JEE Main 6 April 2026 Shift 2 question paper with solutions

JEE Main 6 April 2026 Shift 2: all 75 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Relations and Functions · Single correct

Let $f : \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = \frac{2x^2 - 3x + 2}{3x^2 + x + 3}$. Then $f$ is:

  1. both one-one and onto
  2. one-one but not onto
  3. onto but not one-one
  4. neither one-one nor onto

Answer: (d)

Solution

Given $\frac{2x^2 - 3x + 2}{3x^2 + x + 3} = y$. $(3y - 2)x^2 + (y + 3)x + (3y - 2) = 0$ $D \geq 0$ $(7y - 1)(5y - 7) \leq 0$ $y \in \left[ \frac{1}{7}, \frac{7}{5} \right]$ Hence $f(x)$ is into and $f(x)$ is many-one because $f(x)$ is non-monotonic.

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

Consider the quadratic equation $(n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0, \ n \in \mathbb{R}$. Let $\alpha$ be the minimum value of the product of its roots and $\beta$ be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is $\alpha$ and the common ratio is $\frac{\alpha}{\beta}$, is:

  1. $\frac{61}{37}$
  2. $\frac{121}{81}$
  3. $\frac{364}{243}$
  4. $\frac{1093}{729}$

Answer: (c)

Solution

Given $\alpha = n^2 - 2n + 2$. $\alpha = (n-1)^2 + 1$. Therefore, the minimum value of $\alpha$ is 1. Similarly, $\beta = \frac{3}{(n-1)^2 + 1}$. Therefore, the maximum value of $\beta$ is 3. The required G.P. is 1, $\frac{1}{3}$, $\frac{1}{3^2}$, $\ldots$ $$S_6 = \frac{1 \left( 1 - \left( \frac{1}{3} \right)^6 \right)}{1 - \frac{1}{3}} = \frac{3}{2} \left[ 1 - \left( \frac{1}{3} \right)^6 \right] = \frac{364}{243}$$

Question 3

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $S = \{ z \in \mathbb{C} : z^2 + \sqrt{6} \, iz - 3 = 0 \}$. Then $\sum_{z \in S} z^8$ is equal to:

  1. 162
  2. 184
  3. 262
  4. 324

Answer: (a)

Solution

Given $$z = \frac{-\sqrt{6}i \pm \sqrt{-6 + 12}}{2}$$ $$= \frac{-\sqrt{6}i \pm \sqrt{6}}{2}$$ $$= \frac{\sqrt{6}}{2}(\pm 1 - i)$$ $$\alpha = \frac{-\sqrt{6}}{2}(1+i), \beta = \frac{\sqrt{6}}{2}(1-i)$$ $$\alpha = \frac{-\sqrt{6}}{2} \times \sqrt{2} e^{i\pi/4}, \beta = +\frac{\sqrt{6}}{2} \times \sqrt{2} e^{-i\pi/4}$$ $$\alpha = -\sqrt{3} e^{i\pi/4}, \beta = +\sqrt{3} e^{-i\pi/4}$$ $$\alpha^8 + \beta^8 = 81 \cdot e^{i2\pi} + 81 \cdot e^{i(-2\pi)}$$ $$= 81 (1 + 1) = 162$$

Question 4

Maths · Determinants · Single correct

The sum of all possible values of $\theta \in [0, 2\pi]$, for which the system of equations: $$x \cos 3\theta - 8y - 12z = 0$$ $$x \cos 2\theta + 3y + 3z = 0$$ $$x + y + 3z = 0$$ has a non-trivial solution, is equal to:

  1. $\pi$
  2. $2\pi$
  3. $3\pi$
  4. $4\pi$

Answer: (d)

Solution

Given the determinant: $$\begin{vmatrix} \cos 3\theta & -8 & -12 \\ \cos 2\theta & 3 & 3 \\ 1 & 1 & 3 \end{vmatrix} = 0$$ Perform the column operations $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$: $$\begin{vmatrix} \cos 3\theta + 8 & -4 & -3 \\ \cos 2\theta - 3 & 2 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0$$ Expanding the determinant: $$2\cos 3\theta + 16 + 4\cos 2\theta - 12 = 0$$ Simplifying: $$(4\cos^3\theta - 3\cos\theta) + 2(2\cos^2\theta - 1) + 2 = 0$$ $$4\cos^3\theta + 4\cos^2\theta - 3\cos\theta = 0$$ Factorizing: $$\cos\theta (4\cos^2\theta + 4\cos\theta - 3) = 0$$ $$\cos\theta (2\cos\theta + 3)(2\cos\theta - 1) = 0$$ Solutions for $\cos\theta$ are $0$, $\frac{1}{2}$, $-\frac{3}{2}$ (rejected). Thus, $\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{3}, \frac{5\pi}{3}$. Sum = $4\pi$

Question 5

Maths · Matrices · Single correct

Let $A = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{pmatrix}$ and $B = [b_{ij}], 1 \leq i, j \leq 3$. If $B = A^{99} - I$, then the value of $\frac{b_{31} - b_{21}}{b_{32}}$ is:

  1. 99
  2. 199
  3. 149
  4. 159

Answer: (c)

Solution

Given $A = P + I$. $$A = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $A = P + I$ where $P^3 = 0$. $$A^{99} = I + 99P + \binom{99}{2} P^2$$ $$A^{99} - I = 99 \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} + \binom{99}{2} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$$ $$= \begin{bmatrix} 0 & 0 & 0 \\ 99 \times 3 & 0 & 0 \\ 44550 & 99 \times 3 & 0 \end{bmatrix}$$ $b_{31} = 44550$, $b_{21} = 99 \times 3$, $b_{32} = 99 \times 3$ $$\frac{44550 - 297}{297} = 149$$

Question 6

Maths · Sequences and Series · Single correct

The sum $1 + \frac{1}{2}(1^2 + 2^2) + \frac{1}{3}(1^2 + 2^2 + 3^2) + \ldots$ upto 10 terms is equal to:

  1. 130
  2. 155
  3. $\frac{315}{2}$
  4. $\frac{325}{2}$

Answer: (c)

Solution

Given $T_n = \frac{1}{n}(1^2 + 2^2 + \ldots + n^2)$. $$T_n = \frac{n(n+1)(2n+1)}{6n}$$ $$T_n = \frac{1}{6}[2n^2 + 3n + 1]$$ $S_n = \Sigma T_n = \frac{1}{6}[2\Sigma n^2 + 3\Sigma n + \Sigma 1]$ $$\Sigma T_n = \frac{1}{6}\left[2\cdot \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} + n\right]$$ $$S_{10} = \frac{1}{6}\left[\frac{2 \times 10 \times 11 \times 21}{6} + \frac{3}{2} \times 10 \times 11 + 10\right]$$ $$= \frac{10}{6}\left[77 + \frac{33}{2} + 1\right] = \frac{315}{2}$$

Question 7

Maths · Permutations and Combinations · Single correct

A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the $10^{th}$ floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to:

  1. 2184
  2. 3064
  3. 7056
  4. 11340

Answer: (c)

Solution

Form group of 4 and 5 person $\frac{9!}{4!5!} = 126$ choose two floor out of 8 floor $^8C_2 = 28$ Total ways in which 4 $\&$ 5 persons can leave at different floor $126 \times 28 \times 2 = 7056$

Question 8

Maths · Complex Numbers and Quadratic Equations · Single correct

Let the mean and the variance of seven observations 2, 4, $\alpha$, 8, $\beta$, 12, 14, $\alpha < \beta$, be 8 and 16 respectively. Then the quadratic equation whose roots are $3\alpha + 2$ and $2\beta + 1$ is:

  1. $x^2 - 35x + 306 = 0$
  2. $x^2 - 41x + 420 = 0$
  3. $x^2 - 45x + 506 = 0$
  4. $x^2 - 37x + 342 = 0$

Answer: (b)

Solution

Given $\frac{2 + 4 + \alpha + \beta + 8 + 12 + 14}{7} = 8$ $\alpha + \beta = 56 - 40 = 16 \Rightarrow \alpha + \beta = 16$ .....(i) Variance = $\frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2$ $16 = \frac{2^2 + 4^2 + \alpha^2 + \beta^2 + 8^2 + 12^2 + 14^2}{7} - 64$ $560 = \alpha^2 + \beta^2 + 424$ $\Rightarrow \alpha^2 + \beta^2 = 136$ ....(ii) From (i) $\&$ (ii) $\alpha = 6 \& \beta = 10$ $x_1 = 3\alpha + 2 = 20 \; ; \; x_2 = 2\beta + 1 = 21$ Q.E. : $x^2 - 41x + 420 = 0$

Question 9

Maths · Probability · Single correct

A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is:

  1. $\frac{63}{925}$
  2. $\frac{17}{231}$
  3. $\frac{16}{231}$
  4. $\frac{64}{925}$

Answer: (c)

Solution

Probability is given by $$\frac{6 \times 6}{\left(\frac{12}{(2)^6} \right) \times 6} = \frac{16}{231}.$$

Question 10

Maths · Conic Sections · Single correct

Let C be a circle having centre in the first quadrant and touching the x-axis at a distance 3 units from the origin. If the circle C has an intercept of length $6\sqrt{3}$ on y-axis, then the length of the chord of the circle C on the line $x - y = 3$ is :

  1. 8
  2. 6
  3. 6$\sqrt{2}$
  4. 8$\sqrt{2}$

Answer: (c)

Solution

Centre of circle is (3, 6) and radius = 6 units. $$PQ = 2 \sqrt{6^2 - p^2} \therefore p = \frac{6}{\sqrt{2}}$$ $$p = \frac{|3 - 6 - 3|}{\sqrt{2}}$$ $$PQ = 2 \sqrt{36 - \frac{36}{2}}$$ $$= 2 \times \frac{6}{\sqrt{2}} = 6 \sqrt{2}$$

Question 11

Maths · Conic Sections · Single correct

The eccentricity of an ellipse E with centre at the origin O is $\frac{\sqrt{3}}{2}$ and its directrices are $x = \pm \frac{4\sqrt{6}}{3}$. Let H : $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of E, and whose length of latus rectum is equal to the length of minor axis of E. Then the distance between the foci of H is :

  1. $\frac{4\sqrt{2}}{\sqrt{7}}$
  2. $\frac{4\sqrt{2}}{7}$
  3. $\frac{4}{\sqrt{7}}$
  4. $\frac{8}{7}$

Answer: (d)

Solution

Given $e = \frac{\sqrt{3}}{2}$. $$\frac{a}{e} = \frac{4\sqrt{6}}{3}$$ Thus, $a = 2\sqrt{2}$. $$(ae)^2 = a^2 - b^2$$ $$b^2 = 2$$ So, $E : \frac{x^2}{8} + \frac{y^2}{2} = 1$. Now, $e_H = 2\sqrt{2} \Rightarrow b^2 = 7a^2$. $$\frac{2b^2}{a} = 2\sqrt{2} \Rightarrow a = \frac{\sqrt{2}}{7}$$ $$(ae_H)^2 = a^2 + b^2 = 8a^2 = \frac{16}{49}$$ Thus, $ae_H = \frac{4}{7}$. The distance between focii is $2ae_H = \frac{8}{7}$.

Question 12

Maths · Conic Sections · Single correct

Let $x = 9$ be a directrix of an ellipse $E$, whose centre is at the origin and eccentricity is $\frac{1}{3}$. Let $P (\alpha, 0)$, $\alpha > 0$, be a focus of $E$ and $AB$ be a chord passing through $P$. Then the locus of the mid point of $AB$ is :

  1. $9y^2 = 8x(1-x)$
  2. $3y^2 = 4x(1-x)$
  3. $9y^2 = 8x(x-1)$
  4. $3y^2 = 4x(x-1)$

Answer: (a)

Solution

Given $\frac{a}{e} = 9 \Rightarrow a = 3$. $b^2 = a^2 (1 - e^2) = 9 \left( 1 - \frac{1}{9} \right) = 8$. The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{8} = 1$. The focus $S(\alpha, 0) \equiv S(1, 0)$. $T = S_1 \Rightarrow \frac{hx}{9} + \frac{ky}{8} = \frac{h^2}{9} + \frac{k^2}{8}$. For $(1, 0) \Rightarrow \frac{h}{9} + 0 = \frac{h^2}{9} + \frac{k^2}{8}$. Thus, $h = \frac{h^2}{9} + \frac{k^2}{8}$. Finally, $9y^2 = 8x(1-x)$.

Question 13

Maths · Inverse Trigonometric Functions · Single correct

If $\sin\left(\tan^{-1}(x\sqrt{2})\right) = \cot\left(\sin^{-1}\sqrt{1-x^2}\right)$, $x \in (0, 1)$, then the value of $x$ is :

  1. $\frac{1}{2}$
  2. $\frac{1}{3}$
  3. $\frac{2}{3}$
  4. $\frac{5}{8}$

Answer: (a)

Solution

Given $\frac{x \sqrt{2}}{\sqrt{2x^2 + 1}} = \frac{1}{\sqrt{1-x^2}}$. $2(1-x^2) = (2x^2 + 1)$ $2 - 2x^2 = 2x^2 + 1$ $x^2 = \frac{1}{4}$ $x = \frac{1}{2}$

Question 14

Maths · Three Dimensional Geometry · Single correct

The shortest distance between the lines $\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$ is:

  1. $\frac{5\sqrt{6}}{6}$
  2. $2\sqrt{5}$
  3. $3\sqrt{5}$
  4. $4\sqrt{5}$

Answer: (c)

Solution

Given the determinant, we have: $$ SD = \begin{vmatrix} 4 - (-2) & 3 - 6 & 2 - 5 \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} $$ This simplifies to: $$ SD = \frac{|6(-10 + 12) - (-3)(-5 + 6) - 3(4 - 4)|}{|2\hat{i} - \hat{j}|} $$ Further simplification gives: $$ SD = \left| \frac{12 + 3}{\sqrt{5}} \right| $$ Which results in: $$ = \frac{15}{\sqrt{5}} $$ Finally, we have: $$ SD = 3\sqrt{5} $$

Question 15

Maths · Vector Algebra · Single correct

Let $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2\vec{a} + 3\vec{b})$ and $(\vec{a} - \vec{b})$ is:

  1. 450
  2. 900
  3. 1800
  4. 2400

Answer: (c)

Solution

Given $2\bar{a} + 3\bar{b} = 2(2\hat{i} + 3\hat{j} + 3\hat{k}) + 3(6\hat{i} + 3\hat{j} + 3\hat{k})$. This simplifies to $22\hat{i} + 15\hat{j} + 15\hat{k}$. We have $\bar{a} - \bar{b} = -4\hat{i}$. The area is given by $\frac{1}{2} |(2\bar{a} + 3\bar{b}) \times (\bar{a} - \bar{b})|$. This equals $\frac{1}{2} |-60\hat{j} + 60\hat{k}| = \frac{1}{2} \sqrt{(60)^2 \times 2}$. Thus, $A = \frac{60}{\sqrt{2}}$. The square of the area is $A^2 = 1800$.

Question 16

Maths · Limits and Derivatives · Single correct

Let $\lim_{x \to 2} \frac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2}$ = 5 for some r, p $\in \mathbb{R}$. If the set of all possible values of q, such that the roots of the equation $rx^2 - px + q = 0$ lie in (0, 2), be the interval ($\alpha$, $\beta$], then 4($\alpha$ + $\beta$) equals.

  1. 11
  2. 13
  3. 17
  4. 21

Answer: (c)

Solution

Given $\lim_{x \to 2} \frac{\tan(x-2) \cdot [rx^2 + (p-2)x - 2p]}{x-2} = 5$. $$\Rightarrow \lim_{x \to 2} \frac{rx^2 - 2x + p(x-2)}{x-2} = 5$$ Since $D^r \to 0$ and $N^r \to 0$, $r = 1$. $$\Rightarrow \lim_{x \to 2} \frac{x(x-2) + p(x-2)}{x-2} = 5$$ $$\Rightarrow 2 + p = 5$$ $$\Rightarrow p = 3$$ Now, the quadratic equation is $x^2 - 3x + q = 0$ and its both roots lie in $(0, 2)$. $D \geq 0$, $0 0$, $f(2) > 0$. $$\Rightarrow 9 - 4q \geq 0 and q > 2$$ $$\Rightarrow q \in \left( 2, \frac{9}{4} \right]$$ $\alpha = 2$, $\beta = \frac{9}{4} \Rightarrow \left( 2 + \frac{9}{4} \right) = 17$

Question 17

Maths · Determinants · Single correct

Let $A = \begin{vmatrix} 1 & 3 & -1 \\ 2 & 1 & \alpha \\ 0 & 1 & -1 \end{vmatrix}$ be a singular matrix. Let $$f(x) = \int_{0}^{x} (t^2 + 2t + 3) \, dt , \ x \in [1, \alpha].$$ If $M$ and $m$ are respectively the maximum and the minimum values of $f$ in $[1, \alpha]$ then $3(M - m)$ is equal to :

  1. 64
  2. 68
  3. 72
  4. 76

Answer: (b)

Question 18

Maths · Integrals · Single correct

Let $f : \mathbb{R} \to \mathbb{R}$ be such that $f(xy) = f(x)f(y)$, for all $x, y \in \mathbb{R}$ and $f(0) \neq 0$. Let $g: [1, \infty) \to \mathbb{R}$ be a differentiable function such that $$x^2 g(x) = \int_{1}^{x} (t^2 f(t) - t g(t)) \, dt$$ Then $g(2)$ is equal to:

  1. $\frac{13}{8}$
  2. $\frac{11}{16}$
  3. $\frac{15}{32}$
  4. $\frac{17}{64}$

Answer: (c)

Solution

Since $f(xy) = f(x)f(y)$, it follows that $f(x) = x^n$ or $f(x) = k$. But $f(0) \neq 0 \Rightarrow f(x) \neq x^n$. For $f(x) = k \Rightarrow k = k.k \Rightarrow k = 1$ (since $f(0) \neq 0$). Therefore, $f(x) = 1$. $$x^2 g(x) = \int_1^x \left(t^2 (1) - t g(t) \right) \, dt$$ Differentiate with respect to $x$: $$x^2 g'(x) + g(x) \cdot 2x = x^2 - x g(x)$$ $$\Rightarrow x g'(x) + 3 g(x) = x$$ $$\Rightarrow g'(x) + \frac{3}{x} g(x) = 1$$ If the integrating factor is $e^{\int \frac{3}{x} \, dx} = e^{3 \ln(x)} = x^3$. Solution of the differential equation: $$\Rightarrow g(x) \cdot x^3 = \int x^3 \, dx$$ $$\Rightarrow x^3 g(x) = \frac{x^4}{4} + C$$ $$\Rightarrow C = -\frac{1}{4} \{ since g(1) = 0 \}$$ $$g(x) \cdot x^3 = \frac{x^4}{4} - \frac{1}{4}$$ Put $x = 2$: $$g(2) = 4 - \frac{1}{4}$$ $$g(2) = \frac{15}{32}.$$

Question 19

Maths · Applications of Integrals · Single correct

The area of the region $\{(x, y) : x^2 - 8x \leq y \leq -x \}$ is:

  1. $\frac{343}{6}$
  2. $\frac{637}{6}$
  3. $\frac{437}{6}$
  4. $\frac{523}{6}$

Answer: (a)

Solution

Required area is given by the integral: $$\int_{0}^{7} (7x - x^2) \, dx$$ Differentiating, we have: $$= \left[ \frac{7x^2}{2} - \frac{x^3}{3} \right]_{0}^{7}$$ Evaluating the integral, we get: $$= \frac{343}{6}$$

Question 20

Maths · Integrals · Single correct

The value of the integral $$\int_{-1}^{1} \left( \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \right) dx$$ is equal to

  1. 3 $\log_e$ 2
  2. 2 $\log_e$ 2
  3. 5 $\log_e$ 3
  4. 3 $\log_e$ 3

Answer: (b)

Solution

Given $$I = \int_{-1}^{1} \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \, dx.$$ We can split the integral as follows: $$I = \int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} \, dx + \int_{-1}^{1} \frac{|x| + 1}{x^2 + 2|x| + 1} \, dx.$$ The first integral evaluates to zero, so we have: $$I = 0 + \int_{-1}^{1} \frac{|x| + 1}{|x|^2 + 2|x| + 1} \, dx.$$ This simplifies to: $$2 \int_{0}^{1} \frac{1}{x + 1} \, dx.$$ Evaluating the integral, we get: $$I = 2 \left[ \ln |x + 1| \right]_{0}^{1}.$$ Finally, $$I = 2 \ln(2).$$

Question 21

Maths · Relations and Functions (Advanced) · Numerical

Let $R = \{(x,y) \in \mathbb{N} \times \mathbb{N} : \log_e (x + y) \leq 2\}$. Then the minimum number of elements, required to be added in R to make it a transitive relation, is $\ldots$.

Answer: 15

Solution

Given $x + y \leq e^2$ ($e^2 \simeq 7.29$) $R = \{(1, 1), (1, 2), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (6, 1)\}$ For $R$ to be transitive, we have to add $(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$ $(5, 3), (5, 4), (5, 5), (5, 6)$ $(4, 4), (4, 5), (4, 6)$ $(3, 5), (3, 6)$ $(2, 3)$ $= 15$ elements

Question 22

Maths · Binomial Theorem · Numerical

If $(1-x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r}$, then $\frac{9a_9}{a_{10}}$ is equal to $\ldots$

Answer: 30

Solution

(1-x)^3 = 1 - x^3 - 3x(1-x) (1-x)^3 = 1 - x^3 - 3x + 3x^2 (1-x)^3 + 3x - 3x^2 = 1 - x^3 (1-x^3)^{10} = [(1-x)^3 + 3x(1-x)]^{10} $\Rightarrow \binom{10}{r}$ (3x(1-x))^r ((1-x)^3)^{10-r} (1-x^3)^{10} = $\binom{10}{r}$ 3^r x^r (1-x)^{30-2r} $\therefore$ a_r = $\binom{10}{r}$ 3^r $\frac{a_9}{a_{10}}$ = $\frac{\binom{10}{9} 3^9}{\binom{10}{10} 3^{10}}$ $\frac{a_9}{a_{10}}$ = $\frac{10}{3}$ $\therefore \frac{9a_9}{a_{10}}$ = 30

Question 23

Maths · Conic Sections · Numerical

Let the line $x - y = 4$ intersect the circle $C : (x - 4)^2 + (y + 3)^2 = 9$ at the points $Q$ and $R$. If $P(\alpha, \beta)$ is a point on $C$ such that $PQ = PR$, then $(6\alpha + 8\beta)^2$ is equal to $\ldots$.

Answer: 18

Solution

Given $x - y = 4 \ldots$ (1) $$(x - 4)^2 + (y + 3)^2 = 9$$ $\ldots$ (2) $(\alpha, \beta)$ lies on circle Therefore, $$(\alpha - 4)^2 + (\beta + 3)^2 = 9$$ $\ldots$ (3) Equation (1) and (2) imply $$y^2 + (y + 3)^2 = 9 \implies 2y^2 + 6y = 0$$ This implies $y = 0, -3$ Therefore, $x = 4, 1$ Points $Q(4, 0)$, $R(1, -3)$ Now $$(PQ)^2 = (PR)^2$$ This implies $$(\alpha - 4)^2 + \beta^2 = (\alpha - 1)^2 + (\beta + 3)^2$$ Therefore, $\alpha + \beta = 1 \ldots$ (4) Using equations (3) and (4) This implies $$\alpha = 4 + \frac{3}{\sqrt{2}}, \beta = -3 - \frac{3}{\sqrt{2}}$$ Therefore, $$(6\alpha + 8\beta)^2 = 18$$

Question 24

Maths · Three Dimensional Geometry · Numerical

Let the image of the point P(0, -5, 0) in the line $\frac{x-1}{2}$ = $\frac{y}{1}$ = $\frac{z+1}{-2}$ be the point R and the image of the point Q$\left(0, -\frac{1}{2}, 0\right)$ in the line $\frac{x-1}{-1}$ = $\frac{y+9}{4}$ = $\frac{z+1}{1}$ be the point S. Then the square of the area of the parallelogram PQRS is $\ldots$.

Answer: 162

Solution

Given $P(0, -5, 0)$. The line equation is given by: $$\frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2} = k$$ Let $R(x_1, y_1, z_1)$ be a point on the line: $$2k + 1, k, -2k - 1$$ Substituting into the line equation: $$(2k + 1)(2) + (k + 5)(1) + (-2k - 1)(-2) = 0$$ Simplifying gives: $$9k + 9 = 0 \implies k = -1$$ Thus, $M_1 \Rightarrow (-1, -1, 1)$ and $R \Rightarrow (-2, 3, 2)$. For $Q(0, -1/2, 0)$, the line equation is: $$\frac{x-1}{-1} = \frac{y+9}{4} = \frac{z+1}{1} = \lambda$$ This gives: $$(-\lambda + 1, 4\lambda - 9, \lambda - 1)$$ Let $S(x_2, y_2, z_2)$ be another point: $$(-\lambda + 1)(-1) + \left(4\lambda - \frac{17}{2}\right)(4) + (\lambda - 1)(1) = 0$$ Solving gives: $$18\lambda - 36 = 0 \implies \lambda = 2$$ Thus, $M_2 \Rightarrow (-1, -1, 1)$ and $S \Rightarrow (-2, -3/2, 2)$. Calculating distances: $$PR = \sqrt{4 + 64 + 4} = \sqrt{72} = 6\sqrt{2}$$ $$QS = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$ The area is: $$Ar. = \frac{1}{2} \cdot 6\sqrt{2} \cdot 3 = 9\sqrt{2}$$ Finally, the square of the area is: $$(Ar.)^2 = (9\sqrt{2})^2 = 162 Sq.unit$$

Question 25

Maths · Continuity and Differentiability · Numerical

Let $f(x) = \begin{cases} x^3 + 8 & x < 0 \\ x^2 - 4 & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} (x - 8)^{1/3} & x < 0 \\ (x + 4)^{1/2} & x \geq 0 \end{cases}$ Then the number of points, where the function gof is discontinuous, is $\ldots$

Answer: 3

Solution

Given $$g(f(x)) = \begin{cases} (f(x) - 8)^{1/3}, & f(x) < 0 \\ (f(x) + 4)^{1/2}, & f(x) \geq 0 \end{cases}$$ $$g(f(x)) = \begin{cases} (x^3)^{1/3}, & x < -2 \\ (x^2 - 12)^{1/3}, & 0 < x < 2 \\ (x^3 + 12)^{1/2}, & -2 \leq x \leq 0 \\ (x^2)^{1/2}, & x \geq 2 \end{cases}$$ $$g(f(x)) = \begin{cases} x, & x < -2 \\ (x^3 + 12)^{1/2}, & -2 \leq x \leq 0 \\ (x^2 - 12)^{1/3}, & 0 < x < 2 \\ x, & x \geq 2 \end{cases}$$ Number of points of discontinuity is equal to 3.

Physics

Question 26

Physics · Physical World, Units and Measurements · Single correct

The percentage error in the calculated volume of a sphere, if there is 2$\%$ error in its diameter measurement, is $\ldots$.

  1. 1
  2. 2
  3. 6
  4. 8

Answer: (c)

Solution

The volume $v$ is given by $$v = \frac{4}{3} \pi R^3$$ The relative change in volume is $$\frac{\Delta v}{v} = \frac{3 \Delta R}{R}$$ which equals $6\%$.

Question 27

Physics · Physical World, Units and Measurements · Single correct

Match List-I with List-II.

  1. A-I, B-II, C-III, D-IV
  2. A-IV, B-III, C-II, D-I
  3. A-III, B-IV, C-II, D-I
  4. A-II, B-I, C-IV, D-III

Answer: (c)

Solution

(A) $\rightarrow$ ML^2 T^{-2} K^{-1} ($\mathrm{III}$) (B) $\rightarrow$ MT^{-3} K^{-4} ($\mathrm{IV}$) (C) $\rightarrow$ ML^2 T^{-1} ($\mathrm{II}$) (D) $\rightarrow$ M^{-1} L^3 T^{-2} ($\mathrm{IV}$)

Question 28

Physics · System of Particles and Rotational Motion · Single correct

A solid sphere (A) of mass, $5m$ and spherical shell (B) of mass $m$, both having same radius, are placed on a rough surface. When a force of same magnitude is applied tangentially at the highest points of A and B, they start rolling without slipping with an acceleration of $a_A$ and $a_B$, respectively. The ratio of $a_A$ and $a_B$ is $\ldots$.

  1. 5 : 21
  2. 6 : 10
  3. 21 : 25
  4. 1 : 5

Answer: (a)

Solution

For the solid sphere, we have: $$F + f = 5m \, a_A$$ $$F(R) - f(R) = \frac{2}{5} (5m) R^2 \cdot \alpha$$ Since $\alpha = R \alpha$, we get: $$2F = 7m \, a_A$$ Thus, $$a_A = \frac{2F}{7m}$$ For the hollow sphere, we have: $$F + f = ma_B$$ $$F(R) - f(R) = \frac{2}{3} m R^2 \times \alpha$$ Since $a = R \alpha$, we get: $$2F = \frac{5m a_B}{3}$$ Thus, $$\frac{a_A}{a_B} = \frac{2F}{7m} \times \frac{5m}{6F} = \frac{5}{21}$$

Question 29

Physics · Work, Energy and Power · Single correct

A body of mass 1 kg moves along a straight line with a velocity $v = 2x^2$. The work done by the body during displacement from $x = 0$ to $5\, \mathrm{m}$ is $\ldots$ J.

  1. 0
  2. 250
  3. 1250
  4. 1000

Answer: (c)

Solution

Given $w \cdot \Delta = \Delta K.E$. $$= \frac{1}{2} m \left( v_f^2 - v_i^2 \right)$$ $$= \frac{1}{2} \times 1 \left[ \left( 2(5)^2 \right)^2 - \left( 2(0) \right)^2 \right]^0$$ $$= \frac{1}{2} \times 50 \times 50 = 1250 \, \mathrm{J}$$

Question 30

Physics · Thermodynamics · Single correct

A cylinder with adiabatic walls is closed at both ends and is divided into two compartments by a frictionless adiabatic piston. Ideal gas is filled in both (left and right) the compartments at same P, V, T. Heating is started from left side until pressure changes to $\frac{27P}{8}$. If initial volume of each compartment was 9 litres then the final volume in right-hand side compartment is $\ldots$ litres. (for this ideal gas $\frac{C_p}{C_v} = 1.5$)

  1. 3
  2. 4
  3. 14
  4. 9

Answer: (b)

Solution

Given $P_{Left} = P_{Right}$ (final) for right chamber. $$P_i V_i^\gamma = P_f V_f^\gamma$$ $$P(9)^{3/2} = \frac{275}{8} \times V_f^{3/2}$$ $$27 = \frac{27}{8} V_f^{3/2}$$ Therefore, $V_f = 4$ litres.

Question 31

Physics · Electromagnetic Waves · Single correct

For an electromagnetic wave propagating through vacuum, $\vec{k}$, $\vec{E}$ and $\omega$ represent propagation vector, electric field and angular frequency, respectively. The magnetic field associated with this wave is represented by:

  1. $\frac{\vec{E} \times \vec{k}}{\omega}$
  2. $\frac{\vec{k} \times \vec{E}}{\omega}$
  3. $\omega (\vec{E} \times \vec{k})$
  4. $\omega (\vec{k} \times \vec{E})$

Answer: (b)

Solution

The solution is given by $$\vec{B} = \frac{1}{\omega} (\vec{k} \times \vec{E})$$

Question 32

Physics · System of Particles and Rotational Motion · Single correct

Two identical bodies A and B of equal masses have initial velocities $\vec{v}_1 = 4\hat{i} \, \mathrm{m/s}$ and $\vec{v}_2 = 4\hat{j} \, \mathrm{m/s}$ respectively. The body A has acceleration $\vec{a}_1 = 6\hat{i} + 6\hat{j} \, \mathrm{m/s^2}$ while the acceleration of the other body B is zero. The centre of mass of the two bodies moves in _____path.

  1. circular
  2. parabolic
  3. straight line
  4. elliptical

Answer: (c)

Solution

The velocity of the center of mass $\vec{V}_{cm}$ is given by $$\vec{V}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2} = \frac{m(4\hat{i}) + m(4\hat{j})}{2m} = 2\hat{i} + 2\hat{j}.$$ The acceleration of the center of mass $\vec{a}_{cm}$ is given by $$\vec{a}_{cm} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{m_1 + m_2} = \frac{m(6\hat{i} + 6\hat{j}) + m(0)}{2m} = 3\hat{i} + 3\hat{j}.$$ Since $\dot{\vec{V}}_{cm} \parallel \vec{a}_{cm}$, the path is a straight line.

Question 33

Physics · Mechanical Properties of Solids · Single correct

Figure represents the extension ($\Delta l$) of a wire of length 1 meter, suspended from the ceiling of the room at one end with load $W$ connected to the other end. If the cross-sectional area of the wire is $10^{-5} \, \mathrm{m}^2$ then the Young's modulus of the wire is $\ldots \, \mathrm{N/m}^2$.

  1. $1.0 \times 10^{11}$
  2. $2.0 \times 10^{10}$
  3. $1.0 \times 10^{10}$
  4. $2.0 \times 10^{11}$

Answer: (c)

Solution

Given $\frac{\Delta \ell}{w} = \tan \theta = \frac{\Delta y}{\Delta x} = \frac{1 \times 10^{-4}}{10} = 10^{-5}$ $$\frac{w}{\Delta \ell} = 10^5$$ Therefore, $y = \frac{F/A}{\Delta \ell / \ell} = \frac{F \ell}{A \Delta \ell} = \left( \frac{w}{\Delta \ell} \right) \frac{\ell}{A}$ $$\Rightarrow y = \frac{10^5 \times 1}{10^{-5}} = 10^{10} \, \mathrm{N/m^2}$$

Question 34

Physics · Mechanical Properties of Fluids · Single correct

A cylindrical vessel of $40\,\mathrm{cm}$ radius is completely filled with water and its capacity is $528\,\mathrm{dm^3}$ (dm : decimeter). The vessel is placed on a solid block of exactly same height as vessel. If a small hole is made at $70\,\mathrm{cm}$ below the top of water level, then horizontal range of water falling on the ground in the beginning is $\ldots$ cm.

  1. 120$\sqrt{2}$
  2. 140$\sqrt{2}$
  3. 140$\sqrt{3}$
  4. 120$\sqrt{3}$

Answer: (b)

Solution

Given $v = \pi r^2 h \Rightarrow 528 \times 10^{-3} = \pi (0.4)^2 h$. $$h = \frac{528 \times 10^{-3} \times 7}{(0.4)^2 \times 22} = 1.05 \, \mathrm{m} = 105 \, \mathrm{cm}$$ $$t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times (105 + 35) \times 10^{-2}}{10}}$$ $$= \sqrt{\frac{2(140) \times 10^{-2}}{10}} = \sqrt{28 \times 10^{-1}} \, \mathrm{sec}$$ $$R = ut = \sqrt{2 \times 10 \times 70 \times 10^{-2} \times \sqrt{28 \times 10^{-1}}}$$ $$= (\sqrt{14})(\sqrt{28}) \times 10^{-1} = 14 \sqrt{2} \times 10^{-1} \, \mathrm{m}$$ $$R = 140 \sqrt{2} \, \mathrm{cm}$$

Question 35

Physics · Kinetic Theory · Single correct

If 2 mole of an ideal monoatomic gas at temperature $T$, is mixed with 6 mole of another ideal monoatomic gas at temperature $2T$ then the temperature of mixture is:

  1. $\frac{5}{2} T$
  2. $\frac{5}{4} T$
  3. $\frac{7}{2} T$
  4. $\frac{7}{4} T$

Answer: (d)

Solution

The expression for $T_{mix}$ is given by $$T_{mix} = \frac{n_1 C_1 T_1 + n_2 C_2 T_2}{n_1 C_1 + n_2 C_2}$$ Substituting the values, we have $$= \frac{\left(2 \times \frac{3}{2} R \times T\right) + \left(6 \times \frac{3}{2} R \times 2T\right)}{\left(2 \times \frac{3}{2} R\right) + \left(6 \times \frac{3}{2} R\right)}$$ Simplifying further, we get $$T_{mix} = \frac{2T + 12T}{8} = \frac{7}{4} T$$

Question 36

Physics · Oscillations · Single correct

A spring stretches by $2\,\mathrm{mm}$ when it is loaded with a mass of $200\,\mathrm{g}$. From equilibrium position the mass is further pulled down by $2\,\mathrm{mm}$ and released. The frequency associated with the system and maximum energy in the spring are $\ldots \mathrm{Hz}$ and $\ldots \mathrm{J}$, respectively. (Take g = $10\,\mathrm{m/s^2}$)

  1. $\frac{5\sqrt{50}}{\pi}$ and $8 \times 10^{-3}$
  2. $\frac{5\sqrt{50}}{\pi}$ and $8$
  3. $10\sqrt{50}$ and $2 \times 10^{-3}$
  4. $\frac{5\sqrt{50}}{\pi}$ and $16 \times 10^{-3}$

Answer: (a)

Solution

Given $kx = 0.2g$; $k \times 2 \times 10^{-3} = 2$ $k = 1000 \, \mathrm{N/m}$ $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$ $f = \frac{1}{2\pi} \sqrt{\frac{1000}{0.2}}$ $= \frac{1}{2\pi} \sqrt{5000} = \frac{5}{\pi} \sqrt{50}$ $E_{\max} = \frac{1}{2} k \, x_{\max}^2$ $\frac{1}{2} \times 1000 \times (4 \times 10^{-3})^2 = 8 \times 10^{-3} \, \mathrm{Joule}$

Question 37

Physics · Electrostatic Potential and Capacitance · Single correct

The electric potential as a function of $x$, $y$ is given by $V = 5(x^2 - y^2) \, \mathrm{V}$. The electric field at a point $(2, 3) \, \mathrm{m}$ is $\ldots$ V/m.

  1. $-20\hat{i} + 30\hat{j}$
  2. $20\hat{i} - 30\hat{j}$
  3. $20\hat{i} + 45\hat{j}$
  4. $-4\hat{i} + 6\hat{j}$

Answer: (a)

Solution

Given $V = 5(x^2 - y^2)V$. $$\varepsilon_x = -\frac{dv}{dx} = -10x\hat{i}$$ $$\varepsilon_y = -\frac{dv}{dx} = +10y\hat{i} \Rightarrow \varepsilon(2,3) = -20\hat{i} + 30\hat{j}$$

Question 38

Physics · Moving Charges and Magnetism · Single correct

A current of $30\,\mathrm{A}$ each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm. The magnetic field at the mid point between the two wires is $\ldots \mu \mathrm{T}$. $\frac{\mu_0}{4\pi} = 10^{-7} \, \mathrm{N/A^2}$

  1. 30
  2. 300
  3. 150
  4. 0.1

Answer: (b)

Solution

The magnetic field $B$ is given by the formula: $$B = \frac{\mu_0 I}{2 \pi (4) \times 10^{-2}} \times 2$$ Calculating further: $$= 10^{-7} \times 30 \times 10^2$$ $$= 30 \times 10^{-5} \, \mathrm{T}$$ $$= 300 \, \mu \mathrm{T}$$

Question 39

Physics · Electromagnetic Induction · Single correct

A square loop of side $2\,\mathrm{cm}$ is placed in a time varying magnetic field with magnitude as $B = 0.4 \sin(300t)$ Tesla. The normal to the plane of loop makes an angle of $60^\circ$ with the field. The maximum induced emf produced in the loop is $\ldots$ mV.

  1. 12
  2. 18
  3. 21
  4. 24

Answer: (d)

Solution

Given $B = 0.4 \sin 300t$ and $\theta = 60^\circ$. The area $A = (0.02)^2 = 4 \times 10^{-4} \, \mathrm{m^2}$. The magnetic flux $\phi = BA \cos \theta$. Therefore, $\phi = 0.4 \sin (300t) A \cos \theta$. The emf is given by $\mathrm{emf} = \frac{d\phi}{dt} = 0.4 \, A \cos \theta \cos (300t) \times 300$. The maximum emf is $0.4 \times 4 \times 10^{-4} \times 300 \times \frac{1}{2}$. This equals $240 \times 10^{-4} \, \mathrm{V}$, which is $24 \, \mathrm{mV}$.

Question 40

Physics · Electrostatic Potential and Capacitance · Single correct

A sphere of capacitance $100\,\mathrm{pF}$ is charged to a potential of $100\,\mathrm{V}$. Another identical undercharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is $\alpha \times 10^{-7} \mathrm{J}$. The value of $\alpha$ is $\ldots$. (combined capacitance of spheres is $200\,\mathrm{pF}$).

  1. 5
  2. $\frac{5}{2}$
  3. $\frac{7}{2}$
  4. $\frac{9}{2}$

Answer: (b)

Solution

Given $Q_1 = CV = 100 \times 100 \, \mathrm{pC} = 10^4 \, \mathrm{pC}$. $V_1 = V_2 = \frac{V}{2} = 50$. $\varepsilon_i = \frac{1}{2} \times 100 \times (100)^2 \, \mathrm{pJ} = \varepsilon$. $\varepsilon_f = 2 \times \frac{1}{2} \times 100 \times (50)^2 = \frac{\varepsilon}{2}$. $\Delta \varepsilon = \frac{\varepsilon}{2} = 100 \times (50)^2 \, \mathrm{pJ}$. $= 25 \times 10^4 \times 10^{-12} = 2.5 \times 10^{-7}$.

Question 41

Physics · Nuclei · Single correct

The energy released if hydrogen atoms are combined to form $^4_2\mathrm{He}$ is $\ldots$. (Take binding energies per nucleon of $^2_1\mathrm{H}$ and $^4_2\mathrm{He}$ as $1.1 \, \mathrm{MeV}$ and $7.2 \, \mathrm{MeV}$, respectively).

  1. 6.1
  2. 24.4
  3. 26.6
  4. 5

Answer: (b)

Solution

Sol. $^2_1\mathrm{H} + ^2_1\mathrm{H} \rightarrow ^4_2\mathrm{He}$ $Q = \mathrm{BE} (RHS) - \mathrm{BE} (LHS)$ $= 4 \times 7.2 - 4 \times 1.1$ $= 4 \times 6.1$ $Q = 24.4 \, \mathrm{MeV}$

Question 42

Physics · Ray Optics and Optical Instruments · Single correct

Angle of minimum deviation is equal to the half of the angle of prism is an equilateral prism. The refractive index of the prism is $\ldots$.

  1. 1.5
  2. $\sqrt{3}$
  3. $\sqrt{2}$
  4. 1.65

Answer: (c)

Solution

Given $\mu_p = \mu_s \frac{\sin \left( \frac{A + \delta_{\min}}{2} \right)}{\sin \left( \frac{A}{2} \right)} = \frac{\sin \frac{3A}{4}}{\sin \frac{A}{2}}$. $$\mu_p = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} = \sqrt{2}$$

Question 43

Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct

Refer to the logic circuit given below. For two inputs (A = 1, B = 1) and (A = 0, B = 1), output (Y) will be $\ldots$.

  1. 1, 0 respectively
  2. 0, 1 respectively
  3. 0, 0 respectively
  4. 1, 1 respectively
  5. 0, 0

Answer: (c)

Solution

The expression for the output is given by $$y = \overline{\overline{A} + B + \overline{A}B}$$ This simplifies to $$= \overline{\overline{A} + B} = \overline{A} \cdot \overline{B}$$ Thus, $$= \overline{A} \overline{B}$$ For $A = 1$, $B = 1$, $y = 0$. For $A = 0$, $B = 1$, $y = 0$.

Question 44

Physics · Motion in a Straight Line · Single correct

The velocity at which 6 kg mass (shown in figure) strikes the ground when it is released from a height of 6 m above the ground is $\ldots$ m/s. Assume pulley is massless and string is light and inextensible (Take $g = 10 \, \mathrm{m/s^2}$).

  1. 7.74
  2. 7.20
  3. 6.55
  4. 4.50

Answer: (a)

Solution

The work done by gravity $W_G$ is equal to the change in kinetic energy $\Delta K$. The equation is given by: $$m_B gh - m_A gh = \frac{1}{2} (m_A + m_B) V^2$$ Substituting the values: $$(6 - 2) \times 10 \times 6 = \frac{1}{2} (8) v^2$$ Solving for $v$: $$v = \sqrt{60} \, \mathrm{m/s}$$ $$v = 7.746 \, \mathrm{m/s}$$

Question 45

Physics · Wave Optics · Single correct

In a Young double slit experiment, the wavelength of incident light is $6000\,\mathrm{\AA}$, the separation between slits $S_1$ and $S_2$ is 5 cm, and the distance between slits plane and screen is 50 cm, as shown in the figure below. If the resultant intensity at $P$ is equal to the intensity due to individual slits, the path difference between interfering waves is $\ldots \mathrm{\AA}$.

  1. 4000
  2. 3000
  3. 2000
  4. 1000

Answer: (c)

Solution

Given $I_P = I_0 = 4I_0 \cos^2 \frac{\Delta \phi}{2}$. $\cos \frac{\Delta \phi}{2} = \frac{1}{2}$. $\frac{\Delta \phi}{2} = \frac{\pi}{3}$. $\frac{2\pi}{\lambda} = \Delta x = \Delta \phi = \frac{2\pi}{3}$. $\Delta x = \frac{\lambda}{3} = 2000 A^\circ$.

Question 46

Physics · Laws of Motion · Numerical

A block takes $t$ time to slide down a plane inclined at $45^\circ$ to the horizontal. If the surface is made smooth (frictionless), the block takes time $\frac{t}{2}$ to slide down the plane. The coefficient of friction between the block and the inclined plane is $\left( \frac{\alpha}{100} \right)$. The value of $\alpha$ is $\ldots$.

Answer: 75

Solution

For a frictionless surface, $a_2 = g \sin \theta$. $$ \ell = \frac{1}{2} a_2 \frac{t^2}{4} $$ $$ a_1 t^2 = a_2 \frac{t^2}{4} $$ $$ g(\sin 45^\circ - \mu \cos 45^\circ) = \frac{g \sin 45^\circ}{4} $$ $$ 1 - \mu = \frac{1}{4} $$ $$ \mu = \frac{3}{4} = \frac{\alpha}{100} $$ $$ \alpha = 75 $$

Question 47

Physics · Dual Nature of Radiation and Matter · Numerical

The de Broglie wave length for an electron accelerated through the potential difference $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by 50$\%$. If $(V_1/V_2) = (9/\alpha)$, then the value of $\alpha$ is $\ldots$.

Answer: 4

Solution

Given $\lambda_1 = \frac{h}{mv} = \frac{h}{\sqrt{2meV_1}} = \lambda$. $\lambda_2 = \frac{h}{\sqrt{2meV_2}} = \frac{3\lambda}{2} = \frac{3}{2} \frac{h}{\sqrt{2meV_1}}$. $V_2 = \frac{4}{9} \cdot V_1$. $\frac{V_1}{V_2} = \frac{9}{4} \Rightarrow \alpha = 4$.

Question 48

Physics · Current Electricity · Numerical

A moving coil of a galvanometer when shunted with $2\,\Omega$ resistance gives a full scale deflection for a current of $500\,\mathrm{mA}$. When a resistance of $470\,\Omega$ is connected in series it gives a full scale deflection for $10\,\mathrm{V}$ potential applied on it. The value of resistance of galvanometer coil is $\ldots\,\Omega$.

Answer: 50

Solution

Given $i_g R_g = 2 \left( \frac{1}{2} - i_g \right)$. $i_g = \frac{1}{R_g + 2}$. $10 = (R_g + 470) i_g$. $10 = \frac{R_g + 470}{R_g + 2}$. $10 R_g + 20 = R_g + 470$. $R_g = 50 \, \Omega$.

Question 49

Physics · Current Electricity · Numerical

Two cells of emfs 1V and 2V and internal resistance $2\Omega$ and $1\Omega$, respectively connected in parallel, gave current of 1A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be $\frac{\alpha}{5}$ A. The value of $\alpha$ is $\ldots$.

Answer: 3

Solution

The equivalent circuit is shown with $E_{eq}$ and $r_{eq}$. For the first circuit, $$E_{eq} = \frac{1}{2} + \frac{2}{1} = \frac{5}{3}$$ The equivalent resistance is $$r_Q = \frac{3}{2}$$ The current $i$ is given by $$i = \frac{E_{eq}}{r_{eq} + R} = \frac{\frac{5}{3}}{\frac{3}{2} + R} = 1$$ Solving for $R$, $$\frac{5}{3} = \frac{3}{2} + R$$ Thus, $$R = \frac{1}{6} \, \Omega$$ For the second circuit, $$r_{eq} = \frac{3}{2}$$ The equivalent voltage is $$E_{eq} = \frac{2}{\frac{1}{1} + \frac{1}{2}} = 1$$ The current $i$ is given by $$i = \frac{E_{eq}}{r_{eq} + R} = \frac{1}{\frac{3}{2} + \frac{1}{6}} = \frac{3}{5}$$ Therefore, $$\alpha = 3$$

Question 50

Physics · Ray Optics and Optical Instruments · Fill in the blank

A concave mirror of focal length $10\,\mathrm{\ cm}$ froms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is ____ $\mathrm{\ cm}$.

Answer: 10

Solution

Case I: $v = 2u$ $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{-2u} + \frac{1}{-u} = \frac{1}{-f}$$ $$\frac{3}{2u} = \frac{1}{f}$$ $$u_1 = \frac{3f}{2}$$ Case 2: $v = -2u$ $$\frac{1}{+2u} + \frac{1}{-u} = \frac{1}{-f}$$ $$\frac{-1}{2u} = \frac{1}{-f}$$ $$u_2 = \frac{f}{2}$$ $\Delta u = f = 10 \, \mathrm{cm}$

Chemistry

Question 51

Chemistry · Some Basic Concepts of Chemistry · Single correct

Which of the following contain the same number of atoms? (Given: Molar mass in $\mathrm{g\,mol^{-1}}$ of H, He, O and S are 1, 4, 16 and 32 respectively) A. $2\,\mathrm{g}$ of $\mathrm{O_2}$ gas \ B. $4\,\mathrm{g}$ of $\mathrm{SO_2}$ gas \ C. $1400\,\mathrm{mL}$ of $\mathrm{O_2}$ at STP \ D. $0.05\,\mathrm{L}$ of $\mathrm{He}$ at STP \ E. $0.0625\,\mathrm{mol}$ of $\mathrm{H_2}$ gas Choose the correct answer from the options given below:

  1. A and B only
  2. B and C only
  3. C and D only
  4. A, C and E only

Answer: (d)

Solution

The number of atoms in $\mathrm{O_2}$ is given by $$N_{atoms in \mathrm{O_2}} = \frac{m_{\mathrm{O_2}}}{M_{\mathrm{O_2}}} N_A = \frac{2}{32} \times 2N_A = \frac{N_A}{8} = 0.125N_A$$ The number of atoms in $\mathrm{SO_2}$ is given by $$N_{atoms in \mathrm{SO_2}} = \frac{m_{\mathrm{SO_2}}}{M_{\mathrm{SO_2}}} = 3N_A = \frac{4}{64} \times 3N_A = \frac{3N_A}{16} = 0.1875N_A$$ The number of atoms in $\mathrm{O_2}$ at STP is given by $$N_{atoms in \mathrm{O_2}} = \frac{V_{\mathrm{O_2}}}{22.7 lit} N_A = \frac{1400 ml}{22400 ml} \times 2N_A = 0.125 \, N_A$$ The number of atoms in $\mathrm{He(g)}$ at STP is given by $$N_{atoms in \mathrm{He(g)}} = \frac{V_{\mathrm{He}}}{22.4 lit} N_A = \frac{0.05}{22.4} N_A = 0.002N_A$$ The number of atoms in $\mathrm{H_2(g)}$ is given by $$N_{atoms in \mathrm{H_2(g)}} = moles \times atomicity \times N_A = 0.0625 \times 2N_A = 0.125 \, N_A$$

Question 52

Chemistry · Structure of Atom · Single correct

The Bohr radius of a hydrogen like species is $70.53\,\mathrm{pm}$. The species and the stationary state(n) are respectively (Given: Hydrogen atom Bohr radius is $52.9\,\mathrm{pm}$)

  1. $\mathrm{Li}^{2+}$, 3
  2. $\mathrm{He}^{+}$, 3
  3. $\mathrm{He}^{+}$, 2
  4. $\mathrm{Li}^{2+}$, 2

Answer: (d)

Solution

Given $r_{species} = a_0 \frac{n^2}{Z}$. $$70.53 = 52.9 \left( \frac{n^2}{Z} \right)$$ $$\frac{n^2}{Z} = \frac{70.53}{52.9} = 1.33 = \frac{4}{3}$$ Data will satisfy if $n = 2$ and $z = 3$. $$Li^{+2}, 2 \Rightarrow \frac{4}{3} = 1.33$$

Question 53

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Given below are two statements: Statement I : The number of compounds among $\mathrm{SO_2}$, $\mathrm{SO_3}$, $\mathrm{SF_4}$, $\mathrm{SF_6}$ and $\mathrm{H_2S}$ in which sulphur does not obey the Octet rule is 3. Statement II : Among $[\mathrm{H_2O}, \mathrm{ClF_3}, \mathrm{SF_4}], [\mathrm{NH_3}, \mathrm{BrF_5}, \mathrm{SF_4}], [\mathrm{BrF_5}, \mathrm{ClF_3}, \mathrm{XeF_4}]$ and $[\mathrm{XeF_4}, \mathrm{ClF_3}, \mathrm{H_2O}]$, the number of sets in which all the molecules have one lone pair of electrons on the central atom is 1. In the light of the above statements, choose the correct answer from the options given below:

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true Statement II is false
  4. Statement I is false Statement II is true

Answer: (d)

Solution

SO_2, SO_3, SF_6 $\&$ SF_4 do not follow octet rule. NH_3, $\mathrm{BrF_5}$, SF_4 have one lone pair on central atom. $\mathrm{ClF_3}$, $\mathrm{XeF_4} \& \mathrm{H_2O}$ have two lone pair on central atom.

Question 54

Chemistry · Thermodynamics · Single correct

Match List-I with List-II Given $V_1$ and $V_2$ are initial and final volumes respectively: Choose the correct answer from the options given below :

  1. A-II, B-III, C-I, D-IV
  2. A-II, B-I, C-IV, D-III
  3. A-II, B-I, C-III, D-IV
  4. A-I, B-II, C-III, D-IV

Answer: (c)

Solution

(A) For isothermal process, $\Delta U = 0$ $$W = -nRT \ln \left( \frac{V_2}{V_1} \right)$$ so $q = nRT \ln \left( \frac{V_2}{V_1} \right)$ (B) Free expansion which is isothermal must be adiabatic, so $q = 0$ (C) Irreversible process $W = -P_{ext} \left[ V_2 - V_1 \right]$ (D) Cyclic process, change in state function (S) so change in entropy must be zero. $$dS = \frac{dq_{rev}}{T}$$ $$\oint dS = 0$$

Question 55

Chemistry · Solutions · Single correct

Given below are two statements : Statement I : $\mathrm{H_2O}$ molecules move from the chamber 1 to chamber 2. Statement II : The osmotic pressure of a solution prepared by dissolving 50 mg of potassium sulphate (molar mass = 174 g/mol) in 2 L of water (at 27°C) is 0.0107 bar. (Given : $R = 0.083 \, \mathrm{dm^3 \, bar \, K^{-1} \, mol^{-1}}$ and assume complete dissociation of electrolyte) In the light of the above statements, choose the correct answer from the options given below :

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is false but Statement II is true

Answer: (d)

Solution

On side (I) $C_1 = \frac{18/180}{0.1} = 1\, \mathrm{M}$. On side (II) $C_2 = \frac{30/180}{250/1000} = \frac{2}{3}\, \mathrm{M}$; $\{ \pi_1 > \pi_2 \}$ Hence water will flow from chamber-II to Chamber-I. $\pi = iCRT$; $i = 3$ for $\mathrm{K_2SO_4}$ And $$\left( \pi = 3 \times \frac{50 \times 10^{-3} / 174}{2} \right) \times 0.083 \times 300$$ $= 0.0107\, \mathrm{bar}$.

Question 56

Chemistry · Equilibrium · Single correct

Given is a concentrated solution of a weak electrolyte $\mathrm{A}_x\mathrm{B}_y$ of concentration 'c' and dissociation constant 'K'. the degree of dissociation is given by:

  1. $\left[ K \times c^{x+y-1} x^x y^y \right]^{x+y}$
  2. $\left( \frac{K}{c^{x+y-1} x^x y^y} \right)^{\frac{1}{x+y}}$
  3. $\left( \frac{c^{x+y-1} x^x y^y}{K} \right)^{x+y}$
  4. $\left( \frac{c^{x+y-1} x^x y^y}{K} \right)^{\frac{1}{x+y}}$

Answer: (b)

Solution

The reaction is given by $\mathrm{A_xB_y} \rightleftharpoons \mathrm{xA^{y+} + yB^{x-}}$. The concentrations are $\mathrm{C(1 - \alpha)}$, $\mathrm{xC\alpha}$, and $\mathrm{yC\alpha}$. The equilibrium constant $K$ is given by $$K = \frac{(xC\alpha)^x (xC\alpha)^y}{C(1 - \alpha)}.$$ As $\alpha \ll 1$, we have $(1 - \alpha) \simeq 1$. Therefore, $$K = x^x y^y C^{x+y-1} \alpha^{x+y}.$$ Solving for $\alpha$, we get $$\alpha = \left( \frac{K}{C^{x+y-1} x^x y^y} \right)^{\frac{1}{x+y}}.$$

Question 57

Chemistry · Redox Reactions · Single correct

For a general redox reaction Anode $\mathrm{Red}_1 \rightarrow \mathrm{Ox}_1^{n_1^+} + n_1 e^-$ Cathode $\mathrm{Ox}_2 + n_2 e^- \rightarrow \mathrm{Red}_2^{n_2^-}$ Which of the following statement is incorrect ?

  1. The overall reaction can be written as $n_2 \mathrm{Red}_1 + n_1 \mathrm{Ox}_2 \rightleftharpoons n_2 \mathrm{Ox}_1^{n_1^+} + n_1 \mathrm{Red}_2^{n_2^-}$
  2. The electrons do not appear in the overall reaction because electrons produced at the anode are consumed at the cathode.
  3. Here n is the number of electrons transferred in redox reaction.
  4. If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.

Answer: (d)

Solution

Anode: $$\left[ \mathrm{Red}_1 \rightarrow \mathrm{Ox}_1^{n_1^+} + n_1 e^- \right] \times n_2$$ Cathode: $$\left[ \mathrm{Ox}_2 + n_2 e^- \rightarrow \mathrm{Red}_2^{n_2^-} \right] \times n_1$$ Overall reaction: $$n_2 \mathrm{Red}_1 + n_1 \mathrm{Ox}_2 \rightarrow n_2 \mathrm{Ox}_1^{n_1^+} + n_1 \mathrm{Red}_2^{n_2^-}$$ The electrons does not appear in overall reaction. $$E = E^\circ - 2.303 \frac{RT}{nF} \log Q$$ $$\frac{(E - E^\circ)}{RT/F} = -\frac{2.303}{n} \log Q$$ $$Slope = -\frac{2.303}{n}$$ Straight line passing through origin. Electrical work = Charge $\times$ Potential difference

Question 58

Chemistry · Classification of Elements and Periodicity in Properties · Single correct

In a period, the first ionisation enthalpy of the element at extreme left and the negative electron gain enthalpy of the extreme right element, except noble gases, are respectively.

  1. lowest and lowest
  2. highest and lowest
  3. lowest and highest
  4. highest and highest

Answer: (c)

Solution

Left most element in a period will be of group 1 which will have lowest IE, while right most element will be of group 17 which will have most negative $\Delta H_{eg}$.

Question 59

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Given below are two statements : Statement I : $\mathrm{F_2O < H_2O < Cl_2O}$ is the correct trend in terms of bond angle. Statement II : $\mathrm{SiF_4, SnF_4}$ and $\mathrm{PbF_4}$ are ionic in nature. In the light of the above statements, choose the correct answer from the options given below :

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is false but Statement II is true

Answer: (c)

Solution

Specie Bond angle $\mathrm{OCl_2} 111^\circ$ $\mathrm{H_2O} 104.5^\circ$ $\mathrm{OF_2} 103^\circ$ $\mathrm{SiF_4}$ is covalent.

Question 60

Chemistry · The d-and f-Block Elements · Single correct

The correct order of first ($\Delta_{i}H_{1}$) and second ($\Delta_{i}H_{2}$) ionisation enthalpy values of Cr and Mn are: A. $\Delta_{i}H_{1}$: Cr $>$ Mn B. $\Delta_{i}H_{2}$: Cr $>$ Mn C. $\Delta_{i}H_{1}$: Mn $>$ Cr D. $\Delta_{i}H_{2}$: Mn $>$ Cr Choose the \textbf{correct answer} from the options given below:

  1. A and B only
  2. B and C only
  3. A and D only
  4. C and D only

Answer: (b)

Solution

Cr - [$\mathrm{Ar}$]3d^5 4s^1 $\newline$ Mn - [$\mathrm{Ar}$]3d^5 4s^2 $\newline \mathrm{IE}_1$ - $\mathrm{Mn}$ > $\mathrm{Cr} \newline \mathrm{IE}_2$ - $\mathrm{Mn}$ < $\mathrm{Cr}$

Question 61

Chemistry · Co-ordination Compounds · Single correct

Which of the following sequence of hybridisation, geometry and magnetic nature are correct for the given coordination compounds? A. $[NiCl_4]^{2-}$ – $sp^3$, tetrahedral, paramagnetic B. $[Ni(NH_3)_6]^{2+}$ – $sp^3d^2$, octahedral, paramagnetic C. $[Ni(CO)_4]$ – $sp^3$, tetrahedral, paramagnetic D. $[Ni(CN)_4]^{2-}$ – $dsp^2$, square planar diamagnetic Choose the correct answer from the options given below:

  1. A, B, C and D
  2. B, C and D only
  3. A, C and D only
  4. A, B and D only

Answer: (d)

Solution

[$\mathrm{Ni(CO)_4}$] is sp^3, tetrahedral and diamagnetic

Question 62

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Given below are two statements: Statement I: A mixture of $C_{12}H_{22}O_{11}$ (sugar) and NaCl can be separated by dissolving sugar in alcohol, due to differential solubility. Statement II: Rose essence from rose petals is separated by steam distillation due to its high volatility and insolubility in $H_2O$. In the light of the above statements, choose the correct answer from the options given below:

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true Statement II is false
  4. Statement I is false Statement II is true

Answer: (a)

Solution

Sugar is soluble in alcohol not NaCl. So they are separated by solvent extraction (or differential extraction). Rose essence oil is steam volatile so it is separated by steam distillation.

Question 63

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Shown below is the structure of methyl acetate with three different $\alpha$, $\beta$ and $\gamma$ carbon-oxygen bonds. The correct order of bond length of these bonds is:

  1. $\alpha > \beta > \gamma$
  2. $\alpha < \beta < \gamma$
  3. $\alpha = \beta = \gamma$
  4. $\alpha < \beta = \gamma$

Answer: (b)

Solution

RS1 is more stable than RS2. Therefore, the contribution of RS1 is more than RS2 in its R.H. Therefore, order of bond length $\alpha < \beta < \gamma$.

Question 64

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

'x' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at $333\,\mathrm{K}$. 'y' is the product which is obtained by the reaction of ethane nitrile with methyl magnesium bromide in dry ether followed by hydrolysis. IUPAC name of product obtained from 'x' and 'y' in the presence of barium hydroxide followed by heating is:

  1. 2-Methylpent-4-en-3-one
  2. 4-Methylpent-3-en-2-one
  3. 4-Methylpent-1-ene
  4. 2-Methylpent-3-one

Answer: (b)

Solution

Propyne is treated with $\mathrm{HgSO_4}$ and dilute $\mathrm{H_2SO_4}$ (Kucherov reaction) to form compound $[x]$. Ethanenitrile is treated with (1) $\mathrm{MeMgBr}$ and (2) $\mathrm{H_3O^+}$ to form compound $[y]$. Compounds $[x]$ and $[y]$ undergo aldol condensation in the presence of $\mathrm{Ba(OH)_2}$ and heat $\Delta$ to form compound $[z]$, which is 4-Methylpent-3-en-2-one.

Question 65

Chemistry · Haloalkanes and Haloarenes · Single correct

An optically active alkyl bromide $\mathrm{C_4H_9Br}$, reacts with ethanolic KOH to form major compound $[A]$ which reacts with bromine to give compound $[B]$. Compound $[B]$ reacts with ethanolic KOH and sodamide to give compound $[C]$. One molecule of water adds to compound $[C]$ on warming with mercuric sulphate and dilute sulphuric acid at $333 \, \mathrm{K}$ to form compound $[D]$. The functional group in compound $D$ will be confirmed by:

  1. Haloform test
  2. Lucas test
  3. Silver mirror test
  4. Benedict test

Answer: (a)

Solution

The reaction starts with the given compound reacting with alcoholic KOH under heat to form compound (A). Compound (A) is then treated with $\mathrm{Br_2/CCL_4}$ to form compound (B). Compound (B) undergoes two steps: (i) treatment with alcoholic KOH and (ii) treatment with $\mathrm{NaNH_2/\Delta}$ to form compound (C). Compound (C) is then reacted with $\mathrm{HgSO_4}$ and dilute $\mathrm{H_2SO_4}$ to form the final product, $\mathrm{CH_3-CH_2-C(=O)-CH_3}$. The final product gives a positive haloform test.

Question 66

Chemistry · Haloalkanes and Haloarenes · Single correct

Consider the following reaction. Statement I: In the above reaction, product formed will be a mixture of benzyl alcohol and iodobenzene. Statement II: In the above reaction, the $-\mathrm{O}-\mathrm{CH}_2-$ bond is cleaved to give the product. In the light of the above statements, choose the correct answer from the options given below:

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is false but Statement II is true

Answer: (d)

Solution

The reaction is as follows: $$Ph–O–CH_2–Ph \xrightarrow{HI} Ph–OH + I–CH_2–Ph$$ This results in phenol and benzyl iodide.

Question 67

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Consider the following organic reaction sequence. Choose the final product (X) from the following (consider the major product in all intermediate reactions)

  1. Benzene
  2. Phenol
  3. Propanol
  4. Chlorobenzene

Answer: (a)

Solution

The reaction sequence is as follows: $$CH_3CHCOOH \xrightarrow{NH_3, \Delta} CH_3CH_2C \equiv NH_2$$ Then, with NaOH and Br$_2$: $$CH_3CH_2NH_2 \xrightarrow{HNO_2, H_2O/0^\circC}$$ Next, the reaction with Ph-$N_2^+Cl^-$ at $0^\circC$: $$CH_3CH_2OH \xrightarrow{Ph-N_2^+Cl^-} Benzene$$ The final products are: $$CH_3C \equiv H + N_2 + HCl$$

Question 68

Chemistry · Amines · Single correct

The number of compounds from the following which can undergo reaction with $\mathrm{Br_2/KOH/(alcoholic)}$ to give respective products and these respective products can also be obtained separately by Gabriel phthalimide reaction is:

  1. 5
  2. 4
  3. 3
  4. 6

Answer: (c)

Solution

Aniline is not formed by Gabriel phthalimide synthesis. (ii) $$Ph - CH_2 - C - NH_2 \xrightarrow{Br_2 + KOH \, \Delta} Ph - CH_2 - NH_2$$ (iii) $$CH_3 - C - NH_2 \xrightarrow{Br_2 + KOH \, \Delta} CH_3 - NH_2$$ (iv) $$\begin{array}{c} O \\ || \\ C - NH - CH_2 - CH_3 \xrightarrow{Br_2 + KOH \, \Delta} \end{array}$$ No 1° amine is formed. (v) $$\begin{array}{c} O \\ || \\ C - NH - CH_3 \xrightarrow{Br_2 + KOH \, \Delta} \end{array}$$ No 1° amine is formed. (vi) $$\begin{array}{c} O \\ || \\ C - NH_2 \xrightarrow{Br_2 + KOH \, \Delta} \end{array} NH_2$$

Question 69

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Consider the following reactions. Total number of electrons in the $\pi$ bonds and lone pair of electrons in the product (X) is:

  1. 12
  2. 16
  3. 14
  4. 18

Answer: (d)

Solution

The reaction sequence involves the conversion of the given compound using Red P and HI with heat, followed by treatment with $\mathrm{V_2O_5}$ at $773 \, \mathrm{K}$ and $10-20 \, \mathrm{atm}$. The product is then reacted with $\mathrm{Ph-C(O)Cl}$ in the presence of $\mathrm{AlCl_3}$. The final product contains a benzophenone structure with a carbonyl group and lone pairs on the oxygen atom. Total number of $\pi$ and lone pair electrons in final product $= 18$

Question 70

Chemistry · Analytical Chemistry · Single correct

Treatment of a gas 'X' with a freshly prepared ferrous sulphate solution gives a compound 'Y' as a brown ring. The compounds X and Y are.

  1. NO and [$\mathrm{Fe(NO)}$]$\mathrm{SO}_4$
  2. $\mathrm{NO}_2$ and [$\mathrm{Fe(NO}_2$)]$\mathrm{SO}_4$
  3. $\mathrm{N}_2 \mathrm{O}$ and [$\mathrm{Fe(N}_2 \mathrm{O}$)]$\mathrm{SO}_4$
  4. $\mathrm{N}_2 \mathrm{O}_4$ and [$\mathrm{Fe(N}_2 \mathrm{O}_4$)]$\mathrm{SO}_4$

Answer: (a)

Solution

NO + $\mathrm{FeSO_4} \xrightarrow{dil. \mathrm{H_2SO_4}}$ [$\mathrm{Fe^{I}(NO)}$]$\mathrm{SO_4}$

Question 71

Chemistry · Co-ordination Compounds · Numerical

An excess of $\mathrm{AgNO_3}$ is added to $100 \, \mathrm{mL}$ of a $0.05 \, \mathrm{M}$ solution of tetraaquadichloridochromium (III) chloride. The number of moles of $\mathrm{AgCl}$ precipitated will be $\ldots \times 10^{-3}$. (Nearest integer)

Answer: 5

Solution

[$\mathrm{Cr(H_2O)_4Cl_2}$]$\mathrm{Cl}$ + $\mathrm{AgNO_3} \rightarrow \mathrm{AgCl} \downarrow$ + [$\mathrm{Cr(H_2O)_4Cl_2}$]$\mathrm{NO_3} \newline \Rightarrow$ 1 mole of complex will precipitate 1 mole of $\mathrm{AgCl}$. $\newline \Rightarrow$ Given moles of complex = $\newline$ M $\times$ V = (0.05 $\times$ 100 $\times$ 10^{-3}) moles $\newline \Rightarrow$ Moles of $\mathrm{AgCl}$ = 5 $\times$ 10^{-3} moles $\newline \Rightarrow$ x = 5

Question 72

Chemistry · Hydrocarbons · Numerical

An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with $\mathrm{Cl}_2/h\nu$, (Y) gives only one monochlorinated product (Z). The total number of primary carbon atoms in (Y) is$\ldots$.

Answer: 4

Solution

Given $\mathrm{C}_n\mathrm{H}_{2n+2} + \left( \frac{3n+1}{2} \right) \mathrm{O}_2 \rightarrow n\mathrm{CO}_2 + (n+1)\mathrm{H}_2\mathrm{O}$. $$\frac{3n+1}{2} = 8$$ $$3n + 1 = 8 \times 2$$ $$n = 5 (\therefore \mathrm{C}_5\mathrm{H}_{12})$$ Neopentane (X): $$\begin{array}{c} \mathrm{CH}_3 \\ | \\ \mathrm{CH}_3 - \mathrm{C} - \mathrm{CH}_3 \\ | \\ \mathrm{CH}_3 \end{array} \xrightarrow{\mathrm{Br}_2/\mathrm{hv}} \begin{array}{c} \mathrm{CH}_3 \\ | \\ \mathrm{CH}_3 - \mathrm{C} - \mathrm{CH}_2 - \mathrm{Br} \\ | \\ \mathrm{CH}_3 \end{array}$$ Only one mono-bromo product. Number of primary carbons = 4

Question 73

Chemistry · Redox Reactions · Numerical

$500\,\mathrm{mL}$ of $0.2\,\mathrm{M} \mathrm{MnO}_4^-$ solution in basic medium when mixed with $500\,\mathrm{mL}$ of $1.5\,\mathrm{M}$ KI solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard x $\mathrm{M}$ thiosulphate solution in presence of starch till the end point. If $300\,\mathrm{mL}$ of thiosulphate was consumed, then the value of x is $\ldots$

Answer: 1

Solution

The reaction is given by: $$3e^- + \mathrm{MnO_4^-} + 2\mathrm{H_2O} \rightarrow \mathrm{MnO_2} + 4\mathrm{OH^-}$$ $$2\mathrm{I^-} \rightarrow \mathrm{I_2} + 2e^-$$ $$\mathrm{I_2} + 2\mathrm{S_2O_3^{2-}} \rightarrow 2\mathrm{I^-} + \mathrm{S_4O_6^{2-}}$$ Gram equivalent of $\mathrm{MnO_4^-}$: $$= 0.2 \times \frac{500}{1000} \times 3 = 0.3 (Limiting reagent)$$ Gram equivalent of KI: $$= 1.5 \times \frac{500}{1000} \times 1 = 0.75$$ Gram equivalent of $\mathrm{MnO_4^-}$ is equal to gram equivalent of hypo: $$0.3 = x \times \frac{300}{1000} \times 1$$ Solving for $x$ gives: $$x = 1$$

Question 74

Chemistry · Equilibrium · Numerical

In a closed flask at $600\,\mathrm{K}$, one mole of $\mathrm{X}_2\mathrm{Y}_4(g)$ attains equilibrium as given below: $$ \mathrm{X}_2\mathrm{Y}_4(g) \rightleftharpoons 2\,\mathrm{XY}_2(g) $$ At equilibrium, $75\%$ $\mathrm{X}_2\mathrm{Y}_4(g)$ was dissociated and the total pressure is $1\,\mathrm{atm}$. The magnitude of $\Delta_r G^\Theta$ (in $\mathrm{kJ\,mol^{-1}}$) at this temperature is $\ldots$ (Nearest Integer). (Given: $R = 8.3\,\mathrm{J\,mol^{-1}\,K^{-1}}$; $\ln 10 = 2.3$, $\log 2 = 0.3$, $\log 3 = 0.48$, $\log 5 = 0.69$, $\log 7 = 0.84$)

Answer: 8

Solution

The reaction is $X_2Y_4(g) \rightleftharpoons 2XY_2(g)$. At $t=0$, the pressure is $P^\circ$. At $t=t$, the pressures are $P^\circ (1 - \alpha)$ and $2P^\circ \alpha$. The equation $P^\circ (1 + \alpha) = 1$ gives $\alpha = \frac{3}{4}$ and $P^\circ = \frac{4}{7}$. The equilibrium constant $k_P$ is given by $$k_P = \frac{4P^\circ \alpha^2}{1 - \alpha^2} = \frac{36}{7}.$$ The change in Gibbs free energy is $$\Delta G^\circ = -RT \ln k_P.$$ The magnitude of the change is $$|\Delta G^\circ| = 8.314 \times 600 \ln \left( \frac{36}{7} \right)$$ which equals $8169.5 \, \mathrm{J \, mol^{-1}}$ or $8.169 \, \mathrm{kJ/mol}$.

Question 75

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11} \, \mathrm{s}^{-1}) \, e^{-\frac{28000 \, K}{T}}$. The activation energy of reaction is $\ldots$ kJ mol$^{-1}$. (Nearest Integer) Given : $R = 8.3 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$

Answer: 232

Solution

Given $K = (5.5 \times 10^{11} \, \mathrm{S^{-1}}) \exp\left(\frac{28000 \, \mathrm{K}}{T}\right)$. $K = A e^{-E_a/RT}$ $\frac{E_a}{R} = 28000$ $$E_a = \frac{28000 \times 8.3}{1000} \, \mathrm{kJ/mol} = 232.4 \, \mathrm{kJ/mol}$$