Previous Year Paper

JEE Mains 2 Apr 2026 Shift 1 (Previous Year Paper)

Held in 2026

75 questions180 minutes

With answers

Free

Paper last updated: 15 October 2026

Practise under exam conditions

Take this paper as a timed mock test. Review your answers after you submit.

Attempt timed test

Questions & answers

Read the paper at your pace. Open an explanation when you need it.

Mathematics Section A (20 questions)

Question 1

Let $\alpha, \alpha + 2, \alpha \in \mathbf{Z}$ , be the roots of the quadratic

equation $x(x + 2) + (x + 1)(x + 3) + (x + 2) (x + 4)$

$+.....+ (x + n - 1) (x + n + 1) = 4n$ for some $n \in \mathbf{N}$ .

Then $(n + \alpha)$ is equal to :

AOption (a)
0
BOption (b)
1
COption (c)
2
DOption (d)
3

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$\alpha, \alpha + 2, \alpha \in \mathbf{Z}$ , are the roots of the quadratic equation:

$x(x + 2) + (x + 1)(x + 3) + (x + 2) (x + 4) + \ldots + (x + n - 1) (x + n + 1) = 4n$ , where $n \in \mathbf{N}$ .

Calculation:

$nx^2 + x(2 + 4 + 6 + \ldots + 2n) + (1 \cdot 3 + \ldots + (n - 1)(n + 1)) = 4n$

⇒ $nx^2 + n(n + 1)x + \frac{n(n - 1)(2n + 5)}{6} = 4n$

⇒ $x^2 + (n + 1)x + \frac{(n - 1)(2n + 5)}{6} = 4$

Discriminant must be a perfect square:

$D = (n + 1)^2 - 4 \times \frac{(n - 1)(2n + 5)}{6}$

⇒ $D = \frac{122 - 2n^2}{6} = 20 - \left( \frac{n^2 - 1}{3} \right)$

For a perfect square:

$\frac{n^2 - 1}{3} = 16$

⇒ $n = 7$

Substitute $n = 7$ into the equation:

$x^2 + 8x + \frac{8 \times 15}{6} - 5 = 0$

⇒ $x^2 + 8x + 15 = 0$

Roots: $x = -3, -5$

$\alpha = -5, \alpha + 2 = -3$

$\alpha + n = 7 - 5 = 2$

∴ The correct answer is option (3).

Question 2

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 :

AOption (a)
20
BOption (b)
31
COption (c)
35
DOption (d)
75

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$50\left(\frac{2x}{1+3i}-\frac{y}{1-2i}\right)=31+17i$

Calculation:

$50\left(\frac{2x}{1+3i}-\frac{y}{1-2i}\right)=31+17i$

⇒ $\frac{1}{1+3i}=\frac{1-3i}{10}$ and $\frac{1}{1-2i}=\frac{1+2i}{5}$

⇒ $10x-30xi-10y-20yi=31+17i$

⇒ Real part: $10(x-y)=31$ , Imaginary part: $-30x-20y=17$

⇒ $x=0.9,;y=-2.2$

⇒ $10(x-3y)=75$

∴ The correct answer is option (4).

Question 3

Let $\alpha, \beta \in \mathbf{R}$ 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 :

AOption (a)
-4
BOption (b)
4
COption (c)
8
DOption (d)
-8

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$x + 2y + z = 5$

$2x + y + \alpha z = 5$

$8x + 4y + \beta z = 18$

Calculation:

$\Delta = \begin{vmatrix} 1 & 2 & 1 \ 2 & 1 & \alpha \ 8 & 4 & \beta \end{vmatrix} = 0$

$\Delta = 1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(0) = 0$

$\beta - 4\alpha - 4\beta + 16\alpha = 0$

$-3\beta + 12\alpha = 0$

$4\alpha = \beta$

$\frac{\beta}{\alpha} = 4$

∴ The correct answer is option (2).

Question 4

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) : $\text{det (adj(A + B))} = -5$ ,

AOption (a)
only (S1) is correct
BOption (b)
only (S2) is correct
COption (c)
both (S1) and (S2) are correct
DOption (d)
both (S1) and (S2) are wrong

View answer & explanation

Correct answer

only (S2) is correct

Why this answer is correct

Given that:

$A = \begin{bmatrix} 1 & 2 \ 1 & \alpha \end{bmatrix}, ; A^2 - 4A + I = O$

$B = \begin{bmatrix} 3 & 3 \ \beta & 2 \end{bmatrix}, ; B^2 - 5B - 6I = O$

Calculation:

$\text{Trace of } A:$ ; $\alpha + 1 = 4 \Rightarrow \alpha = 3$

$\text{Determinant of } B:$ ; $\text{det(B)} = 6 - 3\beta = -6 \Rightarrow 3\beta = 12 \Rightarrow \beta = 4$

B - $A = \begin{bmatrix} 2$ & 1 \ 3 & -1 $\end{bmatrix},$ ; B + $A = \begin{bmatrix} 4$ & 5 \ 5 & 5 $\end{bmatrix}$

(B - A)(B + A) = $\begin{bmatrix} 2$ & 1 \ 3 & -1 $\end{bmatrix} \begin{bmatrix} 4$ & 5 \ 5 & 5 $\end{bmatrix} \Rightarrow \begin{bmatrix} 13$ & 15 \ 7 & 10 $\end{bmatrix}$

$\text{adj(A + B)} = \begin{bmatrix} 5$ & -5 \ -5 & 4 $\end{bmatrix},$ ; $\text{det(adj(A + B))} =$ (5)(4) - (-5)(-5) = 20 - 25 = -5

∴ The correct answer is option (3).

Question 5

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 :

AOption (a)
4
BOption (b)
5
COption (c)
6
DOption (d)
7

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$\text{AP}_1 \text{ (Set A)} = { 1, 6, 11, 16, \dots, 101 \text{ terms} }$

$\text{AP}_2 \text{ (Set B)} = { 9, 16, \dots, 71 \text{ terms} }$

$D = \text{L.C.M. } { d_1, d_2 } = 35$

$1^{\text{st}} \text{ Common term} = 16$

Calculation:

$16 + (n - 1) \cdot 35 \leq 499$

⇒ $n \leq 14.8$

⇒ $n = 14$

$A \cap B = { 16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471 }$

Divisible by 3:

$= { 51, 156, 261, 366, 471 }$

∴ The correct answer is option (2).

Question 6

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 ;

AOption (a)
15400
BOption (b)
17800
COption (c)
16800
DOption (d)
29400

View answer & explanation

Correct answer

16800

Why this answer is correct

Given that:

${}^5C_1 \times \frac{7!}{3!} + {}^5C_2 \times \frac{7!}{2!2!}$

Calculation:

${}^5C_1 \times \frac{7!}{3!} = 5 \times \frac{5040}{6} = 5 \times 840 = 4200$

${}^5C_2 \times \frac{7!}{2!2!} = 10 \times \frac{5040}{4} = 10 \times 1260 = 12600$

⇒ $4200 + 12600 = 16800$

∴ The correct answer is option (3).

Question 7

The number of elements in the set

$S = \left{ (r, k) : k \in \mathbf{Z} \text{ and } {}^{36}C_{r+1} = \frac{6({}^{35}C_r)}{(k^2 - 3)} \right}$ , is

AOption (a)
2
BOption (b)
4
COption (c)
8
DOption (d)
16

View answer & explanation

Correct answer

Why this answer is correct

Given that:

${}^{36}C_{r+1} = \frac{6({}^{35}C_r)}{k^2 - 3}$

Calculation:

$\frac{36}{r+1} \cdot {}^{35}C_r = 6 \cdot \frac{{}^{35}C_r}{k^2 - 3}$

⇒ $k^2 - 3 = \frac{r+1}{6}$

⇒ $k^2 = \frac{r+19}{6}$

$0 \leq r \leq 35$

⇒ $3.1 \leq \frac{r+19}{6} \leq 9$

⇒ $k^2 = 4, 9$

⇒ $k = \pm 2, \pm 3$

⇒ $(r, k) = (5, \pm 2), (35, \pm 3)$

∴ The correct answer is option (2).

Question 8

If the mean of the data

Class	5-10	10-15	15-20	20-25	25-30	30-35
Frequency	2	k	28	54	k+1	5

is 21, then k is one of the roots of the equation :

AOption (a)
$2x^2 - 23x - 10 = 0$
BOption (b)
$4x^2 - 35x + 24 = 0$
COption (c)
$2x^2 - 19x - 10 = 0$
DOption (d)
$2x^2 - 35x + 98 = 0$

View answer & explanation

Correct answer

$2x^2 - 19x - 10 = 0$

Why this answer is correct

Given that:

$21 = \frac{15 + 12.5k + 490 + 1215 + 27.5 + 27.5k + 162.5}{90 + 2k}$

Calculation:

$21(90 + 2k) = 15 + 12.5k + 490 + 1215 + 27.5 + 27.5k + 162.5$

⇒ $1890 + 42k = 1910 + 40k$

⇒ $42k - 40k = 1910 - 1890$

⇒ $2k = 20$

⇒ $k = 10$

$k$ satisfies the equation $2x^2 - 23x - 10 = 0$ .

∴ The correct answer is option (1).

Question 9

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 :

AOption (a)
11
BOption (b)
12
COption (c)
13
DOption (d)
14

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$\text{Midpoints: } \left( \frac{5}{2}, 7 \right), \left( \frac{5}{2}, 3 \right), (4, 5)$

$\text{Incentre: } (h, k)$

Calculation:

$I = \left( \frac{24}{9}, 5 \right)$

⇒ $h = \frac{24}{9} = \frac{8}{3}, ; k = 5$

⇒ $3h + k = 3 \times \frac{8}{3} + 5 = 8 + 5 = 13$

∴ The correct answer is option (3).

Question 10

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 :

AOption (a)
$\frac{4\sqrt{5}}{3}$
BOption (b)
$2\sqrt{5}$
COption (c)
$\frac{7\sqrt{5}}{3}$
DOption (d)
$\frac{8\sqrt{5}}{3}$

View answer & explanation

Correct answer

$\frac{8\sqrt{5}}{3}$

Why this answer is correct

Given that:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,; a < b$

$(4, 3)$ lies on the ellipse

$e = \frac{\sqrt{5}}{3}$

Calculation:

$e^2 = 1 - \frac{a^2}{b^2} \Rightarrow \frac{a^2}{b^2} = 1 - e^2 = 1 - \frac{5}{9} = \frac{4}{9}$

$\Rightarrow \frac{a^2}{b^2} = \frac{4}{9} \dots (1)$

$\frac{16}{a^2} + \frac{9}{b^2} = 1 \dots (2)$

From (1) and (2):

$a^2 = 20,; b^2 = 45$

$LR = \frac{2a^2}{b} = \frac{2(20)}{3\sqrt{5}} = \frac{8\sqrt{5}}{3}$

∴ The correct answer is option (4).

Question 11

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 :

AOption (a)
$\frac{\sqrt{3} + 1}{2\sqrt{2}}$
BOption (b)
$\frac{\sqrt{3} - 1}{\sqrt{2}}$
COption (c)
$\frac{\sqrt{3}}{2}$
DOption (d)
$\frac{1}{2}$

View answer & explanation

Correct answer

$\frac{\sqrt{3} + 1}{2\sqrt{2}}$

Why this answer is correct

Given that:

$K = \sin 10^\circ \sin 50^\circ \sin 70^\circ$

Calculation:

$K = \sin 10^\circ \sin 50^\circ \sin 70^\circ$

⇒ $K = \frac{1}{4} \sin 30^\circ$

⇒ $K = \frac{1}{8}$

$\sin \frac{10K\pi}{3} = \sin\left( 10 \times \frac{1}{8} \times \frac{\pi}{3} \right)$

⇒ $\sin \frac{10K\pi}{3} = \sin \frac{5\pi}{12}$

⇒ $\sin \frac{5\pi}{12} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

∴ The correct answer is option (1).

Question 12

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

AOption (a)
3
BOption (b)
6
COption (c)
7
DOption (d)
9

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$S = { x \in [-\pi, \pi] : \sin x (\sin x + \cos x) = a, a \in \mathbf{Z} }$

Calculation:

$\sin x (\sin x + \cos x) \in \left[ \frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2} \right]$

⇒ $a = 0, 1$ (since $a \in \mathbf{Z}$ )

Case 1: $a = 0$

$\sin x (\sin x + \cos x) = 0$

⇒ $\sin x = 0$ or $\sin x + \cos x = 0$

$\sin x = 0 \Rightarrow x = -\pi, 0, \pi$ (3 solutions)

$\sin x + \cos x = 0 \Rightarrow \tan x = -1 \Rightarrow x = -\frac{\pi}{4}, \frac{3\pi}{4}$ (2 solutions)

Total solutions for $a = 0$ : 3 + 2 = 5

Case 2: $a = 1$

$\sin x (\sin x + \cos x) = 1$

⇒ $\sin^2 x + \sin x \cos x = 1$

⇒ $2\sin^2 x + 2\sin x \cos x = 2$

⇒ $1 - \cos 2x + \sin 2x = 2$

⇒ $\sin 2x - \cos 2x = 1$

Squaring both sides:

$\sin^2 2x - 2\sin 2x \cos 2x + \cos^2 2x = 1$

⇒ $1 - \sin 4x = 1$

⇒ $\sin 4x = 0 \Rightarrow 4x = -4\pi, -3\pi, \dots, 3\pi, 4\pi$

⇒ $x = -\pi, -\frac{3\pi}{4}, -\frac{\pi}{2}, -\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$

Total solutions for $a = 1$ : 4

Total solutions: $5 + 4 = 9$

∴ The correct answer is option (4).

Question 13

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 :

AOption (a)
2
BOption (b)
5
COption (c)
7
DOption (d)
9

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7}$

$\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7}$

Calculation:

$3r_1 - 1 = r_2 + 2 \Rightarrow r_2 = 3r_1 - 3$

$7r_1 - b - 1 = 0 \Rightarrow b = 7r_1 - 1$

$7r_2 + 2a = 0 \Rightarrow 2a = -7r_2 \Rightarrow a = -\frac{7}{2}r_2$

$r_2 = 3r_1 - 3 \Rightarrow a = -\frac{7}{2}(3r_1 - 3) \Rightarrow a = -\frac{21}{2}r_1 + \frac{21}{2}$

$a + b = -\frac{21}{2}r_1 + \frac{21}{2} + 7r_1 - 1$

$a + b = \left(-\frac{21}{2} + 7\right)r_1 + \left(\frac{21}{2} - 1\right)$

$a + b = \frac{14}{2}r_1 + \frac{19}{2} \Rightarrow r_1 = 1$

$a + b = 7 + 2 = 9$

∴ The correct answer is option (4).

Question 14

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

AOption (a)
30
BOption (b)
36
COption (c)
60
DOption (d)
72

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$|\vec{a}| = 2, ; |\vec{b}| = 3$

$E = 3\left| \left( 3\vec{a} + 2\vec{b} \right) \right| + 4\left| \left( 3\vec{a} - 2\vec{b} \right) \right|$

Calculation:

$E = 3\sqrt{9a^2 + 4b^2 + 12\vec{a}.\vec{b}} + 4\sqrt{9a^2 + 4b^2 - 12\vec{a}.\vec{b}}$

⇒ $a = 2, ; b = 3, ; \vec{a}.\vec{b} = 6 \cos \theta$

⇒ $E = 3\sqrt{36 + 36 + 12 \times 6 \cos \theta} + 4\sqrt{36 + 36 - 72 \cos \theta}$

⇒ $E = 3\sqrt{72 + 72 \cos \theta} + 4\sqrt{72 - 72 \cos \theta}$

⇒ $E = 18\sqrt{2} \sqrt{1 + \cos \theta} + 24\sqrt{2} \sqrt{1 - \cos \theta}$

⇒ $E = 36 \cos \frac{\theta}{2} + 48 \sin \frac{\theta}{2}$

⇒ $E_{\text{max}} = \sqrt{(36)^2 + (48)^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60$

∴ The correct answer is option (3).

Question 15

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 :

AOption (a)
50
BOption (b)
80
COption (c)
100
DOption (d)
120

View answer & explanation

Correct answer

100

Why this answer is correct

Given that:

$\vec{L}$ passes through point (1, 1, 1) and is perpendicular to $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + \hat{k}$ .

Calculation:

$\text{Direction vector of } \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 2 & 2 & 1 \end{vmatrix} = -2\hat{i} + 3\hat{j} - 2\hat{k}$

Equation of line: $\frac{x-1}{-2} = \frac{y-1}{3} = \frac{z-1}{-2} = \lambda$

Let $P(a, b, c) \equiv (2\lambda + 1, -3\lambda + 1, 2\lambda + 1)$ be the foot of perpendicular from (0, 0, 0).

$\vec{OM} \cdot (-2\hat{i} + 3\hat{j} - 2\hat{k}) = 0$

⇒ $-2(2\lambda + 1) + 3(-3\lambda + 1) - 2(2\lambda + 1) = 0$

⇒ $\lambda = -\frac{1}{17}$

Substitute $\lambda = -\frac{1}{17}$ :

$a = \frac{-2}{17} + 1, ; b = \frac{3}{17} + 1, ; c = \frac{-2}{17} + 1$

⇒ $34(a+b+c) = 100$

∴ The correct answer is option (3).

Question 16

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:

AOption (a)
5
BOption (b)
6
COption (c)
8
DOption (d)
10

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$\lim_{x \to 2} \frac{\sin(x^3 - 5x^2 + ax + b)}{(\sqrt{x-1} - 1)\log_e(x-1)} = m$

Calculation:

$\because$ Denominator = 0 at $x = 2$

$\Rightarrow$ Numerator = 0 at $x = 2$

$\Rightarrow 2^3 - 5(2)^2 + a(2) + b = 0$

$\Rightarrow 2a + b = 12$ ... (1)

$m = \lim_{x \to 2} \frac{\frac{\sin(x^3 - 5x^2 + ax + b)}{(x^3 - 5x^2 + ax + b)} \cdot (x^3 - 5x^2 + ax + b)}{\frac{\sqrt{x-1} - 1}{\sqrt{x-1} + 1} \cdot \frac{\log_e(1 + (x-2))}{(x-2)} \cdot (x-2)}$

$\Rightarrow m = \lim_{x \to 2} 2 \frac{x^3 - 5x^2 + ax + b}{(x-2)^2}$

$\Rightarrow m = \lim_{x \to 2} 2 \frac{3x^2 - 10x + a}{2(x-2)}$

$\because$ Denominator = 0 at $x = 2$

$\Rightarrow$ Numerator = 0 at $x = 2$

$\Rightarrow 3(2)^2 - 10(2) + a = 0$

$\Rightarrow a = 8$

Substituting $a = 8$ in (1):

$2(8) + b = 12$

$\Rightarrow b = -4$

$\therefore m = \lim_{x \to 2} 2 \left( \frac{6x - 10}{2} \right)$

$\Rightarrow m = 2$

$\therefore a + b + m = 8 - 4 + 2 = 6$

∴ The correct answer is option (2).

Question 17

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

AOption (a)
$e^e$
BOption (b)
$e^{e^2}$
COption (c)
$e^{2e}$
DOption (d)
$e^{2^e}$

View answer & explanation

Correct answer

$e^{2e}$

Why this answer is correct

Given that:

$\frac{dy}{dx} = y(2 + \ln x)$

Calculation:

$\int \frac{1}{y} , dy = \int (2 + \ln x) , dx$

⇒ $\ln y = 2x + x \ln x - x + C$

⇒ $\ln y = x + x \ln x + C$

Given point (1, e): $\ln e = 1 + 1 \ln 1 + C$

⇒ $1 = 1 + 0 + C \Rightarrow C = 0$

⇒ $\ln y = x + x \ln x$

⇒ $f(x) = y = e^{x + x \ln x}$

⇒ $f(e) = e^{e + e} = e^{2e}$

∴ The correct answer is option (3).

Question 18

The number of critical points of the function

$f(x) = \begin{cases} \left| \frac{\sin x}{x} \right| & , \quad x \neq 0 \ 1 & , \quad x = 0 \end{cases}$ in the interval $(-2\pi, 2\pi)$ is equal to:

AOption (a)
1
BOption (b)
3
COption (c)
5
DOption (d)
7

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$f(x) = \begin{cases} \left| \frac{\sin x}{x} \right| & , \quad x \neq 0 \ 1 & , \quad x = 0 \end{cases}$ , $x \in (-2\pi, 2\pi)$

Calculation:

$\lim_{x \to 0} \left| \frac{\sin x}{x} \right| = 1 = f(0)$ , hence $f(x)$ is continuous.

$f'(x) = \frac{d}{dx} \left| \frac{\sin x}{x} \right| = \frac{x \cos x - \sin x}{x^2}$

$f'(x) = 0 \Rightarrow \tan x = x$

From the graph, there are 3 solutions to $\tan x = x$ in $(-2\pi, 2\pi)$ .

Additionally, $f'(x)$ does not exist at $x = -\pi, \pi$ .

Total critical points = 3 (from $\tan x = x$ ) + 2 (where $f'(x)$ does not exist) = 5.

∴ The correct answer is option (3).

Question 19

Let [.] denote the greatest integer function. Then the value of $\int_0^3 \frac{e^x + e^{-x}}{[x]!} dx$ is:

AOption (a)
$e^2 + e^3 - \frac{1}{e^2} - \frac{1}{e^3}$
BOption (b)
$\frac{1}{2} \left( e^2 + e^3 - \frac{1}{e^2} - \frac{1}{e^3} \right)$
COption (c)
$e^2 + e^3 - \frac{1}{2e^2} - \frac{1}{2e^3}$
DOption (d)
$\frac{1}{2} (e^2 + e^3) - \frac{1}{e^2} - \frac{1}{e^3}$

View answer & explanation

Correct answer

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

Why this answer is correct

Given that:

$\int_{0}^{3}\dfrac{e^x+e^{-x}}{[x]!},dx$

Calculation:

$\int_{0}^{3}\dfrac{e^x+e^{-x}}{[x]!},dx$

⇒ $\int_{0}^{1}(e^x+e^{-x}),dx+\int_{1}^{2}(e^x+e^{-x}),dx+\dfrac{1}{2}\int_{2}^{3}(e^x+e^{-x}),dx$

⇒ $\left[e^x-e^{-x}\right]{0}^{1}+\left[e^x-e^{-x}\right]{1}^{2}+\dfrac{1}{2}\left[e^x-e^{-x}\right]_{2}^{3}$

⇒ $\left(e-\dfrac{1}{e}\right)+\left(e^2-e-\dfrac{1}{e^2}+\dfrac{1}{e}\right)+\dfrac{1}{2}\left(e^3-e^2-\dfrac{1}{e^3}+\dfrac{1}{e^2}\right)$

⇒ $\dfrac{1}{2}\left(e^2+e^3-\dfrac{1}{e^2}-\dfrac{1}{e^3}\right)$

∴ The correct answer is option (2).

Question 20

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 :

AOption (a)
$\frac{\pi}{6}$
BOption (b)
$\frac{\pi}{4}$
COption (c)
$\frac{\pi}{3}$
DOption (d)
$\frac{\pi}{2}$

View answer & explanation

Correct answer

$\frac{\pi}{2}$

Why this answer is correct

Given that:

$(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0, ; y(0) = 0$

Calculation:

$\int \frac{dy}{y+1} = -\int \frac{\cos x}{1+\sin x} dx$

⇒ $\ln(y+1) = -\ln(1+\sin x) + C$

⇒ $y+1 = \frac{e^{C}}{1+\sin x}$

Using $y(0) = 0$ :

⇒ $1+1 = \frac{e^{C}}{1+\sin 0}$

⇒ $C = 0$

⇒ $y+1 = \frac{1}{1+\sin x}$

Given $y = -\frac{1}{2}, ; x = \alpha$ :

⇒ $-\frac{1}{2} + 1 = \frac{1}{1+\sin \alpha}$

⇒ $\frac{1}{2} = \frac{1}{1+\sin \alpha}$

⇒ $1+\sin \alpha = 2$

⇒ $\sin \alpha = 1$

⇒ $\alpha = \frac{\pi}{2}$

∴ The correct answer is option (4).

Mathematics Section B (5 questions)

Question 21

If the domain of the function $f(x) = \sqrt{\log_{(0.6)} \left( \left| \frac{2x - 5}{x^2 - 4} \right| \right)}$

is $(-\infty, a] \cup {b} \cup [c, d) \cup (e, \infty)$ , then the value of

a + b + c + d + e is ____________.

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$\log_{0.6} \left| \frac{2x - 5}{x^2 - 4} \right| \geq 0$

$\left| \frac{2x - 5}{x^2 - 4} \right| \leq 1$

$x \neq \frac{5}{2}$

$-1 \leq \frac{2x - 5}{x^2 - 4} \leq 1$

$\frac{(x+1)^2 - 10}{(x-2)(x+2)} \geq 0$

$x \in (-\infty, -1-\sqrt{10}] \cup (-2, 2) \cup [-1+\sqrt{10}, \infty)$

$\frac{(x-1)^2}{(x-2)(x+2)} \geq 0$

$x \in (-\infty, -2) \cup (2, \infty) \cup {1}$

Calculation:

$(1) \cap (2) \cap (3)$

$x \in (-\infty, -1-\sqrt{10}] \cup {1} \cup \left[ -1+\sqrt{10}, \frac{5}{2} \right) \cup \left( \frac{5}{2}, \infty \right)$

$a + b + c + d + e = -2 + 1 + 5 = 4$

Question 22

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 ________.

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$a_1 + a_2 + \ldots + a_n = 6n^3$

$a_1 + a_2 + \ldots + a_n + a_{n+1} = 6(n+1)^3$

Calculation:

$a_{n+1} = 6(n+1)^3 - 6n^3$

⇒ $a_{n+1} = 6((n+1)-n)((n+1)^2 + n^2 + n(n+1))$

⇒ $a_{n+1} = 6(1)(3n^2 + 3n + 1)$

⇒ $a_n = 6(3(n-1)^2 + 3(n-1) + 1)$

⇒ $a_n = 6(3n^2 - 3n + 1)$

⇒ $\sum_{k=1}^6 \left( \frac{6(3k^2 + 3k + 1) - 6(3k^2 - 3k + 1)}{36} \right)^2$

⇒ $\sum_{k=1}^6 \left( \frac{36k}{36} \right)^2 = \sum_{k=1}^6 k^2 = \frac{6 × 7 × 13}{6} = 91$

Question 23

Let a, b, c $\in {1, 2, 3, 4}$ . If the probability, that

$ax^2 + 2\sqrt{2} bx + c > 0$ for all x $\in \mathbf{R}$ , is $\frac{m}{n}$ ,

gcd (m, n) = 1, then m + n is equal to ________.

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$8b^2 - 4ac < 0 \Rightarrow 2b^2 < ac$

Calculation:

Total combinations of $b, a, c$ :

$1 \times 4 + 1 \times 3 + 1 \times 4 + 1 \times 4 + 1 \times 2 + 1 \times 2 = 19$

Valid combinations satisfying $2b^2 < ac$ :

$2 + 3 + 4 + 4 + 2 + 2 = 17$

Required probability:

$\frac{17}{64} = \frac{m}{n}$ , where $m = 17, n = 64$

$m + n = 17 + 64 = 81$

Question 24

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 ________.

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$r^2 = \left( \frac{\sqrt{14}}{2} \right)^2 + \left( r - \frac{1}{\sqrt{2}} \right)^2$

Calculation:

$r^2 = \frac{7}{2} + \left( r - \frac{1}{\sqrt{2}} \right)^2$

⇒ $r^2 = \frac{7}{2} + r^2 - \sqrt{2}r + \frac{1}{2}$

⇒ $\sqrt{2}r = 4$

⇒ $r = 2\sqrt{2}$

⇒ $r^2 = 8$

Question 25

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 ________.

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

Given that:

$\int_a^b f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(x) dx = bf(b) - af(a)$

$\alpha = (2\sqrt{3})(4) - (0)(2)$

Calculation:

$\alpha = 8\sqrt{3}$

$\alpha^2 = 192$

Chemistry Section A (20 questions)

Question 26

The mass of iron converted into $\text{Fe}_3\text{O}_4$ by the action of 18 g of steam is :

(Given : Molar mass of H, O and Fe are 1, 16 and 56 $\text{g mol}^{-1}$ respectively)

Assume iron is present in excess :

AOption (a)
2.1 g
BOption (b)
4.2 g
COption (c)
21 g
DOption (d)
42 g

View answer & explanation

Correct answer

42 g

Why this answer is correct

CONCEPT:

Stoichiometry of the Reaction

Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction.
In the given reaction:

3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

From the balanced chemical equation:
3 moles of Fe react with 4 moles of H₂O.
The molar ratio of Fe to H₂O is 3:4.
Molar masses:
Molar mass of Fe = 56 g/mol
Molar mass of H₂O = 18 g/mol

EXPLANATION:

Given mass of steam (H₂O) = 18 g
Number of moles of steam:

n_steam = ⅛18 = 1 mol

From the stoichiometric ratio, moles of Fe required:

n_Fe = ¾

Mass of Fe required:

Mass of Fe = n_Fe × Molar mass of Fe

= ¾ × 56

= 42 g

Therefore, the mass of iron converted into Fe₃O₄ by the action of 18 g of steam is 42 g.

Question 27

What is the energy (in $\text{J atom}^{-1}$ ) required for the following process?

$\text{Li}^{2+}\text{(g)} \rightarrow \text{Li}^{3+}\text{(g)} + \text{e}^-$

(Take the ionization energy for the H atom in the ground state as $2.18 \times 10^{-18} \text{ J atom}^{-1}$ )

AOption (a)
$8.72 \times 10^{-18}$
BOption (b)
$1.962 \times 10^{-18}$
COption (c)
$1.962 \times 10^{-17}$
DOption (d)
$6.54 \times 10^{-17}$

View answer & explanation

Correct answer

$1.962 \times 10^{-17}$

Why this answer is correct

CONCEPT:

Ionization Energy and Its Dependence on Nuclear Charge (Z)

The ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state.
For hydrogen-like ions (single electron systems like $\text{Li}^{2+}$ ), the ionization energy can be calculated using the formula:

$\text{E} = \text{E}_{\text{H}} \times \text{Z}^2$

Here:
$\text{E}_{\text{H}}$ is the ionization energy of a hydrogen atom in its ground state.
$\text{Z}$ is the atomic number of the element.

EXPLANATION:

The given reaction is:

$\text{Li}^{2+}\text{(g)} \rightarrow \text{Li}^{3+}\text{(g)} + \text{e}^-$

Here, $\text{Li}^{2+}$ is a hydrogen-like ion with a single electron in its orbit. Its ionization energy can be calculated using the formula:

$\text{E} = \text{E}_{\text{H}} \times \text{Z}^2$

The ionization energy of hydrogen, $\text{E}_{\text{H}} = 2.18 \times 10^{-18} \text{ J atom}^{-1}$ .
The atomic number of lithium, $\text{Z} = 3$ .
Substitute the values into the formula:

$\text{E}_{\text{Li}^{2+}} = 2.18 \times 10^{-18} \times (3)^2$

$\text{E}_{\text{Li}^{2+}} = 2.18 \times 10^{-18} \times 9$

$\text{E}_{\text{Li}^{2+}} = 1.962 \times 10^{-17} \text{ J atom}^{-1}$

Therefore, the energy required for the process is $1.962 \times 10^{-17} \text{ J atom}^{-1}$ .

Question 28

Given below are two statements :

Statement (I) : The correct sequence of bond lengths in the following species is: $\text{O}_2^+ < \text{O}_2 < \text{O}_2^- < \text{O}_2^{2-}$

Statement (II) : The correct sequence of number of unpaired electrons in the following species is: $\text{O}_2 > \text{O}_2^+ > \text{O}_2^- > \text{O}_2^{2-}$

In the light of the above statements, choose the correct answer from the options given below :

AOption (a)
Both Statement I and Statement II are true
BOption (b)
Both Statement I and Statement II are false
COption (c)
Statement I is true but Statement II is false
DOption (d)
Statement I is false but Statement II is true

View answer & explanation

Correct answer

Statement I is true but Statement II is false

Why this answer is correct

CONCEPT:

Bond Order and Unpaired Electrons in Molecular Orbital Theory

Bond Order:

Bond Order = ^{(Number of bonding electrons - Number of antibonding electrons)}⁄₂

The bond order is a measure of the number of chemical bonds between a pair of atoms. It is given by the formula:
A higher bond order corresponds to a shorter and stronger bond, while a lower bond order corresponds to a longer and weaker bond.
Unpaired Electrons:
The number of unpaired electrons in a molecule is determined based on the electronic configuration of its molecular orbitals.
Molecules with unpaired electrons are paramagnetic, while those without unpaired electrons are diamagnetic.

EXPLANATION:

The electronic configurations of the molecular orbitals for the given oxygen species are as follows:
O+2O2+

: Bond Order = 2.5, Unpaired Electrons = 1

O2O2

: Bond Order = 2, Unpaired Electrons = 2

O−2O2−

: Bond Order = 1.5, Unpaired Electrons = 1

O2−2O22−

: Bond Order = 1, Unpaired Electrons = 0

Statement I: The bond length is inversely proportional to the bond order. Therefore, the correct sequence of bond lengths is:

O+2<O2<O−2<O2−2O2+<O2<O2−<O22−

This makes Statement I true.

Statement II: The number of unpaired electrons decreases in the following order:

O2>O+2>O−2>O2−2O2>O2+>O2−>O22−

This makes Statement II true.

Conclusion: Both Statement I and Statement II are true.

Question 29

Consider the following data

(i) $2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 3\text{H}_2\text{(g)} +$ 1200 kJ/mol.

(ii) $\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)} + 164 \text{ kJ/mol.}$

(iii) $\text{HCl(g)} + \text{aq} \rightarrow \text{HCl(aq)} + 83 \text{ kJ/mol.}$

(iv) $\text{Al}_2\text{Cl}_6\text{(s)} + \text{aq} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 663 \text{ kJ/mol.}$

The enthalpy of formation of anhydrous solid

$\text{Al}_2\text{Cl}_6$ is :

AOption (a)
$-648 \text{ kJ mol}^{-1}$
BOption (b)
$-1350 \text{ kJ mol}^{-1}$
COption (c)
$-2002 \text{ kJ mol}^{-1}$
DOption (d)
$-1527 \text{ kJ mol}^{-1}$

View answer & explanation

Correct answer

$-1527 \text{ kJ mol}^{-1}$

Why this answer is correct

CONCEPT:

Enthalpy of Formation ( $\Delta_{\text{f}}H$ )

The enthalpy of formation ( $\Delta_{\text{f}}H$ ) of a compound is the change in enthalpy when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions.
To calculate the enthalpy of formation for a reaction, the enthalpy changes ( $\Delta_{\text{r}}H$ ) from individual steps involved in the reaction can be summed up using Hess's Law.
Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps that lead to the reaction.

EXPLANATION:

Given data:
Step (i): $ 2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 3\text{H}2\text{(g)} $, $ \Delta{\text{r}}H_1 = -1200 , \text{kJ/mol} $
Step (ii): $ \text{H}_2\text{(g)} + \text{Cl}2\text{(g)} \rightarrow 2\text{HCl(g)} $, $ \Delta{\text{r}}H_2 = -164 , \text{kJ/mol} $ (multiplied by 3: $-164 \times 3 , \text{kJ/mol}$ )
Step (iii): $\text{HCl(g)} + \text{aq} \rightarrow \text{HCl(aq)}$ , $\Delta_{\text{r}}H_3 = -83 , \text{kJ/mol}$ (multiplied by 6: $-83 \times 6 , \text{kJ/mol}$ )
Step (iv): $ \text{Al}_2\text{Cl}_6\text{(aq)} \rightarrow \text{Al}_2\text{Cl}6\text{(s)} + \text{aq} $, $ \Delta{\text{r}}H_4 = +663 , \text{kJ/mol} $
Target reaction: $2\text{Al(s)} + 3\text{Cl}_2 \rightarrow \text{Al}_2\text{Cl}_6\text{(s)}$
Using Hess's Law:
$\Delta_{\text{f}}H = \Delta_{\text{r}}H_1 + \Delta_{\text{r}}H_2 + \Delta_{\text{r}}H_3 + \Delta_{\text{r}}H_4$
$\Delta_{\text{f}}H = -1200 - (164 \times 3) - (83 \times 6) + 663$
$\Delta_{\text{f}}H = -1200 - 492 - 498 + 663$
$\Delta_{\text{f}}H = -1527 , \text{kJ/mol}$

Therefore, the enthalpy of formation ( $\Delta_{\text{f}}H$ ) of anhydrous solid $\text{Al}_2\text{Cl}_6$ is $-1527 , \text{kJ/mol}$ .

Question 30

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

AOption (a)
$10^{-6}$
BOption (b)
$4 \times 10^{-4}$
COption (c)
$3 \times 10^{-5}$
DOption (d)
$3 \times 10^{-3}$

View answer & explanation

Correct answer

$3 \times 10^{-3}$

Why this answer is correct

CONCEPT:

Freezing Point Depression and Degree of Dissociation

The depression in the freezing point of a solution is given by the equation:

ΔT_f = iK_fm

where:

ΔT_f = depression in freezing point
i = van't Hoff factor
K_f = molal freezing point depression constant
m = molality of the solution
The van't Hoff factor, i, is related to the degree of dissociation (α) by the formula:

i = 1 + (n - 1)α

where:

n = number of ions produced per molecule upon dissociation
α = degree of dissociation
The dissociation constant (K_a) is given by the formula:

K_a = (Cα²) / (1 - α)

where:

C = concentration of the solute
α = degree of dissociation

EXPLANATION:

Given data:
Mass of fluoroacetic acid = 19.5 g
Molar mass of fluoroacetic acid = 78 g mol^-1
Mass of water = 500 g = 0.5 kg
ΔT_f = 1°C
K_f = 1.86 K kg mol^-1
Calculate the number of moles of fluoroacetic acid:

= ^19.5 / ₇₈ = 0.25 mol

Calculate the molality (m):

m = ^{Number of moles of solute} / _{Mass of solvent in kg}

= ^0.25 / _0.5 = 0.5 mol kg^-1

ΔT_f = iK_fm

1 = i × 1.86 × 0.5

i = ¹ / _{1.86 × 0.5} = ² / _1.86

i = 1 + α (since fluoroacetic acid dissociates into 2 particles: H⁺ and CH₂FCOO^-)

² / _1.86 = 1 + α

α = ² / _1.86 - 1 = 0.075

K_a = ^Cα² / _{1 - α}

Here, C = molality = 0.5 mol kg^-1

K_a = ^{0.5 × (0.075)²} / _{1 - 0.075}

K_a = ^{0.5 × 0.005625} / _0.925 = ^0.0028125 / _0.925

K_a ≈ 3 × 10^-3

Therefore, the dissociation constant (K_a) of fluoroacetic acid is approximately 3 × 10^-3.

Question 31

The solubility product constants of $\text{Ag}_2\text{CrO}_4$ and $\text{AgBr}$ are 32x and 4y respectively at 298 K. The value of $\left( \frac{\text{molarity of Ag}_2\text{CrO}_4}{\text{molarity of AgBr}} \right)$ can be expressed as :

AOption (a)
$\frac{2\sqrt[3]{\text{x}}}{\text{y}}$
BOption (b)
$2\sqrt{\frac{\text{x}}{\text{y}}}$
COption (c)
$\sqrt{\frac{\text{x}}{\text{y}}}$
DOption (d)
$\frac{\sqrt[3]{\text{x}}}{\sqrt{\text{y}}}$

View answer & explanation

Correct answer

$\frac{\sqrt[3]{\text{x}}}{\sqrt{\text{y}}}$

Why this answer is correct

CONCEPT:

Solubility Product and Molarity Relation

The solubility product constant ( $K_{\text{sp}}$ ) represents the equilibrium constant for the dissolution of a sparingly soluble salt.
The molarity of ions in solution can be derived from the solubility product constant using the stoichiometric relationship between the ions and the salt.

EXPLANATION:

For $\text{Ag}_2\text{CrO}_4$ :
The dissociation equation is:

$\text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}$

The solubility product is given by:

$K_{\text{sp}} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}]$

If the molarity of $\text{Ag}_2\text{CrO}_4$ is $S_1$ , then:
$[\text{Ag}^+] = 2S_1$
$[\text{CrO}_4^{2-}] = S_1$
Substitute these values into the solubility product equation:

$K_{\text{sp}} = (2S_1)^2(S_1)$

$K_{\text{sp}} = 4S_1^3$

Given $K_{\text{sp}} = 32x$ , solve for $S_1$ :

$4S_1^3 = 32x$

$S_1 = 2x^{1/3}$

For $\text{AgBr}$ :
The dissociation equation is:

$\text{AgBr} \rightleftharpoons \text{Ag}^+ + \text{Br}^-$

The solubility product is given by:

$K_{\text{sp}} = [\text{Ag}^+][\text{Br}^-]$

If the molarity of $\text{AgBr}$ is $S_2$ , then:
$[\text{Ag}^+] = S_2$
$[\text{Br}^-] = S_2$
Substitute these values into the solubility product equation:

$K_{\text{sp}} = S_2^2$

Given $K_{\text{sp}} = 4y$ , solve for $S_2$ :

$S_2^2 = 4y$

$S_2 = 2y^{1/2}$

Calculate the ratio $\left( \frac{\text{S}_1}{\text{S}_2} \right)$ :

$\frac{\text{S}_1}{\text{S}_2} = \frac{2x^{1/3}}{2y^{1/2}}$

$\frac{\text{S}_1}{\text{S}_2} = \frac{\sqrt[3]{x}}{\sqrt{y}}$

Therefore, the ratio $\left( \frac{\text{molarity of Ag}_2\text{CrO}_4}{\text{molarity of AgBr}} \right)$ is $\frac{\sqrt[3]{x}}{\sqrt{y}}$ .

Question 32

An electrochemical cell is constructed using half cells in the direction of spontaneous change $\text{Fe(OH)}_2(\text{s}) + 2\text{e}^- \rightarrow \text{Fe(s)} + 2\text{OH}^-(\text{aq}) \text{ E}^0 = -0.88\text{V}$ and $\text{AgBr(s)} + \text{e}^- \rightarrow \text{Ag(s)} + \text{Br}^-(\text{aq}) \text{ E}^0 = +0.07\text{V}$ Which of the following option is correct ?

AOption (a)
Overall reaction $\text{Fe(s)} + 2\text{OH}^-(\text{aq}) + 2\text{AgBr(s)} \rightleftharpoons \text{Fe(OH)}_2(\text{s}) + 2\text{Ag(s)} + 2\text{Br}^-(\text{aq})$
BOption (b)
$\text{E}^0_{\text{cell}} = -0.95\text{V}$
COption (c)
Fe is reduced in the electrochemical cell
DOption (d)
$\text{E}^0_{\text{cell}}$ is an extensive property

View answer & explanation

Correct answer

Overall reaction $\text{Fe(s)} + 2\text{OH}^-(\text{aq}) + 2\text{AgBr(s)} \rightleftharpoons \text{Fe(OH)}_2(\text{s}) + 2\text{Ag(s)} + 2\text{Br}^-(\text{aq})$

Why this answer is correct

CONCEPT:

Standard Cell Potential (E⁰_cell)

The standard cell potential (E⁰_cell) is the difference in the standard reduction potentials of the cathode and the anode in an electrochemical cell.
The standard cell potential can be calculated using the equation:

E⁰_cell = E⁰_cathode - E⁰_anode

EXPLANATION:

In the given reaction:

Fe(s) + 2OH^-(aq) + 2AgBr(s) → Fe(OH)₂(s) + 2Ag(s) + 2Br^-(aq)

Iron (Fe) is oxidized to Fe(OH)₂, so Fe is the anode.
Silver bromide (AgBr) is reduced to Ag, so AgBr is the cathode.
Given standard reduction potentials:
E⁰_AgBr = +0.07 V (for AgBr(s) + e^- → Ag(s) + Br^-(aq))
E⁰_Fe = -0.88 V (for Fe(OH)₂(s) + 2e^- → Fe(s) + 2OH^-(aq))
Using the standard cell potential equation:
E⁰_cell = E⁰_cathode - E⁰_anode
= 0.07 V - (-0.88 V)
= 0.07 V + 0.88 V
= 0.95 V
Key points to note:
Fe is oxidized in the reaction, not reduced.
E⁰_cell is not an extensive property; it is an intensive property, as it does not depend on the amount of material.

Therefore, the correct overall reaction is:

Fe(s) + 2OH^-(aq) + 2AgBr(s) → Fe(OH)₂(s) + 2Ag(s) + 2Br^-(aq)

The standard cell potential (E⁰_cell) is 0.95 V.

Question 33

$ \text{t}{100%} $ is the time required for the 100% completion of the reaction while $ \text{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 $ \text{t}{100%} $ and $ \text{t}{1/2} $ for zero and first order reactions respectively ?

AOption (a)
$ \text{t}{100%} = (\text{t}{1/2})^2 $ and $ \text{t}{100%} = (\text{t}{1/2})^\infty $
BOption (b)
$ \text{t}{100%} = 2\text{t}{1/2} $ and $ \text{t}{100%} = (\text{t}{1/2})^\infty $
COption (c)
$ \text{t}{100%} = 2\text{t}{1/2} $ and $ \text{t}{100%} = (2\text{t}{1/2})^2 $
DOption (d)
$ \text{t}{100%} = (\text{t}{1/2})^\infty $ and $ \text{t}{100%} = 2\text{t}{1/2} $

View answer & explanation

Correct answer

$ \text{t}{100%} = 2\text{t}{1/2} $ and $ \text{t}{100%} = (\text{t}{1/2})^\infty $

Why this answer is correct

CONCEPT:

Zero-order and First-order Reaction Time Relationships

For a zero-order reaction, the rate of reaction is constant and independent of the concentration of the reactant.
For a first-order reaction, the rate of reaction is proportional to the concentration of the reactant.
The times required for 50% completion ($ \text{t}{1/2} $) and 100% completion ($ \text{t}{100%} $) of the reaction differ for zero-order and first-order reactions.

EXPLANATION:

Zero-order Reaction:
For a zero-order reaction, the time required for the reaction to complete 100% is given by:

$\text{t}_{100\%} = \frac{\text{a}}{\text{k}}$

The time required for 50% completion is:

$\text{t}_{1/2} = \frac{\text{a}}{2\text{k}}$

By dividing $ \text{t}{100%} $ by $ \text{t}{1/2} $, we get:

$ \text{t}{100%} = 2\text{t}{1/2} $

First-order Reaction:
For a first-order reaction, the time required for 100% completion is theoretically infinite, because the reactant concentration never truly reaches zero.
The time required for 50% completion is:

$\text{t}_{1/2} = \frac{\ell\text{n } 2}{\text{k}}$

Since $\text{t}_{100\%}$ is infinite, we represent the relationship as:

$ \text{t}{100%} = (\text{t}{1/2})^\infty $

So, For a zero-order reaction, $ \text{t}{100%} = 2\text{t}{1/2} $.

For a first-order reaction, $ \text{t}{100%} = (\text{t}{1/2})^\infty $.

Therefore, the correct relationship is:

$ \text{t}{100%} = 2\text{t}{1/2} $ for zero-order reactions and $ \text{t}{100%} = (\text{t}{1/2})^\infty $ for first-order reactions.

Question 34

Given below are two statements :

Statement (I) : The first ionisation enthalpy of the elements Na, Mg, Cl and Ar follows the order 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 :

AOption (a)
Both Statement I and Statement II are true
BOption (b)
Both Statement I and Statement II are false
COption (c)
Statement I is true but Statement II is false
DOption (d)
Statement I is false but Statement II is true

View answer & explanation

Correct answer

Statement I is false but Statement II is true

Why this answer is correct

CONCEPT:

Ionization Enthalpy

The ionization enthalpy is the energy required to remove an electron from the outermost shell of an isolated gaseous atom.
The first ionization enthalpy corresponds to the removal of the first electron, the second ionization enthalpy corresponds to the removal of the second electron, and so on.
Ionization enthalpy is influenced by factors such as atomic size, nuclear charge, shielding effect, and electron configuration.

EXPLANATION:

Statement (I): The first ionization enthalpy of the elements Na, Mg, Cl, and Ar follows the order Na > Mg > Cl > Ar.
This statement is false because the correct order of first ionization enthalpy is Na < Mg < Cl < Ar.
The trend in ionization energy across a period increases from left to right due to increasing nuclear charge and decreasing atomic size.
Na has the lowest ionization enthalpy, followed by Mg, Cl, and Ar, which has the highest due to its stable noble gas configuration.
Statement (II): Among Ca, Al, Fe, and B, the third ionization enthalpy is very high for Ca.
This statement is true.
Calcium (Ca) has a very high third ionization enthalpy because after losing two electrons, it attains a stable noble gas configuration (similar to Argon: [Ar]).
Removing a third electron from this stable configuration requires a significantly higher amount of energy.

So, Statement I is false but Statement II is true.

Question 35

Given below are two statements :

Statement (I) : Oxidising power of halogens decreases in the order $\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2$ , which is the basis of “Layer test”.

Statement (II) : “Layer test” to identify $\text{Br}_2$ and $\text{I}_2$ in aqueous solution involves the oxidation of bromide or iodide into $\text{Br}_2$ or $\text{I}_2$ respectively with , which is a type of displacement redox reaction.

In the light of the above statements, choose the correct answer from the options given below :

AOption (a)
Both Statement I and Statement II are true
BOption (b)
Both Statement I and Statement II are false
COption (c)
Statement I is true but Statement II is false
DOption (d)
Statement I is false but Statement II is true

View answer & explanation

Correct answer

Both Statement I and Statement II are true

Why this answer is correct

CONCEPT:

Oxidizing Power of Halogens

Oxidizing power refers to the ability of a substance to oxidize another substance, i.e., to accept electrons.
Among the halogens, the oxidizing power decreases down the group in the periodic table in the following order: F₂ > Cl₂ > Br₂ > I₂.
This trend is due to the decreasing standard reduction potentials of the halogens as we move down the group, making them less effective oxidizing agents.

Layer Test

The "Layer Test" is a chemical test used to identify halogens in aqueous solutions based on their relative oxidizing powers.
In this test, a more powerful oxidizing agent (e.g., Cl₂) is used to oxidize halide ions (e.g., Br^- or I^-) to their molecular forms (Br₂ or I₂), which can then be detected by their characteristic colors.
For example:
Cl₂ oxidizes bromide ions (Br^-) to bromine (Br₂), which appears as an orange-brown color in aqueous solution:

NaBr + Cl₂-water → Br₂ + NaCl

Cl₂ also oxidizes iodide ions (I^-) to iodine (I₂), which appears as a violet color in organic solvents or a brown color in aqueous solution:

NaI + Cl₂-water → I₂ + NaCl

EXPLANATION:

Statement (I): "Oxidizing power of halogens decreases in the order F₂ > Cl₂ > Br₂ > I₂, which is the basis of 'Layer test'" is true. This is because fluorine has the highest oxidizing power, followed by chlorine, bromine, and iodine, consistent with their standard reduction potentials.
Statement (II): "'Layer test' to identify Br₂ and I₂ in aqueous solution involves the oxidation of bromide or iodide into Br₂ or I₂ respectively with Cl₂, which is a type of displacement redox reaction" is also true. This statement correctly describes the chemical basis of the layer test, where Cl₂, being a stronger oxidizing agent, displaces Br₂ and I₂ from their respective halide ions.
Since both statements are correct and Statement (II) explains the process mentioned in Statement (I), the correct interpretation is that both Statement I and Statement II are true.

Therefore, the answer is that both Statement I and Statement II are true.

Question 36

Which of the following sets includes all the species that will change the orange colour of $\text{K}_2\text{Cr}_2\text{O}_7$ in acidic medium ?

AOption (a)
$\text{Fe}^{2+}, \text{Sn}^{2+}, \text{I}^-, \text{S}^{2-}$
BOption (b)
$\text{S}^{2-}, \text{Fe}^{3+}, \text{I}^-, \text{C}_2\text{O}_4^{2-}$
COption (c)
$\text{Fe}^{2+}, \text{NO}_2^-, \text{SO}_2, \text{Sn}^{4+}$
DOption (d)
$\text{Fe}^{3+}, \text{SO}_4^{2-}, \text{S}^{2-}, \text{Sn}^{4+}$

View answer & explanation

Correct answer

$\text{Fe}^{2+}, \text{Sn}^{2+}, \text{I}^-, \text{S}^{2-}$

Why this answer is correct

CONCEPT:

Oxidation-Reduction Reactions and Changes in Color

In an acidic medium, $\text{K}_2\text{Cr}_2\text{O}_7$ (potassium dichromate) undergoes a reduction reaction. The dichromate ion $\text{Cr}_2\text{O}_7^{2-}$ is reduced to $\text{Cr}^{3+}$ , causing a color change from orange to green.
This color change occurs when the dichromate ion reacts with a reducing agent, which donates electrons to reduce $\text{Cr}_2\text{O}_7^{2-}$ .
Common reducing agents include species that can themselves undergo oxidation, such as $\text{Fe}^{2+}, \text{Sn}^{2+}, \text{I}^-, \text{S}^{2-}$ , etc.

EXPLANATION:

To determine which species will cause the color change of $\text{K}_2\text{Cr}_2\text{O}_7$ in acidic medium, we need to identify species that act as reducing agents and undergo oxidation.
For the species in the given options:
$\text{Fe}^{2+}$ : Oxidized to $\text{Fe}^{3+}$ , donating electrons and reducing $\text{Cr}_2\text{O}_7^{2-}$ . Thus, it causes the color change.
$\text{Sn}^{2+}$ : Oxidized to $\text{Sn}^{4+}$ , donating electrons and reducing $\text{Cr}_2\text{O}_7^{2-}$ . Thus, it causes the color change.
$\text{I}^-$ : Oxidized to $\text{I}_2$ , donating electrons and reducing $\text{Cr}_2\text{O}_7^{2-}$ . Thus, it causes the color change.
$\text{S}^{2-}$ : Oxidized to sulfur ( $\text{S}$ ), donating electrons and reducing $\text{Cr}_2\text{O}_7^{2-}$ . Thus, it causes the color change.
Other species, such as $\text{Fe}^{3+}$ , $\text{Sn}^{4+}$ , or $\text{SO}_4^{2-}$ , cannot act as reducing agents in this context, as they are already in their oxidized states.

Conclusion: The species $\text{Fe}^{2+}, \text{Sn}^{2+}, \text{I}^-, \text{S}^{2-}$ will change the orange color of $\text{K}_2\text{Cr}_2\text{O}_7$ to green in acidic medium, as they are all capable of acting as reducing agents.

Question 37

Match List – I with List – II.

List – I	List – II
Chromium (III) Complexes	$\Delta_0(\text{cm}^{-1})$
(en = ethylene diamine)
A. $[\text{Cr(CN)}_6]^{3-}$	I. 15,060
B. $[\text{CrF}_6]^{3-}$	II. 17,400
C. $[\text{Cr(H}_2\text{O)}_6]^{3+}$	III. 22,300
D. $[\text{Cr(en)}_3]^{3+}$	IV. 26,600

Choose the correct answer from the options given

below :

AOption (a)
a-I, b-II, c-III, d-IV
BOption (b)
a-II, b-III, c-IV, d-I
COption (c)
a-III, b-IV, c-I, d-II
DOption (d)
a-IV, b-I, c-II, d-III

View answer & explanation

Correct answer

a-IV, b-I, c-II, d-III

Why this answer is correct

CONCEPT:

Crystal Field Splitting Energy (CFSE) and Ligand Strength

The crystal field splitting energy ( $\Delta_0$ ) is directly proportional to the strength of the ligand in the spectrochemical series.
Stronger ligands cause greater splitting of the d-orbitals, resulting in higher $\Delta_0$ values.
The spectrochemical series for the given ligands is as follows:

CN^- > en (ethylene diamine) > H₂O > F^-

Thus, CN^- is the strongest ligand, and F^- is the weakest ligand.

Additionally, chelation (e.g., by ethylene diamine) increases the ligand strength compared to monodentate ligands of similar type.

EXPLANATION:

The given complexes are:
A. $[\text{Cr(CN)}_6]^{3-}$
B. $[\text{CrF}_6]^{3-}$
C. $[\text{Cr(H}_2\text{O)}_6]^{3+}$
D. $[\text{Cr(en)}_3]^{3+}$
The crystal field splitting energy ( $\Delta_0$ ) values are:
I. 15,060 cm^-1
II. 17,400 cm^-1
III. 22,300 cm^-1
IV. 26,600 cm^-1
Based on the ligand strength:
$[\text{Cr(CN)}_6]^{3-}$ (CN^-, strongest ligand) should have the highest splitting energy, i.e., $\Delta_0$ $= 26,600 , \text{cm}^{-1} \) (IV). -$ $[\text{Cr(en)}_3]^{3+}$ (chelating ligand, moderate strength) should have the second-highest splitting energy, i.e., $\Delta_0$ $= 22,300 , \text{cm}^{-1} \) (III). -$ $[\text{Cr(H}_2\text{O)}_6]^{3+}$ (H₂O, weaker ligand) should have the second-lowest splitting energy, i.e., $\Delta_0$ $= 17,400 , \text{cm}^{-1} \) (II). -$ $[\text{CrF}_6]^{3-}$ (F^-, weakest ligand) should have the lowest splitting energy, i.e., $\Delta_0$ $= 15,060 , \text{cm}^{-1} \) (I).

MATCHING:

Complex	$ $\Delta_0 , (\text{cm}^{-1})$
A. $[\text{Cr(CN)}_6]^{3-}$	IV. 26,600
B. $[\text{CrF}_6]^{3-}$	I. 15,060
C. $[\text{Cr(H}_2\text{O)}_6]^{3+}$	II. 17,400
D. $[\text{Cr(en)}_3]^{3+}$	III. 22,300

FINAL ANSWER:

A-IV, B-I, C-II, D-III

Question 38

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 :

AOption (a)
Both Statement I and Statement II are true
BOption (b)
Both Statement I and Statement II are false
COption (c)
Statement I is true but Statement II is false
DOption (d)
Statement I is false but Statement II is true

View answer & explanation

Correct answer

Statement I is false but Statement II is true

Why this answer is correct

CONCEPT:

Distillation and Separation of Mixtures

Distillation is a separation technique based on the difference in boiling points of substances.
Simple distillation is effective when the boiling points of the components differ significantly, while fractional distillation is used for closer boiling points.
An azeotropic mixture is a mixture of two or more liquids that boils at a constant temperature and retains its composition in the vapor phase, making it impossible to separate by conventional fractional distillation.

EXPLANATION:

Statement (I): 1,2,3-Trihydroxypropane (commonly known as glycerol) cannot be separated from water by simple distillation because glycerol has a very high boiling point (approximately 290°C) compared to water (100°C). Moreover, glycerol decomposes at high temperatures, so it is typically separated by distillation under reduced pressure to avoid decomposition.
Statement (II): An azeotropic mixture cannot be separated by fractional distillation because the composition of the liquid and vapor phases remains constant during boiling, preventing further separation. This statement is true.
Based on the above analysis:
Statement (I) is false because glycerol cannot be separated from water by simple distillation.
Statement (II) is true because azeotropic mixtures cannot be separated by fractional distillation.

So, Statement I is false but Statement II is true.

Question 39

Given below are two statements :

Statement (I) : Benzyl chloride reacts faster in $\text{S}_{\text{N}}1$ 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 :

AOption (a)
Both Statement I and Statement II are true
BOption (b)
Both Statement I and Statement II are false
COption (c)
Statement I is true but Statement II is false
DOption (d)
Statement I is false but Statement II is true

View answer & explanation

Correct answer

Both Statement I and Statement II are true

Why this answer is correct

CONCEPT:

Rate of S_N1 Reaction

The rate of an S_N1 reaction is directly proportional to the stability of the intermediate carbocation.
Carbocation stability is influenced by factors like resonance, hyperconjugation, and inductive effects.
Order of carbocation stability:

Ph-⁺C_H₂ > CH₃-⁺C_H₂

Resonance stabilization is more effective than hyperconjugation in stabilizing carbocations.

EXPLANATION:

Statement (I): Benzyl chloride reacts faster in an S_N1 mechanism than ethyl chloride.
This is true because the benzyl carbocation (Ph-⁺C_H₂) formed during the reaction is stabilized by resonance with the aromatic ring.
In contrast, the ethyl carbocation (CH₃-⁺C_H₂) is stabilized only by hyperconjugation, which is less effective compared to resonance.
Statement (II): Ethyl carbocation intermediate is less stabilized by hyperconjugation than benzyl carbocation by resonance.
This is also true because the benzyl carbocation has multiple resonance structures, allowing delocalization of the positive charge over the aromatic ring, which provides significant stabilization.
On the other hand, the ethyl carbocation relies only on hyperconjugation, which provides limited stabilization compared to resonance.
Since both statements are true and align with the principles of carbocation stability, the correct answer is that both Statement (I) and Statement (II) are true.

Therefore, the stability of the benzyl carbocation due to resonance explains the faster reaction rate of benzyl chloride in the S_N1 mechanism compared to ethyl chloride.

Question 40

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

AOption (a)
$-\text{CONH}_2, >\text{C} = \text{O}, -\text{CHO}, -\text{NH}_2, -\text{C} \equiv \text{C} -$
BOption (b)
$-\text{CONH}_2, -\text{COOCH}_3, -\text{CHO}, -\text{NH}_2, -\text{OH}$
COption (c)
$-\text{CONH}_2, -\text{CHO}, >\text{C} = \text{O}, -\text{NH}_2, -\text{C} \equiv \text{C} -$
DOption (d)
$-\text{CONH}_2, -\text{CHO}, -\text{CN}, -\text{NH}_2, -\text{C} \equiv \text{C} -$

View answer & explanation

Correct answer

$-\text{CONH}_2, -\text{CHO}, >\text{C} = \text{O}, -\text{NH}_2, -\text{C} \equiv \text{C} -$

Why this answer is correct

CONCEPT:

IUPAC Nomenclature Priority of Functional Groups

In IUPAC nomenclature, the priority of functional groups is based on the specific rules established by IUPAC. The priority order is used to determine the principal functional group, which dictates the suffix of the compound name.
The priority order is determined based on the oxidation state of the carbon atom to which the functional group is attached. Functional groups with higher oxidation states have higher priority.
In general, the priority order from highest to lowest is:
Carboxylic acids (-COOH)
Esters (-COOR)
Amides (-CONH₂)
Aldehydes (-CHO)
Ketones (>C=O)
Nitriles (-CN)
Alcohols (-OH)
Amines (-NH₂)
Alkenes (C=C)
Alkynes (C≡C)

EXPLANATION:

Memonics: "Crazy Elephants Always Ask Kind Neighbors About Amazing Airplanes Always."
Functional groups provided in the question:
-CONH₂ (Amide)
-CHO (Aldehyde)
C=O (Ketone)
-NH₂ (Amine)
-C≡C- (Alkyne)
Based on the priority order of functional groups:
Amides (-CONH₂) have the highest priority among the groups provided.
Aldehydes (-CHO) have the next highest priority.
Ketones (>C=O) follow aldehydes in priority.
Amines (-NH₂) come after ketones.
Alkynes (-C≡C-) have the lowest priority among the given groups.
Thus, the correct decreasing order of priority is:

-CONH₂, -CHO, >C=O, -NH₂, -C≡C-

Therefore, the correct answer is Option 3.

Question 41

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

AOption (a)
Predominantly at “r”
BOption (b)
“r” and “u”
COption (c)
“p” and “s”
DOption (d)
Predominantly at “u”

View answer & explanation

Correct answer

Predominantly at “u”

Why this answer is correct

CONCEPT:

Electrophilic Attack on Aromatic Compounds

Electrophilic substitution reactions occur when an electrophile replaces a substituent (usually a hydrogen atom) in an aromatic ring.
The position at which the electrophile attacks depends on the electron density of the aromatic ring and the nature of the substituents already present.
Electron-donating groups (+M or +I effect) increase electron density at certain positions, making them more favorable for electrophilic attack.
Electron-withdrawing groups (-M or -I effect) decrease electron density at specific positions, making them less favorable for electrophilic attack.

EXPLANATION:

In the given molecule, the nitrogen atom is attached to the aromatic ring and exhibits a +M (resonance donating) effect.
The +M effect of nitrogen increases the electron density at the ortho and para positions of the aromatic ring.
Among the labeled positions ("r", "u", "p", and "s"), position "u" is predominantly favored for electrophilic attack because it is para to the nitrogen atom, and the +M effect enhances the electron density at this position.
Position "r" also experiences an increase in electron density due to the +M effect, but it is less favorable than "u" because steric hindrance near the nitrogen may play a minor role in reducing reactivity at the ortho position.

Therefore, the preferred site for the attack of the electrophile is predominantly at "u" because of the +M nature of nitrogen.

Question 42

Match List-I with List-II.

List-I	List-II
Mixture of Compounds	Reagent used to distinguish
A. Diethyl amine + Ethyl amine	I. Bromine water
B. Acetaldehyde + Acetone	II. $CHCl_3 + KOH, \Delta$
C. Ethanol + Phenol	III. Neutral $FeCl_3$
D. Benzoic acid + Cinnamic acid	IV. Ammonical silver nitrate

Choose the correct answer from the options given below :

AOption (a)
a-IV, b-II, c-I, d-III
BOption (b)
a-IV, b-II, c-III, d-I
COption (c)
a-II, b-IV, c-I, d-III
DOption (d)
a-II, b-IV, c-III, d-I

View answer & explanation

Correct answer

a-II, b-IV, c-III, d-I

Why this answer is correct

CONCEPT:

The question involves distinguishing between different chemical compounds in mixtures using specific reagents. The concept relies on the unique chemical reactions of these compounds with the reagents provided.

Phenol vs Ethanol: Phenol reacts with neutral $FeCl_3$ to give a violet color, whereas ethanol does not react with neutral $FeCl_3$ .
Diethyl amine vs Ethyl amine: Ethyl amine (a primary amine) reacts with $CHCl_3/KOH$ to give a carbylamine (foul-smelling test), whereas diethyl amine (a secondary amine) does not give this test.
Acetaldehyde vs Acetone: Acetaldehyde (an aldehyde) gives a positive test with ammonical silver nitrate (Tollens' test), forming a silver mirror, whereas acetone (a ketone) does not.
Benzoic acid vs Cinnamic acid: Cinnamic acid reacts with bromine water due to the presence of a double bond in its structure, whereas benzoic acid does not react with bromine water.

EXPLANATION:

A. Diethyl amine + Ethyl amine: Ethyl amine (primary amine) reacts with $CHCl_3/KOH$ , producing a foul smell (carbylamine test). Diethyl amine (secondary amine) does not react. Therefore, the reagent used is II. $CHCl_3 + KOH, \Delta$ .
B. Acetaldehyde + Acetone: Acetaldehyde gives a positive test with ammonical silver nitrate (Tollens' test), forming a silver mirror, whereas acetone does not. Therefore, the reagent used is IV. Ammonical silver nitrate.
C. Ethanol + Phenol: Phenol reacts with neutral $FeCl_3$ to produce a violet color, whereas ethanol does not. Therefore, the reagent used is III. Neutral $FeCl_3$ .
D. Benzoic acid + Cinnamic acid: Cinnamic acid reacts with bromine water due to the presence of a double bond, whereas benzoic acid does not. Therefore, the reagent used is I. Bromine water.

So the correct matching is A-II, B-IV, C-III, D-I

Mixture of Compounds (List-I)	Reagent used to distinguish (List-II)
A. Diethyl amine + Ethyl amine	II. $CHCl_3 + KOH, \Delta$
B. Acetaldehyde + Acetone	IV. Ammonical silver nitrate
C. Ethanol + Phenol	III. Neutral $FeCl_3$
D. Benzoic acid + Cinnamic acid	I. Bromine water

Question 43

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 $Ph-N \equiv N \ Cl^{(-)}$ under optimum but slightly acidic medium is :

AOption (a)
P > Q > R
BOption (b)
R > P > Q
COption (c)
R > Q > P
DOption (d)
P > R > Q

View answer & explanation

Correct answer

P > Q > R

Why this answer is correct

CONCEPT:

Electrophilic Substitution Reaction (ESR) Reactivity

The reactivity of an aromatic compound towards electrophilic substitution reactions (ESR) depends on the electron density of the benzene ring.
Electron-donating groups increase the nucleophilicity of the benzene ring, enhancing its reactivity with electrophiles.
Electron-withdrawing groups decrease the nucleophilicity of the benzene ring, reducing its reactivity with electrophiles.

EXPLANATION:

Rate of ESR $\propto$ nucleophilicity of benzene ring.
The nucleophilicity of the benzene rings follows the order is P > R > Q
Thus, the reactivity of these compounds towards the diazonium salt in a slightly acidic medium is P > R > Q

Therefore, the correct order regarding the reactivity of these compounds with $Ph-N \equiv N \ Cl^{(-)}$ is P > R > Q.

Question 44

Match List-I with List-II.

List-I	List-II
Vitamin	Name
A. Vitamin $B_1$	I. Pyridoxine
B. Vitamin $B_2$	II. Ascorbic acid
C. Vitamin $B_6$	III. Thiamine
D. Vitamin C	IV. Riboflavin

Choose the correct answer from the options given below :

AOption (a)
a-II, b-I, c-III, d-IV
BOption (b)
a-IV, b-III, c-II, d-I
COption (c)
a-III, b-IV, c-I, d-II
DOption (d)
a-I, b-III, c-II, d-IV

View answer & explanation

Correct answer

a-III, b-IV, c-I, d-II

Why this answer is correct

CONCEPT:

Vitamins and Their Chemical Names

Vitamins are organic compounds essential for various metabolic processes in the human body.
Each vitamin has a specific chemical name that represents its structure and function.
Matching vitamins with their chemical names helps in understanding their roles in human nutrition.

EXPLANATION:

The matching of vitamins with their chemical names is as follows:
Vitamin B₁ (A): Its chemical name is Thiamine. Thiamine plays a crucial role in energy metabolism and nerve function.
Vitamin B₂ (B): Its chemical name is Riboflavin. Riboflavin is involved in energy production and cellular function.
Vitamin B₆ (C): Its chemical name is Pyridoxine. Pyridoxine is important for amino acid metabolism and neurotransmitter synthesis.
Vitamin C (D): Its chemical name is Ascorbic acid. Ascorbic acid is an antioxidant and is essential for collagen synthesis and immune function.

Therefore, the correct matching is: A - III, B - IV, C - I, D - II.

Question 45

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 $K_2CrO_4$ to give yellow precipitate. When the sodium carbonate extract of the salt solution is heated with conc. $HNO_3$ and ammonium molybdate, it resulted a canary yellow precipitate. The cation and anion present in the salt are respectively.

AOption (a)
$Ca^{2+}$ and $SO_4^{2-}$
BOption (b)
$Ba^{2+}$ and $PO_4^{3-}$
COption (c)
$Mn^{2+}$ and $PO_4^{3-}$
DOption (d)
$Ba^{2+}$ and $SO_4^{2-}$

View answer & explanation

Correct answer

$Ba^{2+}$ and $PO_4^{3-}$

Why this answer is correct

CONCEPT:

Flame Test and Identification of Ions

The flame test is used to identify certain metal ions based on the characteristic color they produce when heated in a flame.
Flame Colors:
Ba+2Ba+2

produces an apple green color.
Ca+2Ca+2

produces a brick red color.
Sr+2Sr+2

produces a crimson red color.
Reactions for Identification:
Ba+2+CrO2−4→BaCrO4Ba+2+CrO42−→BaCrO4

: Yellow precipitate indicates the presence of Ba+2Ba+2

The sodium carbonate extract test with ammonium molybdate and concentrated nitric acid produces a canary yellow precipitate if PO3−4PO43−

is present.

EXPLANATION:

From the flame test:
The apple green color in the flame test confirms the presence of Ba+2Ba+2

From the reaction with K2CrO4K2CrO4

The formation of a yellow precipitate confirms the presence of Ba+2Ba+2

as BaCrO4BaCrO4

From the sodium carbonate extract test:
Heating the sodium carbonate extract with concentrated HNO3HNO3

and ammonium molybdate produces a canary yellow precipitate, confirming the presence of PO3−4PO43−

Based on the observations:
The cation present in the salt is Ba+2Ba+2

The anion present in the salt is PO3−4PO43−

Therefore, the cation and anion present in the salt are Ba2+Ba2+

and PO3−4PO43−

respectively.

Chemistry Section B (5 questions)

Question 46

5.33 g of $CrCl_3 \cdot 6H_2O$ , 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 $AgNO_3$ results in 8.61 g of AgCl. The ratio of moles of complex reacted and moles of AgCl formed is ________ $\times 10^{-2}$ . (Nearest integer)

[Molar mass in g $mol^{-1}$ Cr : 52, Ag : 108, Cl : 35.5, H : 1, O : 16]

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

CONCEPT:

Coordination Chemistry and Precipitation Reaction

The given reaction involves a coordination complex $[Cr(H_2O)_6]Cl_3$ , which is a hexaaquachromium(III) chloride complex.
When this complex reacts with an excess of AgNO₃ (silver nitrate), the chloride ions present outside the coordination sphere precipitate as AgCl (silver chloride).
The number of moles of AgCl precipitated depends on the number of chloride ions outside the coordination sphere of the complex.

EXPLANATION:

The reaction can be written as:

$ [Cr(H_2O)_6]Cl_3 + 3AgNO_3 \xrightarrow{excess} 3AgCl(s) + Cr(H_2O)_6_3 $

In the given coordination complex $[Cr(H_2O)_6]Cl_3$ , all six water molecules are within the coordination sphere of the chromium ion, while the three chloride ions are outside the coordination sphere.
When reacted with excess AgNO₃, each chloride ion forms a precipitate of AgCl:

$Cl^- + Ag^+ \rightarrow AgCl(s)$

Thus, one mole of $[Cr(H_2O)_6]Cl_3$ reacts to produce 3 moles of AgCl precipitate.
The ratio of the moles of the complex reacted to the moles of AgCl precipitated is given as:

$\frac{\text{no. of moles of complex reacted}}{\text{no. of moles of AgCl precipitated}} \times 100$

= $\frac{1}{3} \times 100$

= 33.3%

Therefore, the percentage of the moles of the complex reacted with respect to the moles of AgCl precipitated is 33.3%.

Question 47

Consider the isomers of hydrocarbon with molecular formula $C_5H_{10}$ . These isomers do not decolourise $KMnO_4$ 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 ________.

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

CONCEPT:

Mathematical Addition

The addition of numbers is a fundamental arithmetic operation.
It involves combining two or more numbers to find their total.
The general form for addition is:

a + b = c, where c is the sum of a and b.

EXPLANATION:

Therefore, the sum of 1 + 4 + 4 + 2 + 3 is 14.

Question 48

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 ________.

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

CONCEPT:

Combustion of Alkanes

Alkanes are hydrocarbons with the general formula $C_nH_{2n+2}$ .
When alkanes undergo combustion in the presence of oxygen ( $O_2$ ), they produce carbon dioxide ( $CO_2$ ) and water ( $H_2O$ ) as products.
The balanced chemical equation for the combustion of an alkane is:

$C_nH_{2n+2} + \left( \frac{3n+1}{2} \right) O_2 \rightarrow nCO_2 + (n+1) H_2O$

EXPLANATION:

In the given question, we are tasked with identifying the alkane based on the combustion equation.
The given combustion equation includes the term $\left( \frac{3n+1}{2} \right)$ for the oxygen ( $O_2$ ) coefficient.
We are also provided with the condition $\frac{3n+1}{2} = 8$ .
Solving this equation:
$\frac{3n+1}{2} = 8$
$3n+1 = 16$ (multiplying both sides by 2)
$3n = 15$ (subtracting 1 from both sides)
$n = 5$ (dividing both sides by 3)
Since $n = 5$ , the alkane is $C_5H_{12}$ (pentane).
Substituting $n = 5$ into the combustion equation:

$C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O$

Therefore, the alkane is pentane ( $C_5H_{12}$ ), and the number of oxygen molecules required for complete combustion is 8.

Question 49

For reaction A $\rightarrow$ P, rate constant k = 1.5 $\times 10^3 \ s^{-1}$ at 27°C. If activation energy for the above reaction is 60 kJ $mol^{-1}$ , then the temperature (in °C) at which rate constant, k = 4.5 $\times 10^3 \ s^{-1}$ is ________. (Nearest integer)

Given : log 2 = 0.30, log 3 = 0.48, R = 8.3 J $K^{-1} \ mol^{-1}$ , ln 10 = 2.3

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

CONCEPT:

Arrhenius Equation and Activation Energy

The Arrhenius equation relates the rate constants of a reaction at two different temperatures with the activation energy of the reaction.
The equation is given as:

ℓn(K2K1)=EaR[T2−T1T1T2]ℓn(K2K1)=EaR[T2−T1T1T2]

Here:
K1K1

and K2

are the rate constants at temperatures T1

and T2

respectively.

is the activation energy of the reaction.

is the universal gas constant (8.3 J mol−1 K−1

and T2

are the absolute temperatures in Kelvin.

EXPLANATION:

In the given question:
K1=1.5×103

, K2=4.5×103

Ea=60×103 J mol−1

, T1=300 K

Substitute these values into the Arrhenius equation:

ℓn(4.5×1031.5×103)=60×1038.3[T2−300300.T2]

Simplify the natural logarithm:
ℓn(4.5×1031.5×103)=ℓn 3
ℓn 3=1.0986
Rearrange the equation:

1.0986=60×1038.3[T2−300300.T2]

Calculate:
60×1038.3=7228.92
1.0986=7228.92[T2−300300.T2]
T2−300300.T2=1.09867228.92
T2−300300.T2=0.000152
T2−300=0.000152⋅300.T2
T2−300=0.0456⋅T2
T2−0.0456⋅T2=300
0.9544⋅T2=300
T2=3000.9544
T2=314.4 K
Convert T2

to Celsius:

t=T2−273
t=314.4−273
t=41.4∘C

Therefore, the temperature T2

is 314.4 K, which corresponds to 41.4∘C

Question 50

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

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

CONCEPT:

Equilibrium and Gibbs Free Energy

At equilibrium under standard pressure conditions, the Gibbs free energy change ( $\Delta G^\circ$ ) is zero.
The relationship between Gibbs free energy and equilibrium temperature can be expressed using thermodynamic principles.

EXPLANATION:

The given equation is:

105 – 35 log T = 0

We solve this equation to find the value of temperature ( $T$ ):
Rearranging the equation:

35 log T = 105

Dividing both sides by 35:

log T = 105 / 35

log T = 3

Converting from logarithmic to exponential form:

$T = 10^{3}$

$T = 1000 , \text{K}$

Converting temperature from Kelvin to Celsius:

$T (\text{in } ^\circ \text{C}) = T (\text{in K}) - 273$

$T (\text{in } ^\circ \text{C}) = 1000 - 273$

$T (\text{in } ^\circ \text{C}) = 727^\circ \text{C}$

Therefore, the temperature at equilibrium is $T = 1000 , \text{K}$ or $T = 727^\circ \text{C}$ .

Physics Section A (20 questions)

Question 51

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 ________ $rad/s^2$ .

AOption (a)
150
BOption (b)
120
COption (c)
130
DOption (d)
170

View answer & explanation

Correct answer

130

Why this answer is correct

$\theta = \frac{5t^4}{40} - \frac{t^3}{3}$

$\frac{d\theta}{dt} = \frac{5}{40}(4t^3) - \frac{3t^2}{3} = \frac{t^3}{2} - t^2$

$\alpha = \frac{d^2\theta}{dt^2} = \frac{3t^2}{2} - 2t$

at t = 10

$\alpha = \frac{3}{2}(10)^2 - 2 \times 10 = 130 rad/sec^2$

Question 52

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

AOption (a)
33.34
BOption (b)
66.67
COption (c)
200
DOption (d)
400

View answer & explanation

Correct answer

66.67

Why this answer is correct

Initial capacitance

$C_i = \frac{A \epsilon_0}{d} = C$

Final capacitance

$C_1 = \frac{A \epsilon_0 \times 5}{\frac{d}{2}} = 10C$

$C_2 = \frac{A \epsilon_0}{\frac{d}{2}} = 2C$

$C_f = \frac{C_1 C_2}{C_1 + C_2} = \frac{10 \times 2}{10 + 2} C = \frac{5}{3} C$

% increase = $\frac{\frac{5}{3} C - C}{C} \times 100 = 66.67$

Question 53

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

AOption (a)
2 : 3 : 5
BOption (b)
5 : 3 : 2
COption (c)
2 : 5 : 7
DOption (d)
7 : 5 : 2

View answer & explanation

Correct answer

7 : 5 : 2

Why this answer is correct

$\Delta Q_p = nC_p \Delta T = \frac{7}{2} nR \Delta T$

$\Delta U = nC_v \Delta T = \frac{5}{2} nR \Delta T$

W = $nR \Delta T$

$\Delta Q : \Delta U : W :: 7 : 5 : 2$

Question 54

Two charged conducting spheres S₁ and S₂ of radii 8 cm and 18 cm are connected to each other by a wire. After equilibrium is established, the ratio of

electric fields on S₁ and S₂ spheres are E_S1 and E_S2respectively. The value of $\frac{E_{S1}}{E_{S2}}$ is ________.

AOption (a)
$\frac{3}{2}$
BOption (b)
$\frac{2}{3}$
COption (c)
$\frac{4}{9}$
DOption (d)
$\frac{9}{4}$

View answer & explanation

Correct answer

$\frac{9}{4}$

Why this answer is correct

$V_1 = V_2$

$\frac{kQ_1}{8} = \frac{kQ_2}{18} \quad \left{ V_{\text{surface}} = \frac{kQ}{R} \right}$

$\frac{Q_1}{Q_2} = \frac{4}{9}$

$\frac{E_{S_1}}{E_{S_2}} = \frac{kQ_1}{r_1^2} \frac{r_2^2}{kQ_2} \quad \left{ E_{\text{surface}} = \frac{kQ}{R^2} \right}$

$= \frac{Q_1}{Q_2} \left( \frac{r_2}{r_1} \right)^2 = \frac{4}{9} \left( \frac{9}{4} \right)^2 = \frac{9}{4}$

Question 55

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 ________ m/s

AOption (a)
120
BOption (b)
150
COption (c)
200
DOption (d)
300

View answer & explanation

Correct answer

300

Why this answer is correct

$v = \frac{\text{coeff of t}}{\text{coeff of x}} = \frac{200}{\frac{1}{150} \times 100} = 300\text{m/s}$

Question 56

Two short electric dipoles A and B having dipole moment p₁ and p₂ 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° with the line joining points O and x. The ratio of the dipole moments p₂ / p₁ is ________.

AOption (a)
$\frac{\sqrt{3}}{2}$
BOption (b)
$2\sqrt{3}$
COption (c)
$\frac{1}{\sqrt{3}}$
DOption (d)
$\sqrt{3}$

View answer & explanation

Correct answer

$2\sqrt{3}$

Why this answer is correct

$\vec{E}_A = \frac{2kp_1}{x^3} (-\hat{i})$ {Axial point}

$\vec{E}_B = \frac{kp_2}{x^3} (-\hat{j})$ {Equatorial point}

Now : $\tan \theta = \left| \frac{\vec{E}_B}{\vec{E}_A} \right|$

$\Rightarrow \tan 60^\circ = \frac{p_2}{2p_1}$

$\Rightarrow \frac{p_2}{p_1} = 2\sqrt{3}$

Question 57

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

(A) Circuit used

Part(B) : Input voltage

Part(C) : Output voltage

AOption (a)
BOption (b)
COption (c)
DOption (d)

View answer & explanation

Correct answer

Why this answer is correct

will be updated soon.

Question 58

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 coils 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 ________.

AOption (a)
36
BOption (b)
$128\sqrt{2}$
COption (c)
16
DOption (d)
64

View answer & explanation

Correct answer

Why this answer is correct

Sol. 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 : $A = \pi R_{coil}^2 \Rightarrow R_{coil} = \sqrt{\frac{A}{\pi}}$

Thus ; $R = \frac{\rho N(2\pi) \sqrt{\frac{A}{\pi}}}{\pi r^2}$

$P = \frac{N^2 A^2 \left( \frac{dB}{dt} \right)^2}{\frac{2\rho N}{r^2} \sqrt{\frac{A}{\pi}}} \Rightarrow P \propto NA^{3/2}r^2 ......(1)$

$P' \propto (2N)(2A)^{3/2}(3r)^2 ......(2)$

$\frac{(2)}{(1)} : \frac{P'}{P} = 36\sqrt{2} \Rightarrow P' = (\sqrt{2})(36)P$

Thus : $\alpha = 36$

Question 59

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 ________.

AOption (a)
$\frac{2 + \pi}{1 + \pi}$
BOption (b)
$\frac{1 + \pi}{1 - \pi}$
COption (c)
$\frac{2 + \pi}{1 - \pi}$
DOption (d)
$\frac{1 + \pi}{2 - \pi}$

View answer & explanation

Correct answer

$\frac{2 + \pi}{1 + \pi}$

Why this answer is correct

Sol. $B_1 = \frac{\mu_0 I}{4\pi r} \times 2 + \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r} \left( \frac{2}{\pi} + 1 \right)$

$B_2 = \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r} \left( \frac{1}{\pi} + 1 \right)$

$\frac{B_1}{B_2} = \frac{2 + \pi}{1 + \pi}$

Question 60

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

AOption (a)
3 : 4
BOption (b)
3 : 2
COption (c)
2 : 1
DOption (d)
1 : 2

View answer & explanation

Correct answer

3 : 2

Why this answer is correct

Sol. We have for thin prism minimum deviation :

$\delta_m = (\mu - 1)A$

$\Rightarrow \delta_m = (1.5 - 1)A \Rightarrow \delta_m = \frac{A}{2}$

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 61

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

AOption (a)
1
BOption (b)
0.5
COption (c)
1.2
DOption (d)
0.25

View answer & explanation

Correct answer

Why this answer is correct

Sol. $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$

$\frac{1.54}{v} - \frac{1}{(-40)} = \frac{1.54 - 1}{-20}$

$\frac{1.54}{v} = -\frac{1}{40} - \frac{0.54}{20}$

$\frac{1.54}{v} = \frac{-2.08}{40}$

$v = -29.61 \text{ cm}$

$\frac{h_i}{h_o} = \frac{\mu_1 v}{\mu_2 u} = \frac{1(-29.61)}{1.54(-40)} = 0.48$

$h_i = 0.48 \times 2 = 0.96 \approx 1 \text{cm}$

Question 62

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 ________.

AOption (a)
1
BOption (b)
4
COption (c)
2
DOption (d)
3

View answer & explanation

Correct answer

Why this answer is correct

Sol. $\frac{hc}{\lambda} - \phi = 3eV_0 \dots (i)$

$\frac{hc}{2\lambda} - \phi = eV_0 \dots (ii)$

$\frac{hc}{\lambda} - \phi = 3 \left[ \frac{hc}{2\lambda} - \phi \right]$

$\phi = \frac{hc}{4\lambda}$

$\frac{hc}{\lambda_0} = \frac{hc}{4\lambda}$

$\lambda_0 = 4\lambda$

$\alpha = 4$

Question 63

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 $1.5 \times 10^6 \text{ m/s}$ then ratio of maximum electric and magnetic forces acting on the electron is ________.

AOption (a)
200
BOption (b)
150
COption (c)
400
DOption (d)
300

View answer & explanation

Correct answer

200

Why this answer is correct

Sol. $E_y = 300 \sin \omega \left( t - \frac{x}{v} \right)$

$F_M = qvB \dots (i)$

$F_e = qE \dots (ii)$

$\frac{F_e}{F_m} = \frac{qE}{qvB} = \frac{E}{vB} = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^6} = 200$

Question 64

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

AOption (a)
- 1.51
BOption (b)
- 0.85
COption (c)
- 0.38
DOption (d)
- 0.28

View answer & explanation

Correct answer

0.38

Why this answer is correct

Sol. 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 65

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

AOption (a)
10
BOption (b)
7
COption (c)
8
DOption (d)
11

View answer & explanation

Correct answer

Why this answer is correct

Sol. $V_1 = V_2$

$\frac{4}{3} \pi R^3 = 512 \left( \frac{4}{3} \pi r^3 \right)$

$R = (512)^{1/3} r$

1 = 8r

r = $\frac{1}{8}$ mm

$\Delta SE = SE_2 - SE_1$

= $512 \left[ T \times 4\pi r^2 \right] - \left[ T \times 4\pi R^2 \right]$

= $4\pi T [512r^2 - R^2]$

= $4\pi (0.08) \left[ 512 \left( \frac{1}{8} \right)^2 - 1^2 \right] \times 10^{-6} \approx 7 \times 10^{-6}$

$\alpha = 7$

Question 66

The dimensional formula of $\frac{1}{2} \epsilon_0 E^2$ ( $\epsilon_0$ = permittivity of vacuum and E = electric field) is $M^a L^b T^c$ . The value of 2a - b + c = ________.

AOption (a)
0
BOption (b)
1
COption (c)
-1
DOption (d)
2

View answer & explanation

Correct answer

Why this answer is correct

$\frac{1}{2} \epsilon_0 E^2$ = Energy density

$\therefore \left[ \frac{1}{2} \epsilon_0 E^2 \right] = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$

$\therefore$ a = 1, b = -1, c = -2

$\therefore$ 2a - b + c = 2 + 1 - 2 = 1

Question 67

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 $\alpha \times 10^9 N/m^2$ . The value of $\alpha$ is ________.

(Ignore the contribution of the load to Young's modulus error calculation)

AOption (a)
1.3
BOption (b)
1.65
COption (c)
0.13
DOption (d)
0.25

View answer & explanation

Correct answer

1.65

Why this answer is correct

Y = $\frac{F\ell}{A\Delta\ell}$

$\therefore \frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta \ell}{\ell} + 2\frac{\Delta d}{d} + \frac{\Delta(\Delta\ell)}{\Delta\ell}$

$\left[ A = \pi \frac{d^2}{4} \Rightarrow \frac{\Delta A}{A} = \frac{2\Delta d}{d} \right]$

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 = $\frac{100 \times 1.5}{\frac{\pi}{4} \left( \frac{0.08}{100} \right)^2 \times 0.5 \times 10^{-2}} = 5.97 \times 10^{10} N/m^2$

$\therefore \Delta Y = 0.0277 \times 5.97 \times 10^{10} N/m^2 = 1.65 \times 10^9 N/m^2$

Question 68

The velocity of a particle is given as

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

AOption (a)
$\sqrt{6}$
BOption (b)
9
COption (c)
$\sqrt{33}$
DOption (d)
0

View answer & explanation

Correct answer

Why this answer is correct

$\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$

$\vec{a} = \frac{d\vec{v}}{dt} = -\frac{dx}{dt}\hat{i} + 2\frac{dy}{dt}\hat{j} - \frac{dz}{dt}\hat{k}$

= $-(-x)\hat{i} + 2(2y)\hat{j} - (-z)\hat{k} = x\hat{i} + 4y\hat{j} + z\hat{k}$

$\therefore \vec{a}$ at (1,2,4) = $\hat{i} + 8\hat{j} + 4\hat{k}$

$\therefore |\vec{a}| = 9 m/s^2$

Question 69

The position of an object having mass 0.1 kg as a

function of time t is given as $\vec{r} = (10t^2\hat{i} + 5t^3\hat{j}) m$ .

At t = 1 s, which of the following statements are

correct?

A. The linear momentum $\vec{p} = (2\hat{i} + 1.5\hat{j}) kg.m/s$ .

B. The force acting on the object $\vec{F} = (2\hat{i} + 3\hat{j}) N$ .

C. The angular momentum of the object about its

origin $\vec{L} = 15\hat{k} Js$ .

D. The torque acting on the object about its origin

$\vec{\tau} = 20\hat{k} Nm$ .

Choose the correct answer from the option given

below.

AOption (a)
A, B and C only
BOption (b)
B, C and D only
COption (c)
A, C and D only
DOption (d)
A, B and D only

View answer & explanation

Correct answer

A, B and D only

Why this answer is correct

Given $\vec{r} = 10t^2\hat{i} + 5t^3\hat{j} \Rightarrow \vec{r}_{t=1} = 10\hat{i} + 5\hat{j}$

$\vec{v} = \frac{d\vec{r}}{dt} = 20t\hat{i} + 15t^2\hat{j} \Rightarrow \vec{v}_{t=1sec} = 20\hat{i} + 15\hat{j}$

$\vec{a} = \frac{d^2\vec{r}}{dt^2} = 20\hat{i} + 30t\hat{j} \Rightarrow \vec{a}_{t=1sec} = 20\hat{i} + 30\hat{j}$

$\therefore \vec{P} = m\vec{v} = 2\hat{i} + 1.5\hat{j}$

$\vec{F} = m\vec{a} = 2\hat{i} + 3\hat{j}$

$\vec{\tau} = \vec{r} \times \vec{F} = 20\hat{k}$

$\vec{L} = \vec{r} \times \vec{P} = 5\hat{k}$

Question 70

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 ________.

AOption (a)
$\frac{1}{2}$
BOption (b)
2
COption (c)
4
DOption (d)
$\frac{1}{4}$

View answer & explanation

Correct answer

Why this answer is correct

From Kepler's law,

$T^2 = \frac{4\pi^2}{GM_s}r^3$

$\therefore T_{P_1} = \frac{2\pi}{\sqrt{G(2M)}} R^{3/2}$

$T_{P_2} = \frac{2\pi}{\sqrt{G(4M)}} (2R)^{3/2}$

$\therefore \frac{T_{P_2}}{T_{P_1}} = \frac{(2)^{3/2}}{\sqrt{4}} = \frac{\sqrt{2}}{1} \times \frac{2\sqrt{2} \times \sqrt{2}}{2} = 2$

Physics Section B (5 questions)

Question 71

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

degrees. The value of $\alpha$ is______.

View answer & explanation

Correct answer

Answer: 36

Why this answer is correct

Sol. 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 72

A vessel contains 0.15 m³ of a gas at pressure 8 bar and temperature 140°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 ____ kJ. (R is gas constant)

View answer & explanation

Correct answer

Answer: 120

Why this answer is correct

Sol. $\gamma = \frac{C_p}{C_v} = \frac{3}{2}$

$\because P_1 V_1^\gamma = P_2 V_2^\gamma$

$\Rightarrow V_2 = \left( \frac{P_1}{P_2} \right)^{1/\gamma} V_1$

$\Rightarrow V_2 = \left( \frac{8}{1} \right)^{2/3} \times 0.15$

$\Rightarrow V_2 = 0.6 \text{ m}^3$

$\because W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$

= $\left( \frac{8 \times 0.15 - 1 \times 0.6}{1.5 - 1} \right) \times 10^5 = 1.2 \times 10^5$

$\Rightarrow W = 120 \text{ kJ}$

Question 73

1 $\mu$ C charge moving with velocity

$\vec{v} = (\hat{i} - 2\hat{j} + 3\hat{k}) \text{ m/s}$ in the region of magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \text{ T}$ . The magnitude of force acting on it is $\sqrt{\alpha} \times 10^{-6} \text{ N}$ . The value of $\alpha$ is ____.

View answer & explanation

Correct answer

Answer: 171

Why this answer is correct

Sol. $\vec{F} = q(\vec{v} \times \vec{B})$

= $1 \times 10^{-6} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & 3 & -5 \end{vmatrix}$

= $10^{-6} [\hat{i} + 11\hat{j} + 7\hat{k}]$

$\Rightarrow \vec{F} = \sqrt{171} \times 10^{-6} \text{ N}$

$\therefore \alpha = 171$

Question 74

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 $\left( \frac{W}{A} + \frac{2}{\gamma} \frac{w}{A} \right)$ , where, A is the cross sectional area of the wire. The value of $\gamma$ is______.

View answer & explanation

Correct answer

Answer: 6

Why this answer is correct

Sol.

$\sigma = \frac{T}{A} = \frac{W + \frac{w}{3}}{A}$

$\Rightarrow \sigma = \frac{W}{A} + \frac{2}{6} \frac{w}{A}$

$\therefore \gamma = 6$

Question 75

A tube is filled with water and a wooden cube 10 cm $\times$ 10cm $\times$ 10cm 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 cm. The mass of the metal coin

is ____ gram. (Take water density as 1g/cm³ and density of wood as 0.4 g/cm³)

View answer & explanation

Correct answer

Answer: 387

Why this answer is correct

Sol.

Weight of coin = Additional upthrust

$\Rightarrow m_c g = \rho_l V_{add} g$

$\Rightarrow m_c = 1 \times 3.87 \times (10)^2$

$\Rightarrow m_c = 387 \text{ gram}$

75 questions in the full paper

SSC Exams

Banking

Railways

Defence & Police

Teaching

State & Civil Services

Entrance Exams

Other Govt Exams

Quantitative Aptitude

Logical Reasoning

English Language

General Awareness

SSC CGL

Short sets.Sharper decisions.

Reasoning

Quantitative Aptitude

English Comprehension

General Awareness

Daily Current Affairs

Revisit earlier editions

Put your knowledge into practice

JEE Mains 2 Apr 2026 Shift 1 (Previous Year Paper)

Questions & answers

Mathematics Section A (20 questions)

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Mathematics Section B (5 questions)

Question 21

Question 22

Question 23

Question 24

Question 25

Chemistry Section A (20 questions)

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Question 35

Question 36

Question 37

Question 38

Question 39

Question 40

Question 41

Question 42

Question 43

Question 44

Question 45

Chemistry Section B (5 questions)

Question 46

Question 47

Question 48

Question 49

Question 50

Physics Section A (20 questions)

Question 51

Question 52

Short sets.
Sharper decisions.