas \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1\) is ________.
(a) 4(a² + b²)
(b) 2(a² + b²)
(c) a² + b²
(d) \(\frac{1}{2}\) (a² + b²)
Answer:
(b) 2(a² + b²)

Question 9.
The directrix of the parabola x² = -4y is ________.
(a) x = 1
(b) x = 0
(c) y = 1
(d) y = 0
Answer:
(c) y = 1

Question 10.
If the line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane x + 3y – αz + b = β, then (α, β) is ______
(a) (-5, 5)
(b) (-6, 7)
(c) (5, -5)
(d) (6, -7)
Answer:
(b) (-6, 7)

Question 11.
If \(\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \times \vec{b}) \times \vec{c}\) for non-coplanar vectors \(\vec{a}, \vec{b}, \vec{c}\) then ________.
(a) \(\vec{a}\) parallel to \(\vec{b}\)
(b) \(\vec{b}\) parallel to \(\vec{c}\)
(c) \(\vec{c}\) parallel to \(\vec{a}\)
(d) \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)
Answer:
(c) \(\vec{c}\) parallel to \(\vec{a}\)

Question 12.
The maximum product of two positive numbers, when their sum of the squares is 200, is ________.
(a) 100
(b) \(25 \sqrt{7}\)
(c) 28
(d) \(24 \sqrt{14}\)
Answer:
(a) 100

Question 13.
If w(x,y, z) = x² (y – z) + y² (z – x) + z² (x – y), then \(\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}+\frac{\partial w}{\partial z}\) is ________.
(a) xy + yz + zx
(b) x (y + z)
(c) y (z + x)
(d) 0
Answer:
(d) 0

Question 14.
If u (x, y) = x² + 3xy + y – 2019, then \(\left.\frac{\partial u}{\partial x}\right|_{(4,-5)}\) is equal to ______.
(a) -4
(b) -3
(c) -7
(d) 13
Answer:
(c) -7

Question 15.
The volume of solid of revolution of the region bounded by y² = x (a – x) about x-axis is ________.

Answer:
(d) \(\frac{\pi a^{3}}{6}\)

Question 16.
\(\int_{0}^{a} f(x) d x+\int_{0}^{a} f(2 a-x) d x\) = _________.

Answer:
(c) \(\int_{0}^{2 a} f(x) d x\)

Question 17.
The differential equation of the family of curves y = Ae^x + Be^-x, where A and B are arbitrary constants is _______.

Answer:
(b) \(\frac{d^{2} y}{d x^{2}}-y=0\)

Question 18.
The differential equation corresponding to xy = c² where c is an arbitrary constant, is ______.
(a) xy”+ x = 0
(b) y” = 0
(c) xy’ + y = 0
(d) xy”- x = 0
Answer:
(c) xy’ + y = 0

Question 19.
If \(f(x)=\left\{\begin{array}{ll}
2 x, & 0 \leq x \leq a \\
0 & , \text { otherwise }
\end{array}\right.\)
is a probability density function of a random variable, then the value of a is _________.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Question 20.
The proposition \(p \wedge(\neg p \vee q)\) is
(a) a tautology
(b) a contradiction
(c) logically equivalent to p ∧ q
(d) logically equivalent to p ∨ q
Answer:
(c) logically equivalent to p ∧ q

Part – II

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

Question 21.
A 12 metre tall tree was broken into two parts. It was found that the height of the part which was left standing was the cube root of the length of the part that was cut away. Formulate this into a mathematical problem to find the height of the part which was cut away.
Answer:
Let the two parts be x and (12 – x)
Given that x = \(\sqrt[3]{12-x}\)
Cubing on both side, x³ = 12 – x
x³ + x – 12 = 0

Question 22.
Find the value of \(\sin ^{-1}\left(\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} \sin \frac{\pi}{9}\right)\)
Answer:

Question 23.
Obtain the equation of the circles with radius 5 cm and touching x-axis at the origin in general form.
Answer:

Given radius = 5 cm and the circle is touching x axis
So centre will be (0, ± 5) and radius = 5
The equation of the circle with centre (0, ± 5) and radius 5 units is
(x – 0)² + (y ± 5)² = 5²
(i.e) x² + y² ± 10y + 25 – 25 = 0
(i.e) x² + y² ± 10y = 0

Question 24.
Find the length of the perpendicular from the origin to the plane.
\(\bar{r} \cdot(3 \vec{i}+4 \bar{j}+12 \vec{k})=26\).
Answer:
Taking the equation of the plane in cartesian form we get,
\((x \vec{i}+y \vec{j}+z \vec{k}) \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})=26\)
i.e. 3x + 4y + 12z – 26 = 0
The length of the perpendicular from (0, 0, 0) to the above plane is

Question 25.
Evaluate \(\lim _{x \rightarrow \pi / 2} \frac{\log (\sin x)}{(\pi-2 x)^{2}}\)
Answer:

Note that here l’ Hospitals rule, applied twice yields the result.

Question 26.
Evaluate \(\int_{-1}^{1} e^{-\lambda x}\left(1-x^{2}\right) d x\)
Answer:
Taking u = 1 – x² and v= e^-λx, and applying the Bernoulli’s formula, we get

Question 27.
Solve: \(\frac{d y}{d x}+2 y=e^{-x}\)
Answer:
Given that \(\frac{d y}{d x}+2 y=e^{-x}\)
This is a linear differential equation.
Here P = 2; Q = e
∫P dx = ∫2 dx = 2x
Thus, I.F = e^∫Pdx = e^2x
Hence the solution of (1) is \(y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c\)
That is, ye^2x = ∫e^-x e^2x dx + c (or) ye^2x = e^x + c (or) y = e^-x + ce^-2x is the required solution.

Question 28.
Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
Answer:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3
Hence, the probabilities
P(X = 0) = P (No heads) = \(\frac{1}{8}\);
P(X = 1) = P (1 head) = \(\frac{3}{8}\);
P(X = 2) = P (2 heads) = \(\frac{3}{8}\);
P(X = 3) = P (3 heads)= \(\frac{1}{8}\);
∴ The probability mass function is
\(f(x)=\left\{\begin{array}{lll}
1 / 8 & \text { for } & x=0,3 \\
3 / 8 & \text { for } & x=1,2
\end{array}\right.\)

Question 29.
Construct the truth table for \((p \vee q) \vee \neg q\)
Answer:
truth table for \((p \vee q) \vee \neg q\)

Question 30.
Show that f(x, y) = \(\frac{x^{2}-y^{2}}{y^{2}+1}\) is continuous at every (x, y) ∈ R².
Answer:

Here, f satisfies all the three conditions of continuity at (a, b). Hence, f is continuous at every point of R² as (a, b) ∈ R².

Part – III

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

Question 31.
Form a polynomial equation with integer coefficients with \(\sqrt{\frac{\sqrt{2}}{\sqrt{3}}}\) as a root.

Question 32.
Find the equation of the tangents from the point (2, -3) to the parabola y² = 4x

Question 33.
Find the vector and cartesian equations of the straight line passing through the points (-5, 2, 3) and (4,-3, 6).

Question 34.
Find the points on the curve y = x³ – 6x² + x + 3 where the normal is parallel to the line x + y =1729.

Question 35.
Assuming log₁₀ e = 0.4343, find an approximate value of log₁₀ 1003.

Question 36.
Evaluate: \(\int_{0}^{\frac{\pi}{2}} \sin ^{2} x \cos ^{4} x d x\)

Question 37.
Solve the differential equation: \(\frac{d y}{d x}\) = e^x+y + x³ e^y

Question 38.
Using binomial distribution find the mean and variance of X for the following experiments
(i) A fair coin is tossed 100 times, and X denote the number of heads.
(ii) A fair die is tossed 240 times, and X denote the number of times that four appeared.

Question 39.
Let

be any three Boolean matrices of the same type. Find (A ∨ B) ∧C

Question 40.
Sketch the graph of y = sin\(\left(\frac{1}{3} x\right)\) for 0 ≤ x < 6π.

Part – IV

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

Question 41.
(a) A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself ? (Use Cramer’s rule to solve the problem).
[OR]
(b) Using elementary transformations find the inverse of the matrix \(\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)

Question 42.
42. (a) Let z₁, z₂, and z₃ be complex numbers such that |z₁| = |z₂| = |z₃| = r > 0 and z₁ +z₂ + z₃ ≠ 0.
Prove that \(\left|\frac{z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}}{z_{1}+z_{2}+z_{3}}\right|=r\)
[OR]
(b) Find all cube roots of \(\sqrt{3}\) + i.

Question 43.
(a) Find all zeros of the polynomial x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135, if it is known that 1 + 2i and \(\sqrt{3}\) are two of its zeros.
[OR]
(b) Let W(x, y, z) = x² – xy + 3 sin z, x, y, z∈R. Find the linear approximation at (2, -1, 0)

Question 44.
(a) Prove p → (q → r) ≡ (p ∧ q) → r without using truth table.
[OR]
(b) Evaluate \(\int_{0}^{\frac{\pi}{4}} \frac{1}{\sin x+\cos x} d x\)

Question 45.
(a) Prove that \(\text { (i) } \tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2} \text { (ii) } \sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{12}{13}=\sin ^{-1} \frac{16}{65}\)
[OR]
(b) Find two positive numbers whose product is 100 and whose sum is minimum.

Question 46.
(a) If X is the random variable with distribution function F (x) given by,
\(\mathrm{F}(x)=\left\{\begin{array}{ll}
0, & x<0 \\
\frac{1}{2}\left(x^{2}+x\right) & 0 \leq x<1 \\
1, & x \geq 1
\end{array}\right.\)
then find (i) the probability density function f(x) (ii) P(0.3 ≤ X ≤ 0.6)
[OR]
(b) Find the vertex, focus, equation of directrix and length of the latus rectum of the following: y² – 4y – 8x + 12 = 0

Question 47.
(a) The rate at which the population of a city increases at any time is proportional to the population at that time. If there were 1,30,000 people in the city in 1960 and 1,60,000 in 1990, what population may be anticipated in 2020? [log_e \(\left(\frac{16}{3}\right)\) = 0.2070; e^-0.42 = 1.52]
[OR]
(b) Find the parametric vector, non-parametric vector and Cartesian form of the equation of the plane passing through the point (3, 6, -2), (-1, -2, 6) and (6, 4, -2).

Students can Download Tamil Nadu 12th Maths Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th 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 12th Maths Model Question Paper 2 English Medium

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. questions of Part I, II. III and IV are to be attempted separately
4. Question numbers 1 to 20 in Part I are objective type 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-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part – I

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

Question 1.
If A = \(\left[\begin{array}{cc}
2 & 3 \\
5 & -2
\end{array}\right]\) be such that λA^-1 = A, then λ is _______.
(a) 17
(b) 14
(c) 19
(d) 21
Answer:
(c) 19

Question 2.
If ω ≠ 1 is a cubic root of unity and (1+ ω)⁷ = A+ B ω, then (A, B) equals to _______.
(a) (1,0)
(b) (-1, 1)
(c) (0, 1)
(d) (1, 1)
Answer:
(d) (1, 1)

Question 3.
The value of z – \(\bar{Z}\) is ______.
(a) 2 Im (z)
(b) 2 i Im (z)
(c) Im (z)
(d) i Im (z)
Answer:
(b) 2 i Im (z)

Question 4.
If x³ + 12x² + 10ax + 1999 definitely has a positive zero, if and only if ________.
(a) a ≥ 0
(b) a > 0
(c) a < 0
(d) a < 0
Answer:
(c) a < 0

Question 5.
sin(tan^-1 x), |x| < 1 is equal to _______.

Answer:
(d) \(\frac{x}{\sqrt{1+x^{2}}}\)

Question 6.
The centre of the circle inscribed in a square formed by the lines x² – 8x – 12 = 0 and y² – 14y + 45 = 0 is _____.
(a) (4, 7)
(b) (7, 4)
(c) (9, 4)
(d) (4, 9)
Answer:
(a) (4, 7)

Question 7.
The axis of the parabola x² = – 4y is ______.
(a) y= 1
(b) x = 0
(c) y = 0
(d) x = 1
Answer:
(b) x = 0

Question 8.
The coordinates of the point where the line \(\vec{r}=(6 \hat{i}-\hat{j}-3 \hat{k})+t(-\hat{i}+4 \hat{k})\) meets the plane \(\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=3\) are _______.
(a) (2, 1, 0)
(b) (7, -1, -7)
(c) (1, 2, -6)
(d) (5, -1, 1)
Answer:
(d) (5, -1, 1)

Question 9.
If the vectors \(\vec{a}=3 \vec{i}+2 \vec{j}+9 \vec{k}\) and \(\vec{b}=\vec{i}+m \vec{j}+3 \vec{k}\) are parallel then m is _________.

Answer:
(b) \(\frac{2}{3}\)

Question 10.
The minimum value of the function |3 – x | + 9 is ________.
(a) 0
(b) 3
(c) 6
(d) 9
Answer:
(d) 9

Question 11.
The curve y² = x² (1 – x²) has ______.
(a) an asymptote x = -1
(b) an asymptote x = 1
(c) two asymptotes x = 1 and x = -1
(d) no asymptote
Answer:
(d) no asymptote

Question 12.
If /(x, y, z) = xy + yz + zx, then f_x – f_z is equal to _______.
(a) z – x
(b) y – z
(c) x – z
(d) y – x
Answer:
(a) z – x

Question 13.
A circular template has a radius of 10 cm. The measurement of radius has an approximate error of 0.02 cm. Then the percentage error in calculating area of this template is _______.
(a) 0.2%
(b) 0.4%
(c) 0.04%
(d) 0.08%
Answer:
(b) 0.4%

Question 14.
The value of \(\int_{0}^{\pi} \sin ^{4} x d x\) is _______.

Answer:
(b) \(\frac{3 \pi}{8}\)

Question 15.
\(\int_{a}^{b} f(x) d x\) is _______.

Answer:
(d) \(\int_{a}^{b} f(a+b-x) d x\)

Question 16.
The degree of the differential equation \(y(x)=1+\frac{d y}{d x}+\frac{1}{1.2}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{1.2 .3}\left(\frac{d y}{d x}\right)^{3}+\ldots\) is ________.
(a) 2
(b) 3
(c) 1
(d) 4
Answer:
(c) 1

Question 17.
In finding the differential equation corresponding toy = e^mx where m is the arbitrary constant, then m is ____.
(a) \(\frac{y}{y^{\prime}}\)
(b) \(\frac{y^{\prime}}{y}\)
(c) y’
(d) y
Answer:
(b) \(\frac{y^{\prime}}{y}\)

Question 18.
Let X be random variable with probability density function f(x) = \(\left\{\begin{array}{ll}
2 / x^{3} & x \geq 1 \\
0 & x<1
\end{array}\right.\)
Which of the following statement is correct
(a) both mean and variance exist
(b) mean exists but variance does not exist
(c) both mean and variance do not exist
(d) variance exists but mean does not exist
Answer:
(b) mean exists but variance does not exist

Question 19.
The random variable X has the probability density function f(x) = \(\left\{\begin{array}{cc}
a x+b, & 0<x<1 \\
0, & \text { otherwise }
\end{array}\right.\)
and E(X) = \(\frac{7}{12}\), then a and b are respectively _______.
(a) 1 and \(\frac{1}{2}\)
(b) \(\frac{1}{2}\) and 1
(c) 2 and 1
(d) 1 and 2
Answer:
(a) 1 and \(\frac{1}{2}\)

Question 20.
A binary operation on a set S is a function from ________.
(a) S → S
(b)(S x S) → S
(c) S → (S x S)
(d) (S x S) → (S x S)
Answer:
(b)( S x S) → S

Part – II

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

Question 21.
Solve the following system of homogeneous equations.
3x + 2y + 7z = 0, 4x – 3y – 2z = 0, 5x + 9y + 23z = 0
Answer:
The matrix form of the above equation is

The augmented matrix [A, B] is

The above matrix is in echelon form. Here ρ(A, B) = ρ( A) < number of unknowns.
⇒ The system is consistent with infinite number of solutions. To find the solutions.
Writing the equivalent equations.
We get 3x + 2y + 7z = 0 ……..(1)
-17y – 34z = 0 …….(2)
Taking z = t in (2) we get -17y – 34t = 0
⇒ -17y = 34t
⇒ y= \(\frac{34 t}{-17}\) = -2t
Taking z = t; y = -2t in (1) we get
3x + 2 (-2t) + 7t = 0
3x – 4t + 7t = 0
⇒ 3x = -3t ⇒ x = -t
So the solution is x = -t; y = -2t; and z = t, t∈R

Question 22.
Show that |3z – 5 + i| = 4 represents a circle, and, find its centre and radius.
Answer:
The given equation |3z – 5 + i| = 4 can be written as

It is of the form |z – z| = r and so it represents a circle, whose center and radius are \(\left(\frac{5}{3},-\frac{1}{3}\right)\) and 4/3 respectively.

Question 23.
Find the equation of the circle whose centre is (2, -3) and passing through the intersection of the line 3x – 2y = 1 and 4x + y = 27.
Answer:
Solving 3x – 2y = 1 and 4x + y = 27
Simultaneously, we get x = 5 and y = 7
∴ The point of intersection of the lines is (5, 7)
Now we have to find the equation of a circle whose centre is
(2, -3) and which passes through (5, 7)

∴ Required equation of the circle is
(x – 2)² + (y + 3)² = \((\sqrt{109})^{2}\)
⇒ x² + y² – 4x + 6y – 96 = 0

Question 24.
Find the intercepts cut off by the plane \(\vec{r} \cdot(6 \hat{i}+4 \hat{j}-3 \hat{k})=12\) on the coordinate axes.
Answer:
\(\vec{r} \cdot(6 \vec{i}+4 \vec{j}-3 \vec{k})=12\)
Compare the above equations into \(\vec{r} \cdot \vec{n}=q\) so q = 12
Let a, b, c are intercepts of x-axis, y-axis and z-axis respectively.
Clearly

x – intercept = 2; y – intercept = 3; z – intercept = -4

Question 25.
Find the values in the interval (1, 2) of the mean value theorem satisfied by the function f(x) = x – x² for 1 ≤ x ≤ 2.
Answer:
f(1) = 0 and f(2) = -2. Clearly f(x) is defined and differentiable in 1 < x < 2. Therefore, by the Mean Value Theorem, there exists a c ∈(1, 2) such that
f'(c) = \(\frac{f(2)-f(1)}{2-1}\) = 1 – 2c
That is, 1 – 2c = -2 ⇒ c = \(\frac{3}{2}\)

Question 26.
Show that the percentage error in the n^th root of a number is approximately \(\frac{1}{n}\) times the percentage error in the number.
Answer:

Question 27.
Solve the differential equation: tany \(\frac{d y}{d x}\) = cos (x + y) + cos (x -y)
Answer:
tan y \(\frac{d y}{d x}\) = cos (x + y) + cos(x – y)
tan y \(\frac{d y}{d x}\) = cos x cos y – sin x sin y + cos x cos y + sin x sin y
tan y \(\frac{d y}{d x}\) = 2 cos x cos y
seperating the variables
\(\int \frac{\tan y}{\cos y}\) dy = 2∫cos x dx ⇒ ∫sec y tan y dy = 2∫cos x dx
sec y = 2 sin x + c

Question 28.
The probability density function of X is given by \(f(x)=\left\{\begin{array}{cc}
k e^{-\frac{x}{3}} & \text { for } x>0 \\
0 & \text { for } x \leq 0
\end{array}\right.\)
Find the value of k.
Answer:

Question 29.
Construct the truth table for the following statement. \(\neg(p \wedge \neg q)\).
Answer:

Question 30.
Find an approximate value of \(\int_{1}^{1.5} x^{2} d x\) by applying the right-hand rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}.
Answer:
Here a = 1; b = 1.5; n = 5; f(x) = x²
So, the width of each subinterval is

x₀ = 1; x₁ = 1.1; x₂ = 1.2; x₃ = 1.3; x₄ = 1.4; x₅ = 1.5
The Right hand rule for Riemann sum,
S = [f(x₁) + f(x₂) + f(x₃) + f(x₄) + f(x₅)] Δx
= [f(1.1) + f(1.2) + f(1.3) + f(1.4) + f(1.5)] (0.1)
= [1.21 + 1.44 + 1.69 + 1.96 + 2.25] (0.1)
= [8.55] (0.1)
= 0.855.

Part – III

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

Question 31.
Find a matrix A if adj (A) = \(\left[\begin{array}{ccc}
7 & 7 & -7 \\
-1 & 11 & 7 \\
11 & 5 & 7
\end{array}\right]\)

Question 32.
Obtain the Cartesian form of the locus of z = x + iy in the following case Im[(1 – i)z +1] = 0

Question 33.
If \(\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}, \vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}\), show that \([\vec{a} \vec{b} \vec{c}]\) depends on neither x nor y.

Question 34.
The Taylor’s series expansion of f(x) = sin x about x = \(\frac{\pi}{2}\) is obtained by the following way.

Question 35.
The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error in computing (i) the volume of the cube and (ii) the surface area of cube.

Question 36.
Evaluate \(\int_{0}^{1} \frac{\sin \left(3 \tan ^{-1} x\right) \tan ^{-1} x}{1+x^{2}} d x\)

Question 37.
Find the particular solution of (1 + x³) dy – x² ydx = 0 satisfying the condition y(1) = 2.

Question 38.
If X is the random variable with distribution function F(x) given by,
\(\mathrm{F}(x)=\left\{\begin{array}{ll}
0, & x<0 \\
x, & 0 \leq x<1 \\
1, & 1 \leq x
\end{array}\right.\)
then find (z) the Probability density function f(x)

Question 39.
Show that \(((\neg q) \wedge p) \wedge q\) is a contradiction.

Question 40.
Show that the absolute value of difference of the focal distances of any point P on the hyperbola is the length of its transverse axis.

Part – IV

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

Question 41.
(a) By using Gaussian elimination method, balance the chemical reaction equation:
C₂H₆ + O₂ → H₂O + CO₂.
[OR]
(b) \(\frac{d y}{d x}+\frac{3 y}{x}=\frac{1}{x^{2}}\), given that y = 2 when x = 1

Question 42.
(a) Find the real values of x and y for the equation \(\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}=i\)
[OR]
(b) Find the area between the line y = x + 1 and the curve y = x² – 1.

Question 43.
(a) Determine k and solve the equation 2x³ – 6x² + 3x + k = 0 if one of its roots is twice the sum of the other two roots.
[OR]
(b) Evaluate: \(\int_{0}^{\frac{\pi}{2}} \frac{d x}{5+4 \sin ^{2} x}\)

Question 44.
(a) A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be 16 m, and the height at the edge of the road must be sufficient for a truck 4 m high to clear if the highest point of the opening is to be 5 m approximately. How wide must the opening be?
[OR]
(b) Using truth table check whether the statements \(\neg(p \vee q) \vee(\neg p \wedge q)\) and \(\neg p\) are logically equivalent.

Question 45.
(a) Find the value of cot^-1 x – cot^-1 (x + 2) = \(\frac{\pi}{12}\), x > 0
[OR]
(b) Verify Euler’s theorem for f(x, y) = \(\frac{1}{\sqrt{x^{2}+y^{2}}}\)

Question 46.
(a) Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz planes.
[OR]
(b) If X is the random variable with probability density function f(x) given by,
\(f(x)=\left\{\begin{array}{rc}
x+1, & -1 \leq x<0 \\
-x+1, & 0 \leq x<1 \\
0, & \text { otherwise }
\end{array}\right.\)
then find (z) the distribution function f(x) (ii) P (-0.5 ≤ X ≤ 0.5)

Question 47.
(a) Sketch the graph of the function: y = \(x \sqrt{4-x}\)
(b) The velocity v, of a parachute falling vertically satisfies the equation, \(v \frac{d v}{d x}=g\left(1-\frac{v^{2}}{k^{2}}\right)\)
where g and k are constants. If v and x are both initially zero, find v in terms of x.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Chemistry Model Question Paper 5 English Medium

Time: 3 Hours
Maximum Marks: 70

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 lo be answered in about one or two sentences
6. Question numbers 25 to 33 in Part III are three-marks 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 answered in detail. Draw diagrams wherever necessary

Part – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
In the extraction of aluminium from alumina by electrolysis, cryolite is added to …………..
(a) Lower the melting point of alumina
(b) Remove impurities from alumina
(c) Decrease the electrical conductivity
(d) Increase the rate of reduction
Answer:
(a) Lower the melting point of alumina

Question 2.
Compound used for propellant is
(a) BN
(b) H₂B₄O₇
(c) B₂H₆
(d) Borax
Answer:
(c) B₂H₆

Question 3.
P₄O₆ reacts with cold water to give …………..
(a) H₃PO₃
(b) H₄P₂O₇
(c) HPO₃
(d) H₃PO₄
Answer:
(a) H₃PO₃

Question 4.
Which one of the following elements show high positive electrode potential?
(a) Ti²⁺
(b) Mn²⁺
(c) CO²⁺
(d) Cr²⁺
Answer:
(c) CO²⁺

Question 5.
As per IUPAC guidelines, the name of the complex [Co(en)₂(ONO)Cl]Cl is …………..
(a) chlorobisethylenediaminenitritocobalt(III) chloride
(b) chloridobis(ethane-l,2-diamine)nitro k-Ocobaltate(III) chloride
(c) chloridobis(ethane-l,2-diammine)nitrito k-Ocobalt(II) chloride
(d) chloridobis(ethane-l,2-diamine)nitro k-Ocobalt(III) chloride
Answer:
(d) chloridobis(ethane-l,2-diamine)nitro k-Ocobalt(III) chloride

Question 6.
Solid NH₃ solid CO₂ are examples of …………………..
(a) Covalent solids
(b) polar molecular solids
(c) molecular solids
(d) ionic solids
Answer:
(b) polar molecular solids

Question 7.
After 2 hours, a radioactive substance becomes\(\left(\frac{1}{16}\right)^{4}\) of original amount. Then the half life ( in min ) is ………
(a) 60 minutes
(b) 120 minutes
(c) 30 minutes
(d) 15 minutes
Answer:
(c) 30 minutes
Solution:

Question 8.
What is the decreasing order of strength of bases?

Answer:

Question 9.
Which electrolyte is used in Leclanche cell?
(a) ZnSO₄ + CuSO₄
(6) NH₄Cl + ZnCl₂
(e) NaCl + CuSO₄
(d) MnSO₄ + MnO₂
Answer:
(6) NH₄Cl + ZnCl₂

Question 10.
The phenomenon observed when a beam of light is passed through a colloidal solution is ………………
(a) Cataphoresis
(b) Eleætrophoresis
(c) Coagulation
(d) Tyndall effect
Answer:
(d) Tyndall effect

Question 11.
In the following sequence of reactions,

(a) Butanal
(b) n-butyl alcohol
(c) propan-1-ol
(d) Propanal
Answer:
(c) propan-1-ol

Question 12.
Of the following, which is the product formed when cyclohexanone undergoe’s aldol condensation followed by heating?

Answer:
(a)

Question 13.

Answer:
(a) A – 2, B – 1, C – 4, D – 3

Question 14.
Insulin, a hormone chemically is ………………..
(a) Fat
(b) Steroid
(c) Protein
(d) Carbohydrates
Answer:
(c) Protein

Question 15.
The role of phosphate in detergent powder is
(a) control pH level of the detergent water mixture
(b) remove Ca²⁺ and Mg²⁺ ions from water that causes hardness of water
(c) provide whiteness to the fabric
(d) more soluble in soft water
Answer:
(b) remove Ca²⁺ and Mg²⁺ ions from water that causes hardness of water

Part – II

Answer any six questions. Question No. 23 is compulsory. [6 x 2 = 12]

Question 16.
What are all the steps involved in metallurgical process?
Answer:
The extraction of a metal from its ore consists the following metallurgical process.

• Concentration of the ore
• Extraction of crude metal
• Refining of crude metal

Question 17.
Give the uses of Borax.
Answer:

• Borax is used for the identification of coloured metal ions.
• In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
• It is also used as a flux in metallurgy and also acts as a good preservative.

Question 18.
Differentiate primary valency and secondary valency.
Answer:
Primary Valency :

1. The primary valence of a metal ion positive in most of the cases and zero in certain cases.
2. The primary valence is always satisfied by negative ions.
3. The primary valences are non directional
4. Example: In COCl₃.6NH₃, the primary valence of cobalt is +3 Example: In CoCl3.6NH3, the secondary valence of cobalt +3

Secondary Valency :

1. The secondary valence as the coordination number.
2. The secondary valence is satisfied by negative ions, neutral molecular or positive ions.
3. The secondary valences are directional
4. Example: In COCl₃.6NH₃, the secondary valence of cobalt is 6

Question 19.
Write a note about molecular solids.
Answer:

• In molecular solids, the constituents are neutral molecules. They are held together by weak vander waals forces.
• Molecular solids are soft and they do not conduct electricity. Eg., Solid CO₂

Question 20.
What are the limitations of Arrhenius concept?
Answer:

• Arrhenius theory does not explain the behaviour of acids and base in non-aqueous solvents such as acetone, tetrahydro furan.
• This theory does not account for the basicity of the substances like ammonia which do not possess hydroxyl group.

Question 21.
Write a note on catalytic poison.
Answer:
Catalytic poison: Certain substances when added to a catalysed reaction, decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons.
For example, in the reaction, 2SO₂ + O₂ → 2SO₃ with a Pt catalyst, the poison is AS₂O₃. i. e., AS₂O₃ destroys the activity of pt. AS₂O₃ blocks the activity of the catalyst. So, the activity is lost.

Question 22.
Convert phenyl magnesium bromide to phenyl methanol (or) How would you prepare phenyl methanol from Grignard reagent?
Answer:

Question 23.
Identify compounds A,B and C in the following sequence of reactions.

Answer:

Question 24.
Why cannot Vitamin C be stored in our body?
Answer:
Vitamin C is water soluble, therefore it is readily excreted in urine and hence cannot be stored in the body.

Part – III

Answer any six questions. Question No. 29 is compulsory. [6 x 3 = 18]

Question 25.
All ores are minerals, but all minerals are not ores. Explain.
Answer:
A naturally occurring substance obtained by mining which contains the metal in free state or in the form of compounds is called a mineral. In most of the minerals, the metal of interest is present only in small amounts and some of them contains a reasonable percentage of metal. Such minerals that – contains a high percentage of metal, from which it can be extracted conveniently and economically are called ores. Hence all ores are minerals but all minerals are not ores.

Question 26.
Discuss the Commercial method to prepare Nitric acid.
[OR]
How will you prepare nitric acid by Ostwald’s process?
Answer:
Nitric acid prepared in large scales using Ostwald’s process. In this method ammonia from Haber’s process is mixed about 10 times of air. This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze. The temperature rises to about 1275 K and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO, which then oxidised to nitrogen dioxide.

4NH₃ + 5O₂ → 4NO + 6H₂O + 120 kJ
2NO + O₂ → 2NO₂

The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.

6NO₂ + 3H₂O → 4HNO₃ + 2NO + H₂O

Question 27.
Actinoid contraction is greater from element to element than the lanthanoid contraction, why?
Answer:

• Actinoid contraction is greater from element to element than lanthanoid cintraction. The 5f orbitals in Actinoids have a very poorer shielding effect than 4f orbitals in lanthanoids.
• Thus, the effective nuclear charge experienced by electron in valence shells in case of actinoids is much more than that experienced by lanthanoids.
• In actinoids, electrons are shielded by 5d, 4f, 4d and 3d whereas in lanthanoids, electrons are shielded by 4d, 4f only. ,
• Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.

Question 28.
What are the examples of first order reaction?
Answer:
(i) Decompostion of dinitrogen pentoxide
2N₂O_5(g) → 2NO_2(g) + 1/2 O_2(g)

(ii) Decomposition of thionylchloride
SO₂Cl_2(g) → SO_2(g) + Cl_2(g)

(iii) Decompostion of H₂O₂ in aqueous solution.
H₂O_2(aq) → H₂O_(l) + 1/2 O_2(g)

(iv) Isomerisation of cyclopropane to propene

Question 29.
Complete the following reaction.

Answer:

Question 30.
How will you calculate degree of dissociation of weak electrolytes and dissociation constant using Kohlrausch’s law?
Answer:
(i) The degree of dissociation of weak electrolyte can be calculated from the molar conductivity at a given concentration and the molar conductivity in infinite dilution using the formula \(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\)

(ii) According to Ostwald’s dilution law \(\mathrm{K}_{\mathrm{a}}=\frac{\alpha^{2} \mathrm{C}}{1-\alpha}\)
Substituting α value in the above equation

Question 31.
What are the characteristics of adsorption?
Answer:

• Adsorption can occur in all interfacial faces i.e., the adsorption can occur in between gas – solid, liquid – solid, liquid – liquid, solid – solid and gas – liquid. .
• Adsorption is always accompanied by decrease in free energy. When AG reaches zero, the equilibrium is attained.
• Adsorption is a spontaneous process.
• When molecules are getting adsorbed, there is always decrease in randomness of the molecules.
  ∆G = ∆H – T ∆S where ∆G = change in free energy
  ∆H = change in enthalpy
  ∆S = change in entropy
  .’. ∆H = ∆G + T∆S
• Adsorption is exothermic and it is a quick process.
• If simultaneous adsorption and absorption take place, it is termed as ‘sorption’ and sorption of gases on metal surface is called occlusion.

Question 32.
What are the uses of cellulose?
Answer:

• Cellulose is used extensively in manufacturing paper, cellulose fibres and rayon explosive.
• Gun cotton – nitrated ester of cellulose an explosive is prepared from cellulose.
• Cellulose act as food for animals.

Question 33.
How will you prepare PHBV? Give its use?
Answer:
(i) The biodegrable polymer PHBV (Poly hydroxy butyrate-co hydroxyl valerate) is prepared by the polymerisation of monomers 3 – hydroxy butanoic acid and 3 – hydroxy pentanoic acid.

(ii) It is used in orthopaedic devices and in controlled release of drugs.

Part – IV

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

Question 34.
(a) (i) Which type of ores can be concentrated by froth floatation method?
Give two examples for such ores. (2)
(ii) Explain the variation In E°_M3+/M2+ series. (3)
[OR]
(b) (i) Mention the uses of silicon tetrachtoride. (2)
(ii) What are all the conditions that are necessary for catenation? (3)
Answer:
(a) (i) Suiphide ores can be concentrated by froth floatation method.
e.g., (i) Copper pyrites (CuFeS₂) (ii) Zinc blende (ZnS) (iii) Galena (PbS)

(ii)

• In transition series, as we move down from Ti to Zn, the standard reduction potential E° _M3+/M2+value is approaching towards less negative value and copper has a positive reduction potential, i.e. elemental copper is more stable than Cu²⁺
• E°_M3+/M2+ value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d⁵ configuration in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺.
• The standard electrode potential for the M³⁺ /M²⁺ half cell gives’the relative stability between M³⁺ and M²⁺.
• The high reduction potential of Mn³⁺ / Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺.
• For Fe³⁺/Fe²⁺ the reduction potential is 0.77 V, and this low value indicates that both Fe³⁺ and Fe²⁺ can exist under normal condition.
• Mn³⁺ has a 3d⁴ configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible [E° = +1.5 IV]

[OR]

(b) (i) Silicon tetrachloride is used in the production of semiconducting silicon.
It is used as a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

(ii) Essential condition for catenation:

• The valency of elements is greater than or equal to two.
• Element should have an ability to bond with itself.
• The self bond must be as strong as its bond with other elements.
• Kinetic inertness of catenated compound towards other molecules.

Question 35.
(a) (i) Discuss the manufacture of chlorine. (3)
(ii) What is inert pair effect? (2)
[OR]
(b) (i) Calculate the magnetic moment of Ti³⁺ and V⁴⁺. (2)
(ii) Draw all possible geometrical isomers of the complex [CO(en)₂Cl₂]⁺ and identify the optically active isomer. (3)
Answer:
(a) (i) Electrolytic process: When a solution of brine (NaCl) is electrolysed, Na⁺ and Cl^– ions are formed. Na⁺ ion reacts with OH^– ions of water and forms sodium hydroxide.

Hydrogen and chlorine are liberated as gases.

Deacon’s process: In this process a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed. Hot gases at about 723 K are passed through a jacket that surrounds the chamber.

The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is given below,

(ii) In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

[OR]

(b) (i) Ti (Z = 22) Ti³⁺3d¹
V (Z = 23) V⁴⁺ 3d¹
.’. \(\mu=\sqrt{1(1+2)}=\sqrt{3}=1.73 \mu_{\mathrm{B}}\) So they are paramagnetic.

(ii) 1. Cis – [Co(en)₂Cl₂]⁺

The coordination complex [Co(en)₂Cl₂]⁺ has three isõmers two optically active cis forms and the optically inactive trans form.

Question 36.
(a) (i) Calculate the number of atoms in a fee unit cell.
(ii) How do nature of the reactant influence rate of reaction?
[OR]
(b) (i) Account for the acidic nature of HClO₄
In terms of Bronsted – Lowry theory, identify its conjugate base.
(ii) IS it possible to store copper sulphate in an iron vessel for a long time?
Given \(\mathbf{E}_{\mathrm{Cu}^{2+} \mathrm{Cu}}^{\circ}=\mathbf{0 . 3 4 V}\) and \(\mathbf{E}_{\mathbf{F e}^{2+} \mathbf{F e}}^{s}=+\mathbf{0 . 4 4 V}\)
Answer:
(a) (i) Number of atoms in a fcc unit cell,

(ii) Nature and state of the reactant:
We know that a chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product. The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.
Let us compare the following two reactions that we carried out in volumetric analysis.

1. Redox reaction between ferrous ammonium sulphate (FAS) and KMnO₄
2. Redox reaction between oxalic acid and KMnO₄

The oxidation of oxalate ion by KMnO₄ is relatively slow compared to the reaction between KMnO₄ and Fe²⁺ . In fact heating is required for the reaction between KMnO₄ and Oxalate ion and is carried out at around 60°C.

The physical state of the reactant also plays an important role to influence the rate of reactions. Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants. For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine. Let us consider another example that we carried out in inorganic qualitative analysis of lead salts. If we mix the aqueous solution of colorless potassium iodide with the colorless solution of lead nitrate, precipitation of yellow lead iodide take place instantaneously, whereas if we mix the solid lead nitrate with solid potassium iodide, yellow coloration will appear slowly.

[OR]

(b) (i) HClO₄ ⇌ H⁺ + ClO₄^–

According to Lowry – Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.

Let us consider the stabilities of the conjugate bases ClO₄^– , CIO₃^– , ClO₂^– and CIO^– formed from these acid HClO₄, HClO₃ , HClO₂, HOCl respectively. These anions are stabilized to greater extent, it has lesser attraction for proton and therefore, will behave as weak base. Consequently the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjugate base has weak acid.

The charge stabilization increases in the order, ClO^– < ClO^–₂ < ClO^–₃ < CIO^–₄ . This means ClO^–₄ will have maximum stability and therefore will have minimum attraction for H⁺. Thus ClO^–₄ will be weakest base and its conjugate acid HClO₄ is the strongest acid.

ClO^–₄ is the conjugate base of the acid HClO₄.

(ii) E⁰_cell = E⁰_ox + E⁰_red = 0.44 V +0.34 V = 0.78 V
These +ve E⁰_cell values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

Question 37.
(a) (i) Explain the formation of water with copper catalyst by intermediate compound formation theory. (3)
(ii) O-nitro phenol is slightly soluble in water where as P-nitro phenol is more soluble. Give reason. (2)
[OR]
(b) (i) What happens when the following alkenes are subjected to reductive ozonolysis.
1. propene
2.1-Butene
3. Isobutylene (3)
(ii) What are reducing and non – reducing sugars? (2)
Answer:
(a) (i) Formation of water due to the reaction of H₂ and O₂ in the presence of Cu can be given as

(ii) O-nitro phenol is slightly soluble in water and more volatile due to intra molecular hydrogen bonding, whereas P-nitro phenol is more soluble in water and less volatile due to intermolecular hydrogen bonding.

[OR]

(b)

(ii) Reducing sugars: Those carbonhydrates which contain free aldehyde or ketonic group and reduces Fehling’s solution and Tollen’s reagent are called reducing sugars. All monosacchaides whether aldose or ketone are reducing sugars.

Non-reducing sugars: Cabohydrates which do not reduce Tollen’s reagent and Fehling’s solution are called non-reducing sugars. Example Sucrose. They do not have free aldehyde group.

Question 38.
(a) A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO₂ (B) on heating with liquid ammonia followed by treating with Br₂ /KOH gives (C) which on treating with NaNO₂ and HCl at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D. (5)
[OR]

(b) Explain the mechanism of cleansing action of soaps and detergents.(5)
Answer:

[OR]

(b) Mechanism of cleansing action of soaps and detergents:

• The cleansing action of both soaps and detergents from their ability to lower the surface tension of water, to emulsify oil or grease and to hold them in a suspension in water.
• This ability is due to the structure of soaps and detergents.
• In water a sodium soap dissolves to form soap anions and sodium cations. For example, the following chemical equation shows the ionisation of sodium palmitate.
  
• A soap anion consists of a long hydrocarbon chain with a carboxylate group on one end. The hydrocarbon chain, which is hydrophobic, is soluble in oils or grease. The ionic part is the carboxylate group which is hydrophilic, is soluble in water.
  
• In water, detergent dissolves to form detergent anions and sodium cations. For example the following chemical equations show the ionisation of sodium alkyl sulphate and sodium alkyl benzene sulphate.
  

6. The following explains the cleansing action of a soap or detergent on a piece of cloth with a greasy stain:

• A soap or detergent anion consists of a hydrophobic part and a hydrophilic part.
• Soap or detergent reduces the surface tension of water. Therefore the surface of the cloth is wetted thoroughly.
• The hydrophobic parts of the soap or detergents anions are soluble in grease.
• The hydrophilic parts of the anions are soluble in water.
• Scrubbing or mechanical agitation helps to pull the grease away from the cloth and the
• grease is broken into smaller droplets.
• Repulsion between the droplets causes the droplets to be suspended in water, forming an emulsion.
• Thus the droplets do not coagulate or redeposit on the cloth. Rinsing washes away the droplets.

Students can Download Tamil Nadu 12th Maths Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th 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 12th Maths Model Question Paper 4 English Medium

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. questions of Part I, II. III and IV are to be attempted separately
4. Question numbers 1 to 20 in Part I are objective type 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-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part – I

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

Question 1.
If A^T A^-1 is symmetric, then A² = _______
(a) A^-1
(b) (A^T)²
(c) A^T
(d) (A^-1)²
Answer:
(b) (AT)2

Question 2.
If p + iq = \(\frac{a+i b}{a-i b}\), then p² + q² = ________.
(a) 0
(b) 2
(c) 1
(d) -1
Answer:
(c) 1

Question 3.
If ω ≠ 1 is a cubic root of unity and \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -\omega^{2}-1 & \omega^{2} \\
1 & \omega^{2} & \omega^{7}
\end{array}\right|\) = 3k, then k is equal to _______.
(a) 1
(b) -l
(c) \(\sqrt{3} i\)
(d) \(-\sqrt{3} i\)
Answer:
(d) \(-\sqrt{3} i\)

Question 4.
The value of sin^-1 (cos x), 0 ≤ x ≤ π is _______.
(a) π – x
(b) x – \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{2} – x\)
(d) π – x
Answer:
(c) \(\frac{\pi}{2} – x\)

Question 5.
The radius of the circle 3x² + by² + 4bx – 6by + b² = 0 is ________.
(a) 1
(b) 3
(c) \(\sqrt{10}\)
(d) \(\sqrt{11}\)
Answer:
(c) \(\sqrt{10}\)

Question 6.
The equation of the directrix of the parabola y² = -8x is ______.
(a) y + 2 = 0
(b) x – 2 = 0
(c) y – 2 = 0
(d) x + 2 = 0
Answer:
(b) x – 2 = 0

Question 7.
If \(\vec{a}\) and \(\vec{b}\) are parallel vector, then \([\vec{a}, \vec{c}, \vec{b}]\) is equal to _____
(a) 2
(b) -1
(c) 1
(d) 0
Answer:
(d) 0

Question 8.
The length of the perpendicular from the origin to the plane \(\vec{r} \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})=26\) is _______.
(a) 26
(b) \(\frac{26}{169}\)
(c) 2
(d) \(\frac{1}{2}\)
Answer:
(c) 2

Question 9.
The curve y = ax⁴ + bx² with ab > 0
(a) has no horizontal tangent
(b) is concave up
(c) is concave down
(d) has no points of inflection
Answer:
(d) has no points of inflection

Question 10.
The asymptote to the curve y² (1 + x) = x² (1 – x) is _______.
(a) x = 1
(b) y = 1
(c) y = -1
(d) x = -1
Answer:
(d) x = -1

Question 11.
If f(x, y, z) = xy + yz + zx, then f_x – f_z is equal to ________.
(a) z – x
(b) y – z
(c) x – z
(d) y – x
Answer:
(a) z – x

Question 12.
If f(x, y) = e^xy , then \(\frac{\partial^{2} f}{\partial x \partial y}\) is equal to ________.
(a) xye^xy
(b) (1 + xy) e^xy
(c) (1 + y) e^xy
(d) (1 + x) e^xy
Answer:
(b) (1 + xy) e^xy

Question 13.
The value of \(\int_{0}^{\frac{\pi}{6}} \cos ^{3} 3 x d x\) is _______.
(a) \(\frac{2}{3}\)
(b) \(\frac{2}{9}\)
( c) \(\frac{1}{9}\)
(d) \(\frac{1}{3}\)
Answer:
(b) \(\frac{2}{9}\)

Question 14.
If f(x) is even then \(\int_{-a}^{a} f(x) d x \) _______.

Answer:
(b) \(2 \int_{0}^{a} f(x) d x\)

Question 15.
The order and degree of the differential equation \(\sqrt{\sin x}\)(dx + dy) = \(\sqrt{\sin x}\) (dx- dy) is ________.
(a) 1, 2
(b) 2, 2
(c) 1, 1
(d) 2, 1
Answer:
(c) 1, 1

Question 16.
The solution of the differential equation \(\frac{d y}{d x}\) = 2xy is _______.
(a) y = c e^x2
(b) y = 2x² + c
(c) = ce^-x2 + c
(d) y = x² + c
Answer:
(a) y = c e^x2

Question 17.
If P{X = 0} = 1 – P{X = 1}. If E[X] = 3Var(X), then P{X = 0} ________.
(a) \(\frac{2}{3}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{1}{3}\)
Answer:
(d) \(\frac{1}{3}\)

Question 18.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times. Then the possible values of X are _________.
(a) i + 2n, i = 0, 1, 2 … n
(b) 2i – n, i = 0, 1, 2 … n
(c) n – i, i = 0, 1, 2 … n
(d) 2i + 2n, i = 0, 1, 2 … n
Answer:
(b) 2i – n, i = 0, 1, 2 … n

Question 19.
In the set Q define a Θ b= a + b + ab. For what value of y, 3 Θ (y Θ 5) = 7 ?

Answer:
(b) y = \(\frac{-2}{3}\)

Question 20.
If X is a continuous random variable then P(X > a) =
(a) P (X < a)
(b) 1 – P (X > a)
(c) P (X > a)
(d) 1 – P (x ≥ a)
Answer:
(c) P (X > a)

Part – II

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

Question 21.
Reduce the matrix \(\left[\begin{array}{ccc}
3 & -1 & 2 \\
-6 & 2 & 4 \\
-3 & 1 & 2
\end{array}\right]\) to a row-echelon form.
Answer:

Question 22.
Find the least positive integer n such that \(\left(\frac{1+i}{1-i}\right)^{n}=1\)
Answer:

Question 23.
Find the value of \(\sin ^{-1}\left(\sin \left(\frac{5 \pi}{4}\right)\right)\)
Answer:

Question 24.
Identify the type of conic section for the equation 3x² + 3y² – 4x + 3y + 10 = 0
Answer:
Comparing this equation with the general equation of the conic
Ax² + Bxy + cy² + Dx + Ey +F = 0
We get A = C also B = 0
So the given conic is a circle.

Question 25.
If U(x, y, z) = log(x³ + y³ + z³), find \(\frac{\partial \mathrm{U}}{\partial x}+\frac{\partial \mathrm{U}}{\partial y}\) and \(\frac{\partial U}{\partial z}\)
Answer:

Question 26.
Find, by integration, the volume of the solid generated by revolving about the x-axis, the region enclosed by y = 2x², y = 0 and x = 1.
Answer:

Question 27.
Solve the differential equation: \(\frac{d y}{d x}-x \sqrt{25-x^{2}}=0\)
Answer:

Question 28.
Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
Answer:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3
Hence, the probabilities
P(X = 0) = P (No heads) = \(\frac{1}{8}\);
P(X = 1) = P (1 head) = \(\frac{3}{8}\);
P(X = 2) = P (2 heads) = \(\frac{3}{8}\);
P(X = 3) = P (3 heads)= \(\frac{1}{8}\);
∴ The probability mass function is
\(f(x)=\left\{\begin{array}{lll}
1 / 8 & \text { for } & x=0,3 \\
3 / 8 & \text { for } & x=1,2
\end{array}\right.\)

Question 29.
Construct the truth table for the following statements. \(\neg p \wedge \neg q\)
Answer:
Truth table for \(\neg p \wedge \neg q\)

Question 30.
Write the Maclaurin series expansion of the function: ex
Answer:
f (x) = e^x; f (0) = e⁰ = 1
f’ (x) = e^x; f’ (0) = 1
f”(x) = e^x; f”(0) = 1

Part – III

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

Question 31.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\)show that A^-1 = \(\frac{1}{2}\) (A² – 3I).

Question 32.
Find the values of the real numbers x and y, if the complex numbers.
(3 – i)x – (2 – i) y + 2i + 5 and 2x + (-1 + 2i) y + 3 + 2i are equal.

Question 33.
It is known that the roots of the equation x³ – 6x² – 4x + 24 = 0 are in arithmetic progression. Find its roots.

Question 34.
Prove that: \(\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}\)

Question 35.
Find the equation of a circle of radius 5 whose centre lies on x-axis and which passes through the point (2, 3).

Question 36.
Using the l’ Hopital Rule prove that, \(\lim _{x \rightarrow 0^{+}}(1+x)^{\frac{1}{x}}=e\)

Question 37.
If v(x, y) = x² – xy + \(\frac{1}{4}\) y² + 7, x, y ∈ R, find the differential dv.

Question 38.
Find the area of the region bounded by 2x – y + 1 =0, y = – 1, y = 3 and y-axis..

Question 39.
Solve: \(\frac{d y}{d x}\) + 2y cot x = 3x² cosec²x

Question 40.
If the straight lines \(\frac{x-5}{5 m+2}=\frac{2-y}{5}=\frac{1-z}{-1}\) and x = \(\frac{2 y+1}{4 m}=\frac{1-z}{-3}\) are perpendicular to each other, find the value of m.

Part – IV

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

Question 41.
(a) Investigate the values of X and p the system of linear equations.
2x + 3y + 5z = 9, 7x + 3y – 5z = 8, 2x + 3y + λz = µ, have
(i) no solution (ii) a unique solution (iii) an infinite number of solutions.
[OR]
(b) If z(x, y) = x tan^-1 (xy), x = t², y = s e^t, s, t ∈ R, Find \(\frac{\partial z}{\partial t}\) and \(\frac{\partial z}{\partial t}\) at s = t = 1.

Question 42.
(a) Form the equation whose roots are the squares of the roots of the cubic equation
x³ + ax² + bx + c = 0.
[OR]
(b) Find the intervals of concavity and the points of inflection of the function.
f(θ) = sin 2θ in (0, π)

Question 43.
(a) If a = cos 2α + i sin 2α, b = cos 2β + i sin 2β and c = cos 2γ + i sin 2γ, prove that.

(b) A closed (cuboid) box with a square base is to have a volume of 2000 c.c. The material for the top and bottom of the box is to cost Rs. 3 per square cm and the material for the sides is to cost Rs. 1.50 per square cm. If the cost of the materials is to be the least, find the dimensions of the box.

Question 44.
(a) Prove that a straight line and parabola cannot intersect at more than two points.
[OR]
(b) Solve \(\left(y-e^{\sin ^{-1} x}\right) \frac{d x}{d y}+\sqrt{1-x^{2}}=0\)

Question 45.
(a) Solve \(\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)
[OR]
(b) Show that the lines \(\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\) and \(\frac{x-4}{2}=\frac{y}{0}=\frac{z+1}{3}\) the point of intersection.

Question 46.
(a) A tank initially contains 50 liters of pure water. Starting at time t = 0 a brine containing with 2 grams of dissolved salt per litre flows into the tank at the rate of 3 liters per minute. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the amount of salt present in the tank at any time t > 0.
[OR]
(b) If X ~ B(n, p) such that 4P (X = 4) = P (x = 2) and n = 6 . Find the distribution, mean and standard deviation.

Question 47.
(a) Find the centre, foci, and eccentricity of the hyperbola 11x² – 25y² – 44x + 50y – 256 = 0
[OR]
(b) Verify (i) closure property (ii) commutative property (iii) associative property (iv) existence of identity and (v) existence of inverse for the operation +5 on Z5 using table corresponding to addition modulo 5.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Chemistry Model Question Paper 3 English Medium

Time: 3 Hours
Maximum Marks: 70

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 lo be answered in about one or two sentences
6. Question numbers 25 to 33 in Part III are three-marks 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 answered in detail. Draw diagrams wherever necessary

Part – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Wolframite ore is separated from tinstone by the process of
(a) Smelting
(b) Calcination
(c) Roasting
(d) Electromagnetic separation
Answer:
(d) Electromagnetic separation

Question 2.

Identify A
(a) BN₃
(b) B₃N
(c) (BN)₃
(d) BN
Answer:
(d) BN

Question 3.
Among the following the correct order of acidity is
(a) HClO₂ < HCIO < HClO₃ < HClO₄
(b) HClO₄ < HClO₂ < HClO < HClO₃
(c) HClO₃ < HClO₄ < HClO₂ < HClO
(d) HClO < HClO₂ < HClO₃ < HClO₄
Answer:
(d) HClO < HClO₂ < HClO₃ < HClO₄

Question 4.
Which of the following transition metal is present in Vitamin B₁₂ ?
(a) Cobalt
(b) Platinum
(c) Copper
(d) Iron
Answer:
(a) Cobalt

Question 5.
A magnetic moment of 1.73BM will be shown by one among the following.
(a)TiCl₄
(b) [COCl₆]^4-
(c) [CU(NH₃)₄]²⁺
(d) [Ni(CN)₄]^2-
Answer:
(c) [CU(NH₃)₄]²⁺

Question 6.
Consider the following statements.
(,i) metallic solids possess high electrical and thermal conductivity
(ii) solid ice are soft solids under room temperature
(iii) In non polar molecular solids constituent molecules are held together by strong electrostatic forces of attraction
Which of the above statements is./ are not correct?
(a) (i) & (ii)only
(b) (iii) only
(c) (ii) only
(d) (i) only
Answer:
(b) (iii) only

Question 7.
For a first order reaction, the rate constant is 6.909 min^-1.The time taken for 75% conversion
in minutes is …………………………..

Answer:
(b) \(\left(\frac{2}{3}\right) \log 2\)
Solution:

Question 8.
The pH of 10^-5 M KOH solution will be ……………….
(a) 9
(b) 5
(c) 19
(d) none of these
Answer:
(a) 9

Solution:
KOH → K⁺ + OH^–
10^-5M 10^-5M 10^-5M
[OH^–]= 10^-5M
pH = 14 – pOH
pH = 14 – (-log [OH^–])
= 14 + log [OH^– ] = 14 + log10^-5
= 14 – 5 = 9

Question 9.
The Lead storage battery is used in
(a) pacemakers
(b) automobiles
(c) electronic watches
(d) flash light
Answer:
(b) automobiles

Question 10.
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of AS2S3 are given below
(I) (NaCl) = 52
(II) (BaCl) = 0.69
(III) (MgSO₄) = 0.22
The correct order of their coagulating power is
(a) III > II > I
(b) I > II > III
(c) I > III > II
(d) II > III> I
Hint: coagulating power ± \(\frac{1}{\text { coagulation value }}\)
Answer:
(a) III > II > I

Question 11.
Oxygen atom in ether is
(a) very active
(b) replacable
(c) comparatively inert
(d) less active
Answer:
(c) comparatively inert

Question 12.
During nucleophilic addition reaction, the hybridisation of carbon changes from
(a) sp² to sp³
(b) sp³ to sp²
(c) sp to sp³
(d) dsp² to sp³
Answer:
(a) sp² to sp³

Question 13.
Match the following:

Answer:
(a) A – 3, B – 4, C – 1, D – 2

Question 14.
Which one of the following is levorotatory?

Answer:

Question 15.
Which one of the following structures represents nylon 6,6 polymer?

Answer:

Part – II

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

Question 16.
Carbon monoxide is more effective reducing agent than carbon below 983 K but, above this temperature, the reverse is true -Explain.
Answer:
From the Ellingham diagram, we find that at 983 K, the curves intersect.

The value of ∆G° for change of C to CO₂ is less than the value of ∆G° for change of CO to CO₂ Therefore, coke (C) is a better reducing agent than CO at 983K or above temperature. However below this temperature (e.g. at 673K), CO is more effective reducing agent than C.

Question 17.
Mention the uses of the potash alum.
Answer:

• It is used for purification of water
• It is also used for water proofing and textiles
• It is used in dyeing, paper and leather tanning industries
• It is employed as a styptic agent to arrest bleeding.

Question 18.
Why Gd³⁺ is colourless?
Answer:
Gd – Electronic Configuration : [Xe] 4f⁷5d¹ 6s²
Gd³⁺ Electronic Configuration : [Xe] 4f⁷
In Gd³⁺ , no electrons are there in outer d-orbitals. d-d transition is not possible. So it is colourless.

Question 19.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
In the molten state, ionic solids dissociate to give free ions and hence can conduct electricity. However in the solid state, as the ions are not free to move but remain held together by strong electrostatic forces of attraction, so they cannot conduct electricity in the solid state.

Question 20.
What are Lewis acids and bases? Give two example for each.
Answer:
I. Lewis acids:
(i) Lewis acid is a species that accepts an electron pair.
(ii) Lewis acid is a positive ion (or) an electron deficient molecule.
(iii) Example, Fe²⁺, CO₂, BF₃ , SiF₄ etc…

II. Lewis bases:
(i) Lewis base is a species that donates an electron pair.
(ii) Lewis base is an anion (or) neutral molecule with atleast one lone pair of electrons.
(iii) Example, NH₃, F^–, CH₂= CH₂ CaO etc….

Question 21.
NH₃, CO₂ are readily adsorbed where as H₂, N₂ are slowly adsorbed. Give reason.
Answer:

• The nature of adsorbate can influence the adsorption. Gases like NH, C02 are easily liquefiable as have greater Van der Waals forces of attraction and hence readily adsorbed due to high critical temperature.
• But permanent gases like H₂. N₂ can not be easily liquefied and having low critical temperature and adsorbed slowly.

Question 22.
Alcohol can act as Bronsted base. Prove this statement.
Answer:
Alcohols can also act as a Bronsted bases. It is due to the presence of unshared electron pairs on oxygen which make them to accept proton. So proton acceptor are Bronsted bases. i. e., alcohols are Bronsted bases.

Question 23. Arrange the following in decreasing order of basic strength

Answer:
(i) Aliphatic amines are more basic than aromatic amines. Therefore CH₃CH₂NH₂ and CH₃NH₂ are more basic. Among the ethylamine and methylamine, ethylamine was experienced more +1 effect than methylamine and hence ethylamine is more basic than methylamine.

(ii) Nitrogroup has a powerfol electron withdrawing group and they have both -R effect as well as -I effect. As a result, all the nitro anilines are weaker bases than aniline. In P-nitroaniline

both R effect and -I effect of the NO₂ group decrease the basicity.

(iii) Therefore decreasing order of basic strength is,

Question 24.
How are biopolymers more beneficial than synthetic polymers?
Answer:
Durability of synthetic polymers is advantageous, however it presents a serious waste disposable problem. In renewal of the disposable problem, biodegradable polymers are useful to us.

Biopolymers are safe in use. They disintegrate by themselves in biological system during a certain period of time by enzymatic hydrolysis and to some extent by oxidation and hence, are biodegradable. As a result, they do not cause any pollution.

Part – III

Answer any six questions. Question No. 26 is compulsory. [6 × 3 = 18]

Question 25.
Explain Aluminothermic process.
Answer:
Metallic oxides such as Cr₂O₃ can be reduced by an aluminothermic process. In this process, the metal oxide is mixed with aluminium powder and placed in a fire clay crucible. To initiate the reduction process, an ignition mixture (usually magnesium and barium peroxide) is used.

BaO₂ + Mg → BaO + Mgo

During the above reaction a large amount of heat is evolved (temperature upto 2400°C, is generated and the reaction enthalpy is : 852 kJ-moF1) which facilitates the reduction of Cr₂O₃ by aluminium power.

Question 26.
Give a reason to support that sulphuric acid is a dehydrating agent.
A double salt which contains fourth period alkali metal (A) on heating at 500K gives (B). Aqueous solution of (B) gives white precipitate with BaCl₂ and gives a red colour compound with alizarin. Identify A and B.
Answer:
(i) A double salt which contains fourth period alkali metal (A) is potash alum
K₂SO₄ Al₂ (SO₄ )₃ .24H₂O

(ii) On heating potash alum (A) 500 K give anhydrous potash alum (or) burnt alum (B).

(iii) Aqueous solution of burnt alum, has sulphates ion, potassium ion and aluminium ion.
Sulphate ion reacts with BaCl2 to form white precipitate of Barium Sulphate:
(SO₄ )^2- + BaCl₂ → BaSO₄ + 2Cl^–
Aluminium ion reacts with alizarin solution to give a red colour compound.

Question 27.
[Ti (H₂O)₆ ]²⁺ is purple in colour. Prove this statement.
Answer:
(i) In [Ti (H₂O)₆ ]²⁺, the central metal ion is Ti³⁺ which has d¹ configuration. This single electron occupies one of the t_tgorbitals in the octahedral aqua ligand field.

(ii) When white light falls on this complex, the d electron absorbs light and promotes itself to eg level.

(iii) The spectral data show the absorption maximum is at 20000 cm^-1 corresponding to the crystal field splitting energy (∆₀) 239.7 kJ mol^-1. The transmitted colour associated with this absorption is purple and hence the complex appears in purple in colour.

Question 28.
Define half life of a reaction. Show that for a first order reaction half life is independent of initial concentration.
Answer:
Half life of a reaction is defined as the time required for the reactant concentration to reach one half of its initial value.
For a first order reaction, the half life is a constant i.e., it does not depend on the initial concentration.
The rate constant for a first order reaction is given by,

Question 29.
State Kohlrausch Law. How is it useful to determine the molar conductivity of weak electrolyte at infinite dilution.
Answer:
Kohlrausch’s law: It is defined as, at infinite dilution the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions.

Determination of the molar conductivity of weak electrolyte at infinite dilution:
It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally. However, the same can be calculated using Kohlraush’s Law. For example, the molar conductance of CH3COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH₃COONa .

Question 30.
Explain the following observations.
(a) Lyophilic colloid is more stable than lyophobic colloid.
(b) Coagulation takes place when sodium chloride solution added to a colloidal solution of ferric hydroxide.
(c) Sky appears blue in colour.
AnsweR:
(a) A lyophilic sol is stable due to the charge and the hydration of the sol particles. Such a sol can only be coagulatd by removing the water and adding solvents like alcohol, acetone, etc. and then an electrolyte. On the other hand, a lyophobic sol is stable due to charge only and hence it can be easily coagulated by adding small amount of an electrolyte.

(b) The colloidal particles get precipitated, i.e., ferric hydroxide is precipitated.

(c) The atmospheric particles of colloidal range scatter blue component of the white sunlight preferentially. That is why the sky appears blue.

Question 31.
Mention the uses of formic acid?
Answer:
Formic acid. It is used
(i) for the dehydration of hides.
(ii) as a coagulating agent for rubber latex
(iii) in medicine for treatment of gout
(iv) as an antiseptic in the preservation of fruit juice

Question 32.
Write the structure of all possible dipeptides which can be obtained from glycine and alanine.
Answer:

Therefore two dipeptides structures are possible from glycine and alanine. They are (i) glycyl alanine and (ii) Alanyl glycine

Question 33.
How the tranquilizers work in body?
Answer:

• They are neurologically active drugs.
• Tranquilizer acts on the central nervous system by blocking the neurotransitter dopamine in the brain.
• This drug is used for treatment of stress anxiety, depression, sleep disorders and severe mental diseases like schizophrenia.

Part – IV

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

Question 34.
(a) (i) How will you purify metals by using iodine?(3)
(ii) Boron does not react directly with hydrogen. Suggest one method to
prepare diborane from BF₃(2)
[OR]
(b) (i) Write the reason for the anomalous behaviour of Nitrogen. (3)
(ii) Mn²⁺ is more stable than Mn⁴⁺. Why? (2)
Answer:
(a) (i) This method is based on the thermal decomposition of metal compounds which lead to the formation of pure metals. Titanium and zirconium can be purified using this method. For example, the impure titanium metal is heated in an evacuated vessel with iodine at a temperature of 550 K to form the volatile titanium tetra-iodide.( TiI₄ ). The impurities are left behind, as they do not react with iodine.
Ti_(s) + 2I_2(s) → TiI₄ (vapour)

The volatile titanium tetraiodide vapour is passed over a tungsten filament at a
temperature around I 800 K. The titanium tetraiodide is decomposed and pure titanium
is deposited on the filament. The iodine is reused.
TiI₄ (vapour) ) → Ti_(s) +2I_2(s)

(ii) Boron does not react directly with hydrogen. However it forms a variety of hydrides
called boranes. Treatment of gaseous boron triuluoride with sodium hydride around
450 K gives diborane.

(b) (i) 1. Due to its small size, high electro negativity, high ionisation enthalpy and absence of d-orbitals.

2. N₂ has a unique ability to form pπ- pπ multiple bond whereas the heavier members of this group (15) do not form pπ- pπ bond, because their atomic orbitals are so large and diffused that they cannot have effective overlapping.

3. Nitrogen exists a diatomic molecule with triple bond between the two atoms whereas other elements form single bond in the elemental state.

4. N cannot form dπ- dπ bond due to the absence of d-orbitals whereas other elements can.

(ii) The relative stability of different oxidation states of 3d metals is correlated with the extra stability of half-filled and fully filled electronic configurations.
Example: Mn²⁺ (3d⁵) is more stable than Mn⁴⁺ (3d³)

Question 35.
(a) (i) Draw all possible stereo isomers of a complex Ca[CO(NH₃)Cl(Ox)₂] (3)
(ii) What is Bragg’s equation? (2)
(b) (i) What is an elementary reaction? Give the differences between order and molecularity of a reaction. (3)
(ii) In a first order reaction A → products, 60% of the given sample of A decomposes in 40 min. what is the half life of the reaction? (2)
Answer:
(a) (i) Possible stereo isomers of a complex Ca[CO(NH₃)Cl(Ox)₂]

(ii) 1. X-ray diffraction analysis is the most powerful tool for the determination of crystal structure.

2. The interplanar distance (d) between two successive planes of atoms can be calculated using the following equation form the X-ray diffraction data 2d sin θ = nλ. The equation is known as Bragg’s equation
Where λ = wavelength of X-ray; d = Interplanar distance
θ = The angle of diffraction n = order of reflection
By knowing the values of λ, λ and n, we can calculate the value of d
\(d=\frac{n \lambda}{2 \sin \theta}\)

Using these values, the edge of the unit cell can be calculated.

[OR]

(b) (i) Elementary reaction: Each and every single step in a reaction mechanism is called an elementary reaction.

Differences between order and molecularity:

Order of a reaction :

• It is the sum of the powers of concentration terms involved in the experimentally determined rate law.
• It can be zero (or) fractional (or) integer.
• It is assigned for a overall reaction.

Molecularity of a reaction :

• It is the total number of reactant species that are involved in an elementary step.
• It is always a whole number, cannot be zero or a fractional number.
• It is assigned for each elementary step of mechanism.

(ii)

Question 36.
(a) (i) Derive the relation between pH and pOH (3)
(ii) Give three uses of emulsions.(2)
[OR]
(b) How would you measure the conductivity of ionic solutions? (5)
Answer:
(a) (i) pH = -log_1o [H₃O⁺] …………..(1)
pOH = -log_1o [OH^–] ……..(2)
Adding equations (1) and (2),
pH + pOH = (-log_1o[H₃O⁺]) + (-log_1o [OH^–])
= -[(log10tHO+]) + (logio [OfT])]
pH + pOH = -log₁₀[H₃O⁺] [OH^– ]
[H₃O⁺] [OH^–] = K_w
∴ pH + pOH = – log_1oK_w
pH + pOH = pKw [pk_w = -log_1oK_w]

At 25°C, the ionic product of water K_w =1 x 10^-14.
pK_w = -log_1o10^-14 = 14log_1o10= 14
pK_w = 14
∴ pH + pOH = 14 at 25°C.

(ii)

1. The cleansing action of soap is due to emulsions.
2. It is used in the preparation of vanishing cream.
3. It is used in the preparation of cold liver oil.

[OR]

(b) The conductivity of an electrolytic solution is determined by using a wheatstone bridge arrangement in which one resistance is replaced by a conductivity cell filled with the electrolytic solution of unknown conductivity.

In the measurement of specific resistance of a metallic wire, a DC power supply is used. Here AC is used for this measurement to prevent electrolysis. Because DC current through the conductivity cell leads to the electrolysis of the solution taken in the cell.

A wheatstone bridge is constituted using known resistances P,Q, a variable resistance S and conductivity cell. An AC source (550 Hz to 5 KHz) is connected between the junctions A and C.
A suitable detector is connected between the junctions B and D.
The variable resistance S is adjusted until the bridge is balanced and in this conditions, there is no current flow through the detector.

Under balanced condition
P/Q = R/S
∴ R = P/Q x S …………(1)
The resistance of the electrolytic solution (R) is calculated from the known resistance values P, Q and the measured S value using the equation (1).

Specific conductance (or) conductivity of an electrolyte can be calculated from the resistance value (R) using the following expression
\(\kappa=\frac{1}{R}\left[\frac{l}{A}\right]\)

The value of cell constant \(\frac{l}{A}\) is usually provided by the cell manufacturer. Alternatively the cell constant may be determined using KC1 solution whose concentration and specific conductance are known.

Question 37.
(a) (1) What is metamerism? Give the structure and ¡UPAC name of metamers of 2-methoxy propane (2)
(ii) Explain the following reactions.
(2)
(OR]
(b) An organic compound (A) of molecular formula C₇H₈O on oxidation with alkaline KMnO₄ gives (B) of formula C₇H₆O.
(B) on reaction with Cl₂ in the presence of catalyst FeCl₃ gives
(C) of formula C₇H₅OCl. (B) on reaction with Cl in the absence
of catalyst gives C₇H₅OCl. Identify A,B,C,D and explain the reaction involved.
Answer:
(a) (i) Metamerism: It is a special type of isomerism in which molecules with same formula, same functional group, but different only in the nature of the alkyl group attached to oxygen.

Ethoxy ethane and 1-methoxy propane are metamers of 2-methoxy propane.

(ii)

(b)

Question 38.
(a) (0 Account for the following:
1. Primary amines (R – NH2)have higher boiling point than tertiary amines (R₃N).
2. Aniline does not undergo Friedel-Crafts reaction.
3. (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution. (3)
(H) Name two fat soluble vitamins, their sources and the diseases caused due to
their deficiency in diet. (2)
[OR]
(b) (i) Why ranitidine is a better antacid than magnesium hydroxide? (2)
(ii) What is bakelite? How is it prepared? Give its uses. (3)

(a) (i) 1. Due to maximum intermolecular hydrogen bonding in primary amines (due to presence of more number of H-atoms), primary amines have higher boiling point in comparison to tertiary amines.

2. Aniline does not undergo Friedel-Crafts reaction due to acid-base reaction. Aniline and a Lewis Acid / Protic Acid, which is used in Friedel-crafts reaction.

3. In (CH₃)₃N there is maximum steric hindrance and least solvation but in (CH₃)₂NH the solvation is more group; di-methyl amine is still a stronger base than trimethyl amine.

(ii)

Vitamins	Sources Deficiency	Diseases
Vitamin A	Fish, liver oil, carrot	Night blindness
Vitamin D	Sunlight, milk, egg yolk	Rickets and osteomalacia

[OR]

(b) (i) To treat acidity, weak base such as magnesium hydroxide is used. But this weak base make the stomach alkaline and trigger the production of much acid. This treatment only relieves the symptoms and does not control the cause. But ranitine stimulate the secretion of HC1 by activating the receptor in the stomach wall which binds the receptor and inactivate them. So ranitine is a better antacid than magnesium hydroxide.

(ii) 1. Bakelite is a thermo setting plastic. It is prepared from the monomers such as phenol and formaldehyde. The condensation polymerisation take place in the presence of acid or base catalyst.

2. Phenol reacts with methanal to form ortho or para hydroxyl methyl phenols which on further reaction with phenol gives linear polymer called novolac. Novolac on further healing with formaldehyde undergoes cross linkages to form bakelite.

• Novolac is used in paints ‘Soft bakelites are used in making glue for binding laminated
  wooden planks and in varnishes
• Hard bakelites are used to prepare combs, pens.

Students can Download Tamil Nadu 12th Biology Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 5 English Medium

General Instructions:

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

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Choose the correct statement(s) about tenuinucellate ovule
(a) Sporogenous cell is hypodermal
(b) Ovules have fairly large nucellus
(c) Sporogenous cell is epidermal
(d) Ovules have single layer of nucellus tissue
Answer:
(a) Sporogenous cell is hypodermal

Question 2.
The genotype of a plant showing the dominant phenotype can be determined by _______.
(a) Back cross
(b) Test cross
(c) Dihybrid cross
(d) Pedigree analysis
Answer:
(b) Test cross

Question 3.
An allo-hexaploidy contains ________.
(a) Six different genomes
(b) Six copies of three different genomes
(c) Two copies of three different genomes
(d) Six copies of one genome
Answer:
(b) Six copies of three different genomes

Question 4.
Match the following column I with column II


Answer:
(D) 1 – c, 2 – d, 3 – a, 4 – b

Question 5.
Assertion (A): Incubation is followed by Inoculation.
Reason (R): Explant is inoculated to media.
(a) Both A and R are correct but R is not a correct explanation to A
(b) R explains A
(c) A is correct R is incorrect
(d) Both A and R are incorrect
Answer:
(b) R explains A

Question 6.
Environment of any community is called
(a) Paratope
(b) Biotope
(c) Ecotope
(d) Epitope
Answer:
(b) Biotope

Question 7.
Which of the following is not a sedimentary cycle?
(a) Nitrogen cycle
(b) Phosphorus cycle
(c) Sulphur cycle
(d) Calcium cycle
Answer:
(a) Nitrogen cycle

Question 8.
The plants which are grown in silvopasture system are
(a) Sesbania and Acacia
(b) Solanum and Crotolaria
(c) Clitoria and Begonia
(d) Teak and sandal
Answer:
(a) Sesbania and Acacia

Part – II

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

Question 9.
What are multiple alleles? Give an example.
Answer:
Three or more alternative forms of a gene that occupy the same locus and control the expression of a single trait.
E.g: ABO blood group

Question 10.
List out the benefits of herbicide tolerant crops.
Answer:
Advantages of Herbicide Tolerant Crops:

• Weed control improves higher crop yields.
• Reduces spray of herbicide.
• Reduces competition between crop plant and weed.
• Use of low toxicity compounds which do not remain active in the soil.
• The ability to conserve soil structure and microbes.

Question 11.
Soil formation can be initiated by biological organisms. How?
Answer:
Soil formation is initiated by the biological weathering process. Biological weathering takes place when organisms like bacteria, fungi, lichens and plants help in the breakdown of rocks through the production of acids and certain chemical substances.

Question 12.
Where did Montreal Protocol was held? State its objectives.
Answer:
The International treaty called the Montreal Protocol (1987) was held in Canada on substances that deplete ozone layer and the main goal of it is gradually eliminating the production and consumption of ozone depleting substances and to limit their damage on the Earth’s ozone layer.

Question 13.
What is Heterosis?
Answer:
The superiority of the F₁ hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to increase in growth, yield, greater adaptability of resistance to diseases, pest and drought.

Question 14.
Write the cosmetic uses of Aloe.
Answer:
Aloe gel are used as skin tonic. It has a cooling effect and moisturising characteristics and hence used in preparation of creams, lotions, shampoos, shaving creams, after shave lotions and allied products. It is used in gerontological applications for rejuvenation of aging skin. Products prepared from aloe leaves have multiple properties such as emollient, antibacterial, antioxidant, antifungal and antiseptic. Aloe vera gel is used in skin care cosmetics.

Part – III

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

Question 15.
List out the functions of tapetum.
Answer:

• It supplies nutrition to the developing microspores.
• It contributes sporopollenin through ubisch bodies thus plays an important role in pollen wall formation.
• The pollenkitt material is contributed by tapetal cells and is later transferred to the pollen surface.
• Exine proteins responsible for ‘rejection reaction’ of the stigma are present in the cavities of the exine. These proteins are derived from tapetal cells.

Question 16.
What do you mean by Embryoids? Write its application.
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids or from the in vitro cells directly form pre-embryonic cells which differentiate into embryoids.

Applications:

• Somatic embryogenesis provides potential plantlets which after hardening period can establish into plants.
• Somatic embryoids can be used for the production of synthetic seeds.
• Somatic embryogenesis is now reported in many plants such as Allium sativum, Hordeum vulgare, Oryza sativa, Zea mays and this is possible in any plant.

Question 17.
Give a comparative account on Seaweed liquid fertilizer.
Answer:
Seaweed liquid fertilizer (SLF) contains cytokinin, gibberellins and auxin apart from macro and micro nutrients. Most seaweed based fertilizers are made from kelp (brown algae) which grows to length of 150 metres. Liquid seaweed fertilizer is not only organic but also ecofriendly. The alginates in the seaweed that reacts with metals in the soil and form long, crosslinked polymers in the soil.

These polymers improve the crumbing in the soil, swell up when they get wet and retain moisture for a long time. They are especially useful in organic gardening which provides carbohydrates for plants. Seaweed has more than 70 minerals, vitamins and enzymes. It promotes vigorous growth. Improves resistance of plants to frost and disease. Seeds soaked in seaweed extract germinate much rapidly and develop a better root system.

Question 18.
Write a brief note on Detritus food chain.
Answer:
Detritus food chain is a type of food chain which begins with dead organic matter which is an important source of energy. A large amount of organic matter is derived from the dead plants, animals and their excreta. This type of food chain is present in all ecosystems. The transfer of energy from the dead organic matter, is transferred through a series of organisms called detritus consumers (detritivores)- small carnivores – large (top) carnivores with repeated eating and being eaten respectively. This is called the detritus food chain.

Question 19.
Mention the name of man-made cereal. How it is developed?
Answer:

Triticale, the successful first man made cereal. Depending on the ploidy level Triticale can be divided into three main groups:

1. Tetraploidy: Crosses between diploid wheat and rye.
2. Hexaploidy: Crosses between tetraploid wheat Triticum durum (macaroni wheat) and rye.
3. Octoploidy: Crosses between hexaploid wheat T. aestivum (bread wheat) and rye.

Hexaploidy Triticale hybrid plants demonstrate characteristics of both macaroni wheat and rye. For example, they combine the high-protein content of wheat with rye’s high content of the amino acid lysine, which is low in wheat.

Part – IV

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

Question 20.
(a) Describe the structure of a Cicer seed (dicot seed) with labelled diagram.
Answer:
Structure of a Cicer seed as an example for Dicot seed The mature seeds are attached to the fruit wall by a stalk called funiculus. The funiculus disappears leaving a scar called hilum. Below the hilum a small pore called micropyle is present. It facilitates entry of oxygen and water into the seeds during germination.

Each seed has a thick outer covering called seed coat. The seed coat is developed from integuments of the ovule. The outer coat is called testa and is hard whereas the inner coat is thin, membranous and is called tegmen.

In Pea plant the tegmen and testa are fused. Two cotyledons laterally attached to the embryonic axis are present. It stores the food materials in pea whereas in other seeds like castor the endosperm contains reserve food and the cotyledons are thin. The portion of embryonal axis projecting beyond the cotyledons is called radicle or embryonic root.

The other end of the axis called embryonic shoot is the plumule. Embryonal axis above the level of cotyledon is called epicotyl whereas the cylindrical region between the level of cotyledon is called hypocotyl. The epicotyl terminates in plumule whereas the hypocotyl ends in radicle.

[OR]

(b) (i) Explain the three types of hyperploidy.
(ii) List out the significance of ploidy.
Answer:
(i) (a) Trisomy: Addition of single chromosome to diploid set is called Simple trisomy (2n+l). Trisomics were first reported by Blackeslee (1910) in Datura stramonium (Jimson weed). But later it was reported in Nicotiana, Pisum and Oenothera. Sometimes addition of two individual chromosome from different chromosomal pairs to normal diploid sets are called Double trisomy (2n + 1 + 1).
(b) Tetrasomy: Addition of a pair or two individual pairs of chromosomes to diploid set is called tetrasomy (2n+2) and Double tetrasomy (2n+2+2) respectively. All possible tetrasomics are available in Wheat.
(c) Pentasomy: Addition of three individual chromosome from different chromosomal pairs to normal diploid set are called pentasomy (2n+3).

(ii)

• Many polyploids are more vigorous and more adaptable than diploids.
• Many ornamental plants are autotetraploids and have larger flower and longer flowering duration than diploids.
• Autopolyploids usually have increase in fresh weight due to more water content.
• Aneuploids are useful to determine the phenotypic effects of loss or gain of different chromosomes.
• Many angiosperms are allopolyploids and they play a role in an evolution of plants.

Question 21.
a. (i) Define tissue culture.
(ii) Explain the basic concepts involved in plant tissue culture.
Answer:
(i) Growing plant protoplasts, cells, tissues or organs away from their natural or normal environment, under artificial condition, is known as Tissue Culture.

(ii) Basic concepts of plant tissue culture are totipotency, differentiation, dedifferentiation and redifferentiation.
Totipotency: The property of live plant cells that they have the genetic potential when cultured in nutrient medium to give rise to a complete individual plant.

Differentiation: The process of biochemical and structural changes by which cells become specialized in form and function.

Redifferentiation: The further differentiation of already differentiated cell into another type of cell. For example, when the component cells of callus have the ability to form a whole plant in a nutrient medium, the phenomenon is called redifferentiation.

Dedifferentiation: The phenomenon of the reversion of mature cells to the meristematic state leading to the formation of callus is called dedifferentiation. These two phenomena of redifferentiation and dedifferentiation are the inherent capacities of living plant cells or tissue. This is described as totipotency.

[OR]

(b) What is soil profile? Explain the characters of different soil horizons.
Answer:
Soil is commonly stratified into horizons at different depth. These layers differ in their physical, chemical and biological properties. This succession of super-imposed horizons is called soil profile.

Horizon	Description
O-Horizon (Organic horizon) Humus	It consists of fresh or partially decomposed organic matter. O1 – Freshly fallen leaves, twigs, flowers and fruits. O2 – Dead plants, animals and their excreta decomposed by micro-organisms. Usually absent in agricultural and deserts.
A-Horizon (Leached horizon) Topsoil – Often rich in humus and minerals.	It consists of top soil with humus, living creatures and in-organic minerals. A1 – Dark and rich in organic matter because of mixture of organic and mineral matters. A2 – Light coloured layer with large sized mineral particles.
B-Horizon (Accumulation horizon) (Subsoil – Poor in humus, rich in minerals)	It consists of iron, aluminium and silica rich clay organic compounds.
C – Horizon (Partially weathered horizon) Weathered rock Fragments – Little or no plant or animal life.	It consists of parent materials of soil, composed of little amount of organic matters without life forms.
R – Horizon (Parent material) Bedrock	It is a parent bed rock upon which underground water is found.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Assertion (A): Asexual reproduction is called blastogenic reproduction.
Reason (R): It is accomplished by mitotic and meiotic divisions.
(a) A and R are correct
(b) A is correct but R is incorrect
(c) Both A and R are incorrect
(d) R is the correct explanation for A
Answer:
(b) A is correct but R is incorrect

Question 2.
The mature sperms are stored in the ________.
(a) Seminiferous tubules
(b) Vas deferens
(c) Epididymis
(d) Seminal vesicle
Answer:
(c) Epididymis

Question 3.
A contraceptive pill prevents ovulation by _______.
(a) blocking fallopian tube
(b) inhibiting the release of FSH and LH
(c) stimulating the release of FSH and LH
(d) causing immediate degeneration of released ovum.
Answer:
(b) inhibiting the release of FSH and LH

Question 4.
Patau’s syndrome is also referred to as ________.
(a) 13-Trisomy
(b) 18-Trisormy
(c) 21-Trisormy
(d) None of these
Answer:
(a) 13-Trisomy

Question 5.
Cyclosporin – A is an immunosuppressive drug produced from ________.
(a) Aspergillus niger
(b) Manascus purpureus
(c) Penicillium notatum
(d) Trichoderma polysporum
Answer:
(d) Trichoderma polysporum

Question 6.
PCR proceeds in three distinct steps governed by temperature, they are in order of ________.
(a) Denaturation, Annealing, Synthesis
(b) Synthesis, Annealing, Denaturation
(c) Annealing, Synthesis, Denaturation
(d) Denaturation, Synthesis, Annealing
Answer:
(a) Denaturation, Annealing, Synthesis

Question 7.
Match column I with column II

(a) A – 4, B – 5, C – 2, D – 3, E – 1
(b) A – 3, B – 1, C – 4, D – 2, E – 5
(c) A – 2, B – 3,C – 1, D – 5, E – 4
(d) A – 5, B – 4, C – 2, D – 3, E – 1
Answer:
(a) A – 4, B – 5, C – 2, D – 3, E – 1

Question 8.
Total number of mega biodiversity countries in the world is _______.
(a) 12
(b) 15
(c) 17
(d) 19
Answer:
(c) 17

Part – II

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

Question 9.
Why are the offsprings of oviparous animals are at a greater risk as compared to offsprings of viviparous organisms?
Answer:
Oviparous animals are egg-layers. The eggs containing embryo are laid out of their body and are highly susceptible to environmental factors (temperature, moisture etc.) and predators. Whereas, in viviparous animals, the embryo develops inside the body of female and comes out as young ones. Hence offsprings of oviparous animals are at risk compared to viviparous animal.

Question 10.
What is “let-down reflex”?
Answer:
Oxytocin causes the “Let-Down” reflex the actual ejection of milk from the alveoli of the mammary glands. During lactation, oxytocin also stimulates the recently emptied uterus to contract, helping it to return to pre – pregnancy size.

Question 11.
In E.coli, three enzymes β- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 12.
Compare relative dating with absolute dating.
Answer:
Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine.the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 13.
Write the name of causative agent, infection site, mode of transmission and any two symptoms of Chikungunya.
Answer:
Causative agent – Alpha virus
Infection site – Nervous system
Mode of transmission – Aedes aegypti (Mosquito)
Symptoms – Fever, headache, joint pain and swelling.

Question 14.
Define the following terms.
(a) Eutrophication (b) Algal Bloom
Answer:
Eutrophication refers to the nutrient enrichment in water bodies leading to lack of oxygen . and will end up in the death of aquatic organisms. Algal Bloom is an excess growth of algae due to abundant excess nutrients imparting distinct color to water.

Part – III

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

Question 15.
Write a short note on foeto scope.
Answer:
Foetoscope is used to monitor the foetal heart rate and other functions during late pregnancy and labour. The average foetal heart rate is between 120 and 160 beats per minute. An abnormal foetal heart rate or pattern may mean that the foetus is not getting enough oxygen and it indicates other problems. A hand-held doppler device is often used during prenatal visits to count the foetal heart rate. During labour, continuous electronic foetal monitoring is often used.

Question 16.
Under which condition does a microbe gains resistance against antibiotic?
Answer:
Antibiotic resistance occurs when bacteria develop the ability to defeat the drug designed to kill or inhibit their growth. It is one of the most acute threat to public health. Antibiotic resistance is accelerated by the misuse and over use of antibiotics, as well as poor infection prevention control. Antibiotics should be used only when prescribed by a certified health professional.

When the bacteria become resistant, antibiotics cannot fight against them and the bacteria multiply. Narrow spectrum antibiotics are preferred over broad spectrum antibiotics. They effectively and accurately target specific pathogenic organisms and are less likely to cause resistance.

Question 17.
State Fisher and Race hypothesis.
Answer:
Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature.

In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1. The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh⁺positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

Question 18.
Extinction of Dodo bird led to the danger of Calvaria tree – Support your answer.
Answer:
An example for co-extinction is the connection between Calvaria tree and the extinct bird of Mauritius Island, the Dodo. The Calvaria tree is dependent on the Dodo bird for completion of its life cycle. The mutualistic association is that the tough homy endocarp of the seeds of Calvaria tree are made permeable by the actions of the large stones in birds gizzard and digestive juices thereby facilitating easier germination. The extinction of the Dodo bird led to the imminent danger of the Calvaria tree coextinction.

Question 19.
Whether PCR can be done for RNA molecules? Yes or No? Explain.
Answer:
The PCR technique can also be used for amplifications of RNA in which case it is referred to as reverse transcription PCR (RT-PCR). In this process the RNA molecules (mRNA) must be converted to complementary DNA by the enzyme reverse transcriptase. The cDNA then serves as the template for PCR.

Part – IV

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

Question 20.
(a) Define the following terms with an example
(i) Hologamy
(ii) Isogamy
(iii) Anisogamy
(iv) Merogamy
(v) Paedogamy
Answer:
(i) Hologamy: In Hologamy, the adult individuals do not produce gametes, but they themselves act as gametes and fuse to form new individuals.
E.g.: Trichonympha

(ii) Isogamy : Fusion of morphologically and physiologically similar gametes.
E.g.: Monocystis

(iii) Anisogamy : Fusion of morphologically and physiologically dissimilar gametes.
E.g.: Vertebrates.

(iv) Merogamy : Fusion of small sized morphologically different gametes (merogametes)

(v) Paedogamy : Fusion of young individuals produced immediately after the mitotic division of adult parent cell.

[OR]

(b) Write in detail about cervical cancer.
Answer:
Cervical cancer is caused by a sexually transmitted virus called Human Papilloma virus (HPV). HPV may cause abnormal growth of cervical cells or cervical dysplasia. The most common symptoms and signs of cervical cancer are pelvic pain, increased vaginal discharge and abnormal vaginal bleeding. The risk factors for cervical cancer include

1. Having multiple sexual partners
2. Prolonged use of contraceptive pills

Cervical cancer can be diagnosed by a Papanicolaou smear (PAP smear) combined with an HPV test. X-Ray, CT scan, MRI and a PET scan may also be used to determine the stage of cancer. The treatment options for cervical cancer include radiation therapy, surgery and chemotherapy.

Modem screening techniques can detect precancerous changes in the cervix. Therefore screening is recommended for women above 30 years once in a year. Cervical cancer can be prevented with vaccination. Primary prevention begins with HPV vaccination of girls aged 9-13 years, before they become sexually active. Modification in lifestyle can also help in preventing cervical cancer. Healthy diet, avoiding tobacco usage, preventing early marriages, practicing monogamy and regular exercise minimize the risk of cervical cancer.

Question 21.
(a) Point out the differences between active and passive immunity.
Answer:
Active Immunity:

1. Active immunity is produced actively by host’s immune system.
2. It is produced due to contact with pathogen or by its antigen.
3. It is durable and effective in protection.
4. Immunological memory is present.
5. Booster effect on subsequent dose is possible.
6. Immunity is effective only after a short period.

Passive Immunity:

1. Passive immunity is received passively and there is no active host participation.
2. It is produced due to antibodies obtained from outside.
3. It is transient and less effective.
4. No memory.
5. Subsequent dose is less effective.
6. Immunity develops immediately.

[OR]

(b) How is the amplification of a gene sample of interest carried out using PCR?
Answer:
Denaturation, renaturation or primer annealing and synthesis or primer extension, are the three steps involved in PCR. The double stranded DNA of interest is denatured to separate into two individual strands by high temperature . This is called denaturation. Each strand is allowed to hybridize with a primer (renaturation or primer annealing). The primer template is used to synthesize DNA by using Taq – DNA polymerase.

During denaturation the reaction mixture is heated to 95°C for a short time to denature the target DNA into single strands that will act as a template for DNA synthesis. Annealing is done by rapid cooling of the mixture, allowing the primers to bind to the sequences on each of the two strands flanking the target DNA. During primer extension or synthesis the temperature of the mixture is increased to 75°C for a sufficient period of time to allow Taq DNA polymerase to extend each primer by copying the single stranded template.

At the end of incubation both single template strands will be made partially double stranded. The new strand of each double stranded DNA extends to a variable distance downstream. These steps are repeated again and again to generate multiple forms of the desired DNA, This process is also called DNA amplification.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Chemistry Model Question Paper 2 English Medium

Time: 3 Hours
Maximum Marks: 70

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 lo be answered in about one or two sentences
6. Question numbers 25 to 33 in Part III are three-marks 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 answered in detail. Draw diagrams wherever necessary

Part – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Blistered copper is .
(a) 98% pure copper
(b) 96% pure copper
(c) 97% pure copper
(d) 88% pure copper
Answer:
(a) 98% pure copper

Question 2.
Which of the following statements is not correct?
(a) Beryl is a cyclic silicate
(b) Mg₂SiO₄ is an orthosilicate
(c) SiO₄ is the basic structural unit of silicates
(d) Feldspar is not aluminosilicate
Answer:
(d) Feldspar is not aluminosilicate

Question 3.
Consider the following statements.
(i) phosphine is the most important hydride of phosphorous
(ii) phosphine is a poisonous gas with rotten egg smell.
(iii) phosphine is a powerful reducing agent
Which of the above statement(s) is / are correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) only
Answer:
(c) (i) and (iii)

Question 4.
The correct order of increasing oxidizing power in the series


Answer:

Hint: VO₂⁺ < Cr₂O₇^2- < MnO₄^– greater the oxidation state, higher is the oxidising power.

Question 5.
The geometry possible in [Fe F₆]^4- and [COF₊₆]^4- is ……………
(a) Trigonal bipyramidal
(b) Square planar
(c) Octahedral
(d) Tetrahedral
Answer:
(c) Octahedral

Question 6.
The number of carbon atoms per unit cell of diamond is
(a) 8
(b) 6
(c) 1
(d) 4
Answer:
(a) 8

Hint: In diamond carbon forming fee. Carbon occupies comers and face centres and also occupying half of the tetrahedral voids.

Question 7.
In a homogeneous reaction
A → B + C + D, the initial pressure was P₀ and after time t it was P. Expression for rate constant in terms of P₀ , P and t will be ……..

Answer:

Solution:

Question 8.
The solubility of an aqueous solution of Mg(OH)₂ be x then its K_sp is
(a) 4x³
(b) 108x⁵
(c) 27x⁴
(d) 9x
Answer:
(a) 4x³
Solution:
Mg(OH)₂ Mg²⁺(x) + 2OH^– ((2x)²)
K_sp = 4x³

Question 9.
During electrolysis of molten sodium chloride, the time required to produce 0. lmol of chlorine gas using a current of 3 A is ……………..
(a) 55 minutes
(b) 107.2 minutes
(c) 220 minutes
(d) 330 minutes
Hint: m = ZIt (mass of 1 mole of Cl₂ gas = 71)
Answer:
(b) 107.2 minutes

Question 10:
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of AS₂S₃ are given below
(I) (NaCl) = 52
(II) (BaCl) = 0.69
(III) (MgSO₄) = 0.22
The correct order of their coagulating power is
(a) III > II > I
(b) I > II > III
(c) I > III > II
(d) II > III> I
coagulating power ± \(\frac{1}{\text { coagulation value }}\)
Answer:
(a) III > II > I

Question 10.
Assertion : Coagulation power of Al³⁺ is more than Na⁺ .
Reason : greater the valency of the flocculating ion added, greater is its power to cause precipitation
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechulze rule)
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer:
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechulze rule)

Question 11.


Answer:
(a) A- 2, B – 3, C – 4, D – 1

Question 12.
Which of the following represents the correct order of acidity in the given compounds (
(a) FCH₂COOH > CH₃COOH > BrCH₂COOH > ClCH₃COOH
(b) FCH₂COOH > ClCH2COOH > BrCH COOH > CH COOH
(c) CH₃COOH > ClCH₂COOH > FCH₂COOH > Br – CH₂COOH
(d) Cl CH₂COOH > CH₃COOH > BrCH₂COOH > ICH₂COOH
Hint. -I effect increases the acidity. If electronegativity is high, -I effect is also high.
Answer:
(b) FCH₂COOH > ClCH2COOH > BrCH COOH > CH COOH

Question 13.
1 -nitrobutane and 2-methyl-1 -nitropropane are belong to
(a) position isomerism
(b) functional isomerism
(c) Tautomerism
(d) chain isomerism
Answer:
(d) chain isomerism

Question 14.
Haemoglobin is
(a) an enzyme
(b) a globular protein
(c) a vitamin
(d) carbohydrate
Answer:
(b) a globular protein

Question 15.
An example of antifertility drug is
(a) novestrol
(b) seldane
(c) salvarsan
(d) Chloramphenicol
Answer:
(a) novestrol

Part – II

Answer any six questions. Question No. 20 is compulsory. [6 x 2 = 12]

Question 16.
Name the method used for the refining of (i) Nickel (ii) Zirconium
Answer:
(i) Mond’s process
(ii) Van Arkel’s method

Question 17.
Complete the following reactions:
(a) B(OH)₃ + NH₃ →
(b) Na₂B₄O₇ + H₂SO₄ + H₂O
Answer:

Question 18.
KMnO₄ does not act as oxidising agent in the presence of HC1. Why?
Answer:
HCl cannot be used for making acidified KMnO₄ as oxidising agent, since it reacts with KMnO₄ as follows.
2MnO₄^– + 10 Cl^– + 16H⁺ → 2Mn²⁺ + 5Cl₂ + 8H₂O

Question 19.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled? ,
Answer:
(i) \(\frac{d x}{d t}=k[\mathrm{A}]^{1}[\mathrm{B}]^{2}\)
(ii) If concentration of ‘B ’ is tripled, then the rate will become 9 times.
(iii) When concentration of both A and B are doubled, then the rate will become 8 times.

Question 20.
Calculate the pH of 0.04 M HNO₃ Solution.
Answer:
Concentration of HNO₃ = 0.04M
[H₃O⁺] = 0.04 mol dm^-3
pH = -log[H₃O⁺]
= – log(0.04)
= – log(4 x 10^-2)
= 2 -log 4 = 2-0.6021
= 1.3979= 1.40

Question 21.
Write a note about Helmoholtz double layer.
Answer:
The surface of a colloidal particle adsorbs one type of ion due to preferential adsorption. This layer attracts the oppositely charged ions in the medium and hence the boundary separating the two electrical double layers are set up. This is called Helmholtz electrical double layer. As the particles nearby are having similar charges, they cannot come close and condense. Hence this helps to explain the stability of the colloid.

Question 22.
Why HCOOH does not give Hell-Volhard Zelinsky reaction but CH₃3COOH does?
Answer:
CH₃COOH contains α – hydrogen atom and hence gives HVZ reaction but HCOOH does not contain an α – hydrogen atom and hence does not give HVZ reaction.

Question 23.
What happens when
Oxidation of acetoneoxime with trifluoroperoxy acetic acid.
Answer:
Oxidation of acetoneoxime with trifluoroperoxy acetic acid gives 2-nitropropane.

Question 24.
What are food preservatives?
Answer:
Food preservatives are chemical substances are capable of inhibiting, retarding or arresting the process of fermentation, acidification or other decomposition of food by growth of micro organisms.
Examples:

• Acetic acid is used mainly as a preservative for preparation of pickles.
• Sodium meta sulphite is used as preservative for fresh vegetables and fruits.
• Sodium benzoate is used as preservative for juices.

Part – III

Answer any six questions. Question No. 31 is compulsory. [6 x 3 = 18]

Question 25.
Explain the principle of electrolytic refining with an example.
Answer:
The crude metal is refined by electrolysis. It is carried out in an electrolytic cell containing aqueous solution of the salts of the metal of interest. The rods of impure metal are used as anode and thin strips of pure metal are used as cathode. The metal of interest dissolves from the anode, pass into the solution while the same amount of metal ions from the solution will be deposited at the cathode. During electrolysis, the less electropositive impurities in the anode, settle down at the bottom and are removed as anode mud.
Let us understand this process by considering electrolytic refining of silver as an example.
Cathode: Pure silver
Anode: Impure silver rods

Electrolyte: Acidified aqueous solution of silver nitrate.
When a current is passed through the electrodes the following reactions will take place
Reaction at anode: 2Ag(s) → Ag⁺ (aq) + le^–
Reaction at cathode: Ag⁺(aq)+ le^– → Ag (s)

During electrolysis, at the anode the silver atoms lose electrons and enter the solution. The positively charged silver cations migrate towards the cathode and get discharged by gaining electrons and deposited on the cathode. Other metals such as copper, zinc etc.,can also be refined by this process in a similar manner.

Question 26.
How will you prepare phosphine and explain the purification of phosphine?
Phosphine’is prepared by action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide or hydrogen.

Phosphine is freed from phosphine dihydride (P2H4) by passing through a freezing mixture. The dihydride condenses while phosphine does not.
Phosphine can also prepared by the hydrolysis of metallic phosphides with water or dilute mineral acids.

Phosphine is prepared in pure form by heating phosphorous acid.

A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda S0lution.

Question 27.
Justify the position of lanthanides and actinides in the periodic table.
Answer:
(i) In sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these 14 elements following lanthanum show similar chemical properties. Therefore these elements are grouped together and placed at the bottom of the periodic table. This position can be justified as follows.
(a) Lanthanoids have general electronic configuration [Xe] 4f2 14 5d° 1 6s2
(b) The common oxidation state of lanthanoids is +3
(c) All these elements have similar physical and chemical properties.

(ii) Similarly the fourteen elements following actinium resemble in their physical and chemical properties.

(iii) If we place these elements after Lanthanum in the periodic table below 4d series and actinides below 5d series, the properties of the elements belongs to a group would be different and it would affect the proper structure of the periodic table.

(iv) Hence a separate position is provided to the inner transition elements at the bottom of the periodic table.

Question 28.
Distinguish between isotropy and anisotropy?
Answer:
Isotropy:

1. Isotropy means uniformity in all directions.
2. Isotropy means having identical values of physical properties such as refractive index, electrical conductance in all directions.
3. Isotropy is the property of amorphous solids.

Anisotropy :

1. Anisotropy means non-uniformity in all directions.
2. Anisotropy is the property which depends on the direction of measurement. They show different values of physical properties when measured along different directions.
3. Anisotropy is the properly of crystalline solids.

Question 29.
What are the merits and limitations of the intermediate compound theory?
Answer:
Merits:
(i) The specificity of a catalyst
(ii) The increase in the rate of the reaction with increase in the concentration of catalyst.

Limitations:
(i) This theory fails to explain the action of catalytic poison and promoters. .
(ii) This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 30.
Write a note on sacrificial protection.
Answer:
In this method, the metallic structure to be protected is made cathode by connecting it with more active metal (anodic metal). So that all the corrosion will concentrate only on the active metal. The artificially made anode thus gradually gets corroded protecting the original metallic structure. Hence this process is otherwise known as sacrificial anodic protection.
Al, Zn and Mg are used as sacrificial anodes.

Question 31.
1mole of Hlis allowed to react with t-butyl methylether. Identify the product and write down the mechanism of the reaction.
Answer:

Question 32.
How will you prove the presence of aldehyde group in glucose?
Answer:
(i) Glucose is oxidised to gluconic acid with ammoniacal silver nitrate (Tollen’s reagent) and alkaline copper sulphate (Fehling’s solution). Tollen’s reagent is reduced to metallic silver and Fehling’s solution to cuprous oxide (red precipitate). These reactions confirm the presence of an aldehye group in glucose.

Question 33.
Explain about anaesthetics with their types.
(i) Local anaesthetics: It causes loss of sensation in the area in which it is applied without losing consciousness. They block pain perception that is transmitted via peripheral nerve fibres to the brain.
Example: Procaine, Li do caine.
They are often used during minor surgical procedures.

(ii) General anaesthetics: They cause a controlled and reversible loss of consciousness by affecting central nervous system.
Example: Propofol, Isoflurane.
They are often used for major surgical procedures.

Part – IV

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

Question 34.
(a) (i) CO is more stable at higher temperature. Why? (2)
(ii) How will you prepare potash alum? (3)
[OR]
(b) (i) Give the balanced equation for the reaction between chlorine with cold NaOH. (3)
(ii) Nitrogen exists as diatomic molecule and Phosphrus as P₄ Why? (2)
Answer:
(a) (i) In the Ellingham diagram, the formation of carbon monoxide is a straight line with negative slope. In this case AS is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen. Hence, CO is more stable at higher temperature.

(ii) The alunite the alum stone is the naturally occurring form and it is K₂SO₄ . Al₂(S0₄)₃.4Al(0H)₃. When alum stone is treated with excess of sulphuric acid, the aluminium hydroxide is converted to aluminium sulphate. A calculated quality of – potassium sulphate is added and the solution is crystallised to generate potash alum. It is purified by recrystallisation.
(2)

(b)(i) Reaction between chlorine with cold NaoH

Chlorine reacts with cold NaOH to give sodium chloride and sodium hypochlorite.

(ii) Nitrogen has a triple bond between its two atoms because of its small size and high electronegativity. Phosphorus P₄ has single bond, that is why it is tetra-atomic.

Question 35.
(a) (i) Discuss the ortho and pyro silicates. (2)
(ii) Compare lanthanides and actinides. (3)
[OR]
(b) (i) Ni²⁺ is identified using alcoholic solution of dimethyl glyoxime. Write the structural formula for the rosy red precipitate of a complex formed in the reaction. (3)
Cu⁺, Zn²⁺, Sc³⁺, Ti⁴⁺ are colourless. Prove this statement. (2)
Answer:
Ortho silicates: The simplest silicates which contain discrete [SiO₄]^4- tetrahedral units are called ortho silicates or neso silicates.

Examples: Phenacite – Be₂SiO₄(Be²⁺ ions are tetrahedrally surrounded by O^2- ions), Olivine – (Fe/Mg)₂ SiO₄ (‘Fe²⁺ and Mg²⁺ cations are octahedrally surrounded by O

Pyro silicates: Silicates which contain [Si₂O₇]^6- ions are called pyro silicates (or) Soro silicates. They are formed by joining two [SiO₄]^4-

tetrahedral units by sharing one oxygen atom at one comer, (one oxygen is removed while joining). Example: Thortveitite – Sc₂Si₂O₇

(ii)
Lanthanoids

1. Differentiating electron enters in 4f orbital
2. Binding energy of 4f orbitals are higher
3. They show less tendency to form complexes
4. Most of the lanthanoids are colourless
5. They do not form oxo cations
6. Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases.

Actinoids

1. Differentiating electron enters in 5f orbital
2. Binding energy of 5f orbitals are lower
3. They show greater tendency to form complexes
4. Most of the actinoids are coloured.
5. E.g : U³⁺ (red), U⁴⁺ (green), UO₂²⁺ (yellow)
6. They do form oxo cations such as UO₂²⁺ NpO₂²⁺ etc.
7. Besides +3 oxidation states actinoids show higher oxidation states such as +4, +5, +6 and +7.

[OR]

(b) (i) Ni²⁺ ions present in Nickel chloride solution is estimated accurately for forming an insoluble complex called [Ni(DMG)₂]
Nickel ion reacts with alcoholic solution of DMG in the presence of ammonical medium, to give rosy red precipitate of [Ni(DMG)₂] complex.

(ii) 1. Cu⁺, Zn²⁺ have d10 configuration and Sc³⁺, Ti⁴⁺have d¹ configuration.
2. d-d transition is not possible in the above complexes. So they are colourless.

Question 36.
(a) i)0 What is meant by the term “coordination number”? What is the coordination number of atoms in a bcc structure? (2)
(ii) Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation
\(\log \mathbf{k}=\log \mathbf{A} \frac{\mathbf{E}_{\mathbf{a}}}{2.303 \mathbf{R}}\left(\frac{1}{\mathbf{T}}\right)\)

Where E_a is the activation energy. When a graph is plotted for logk Vs \(\frac{1}{T}\) a straight line
with a slope of- 4000K Is obtained. Calculate the activation energy. (3)

(OR)

(b) (i) Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of K_h for that reaction. (3)
(ii) Identify the Lewis acid and the Lewis base in the following reactions. (2)

Answer:
(a) (i) 1. Let us consider the reaction between sodium hydroxide and acetic acid to give sodium acetate and water

NaOH_(aq) + CH₃COOH_(aq) ⇌ CH₃COONa_(aq) + H₂O

2. Coordination number of atoms in a bcc structure is 8

(ii) logk = logA – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{\mathrm{T}}\right)\)
y = c + mx
m = \(-\frac{E_{a}}{2.303 R}\)
E_a =—2.303Rm
E_a = – 2.303 x 8.314 J K^-1 mol^-1 x (— 4000 K)
E_a = 76,589 J mol^-1
E_a =76589kJmol^-1

[OR]

(b) (i) 1. Coordination number: The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.
2. In aqueous solution, CH₃COONa is completely dissociated as follows.
CH₃COONa_(aq), →  CH₃COO^–_(aq) + Na⁺_(aq)

3. CH₃COO^– is a conjugate base of the weak acid CHCOOH and it has a tendency to react with H⁺ from water to produce unionised acid. But there is no such tendency
for Na⁺ to react with OH^–

4.

such cases, the solution is basic due to the hydrolysis and pH is greater than 7.

5. Relationship between equilibrium constant, hydrolysis constant and the dissociation constant of acid is derived as follows:

K_h value in terms of degree of hydrolysis (h) and the concentration of salt (c) for the equilibrium can be obtained as in the case of Ostwald’s dilution law K_h = h²C and \(\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{c}}\)

Question 37.
(a) (i) What are lyophilic and lyophobic sols? Give one example of each type. Why are
Answer:
(a) (i) Lyophilic Sols: Colloidal sols directly formed by mixing substances like gums, gelatin, starch, rubber, etc. with a suitable liquid (The dispersion medium) are lyophilic sols.
An important characteristic of these sols is that if the dispersion medium is separated from the dispersed phase (say by evaporation) the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols. These sols are quite stable and cannot be easily coagulated.

Lyophobic sols: These colloidal sols can only be prepared by some special methods. These sols are readily precipitated on the addition of small amount of electrolytes, by heating or by shaking and hence are not stable.
Hydrophobic sols are water hating. They are formed by indirect method. These sols are irreversible sols. These sols are readily precipitated by the addition of small amount of electrolytes, by heating or by shaking and hence are not stable.

(ii) 1. The substances when added to a catalysed reaction decreases or completely destroys the activity of a catalyst are often known as catalytic poisons.
2. In the reaction 2SO₂ + O₂ → 2SO₃ with Pt catalyst, the catalyst poison is AS₂O₃

[OR]

(b) (i) 1. Substances like AgCl, PbS04 are sparingly soluble in water. The solubility product can be determined using conductivity experiments.
2. Let us consider AgCl as an example
Agci_(s)) ⇌ Ag⁺ + Cl^–
K_Sp = [Ag ⁺] [Cl^–]

3. Let the concentration of [Ag⁺] be ‘C’ mol L^-1
If [Ag⁺] = C, then [Cl^–] is also equal to C mol L^-1.
.’. K_sp = C.C
K_sp = C²

4. The relationship between molar conductance and equivalent conductance is

(ii) 1. Formic acid contains both an aldehyde as well as an acid group. Hence, like other aldehydes, formic acid can easily be oxidised and therefore acts as a strong reducing agent.

2. Formic acid reduces Tollen’s reagent (ammonical silver nitrate solution) to metallic silver.

3. Formic acid reduces Fehling’s solution. It reduces blue coloured cupric ions to red coloured cuprous ions.

Question 38.
(a) (i) What are the uses of nitrobenzene? (2)
(ii) Write a note on formation of -helix. (3)
[OR]
(b) (i) Define TFM value. (2)
(ii) Differentiate thermoplastic and thermosetting. (3)
Answer:
(a) (i) 1. Nitro benzene is used to produce lubricating oils in motors and machinery.
2. It is used in the manufacture of dyes, drugs, pesticides, synthelic rubber, aniline and explosives like TNT, TNB.
(ii) 1. In the a-helix sub-structure, the aminoacids are arranged in a right handed helical (spiral) structure and are stabilised by the hydrogen bond between the carbonyl oxygen one aminoacid (nth residue) with amino hydrogen of the fifth residue (n+4th residue).

2. The side chains of the residues protrude outside of the helix. Each turn of an a-helix contains about 3.6 residues and is about 5.4 A long.

3. The amino acid pro line produces a line in the helical structure and often called as a helical breaker due to its rigid cyclic structure.

4. Many fibrous proteins such as a-Keratin in hair, nails, wool, skin and myosin in muscles have a-helix structure. Stretching property of human hair is due to the helical structure of a-keratin in hair.

5. Structure of α – helix

[OR]

(b) (i) 1. The quality of a soap is described in terms of total fatty matter (TFM value). It is defined as the total amount of fatty matter that can be separated from a sample after spliting with mineral acids. Higher the TFM value in the soap, better is its quality.
2. As per BIS standards, Grade I soaps should have 76% minimum TFM value (ii) Difference between thermoplastic and thermosetting:

Thermoplastic :

1. They soften on heating and harden on cooling, and they can be remoulded.
2. They consists of linear long chain polymers and low molecular weights polymers.
3. All the polymer chains are held together by weak Vanderwaals forces.
4. They are weak, soft and less brittle.
5. They are formed by addition polymerisation.
6. They are soluble in organic solvents.
7.  Example: PVC, polythene, polystrene etc.

Thermosetting

1. They do not soften on heating and they cannot be remoulded.
2. The consist of three dimensional network structure and high molecular weight polymers.
3. All the polymer chains are linked by strong covalent.
4. They are strong, hard and more brittle.
5. They are formed by condensation polymerisation.
6. They are insoluble in organic solvents.
7. Example: Bakelite, melamine etc.

Students can Download Tamil Nadu 12th Biology Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 4 English Medium

General Instructions:

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

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
A Plant called ‘X’ possesses small flower with reduced perianth and versatile anther. The probable agent for pollination would be _________.
(a) water
(b) air
(c) butterflies
(d) beetles
Answer:
(b) air

Question 2.
“Gametes are never hybrid”. This is a statement of _______.
(a) Law of dominance
(b) Law of independent assortment
(c) Law of segregation
(d) Law of random fertilization
Answer:
(c) Law of segregation

Question 3.
In which techniques Ethidium Bromide is used?
(a) Southern Blotting techniques
(b) Western Blotting techniques
(c) Polymerase Chain Reaction
(d) Agarose Gel Electrophoresis
Answer:
(d) Agarose Gel Electrophoresis

Question 4.
Which of the following statement is correct?
(a) Agar is not extracted from marine algae such as seaweeds
(b) Callus undergoes differentiation and produces somatic embryoids
(c) Surface sterilization of explants is done by using mercuric bromide
(d) pH of the culture medium is 5.0 to 6.0
Answer:
(d) pH of the culture medium is 5.0 to 6.0

Question 5.
The term pedogenesis is related to _______.
(a) Fossils
(b) Water
(c) Population
(d) Soil
Answer:
(d) Soil

Question 6.
Depletion of which gas in the atmosphere can lead to an increased incidence of skin cancer?
(a) Ammonia
(b) Methane
(c) Nitrous oxide
(d) Ozone
Answer:
(d) Ozone

Question 7.
A wheat variety, Atlas 66 which has been used as a donor for improving cultivated wheat is rich in _______.
(a) iron
(b) carbohydrates
(c) proteins
(d) vitamins
Answer:
(c) proteins

Question 8.
Statement A: Coffee contains caffeine.
Statement B: Drinking coffee enhances cancer.
(a) A is correct, B is wrong
(b) A and B – Both are correct
(c) A is wrong, B is correct
(d) A and B – Both are wrong
Answer:
(b) A and B – Both are correct

Part – II

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

Question 9.
Draw and label the structure of a typical pollen grain.
Answer:

Question 10.
What is test cross? Why it is done?
Answer:
Test cross is crossing an individual of unknown genotype with a homozygous recessive. Test cross is used to identify whether an individual is homozygous or heterozygous for dominant character.

Question 11.
You are working in a biotechnology lab with a bacterium namely E.coli. How will you cut the nucleotide sequence? Explain it.
Answer:
The DNA nucleotide sequence can be cut using Restriction endonucleases (RE). Restriction

Question 12.
What is ecological hierarchy? Name the levels of ecological hierarchy.
Answer:
The interaction of organisms with their environment results in the establishment of grouping of organisms which is called ecological hierarchy.

Question 13.
Mutagens are the substances that induces mutation. Name any two physical and chemical mutagens.
Answer:
UV short waves, X-rays – Physical mutagens.
Nitromethyl, Urea – Chemical mutagens.

Question 14.
What is pseudo cereal? Give an example.
Answer:
The term pseudo-cereal is used to describe foods that are prepared and eaten as a whole grain, but are botanical outliers from grasses. Example: quinoa. It is actually a seed from the Chenopodium quinoa plant, belongs to the family Amaranthaceae.

Part – III

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

Question 15.
“Endothecium is associated with dehiscence of anther” Justify the statement.
Answer:
The inner tangential wall develops bands (sometimes radial walls also) of α cellulose (sometimes also slightly lignified). The cells are hygroscopic. The cells along the junction of the two sporangia of an anther lobe lack these thickenings. This region is called stomium. This region along with the hygroscopic nature of endothecium helps in the dehiscence of anther at maturity.

Question 16.
What is gene mapping? Write its uses.
Answer:
The diagrammatic representation of position of genes and related distances between the adjacent genes is called genetic mapping. It is directly proportional to the frequency of recombination between them. It is also called as linkage map.

Uses of genetic mapping:

• It is used to determine gene order, identify the locus of a gene and calculate the distances between genes.
• They are useful in predicting results of dihybrid and trihybrid crosses.
• It allows the geneticists to understand the overall genetic complexity of particular organism.

Question 17.
How synthetic seeds are developed?
Answer:
Artificial seeds or synthetic seeds (synseeds) are produced by using embryoids (somatic embryos) obtained through in vitro culture. They may even be derived from single cells from any part of the plant that later divide to form cell mass containing dense cytoplasm, large nucleus, starch grains, proteins, and oils, etc. To prepare the artificial seeds different inert materials are used for coating the somatic embryoids like agrose and sodium alginate.

Question 18.
Discuss the three zones of a lentic ecosystem.
Answer:
There are three zones, littoral, limnetic and profundal. The littoral zone, which is closest to the shore with shallow water region, allows easy penetration of light. It is warm and occupied by rooted plant species. The limnetic zone refers the open water of the pond with an effective penetration of light and domination of planktons.

The deeper region of a pond below the limnetic zone is called profundal zone with no effective light penetration and predominance of heterotrophs. The bottom zone of a pond is termed benthic and is occupied by a community of organisms called benthos (usually decomposers).

Question 19.
Write a short note on clean development mechanism.
Answer:
Clean Development Mechanism (CDM) is defined in the Kyoto protocol (2007) which provides project based mechanisms with two objectives to prevent dangerous climate change and to reduce green house gas emissions. CDM projects helps the countries to reduce or limit emission and stimulate sustainable development.

An example for CDM project activity, is replacement of conventional electrification projects with solar panels or other energy efficient boilers. Such projects can earn Certified Emission Reduction (CER) with credits / scores, each equivalent to one tonne of CO₂, which can be counted towards meeting Kyoto targets.

Part – IV

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

Question 20.
(a) Describe dominant epistasis with an example.
Answer:

Dominant Epistasis – It is a gene interaction in which two alleles of a gene at one locus interfere and suppress or mask the phenotypic expression of a different pair of alleles of another gene at another locus. The gene that suppresses or masks the phenotypic expression of a gene at another locus is known as epistatic. The gene whose expression is interfered by non-allelic genes and prevents from exhibiting its character is known as hypostatic. When both the genes are present together, the phenotype is determined by the epistatic gene and not by the hypostatic gene.

In the summer squash the fruit colour locus has a dominant allele ‘W’ for white colour and a recessive allele ‘w’ for coloured fruit. ‘W’ allele is dominant that masks the expression of any colour. In another locus hypostatic allele ‘G’ is for yellow fruit and its recessive allele ‘g’ for green fruit. In the first locus the white is dominant to colour where as in the second locus yellow is dominant to green. When the white fruit with genotype WWgg is crossed with yellow fruit with genotype wwGG, the F₁ plants have white fruit and are heterozygous (WwGg). When F₁ heterozygous plants are crossed they give rise to F₂ with the phenotypic ratio of 12 white : 3 yellow : 1 green.

Since W is epistatic to the alleles ‘G’ and ‘g’ the white which is dominant, masks the effect of yellow or green. Homozygous recessive ww genotypes only can give the coloured fruits (4/16). Double recessive ‘wwgg’ will give green fruit (1/16). The Plants having only ‘G’ in its genotype (wwGg or wwGG) will give the yellow fruit(3/16).

[OR]

(b) Point out the significance of plant succession.
Answer:
Significance of Plant Succession:

• Succession is a dynamic process. Hence an ecologist can access and study the seral stages of a plant community found in a particular area.
• The knowledge of ecological succession helps to understand the controlled growth of one or more species in a forest.
• Utilizing the knowledge of succession, even dams can be protected by preventing siltation.
• It gives information about the techniques to be used during reforestation and afforestation.
• It helps in the maintenance of pastures.
• Plant succession helps to maintain species diversity in an ecosystem.
• Patterns of diversity during succession are influenced by resource availability and disturbance by various factors.
• Primary succession involves the colonization of habitat of an area devoid of life.
• Secondary succession involves the re-establishment of a plant community in disturbed area or habitat.
• Forests and vegetation that we come across all over the world are the result of plant succession.

Question 21.
(a) Compare the various types of Blotting techniques.
Answer:

[OR]

(b) Explain different types of hybridization.
Answer:
Types of Hybridization:
According to the relationship between plants, the hybridization is divided into.
1. Intravarietal hybridization – The cross between the plants of same variety. Such crosses are Useful only in the self-pollinated crops.

2.. Intervarietal hybridization – The cross between the plants belonging to two different varieties of the same species and is also known as intraspecific hybridization. This technique has been the basis of improving self-pollinated as well as cross pollinated crops.

3. Interspecific hybridization – The cross between the plants belonging to different species belonging to the same genus is also called intragenic hybridization. It is commonly used for transferring the genes of disease, insect, pest and drought resistance from one species to another.
Example: Gossypium hirsutum x Gossypium arboreum – Deviraj.

4. Intergeneric hybridization – The crosses are made between the plants belonging to two different genera. The disadvantages are hybrid sterility, time consuming and expensive procedure. Example: Raphanobrassica and Triticale.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Select the correct production site and action site of Relaxin.
(a) Hypothalamus and pituitary gland
(b) Pituitary gland and Pelvic joints and cervix
(c) Placenta and pelvic joint and cervix
(d) Hypothalamus and placenta
Answer:
(c) Placenta and pelvic joint and cervix

Question 2.
Fusion of young individuals produced immediately after the mitotic division of adult parent cell is called _______.
(a) Merogamy
(b) Anisogamy
(c) Hologamy
(d) Paedogamy
Answer:
(d) Paedogamy

Question 3.
Mangolism is a genetic disorder which is caused by the presence of an extra chromosome number ______.
(a) 20
(b) 21
(c) 23
(d) 19
Answer:
(b) 21

Question 4.
DNA finger printing techniques was developed by _______.
(a) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
(d) Hershey and Chase
Answer:
(b) Alec Jeffreys

Question 5.
Identify the correct sequence of periods from oldest to youngest
(a) Cambrian → Permian → Devonian → Silurian → Ordovician
(b) Permian → Silurian → Devonian → Ordovician → Cambrian
(c) Permian → Devonian → Silurian → Cambrian → Ordovician
(d) Cambrian → Ordovician → Silurian → Devonian → Permian
Answer:
(d) Cambrian → Ordovician → Silurian → Devonian → Permian

Question 6.
Spread of cancerous cells to distant sites is termed as _________.
(a) Metastasis
(b) Oncogenes
(c) Proto-oncogenes
(d) Malignant neoplasm
Answer:
(a) Metastasis

Question 7.
Assertion (A): Streptomycin is an antibiotic.
Reason (R): Antibiotic are microbial chemicals inhibits the growth of pathogenic microbe.
(a) A is right R is wrong
(b) R explains A
(c) A and R are wrong
(d) A is wrong but R is right
Answer:
(b) R explains A

Question 8.
World Ozone Day was observed on _________.
(a) September 16^th
(b) October 12^th
(c) December 1^th
(d) August 18^th
Answer:
(a) September 16^th

Part – II

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

Question 9.
Name the active chemical found in the medicinal plant Rauwolfia vomitoria. What type of diversity does it belongs to?
Answer:
Rauwolfia vomitoria can be cited as an example for genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria.

Question 10.
State any two unique features of ELISA test.
Answer:
ELISA is highly sensitive and can detect antigen even in nanograms.
ELISA test does not require radioisotopes or radiation counting apparatus.

Question 11.
Define Anaphylaxis.
Answer:
Anaphylaxis is the classical immediate hypersensitivity reaction. It is a sudden, systematic, severe and immediate hypersensitivity reaction occurring as a result of rapid generalized mast-cell degranulation

Question 12.
Who is Cro-Magnon?
Answer:
Cro-Magnon was one of the most talked forms of modem human found from the rocks of Cro-Magnon, France and is considered as the ancestor of modem Europeans. They were not only adapted to various environmental conditions, but were also known for their cave paintings, figures on floors and walls.

Question 13.
What is S – D sequence?
Answer:
The 5′ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgamo sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 14.
Expand (a) GIFT (b) ICSI
Answer:
GIFT – Gamete Intra – Fallopian Transfer
ICSI – Intra-cytoplasmic sperm injection

Part – III

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

Question 15.
Write a short note on encystment in amoeba.
Answer:
During unfavorable conditions (increase or decrease in temperature and scarcity of food) Amoeba withdraws its pseudopodia and secretes a three-layered, protective, chitinous cyst wall around it and becomes inactive. This phenomenon is called encystment. When conditions become favourable, the encysted Amoeba divides by multiple fission and produces many minute amoebae called pseudopodiospore or amoebulae.

The cyst wall absorbs water and breaks off liberating the young pseudopodiospores, each with a fine pseudopodia. They feed and grow rapidly to lead an independent life.

Question 16.
Draw a schematic representation of human oogenesis.
Answer:

Question 17.
Comment on the methods of Eugenics.
Answer:
Eugenics refers to the study of the possibility of improving the qualities of human population. Methods of Eugenics:

• Sex-education in school and public forums.
• Promoting the uses of contraception.
• Compulsory sterilization for mentally retarded and criminals.
• Egg donation.
• Artificial insemination by donors.
• Prenatal diagnosis of genetic disorders and performing MTP.
• Gene therapy.
• Cloning.
• Egg/sperm donation of healthy individuals.

Question 18.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.
1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.

2. If two RNA molecules were produced simultaneously, double stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Question 19.
What is Q₁₀ value? How it is calculated?
Answer:
The effect of temperature on the rate of reaction is expressed in terms of temperature coefficient or Q₁₀ value. The Q₁₀ values are estimated taking the ratio between the rate of reaction at X°C and rate of reaction at (X-10°C).

Part – IV

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

Question 20.
(a) Differentiate between r-selected species and k-selected species.
Answer:
r – selected species:

• Smaller sized organisms
• Produce many offspring
• Mature early
• Short life expectancy
• Each individual reproduces only once or few times in their life time
• Only few reach adulthood
• Unstable environment, density independent

k – seleced species:

• Larger sized organisms
• Produce few offspring
• Late maturity with extended parental care
• Long life expectancy
• Can reproduce more than once in lifetime
• Most individuals reach maximum life span
• Stable environment, density dependent

[OR]

(b) Give a detailed account on ethanol production by microbes and the uses of ethanol.
Answer:
Ethanol production:
Saccharomyces cerevisiae (Yeast) is the major product of ethanol.

Since ethanol is used for industrial, laboratory and fuel proposes, it is called as industrial alcohol.

Organism used: Saccharomyces cerevisiae, bacteria like Zymomonas mobilis and Sarcina ventriculi.
Substances used: Molasses, Com, Potatoes, wood waste.

Process of ethanol production:
Step-1: Milling of fees stock.
Step-2: Adding fungal (Aspergillus) amylase to break down starch into sugar.
Step-3: Yeast is added to convert sugar into ethanol.
Step-4: Distillation yield 96% concentrated ethanol.

Uses of Ethanol:
Ethanol and bio-diesel are the two commonly used first generation bio-fuels.
Ethanol is used as fuel, mainly as bio-fuel additive for gasoline.

Question 21.
(a) How DNA is packed in an eukaryotic cell?
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (H1) that is exposed to enzymes. The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an H1 molecule. Chromatin lacking H1 has a beads-on-a-string appearance in which DNA enters and leaves the nucleosomes at random places. H1 of one nucleosome can interact with H1 of the neighbouring nucleosomes resulting in the further folding of the fibre.

The chromatin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucleosome, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different H1 molecules. DNA is a solenoid and packed about 40 folds. The hierarchical nature of chromosome structure is illustrated.

Additional set of proteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In a typical nucleus, some regions of chromatin are loosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

[OR]

(b) Explain Oparin – Haldane hypothesis on evolution.
Answer:
According to the theory of chemical evolution primitive organisms in the primordial environment of the Earth evolved spontaneously from inorganic substances and physical forces such as lightning, UV radiations, volcanic activities, etc. Oparin (1924) suggested that the organic compounds could have undergone a series of reactions leading to more complex molecules. He proposed that the molecules formed colloidal aggregates or ‘coacervates’ in an aqueous environment.

The coacervates were able to absorb and assimilate organic compounds from the environment. Haldane (1929) proposed that the primordial sea served as a vast chemical laboratory powered by solar energy. The atmosphere was oxygen free and the combination of CO₂, NH₃ and UV radiations gave rise to organic compounds. The sea became a ‘hot’ dilute soup containing large populations of organic monomers and polymers.

They envisaged that groups of monomers and polymers acquired lipid membranes and further developed into the first living cell. Haldane coined the term prebiotic soup and this became the powerful symbol of the Oparin-Haldane view on the origin of life (1924-1929). Oparin and Haldane independently suggested that if the primitive ‘ atmosphere was reducing and if there was appropriate supply of energy such as lightning or UV light then a wide range of organic compounds can be synthesized.

Students can Download Tamil Nadu 12th Biology Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 3 English Medium

General Instructions:

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

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Identity the mismatched pair regarding the anther walls.
(a) Epidermal layer – Protective in function
(b) Endothecium layer – Helps in dehiscence of anther
(c) Middle layer – Persistent layer
(d) Tapetum – Nutritive in function
Answer:
(c) Middle layer – Persistent layer

Question 2.
Extra nuclear inheritance is a consequence of presence of genes in _______.
(a) Mitochondria and chloroplasts
(b) Endoplasmic reticulum and mitochondria
(c) Ribosomes and chloroplast
(d) Lysosomes and ribosomes
Answer:
(b) Endoplasmic reticulum and mitochondria

Question 3.
How many map units separate two alleles A and B, if the recombination frequency is 0.09?
(a) 900 cM
(b) 90 cM
(c) 9 cM
(d) 0.9 cM
Answer:
(c) 9 cM

Question 4.
Plasmids are _______.
(a) circular protein molecules
(b) required by bacteria
(c) tiny bacteria
(d) confer resistance to antibiotics
Answer:
(d) confer resistance to antibiotics

Question 5.
Solar energy used by green plants for photosynthesis is only ________.
(a) 2 – 8%
(b) 2 – 10%
(c) 3 – 10%
(d) 2- 9%
Answer:
(b) 2 – 10%

Question 6.
One of the chief reasons among the following for the depletion in the number of species making endangered is _____.
(a) over hunting and poaching
(b) green house effect
(c) competition and predation
(d) habitat destruction
Answer:
(d) habitat destruction

Question 7.
Assertion (A): Genetic variation provides the raw material for selection.
Reason (R): Genetic variations are differences in genotypes of the individuals.
(a) (A) is right and (R) is wrong
(b) (A) is wrong and (R) is right
(c) Both (A) and (R) are right
(d) Both (A) and (R) are wrong
Answer:
(c) Both (A) and (R) are right

Question 8.
Groundnut is native of ______.
(a) Philippines
(b) India
(c) North America
(d) Brazil
Answer:
(d) Brazil

Part – II

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

Question 9.
Give the phenotypic ratio of
(a) Dihybrid cross
(b) Dihybrid test cross
Answer:
(a) Dihybrid cross ratio = 9 : 3 : 3 : 1 ,
(b) Dihybrid test cross ratio = 1 : 1 : 1 : 1

Question 10.
What are the materials used to grow microorganism like Spirulina.
Answer:
Spirulina can be grown easily on materials like waste water from potato processing plants (containing starch), straw, molasses, animal manure and even sewage, to produce large quantities.

Question 11.
Compare Redifferentiation with Dedifferentiation.
Answer:
Redifferentiation:
A process by which an already differentiated cell undergo further differentiation to form another type of cell.

Dedifferentiation:
A process of reversion of cells (differentiated cells) to meristematic cells leading to formation of callus.

Question 12.
Loamy soil is ideal for crop cultivation – Justify.
Answer:
Loamy soil is ideal soil for cultivation, since it consists of 70% sand and 30% clay or silt or both. It ensures good retention and proper drainage of water. The porosity of soil provides adequate aeration and allows the penetration of roots.

Question 13.
Mention any four environmental benefits of Rain Water Harvesting.
Answer:

• Promotes adequacy of underground water and water conservation.
• Mitigates the effect of drought.
• Improves groundwater quality and water table / decreases salinity.
• Reduces soil erosion as surface run-off water is reduced.

Question 14.
If a person drinks a cup of coffee daily it will help him for his health. Is this correct? If it is correct, list out the benefits.
Answer:
Yes, drinking coffee in moderation enhances the health of a person. Caffeine enhances release of acetylcholine in brain, which in turn enhances efficiency. It can lower the incidence of fatty liver diseases, cirrhosis and cancer. It may reduce the risk of type 2 diabetes.

Part – III

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

Question 15.
How the flowers of salvia are adopted for mellitophily?
Answer:
Pollination in Salvia (Lever mechanism): The flower of Salvia is adapted for Bee pollination. The flower is protandrous and the corolla is bilabiate with 2 stamens. A lever mechanism helps in pollination. Each anther has an upper fertile lobe and lower sterile lobe which is separated by a long connective which helps the anthers to swing freely. When a bee visits a flower, it sits on the lower lip which acts as a platform.

It enters the flower to suck the nectar by pushing its head into the corolla. During the entry of the bee into the flower the body strikes against the sterile end of the connective. This makes the fertile part of the stamen to descend and strike at the back of the bee. The pollen gets deposited on the back of the bee. When it visits another flower, the pollen gets rubbed against the stigma and completes the act of pollination in Salvia.

Question 16.
What is the difference between mis-sense mutation and non-sense mutation?
Answer:
Mis-sense Mutation:
The mutation where the codon for one amino acid is changed into a codon for another amino acid is called Missense or non-synonymous mutations.

Non-sense Mutation:
The mutations where codon for one amino acid is changed into a termination or stop codon is called Nonsense mutation.

Question 17.
Give an account on germplasm conservation. .
Answer:
Germplasm conservation refers to the conservation of living genetic resources like pollen, seeds or tissue of plant material maintained for the purpose of selective plant breeding, preservation in live condition and used for many research works.

Germplasm conservation resources is a part of collection of seeds and pollen that are stored in seed or pollen banks, so as to maintain their viability and fertility for any later use such as hybridization and crop improvement. Germplasm conservation may also involve a gene bank and DNA bank of elite breeding lines of plant resources for the maintenance of biological diversity and also for food security.

Question 18.
How are microbial innoculants used to increase the soil fertility?
Answer:
Biofertilizers or microbial innoculants are defined as preparations containing living cells or latent cells of efficient strains of microorganisms that help crop plants uptake of nutrients by their interactions in the rhizosphere when applied through seed or soil. They are efficient in fixing nitrogen, solubilising phosphate and decomposing cellulose.

They are designed to improve the soil fertility, plant growth, and also the number and biological activity of beneficial microorganisms in the soil. They are ecofriendly organic agro inputs and are more efficient and cost effective than chemical fertilizers. .

Question 19.
Shape of pyramid in a particular ecosystem is always different in shape. Explain with example.
Answer:
In a forest ecosystem the pyramid of number is spindle in shape, it is because the base (T₁) of the pyramid occupies large sized trees (Producer) which are lesser in number. Herbivores (T₂) (Fruit eating birds, elephant and deer) occupying second trophic level, are more in number than the producers. In final trophic level (T₄), tertiary consumers (lion) are lesser in number than the secondary consumer (T₃) (fox and snake).

Part – IV

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

Question 20.
(a) Find out the molecular explanation for the wrinkled pea seeds used by Mendel.
Answer:
The protein called starch branching enzyme (SBEI) is encoded by the wild-type allele of the gene (RR) which is dominant. When the seed matures, this enzyme SBEI catalyzes the formation of highly branched starch molecules. Normal gene (R) has become interrupted by the insertion of extra piece of DNA (0.8 kb) into the gene, resulting in r allele. In the homozygous mutant form of the gene (rr) which is recessive, the activity of the enzyme SBEI is lost resulting in wrinkled peas. The wrinkled seed accumulates more sucrose and high water content.

Hence the osmotic pressure inside the seed rises. As a result, the seed absorbs more water and when it matures it loses water as it dries. So it becomes wrinkled at maturation. When the seed has at least one copy of normal dominant gene heterozygous, the dominant allele helps to synthesize starch, amylopectin an insoluble carbohydrate, with the osmotic balance which minimises the loss of water resulting in smooth structured round seed.

[OR]

(b) Write in detail about Remote sensing and its uses.
Answer:
Remote Sensing is the process of detecting and monitoring the physical characteristics of an area by measuring its reflected and emitted radiation at a distance from the targeted area. It is an tool used in conservation practices by giving exact picture and data on identification of even a single tree to large area of vegetation and wild life for classification of land use patterns and studies, identification of biodiversity rich or less areas for futuristic works on conservation and maintenance of various species including commercial crop, medicinal plants and threatened plants. .

Specific uses:

• Helps predicting favourable climate, for the study of spreading of disease and controlling it.
• Mapping of forest fire and species distribution.
• Tracking the patterns of urban area development and the changes in Farmland or forests over several years.
• Mapping ocean bottom and its resources.

Question 21.
(a) Explain various edaphic factors that affect vegetation.
Answer:
The important edaphic factors which affect vegetation are as follows:

1. Soil moisture: Plants absorbs rain water and moisture directly from the air.
2. Soil water: Soil water is more important than any other ecological factors affecting the distribution of plants. Rain is the main source of soil water. Capillary water held between pore spaces of soil particles and angles between them is the most important form of water available to the plants.
3. Soil reactions: Soil may be acidic or alkaline or neutral in their reaction. pH value of the soil solution determines the availability of plant nutrients. The best pH range of the soil for cultivation of crop plants is 5.5 to 6.8.
4. Soil nutrients: Soil fertility and productivity is the ability of soil to provide all essential plant nutrients such as minerals and organic nutrients in the form of ions.
5. Soil temperature: Soil temperature of an area plays an important role in determining the geographical distribution of plants. Low temperature reduces use of water and solute absorption by roots.
6. Soil atmosphere: The spaces left between soil particles are called pore spaces which contains oxygen and carbon-di-oxide.
7. Soil organisms: Many organisms existing in the soil like bacteria, fungi, algae, protozoans, nematodes, insects and earthworms, etc., are called soil organisms.

[OR]

(b) Describe the procedure involved in Blue-White colony selection methods.
Answer:
Blue- White Colony Selection Method is a powerful method used for screening of recombinant plasmid. In this method, a reporter gene lacZ is inserted in the vector. The lacZ encodes the enzyme β-galactosidase and contains several recognition sites for restriction enzyme.

β-galactosidase breaks a synthetic substrates called X-gal (5-bromo-4-chloroindolyl- β-D- galacto-pyranoside) into an insoluble blue coloured product. If a foreign gene is inserted into lacZ, this gene will be inactivated. Therefore, no-blue colour will develop (white) because p-galactosidase is not synthesized due to inactivation of lacZ.

Therefore, the host cell containing r-DNA form white coloured colonies on the medium contain X-gal, whereas the other cells containing non-recombinant DNA will develop the blue coloured colonies. On the basis of colony colour, the recombinants can be selected.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Identify the proper sequence.
(a) juvenile phase, senescent phase, vegetative phase
(b) juvenile phase, maturity phase, senescent phase
(c) vegetative phase, maturity phase, juvenile phase
(d) senescent phase, juvenile phase, vegetative phase
Answer:
(b) juvenile phase, maturity phase, senescent phase

Question 2.
Which of the following symbol is used in pedigree analysis to represent unspecified sex?

Answer:

Question 3.
Darwin’s finches are an excellent example of _______.
(a) connecting links
(b) seasonal migration
(c) adaptive radiation
(d) geographical isolation
Answer:
(c) adaptive radiation

Question 4.
A 30 year old woman has bleedy diarrhoea for the past 14 hours, which one of the following organisms is likely to cause this illness?
(a) Streptococcus pyogenes
(b) Clostridium difficile
(c) Shigella dysenteriae
(d) Salmonella enteritidis
Answer:
(c) Shigella dysenteriae

Question 5.
Which of the following microorganism is used for the production of citric acid in industries?
(a) Lactobacillus bulgaris
(b) Penicillium citrinum
(c) Aspergillus niger
(d) Rhizopus nigricans
Answer:
(c) Aspergillus niger

Question 6.
The interaction in nature, where one gets benefit on the expense of other is __________.
(a) Predation
(b) Mutualism
(c) Amensalism
(d) Commensalism
Answer:
(d) Commensalism

Question 7.
Which of the following region has maximum bio-diversity?
(a) Taiga
(b) Tropical forest
(c) Temperate rain forest
(d) Mangroves
Answer:
(b) Tropical forest

Question 8.
The thickness of stratospheric ozone layer is measured in ____________.
(a) Sieverts units
(b) Melson units
(c) Dobson units
(d) Beaufort Scale
Answer:
(c) Dobson units

Part – II

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

Question 9.
What is parthenogenesis? Give two examples from animals.
Answer:
Development of an egg into a complete individual without fertilization is known as parthenogenesis. It was first discovered by Charles Bonnet in 1745.
E.g. Honey bees, Aphis.

Question 10.
Mention the production site and action site of following hormones.
(a) GnRH (b) Relaxin
Answer:

Hormone	Production Site	Action Site
GnRH	Hypothalamus	Pituitary gland
Relaxin	Placenta	Pelvic joints and cervix

Question 11.
Differentiate foeticide and infanticide.
Answer:
Female foeticide refers to ‘aborting the female in the mother’s womb’.
Female infanticide is ‘killing the female child after her birth’.

Question 12.
State Van’t Hoff’s rule.
Answer:
Van’t Hoff’s rule states that with the increase of every 10°C, the rate of metabolic activity is doubled or the reaction rate is halved with the decrease of 10°C.

Question 13.
Name the active chemical found in the medicinal plant Rauwolfia vomitoria. What type of diversity does it belongs to?
Answer:
Rauwolfia vomitoria can be cited as an example for genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria.

Question 14.
List any four adverse effect of noise.
Answer:

1. High blood pressure
2. Stress related ailments
3. Sleep disruption
4. Hearing impairment

Part – III

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

Question 15.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosome in females so that both males and females have only one functional X chromosome per cell.

Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Ban- body.

Question 16.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional activity of the structural gene.

• The structural gene codes for proteins, rRNA and tRNA required by the cell.
• Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
• The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 17.
Explain the principles of Lamarckian theory.
Answer:

• The theory of use and disuse – Organs that are used often will increase in size and those that are not used will degenerate. Neck in giraffe is an example of use and absence of limbs in snakes is an example for disuse theory.
• The theory of inheritance of acquired characters – Characters that are developed during the life time of an organism are called acquired characters and these are then inherited.

Question 18.
List the causative agent, mode of transmission and symptoms for Diphtheria and Typhoid.
Answer:

Question 19.
ELISA is a technique based on the principles of antigen-antibody reactions. Can this technique be used in the molecular diagnosis of a genetic disorder such as Phenylketonuria?
Answer:
Yes, ELISA test can be done to diagnose phenylketonuria. The affected person does not produce the enzyme phenylalanine hydroxylase. If specific antibodies are developed against the enzyme and ELISA is performed, the unaffected person will show positive result due to antigen and antibody reaction, whereas the affected individual produces negative result.

[Note: phenylketonuria is an inherited metabolic disorder that causes the accumulation of Phenylalanine (an amino acid) in body cells due to defect in the synthesizing of an enzyme phenylalanine hydroxylase]

Part – IV

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

Question 20.
(a) The following is the illustration of the sequence of ovarian events (a-i) in a human female.

(a) Identify the figure that illustrates ovulation and mention the stage of oogenesis it represents.
(b) Name the ovarian hormone and the pituitary hormone that have caused the above- mentioned events.
(c) Explain the changes that occurs in the uterus simultaneously in anticipation.
(d) Write the difference between C and H.
Answer:
(a) A- Primordial follicle; B- Primary follicle; C- Secondary follicle; D-Tertiary follicle; E- Mature graafian follicle; F- Ovulation (release of egg); G- Empty Graafian follicle; H- Corpus luteum; I – Corpus albicans.

(b) Pituitary hormones: Follicle Stimulating Hormones (FSH) and Lutenizing Hormone (LH). Ovarian hormones: Estrogen and Progesterone.

(c) At the start of menstrual cycle, the endometrium of uterus starts regenerating through proliferation of cells induced by FSH and CH. After ovulation, the progesterone secreted by corpus luteum prepares the endometrium (uterine wall) to receive the egg if it is fertilized.

(d) C- Secondary follicle
H – Corpus luteum
During development of ovum, the primary follicle gets surrounded by many layers of granular cells and forms a new layer called secondary follicle.

Corpus luteum is the empty graafian follicle that remains after ovulation. It acts as a transitory endocrine gland secreting progesterone to maintain pregnancy.

[OR]

(b) Write short notes on the following.
(i) Brewer’s yeast
(ii) Ideonella sakaiensis
(iii) Microbial fuel cells
Answer:
(i) Brewer’s yeast – Saccharomyces cerevisiae is a widely used fungal species in preparation and softening of bakery products like dough.

(ii) Ideonella sakaiensis is a bacterium is used to recycle PET plastics. The enzyme PETase and MHETase in the bacterium breakdown the PET plastics into terephthalic acid and ethylene glycol.

(iii) A microbial fuel cell is a bio-electrochemical system that drives an electric current by using bacteria and mimicking bacterial interaction found in nature. Microbial fuel cells work by allowing bacteria to oxidize and reduce organic molecules. Bacterial respiration is basically one big redox reaction in which electrons are being moved around.

A MFC consists of an anode and a cathode separated by a proton exchange membrane. Microbes at the anode oxidize the organic fuel generating protons which pass through the membrane to the cathode and the electrons pass through the anode to the external circuit to generate current.

Question 21.
(a) Write the salient features of Human Genome Project.
Answer:

• Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
• An average gene consists of3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
• The function of 50% of the genome is derived from transposable elements such as LINE and ALU sequence.
• Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosome 13 and Y chromosome have lowest gene densities.
• The chromosomal organization of human genes shows diversity. .
• There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
• Functions for over 50 percent of the discovered genes are unknown.
• Less than 2 percent of the genome codes for proteins.
• Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
• Chromosome 1 has 2968 genes, whereas chromosome ‘Y’ has 231 genes.
• Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotide polymorphism – pronounce as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

[OR]

(b) Tropical regions are rich in biodiversity. Why?
Answer:
The reasons for the richness of biodiversity in the Tropics are:

• Warm tropical regions between the tropic of Cancer and Capricorn on either side of equator possess congenial habitats for living organisms.
• Environmental conditions of the tropics are favourable not only for speciation but also for supporting both variety and number of organisms.
• The temperatures vary between 25°C to 35°C, a range in which most metabolic activities of living organisms occur with ease and efficiency.
• The average rainfall is often more than 200 mm per year.
• Climate, seasons, temperature, humidity, photo periods are more or less stable and
  encourage both variety and numbers. .
• Rich resource and nutrient availability.

Students can Download Tamil Nadu 12th Biology Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 2 English Medium

General Instructions:

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

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
How many different kinds of gametes will be produced by a plant having the genotype AABbCC?
(a) Three
(b) Four
(c) Nine
(d) Two
Answer:
(d) Two

Question 2.
pBR 322, BR stands for ________.
(a) Plasmid Bacterial Recombination
(b) Plasmid Bacterial Replication
(c) Plasmid Boliver and Rodriguez
(d) Plasmid Baltimore and Rodriguez
Answer:
(c) Plasmid Boliver and Rodriguez

Question 3.
Which of the given plant produces cardiac glycosides?
(a) Calotropis
(b) Acacia
(c) Nepenthes
(d) Utricularia
Answer:
(a) Calotropis

Question 4.
Methane is _______ times as effective as CO₂ at trapping heat.
(a) 5
(b) 10
(c) 20
(d) 100
Answer:
(c) 20

Question 5.
Statement 1: Arachis hypogea belongs to Fabaceae
Statement 2: It is a native of Brazil.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the Statements are correct
(d) Both the Statements are incorrect
Answer:
(c) Both the Statements are correct

Question 6.
Which of the following scientist developed world’s first cotton hybrid?
(a) Dr. B.P. Pal
(b) C.T. Patel
(c) Dr. K. Ramiah
(d) N.G.P. Rao
Answer:
(b) C.T. Patel

Question 7.
Ecosystem is the structural and functional unit of ecology. This statement was given by _______.
(a) Tansley
(b) Odum
(c) Charles Elton
(d) Edwin
Answer:
(b) Odum

Question 8.
Name the scientist(s) who rediscovered the Mendelian work? .
(i) Hugo de Vries (ii) Carl Correns (iii) Tschermak (iv) T.H. Morgan
(a) (i), (ii)and (iv) (b) (i), (ii)and (iii) (iii) ii, iii, iv (d) i, iii and iv
Answer:
(b) (i), (ii)and (iii)

Part – II

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

Question 9.
Detached leaf of Bryophyllum produces new plants. How?
Answer:
In Bryophyllum, the leaf is succulent and notched on its margin. Adventious buds develop at these notches and are called epiphyllous buds. They develop into new plants forming a root system and become independent plants when the leaf gets decayed.

Question 10.
Crossing over occurs only in germinal cells. Yes or No? Support your answer.
Answer:
No. Though crossing over is a common process in germinal cells rarely it also occurs in somatic cells during mitosis. Such crossing over is called mitotic crossing over or somatic crossing over.

Question 11.
Why do we use cryoprotectants in conservation process? Name any two cryoprotectant.
Answer:
Cryoprotectants are the protective agents that are used to protect the cells or tissues from the stress of freezing temperature. E.g: Sucrose, dimethyl sulphoxide.

Question 12.
What is Eltonian pyramid?
Answer:
Eltonian pyramid or Ecological pyramid is a graphic representation of the trophic structure and function at successive trophic levels of an ecosystem.

Question 13.
What do you understand by the term good ozone and bad ozone?
Answer:
The ozone layer of troposphere is called bad ozone.
The ozone layer of stratosphere is called good ozone because the layer acts as an shield for absorbing UV rays coming from sun which is harmful for living organisms causing ONA damage.

Question 14.
A person got irritation while applying chemical dye. What would be your suggestion for alternative?
Answer:
If a grey haired person is allergic on using chemical dyes then he can go for natural dyes like Henna. Henna is an organic dye obtained from leaves and young shoots of Lawsonia inermis. The principal colouring matter is Tacosone’ which is harmless and causes no irritation on skin.

Part – III

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

Question 15.
Differentiate between incomplete dominance and codominance.
Answer:
Incomplete Dominance:

• In incomplete dominance, neither of the allele is not completely dominant to another allele rather combine and produce new trait.
• New phenotype is formed due to character blending (not alleles)
• Example : Pink flowers of Mirabilis Jalapa

Co-dominance:

• In co-dominance, both the alleles in heterozygote are dominant and the traits are equally expressed (joint expression)
• No formation of new phenotype rather both dominant traits are expressed, conjointly.
• Example : Red and white flowers of camellia.

Question 16.
What are the features that a vector must possess to facilitate cloning?
Answer:
The following are the features that are required to facilitate cloning into a vector.

• Origin of replication (ori): This is a sequence from where replication starts and piece of DNA when linked to this sequence can be made to replicate within the host cells.
• Selectable marker: In addition to ori the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.
• Cloning sites: In order to link the alien DNA, the vector needs to have very few, preferably single, recognition sites for the commonly used restriction enzymes.

Question 17.
Define somatic embryogenesis? Give any two of its applications.
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids.

Applications:

• Somatic embryogenesis provides potential plantlets which after hardening period can establish into plants.
• Somatic embryoids can be used for the production of synthetic seeds.

Question 18.
What is TSM? How it is classified and what does it focuses on?
Answer:
TSM stands for Traditional Systems of Medicines India has a rich medicinal heritage. A number of Traditional Systems of Medicine (TSM) are practiced in India some of which come from outside India. TSM in India can be broadly classified into institutionalized or documented and non-institutionalized or oral traditions. Institutionalized Indian systems include Siddha and Ayurveda which are practiced for about two thousand years.

These systems have prescribed texts in which the symptoms, disease diagnosis, drugs to cure, preparation of drugs, dosage and diet regimes, daily and seasonal regimens. Non-institutional systems, whereas, do not have such records and or practiced by rural and tribal peoples across India. The knowledge is mostly held in oral form. The TSM focus on healthy lifestyle and healthy diet for maintaining good health and disease reversal

Question 19.
How do sacred groves help in the conservation of biodiversity?
Answer:
These are the patches or grove of cultivated trees which are community protected and are based on strong religious belief systems which usually have a significant religious connotation for protecting community. Each grove is an abode of a deity mostly village God Or Goddesses like Aiyanar or Amman. 448 grooves were documented throughout Tamil Nadu, of which 6 groves (Banagudi shola, Thirukurungudi and Udaiyankudikadu, Sittannnavasal, Puthupet and Devadanam) were taken up for detailed floristic and faunistic studies.

These groves provide a number of ecosystem services to the neighbourhood like protecting watershed, fodder, medicinal plants and micro climate control.

Part – IV

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

Question 20.
(a) Draw a flow chart depicting the various types of ploidy.
Answer:

[OR]

(b) Mention the application of biotechnology.
Answer:

• Biotechnology is one of the most important applied interdisciplinary sciences of the 21st century. It is the trusted area that enables us to find the beneficial way of life.
• Biotechnology has wide applications in various sectors like agriculture, medicine, environment and commercial industries.
• This science has an invaluable outcome like transgenic varieties of plants e.g. transgenic cotton (Bt-cotton), rice, tomato, tobacco, cauliflower, potato and banana.
• The development of transgenics as pesticide resistant, stress resistant and disease resistant varieties of agricultural crops is the immense outcome of biotechnology.
• The synthesis of human insulin and blood protein in E.coli and utilized for insulin deficiency disorder in human is a breakthrough in biotech industries in medicine.
• The synthesis of vaccines, enzymes, antibiotics, dairy products and beverages are the products of biotech industries.
• Biochip based biological computer is one of the successes of biotechnology.
• Genetic engineering involves genetic manipulation, tissue culture involves aseptic cultivation of totipotent plant cell into plant clones under controlled atmospheric conditions.
• Single cell protein from Spirulina is utilized in food industries.
• Production of secondary metabolites, biofertilizers, biopesticides and enzymes.
• Biomass energy, biofuel, bioremediation and phytoremediation for environmental biotechnology.

Question 21.
(a) Differentiate Primary succession and Secondary succession.
Answer:
Primary succession:

• Developing in an barren area.
• Initiated due to a biological or any other external factors.
• No soil, while primary succession starts
• Pioneer species come from outside environment.
• It takes more time to complete.

Secondary succession:

• Developing in disturbed area.
• Starts due to external factors only.
• It starts where soil covers is already present.
• Pioneer species develop from existing environment.
• It takes comparatively less time to complete.

(b) Enumerate the characters of anemophilous flowers
Answer:
Anemophilous plants have the following characteristic features:

• The flowers are produced in pendulous, catkin-like or spike inflorescence.
• The axis of inflorescence elongates so that the flowers are brought well above the leaves.
• The perianth is absent or highly reduced.
• The flowers are small, inconspicuous, colourless, not scented, do not secrete nectar.
• The stamens are numerous, filaments are long, exerted and versatile.
• Anthers produce enormous quantities of pollen grains compared to number of ovules available for pollination. They are minute, light and dry so that they can be carried to long distances by wind.
• In some plants anthers burst violently and release the pollen into the air. Example: Urtica.
• Stigmas are comparatively large, protruding, sometimes branched and feathery, adapted to catch the pollen grains. Generally single ovule is present.
• Plant produces flowers before the new leaves appear, so the pollen can be carried without hindrance of leaves.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Which one of the following groups includes sexually transmitted diseases caused by bacteria only?
(a) Syphilis, gonorrhoea and candidiasis
(b) Syphilis, chlamydiasis and gonorrhoea
(c) Syphilis, gonorrhoea and trichomoniasis
(d) Syphilis, trichomoniasis and pediculosis
Answer:
(b) Syphilis, chlamydiasis and gonorrhoea

Question 2.
ABO blood group in man is controlled by __________.
(a) Multiple alleles
(b) Lethal genes
(c) Sex linked genes
(d) Y-linked genes
Answer:
(a) Multiple alleles

Question 3.
Hershey and Chase experiment with bacteriophage showed that ____________.
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 4.
Modern man belongs to which period?
(a) Quaternary
(b) Cretaceous
(c) Silurian
(d) Cambrian
Answer:
(a) Quaternary

Question 5.
Allergy involves _______.
(a) IgE
(b) IgG
(c) IgA
(d) IgM
Answer:
(a) IgE

Question 6.
Recombinant Factor VIII is produced in the ______ of the Chinese Hamster.
(a) Liver cells
(b) blood cells
(c) ovarian cells
(d) brain cells
Answer:
(c) ovarian cells

Question 7.
Match with correct pair

(a) A – ii, B – iv, C – i, D – iii
(b) A – ii, B – iii, C – i, D – iv
(d) A – iii, B – i, C – iv, D – ii
(d) A – iv, B – ii, C – i, D – iii
Answer:
(b) A – ii, B – iii, C – i, D – iv

Question 8.
Assertion (A): Snake is a stenotherm.
Reason (R): Organism can tolerate narrow temperature fluctuations.
(a) Both A and R are correct R explain A
(b) A is correct R is incorrect
(c) A and R are correct, R doesn’t explains A
(d) Both A and R are incorrect
Answer:
(a) Both A and R are correct R explain A

PART – II

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

Question 9.
Mention the differences between spermiogenesis and spermatogenesis.
Answer:
Spermiogenesis:
Transformation of spermatids into mature sperm.

Spermatogenesis:
Spermatogenesis is the sequence of events in the seminiferous tubules of testes that produces male gametes, the sperms.

Question 10.
Draw and label a gemmule of sponge.
Answer:

Question 11.
State Wiener’s Hypothesis on Rh-factor.
Answer:
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^Z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R², R⁰, R^z) will produce ‘Rh-positive’ phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 12.
Why t-RNA is called an adapter molecule?
Answer:
The transfer RNA (t-RNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule. This term was postulated by Francis Crick.

Question 13.
Why Red list is prepared periodically?
Answer:
The purpose of preparation of Red List are:

• To create awareness on the degree of threat to biodiversity
• Identification and documentation of species at high risk of extinction
• Provide global index on declining biodiversity .
• Preparing conservation priorities and help in conservation of action
• Information on international agreements on conservation of biological diversity

Question 14.
Write notes on (a) Eutrophication (b) Algal Bloom
Answer:
Eutrophication refers to the nutrient enrichment in water bodies leading to lack of oxygen and will end up in the death of aquatic organisms.
Algal Bloom is an excess growth of algae due to abundant excess nutrients imparting distinct color to water.

Part – III

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

Question 15.
Explain Foetal-ejection reflex.
Answer:
As the pregnancy progresses, increase in the oestrogen concentration promotes uterine contractions. These uterine contractions facilitate moulding of the foetus and downward movement of the foetus. The descent of the foetus causes dilation of cervix of the uterus and vaginal canal resulting in a neurohumoral reflex called Foetal ejection reflex or Ferguson reflex.

This initiates the secretion of oxytocin from the neurohypophysis which in turn brings about the powerful contraction of the uterine muscles and leads to the expulsion of the baby through the birth canal.

Question 16.
Role of Y- chromosome is crucial for maleness – Justify.
Answer:
Current analysis of Y chromosomes has revealed numerous genes and regions with potential genetic function; some genes with or without homologous counterparts are seen on the X. Present at both ends of the Y chromosome are the pseudoautosomal regions (PARs) that are similar with regions on the X chromosome which synapse and recombine during meiosis. The remaining 95% of the Y chromosome is referred as the Non-combining Region of the Y (NRY).

The NRY is divided equally into functional genes (euchromatic) and non-functional genes (heterochromatic). Within the euchromatin regions, is a gene called Sex determining region Y (SRY). In humans, absence of Y chromosome inevitably leads to female development and this SRY gene is absent in X chromosome. The gene product of SRY is the testes determining factor (TDF) present in the adult male testis.

Question 17.
State immunological surveillance theory.
Answer:
The concept of immunological surveillance postulates that the primary function of the immune system is to “seek and destroy” malignant cells that arise by somatic mutation. The efficiency of the surveillance mechanism reduces either as a result of ageing or due to congenital or acquired immuno deficiencies, leads to increased incidence of cancer. Thus, if immunological surveillance is effective, cancer should not occur. The development of tumour represents a lapse in surveillance.

Question 18.
Write a short note on Pathaneer.
Answer:
In some parts of South India, a traditional drink called pathaneer is obtained from fermenting sap of palms and coconut trees. A common source is tapping of unopened spadices of coconut. It is a refreshing drink, which on boiling produces jaggery or palm sugar. When pathaneer is left undisturbed for few hours it gets fermented to form toddy with the help of naturally occurring yeast, to form a beverage that contains 4 percent alcohol. After 24 hours toddy becomes unpalatable and is used for the production of vinegar.

Question 19.
Elucidate the methodology of ELISA test.
Answer:
During diagnosis the sample suspected to contain the antigen is immobilized on the surface of an ELISA plate. The antibody specific to this antigen is added and allowed to react with the immobilized antigen. The anti-antibody is linked to an appropriate enzyme like peroxidase.

The unreacted anti-antibody is washed away and the substrate of the enzyme (hydrogen peroxidase) is added with certain reagents such as 4-chloronaphthol. The activity of the enzyme yields a coloured product indicating the presence of the antigen.

Part – IV

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

Question 20.
(a) What are IUD’s? Explain its way of functioning. Also describe their types.
Answer:
Intrauterine Devices (IUDs) are inserted by medical experts in the uterus through the vagina. These devices are available as copper releasing IUDs, hormone releasing IUDs and nonmedicated IUDs. IUDs increase phagocytosis of sperm within the uterus. IUDs are the ideal contraceptives for females who want to delay pregnancy. It is one of the popular methods of contraception in India and has a success rate of 95 to 99%.

Copper releasing IUDs differ from each other by the amount of copper. Copper IUDs such as Cu T-380 A, Nova T, Cu 7, Cu T 380 Ag, Multiload 375, etc. release free copper and copper salts into the uterus and suppress sperm motility. They can remain in the uterus for five to ten years.

Hormone-releasing IUDs such as Progestasert and LNG – 20 are often called as intrauterine systems (IUS). They increase the viscosity of the cervical mucus and thereby prevent sperms from entering the cervix.

Non-medicated IUDs are made of plastic or stainless steel. Lippes loop is a double S-shaped plastic device.

[OR]

(b) Explain how Urey – Miller’s experiment supports the origin of life?
Answer:

Urey and Miller (1953), paved way for understanding the possible synthesis of organic compounds that led to the appearance of living organisms is depicted in the Figure In their experiment, a mixture of gases was allowed to circulate over electric discharge from an tungsten electrode.

A small flask was kept boiling and the steam emanating from it was made to mix with the mixture of gases (ammonia, methane and hydrogen) in the large chamber that was connected to the boiling water. The steam condensed to form water which ran down the ‘U’ Diagrammatic representation of Urey-Miller’s experiment tube. Experiment was conducted continuously for a week and the liquid was analysed. Glycine, alanine, beta alanine and aspartic acid were identified.

Thus Miller’s experiments had an insight as to the possibility of abiogenetic synthesis of large amount of variety of organic compounds in nature from a mixture of sample gases in which the only source of carbon was methane. Later in similar experiments, formation of all types of amino acids, and nitrogen bases were noticed.

Question 21.
(a) Discuss any two Allosomal abnormalities in human beings.
Answer:
Allosomal abnormalities in human beings:
Mitotic or meiotic non-disjunction of sex chromosomes causes allosomal abnormalities.
Several sex chromosomal abnormalities have been detected. E.g. Klinefelter’s syndrome and
Turner’s syndrome.

1. Klinefelter’s Syndrome (XXY Males):
This genetic disorder is due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47, XXY. Persons with this syndrome have 47 chromosomes (44AA+XXY). They are usually sterile males, tall, obese, with long limbs, high pitched voice, under developed genetalia and have feeble breast (gynaecomastia) development.

2. Turner’s Syndrome (XO Females):
This genetic disorder is due to the loss of a X chromosome resulting in a karyotype of 45 ,X. Persons with this syndrome have 45 chromosomes (44 autosomes and one X chromosome) (44AA+XO) and are sterile females. Low stature, webbed neck, under developed breast, rudimentary gonads lack of menstrual cycle during puberty, are the main symptoms of this syndrome:

[OR]

(b) Explain in detail about various types of extinctions.
Answer:
There are three types of Extinctions
1. Natural extinction: It is a slow process of replacement of existing species with better adapted species due to changes in environmental conditions, evolutionary changes, predators and diseases. A small population can get extinct sooner than the large population due to inbreeding depression (less adaptivity and variation)

2. Mass extinction: The Earth has experienced quite a few mass extinctions due to environmental catastrophes. A mass extinction occurred about 225 million years ago during the Permian, where 90% of shallow water marine invertebrates disappeared.

3. Anthropogenic extinctions: These are abetted by human activities like hunting, habitat destruction, over exploitation, urbanization and industrialization. Some examples of extinctions are Dodo of Mauritius and Steller’s sea cow of Russia. Amphibians seem to be at higher risk of extinction because of habitat destruction.

The most serious aspect of the loss of biodiversity is the extinction of species. The unique information contained in its genetic material (DNA) and the niche it possesses are lost forever.