JEE Main 2 April 2026 Shift 2 question paper with solutions

JEE Main 2 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 · Complex Numbers and Quadratic Equations · Single correct

Let $\alpha$, $\beta$ be the roots of the equation $x^2 - 3x + r = 0$, and $\frac{\alpha}{2}$, $2\beta$ be the roots of the equation $x^2 + 3x + r = 0$. If the roots of the equation $x^2 + 6x = m$ are $2\alpha + \beta + 2r$ and $\alpha - 2\beta - \frac{r}{2}$, then $m$ is equal to :-

  1. -135
  2. -567
  3. 135
  4. 567

Answer: (d)

Solution

Given $\alpha + \beta = 3$ and $\frac{\alpha}{2} + 2\beta = -3$. On solving, we get $\alpha = 6$; $\beta = -3$. Product of roots $= \alpha \beta = r \Rightarrow r = -18$. Now for $x^2 + 6x - m = 0$. Product of roots $= -m$. $$= (2\alpha + \beta + 2r)(\alpha - 2\beta - \frac{r}{2})$$ $$\Rightarrow -m = (-27)(21)$$ $$\Rightarrow m = 567$$

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

Let the circle $C_1 : |z| = r$ and $C_2 : |z - 3 - 4i| = 5$, $z \in \mathbb{C}$, be such that $C_2$ lies within $C_1$. If $z_1$ moves on $C_1$, $z_2$ moves on $C_2$ and $\min |z_1 - z_2| = 2$, then $\max |z_1 - z_2|$ is equal to :-

  1. 12
  2. 17
  3. 22
  4. 24

Answer: (c)

Solution

Given $|z| = r$ and $\mathrm{OA} = r$. The minimum $|z_1 - z_2| = 2$. Therefore, $r - 10 = 2 \Rightarrow r = 12$. The maximum value of $\mathrm{CB}$ is $|z_1 - z_2| = r + 10 = 12 + 10 = 22$.

Question 3

Maths · Determinants · Single correct

If the system of equations x + 5y + 6z = 4, 2x + 3y + 4z = 7, x + 6y + az = b has infinitely many solutions, then the point (a, b) lies on the line

  1. y - x = 3
  2. x - y = 3
  3. x + y = 11
  4. x + y = 12

Answer: (b)

Solution

Given $\begin{vmatrix} 1 & 5 & 6 \\ 2 & 3 & 4 \\ 1 & 6 & a \end{vmatrix} = 0$ $\Rightarrow a = \frac{50}{7}$ $D_z = 0 \Rightarrow \begin{vmatrix} 1 & 5 & 4 \\ 2 & 3 & 7 \\ 1 & 6 & b \end{vmatrix} = 0$ $\Rightarrow b = \frac{29}{7}$ Therefore, $(a, b)$ lies on $x - y = 3$

Question 4

Maths · Sequences and Series · Single correct

Let $a_1, a_2, a_3, \ldots$ be an A.P. and $g_1 = a_1, g_2, g_3, \ldots$ be an increasing G.P. If $a_1 = a_2 + g_2 = 1$ and $a_3 + g_3 = 4$, then $a_{10} + g_5$ is equal to:

  1. 81
  2. 76
  3. 62
  4. 55

Answer: (d)

Solution

A.P. : 1, a_2, a_3, $\ldots$ G.P. : 1, g_2, g_3, $\ldots$ $a_2 + g_2 = 1 \Rightarrow 1 + d + r = 1 \Rightarrow d + r = 0$ $1 + 2d + r^2 = 4 \Rightarrow r^2 - 2r - 3 = 0$ $\Rightarrow r = 3, -1$ (reject) Therefore, $d = -3$ Thus, $a_{10} = 1 + 9d = 1 + 9(-3) = -26$ $g_5 = 1 \cdot r^4 = 3^4 = 81$ Therefore, $a_{10} + g_5 = 55$

Question 5

Maths · Sequences and Series · Single correct

The sum $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \ldots$ up to 8 terms, is:

  1. 70
  2. 71
  3. 72
  4. 73

Answer: (b)

Solution

Given $T_r = \frac{1^3 + 2^3 + 3^3 + \ldots r^3}{1 + 3 + 5 + \ldots + (2r-1)} = \frac{\left(\frac{r(r+1)}{2}\right)^2}{r^2}$. This simplifies to $$= \frac{r^2 + 2r + 1}{4}$$. The sum $S_n$ is given by $$S_n = \sum_{r=1}^{n} T_r$$ which becomes $$S_n = \frac{1}{4} \sum (r^2 + 2r + 1)$$. Simplifying further, $$= \frac{1}{4} \left[ \frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} + n \right]$$. For $S_8$, we have $$S_8 = \frac{1}{4} \left[ \frac{8 \times 9 \times 17}{6} + 8 \times 9 + 8 \right]$$ which simplifies to $$= \frac{1}{4} [204 + 72 + 8] = 71$$.

Question 6

Maths · Binomial Theorem · Single correct

If for $3\leq r\leq 30$, \[ ({}^{30}C_{30-r}) +3({}^{30}C_{31-r}) +3({}^{30}C_{32-r}) +({}^{30}C_{33-r}) ={}^{m}C_r, \] then $m$ equals:

  1. 31
  2. 32
  3. 33
  4. 34

Answer: (c)

Solution

The solution is as follows: $$^3C_3 \cdot {}^{30}C_{30-r} + {}^3C_2 \cdot {}^{30}C_{31-r} + {}^3C_1 \cdot {}^{30}C_{32-r} + {}^3C_0 \cdot {}^{30}C_{33-r}$$ This is equivalent to: $$= {}^{33}C_{33-r} \equiv {}^mC_r \equiv {}^mC_{m-r}$$ Thus, we find: $$\Rightarrow m = 33$$

Question 7

Maths · Permutations and Combinations · Single correct

Let $p_n$ denote the total number of triangles formed by joining the vertices of an $n$-side regular polygon. If $p_{n+1} - p_n = 66$, then the sum of all distinct prime divisors of $n$ is:

  1. 7
  2. 8
  3. 5
  4. 6

Answer: (c)

Solution

Given $P_n = \binom{n}{3}$. $P_{n+1} - P_n = 66$ $$\binom{n+1}{3} - \binom{n}{3} = 66$$ $$\Rightarrow \frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6} = 66$$ $$\Rightarrow \frac{n(n-1)}{6} [n+1-n+2] = 66$$ $$\Rightarrow n(n-1) = 132$$ $n = 12$ Prime divisors of 12 are 2 and 3. Sum $= 2 + 3 = 5$

Question 8

Maths · Probability · Single correct

A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is $\frac{m}{n}$, gcd $(m, n) = 1$, then $m + n$ is equal to:

  1. 53
  2. 55
  3. 107
  4. 105

Answer: (c)

Solution

P = (6 tail) + (4 tail + 1 head) + (2 tail + 2 head) + (3 head) $$P = \left( \frac{1}{2} \right)^6 + \frac{5!}{1!4!} \left( \frac{1}{2} \right)^5 + \frac{4!}{2!2!} \left( \frac{1}{2} \right)^4 + \left( \frac{1}{2} \right)^3$$ P = $\frac{1}{64}$ + $\frac{5}{32}$ + $\frac{3}{8}$ + $\frac{1}{8}$ $$P = \frac{1 + 10 + 24 + 8}{64} = \frac{43}{64} = \frac{m}{n}$$ m + n = 107

Question 9

Maths · Statistics · Single correct

The mean and variance of n observations are 8 and 16, respectively. If the sum of the first (n - 1) observations is 48 and the sum of squares of the first (n - 1) observations is 496, then the value of n is:

  1. 21
  2. 16
  3. 13
  4. 7

Answer: (d)

Question 10

Maths · Conic Sections · Single correct

Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines $x + (k - 1)y + 3 = 0$ and $2x + k^2y - 4 = 0$. If the line $x - y + 2 = 0$ intersects the circle at the points $A$ and $B$, then $(AB)^2$ is equal to:

  1. 10
  2. 27
  3. 18
  4. 34

Answer: (c)

Solution

Given the equations: $$x + (k-1)y + 3 = 0$$ $$2x + k^2y - 4 = 0$$ We have: $$\left( \frac{1}{1-k} \right) \left( \frac{2}{k^2} \right) = 1$$ Solving gives: $$2 = k^2 - k^3$$ $$k^3 - k^2 + 2 = 0$$ $$k = -1$$ Solving: $$2(x - 2y + 3) = 0$$ $$2x + y - 4 = 0$$ This simplifies to: $$-5y + 10 = 0$$ $$y = 2$$ The circle has center $(1, 2)$ and radius $\sqrt{5}$. So the circle equation is: $$(x - 1)^2 + (y - 2)^2 = 5$$ The chord is given by $x - y + 2 = 0$. The perpendicular distance $p$ is: $$p = \frac{1}{\sqrt{2}}$$ The length $\ell$ is: $$\ell = \sqrt{5 - \frac{1}{2}} = \frac{3}{\sqrt{2}}$$ The length $AB$ is: $$AB = \frac{6}{\sqrt{2}} \therefore AB^2 = 18$$

Question 11

Maths · Conic Sections · Single correct

Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the mid point of the line segment $PQ$ is $\left( \frac{1}{2}, -\frac{1}{2} \right)$. Then the area of the triangle $OPQ$ equals:

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

Answer: (c)

Solution

Equation of chord whose midpoint is $\left( \frac{1}{2}, -\frac{1}{2} \right)$ $$\frac{x}{x_1} + \frac{y}{y_1} = 2$$ $$2x - 2y = 2$$ $$x - y = 1$$ solving with $xy = 12$ $P(-3, -4)$, $Q(4, 3)$ $$Area (\triangle OPQ) = \frac{1}{2} \left| -3 \times 3 - (-4)(4) \right| = \frac{7}{2}$$

Question 12

Maths · Conic Sections · Single correct

Let the parabola $y = x^2 + px + q$ passing through the point $(1, -1)$ be such that the distance between its vertex and the x-axis is minimum. Then the value of $p^2 + q^2$ is:

  1. 2
  2. 4
  3. 5
  4. 8

Answer: (b)

Solution

The parabola passes through $(1, -1)$ so $$-1 = 1 + p + q$$ $$p + q = -2$$ $$q = -2 - p ...(1)$$ Distance from x-axis $$\frac{-D}{4a} = \frac{-(p^2 - 4q)}{4(1)} = \frac{4q - p^2}{4}$$ $$= \frac{4(-2 - p) - p^2}{4} from (1)$$ $$= \frac{-8 - 4p - p^2}{4} = \frac{-(p^2 + 4p + 8)}{4} = \frac{-((p + 2)^2 + 4)}{4}$$ Minimum at $p = -2 \Rightarrow q = 0$ $$p^2 + q^2 = 4 + 0 = 4$$

Question 13

Maths · Trigonometric Functions · Single correct

Let $\mathrm{P} = \{ \theta \in [0, 4\pi] : \tan^2 \theta \neq 1 \}$ and $\mathrm{S} = \{ a \in \mathbb{Z} : 2(\cos^8 \theta - \sin^8 \theta) \sec 2\theta = a^2, \theta \in \mathrm{P} \}$. Then $n(\mathrm{S})$ is :

  1. 0
  2. 1
  3. 2
  4. 3

Answer: (a)

Solution

Given $2(\cos^8 \theta - \sin^8 \theta) \sec 2\theta = a^2$. $2(\cos^4 \theta + \sin^4 \theta)(\cos^2 \theta + \sin^2 \theta)(\cos^2 \theta - \sin^2 \theta) \sec 2\theta = a^2$. $2(\cos^4 \theta + \sin^4 \theta) = a^2$. $2(1 - 2 \sin^2 \theta \cos^2 \theta) = a^2$. $$2 \left( 1 - \frac{\sin^2 2\theta}{2} \right) = a^2, \ a \in \mathbb{Z}$$ $a^2 = 2 - \sin^2 2\theta \in [1, 2]$. $a^2 = 1$ at $\sin^2 2\theta = 1$. $2\theta = (2n + 1) \frac{\pi}{2}$. $\theta = (2n + 1) \frac{\pi}{4} \notin P$. $n(S) = 0$

Question 14

Maths · Vector Algebra · Single correct

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$. If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to:

  1. 8
  2. 12
  3. 14
  4. 15

Answer: (b)

Solution

Given $\vec{c} = \lambda \vec{a} + \mu \vec{b}$. $\vec{c} = (\mu - \lambda) \hat{i} + (\lambda + 3\mu) \hat{j} + (3\lambda + \mu) \hat{k}$. $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$. $3(\mu - \lambda) - 6(\lambda + 3\mu) + 2(3\lambda + \mu) = 10$. $-3\lambda - 13\mu = 10$ ...(1) $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$. $5\mu + 3\lambda = -2$ ...(2) Solving (1) $\&$ (2) $\lambda = 1, \mu = -1$. $\vec{c} = -2\hat{i} - 2\hat{j} + 2\hat{k}$. $|\vec{c}|^2 = 12$

Question 15

Maths · Three Dimensional Geometry · Single correct

Let the point A be the foot of perpendicular drawn from the point P(a, b, 0) on the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}$. If the midpoint of the line segment PA is $\left(0, \frac{3}{4}, -\frac{1}{4}\right)$, then the value of $a^2 + b^2 + \alpha^2$ is equal to:

  1. 1
  2. 2
  3. 6
  4. 9

Answer: (a)

Solution

Given $\frac{2r + 1 + a}{2} = 0 \Rightarrow 2r + a = -1$ and $\frac{r + 2 + b}{2} = \frac{3}{4} \Rightarrow r + b = -\frac{1}{2}$. The points are $P(a, b, 0)$ and $A(2r + 1, r + 2, 3r + \alpha)$. For the equation $\frac{3r + \alpha}{2} = -\frac{1}{4} \Rightarrow 3r + \alpha = -\frac{1}{2}$. Solving the equations: $$2 \cdot a + 1 \cdot (b - 3/4) + 3 \cdot \frac{1}{4} = 0 \Rightarrow a + b = 0$$ $$\Rightarrow 2(-1 - 2r) + (-r - 1/2) = 0$$ $$\Rightarrow -3/2 = 3r = 1 \Rightarrow r = -1/2, \ a = 0, \ b = 0, \ \alpha = 1$$ Thus, $a^2 + b^2 + \alpha^2 = 1$.

Question 16

Maths · Vector Algebra · Single correct

Two adjacent sides of a parallelogram PQRS are given by $\overrightarrow{PQ} = \hat{j} + \hat{k}$ and $\overrightarrow{PS} = \hat{i} - \hat{j}$. If the side PS is rotated about the point P by an acute angle $\alpha$ in the plane of the parallelogram so that it becomes perpendicular to the side PQ, then $\sin^2 \left( \frac{5\alpha}{2} \right) - \sin^2 \left( \frac{\alpha}{2} \right)$ is equal to:

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

Answer: (b)

Solution

Let angle between $\overrightarrow{PQ}$ and $\overrightarrow{PS}$ is $\theta$. $$\cos \theta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PS}}{|\overrightarrow{PQ}| |\overrightarrow{PS}|}$$ $$\cos \theta = \frac{0 - 1 + 0}{\sqrt{2} \sqrt{2}}$$ $$\cos \theta = -\frac{1}{2}$$ $$\theta = \frac{2\pi}{3}$$ $$\alpha = \frac{2\pi}{3} - \frac{\pi}{2}$$ So $\alpha = \frac{\pi}{6}$ $$\sin^2 \frac{5\alpha}{2} - \sin^2 \frac{\alpha}{2} = \sin \left( \frac{5\alpha}{2} + \frac{\alpha}{2} \right) \sin \left( \frac{5\alpha}{2} - \frac{\alpha}{2} \right)$$ $$= \sin \frac{6\alpha}{2} \times \sin \frac{4\alpha}{2}$$ $$= \sin 3\alpha \times \sin 2\alpha$$ $$= \sin \frac{\pi}{2} \times \sin \frac{\pi}{3}$$ $$= 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$

Question 17

Maths · Integrals · Single correct

The value of $\int_{0}^{20\pi} (\sin^4 x + \cos^4 x) \, dx$ is equal to:

  1. $\frac{15\pi}{2}$
  2. $25\pi$
  3. $15\pi$
  4. $\frac{25\pi}{2}$

Answer: (c)

Solution

Given $$I = \int_0^{20\pi} (1 - 2 \sin^2 x \cos^2 x) \, dx$$ This simplifies to $$= 20\pi - \frac{1}{2} \int_0^{20\pi} \sin^2 2x \, dx$$ Next, we have $$20\pi - \frac{1}{2} (40) \int_0^{\pi/2} \sin^2 2x$$ This becomes $$20\pi - \frac{20}{2} \left( x - \frac{\sin 2x}{2} \right) \bigg|_0^{\pi/2}$$ Finally, we get $$20\pi - 10 \left( \frac{\pi}{2} \right) = 15\pi$$

Question 18

Maths · Limits and Derivatives · Single correct

Let f(x) be a polynomial of degree 5, and have extrema at x = 1 and x = -1. If $\lim_{x \to 0} \left( \frac{f(x)}{x^3} \right) = -5$, then $f(2) - f(-2)$ is equal to :

  1. 0
  2. 50
  3. 92
  4. 112

Answer: (d)

Solution

Given $f(x)$ is a polynomial of degree 5. Also, $f(1) = 0$; $f'(-1) = 0$. Also $\lim_{x \to 0} \frac{f(x)}{x^3} = -5$ (fixed and finite). Therefore, $f(0) = 0$; $f'(0) = 0$; $f''(0) = 0$. $$\frac{f'''(0)}{6} = -5 \implies f'''(0) = -30$$ Hence, let $f'(x) = (ax + b)(x - 0)(x - 1)(x + 1)$. $$\Rightarrow f'(x) = ax^4 + bx^3 - ax^2 - bx$$ $$f''(x) = 4ax^3 + 3bx^2 - 2ax - b$$ Therefore, $f''(0) = 0 \implies b = 0$. $$f'''(x) = 12ax^2 + 6bx - 2a$$ Therefore, $f'''(0) = -2a = -30$. $$\Rightarrow a = 15$$ Therefore, $f'(x) = 15x.x.(x - 1)(x + 1)$. $$\Rightarrow f'(x) = 15x^4 - 15x^2$$ $$\Rightarrow f(x) = 3x^5 - 5x^3 + C$$ Therefore, $f(0) = 0 \implies C = 0$. Therefore, $f(x) = 3x^5 - 5x^3$. Therefore, $f(2) - f(-2) = 112$

Question 19

Maths · Integrals · Single correct

Let $f(x) = \int \left( \frac{16x + 24}{x^2 + 2x - 15} \right) \, dx$. If $f(4) = 14 \log_e(3)$ and $f(7) = \log_e(2^\alpha \cdot 3^\beta)$, $\alpha, \beta \in \mathbb{N}$, then $\alpha + \beta$ is equal to:

  1. 31
  2. 37
  3. 39
  4. 41

Answer: (c)

Solution

Given $f(x) = \int \frac{8(2x+2) + 8}{x^2 + 2x - 15} \, dx$. Therefore, $f(x) = 8 \ln |x^2 + 2x - 15| + \ln \left| \frac{x-3}{x+5} \right| + C$. Thus, $f(4) = 14 \ln 3 + C$. So, $14 \ln 3 = 14 \ln 3 + C$. Therefore, $C = 0$. Now, $f(7) = 8 \ln 48 - \ln 3$. This equals $\ln \left( \frac{48^8}{3} \right)$. Thus, $f(7) = \ln (2^{32} \times 3^7)$. Given $\alpha = 32$, $\beta = 7$ implies $\alpha + \beta = 39$.

Question 20

Maths · Differential Equations · Single correct

Let $x = x(y)$ be the solution of the differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$, $y > 1$, $x(e) = e$. Then $x(e^2)$ is equal to:

  1. $\frac{3}{2} e^2$
  2. $\frac{2}{3} e^2$
  3. $e^2$
  4. $2e^2$

Answer: (b)

Solution

Given $2y(ydx - xdy) + x^2 dy = 0$. $$\Rightarrow -2yx^2 \frac{d\left(\frac{y}{x}\right)}{dx} + x^2 dy = 0$$ $$\Rightarrow -2y \frac{d\left(\frac{y}{x}\right)}{dx} + dy = 0$$ $$\Rightarrow -2d\left(\frac{y}{x}\right) + \frac{1}{y} dy = 0$$ $$\Rightarrow \frac{-2y}{x} + \log_e y = C$$ Given $x(e) = e$ $$\Rightarrow C = -1$$ Thus, $$\frac{-2y}{x} + \log_e y = -1$$ $$\Rightarrow \frac{2y}{x} - \log_e y = 1$$ Put $y = e^2$, we get $$\Rightarrow \frac{2e^2}{x} - 2 = 1 \Rightarrow \frac{2e^2}{x} = 3 \Rightarrow x = \frac{2e^2}{3}$$ $$\Rightarrow x(e^2) = \frac{2e^2}{3}$$

Question 21

Maths · Relations and Functions · Numerical

Let A = {$2, 3, 4, 5, 6$\}. Let R be a relation on the set A $\times$ A given by (x, y)R(z, w) if and only if x divides z and y $\leq$ w. Then the number of elements in R is $\ldots$.

Answer: 120

Solution

For $x = 2$, $z = 2, 4, 6$. $x = 3$, $z = 3, 6$. $x = 4$, $z = 4$. $x = 5$, $z = 5$. $x = 6$, $z = 6$. Total case $= 8$. And number of combinations of $y$ and $w$ satisfying $y \leq w$ is $1 + 2 + 3 + 4 + 5 = 15$. Number of relations satisfying $(x, y) R(z, w)$ is $8 \times 15 = 120$.

Question 22

Maths · Matrices · Numerical

Consider the matrices $A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}$. If matrices $P$ and $Q$ are such that $PA = B$ and $AQ = B$, then the absolute value of the sum of the diagonal elements of $2(P + Q)$ is $\ldots$.

Answer: 34

Solution

Given $$A^{-1} = \frac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix}$$ Therefore, $PA = B$. Thus, $$P = BA^{-1}$$ $$= \frac{1}{2} \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -21 & 12 \\ -7 & 4 \end{bmatrix}$$ Therefore, $AQ = B$. Thus, $$Q = A^{-1}B$$ $$= \frac{1}{2} \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -2 & -6 \\ -5 & -15 \end{bmatrix}$$ Therefore, $$|\mathrm{tr} \left( 2(P + Q) \right)| = |(-21) + (4) + (-2) + (-15)|$$ $$= |-34| = 34$$

Question 23

Maths · Conic Sections · Numerical

Let A be the point (3, 0) and circles with variable diameter AB touch the circle $x^2 + y^2 = 36$ internally. Let the curve C be the locus of the point B. If the eccentricity of C is e, then $72e^2$ is equal to ____.

Answer: 18

Solution

Let $B(h, k)$. Equation of circle with $AB$ as diameter: $$(x-h)(x-3) + (y-k)(y-0) = 0$$ $$x^2 + y^2 - (h+3)x - ky + 3h = 0$$ Centre: $$\left( \frac{h+3}{2}, \frac{k}{2} \right)$$ Circle touches internally. Therefore, $C_1C_2 = |R - r|$: $$\sqrt{\left( \frac{h+3}{2} \right)^2 + \left( \frac{k}{2} \right)^2} = \left| 6 - \frac{1}{2} \left( \sqrt{(h-3)^2 + k^2} \right) \right|$$ $$\sqrt{(x+3)^2 + y^2} + \sqrt{(x-3)^2 + y^2} = 12$$ $$2a = 12 \therefore a = 6$$ $(-3, 0)$ and $(3, 0)$ are foci. $$2ae = 6$$ $$12e = 6$$ $$e = \frac{1}{2}$$ Therefore, $$72e^2 = 72 \left( \frac{1}{4} \right) = 18$$

Question 24

Maths · Applications of Integrals · Numerical

If the area of the region bounded by $16x^2 - 9y^2 = 144$ and $8x - 3y = 24$ is $A$, then $3(A + 6 \log_e(3))$ is equal to $\ldots$.

Answer: 24

Solution

Given the equation $16x^2 - 9y^2 = 144$. Rewriting it as $16x^2 - (8x - 24)^2 = 144$. This simplifies to $16x^2 - 64(x - 3)^2 = 144 \Rightarrow x^2 - 4(x - 3)^2 = 9$. Expanding gives $3x^2 - 24x + 45 \Rightarrow x^2 - 8x + 15 = 0$. Solving for $x$ gives $x = 3, 5$. The area is given by: $$Area = \int_3^5 \sqrt{\frac{16x^2 - 144}{9}} - \frac{1}{2} \cdot 2 \cdot \frac{16}{3}$$ Simplifying further: $$= \frac{4}{3} \int_3^5 \sqrt{x^2 - 9} - \frac{16}{3}$$ $$= \frac{4}{3} \left( \frac{x}{2} \sqrt{x^2 - 9} - \frac{9}{2} \log_e(x + \sqrt{x^2 - 9}) \right)_3^5 - \frac{16}{3}$$ Evaluating the integral: $$= \frac{4}{3} \left( \frac{5}{2} \cdot 4 - \frac{9}{2} \log_e 9 - \frac{3}{2} \cdot 0 + \frac{9}{2} \log_e 3 \right) - \frac{16}{3}$$ The area is $8 - 6 \log_e 3 = A$. Therefore, $A + 6 \ln 3 = 8$. Thus, $3(A + 6 \ln 3) = 24$.

Question 25

Maths · Continuity and Differentiability · Numerical

The number of points in the interval $[2, 4]$, at which the function $f(x) = \left[ x^2 - x - \frac{1}{2} \right]$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is$\ldots$.

Answer: 10

Physics

Question 26

Physics · Physical World, Units and Measurements · Single correct

Dimensions of universal gravitational constant (G) in terms of Planck's constant (h), distance (L), mass (M) and time (T) are $\ldots$.

  1. [hTL M^{-2}]
  2. [hT^{-1}LM^{-2}]
  3. [hTL^{2}M^{-2}]
  4. [h^{-1}T^{-1}LM^{-2}]

Answer: (b)

Solution

Energy = $\frac{hc}{\lambda}$ = $\frac{GM_1M_2}{r}$ $$[ML^2T^{-2}] = [h][T^{-1}] = [G][M^2L^{-1}]$$ $$[G] = [M^{-2}L^{+1}T^{-1}h^{+1}]$$

Question 27

Physics · Laws of Motion · Single correct

A $0.5\,\mathrm{kg}$ mass is in contact against the inner wall of a cylindrical drum of radius $4\,\mathrm{m}$ rotating about its vertical axis. The minimum rotational speed of the drum to enable the mass to remain stuck to the wall (without falling) is $5\,\mathrm{rad/s}$. The coefficient of friction between the drum's inner wall surface and mass is $\ldots$ (Take g = $10\,\mathrm{m/s^2}$)

  1. 0.1
  2. 0.5
  3. 0.7
  4. 0.3

Answer: (a)

Solution

The normal force is given by $N = m \omega^2 R$. The static friction force is $f_s = mg$. The static friction force must be less than or equal to the limiting friction force $f_s \leq f_\ell$. Therefore, $mg \leq \mu N$. Substituting for $N$, we have $mg \leq \mu m \omega^2 R$. This implies $\mu \geq \frac{g}{\omega^2 R}$. Calculating the minimum coefficient of friction, $\mu_{\min} = \frac{g}{\omega^2 R} = \frac{10}{10^2 \times 1} = 0.1$.

Question 28

Physics · Laws of Motion · Single correct

Two blocks of masses $2\,\mathrm{\,kg}$ and $1\,\mathrm{\,kg}$ respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in 2 s is $\ldots$ m. (Take $g = 10 \mathrm{\,m/s^2}$)

  1. 3.33
  2. 3.12
  3. 2.22
  4. 1.42

Answer: (c)

Solution

Given $a = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g = \frac{(2 - 1)}{3} \times 10 = \frac{10}{3} \, \mathrm{m/s^2}$. $a_{\mathrm{cm}} = \frac{(1)(10/3) - (2)(10/3)}{3} = \frac{-10}{9} \, \mathrm{m/s^2}(\downarrow)$ $S_{\mathrm{cm}} = u_{\mathrm{cm}} t + \frac{1}{2} a_{\mathrm{cm}} t^2$ $= 0 + \frac{1}{2} \left( \frac{10}{9} \right) (4) = \frac{20}{9} \, \mathrm{m} (downward)$ $S_{\mathrm{COM}} = \frac{20}{9} (Downward) = 2.22$

Question 29

Physics · Moving Charges and Magnetism · Single correct

A particle having charge $10^{-9} \, \mathrm{C}$ moving in $x-y$ plane in fields of $0.4 \, \hat{j} \, \mathrm{N/C}$ and $4 \times 10^{-3} \, \hat{k} \, \mathrm{T}$ experiences a force of $(4\hat{i} + 2\hat{j}) \times 10^{-10} \, \mathrm{N}$. The velocity of the particle at that instant is $\ldots$ m/s.

  1. $50\hat{i} + 100\hat{j}$
  2. $100\hat{i} + 50\hat{j}$
  3. $-50\hat{i} + 100\hat{j}$
  4. $50\hat{i} - 100\hat{j}$

Answer: (a)

Solution

Given $\mathbf{F} = q \mathbf{E} + q (\mathbf{\vec{V}} \times \mathbf{B})$ $$(4 \hat{i} + 2 \hat{j}) \times 10^{-10}$$ $$= 10^{-9} [0.4 \hat{j} + (v_x \hat{i} + v_y \hat{j}) \times (4 \times 10^{-3}) \hat{k}]$$ $$(4 \hat{i} + 2 \hat{j}) \times 10^{-10}$$ $$= 10^{-9} (0.4 \hat{j} - 4 \times 10^{-3} v_x \hat{j} + 4 \times 10^{-3} v_y \hat{i})$$ On comparing $v_y = 100, \; v_x = 50$ $$\mathbf{\vec{V}} = 50 \hat{i} + 100 \hat{j}$$

Question 30

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

If X and Y are the inputs, the given circuit works as $\ldots$.

  1. OR gate
  2. AND gate
  3. NAND gate
  4. NOR gate

Answer: (d)

Solution

The circuit diagram shows the logic gates used to derive the output C. The inputs are x and y. The NOT gates invert the inputs to produce $\overline{x}$ and $\overline{y}$. These are then combined using an OR gate to produce $z = x + y$. Finally, the output C is given by the expression: $$C = \overline{x + y}$$

Question 31

Physics · Gravitation · Single correct

If a body of mass 1 kg falls on the earth from infinity, it attains velocity (v) and kinetic energy (k) on reaching the surface of earth. The values of v and k respectively are______. (Take radius of earth to be 6400 km and $g = 9.8 \, \mathrm{m/s^2}$)

  1. $11.2$ km/s; $6.27 \times 10^7$ J
  2. $11.2$ km/s; $12.54 \times 10^7$ J
  3. $8.8$ km/s; $6.27 \times 10^7$ J
  4. $8.8$ km/s; $12.54 \times 10^7$ J

Answer: (a)

Solution

M.E.C $K_i + U_i = K_f + U_f$ $$\Rightarrow O - \frac{GM_e m}{\infty} = \frac{1}{2}mv^2 - \frac{GMm}{R_e}$$ $$V = \sqrt{\frac{2GM}{R_e}} = \sqrt{2gR_e} = \sqrt{2 \times 9.8 \times 6400 \times 1000}$$ $$= 11.2 \, \mathrm{km/s}$$ Kinetic energy $= \frac{1}{2}mv^2$ $$= 6.27 \times 10^7 \, \mathrm{J}$$

Question 32

Physics · Experimental Physics · Single correct

In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is $0.1 \, \mathrm{mm}$. When diameter of a sphere is measured, the reading of main scale is $5 \, \mathrm{mm}$ and $50^{th}$ division of circular scale coincides with the reference line of main scale. The diameter of sphere is $\ldots \, \mathrm{mm}$.

  1. 5.045
  2. 5.055
  3. 5.450
  4. 5.550

Answer: (a)

Solution

Least count: $\frac{0.1}{100} \, \mathrm{mm} = 0.001 \, \mathrm{mm}$ Reading $= \mathrm{MSD} + (\mathrm{CSD}) \times \mathrm{LC} - zero error$ $= 5 \, \mathrm{mm} + 50 \times 0.001 - 5 \times 0.001 = 5.045 \, \mathrm{mm}$

Question 33

Physics · Mechanical Properties of Fluids · Numerical

The surface tension of a soap bubble is $0.03\,\mathrm{N/m}$. The work done in increasing the diameter of bubble from $2\,\mathrm{cm}$ to $6\,\mathrm{cm}$ is $\alpha \pi \times$ $10^{-4}$ $\mathrm{J}$. The value of $\alpha$ is $\ldots$. (Take $\pi$ = 3.14)

  1. 0.86
  2. 0.64
  3. 1.92
  4. 7.68

Answer: (c)

Solution

W = $\Delta$ U = 2T $\left[ 4\pi R_2^2 - 4\pi R_1^2 \right]$ W = 2 $\times$ 3 $\times$ 10^{-2} $\times$ 4$\pi \left[ 9 \times 10^{-4} - 1 \times 10^{-4} \right]$ W = 1.92 $\pi \times$ 10^{-4} $\mathrm{J}$ $\alpha$ = 1.92

Question 34

Physics · Kinetic Theory · Single correct

A mixture of carbon dioxide and oxygen has volume $8310 \, \mathrm{cm}^3$, temperature $300 \, \mathrm{K}$, pressure $100 \, \mathrm{kPa}$ and mass $13.2 \, \mathrm{g}$. The number of moles of carbon dioxide and oxygen gases in the mixture respectively are $\ldots$. (Assume both carbondioxide and oxygen gases behave like ideal gases) $[R = 8.31 \, \mathrm{J/mol.K}]$

  1. 0.15 and 0.18
  2. 0.25 and 0.08
  3. 0.21 and 0.12
  4. 0.13 and 0.20

Answer: (c)

Solution

First, we determine the total number of moles ($n_{total}$) in the mixture using the Ideal Gas Law: $PV = nRT$. Given the values: Pressure (P): $100 \, kPa = 10^5 \, Pa$ Volume (V): $8310 \, cm^3 = 8.31 \times 10^{-3} \, m^3$ Temperature (T): $300 \, K$ Gas Constant (R): $8.31 \, J/mol \cdot K$ $$\frac{PV}{RT} = \frac{10^5 \times 8.31 \times 10^{-3}}{8.31 \times 300} = \frac{100}{300} \approx 0.333 \, mole$$ 1. Mole equation: $n_1 + n_2 = 0.333$ 2. Mass equation: $44n_1 + 32n_2 = 13.2$ (where 44 and 32 are the molar masses of $CO_2$ and $O_2$ respectively) 3. Solve for individual moles Using the first equation to express $n_2$ as $0.333 - n_1$, we substitute this into the mass equation: $$44n_1 + 32(0.333 - n_1) = 13.2$$ $$44n_1 + 10.667 - 32n_1 = 13.2$$ $$12n_1 = 2.533$$ $$n_1 \approx 0.211 \, mol$$ Then, calculate $n_2$: $$n_2 = 0.333 - 0.211 = 0.122 \, mol$$

Question 35

Physics · Mechanical Properties of Fluids · Single correct

If an air bubble of diameter $2\,\mathrm{mm}$ rises steadily through a liquid of density $2000\,\mathrm{kg/m^3}$ at a rate of $0.5\,\mathrm{cm/s}$, then the coefficient of viscosity of liquid is $\ldots \mathrm{Poise}$. (Take g = $10\,\mathrm{m/s^2}$)

  1. 0.88
  2. 8.8
  3. 88.8
  4. 0.088

Answer: (b)

Solution

Given $F_B = \rho_\ell \cdot \frac{4}{3} \pi R^3 g$ and $F_V = 6 \pi \eta R v$. For constant speed, $F_B = F_V$. Therefore, $$\rho_\ell \frac{4}{3} \pi R^3 g = 6 \pi \eta R v$$ $$\frac{2}{9} \frac{\rho_\ell R^2 g}{v} = \eta$$ $$\Rightarrow \frac{2}{9} \times \frac{2000 \times 10^{-6} \times 10}{5 \times 10^{-3}} = \eta$$ $$\Rightarrow \eta = \frac{8}{9} \times 10$$ $$\Rightarrow \eta = \frac{80}{9} Poise = 8.88$$

Question 36

Physics · Work, Energy and Power · Single correct

A spherical ball of mass $2\,\mathrm{kg}$ falls from a height of $10\,\mathrm{m}$ and is brought to rest after penetrating $10\,\mathrm{cm}$ into sand. The average force exerted by sand on the ball is $\ldots \mathrm{N}$.

  1. 1980
  2. 2020
  3. 2000
  4. 1000

Answer: (b)

Solution

Given $(\mathrm{W.D.})_g + (\mathrm{W.D.})_{sand} = \Delta K$. $$2 \times 10 \times \left(10 + \frac{10}{100}\right) - F_{sand} \times \left(\frac{10}{100}\right) = 0 = 0$$ $$F_{sand} = 2(10 + 0.1) \times 100 = 2020$$

Question 37

Physics · Electromagnetic Waves · Single correct

An electromagnetic wave travels in free space along the $x$-direction. At a particular point in space and time, $\mathbf{B} = 2 \times 10^{-7} \hat{\mathbf{j}} \, \mathrm{T}$ is associated with this wave. The value of corresponding electric field $\mathbf{E}$ at this point is $\ldots$ V/m.

  1. $60 \hat{\mathbf{k}}$
  2. $-60 \hat{\mathbf{k}}$
  3. $30 \hat{\mathbf{k}}$
  4. $-600 \hat{\mathbf{k}}$

Answer: (b)

Solution

Given $\vec{E} = \vec{B} \times \vec{C}$. $$= 2 \times 10^{-7} \hat{j} \times 3 \times 10^{8} \hat{i}$$ $$\vec{E} = -60 \hat{k} \, \mathrm{V/m}$$

Question 38

Physics · Current Electricity · Single correct

Two resistors of 200 $\Omega$ and 400 $\Omega$ are connected in series with a battery of $100 \, \mathrm{V}$. A bulb rated at $200 \, \mathrm{V}$, $100 \, \mathrm{W}$ is connected across the $400 \Omega$ resistance. The potential drop across the bulb is $\ldots$ V.

  1. 25
  2. 50
  3. 66.6
  4. 100

Answer: (b)

Solution

For resistance of bulb $$R = \frac{V^2}{P} = \frac{200 \times 200}{100} = 400\, \Omega$$ Now circuit will look like this $V_1$ and $V_2$ will be same as resistances are same $$V_1 = V_2 = 50\, \mathrm{V}$$ $$V_{bulb} = V_2 = 50\, \mathrm{V}$$

Question 39

Physics · Electrostatic Potential and Capacitance · Single correct

Two metal plates (A, B) are kept horizontally with separation of $\frac{12}{\pi}$ cm, with plate A on the top. An atomizer jet sprays oil (density 1.5 g/cm$^3$) droplets of radius 1 mm horizontally. All oil droplets carry a charge 5 nC. The potentials $V_A$ and $V_B$ are required on plates A and B respectively in order to ensure the droplets do not descend. The values of $V_A$ and $V_B$ are $\ldots$ (Neglect the air resistance to the droplets and take $g = 10 \, \mathrm{m/s^2}$)

  1. 100 V and 580 V
  2. 580 V and 100 V
  3. 60 V and 400 V
  4. 0 V and -200 V

Answer: (a)

Solution

Given the setup with plates A and B, the forces acting are $qE$ upwards and $mg$ downwards. $$\frac{\Delta V}{(12/\pi) \times 10^{-2}} \times 5 \times 10^{-9} = \frac{4}{3} \pi (r^3) \rho (g)$$ $$\Delta V \times 5 \times 10^{-7} = 16 (r^3) \rho (g)$$ $$\Delta V = 480, \ V_A < V_B$$ Plate A will be lower potential.

Question 40

Physics · Electric Charges and Fields · Single correct

Two point charges $8\,\mu\mathrm{C}$ and $-2\,\mu\mathrm{C}$ are located at $x = 2\,\mathrm{cm}$ and $x = 4\,\mathrm{cm}$, respectively on the x-axis. The ratio of electric flux due to these charges through two spheres of radii $3\,\mathrm{cm}$ and $5\,\mathrm{cm}$ with their centers at the origin is $\ldots$

  1. 4 : 1
  2. 3 : 4
  3. 4 : 3
  4. 4 : 5

Answer: (c)

Solution

For sphere with radius 3 cm and center origin $$\phi_1 = \frac{q_{en}}{\varepsilon_0} = \frac{8 \, \mu \mathrm{C}}{\varepsilon_0}$$ for sphere of radius 5 cm $$\phi_2 = \frac{q_{en}}{\varepsilon_0} = \frac{(8 - 2) \, \mu \mathrm{C}}{\varepsilon_0} = \frac{6 \, \mu \mathrm{C}}{\varepsilon_0}$$ $$\frac{\phi_1}{\phi_2} = \frac{8}{6} = \frac{4}{3} = 1.33$$ $$\frac{\phi_1}{\phi_2} = 1.33$$

Question 41

Physics · Ray Optics and Optical Instruments · Single correct

One side of an equilateral prism is painted by a transparent material of refractive index $n_2$. The refractive index of prism is $1.6$. The minimum value of $n_2$ required for total internal reflection from painted face is $\ldots$.

  1. $\frac{3\sqrt{3}}{1.6}$
  2. $\sqrt{3}$
  3. $\frac{3.2}{\sqrt{3}}$
  4. $\frac{4\sqrt{3}}{5}$

Answer: (d)

Solution

For surface QR, $i = 60^\circ$. $1.6 \sin 60^\circ = \mu \sin 90^\circ$ $$\mu = 1.6 \times \frac{\sqrt{3}}{2} = \frac{4}{5} \times \sqrt{3}$$

Question 42

Physics · Alternating Current · Single correct

The figure given below shows an LCR series circuit with two switches $S_1$ and $S_2$. When switch $S_1$ is closed keeping $S_2$ open, the phase difference ($\phi$) between the current and source voltage is $30^\circ$ and phase difference is $60^\circ$ when $S_2$ is closed keeping $S_1$ open. The value of $(3L_1 - L_2)$ is $\ldots$ H.

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

Answer: (b)

Solution

For $S_1$ is closed and $S_2$ open. $V = V_0 \sin (300 \, t)$ $$\frac{X_{L_1} - X_C}{R} = \tan (30^\circ) = \frac{1}{\sqrt{3}}$$ $$\sqrt{3} \left( X_{L_1} - X_C \right) = R .....(1)$$ For $S_1$ is open and $S_2$ is closed. $V = V_0 \sin (300t)$ $$\frac{|X_{L_2} - X_C|}{R} = \tan 60^\circ = \sqrt{3}$$ $$\frac{|X_{L_2} - X_C|}{\sqrt{3}} = R ....(2)$$ From Eq. (1) and (2) $$\sqrt{3} \left( X_{L_1} - X_C \right) = \frac{(X_{L_2} - X_C)}{\sqrt{3}}$$ $$3 \left( \omega L_1 - \frac{1}{\omega C} \right) = \left( \omega L_2 - \frac{1}{\omega C} \right)$$ $$|3L_1 - L_2| = \frac{3}{\omega^2 C} - \frac{1}{\omega^2 C} = \frac{2}{9 \times 10^4 \times 100 \times 10^{-6}}$$ $$|3L_1 - L_2| = \frac{2}{9}$$

Question 43

Physics · Electromagnetic Induction · Single correct

A circular current loop of radius $R$ is placed inside square loop of side length $L (L \gg R)$ such that they are co-planar and their centers coincide. The permeability of free space is $\mu_0$. The mutual inductance between circular loop and square loop is____.

  1. 2$\sqrt{2} \frac{\mu_0 L^2}{R}$
  2. $\sqrt{2} \frac{\mu_0 L^2}{R}$
  3. $\sqrt{2} \frac{\mu_0 R^2}{L}$
  4. 2$\sqrt{2} \frac{\mu_0 R^2}{L}$

Answer: (d)

Solution

Magnetic field due to square $$B = 4 \times \frac{\mu_0 I}{4\pi \frac{L}{2}} \{2 \sin 45^\circ\}$$ $$= \frac{4 \mu_0 I}{\pi L} \times \frac{1}{\sqrt{2}} = \frac{2 \sqrt{2} \mu_0 I}{\pi L}$$ $$Q = B \times A_{circle}$$ $$Q = \frac{2 \sqrt{2} \mu_0 I}{\pi L} \times \pi R^2$$ $$Q = \frac{2 \sqrt{2} \mu_0 R^2}{L} I$$ Inductance = $$\frac{2 \sqrt{2} \mu_0 R^2}{L}$$

Question 44

Physics · Nuclei · Single correct

The binding energy per nucleon of $^{209}_{83}\mathrm{Bi}$ is $\ldots$ MeV. [Take $m(^{209}_{83}\mathrm{Bi}) = 208.980388 \, \mathrm{u}$, $m_p = 1.007825 \, \mathrm{u}$, $m_n = 1.008665 \, \mathrm{u}$, $1 \, \mathrm{u} = 931 \, \mathrm{MeV/c^2}$]

  1. 7.48
  2. 7.84
  3. 8.79
  4. 6.94

Answer: (b)

Solution

The reaction is $^{209}_{83}\mathrm{P}^1 + ^{1}_{0}\mathrm{n} \rightarrow ^{209}_{83}\mathrm{Bi}$. The mass defect $\Delta m$ is calculated as $\Delta m = 83m_p + 126m_n - M_{\mathrm{Bi}}$. Substituting the values, we have $$\Delta m = 83 \times 1.007825 + 126 \times 1.008665 - 208.980388$$ $$= 1.760877 \, \mathrm{u}$$ The binding energy (BE) is given by $$\mathrm{BE} = 1.760877 \times 931 \, \mathrm{MeV} = 1639.3764$$ The binding energy per nucleon is $$\frac{\mathrm{BE}}{A} = \frac{1639.3764}{209} \, \mathrm{MeV} = 7.8439 \, \mathrm{MeV}$$

Question 45

Physics · Oscillations · Single correct

The equation of motion of a particle is given by $x = a \sin(50t + \pi/3) \, \mathrm{cm}$. The particle will come to rest at time $t_1$ and it will have zero acceleration at time $t_2$. The $t_1$ and $t_2$ respectively are $\ldots$.

  1. $\frac{\pi}{300} \, \mathrm{s}, \frac{\pi}{75} \, \mathrm{s}$
  2. $\frac{\pi}{75} \, \mathrm{s}, \frac{\pi}{300} \, \mathrm{s}$
  3. $\frac{\pi}{300} \, \mathrm{s}, \frac{\pi}{25} \, \mathrm{s}$
  4. $\frac{\pi}{50} \, \mathrm{s}, \frac{\pi}{100} \, \mathrm{s}$

Answer: (a)

Solution

Phase of the motion of particle is as follows: $V = 0$, when particle crosses A. Therefore, $t_1 = \frac{\pi}{6\omega} = \frac{\pi}{300}$. a = 0, when particle crosses B. Therefore, $t_2 = \frac{2\pi}{3\omega} = \frac{\pi}{75}$.

Question 46

Physics · Wave Optics · Numerical

In a Young's double slit experiment, the intensity at some point on the screen is found to be $\frac{3}{4}$ times of the maximum of the interference pattern. The path difference between the interfering waves at this point is $\frac{\lambda}{x}$ where $\lambda$ is wavelength of the incident light. The value of $x$ is $\ldots$.

Answer: 6

Solution

Given $I_r = I_{\max} \cos^2 \frac{\phi}{2}$. $$\frac{3I_{\max}}{4} = I_{\max} \cos^2 \frac{\phi}{2}$$ $$\cos \frac{\phi}{2} = \frac{\sqrt{3}}{2}$$ $$\frac{\phi}{2} = 30^\circ = \frac{\pi}{6}$$ $$\left( \phi = \frac{\pi}{3} \right)$$ Path difference $\Delta x = \frac{\phi}{2\pi} \times \lambda$ $$\Rightarrow \frac{\pi}{3 \times 2\pi} \times \lambda$$ $$\Delta x = \frac{\lambda}{6}$$ $x = 6$

Question 47

Physics · Atoms · Numerical

Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the $2^{nd}$ and $4^{th}$ orbits of hydrogen atom.

Answer: 64

Solution

The magnetic field $B$ is given by $$B = \frac{\mu_0 I}{2 \pi r}$$ where $I$ is the current. The current $I$ is given by $$I = \frac{ev}{2 \pi r}$$ where $e$ is the charge and $v$ is the velocity. The magnetic field $B$ is proportional to $$B \propto \frac{V}{r^2}$$ where $V$ is the potential. The potential $V$ is proportional to $$V \propto \frac{Z}{n}$$ where $Z$ is the atomic number and $n$ is the principal quantum number. The radius $r$ is proportional to $$r \propto \frac{n^n}{Z}$$ Therefore, $$B \propto \left(\frac{Z}{n}\right) \left(\frac{n^2}{Z}\right)^2$$ Simplifying, $$B \propto \frac{Z^3}{n^5}$$ The ratio of magnetic fields is given by $$\frac{B_1}{B_2} = \frac{n_2^5}{n_1^5} = \frac{4^5}{2^5} = 64$$

Question 48

Physics · Thermodynamics · Numerical

5 moles of unknown gas is heated at constant volume from $10^\circ \mathrm{C}$ to $20^\circ \mathrm{C}$. The molar specific heat of this gas at constant pressure $c_p = 8 \, \mathrm{cal/mol.} \ ^\circ \mathrm{C}$ and $R = 8.36 \, \mathrm{J/mol.} \ ^\circ \mathrm{C}$. The change in the internal energy of the gas is $\ldots$ calorie.

Answer: 300

Solution

Given $C_p - C_V = R \Rightarrow C_V = (C_p - R)$. $C_V = 6 \, cal/mol°C$. $\Delta U = n C_V \Delta T$. $$= 5 \times 6 \times 10 = 300 \, cal$$

Question 49

Physics · Ray Optics and Optical Instruments · Numerical

If sunlight is focused on a paper using convex lens, it starts burning the paper in shortest time when the lens is kept at $30 \, \mathrm{cm}$ above the paper. If the radius of curvature of the lens is $60 \, \mathrm{cm}$ then the refractive index of the lens material is $\frac{\alpha}{10}$. The value of $\alpha$ is $\ldots$.

Answer: 20

Solution

To burn in minimum time distance of paper must be at focus. So $f = 30 \, \mathrm{cm}$ $$\frac{1}{30} = (\mu - 1) \left( \frac{1}{60} - \frac{1}{-60} \right)$$ $$\frac{1}{30} = \frac{\mu - 1}{30}$$ $$\mu = 2$$ $$\alpha = 20$$

Question 50

Physics · System of Particles and Rotational Motion · Numerical

Moment of inertia about an axis AB for a rod of mass $40\,\mathrm{kg}$ and length $3\,\mathrm{m}$ is same as that of a solid sphere of mass $10\,\mathrm{kg}$ and radius R about an axis parallel to AB axis with separation of $3\,\mathrm{m}$ as shown in figure below. The value of R is given as $\sqrt{\frac{\alpha}{2}}$. The value of $\alpha$ is $\ldots$.

Answer: 15

Solution

Given $I_{Rod} = I_{SS}$. $$\frac{M_1 L^2}{3} = \left( \frac{2}{5} m_2 R^2 + m_2 L^2 \right)$$ $$\frac{40 \times (3)^2}{3} = \frac{2}{5} \times 10 \times R^2 + 10 \times (3)^2$$ $$120 = 4R^2 + 90$$ $$4R^2 = 30$$ $$R = \sqrt{\frac{15}{2}} = \sqrt{\frac{\alpha}{2}}$$ $$\alpha = 15$$

Chemistry

Question 51

Chemistry · Hydrocarbons · Single correct

The ratio of mass percentage (w/w) of C : H in a hydrocarbon is 12 : 1. It has two carbon atoms. The weight (in g) of $CO_2(g)$ formed when $3.38\,\mathrm{g}$ of this hydrocarbon is completely burnt in oxygen is : (Given : Molar mass in g $\mathrm{mol}^{-1}$ C : 12, H : 1, O : 16)

  1. 5.68
  2. 11.44
  3. 22.74
  4. 17.05

Answer: (b)

Solution

Molecular formula of hydrocarbon is $\mathrm{C_2H_2}$. Moles of $\mathrm{C_2H_2} = \frac{3.38}{26}$. $$\mathrm{C_2H_2(g) + \frac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(l)}$$ Moles of $\mathrm{CO_2}$ produced $= \left( \frac{3.38}{26} \times 2 \right)$. Mass of $\mathrm{CO_2}$ produced $= \left( \frac{3.38}{26} \times 2 \right) \times 44 = 11.44$.

Question 52

Chemistry · Equilibrium · Single correct

The first and second ionization constants of a weak dibasic acid $\mathrm{H_2A}$ are $8.1 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. $0.1$ mol of $\mathrm{H_2A}$ was dissolved in $1\, \mathrm{L}$ of $0.1\, \mathrm{M}\ \mathrm{HCl}$ solution. The concentration of $\mathrm{HA^-}$ in the resultant solution is:

  1. $0.1\, \mathrm{M}$
  2. $9.53 \times 10^{-6}\, \mathrm{M}$
  3. $8.1 \times 10^{-8}\, \mathrm{M}$
  4. $1.0 \times 10^{-13}\, \mathrm{M}$

Answer: (c)

Solution

HCl $\rightarrow \mathrm{H^+}$ + $\mathrm{Cl^-}$ 0.1 M $\hspace{1cm}$ - $\hspace{1cm}$ - $\hspace{2.5cm}$ 0.1 $\hspace{1cm}$ 0.1 $\mathrm{H_2A} \rightleftharpoons \mathrm{H^+}$ + $\mathrm{HA^-}$ 0.1-x $\hspace{1cm}$ x+0.1 $\hspace{1cm}$ x (Due to common ion effect $x$ is very less) $$K_{a_1} = \frac{[\mathrm{H^+}][\mathrm{HA^-}]}{[\mathrm{H_2A}]}$$ $$K_{a_1} = \frac{0.1 \times [\mathrm{HA^-}]}{0.1}$$ $$[\mathrm{HA^-}] = K_{a_1} = 8.1 \times 10^{-8}$$

Question 53

Chemistry · Chemical Bonding and Molecular Structure · Single correct

$\mathrm{SF_4}$ is isostructural with: \ A. $\mathrm{BrF_4^-}$ \ B. $\mathrm{CH_4}$ \ C. $\mathrm{IF_4^+}$ \ D. $\mathrm{XeF_4}$ \ E. $\mathrm{XeO_2F_2}$ \ Choose the correct answer from the options given below:

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

Answer: (b)

Solution

SF_4 has a see-saw structure. IF_4^+ also has a see-saw structure. BrF_4^- has a square planar structure. XeO_2F_2 has a see-saw structure. CH_4 has a tetrahedral structure. XeF_4 has a square planar structure.

Question 54

Chemistry · Thermodynamics · Single correct

Gas 'A' undergoes change from state 'X' to state 'Y'. In this process, the heat absorbed and work done by the gas is $10\,\mathrm{J}$ and $18\,\mathrm{J}$ respectively. Now gas is brought back to state 'X' by another process during which $6\,\mathrm{J}$ of heat is evolved. In the reverse process of 'Y' to 'X',

  1. $18\,\mathrm{J}$ of the work is done by the gas 'A'.
  2. $2\,\mathrm{J}$ of the work is done by the gas 'A'.
  3. $12\,\mathrm{J}$ of the work is done on the gas 'A' by the surrounding.
  4. $14\,\mathrm{J}$ of the work is done on the gas 'A' by the surrounding.

Answer: (d)

Solution

For process $X \rightarrow Y$: $w = -18 \, \mathrm{J}$ $q = +10 \, \mathrm{J}$ $\Delta U = q + w$ $$\Delta U_{X \rightarrow Y} = 10 + (-18) = -8 \, \mathrm{J}$$ For reverse process: $\Delta U_{Y \rightarrow X} = 8 \, \mathrm{J}$ $q = -6 \, \mathrm{J}$ $\Delta U = q + w$ $$8 = -6 + w$$ $$w = 14 \, \mathrm{J}$$

Question 55

Chemistry · Solutions · Single correct

Solution A is prepared by dissolving 1 g of a protein (molar mass = 50000 $g$ $mol^{-1}$) in 0.5 L of water at 300 K. Its osmotic pressure is x bar. Solution B is made by dissolving 2 g of same protein in 1 L of water at 300 K. Osmotic pressure of solution B is y bar. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is z bar. x, y and z respectively are: (R = 0.083 L bar $mol^{-1} K^{-1}$)

  1. 9.96 $\times$ 10^{-4}; 9.96 $\times$ 10^{-4}; 9.96 $\times$ 10^{-4}
  2. 9.96 $\times$ 10^{-4}; 9.96 $\times$ 10^{-4}; 19.92 $\times$ 10^{-4}
  3. 4.98 $\times$ 10^{-4}; 4.98 $\times$ 10^{-4}; 9.96 $\times$ 10^{-4}
  4. 4.98 $\times$ 10^{-4}; 4.98 $\times$ 10^{-4}; 4.98 $\times$ 10^{-4}

Answer: (a)

Solution

'x' = $\frac{1}{50,000 \times 0.5} \times$ R $\times$ 300 = 9.96 $\times$ 10^{-4} bar 'y' = $\frac{2}{50,000 \times 1} \times$ R $\times$ 300 = 9.96 $\times$ 10^{-4} bar 'z' = $\frac{'x' \times \frac{1}{2} + 'y' \times 1}{\frac{3}{2}}$ = 9.96 $\times$ 10^{-4} bar

Question 56

Chemistry · Equilibrium · Single correct

At $25^\circ\mathrm{C}$, $20.0 \, \mathrm{mL}$ of $0.2 \, \mathrm{M}$ weak monoprotic acid $\mathrm{HX}$ is titrated against $0.2 \, \mathrm{M}$ NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when $10 \, \mathrm{mL}$ of NaOH is added respectively, are: Given: $K_a = 5 \times 10^{-4}$, $\mathrm{p}K_a = 3.3$, $\alpha \ll 1$

  1. 0.7, 2.0
  2. 2.0, 3.3
  3. 1.1, 2.2
  4. 3.0, 2.2

Answer: (b)

Solution

For 0.2 M, 20 ml HX solution, pH of weak acid is given by $$\frac{1}{2} [pK_a - \log c]$$ which simplifies to $$\frac{1}{2} [4 - \log 5 - \log 0.2]$$ and further to $$\frac{1}{2} [4 - \log 5 + \log 5]$$ resulting in $$2$$. The reaction is: $$HX + NaOH \xrightarrow{(LR)} NaA + H_2O$$ For 20 ml, 0.2 M HX and 10 ml, 0.2 M NaOH, we have: 4 m mole of HX and 2 m mole of NaOH. After reaction: 2 m mole of HX remains and 2 m mole of NaOH is used. This forms an acidic buffer: $$[HX + NaA] \Rightarrow acidic buffer$$ The pH is calculated as: $$pH = pK_a + \log \frac{salt}{acid}$$ Substituting the values: $$pH = 3.3 + \log \frac{2}{2}$$ Thus, $$pH = 3.3$$

Question 57

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

Consider the reaction $\mathrm{aX \rightarrow bY}$, for which the rate constant at $30^\circ\mathrm{C}$ is $1\times10^{-3}\,\mathrm{mol^{-1}\,L\,s^{-1}}$. \ Which of the following statements are true? \ A. When concentration of `X' is increased to four times, the rate of reaction becomes 16 times. \ B. The reaction is a second order reaction. \ C. The half-life period is independent of the concentration of X. \ D. Decomposition of $\mathrm{N_2O_5}$ is an example of the above reaction. \ E. is valid for the above reaction. Choose the correct answer from the options given below:

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

Answer: (a)

Solution

Rate of constant (k) = $10^{-3} \, \mathrm{M^{-1} \, sec^{-1}}$ $\therefore \, 1 - n = -1$ [unit of $k = \mathrm{M^{1-n} \, sec^{-1}}$] $n = 2 \Rightarrow 2^{nd}$ order reaction rate = $k[A]^2$ (A) If concentration of A is 4 times then reaction will become 16 times. (B) Order of reaction is 2 (C) Half-life of the reaction is independent of the concentration of the reactant for $1^{st}$ order, not for $2^{nd}$ order reaction. (D) Decomposition of $\mathrm{N_2O_5}$ is example of $1^{st}$ order reaction (E) Graph of $\ln \left( \frac{[R]_0}{[R]_t} \right)$ vs $t$ is straight line for $1^{st}$ order not for $2^{nd}$ order reaction. So statement (A) & (B) are correct

Question 58

Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct

The correct set that contains all kinds (basic, acidic, amphoteric and neutral) of oxides is:

  1. $\mathrm{Na_2O}, \mathrm{K_2O}, \mathrm{Al_2O_3}$ and $\mathrm{As_2O_3}$
  2. $\mathrm{Al_2O_3}, \mathrm{As_2O_3}, \mathrm{CO}$ and $\mathrm{NO}$
  3. $\mathrm{K_2O}, \mathrm{Cl_2O_7}, \mathrm{As_2O_3}$ and $\mathrm{NO}$
  4. $\mathrm{Na_2O}, \mathrm{N_2O}, \mathrm{Al_2O_3}$ and $\mathrm{CO}$

Answer: (c)

Solution

Basic Oxides: $\mathrm{Na_2O}$, $\mathrm{K_2O}$ Acidic Oxides: $\mathrm{Cl_2O_7}$ Neutral Oxides: $\mathrm{N_2O}$, $\mathrm{NO}$, $\mathrm{CO}$ Amphoteric Oxides: $\mathrm{Al_2O_3}$, $\mathrm{As_2O_3}$

Question 59

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

Given below are two statements : Statement-I : The second ionization enthalpy of B, Al and Ga is in the order of B > Al > Ga. Statement-II : The correct order in terms of first ionization enthalpy is Si < Ge < Pb < Sn. 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: (b)

Solution

Order of II ionization energy for boron family elements is: $$\mathrm{B} \,(2427 \, \mathrm{KJ/mol}) > \mathrm{Ga} \,(1979 \, \mathrm{KJ/mol}) > \mathrm{Al} \,(1816 \, \mathrm{KJ/mol})$$ Order of I ionization energy for carbon family elements is: $$\mathrm{Si} \,(786 \, \mathrm{KJ/mol}) > \mathrm{Ge} \,(761 \, \mathrm{KJ/mol}) > \mathrm{Pb} \,(715 \, \mathrm{KJ/mol}) > \mathrm{Sn} \,(708 \, \mathrm{KJ/mol})$$

Question 60

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

Given below are two statements: \ Statement-I: Among Zn, Mn, Sc and Cu, the energy required to remove the third valence electron is highest for Zn and lowest for Sc. \ Statement-II: The correct order of the following complexes in terms of CFSE is $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+} < [\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{3+} < [\mathrm{Co(en)}_3]^{3+}$. \ 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: (a)

Solution

Statement I $\mathrm{IE_3}$ (kJ/mol) Sc: 2393 Mn: 3260 Cu: 3556 Zn: 3837 Sc WFL (en > $\mathrm{H_2O}$) CFSE $\propto$ Charge on metal $\mathrm{Co^{+2} < Co^{+3}}$ $[\mathrm{Co(H_2O)_6}]^{2+} < [\mathrm{Co(H_2O)_6}]^{3+} < [\mathrm{Co(en)_3}]^{3+}$

Question 61

Chemistry · Co-ordination Compounds · Single correct

Which of the following complexes will show coordination isomerism? \ (A) $[\mathrm{Ag(NH_3)_2}][\mathrm{Ag(CN)_2}]$ \ (B) $[\mathrm{Co(NH_3)_6}][\mathrm{Cr(CN)_6}]$ \ (C) $[\mathrm{Co(NH_3)_6}][\mathrm{Co(CN)_6}]$ \ (D) $[\mathrm{Fe(NH_3)_6}][\mathrm{Co(CN)_6}]$ \ (E) $[\mathrm{Co(NH_3)_6}][\mathrm{Fe(CN)_6}]$ \ Choose the correct answer from the options given below:

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

Answer: (a), (b), (d)

Solution

Compound, $[\mathrm{Ag(NH_3)_2}][\mathrm{Ag(CN)_2}]$ cannot show coordination isomerism. All other complexes (B, C, D and E) will show co-ordination isomerism.

Question 62

Chemistry · Some Basic Concepts of Chemistry · Single correct

Complete combustion of $X$ g of an organic compound gave 0.25 g of $\mathrm{CO}_2$ and 0.12 g of $\mathrm{H}_2\mathrm{O}$. If the $\%$ of carbon is 25$\%$ and of hydrogen is 4.89$\%$, then $X = \ldots \times 10^{-3}$ g. (Nearest integer) (Molar mass of C, H and O are 12, 1 and 16 g mol$^{-1}$ respectively.)

  1. 273
  2. 27
  3. 2730
  4. 227

Answer: (a)

Solution

Mass of carbon = $\frac{0.25}{44} \times$ 12 Mass$\%$ of carbon = $\frac{Mass of carbon}{x} \times$ 100 x = $\frac{\frac{0.25 \times 12}{44} \times 100}{25}$ x = $\frac{12}{44}$ x = 0.2727 = 272.7 $\times$ 10^{-3} $\simeq$ 273 $\times$ 10^{-3}

Question 63

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

Given below are two statements 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: (a)

Solution

Both statements are correct.

Question 64

Chemistry · Hydrocarbons · Single correct

The compound (X) on (i) on heating in the presence of anhydrous $\mathrm{AlCl_3}$ and HCl gas gives 2, 4-dimethyl pentane (ii) aromatization gives toluene and (iii) cyclisation gives methyl cyclohexane The correct name of compound (X) is:

  1. Hept-2-ene
  2. Hept-1,3,5-triene
  3. Heptane
  4. Hept-2,4,6-triene

Answer: (c)

Solution

n-Heptane undergoes aromatisation in the presence of $\mathrm{Cr_2O_3}$ at $773 \, \mathrm{K}$ and $10-20 \, \mathrm{atm}$ to form an aromatic compound. It can also undergo isomerisation with anhydrous $\mathrm{AlCl_3}$ and $\mathrm{HCl}$ to form a branched alkane.

Question 65

Chemistry · Haloalkanes and Haloarenes · Single correct

Correct statements regarding alkyl halides (R–X) among the following are: (A) Alcohol being less polar solvent as compared to water, alcoholic KOH favours elimination reaction with R–X. (B) Order of reactivity towards $S_N1$ mechanism is $C_6H_5–CH_2–Cl > C_6H_5–CHCl–C_6H_5$. (C) Non substituted aryl halides exhibit properties similar to alkyl halides. (D) Vinyl chloride is an example of haloalkene and allyl chloride is an example of haloalkyne. (E) R–Cl can be prepared by reacting R–OH with $SOCl_2$ but Ar–Cl cannot be prepared by reacting Ar–OH with $SOCl_2$. Choose the correct answer from the options given below:

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

Answer: (c)

Solution

Ar-Cl and R-Cl shows different chemical properties. Rate of $S_N1 \propto$ stability of first formed carbocation. Water is more polar than alcohol. Vinyl alcohol, allyl alcohol, both are alkene unsaturated alcohols. $$R-OH \xrightarrow{SOCl_2} R-Cl$$ Phenol does not give Ph-Cl with SOCl$_2$.

Question 66

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

An organic compound "x" where molar ratio of C, O and H are equal, on treatment with 50$\%$ KOH under reflux followed by acidification produced "y". The most likely structure of "y" is: [Molar mass of 'x' is $58\,\mathrm{g \, mol^{-1}}$]

  2. CH$_3$–CH=CH–CH=O

Answer: (c)

Solution

The reaction involves an intramolecular Cannizzaro reaction. The starting compound is treated with $\mathrm{KOH}$ to form an intermediate with $\mathrm{CH_2OH}$ and $\mathrm{COO}^-$. Upon protonation, the final product is $\mathrm{CH_2OH}$ and $\mathrm{COOH}$.

Question 67

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Multiple correct

A molecule (X) with following structure under mild acidic condition is hydrolysed to produce (Y) and (Z). Identify the correct statements about (Y) and (Z). \ (A) Both $(Y)$ and $(Z)$ have same molar mass. \ (B) $(Y)$ and $(Z)$ can be distinguished from each other by $\mathrm{NaHCO_3}$. \ (C) $(Y)$ and $(Z)$ react with HCN with same rates. \ (D) $(Y)$ and $(Z)$ undergo addition reaction with 2,4-DNP. \ Choose the correct answer from the options given below:

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

Answer: (d)

Solution

Y will give positive Fehling's test. Molar mass of [Z] $\mathrm{(C_3H_6O)} = 58$. Molar mass of [Y] $\mathrm{(C_3H_6O)} = 58$.

Question 68

Chemistry · Amines · Single correct

Identify compounds A and E in the following reaction sequence.

Answer: (b)

Solution

The solution involves a series of chemical reactions starting with compound (A) which is brominated using $\mathrm{Br_2}$ and $\mathrm{Fe}$ to form compound (B). Compound (B) is then reduced using $\mathrm{Sn}$ and $\mathrm{HCl}$ to form compound (C). Compound (C) undergoes diazotization with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ at $0-5^\circ\mathrm{C}$ to form compound (D) with $\mathrm{N_2^+Cl^-}$. This is followed by an ethanol/heat ($\mathrm{EtOH/\Delta}$) reaction to form compound (E). Finally, compound (E) is oxidized using (i) $\mathrm{KOH/KMnO_4}$ and (ii) $\mathrm{H^+}$ to form compound (F).

Question 69

Chemistry · Biomolecules · Single correct

Identify the correct pair having amino acid (A) and the hormone (B) that is iodinated derivative of the amino acid (A). (T and Y represent one letter code for amino acids)

  1. T Insuline
  2. T Thyroxine
  3. Y Thyroxine
  4. Y Insuline

Answer: (c)

Solution

Thyroxine is derivative of Tyrosine and single letter symbol of Tyrosine amino acid is Y.

Question 70

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

Among $\mathrm{Fe}^{2+}$, $\mathrm{Fe}^{3+}$, $\mathrm{Cr}^{2+}$ and $\mathrm{Zn}^{2+}$, the ion that shows positive borax bead test and with highest ionisation enthalpy is:

  1. $\mathrm{Fe}^{2+}$
  2. $\mathrm{Zn}^{2+}$
  3. $\mathrm{Cr}^{2+}$
  4. $\mathrm{Fe}^{3+}$

Answer: (d)

Solution

Zn^{2+} does not show Borax Bead Test. Among remaining ions; Fe^{3+} has maximum ionization energy.

Question 71

Chemistry · Structure of Atom · Numerical

The surface of sodium metal is irradiated with radiation of wavelength $x$ nm. The kinetic energy of ejected electrons is $2.8 \times 10^{-20}$ J. The work function of sodium is $2.3$ eV. The value of $x$ is $\ldots \times 10^2$ nm. (Nearest integer) (Given : $h = 6.6 \times 10^{-34}$ J s; $1 \mathrm{eV} = 1.6 \times 10^{-19}$ J; $c = 3.0 \times 10^8$ m s$^{-1}$)

Answer: 5

Solution

For photoelectric effect: $$\frac{hc}{\lambda} = \mathrm{W} + \mathrm{K.E.}$$ $$\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{x} = 2.3 \times 1.6 \times 10^{-19} + 2.8 \times 10^{-20}$$ $$\frac{19.8 \times 10^{-26}}{x} = 3.96 \times 10^{-19}$$ $$x = 5 \times 10^{-7}$$ $$x = 500 \times 10^{-9} \, \mathrm{m}$$ $$x = 5 \times 10^2 \, \mathrm{nm}$$

Question 72

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

Consider the following gas phase reaction being carried out in a closed vessel at $25^\circ\mathrm{C}$. \[2\mathrm{A(g)} \rightarrow 4\mathrm{B(g)}+\mathrm{C(g)}\] \begin{tabular}{|c|c|}\hlineTime (min) & Total pressure of the system (mm Hg) \\\hline30 & 300 \\\hline$\infty$ & 600 \\\hline\end{tabular} The pressure of $\mathrm{C(g)}$ at $30$ minutes time interval would be \_\_\_\_\_\_\_\_\_\_ mm Hg. (Nearest integer)

Answer: 20

Question 73

Chemistry · Electrochemistry · Numerical

Consider the following two half-cell reactions along with the standard reduction potential given: \ $\mathrm{CO_2 + 6H^+ + 6e^- \rightarrow CH_3OH + H_2O}\qquad E^\circ_{\mathrm{red}}=0.02\,\mathrm{V}$ \ $\frac{1}{2}\mathrm{O_2}+2\mathrm{H^+}+2e^- \rightarrow \mathrm{H_2O}\qquad E^\circ_{\mathrm{red}}=1.23\,\mathrm{V}$ \ The fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with 80$\%$ efficiency. If the work derived from the cell using 1 mol of $\mathrm{CH_3OH}$ is used to compress an ideal gas isothermally against a constant pressure of 1 kPa, then the change in the volume of the gas, $\Delta V = \ldots\,\mathrm{m^3}$. (Nearest integer) Given: $F=96500\,\mathrm{C\,mol^{-1}}$

Answer: 560

Solution

Given $E^\circ_{cell} = [1.23 - 0.02] = 1.21 \, V$. $\Delta G^\circ = -6 \times 96500 \times 1.21 \, J$. $W = \frac{80}{100} \times \Delta G^\circ = -P \Delta V$. $0.8 \times 6 \times 96500 \times 1.21 = 1 \times 10^3 (\Delta V)$. $\Delta V = 560.472 \, m^3$

Question 74

Chemistry · The d-and f-Block Elements · Numerical

Number of paramagnetic ions among the following d- and f-block metal ions is $\ldots$. \ $\mathrm{Mn}^{2+},\ \mathrm{Cu}^{2+},\ \mathrm{Zn}^{2+},\ \mathrm{Yb}^{2+},\ \mathrm{Sc}^{3+},\ \mathrm{La}^{3+},\ \mathrm{Gd}^{3+},\ \mathrm{Lu}^{3+},\ \mathrm{Ti}^{4+},\ \mathrm{Ce}^{4+}$ \ (Atomic number of $\mathrm{Mn}=25$, $\mathrm{Cu}=29$, $\mathrm{Zn}=30$, $\mathrm{Yb}=70$, $\mathrm{Sc}=21$, $\mathrm{La}=57$, $\mathrm{Gd}=64$, $\mathrm{Lu}=71$, $\mathrm{Ti}=22$, $\mathrm{Ce}=58$)

Answer: 3

Solution

Paramagnetic: $\mathrm{Mn^{+2}}$, $\mathrm{Cu^{+2}}$, $\mathrm{Gd^{+3}}$. Diamagnetic: $\mathrm{Zn^{+2}}$, $\mathrm{Yb^{+2}}$, $\mathrm{Sc^{+3}}$, $\mathrm{La^{+3}}$, $\mathrm{Lu^{+3}}$, $\mathrm{Ti^{+4}}$, $\mathrm{Ce^{+4}}$.

Question 75

Chemistry · Haloalkanes and Haloarenes · Numerical

Consider the following reactions sequence When the product (P) is subjected to Carius analysis using AgNO$_3$, 1.0 g of the product (P) will produce $\ldots$ g of the precipitate of AgBr. (Nearest integer) (Given : molar mass in g mol$^{-1}$ C : 12, H : 1, O : 16, N : 14, Br : 80, Ag : 108)

Answer: 1

Solution

The reaction sequence starts with the reduction of the nitro group to an amine using $\mathrm{Sn + HCl}$. The amine is then acetylated using $\mathrm{Ac_2O}$. Bromination occurs in the presence of $\mathrm{FeBr_3 + Br_2}$. Finally, hydrolysis with $\mathrm{H_3O^+}$ gives the product. Molar mass of product = 186 Mass of $\mathrm{AgBr} = \frac{1}{186} \times 188 \approx 1$