JEE Main 2 April 2026 Shift 1 question paper with solutions

JEE Main 2 April 2026 Shift 1: 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 $\alpha$, $\alpha + 2$, $\alpha \in \mathbb{Z}$, be the roots of the quadratic equation $x(x + 2) + (x + 1)(x + 3) + (x + 2)(x + 4) + \ldots + (x + n - 1)(x + n + 1) = 4n$ for some $n \in \mathbb{N}$. Then $(n + \alpha)$ is equal to:

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

Answer: (c)

Solution

Given $nx^2 + x(2 + 4 + 6 + \ldots + 2n) + (1.3 + \ldots + (n-1)(n+1)) = 4n$. $$nx^2 + n(n+1)x + \frac{n(n-1)(2n+5)}{6} = 4n$$ $$x^2 + (n+1)x + \frac{(n-1)(2n+5)}{6} = 4$$ $D$ must be a perfect square. $$D = \frac{122 - 2n^2}{6} = 20 - \left(\frac{n^2 - 1}{3}\right)$$ If make perfect square $\Rightarrow \frac{n^2 - 1}{3} = 16 \Rightarrow n = 7$ So Equation is $$\Rightarrow x^2 + 8x + \frac{8 \times 15}{6} - 5 = 0$$ $$x^2 + 8x + 15 = 0$$ $$x = -3, -5$$ $$\alpha = -5, \alpha + 2 = -3$$ $$\alpha + n = 7 - 5 = 2$$

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $x$ and $y$ be real number such that $50\left(\frac{2x}{1+3i}-\frac{y}{1-2i}\right)=31+17i$, $i=\sqrt{-1}$. Then the value of $10(x-3y)$ is:

  1. 20
  2. 31
  3. 35
  4. 75

Answer: (d)

Solution

Given the equation: $$50 \left( \frac{2x(1-3i)}{10} - \frac{y(1+2i)}{5} \right) = 31 + 17i$$ Expanding and simplifying, we have: $$10x - (30x)i - 10y - 20y \, i = 31 + 17i$$ This leads to the system of equations: $$\begin{cases} 10(x-y) = 31 \\ -30x - 20y = 17 \end{cases}$$ Solving the system, we get: $$30x - 30y = 93$$ $$-10(3x + 2y) = 17$$ From which: $$x = 0.9, \; y = -2.2$$ Finally, we have: $$10(x - 3y) = 75$$

Question 3

Maths · Determinants · Single correct

Let $\alpha, \beta \in \mathbb{R}$ be such that the system of linear equations \ $x + 2y + z = 5$ \ $2x + y + \alpha z = 5$ \ $8x + 4y + \beta z = 18$ \ has no solution. Then $\frac{\beta}{\alpha}$ is equal to:

  1. -4
  2. 4
  3. 8
  4. -8

Answer: (b)

Solution

Given the system of equations: $$x + 2y + z = 5$$ $$2x + y + \alpha z = 5$$ $$8x + 4y + \beta z = 18$$ For no solution, the determinant is: $$\Delta = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{vmatrix} = 0$$ Calculating the determinant: $$\Delta = 1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1 \cdot 0 = 0$$ Simplifying: $$= \beta - 4\alpha - 4\beta + 16\alpha = 0$$ This implies: $$\Rightarrow 4\alpha = \beta$$

Question 4

Maths · Matrices · Single correct

Let $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$. If $A^2 - 4A + I = O$ and $B^2 - 5B - 6I = O$, then among the two statements: (S1): $[(B - A)(B + A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$ and (S2): $\det(adj(A + B)) = -5$,

  1. only (S1) is correct
  2. only (S2) is correct
  3. both (S1) and (S2) are correct
  4. both (S1) and (S2) are wrong

Answer: (b)

Solution

Given the matrix $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$. The equation $A^2 - 4A + I = 0$ gives the trace of the matrix. From this, $\alpha + 1 = 4$, so $\alpha = 3$. Now consider the matrix $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$. The equation $B^2 - 5B - 6I = 0$ gives the determinant. Calculating, $\det(B) = 6 - 3\beta = -6$. Solving, $3\beta = 12$, so $\beta = 4$. Now, $B - A = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$ and $B + A = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$. The product $(B - A)(B + A) = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$. The adjugate matrices are $adj(A) = \begin{bmatrix} 3 & -2 \\ -1 & 1 \end{bmatrix}$ and $adj(B) = \begin{bmatrix} 2 & -3 \\ -4 & 3 \end{bmatrix}$. Finally, $adj(A + B) = \begin{bmatrix} 5 & -5 \\ -5 & 4 \end{bmatrix}$.

Question 5

Maths · Sets · Single correct

Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let B be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements is A $\cap$ B, which are divisible by 3, is:

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

Answer: (b)

Solution

AP Set A = $\{$1, 6, 11, 16 $\ldots$ 101 terms$\}$ AP Set B = $\{$9, 16 $\ldots$ 71 terms$\}$ D = L.C.M $\{$d_1, d_2$\}$ = 35 1^{st} Common term is 16 16 + (n - 1) 35 $\leq$ 499 n $\leq$ 14.8 $\Rightarrow$ n = 14 A $\cap$ B = $\{$16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471$\}$ terms divisible by '3' = $\{$51, 156, 261, 366, 471$\}$ Ans. 5 terms

Question 6

Maths · Permutations and Combinations · Single correct

The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 5 and 7 such that each digit is used at least once, is ;

  1. 15400
  2. 17800
  3. 16800
  4. 29400

Answer: (c)

Solution

Exactly one digit repeated $\rightarrow \binom{5}{1} \times \frac{7!}{3!} = 4200$ Exactly two digits repeated $\rightarrow \binom{5}{2} \times \frac{7!}{2!2!} = 12600$ Total numbers $= 16800$

Question 7

Maths · Binomial Theorem · Single correct

The number of elements in the set \[ S=\left\{(r,k):k\in\mathbb{Z}\text{ and }{}^{36}C_{r+1} =\frac{6({}^{35}C_r)}{k^2-3}\right\} \] is

  1. 2
  2. 4
  3. 8
  4. 16

Answer: (b)

Solution

Given $\frac{36}{r+1} \cdot \binom{35}{r} = 6 \cdot \frac{\binom{35}{r}}{k^2 - 3}$. $$k^2 - 3 = \frac{r+1}{6}$$ $$k^2 = \frac{r+19}{6}$$ $0 \leq r \leq 35$ $19 \leq r + 19 \leq 54$ $$3.1 \leq \frac{r+19}{6} \leq 9$$ Therefore, $k^2 = 4, 9$ $k = \pm 2, \pm 3$ Thus, $(r, k) \Rightarrow (5, \pm 2) \Rightarrow (35, \pm 3)$ Therefore, $4$

Question 8

Maths · Statistics · Single correct

If the mean of the data \begin{tabular}{|c|c|c|c|c|c|c|} \hline Class & 5-10 & 10-15 & 15-20 & 20-25 & 25-30 & 30-35 \\ \hline Frequency & 2 & $k$ & 28 & 54 & $k+1$ & 5 \\ \hline \end{tabular} is $21$, then $k$ is one of the roots of the equation:

  1. $2x^2 - 23x - 10 = 0$
  2. $4x^2 - 35x + 24 = 0$
  3. $2x^2 - 19x - 10 = 0$
  4. $2x^2 - 35x + 98 = 0$

Answer: (c)

Solution

Given the table: Class: 5-10, 10-15, 15-20, 20-25, 25-30, 30-35 Freq.: 2, k, 28, 54, k+1, 5 The mean is calculated as: $$21 = \frac{15 + 12.5k + 490 + 1215 + 27.5 + 27.5k + 162.5}{90 + 2k}$$ Simplifying, we have: $$1890 + 42k = 1910 + 40k$$ Solving for $k$ gives: $$k = 10$$

Question 9

Maths · Properties of Triangles · Single correct

Let the mid points of the sides of a triangle ABC be $\left( \frac{5}{2}, 7 \right)$, $\left( \frac{5}{2}, 3 \right)$ and (4, 5). If its incentre is (h, k), then 3h + k is equal to:

  1. 11
  2. 12
  3. 13
  4. 14

Answer: (c)

Solution

The coordinates of point I are given by $I = \left( \frac{24}{9}, 5 \right)$. Therefore, $3h + k = 8 + 5 = 13$.

Question 10

Maths · Conic Sections · Single correct

Let an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a < b$, pass through the point $(4, 3)$ and have eccentricity $\frac{\sqrt{5}}{3}$. Then the length of its latus rectum is:

  1. $\frac{4\sqrt{5}}{3}$
  2. $2\sqrt{5}$
  3. $\frac{7\sqrt{5}}{3}$
  4. $\frac{8\sqrt{5}}{3}$

Answer: (d)

Solution

Given $e^2 = 1 - \frac{a^2}{b^2}$, we have $\frac{a^2}{b^2} = 1 - e^2 = 1 - \frac{5}{9} = \frac{4}{9}$. Therefore, $\frac{a^2}{b^2} = \frac{4}{9} \ldots$ (1) The curve passes through $(4, 3)$, so $\frac{16}{a^2} + \frac{9}{b^2} = 1 \ldots$ (2) From equations (1) and (2), we find $a^2 = 20$ and $b^2 = 45$. The length of the latus rectum (LR) is given by: $$LR = \frac{2a^2}{b} = \frac{2(20)}{3\sqrt{5}} = \frac{8\sqrt{5}}{3}$$

Question 11

Maths · Trigonometric Functions · Single correct

If $\sin\left(\frac{\pi}{18}\right) \sin\left(\frac{5\pi}{18}\right) \sin\left(\frac{7\pi}{18}\right) = K$, then the value of $\sin\left(\frac{10K\pi}{3}\right)$ is:

  1. $\frac{\sqrt{3} + 1}{2\sqrt{2}}$
  2. $\frac{\sqrt{3} - 1}{\sqrt{2}}$
  3. $\frac{\sqrt{3}}{2}$
  4. $\frac{1}{2}$

Answer: (a)

Solution

Given $K = \sin 10^\circ \sin 50^\circ \sin 70^\circ$. $$= \frac{1}{4} \sin 30^\circ = \frac{1}{8}$$ $$\sin 10K \frac{\pi}{3} = \sin \left( 10 \times \frac{1}{8} \cdot \frac{\pi}{3} \right) = \sin \frac{5\pi}{12} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$$

Question 12

Maths · Trigonometric Functions · Single correct

Let $S = \{ x \in [-\pi, \pi] : \sin x (\sin x + \cos x) = a, \ a \in \mathbb{Z} \}$. Then $n(S)$ is equal to :

  1. 3
  2. 6
  3. 7
  4. 9

Answer: (d)

Solution

Given $\sin x (\sin x + \cos x) \in \left[ \frac{1 - \sqrt{2}}{2}, \frac{1 + \sqrt{2}}{2} \right]$. 2 integer values will be there $\Rightarrow a = 0, 1$. If $a = 0$, then $\sin x (\sin x + \cos x) = 0$. $\Rightarrow \sin x = 0$ or $\sin x + \cos x = 0$. $\boxed{x = -\pi, 0, \pi}$ $= \tan x = -1$ $\boxed{x = -\frac{\pi}{4}, \frac{3\pi}{4}}$ 2 solutions If $a = 1$, then $\sin x (\sin x + \cos x) = 1$. $\sin^2 x + \sin x \cos x = 1$ $2 \sin^2 x + 2 \sin x \cos x = 2$ $1 - \cos 2x + \sin 2x = 2$ $\sin 2x - \cos 2x = 1 \Rightarrow square$ $\sin 4x = 0$ $4x = -4\pi, -3\pi, \ldots, 3\pi, 4\pi$ $x = -\pi, -\frac{3\pi}{4}, -\frac{\pi}{2}, -\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$ $\Rightarrow x = -\frac{3\pi}{4}, -\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{2}$ 4 solutions Total 9 solutions

Question 13

Maths · Three Dimensional Geometry · Single correct

If the point of intersection of the lines $\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7}$ and $\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7}$ lies on xy-plane, then the value of $a + b$ is:

  1. 2
  2. 5
  3. 7
  4. 9

Answer: (c)

Solution

Line (L_1) $$\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} = r_1$$ General point P on L_1 is $(3r_1 - 1, 5r_1 - a, 7r_1 - b - 1)$. Line (L_2) $$\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} = r_2$$ General point Q on L_2 is $(r_2 + 2, 4r_2 + b, 7r_2 + 2a)$. For point of intersection $$3r_1 - 1 = r_2 + 2 \Rightarrow r_2 = 3r_1 - 3 (1)$$ $$5r_1 - a = 4r_2 + b (2)$$ $$7r_1 - b - 1 = 7r_2 + 2a (3)$$ Since point lies on XY plane, $z$-coordinate = 0. From L_1: $7r_1 - b - 1 = 0 \Rightarrow 7r_1 = b + 1$. From L_2: $7r_2 + 2a = 0 \Rightarrow 2a = -7r_2$. Substitute $r_2 = 3r_1 - 3$ $$2a = -7(3r_1 - 3)$$ $$a = \frac{-21r_1 + 21}{2}$$ Put in equation (2) $$5r_1 - a = 4r_2 + b$$ $$5r_1 - \left( \frac{-21r_1 + 21}{2} \right) = 4(3r_1 - 3) + (7r_1 - 1)$$ $$\Rightarrow r_1 = \frac{5}{7}$$ $$\Rightarrow b = 7r_1 - 1 = 4$$ $$\Rightarrow a = 3 \therefore a + b = 7$$

Question 14

Maths · Vector Algebra · Single correct

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 2$ and $|\vec{b}| = 3$, then the maximum value of $3\left|3\vec{a} + 2\vec{b}\right| + 4\left|3\vec{a} - 2\vec{b}\right|$ is:

  1. 30
  2. 36
  3. 60
  4. 72

Answer: (c)

Solution

Given $$E = 3 \sqrt{9a^2 + 4b^2 + 12 \overline{a} \cdot \overline{b}} + 4 \sqrt{9a^2 + 4b^2 - 12 \overline{a} \cdot \overline{b}}$$ Simplifying, we have: $$= 3 \sqrt{36 + 36 + 12 \times 6 \cos \theta} + 4 \sqrt{36 + 36 - 72 \cos \theta}$$ $$= 3 \sqrt{72 + 72 \cos \theta} + 4 \sqrt{72 - 72 \cos \theta}$$ $$= 18 \sqrt{2} \sqrt{1 + \cos \theta} + 24 \sqrt{2} \sqrt{1 - \cos \theta}$$ $$= 18 \sqrt{2} \sqrt{2} \cos \frac{\theta}{2} + 24 \sqrt{2} \left( \sqrt{2} \sin \frac{\theta}{2} \right)$$ $$= 36 \left( \cos \frac{\theta}{2} \right) + 48 \sin \left( \frac{\theta}{2} \right)$$ The maximum value of $E$ is given by: $$E_{\max} = \sqrt{(36)^2 + (48)^2} = \sqrt{6^4 + 6^2 \times 8^2} = 60$$

Question 15

Maths · Three Dimensional Geometry · Single correct

Let a line L passing through the point (1, 1, 1) be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + \hat{k}$. If $P(a, b, c)$ is the foot of perpendicular from the origin on the line L, then the value of $34(a + b + c)$ is:

  1. 50
  2. 80
  3. 100
  4. 120

Answer: (c)

Solution

Direction vector of line is $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} = -2\hat{i} + 3\hat{j} - 2\hat{k}$$ So equation of line is $$\frac{x-1}{2} = \frac{y-1}{-3} = \frac{z-1}{2} = \lambda let$$ let $(2\lambda + 1, -3\lambda + 1, 2\lambda + 1)$ be the foot of perpendicular from $(0, 0, 0)$ upon given line $$\Rightarrow \overline{OM} \cdot (2\hat{i} - 3\hat{j} + 2\hat{k}) = 0$$ $$\Rightarrow 2(2\lambda + 1) - 3(-3\lambda + 1) + 2(2\lambda + 1) = 0$$ $$\Rightarrow \lambda = -\frac{1}{17}$$ So $(a, b, c) \equiv \left(\frac{-2}{17} + 1, \frac{3}{17} + 1, \frac{-2}{17} + 1\right) \Rightarrow 34(a + b + c) = 100$$

Question 16

Maths · Limits and Derivatives · Single correct

If $\lim_{x \to 2} \frac{\sin(x^3 - 5x^2 + ax + b)}{(\sqrt{x-1} - 1) \log_e(x-1)} = m$, then $a + b + m$ is equal to:

  1. 5
  2. 6
  3. 8
  4. 10

Answer: (b)

Solution

Since the denominator is 0 at $x = 2$, the numerator must also be 0 at $x = 2$. Therefore, $$2^3 - 5(2)^2 + a(2) + b = 0$$ which simplifies to $$2a + b = 12 (1)$$ Now, $$m = \lim_{x \to 2} \frac{\sin(x^3 - 5x^2 + ax + b)}{(x^3 - 5x^2 + ax + b)} \cdot \frac{(x^3 - 5x^2 + ax + b)}{\left(\frac{x-1-1}{\sqrt{x-1}+1}\right) \cdot \frac{\log_e(1+(x-2))}{(x-2)} \cdot (x-2)}$$ This simplifies to $$m = \lim_{x \to 2} \frac{x^3 - 5x^2 + ax + b}{(x-2)^2}$$ Further simplifying, $$m = \lim_{x \to 2} \frac{3x^2 - 10x + a}{2(x-2)}$$ Since the denominator is 0 at $x = 2$, the numerator must also be 0 at $x = 2$. Therefore, $$3(2)^2 - 10(2) + a = 0 \Rightarrow a = 8$$ Substituting $a = 8$ in (1), we get $b = -4$. Thus, $$m = \lim_{x \to 2} \left(\frac{6x - 10}{2}\right) \Rightarrow m = 2$$ Therefore, $a + b + m = 6$.

Question 17

Maths · Differential Equations · Single correct

If the curve $y = f(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y (2 + \log_e x) \, dx$, $x > 0$, then $f(e)$ is equal to:

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

Answer: (c)

Solution

Given $\frac{dy}{dx} = y(2 + \ln x)$. Integrating, $\int \frac{dy}{y} = \int (2 + \ln x) \, dx$. $\ln y = 2x + x \ln x - x + C$. $\ln y = x + x \ln x + C$. Since it passes through $(1, e)$, $1 = 1 + 0 + C \Rightarrow C = 0$. $\ln y = x + x \ln x$. $f(x) = y = e^{x + x \ln x}$. $\Rightarrow f(e) = e^{e + e} = e^{2e}$.

Question 18

Maths · Continuity and Differentiability · Single correct

The number of critical points of the function \[ f(x)= \begin{cases} \left|\dfrac{\sin x}{x}\right|, & x\ne 0,\\ 1, & x=0 \end{cases} \] in the interval $(-2\pi,2\pi)$ is equal to:

  1. 1
  2. 3
  3. 5
  4. 7

Answer: (c)

Solution

Given $\lim_{x \to 0} \left| \frac{\sin x}{x} \right| = 1 = f(0)$, $\Rightarrow f(x)$ is continuous. Now, $\frac{d}{dx} \left( \frac{\sin x}{x} \right) = \frac{x \cos x - \sin x}{x^2}$. $\Rightarrow f'(x) = 0 \Rightarrow \tan x = x$. There are 3 solutions in $(-2\pi, 2\pi)$. Also, $f'(0)$ does not exist at $x = -\pi, \pi$. Total 5 points.

Question 19

Maths · Integrals · Single correct

Let [.] denote the greatest integer function. Then the value of $$\int_{0}^{3} \frac{e^x + e^{-x}}{[x]!} \, dx$$ is:

  1. $e^2 + e^3 - \frac{1}{e^2} - \frac{1}{e^3}$
  2. $\frac{1}{2} \left( e^2 + e^3 - \frac{1}{e^2} - \frac{1}{e^3} \right)$
  3. $e^2 + e^3 - \frac{1}{2e^2} - \frac{1}{2e^3}$
  4. $\frac{1}{2} (e^2 + e^3) - \frac{1}{e^2} - \frac{1}{e^3}$

Answer: (b)

Solution

The solution involves evaluating the integrals: $$\int_0^1 \frac{e^x + e^{-x}}{1} \, dx + \int_1^2 \frac{e^x + e^{-x}}{1} \, dx + \int_2^3 \frac{e^x + e^{-x}}{2} \, dx$$ This can be rewritten as: $$\int_0^2 e^x \, dx + \int_0^2 e^{-x} \, dx + \frac{1}{2} \int_2^3 (e^x + e^{-x}) \, dx$$ Evaluating each integral, we have: $$(e^2 - 1) + \frac{(e^{-2} - 1)}{-1} + \frac{1}{2} \left((e^3 - e^2) - (e^{-3} - e^{-2})\right)$$ Simplifying further: $$\frac{e^3 + e^2}{2} + 1 - 1 - \frac{e^{-2}}{2} - \frac{e^{-3}}{2}$$ Finally, the expression simplifies to: $$\frac{e^3 + e^2 - e^{-2} - e^{-3}}{2}$$

Question 20

Maths · Differential Equations · Single correct

Let y = y(x) be the solution curve of the differential equation (1 + $\sin$ x) $\frac{dy}{dx}$ + (y + 1) $\cos$ x = 0, y(0) = 0. If the curve y = y(x) passes through the point $\left( \alpha, -\frac{1}{2} \right)$, then a value of $\alpha$ is :

  1. $\frac{\pi}{6}$
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{2}$

Answer: (d)

Solution

The solution starts with the integral equation: $$\int \frac{dy}{y+1} = -\int \frac{\cos x}{1+\sin x} \, dx$$ This implies $$\ln(y+1) = -\ln(1+\sin x) + C$$ Given $y(0) = 0$, it follows that $C = 0$. Thus, $$y + 1 = \frac{1}{1+\sin x}$$ The curve passes through the point $$\left( \alpha, -\frac{1}{2} \right)$$ This implies $\sin x = 1$. Therefore, $$x = \frac{\pi}{2} (as per options)$$

Question 21

Maths · Relations and Functions · Numerical

If the domain of the function $f(x) = \sqrt{\log_{(0.6)} \left( \frac{2x - 5}{x^2 - 4} \right)}$ is $(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$, then the value of $a + b + c + d + e$ is $\ldots$.

Answer: 4

Solution

For domain $\log_{0.6} \left| \frac{2x-5}{x^2-4} \right| \geq 0$. $$\left| \frac{2x-5}{x^2-4} \right| \leq 1 \& x \neq \frac{5}{2} \ldots (1)$$ $$-1 \leq \frac{2x-5}{x^2-4} \leq 1$$ $$\frac{2x-5}{x^2-4} + 1 \geq 0$$ $$\frac{x^2 + 2x - 9}{x^2-4} \geq 0$$ $$\frac{(x+1)^2 - 10}{(x-2)(x+2)} \geq 0$$ $$x \in (-\infty, -1-\sqrt{10}] \cup (2,2) \cup [-1+\sqrt{10}, \infty) \ldots (2)$$ $$\frac{2x-5}{x^2-4} - 1 \leq 0$$ $$\frac{2x-5-x^2+4}{x^2-4} \leq 0$$ $$\frac{x^2 - 2x + 1}{x^2-4} \geq 0$$ $$\frac{(x-1)^2}{(x-2)(x+2)} \geq 0$$ $$x \in (-\infty, -2) \cup (2, \infty) \cup \{1\} \ldots (3)$$ $$(1) \cap (2) \cap (3)$$ $$x \in (-\infty, -1-\sqrt{10}] \cup \{1\} \cup \left[-1+\sqrt{10}, \frac{5}{2}\right) \cup \left(\frac{5}{2}, \infty\right)$$ a + b + c + d + e = -2 + 1 + 5 = 4

Question 22

Maths · Sequences and Series · Numerical

If $\sum_{k=1}^{n} a_k = 6n^3$, then $\sum_{k=1}^{6} \left( \frac{a_{k+1} - a_k}{36} \right)^2$ is equal to $\ldots$

Answer: 91

Solution

Given $a_1 + a_2 + \ldots + a_n = 6n^3$. $a_1 + a_2 + \ldots + a_n + a_{n+1} = 6(n+1)^3$ $a_{n+1} = 6(n+1)^3 - 6n^3$ $$= 6 \left( (n+1) - n \right) \left( (n+1)^2 + n^2 + n(n+1) \right)$$ $a_{n+1} = 6(1) \left( 3n^2 + 3n + 1 \right)$ $a_n = 6 \left( 3(n-1)^2 + 3(n-1) + 1 \right)$ $$= 6 \left( 3(n^2 - 2n + 1 + n - 1) + 1 \right)$$ $$= 6 \left( 3(n^2 - n) + 1 \right)$$ $a_n = 6 \left( 3n^2 - 3n + 1 \right)$ $$\sum_{k=1}^{6} \left( \frac{6(3k^2 + 3k + 1) - 6(3k^2 - 3k + 1)}{36} \right)^2$$ $$\sum_{k=1}^{6} \left( \frac{36k}{36} \right)^2 = \sum_{k=1}^{6} k^2 = \frac{6 \times 7 \times 13}{6} = 91$$

Question 23

Maths · Complex Numbers and Quadratic Equations · Numerical

Let $a, b, c \in \{1,2,3,4\}$. If the probability, that $$ax^2 + 2\sqrt{2} bx + c > 0$$ for all $x \in \mathbb{R}$, is $\frac{m}{n}$, gcd $(m,n) = 1$, then m + n is equal to $\ldots$

Answer: 81

Solution

Given $8b^2 - 4ac < 0 \Rightarrow 2b^2 < ac$. For $b = 1$, $a = 1$, $c = 3, 4 \rightarrow 2$. For $b = 1$, $a = 2$, $c = 2, 3, 4 \rightarrow 3$. For $b = 1$, $a = 3$, $c = 1, 2, 3, 4 \rightarrow 4$. For $b = 1$, $a = 4$, $c = 1, 2, 3, 4 \rightarrow 4$. For $b = 2$, $a = 3$, $c = 3, 4 \rightarrow 2$. For $b = 2$, $a = 4$, $c = 3, 4 \rightarrow 2$. Total: 17. Required probability is given by $$\frac{17}{64} \equiv \frac{m}{n}$$ Therefore, $m + n = 17 + 64 = 81$.

Question 24

Maths · Conic Sections · Numerical

Let a circle C have its centre in the first quadrant, intersect the coordinate axes at exactly three points and cut off equal intercepts from the coordinate axes. If the length of the chord of C on the line $x + y = 1$ is $\sqrt{14}$, then the square of the radius of C is $\ldots$.

Answer: 8

Solution

Given the diagram, we have the equation for the radius $r$ as follows: $$r^2 = \left( \frac{\sqrt{14}}{2} \right)^2 + \left( r - \frac{1}{\sqrt{2}} \right)^2$$ Expanding and simplifying the equation: $$r^2 = \frac{7}{2} + r^2 - \sqrt{2}r + \frac{1}{2}$$ Rearranging terms gives: $$\sqrt{2}r = 4$$ Solving for $r$: $$r = 2\sqrt{2}$$ Thus, the square of the radius is: $$r^2 = 8$$

Question 25

Maths · Integrals · Numerical

If $\alpha = \int_{0}^{2\sqrt{3}} \log_{2}(x^2 + 4) \, dx + \int_{2}^{4} \sqrt{2^x - 4} \, dx$, then $\alpha^2$ is equal to $\ldots$

Answer: 192

Solution

As we know that $$\int_{a}^{b} f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1}(x) \, dx = bf(b) - af(a)$$ Therefore, $\alpha = (2\sqrt{3})(4) - (0)(2)$ $$= 8\sqrt{3}$$ $$\Rightarrow \alpha^2 = 192$$

Physics

Question 26

Physics · Physical World, Units and Measurements · Single correct

The dimensional formula of $\frac{1}{2} \varepsilon_0 E^2$ ($\varepsilon_0 =$ permittivity of vacuum and $E =$ electric field) is $M^a L^b T^c$. The value of $2a - b + c = \ldots$

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

Answer: (b)

Solution

Given $\frac{1}{2} \varepsilon_0 E^2 = Energy density$. Therefore, $$\left[ \frac{1}{2} \varepsilon_0 E^2 \right] = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$$ Thus, $a = 1$, $b = -1$, $c = -2$. Therefore, $2a - b + c = 2 + 1 - 2 = 1$

Question 27

Physics · Mechanical Properties of Solids · Single correct

The diameter of a wire measured by a screw gauge of least count 0.001 cm is 0.08 cm. The length measured by a scale of least count 0.1 cm is 150 cm. When a weight of 100 N is applied to the wire, the extension in length is 0.5 cm, measured by a micrometer of least count 0.001 cm. The error in the measured Young's modulus is $\alpha \times 10^9 \, \mathrm{N/m^2}$. The value of $\alpha$ is $\ldots$. (Ignore the contribution of the load to Young's modulus error calculation)

  1. 1.3
  2. 1.65
  3. 0.13
  4. 0.25

Answer: (b)

Solution

Given $Y = \frac{F \ell}{A \Delta \ell}$. Therefore, $$\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta \ell}{\ell} + 2 \frac{\Delta d}{d} + \frac{\Delta (\Delta \ell)}{\Delta \ell}$$ $$\left[ A = \frac{\pi d^2}{4} \Rightarrow \frac{\Delta A}{A} = \frac{2 \Delta d}{d} \right]$$ Since $F$ is known exactly $\Rightarrow \Delta F = 0$ $$\frac{\Delta Y}{Y} = \frac{0.1}{150} + \frac{2 \times 0.001}{0.08} + \frac{0.001}{0.5} = 0.0277$$ Therefore, $\Delta Y = 0.0277 \, Y$ $$Y = \frac{100 \times 1.5}{\left( \frac{\pi}{4} \left( \frac{0.08}{100} \right)^2 \right) \times 0.5 \times 10^{-2}} = 5.97 \times 10^{10} \, \mathrm{N/m^2}$$ Therefore, $\Delta Y = 0.0277 \times 5.97 \times 10^{10} \, \mathrm{N/m^2} = 1.65 \times 10^9 \, \mathrm{N/m^2}$

Question 28

Physics · Motion in a Plane · Single correct

The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k} \, \mathrm{m/s}$. The magnitude of acceleration at point $(1, 2, 4)$ is $\ldots \, \mathrm{m/s^2}$.

  1. $\sqrt{6}$
  2. 9
  3. $\sqrt{33}$
  4. 0

Answer: (b)

Solution

Given $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$. The acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -\frac{dx}{dt}\hat{i} + 2\frac{dy}{dt}\hat{j} - \frac{dz}{dt}\hat{k}$. This simplifies to $= -(-x)\hat{i} + 2(2y)\hat{j} - (-z)\hat{k} = x\hat{i} + 4y\hat{j} + z\hat{k}$. Therefore, $\vec{a}$ at $(1, 2, 4) = \hat{i} + 8\hat{j} + 4\hat{k}$. Thus, $|\vec{a}| = 9 \, \mathrm{m/s^2}$.

Question 29

Physics · System of Particles and Rotational Motion · Single correct

The position of an object having mass $0.1\,\mathrm{kg}$ as a function of time $t$ is given by $\vec{r}=(10t^2\hat{i}+5t^3\hat{j})\,\mathrm{m}$. At $t=1\,\mathrm{s}$, which of the following statements are correct? \ (A) The linear momentum $\vec{p}=(2\hat{i}+1.5\hat{j})\,\mathrm{kg\,m/s}$. \ (B) The force acting on the object $\vec{F}=(2\hat{i}+3\hat{j})\,\mathrm{N}$. \ (C) The angular momentum of the object about its origin $\vec{L}=15\hat{k}\,\mathrm{J\,s}$. \ (D) The torque acting on the object about its origin $\vec{\tau}=20\hat{k}\,\mathrm{N\,m}$. \ Choose the correct answer from the options given below.

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

Answer: (d)

Solution

Given $\vec{r} = 10t^2 \hat{i} + 5t^3 \hat{j}$ implies $\vec{r}_{t=1} = 10 \hat{i} + 5 \hat{j}$. $\vec{v} = \frac{d\vec{r}}{dt} = 20t \hat{i} + 15t^2 \hat{j}$ implies $\vec{v}_{t=1 sec} = 20 \hat{i} + 15 \hat{j}$. $\vec{a} = \frac{d^2\vec{r}}{dt^2} = 20 \hat{i} + 30t \hat{j}$ implies $\vec{a}_{t=1 sec} = 20 \hat{i} + 30 \hat{j}$. Therefore, $\vec{P} = m\vec{v} = 2 \hat{i} + 1.5 \hat{j}$. $\vec{F} = m\vec{a} = 2 \hat{i} + 3 \hat{j}$. $\vec{\tau} = \vec{r} \times \vec{F} = 20 \hat{k}$. $\vec{L} = \vec{r} \times \vec{P} = 5 \hat{k}$.

Question 30

Physics · Gravitation · Single correct

A planet $(P_1)$ is moving around the star of mass $2M$ in the orbit of radius $R$. Another planet $(P_2)$ is moving around another star of mass $4M$ in a orbit of radius $2R$. Ratio of time periods of revolution of $P_2$ and $P_1$ is $\ldots$.

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

Answer: (b)

Solution

From Kepler's law, $$T^2 = \frac{4\pi^2}{GM_S} r^3$$ Therefore, $$T_{p_1} = \frac{2\pi}{\sqrt{G(2M)}} R^{3/2}$$ $$T_{p_2} = \frac{2\pi}{\sqrt{G(4M)}} (2R)^{3/2}$$ Thus, $$\frac{T_{p_2}}{T_{p_1}} = \frac{(2)^{3/2}}{\sqrt{4}} \times \frac{\sqrt{2}}{1} = \frac{2\sqrt{2} \times \sqrt{2}}{2} = 2$$

Question 31

Physics · System of Particles and Rotational Motion · Single correct

A particle is rotating in a circular path and at any instant its motion can be described as $\theta = \frac{5t^4}{40} - \frac{t^3}{3}$. The angular acceleration of the particle after 10 seconds is $\ldots$ rad/s$^2$.

  1. 150
  2. 120
  3. 130
  4. 170

Answer: (c)

Solution

Given $\theta = \frac{5t^4}{40} - \frac{t^3}{3}$. Differentiating with respect to $t$: $$\frac{d\theta}{dt} = \frac{5}{40} (4t^3) - \frac{3t^2}{3} = \frac{t^3}{2} - t^2.$$ Differentiating again: $$\alpha = \frac{d^2\theta}{dt^2} = \frac{3t^2}{2} - 2t.$$ At $t = 10$: $$\alpha = \frac{3}{2} (10)^2 - 2 \times 10 = 130 \, \mathrm{rad/sec^2}.$$

Question 32

Physics · Electrostatic Potential and Capacitance · Single correct

A parallel plate air capacitor has a capacitance C. When it is half filled as shown in figure with a dielectric constant K = 5, the percentage increase in the capacitance is $\ldots$.

  1. 33.34
  2. 66.67
  3. 200
  4. 400

Answer: (b)

Solution

Initial capacitance $$C_i = \frac{A \epsilon_0}{d} = C$$ Final capacitance $$C_1 = \frac{A \epsilon_0 \times 5}{\frac{d}{2}} = 10C$$ $$C_2 = \frac{A \epsilon_0}{\frac{d}{2}} = 2C$$ $$C_f = \frac{C_1 C_2}{C_1 + C_2} = \frac{10 \times 2}{10 + 2} C = \frac{5}{3} C$$ Percentage increase $$\frac{\frac{5}{3} C - C}{C} \times 100 = 66.67$$

Question 33

Physics · Thermodynamics · Single correct

Heat is supplied to a diatomic gas at constant pressure. Then the ratio of $\Delta Q : \Delta U : \Delta W$ is _____.

  1. 2 : 3 : 5
  2. 5 : 3 : 2
  3. 2 : 5 : 7
  4. 7 : 5 : 2

Answer: (d)

Solution

Given $\Delta Q_P = nC_P \Delta T = \frac{7}{2} nR \Delta T$. $\Delta U = nC_V \Delta T = \frac{5}{2} nR \Delta T$. $W = nR \Delta T$. $\Delta Q : \Delta U : W :: 7 : 5 : 2$.

Question 34

Physics · Electrostatic Potential and Capacitance · Single correct

Two charged conducting spheres $S_1$ and $S_2$ of radii $8 \, \mathrm{cm}$ and $18 \, \mathrm{cm}$ are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on $S_1$ and $S_2$ spheres are $E_{S_1}$ and $E_{S_2}$ respectively. The value of $\frac{E_{S_1}}{E_{S_2}}$ is $\ldots$.

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

Answer: (d)

Solution

Assuming very far. Given $V_1 = V_2$, we have $\frac{kQ_1}{8} = \frac{kQ_2}{18}$. This implies $V_{surface} = \frac{kQ}{R}$. Therefore, $\frac{Q_1}{Q_2} = \frac{4}{9}$. For the electric fields, $\frac{E_{S_1}}{E_{S_2}} = \frac{kQ_1}{r_1^2} \frac{r_2^2}{kQ_2}$. This implies $E_{surface} = \frac{kQ}{R^2}$. Thus, $\frac{Q_1}{Q_2} \left( \frac{r_2}{r_1} \right)^2 = \frac{4}{9} \times \left( \frac{9}{4} \right)^2 = \frac{9}{4}$.

Question 35

Physics · Waves · Single correct

The equation of a plane progressive wave is given by $y = 5 \cos \pi \left( 200t - \frac{x}{150} \right)$ where $x$ and $y$ are in cm and $t$ is in second. The velocity of the wave is $\ldots$ m/s

  1. 120
  2. 150
  3. 200
  4. 300

Answer: (d)

Solution

The velocity $v$ is given by the ratio of the coefficient of $t$ to the coefficient of $x$. Therefore, $$v = \frac{coeff of t}{coeff of x} = \frac{200}{\frac{1}{150} \times 100} = 300 \, \mathrm{m/s}.$$

Question 36

Physics · Electric Charges and Fields · Single correct

Two short electric dipoles A and B having dipole moment $p_1$ and $p_2$ respectively are placed with their axis mutually perpendicular as shown in the figure. The resultant electric field at a point x is making an angle of $60^\circ$ with the line joining points O and x. The ratio of the dipole moments $p_2/p_1$ is $\ldots$.

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

Answer: (b)

Solution

Given the vectors $\vec{E}_A$, $\vec{E}_B$, and $\vec{E}_{net}$, we have: $$\vec{E}_A = \frac{2kP_1}{x^3} \left( -\hat{i} \right) \{Axial point\}$$ $$\vec{E}_B = \frac{kP_2}{x^3} \left( -\hat{j} \right) \{Equatorial point\}$$ Now: $\tan \theta = \frac{|\vec{E}_B|}{|\vec{E}_A|}$ $$\Rightarrow \tan 60^\circ = \frac{P_2}{2P_1}$$ $$\Rightarrow \frac{P_2}{P_1} = 2\sqrt{3}$$

Question 37

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

For the given circuit (shown in part (A)) the time dependent input voltage $v_{in}(t)$ and corresponding output $v_0(t)$ are shown in part (B) and part (C), respectively. Identify the components that are used in the circuit between points X and Y.

Answer: (b)

Solution

For the +ve half cycle of input, the middle branch has an ideal diode in forward bias and a Zener diode in reverse bias. Thus, we get a constant 5V output due to Zener breakdown. For this situation, the component between x and y should act as an open circuit. For -ve half cycle of input, the middle branch has an ideal diode in reverse bias, which makes it an open branch without any current. For this situation, the component between x and y should give a constant -5V output due to Zener breakdown. (Note: The input side must contain some resistance else $V_{in}$ is equal to $V_{out}$ for every case.)

Question 38

Physics · Electromagnetic Induction · Single correct

When a coil is placed in a time dependent magnetic field the power dissipated in it is $P$. The number of turns, area of the coil and radius of the coil wire are $N$, $A$ and $r$ respectively. For a second coil number of turns, area of the coil and radius of the coil wire are $2N$, $2A$ and $3r$ respectively. When the first coil is replaced with second coil the power dissipated in it is $\sqrt{2} \alpha P$. The value of $\alpha$ is $\ldots$.

  1. 36
  2. 128$\sqrt{2}$
  3. 16
  4. 64

Answer: (a)

Solution

Power dissipated: $P = \frac{\varepsilon^2}{R}$. Induced emf $\varepsilon = -NA \frac{dB}{dt}$. Resistance of coil: $R = \rho \frac{\ell}{\pi r^2}$. Now: $\ell = N(2\pi R_{coil})$. Also: $A = \pi R_{coil}^2 \implies R_{coil} = \sqrt{\frac{A}{\pi}}$. Thus: $R = \frac{\rho N (2\pi)}{\pi r^2} \sqrt{\frac{A}{\pi}}$. $$P = \frac{N^2 A^2 \left(\frac{dB}{dt}\right)^2}{\frac{2 \rho N}{r^2} \sqrt{\frac{A}{\pi}}} \implies P \propto NA^{3/2} r^2 \ldots (1)$$ $$P' \propto (2N)(2A)^{3/2} (3r)^2 \ldots (2)$$ $$\frac{(2)}{(1)}: \frac{P'}{P} = 36 \sqrt{2} \implies P' = (\sqrt{2})(36)P$$ Thus: $\alpha = 36$

Question 39

Physics · Moving Charges and Magnetism · Single correct

Two identical long current carrying wires are bent into the shapes shown in the following figures. If the magnitude of magnetic fields at the centres P and Q of a semicircular arc are $B_1$ and $B_2$ respectively, then the ratio $\frac{B_1}{B_2}$ is $\ldots$.

  1. $\frac{2+\pi}{1+\pi}$
  2. $\frac{1+\pi}{1-\pi}$
  3. $\frac{2+\pi}{1-\pi}$
  4. $\frac{1+\pi}{2-\pi}$

Answer: (a)

Solution

Given $$B_1 = \frac{\mu_0 I}{4 \pi r} \times 2 + \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r} \left( \frac{2}{\pi} + 1 \right)$$ $$B_2 = \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r} \left( \frac{1}{\pi} + 1 \right)$$ Therefore, $$\frac{B_1}{B_2} = \frac{2 + \pi}{1 + \pi}$$

Question 40

Physics · Ray Optics and Optical Instruments · Single correct

For a thin symmetric prism made of glass (refractive index 1.5), the ratio of incident angle and minimum deviation will be $\ldots$.

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

Answer: (b)

Solution

We have for thin prism minimum deviation: $$\delta_m = (\mu - 1) A$$ Thus, $$\delta_m = (1.5 - 1) A$$ which implies $$\delta_m = \frac{A}{2}$$ Also, $$i = \frac{A + \delta_m}{2} \Rightarrow i = \frac{A + \frac{A}{2}}{2} \Rightarrow i = \frac{3A}{4}$$ Thus, $$\frac{i}{\delta_m} = \frac{3}{2}$$

Question 41

Physics · Ray Optics and Optical Instruments · Single correct

Refer the figure given below. $\mu_1$ and $\mu_2$ are refractive indices of air and lens material. The height of image will be $\ldots$ cm.

  1. 1
  2. 0.5
  3. 1.2
  4. 0.25

Answer: (a)

Solution

Given $\frac{\mu_2}{v} - \frac{\mu_1}{\mu} = \frac{\mu_2 - \mu_1}{R}$ $$\frac{1.54}{v} - \frac{1}{(-40)} = \frac{1.54 - 1}{-20}$$ $$\frac{1.54}{v} = -\frac{1}{40} - \frac{0.54}{20}$$ $$\frac{1.54}{v} = \frac{-2.08}{40}$$ $v = -29.61 \, \mathrm{cm}$ $$\frac{h_i}{h_0} = \frac{\mu_1 v}{\mu_2 u} = \frac{1(-29.61)}{1.54(-40)} = 0.48$$ $h_i = 0.48 \times 2 = 0.96 \approx 1 \, \mathrm{cm}$

Question 42

Physics · Dual Nature of Radiation and Matter · Single correct

For a certain metal, when monochromatic light of wavelength $\lambda$ is incident, the stopping potential for photoelectrons is $3V_0$. When the same metal is illuminated by light of wavelength $2\lambda$, then the stopping potential becomes $V_0$. The threshold wavelength for photoelectric emission for the given metal is $\alpha \lambda$. The value of $\alpha$ is _____.

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

Answer: (b)

Solution

Given the equations: $$\frac{hc}{\lambda} - \phi = 3eV_0 (i)$$ $$\frac{hc}{2\lambda} - \phi = eV_0 (ii)$$ Substituting equation (ii) into equation (i): $$\frac{hc}{\lambda} - \phi = 3 \left[ \frac{hc}{2\lambda} - \phi \right]$$ Solving for $\phi$: $$\phi = \frac{hc}{4\lambda}$$ Equating the expressions: $$\frac{hc}{\lambda_0} = \frac{hc}{4\lambda}$$ Thus, we find: $$\lambda_0 = 4\lambda$$ Therefore, $\alpha = 4$.

Question 43

Physics · Electromagnetic Waves · Single correct

An electromagnetic wave travelling in x-direction is described by field equation $E_y = 300 \sin \omega \left( t - \frac{x}{c} \right)$. If the electron is restricted to move in y-direction only with speed of $1.5 \times 10^6 \, \mathrm{m/s}$ then ratio of maximum electric and magnetic forces acting on the electron is $\ldots$.

  1. 200
  2. 150
  3. 400
  4. 300

Answer: (a)

Solution

Given $E_y = 300 \sin \omega \left( t - \frac{x}{v} \right)$. $F_M = qvB \ldots$ (i) $F_e = qE \ldots$ (ii) $$\frac{F_e}{F_m} = \frac{qE}{qvB} = \frac{E}{vB} = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^6} = 200$$

Question 44

Physics · Atoms · Single correct

Angular momentum of an electron in a hydrogen atom is $\frac{3h}{\pi}$, then the energy of the electron is $\ldots$ eV.

  1. -1.51
  2. -0.85
  3. -0.38
  4. -0.28

Answer: (c)

Solution

Given $L = \frac{nh}{2\pi}$. $$\frac{3h}{\pi} = \frac{nh}{2\pi}$$ $n = 6$ $E = \frac{-13.6}{n^2} \, eV$ $$= \frac{-13.6}{6^2} \approx -0.38 \, eV$$

Question 45

Physics · Mechanical Properties of Fluids · Single correct

A liquid drop of diameter $2\,\mathrm{mm}$ breaks into 512 droplets. The change in surface energy is $\alpha \times 10^{-6} \, \mathrm{J}$. The value of $\alpha$ is $\ldots$ (Take surface tension of liquid $= 0.08 \, \mathrm{N/m}$)

  1. 10
  2. 7
  3. 8
  4. 11

Answer: (b)

Solution

Given $V_1 = V_2$. $$\frac{4}{3} \pi R^3 = 512 \left( \frac{4}{3} \pi r^3 \right)$$ $$R = (512)^{1/3} \, r$$ $$1 = 8r$$ $$r = \frac{1}{8} \, \mathrm{mm}$$ $$\Delta SE = SE_2 - SE_1$$ $$= 512 \left[ T \times 4 \pi r^2 \right] - \left[ T \times 4 \pi R^2 \right]$$ $$= 4 \pi T [512 r^2 - R^2]$$ $$= 4 \pi (0.08) \left[ 512 \left( \frac{1}{8} \right)^2 - 1^2 \right] \times 10^{-6} \approx 7 \times 10^{-6}$$ $$\alpha = 7$$

Question 46

Physics · Wave Optics · Numerical

In single slit diffraction pattern, the wavelength of light used is $628 \, \mathrm{nm}$ and slit width is $0.2 \, \mathrm{mm}$, the angular width of central maximum is $\alpha \times 10^{-2}$ degrees. The value of $\alpha$ is $\ldots$.

Answer: 36

Solution

Angular width of central maxima $\theta_{cm} = \frac{2\lambda}{d}$ $$= \frac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}} rad.$$ $$= (628 \times 10^{-5}) \times \frac{180}{\pi} degrees$$ $$\approx 36 \times 10^{-2} degrees$$ Therefore, Answer: $\alpha = 36$

Question 47

Physics · Thermodynamics · Numerical

A vessel contains $0.15\,\mathrm{m}^3$ of a gas at pressure 8 bar and temperature $140^\circ \mathrm{C}$ with $c_p$ = 3R and $c_v$ = 2R. It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is $\ldots$ k J. (R is gas constant)

Answer: 120

Solution

Given $r = \frac{C_P}{C_V} = \frac{3}{2}$. Therefore, $P_1 V_1^\gamma = P_2 V_2^\gamma$. $$V_2 = \left( \frac{P_1}{P_2} \right)^{1/\gamma} V_1$$ $$V_2 = \left( \frac{8}{1} \right)^{2/3} \times 0.15$$ $$V_2 = 0.6 \, \mathrm{m}^3$$ Therefore, $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$. $$= \left( \frac{8 \times 0.15 - 1 \times 0.6}{1.5 - 1} \right) \times 10^5 = 1.2 \times 10^5$$ $$\Rightarrow W = 120 \, \mathrm{kJ}$$

Question 48

Physics · Moving Charges and Magnetism · Numerical

1 $\mu \mathrm{C}$ charge moving with velocity $\vec{v}$ = ($\hat{i}$ - 2$\hat{j}$ + 3$\hat{k}$) $\mathrm{m/s}$ in the region of magnetic field $\vec{B}$ = (2$\hat{i}$ + 3$\hat{j}$ - 5$\hat{k}$) $\mathrm{T}$. The magnitude of force acting on it is $\sqrt{\alpha} \times 10^{-6}\,\mathrm{N}$. The value of $\alpha$ is $\ldots$

Answer: 171

Solution

Given $\mathbf{F} = q (\vec{v} \times \mathbf{B})$ $$= 1 \times 10^{-6} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}$$ $$= 10^{-6} \left[ \hat{i} + 11 \hat{j} + 7 \hat{k} \right]$$ Therefore, $\mathbf{F} = \sqrt{171} \times 10^{-6} \, \mathrm{N}$ Thus, $\alpha = 171$

Question 49

Physics · Mechanical Properties of Solids · Numerical

A uniform wire of length $l$ of weight $w$ is suspended from the roof with a weight of $W$ at the other end. The stress in the wire at $\frac{l}{3}$ distance from the top is $$\left( \frac{W}{A} + \frac{2w}{\gamma A} \right)$$, where, $A$ is the cross sectional area of the wire. The value of $\gamma$ is $\ldots$.

Answer: 6

Solution

The stress $\sigma$ is given by $\sigma = \frac{T}{A} = \frac{W + \frac{w}{3}}{A}$. Therefore, $\sigma = \frac{W}{A} + \frac{2w}{6A}$. Thus, $\gamma = 6$.

Question 50

Physics · Mechanical Properties of Fluids · Numerical

A tube is filled with water and a wooden cube $10 \, \mathrm{cm} \times 10 \, \mathrm{cm} \times 10 \, \mathrm{cm}$ is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by $3.87 \, \mathrm{cm}$. The mass of the metal coin is $\ldots$ gram. (Take water density as $1 \, \mathrm{g/cm^3}$ and density of wood as $0.4 \, \mathrm{g/cm^3}$)

Answer: 387

Solution

Weight of coin = Additional upthrust $$\Rightarrow m_c g = \rho_f V_{add} g$$ $$\Rightarrow m_c = 1 \times 3.87 \times (10)^2$$ $$\Rightarrow m_c = 387 \, gram$$

Chemistry

Question 51

Chemistry · Some Basic Concepts of Chemistry · Single correct

The mass of iron converted into $\mathrm{Fe}_3\mathrm{O}_4$ by the action of $18 \, \mathrm{g}$ of steam is: (Given: Molar mass of H, O and Fe are $1$, $16$ and $56 \, \mathrm{g \, mol^{-1}}$ respectively) Assume iron is present in excess:

  1. $2.1 \, \mathrm{g}$
  2. $4.2 \, \mathrm{g}$
  3. $21 \, \mathrm{g}$
  4. $42 \, \mathrm{g}$

Answer: (d)

Solution

The reaction is given by: $$3\mathrm{Fe}(s) + 4\mathrm{H_2O}(g) \rightarrow \mathrm{Fe_3O_4}(s) + 4\mathrm{H_2}(g)$$ The number of moles of steam is calculated as: $$n_{steam} = \frac{18}{18} = 1$$ The number of moles of Fe required is: $$n_{\mathrm{Fe}} = \frac{3}{4}$$ The mass of Fe required is: $$\frac{3}{4} \times 56$$ This equals: $$= 42 \, gm$$

Question 52

Chemistry · Structure of Atom · Single correct

What is the energy (in $\mathrm{J\,atom^{-1}}$) required for the following process? $\mathrm{Li}^{2+}(g) \rightarrow \mathrm{Li}^{3+}(g) + e^-$ (Take the ionization energy for the H atom in the ground state as $2.18 \times 10^{-18}\,\mathrm{J\,atom^{-1}}$)

  1. 8.72 $\times$ 10^{-18}
  2. 1.962 $\times$ 10^{-18}
  3. 1.962 $\times$ 10^{-17}
  4. 6.54 $\times$ 10^{-17}

Answer: (c)

Solution

The energy of $\mathrm{Li^{2+}}$ is given by $E_{\mathrm{Li^{2+}}} = E_{\mathrm{H}} \times Z^2$. This equals $2.18 \times 10^{-18} \times (3)^2$. Therefore, $E_{\mathrm{Li^{2+}}} = 1.962 \times 10^{-17} \, \mathrm{J \, atom^{-1}}$.

Question 53

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Given below are two statements : \ Statement (I) : The correct sequence of bond lengths in the following species is : $O_2^+ O_2^+ > O_2^- > O_2^{2-}$ \ 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

Species $\mathrm{O_2^{\oplus}}$ has a bond order of $2.5$ and $1$ unpaired electron. Species $\mathrm{O_2}$ has a bond order of $2$ and $2$ unpaired electrons. Species $\mathrm{O_2^-}$ has a bond order of $1.5$ and $1$ unpaired electron. Species $\mathrm{O_2^{2-}}$ has a bond order of $1$ and $0$ unpaired electrons.

Question 54

Chemistry · Thermodynamics · Single correct

Consider the following data (i) $2\mathrm{Al}(s) + 6\mathrm{HCl}(aq) \rightarrow \mathrm{Al}_2\mathrm{Cl}_6(aq) + 3\mathrm{H}_2(g) + 1200 \, \mathrm{kJ/mol}$. (ii) $\mathrm{H}_2(g) + \mathrm{Cl}_2(g) \rightarrow 2\mathrm{HCl}(g) + 164 \, \mathrm{kJ/mol}$. (iii) $\mathrm{HCl}(g) + aq \rightarrow \mathrm{HCl}(aq) + 83 \, \mathrm{kJ/mol}$. (iv) $\mathrm{Al}_2\mathrm{Cl}_6(s) + aq \rightarrow \mathrm{Al}_2\mathrm{Cl}_6(aq) + 663 \, \mathrm{kJ/mol}$. The enthalpy of formation of anhydrous solid $\mathrm{Al}_2\mathrm{Cl}_6$ is :

  1. $-648 \, \mathrm{kJ \, mol^{-1}}$
  2. $-1350 \, \mathrm{kJ \, mol^{-1}}$
  3. $-2002 \, \mathrm{kJ \, mol^{-1}}$
  4. $-1527 \, \mathrm{kJ \, mol^{-1}}$

Answer: (d)

Solution

Given the reaction: $$2\mathrm{Al}(s) + 6\mathrm{HCl}(aq) \rightarrow \mathrm{Al_2Cl_6}(aq) + 3\mathrm{H_2}(g)$$ $$\Delta_r H_1 = -1200 \, \mathrm{kJ/mol}$$ For the reaction: $$[\mathrm{H_2}(g) + \mathrm{Cl_2}(g) \rightarrow 2\mathrm{HCl}(g)] \times 3$$ $$\Delta_r H_2 = -163 \times 3 \, \mathrm{kJ/mol}$$ For the reaction: $$[\mathrm{HCl}(g) + \mathrm{aq} \rightarrow \mathrm{HCl}(aq)] \times 6$$ $$\Delta_r H_3 = -83 \times 6 \, \mathrm{kJ/mol}$$ For the reaction: $$\mathrm{Al_2Cl_6}(aq) \rightarrow \mathrm{Al_2Cl_6}(s) + \mathrm{aq}$$ $$\Delta_r H_4 = +663 \, \mathrm{kJ/mole}$$ Overall reaction: $$2\mathrm{Al}(s) + 3\mathrm{Cl_2} \rightarrow \mathrm{Al_2Cl_6}(s)$$ $$\Delta_f H$$ The enthalpy change is given by: $$\Delta_f H = \Delta_r H_1 + \Delta_r H_2 + \Delta_r H_3 + \Delta_r H_4$$ $$= -1200 - 163 \times 3 - 83 \times 6 + 663$$ $$= -1527 \, \mathrm{kJ/mole}$$

Question 55

Chemistry · Solutions · Single correct

$19.5\,\mathrm{g}$ of fluoro acetic acid (molar mass = $78\,\mathrm{g \, mol^{-1}}$) is dissolved in $500\,\mathrm{g}$ of water at $298\,\mathrm{K}$. The depression in the freezing point of water was $1^\circ \mathrm{C}$. What is $K_a$ of fluoro acetic acid? (For water, $K_f$ = $1.86\,\mathrm{K \, kg \, mol^{-1}}$). Assume molarity and molality to have same values.

  1. 10^{-6}
  2. 4 $\times$ 10^{-4}
  3. 3 $\times$ 10^{-5}
  4. 3 $\times$ 10^{-3}

Answer: (d)

Solution

Given $\Delta T_f = i K_f m$. The number of moles is calculated as $$\frac{19.5}{78} = \frac{1}{4} mole$$ Then, $$1 = i \times 1.86 \times \frac{1/4}{1/2}$$ Solving for $i$, we get $$i = \frac{2}{1.86}$$ Using the relation $$i = 1 + (n-1) \alpha$$ we have $$i = 1 + \alpha$$ Solving for $\alpha$, $$\alpha = \frac{2}{1.86} - 1 = 0.075$$ The expression for $K_a$ is $$K_a = \frac{C \alpha^2}{1 - \alpha}$$ Substituting the values, $$= \frac{1}{2} \times \frac{(0.075)^2}{(1 - 0.075)}$$ This simplifies to $$= 3 \times 10^{-3}$$

Question 56

Chemistry · Equilibrium · Single correct

The solubility product constants of $\mathrm{Ag_2CrO_4}$ and $\mathrm{AgBr}$ are $32x$ and $4y$ respectively at $298 \, \mathrm{K}$. The value of $\frac{molarity of \mathrm{Ag_2CrO_4}}{molarity of \mathrm{AgBr}}$ can be expressed as:

  1. $\frac{2 \sqrt[3]{x}}{y}$
  2. $2 \frac{\sqrt{x}}{\sqrt{y}}$
  3. $\frac{\sqrt{x}}{\sqrt{y}}$
  4. $\frac{\sqrt[3]{x}}{\sqrt{y}}$

Answer: (d)

Solution

For $\mathrm{Ag_2C_2O_4}$, $4S_1^3 = 32x$. Therefore, $S_1 = 2x^{1/3}$. For $\mathrm{AgBr}$, $S_2^2 = 4y$. Therefore, $S_2 = 2y^{1/2}$. The ratio $\left( \frac{S_1}{S_2} \right) = \frac{\sqrt[3]{x}}{\sqrt{y}}$.

Question 57

Chemistry · Electrochemistry · Single correct

An electrochemical cell is constructed using half-cells in the direction of spontaneous change: $$ \mathrm{Fe(OH)_2(s) + 2e^- \rightarrow Fe(s) + 2OH^-(aq)} \qquad E^\circ = -0.88\,\mathrm{V} $$ \ and $$ \mathrm{AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)} \qquad E^\circ = +0.07\,\mathrm{V} $$ Which of the following option is correct?

  1. Overall reaction $Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightleftharpoons Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)$
  2. $E^0_{cell} = -0.95\, \mathrm{V}$
  3. Fe is reduced in the electrochemical cell
  4. $E^0_{cell}$ is an extensive property

Answer: (a)

Solution

Overall reaction $$\mathrm{Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightleftharpoons Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)}$$ $$E^\circ_{cell} = E^\circ_{R(cathode)} - E^\circ_{R(anode)}$$ $$= 0.07 - (-0.88) = 0.95 \, \mathrm{V}$$

Question 58

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

$t_{100\%}$ is the time required for the 100$\%$ completion of the reaction while $t_{1/2}$ is the time required for 50$\%$ of the reaction to be completed. Which of the following option correctly represents the relation between $t_{100\%}$ and $t_{1/2}$ for zero and first order reactions respectively?

  1. $t_{100\%} = (t_{1/2})^2$ and $t_{100\%} = (t_{1/2})^{-\infty}$
  2. $t_{100\%} = 2t_{1/2}$ and $t_{100\%} = (t_{1/2})^{\infty}$
  3. $t_{100\%} = 2t_{1/2}$ and $t_{100\%} = (2t_{1/2})^2$
  4. $t_{100\%} = (t_{1/2})^{\infty}$ and $t_{100\%} = 2t_{1/2}$

Answer: (b)

Solution

Zero-order $$t_{100\%} = \frac{a}{k}; \; t_{1/2} = \frac{a}{2k}$$ $$t_{100\%} = 2t_{1/2}$$ 1st order $$t_{100\%} = \infty$$ $$t_{1/2} = \frac{\ln 2}{k}$$ $$t_{100\%} = (t_{1/2})^\infty$$

Question 59

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

Given below are two statements: \ Statement (I): The first ionisation enthalpy of the elements Na, Mg, Cl and Ar follows the order $\mathrm{Na > Mg > Cl > Ar}$ \ Statement (II): Among Ca, Al, Fe and B, the third ionisation enthalpy is very high for Ca. \ 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

(i) $\mathrm{Na} < \mathrm{Mg} < \mathrm{Cl} < \mathrm{Ar} \rightarrow 1st ionization energy$ (ii) Among $\mathrm{B}$, $\mathrm{Ca}$, $\mathrm{Al}$, $\mathrm{Fe}$ Ca has maximum 3rd ionization energy among given due to inert gas configuration.

Question 60

Chemistry · Redox Reactions · Single correct

Given below are two statements: \ Statement (I): Oxidising power of halogens decreases in the order $\mathrm{F_2 > Cl_2 > Br_2 > I_2}$, which is the basis of "Layer test". \ Statement (II): "Layer test" to identify $\mathrm{Br_2}$ and $\mathrm{I_2}$ in aqueous solution involves the oxidation of bromide or iodide into $\mathrm{Br_2}$ or $\mathrm{I_2}$ respectively with $\mathrm{Cl_2}$, which is a type of displacement redox reaction. \ 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

(I) $F_2 > Cl_2 > Br_2 > I_2$ (Oxidizing power order) (II) $\mathrm{NaBr} + Cl_2$-water $\rightarrow \mathrm{Br_2} + \mathrm{NaCl}$ $\mathrm{NaI} + Cl_2$-water $\rightarrow \mathrm{I_2} + \mathrm{NaCl}$

Question 61

Chemistry · Redox Reactions · Single correct

Which of the following sets includes all the species that will change the orange colour of $\mathrm{K_2Cr_2O_7}$ in acidic medium?

  1. $\mathrm{Fe^{2+}, Sn^{2+}, I^-, S^{2-}}$
  2. $\mathrm{S^{2-}, Fe^{3+}, I^-, C_2O_4^{2-}}$
  3. $\mathrm{Fe^{2+}, NO_2^-, SO_2, Sn^{4+}}$
  4. $\mathrm{Fe^{3+}, SO_4^{2-}, S^{2-}, Sn^{4+}}$

Answer: (a)

Solution

Reducing agent + $\mathrm{KmNO_4} \xrightarrow{acidic medium} \mathrm{Mn^{+2}}$ + Product Given Reducing Agent $\mathrm{Fe^{+2} \rightarrow Fe^{+3}}$ $\mathrm{Sn^{+2} \rightarrow Sn^{+4}}$ $\mathrm{I^- \rightarrow I_2}$ $\mathrm{S^{-2} \rightarrow S}$

Question 62

Chemistry · Co-ordination Compounds · Single correct

Match List – I with List – II. \begin{tabular}{|c|c|c|c|} \hline \textbf{List-I} & \textbf{Chromium (III) Complexes} & \textbf{List-II} & $\Delta_o\;(\mathrm{cm}^{-1})$ \\ & \textbf{(en = ethylene diamine)} & & \\ \hline A. & $[\mathrm{Cr(CN)_6}]^{3-}$ & I. & $15,060$ \\ \hline B. & $[\mathrm{CrF_6}]^{3-}$ & II. & $17,400$ \\ \hline C. & $[\mathrm{Cr(H_2O)_6}]^{3+}$ & III. & $22,300$ \\ \hline D. & $[\mathrm{Cr(en)_3}]^{3+}$ & IV. & $26,600$ \\ \hline \end{tabular} Choose the correct answer from the options given below:

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

Answer: (d)

Solution

CFSE is proportional to strength of ligand followed by chelation. CN^- is stronger ligand than en. $$\mathrm{CN^- > en > H_2O > F^-}$$

Question 63

Chemistry · Solutions · Single correct

Given below are two statements: \ Statement (I): 1,2,3-Trihydroxypropane can be separated from water by simple distillation. \ Statement (II): An azotropic mixture cannot be separated by fractional distillation. \ 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

Glycerol is separated by distillation under reduced pressure.

Question 64

Chemistry · Haloalkanes and Haloarenes · Single correct

Given below are two statements : \ Statement (I) : Benzyl chloride reacts faster in $S_{N}1$ mechanism than ethyl chloride. \ Statement (II) : Ethyl carbocation intermediate is less stabilized by hyperconjugation than benzyl carbocation by resonance. \ 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

Rate of $\mathrm{S_N^1} \propto$ Stability of carbocation. Order of stability of carbocation: $$Ph - \overset{+}{CH_2} > CH_3 - \overset{+}{CH_2}$$ Hyperconjugation in $CH_3 - \overset{+}{CH_2}$. Resonance in $Ph - \overset{+}{CH_2}$.

Question 65

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

In IUPAC nomenclature, the correct order of decreasing priority of functional group is:

  1. –CONH$_2$, >C=O, –CHO, –NH$_2$, –C≡C–
  2. –CONH$_2$, –COOCH$_3$, –CHO, –NH$_2$, –OH
  3. –CONH$_2$, –CHO, >C=O, –NH$_2$, –C≡C–
  4. –CONH$_2$, –CHO, –CN, –NH$_2$, –C≡C–

Answer: (c)

Solution

Solution: Theory based

Question 66

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

For the given molecule, "x", the preferred site for the attack of the electrophile is :

  1. Predominantly at "r"
  2. "r" and "u"
  3. "p" and "s"
  4. Predominantly at "u"

Answer: (d)

Solution

U is predominant site towards attack of electrophile because of +M nature of nitrogen.

Question 67

Chemistry · Analytical Chemistry · Single correct

Match List-I with List-II. \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{\textbf{List-I}} & \multicolumn{2}{c|}{\textbf{List-II}} \\ \multicolumn{2}{|c|}{\textbf{Mixture of Compounds}} & \multicolumn{2}{c|}{\textbf{Reagent used to distinguish}} \\ \hline A. & Diethyl amine + Ethyl amine & I. & Bromine water \\ \hline B. & Acetaldehyde + Acetone & II. & $\mathrm{CHCl_3 + KOH,\ \Delta}$ \\ \hline C. & Ethanol + Phenol & III. & Neutral $\mathrm{FeCl_3}$ \\ \hline D. & Benzoic acid + Cinnamic acid & IV. & Ammoniacal silver nitrate \\ \hline \end{tabular} Choose the correct answer from the options given below:

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

Answer: (d)

Solution

Phenol gives violet colour with neutral $\mathrm{FeCl_3}$, not ethanol. 1° amine gives positive test with $\mathrm{CHCl_3/KOH}$ not 2° amine. Aldehyde gives positive test with ammoniacal silver nitrate not ketone. Cinnamic acid gives bromine water test.

Question 68

Chemistry · Amines · Single correct

Consider the three aromatic molecules (P, Q and R) whose structures have been given below: The correct order regarding the reactivity of these compounds with $\mathrm{Ph{-}N\equiv N\ Cl^{(-)}}$ under optimum but slightly acidic medium is:

  1. $\mathrm{P > Q > R}$
  2. $\mathrm{R > P > Q}$
  3. $\mathrm{R > Q > P}$
  4. $\mathrm{P > R > Q}$

Answer: (a)

Solution

Rate of ESR $\propto$ nucleophilicity of benzene ring.

Question 69

Chemistry · Biomolecules · Single correct

Match List-I with List-II. \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{\textbf{List-I}} & \multicolumn{2}{c|}{\textbf{List-II}} \\ \multicolumn{2}{|c|}{\textbf{Vitamin}} & \multicolumn{2}{c|}{\textbf{Name}} \\ \hline A. & Vitamin B$_1$ & I. & Pyridoxine \\ \hline B. & Vitamin B$_2$ & II. & Ascorbic acid \\ \hline C. & Vitamin B$_6$ & III. & Thiamine \\ \hline D. & Vitamin C & IV. & Riboflavin \\ \hline \end{tabular} Choose the correct answer from the options given below:

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

Answer: (c)

Solution

Sol. Theory based

Question 70

Chemistry · Analytical Chemistry · Single correct

A salt with few drops of conc. HCl gives apple green colour in flame test. The group precipitate of the salt is dissolved in acetic acid and treated with $\mathrm{K_2CrO_4}$ to give yellow precipitate. When the sodium carbonate extract of the salt solution is heated with conc. $\mathrm{HNO_3}$ and ammonium molybdate, it resulted a canary yellow precipitate. The cation and anion present in the salt are respectively.

  1. $\mathrm{Ca^{2+}}$ and $\mathrm{SO_4^{2-}}$
  2. $\mathrm{Ba^{2+}}$ and $\mathrm{PO_4^{3-}}$
  3. $\mathrm{Mn^{2+}}$ and $\mathrm{PO_4^{3-}}$
  4. $\mathrm{Ba^{2+}}$ and $\mathrm{SO_4^{2-}}$

Answer: (b)

Solution

(i) $\mathrm{Ba^{+2}}$ gives apple green colour while $\mathrm{Ca^{+2}}$ and $\mathrm{Sr^{+2}}$ gives brick red and crimson red in flame test. (ii) $\mathrm{Ba^{+2} + CrO_4^{2-} \rightarrow BaCrO_4}$ Yellow ppt (iii) $\mathrm{PO_4^{3-} + Conc \ HNO_3 + (NH_4)_2MoO_4 \rightarrow (NH_4)_3[P(Mo_3O_{10})_4]}$ Canary yellow ppt

Question 71

Chemistry · Co-ordination Compounds · Numerical

$5.33\,\mathrm{g}$ of $\mathrm{CrCl}_3 \cdot$ $6\,\mathrm{H}_2 \mathrm{O}$, which is a 1:3 electrolyte, is dissolved in water and is passed through a cation exchanger. The chloride ions in the eluted solution, on treatment with $\mathrm{AgNO}_3$ results in $8.61\,\mathrm{g}$ of $\mathrm{AgCl}$. The ratio of moles of complex reacted and moles of $\mathrm{AgCl}$ formed is $\ldots \times$ $10^{\!-2}$. (Nearest integer) [Molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ Cr : 52, Ag : 108, Cl : 35.5, H : 1, O : 16]

Answer: 33

Solution

The reaction is given by: $$[\mathrm{Cr(H_2O)_6}]\mathrm{Cl}_3 + \mathrm{AgNO_3} \xrightarrow{excess} 3\mathrm{AgCl(s)}$$ The percentage of moles of complex reacted is calculated as: $$\frac{no. of moles of complex reacted}{no. of moles of AgCl precipitated} \times 100$$ $$= \frac{1}{3} \times 100 = 33.3$$

Question 72

Chemistry · Hydrocarbons · Numerical

Consider the isomers of hydrocarbon with molecular formula $C_5H_{10}$. These isomers do not decolourise $\mathrm{KMnO_4}$ solution. These isomers are subjected to chlorination with chlorine in presence of light to give monochloro compounds. The total number of monochloro compounds (structural isomers only) formed is $\ldots$.

Answer: 14

Solution

The solution involves a series of chlorination reactions. Each step is labeled with a number indicating the number of chlorine atoms added. The reactions are as follows: 1. The first reaction adds 1 chlorine atom. 2. The second reaction adds 4 chlorine atoms. 3. The third reaction adds 4 chlorine atoms. 4. The fourth reaction adds 2 chlorine atoms. 5. The fifth reaction adds 3 chlorine atoms. The total number of chlorine atoms added is calculated as: $$1 + 4 + 4 + 2 + 3 = 14$$

Question 73

Chemistry · Hydrocarbons · Numerical

One mole of an alkane (x) requires 8 mole oxygen for complete combustion. Sum of number of carbon and hydrogen atoms in the alkane (x) is $\ldots$.

Answer: 17

Solution

Question 74

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

For reaction A $\rightarrow$ P, rate constant $k = 1.5 \times 10^3 \, \mathrm{s}^{-1}$ at $27^\circ \mathrm{C}$. If activation energy for the above reaction is $60 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$, then the temperature (in $^\circ \mathrm{C}$) at which rate constant, $k = 4.5 \times 10^3 \, \mathrm{s}^{-1}$ is $\ldots$. (Nearest integer) Given : $\log 2 = 0.30$, $\log 3 = 0.48$, $R = 8.3 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$, $\ln 10 = 2.3$

Answer: 41

Solution

Given $\ln \left( \frac{K_2}{K_1} \right) = \frac{E_a}{R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$. $$\ln \left( \frac{4.5 \times 10^3}{1.5 \times 10^3} \right) = \frac{60 \times 10^3}{8.3} \left[ \frac{T_2 - 300}{300 \cdot T_2} \right]$$ $$\ln 3 = \frac{60 \times 10^3}{8.3} \left[ \frac{T_2 - 300}{300 \cdot T_2} \right]$$ $$\Rightarrow T_2 = 314.4 \, \mathrm{K}$$ $$\Rightarrow t = 41.4^\circ \mathrm{C}$$

Question 75

Chemistry · Thermodynamics · Numerical

At the transition temperature T, A $\rightleftharpoons$ B and $\Delta G^0 = 105 - 35 \log T$ where A and B are two states of substance X. The transition temperature in $^\circ \mathrm{C}$ when pressure is 1 atm is $\ldots$ (Nearest integer).

Answer: 727

Solution

Given $\Delta G^\circ = 0$ (equilibrium at standard pressure). $$105 - 35 \log T = 0$$ $$\log T = 3$$ $$T = 1000 \, \mathrm{K}$$ $$T = 727^\circ \mathrm{C}$$