JEE Main 6 April 2026 Shift 1 question paper with solutions

JEE Main 6 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 · Inverse Trigonometric Functions · Single correct

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x) = \sin^{-1} \left( \frac{x + [x]}{3} \right)$ is $[\alpha, \beta)$, then $\alpha^2 + \beta^2$ is equal to:

  1. 2
  2. 5
  3. 10
  4. 13

Answer: (b)

Solution

Given $-1 \leq \frac{x + \lfloor x \rfloor}{3} \leq 1$. This implies $-3 \leq x - \lfloor x \rfloor \leq 3$. Therefore, $-3 - \lfloor x \rfloor \leq x \leq 3 - \lfloor x \rfloor$. If $\lfloor x \rfloor = -1$, then $x \in [-1, 0)$. If $\lfloor x \rfloor = 0$, then $x \in [0, 1)$. If $\lfloor x \rfloor = 1$, then $x \in [1, 2)$. Hence, $x \in [-1, 2)$. Thus, $\alpha = -1$ and $\beta = 2$. Therefore, $\alpha^2 + \beta^2 = 5$.

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

Let one root of the quadratic equation in $x$: $\left(k^2 - 15k + 27\right)x^2 + 9(k - 1)x + 18 = 0$ be twice the other. Then the length of the latus rectum of the parabola $y^2 = 6kx$ is equal to:

  1. 4
  2. 6
  3. 8
  4. 12

Answer: (d)

Solution

Given $\alpha + 2\alpha = \frac{-9(k-1)}{k^2 - 15k + 27} = 3\alpha \ldots$ (1) $\alpha \times 2\alpha = \frac{18}{k^2 - 15k + 27} = 2\alpha^2 \ldots$ (2) Solving (1) and (2), $k = 2$ Therefore, the parabola will be $y^2 = 6kx = 12x$. The length of the latus rectum is $12$.

Question 3

Maths · Conic Sections · Single correct

Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2 - ax + 2 = 0$. Let the sets $\{a \in \mathbb{R} : e_1$ and $e_2$ are the eccentricities of hyperbolas$\} = (\alpha, \beta)$, and $\{a \in \mathbb{R} : e_1$ and $e_2$ are the eccentricities of an ellipse and a hyperbola, respectively$\} = (\gamma, \infty)$. The $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

  1. 18
  2. 22
  3. 26
  4. 34

Answer: (c)

Solution

Given $e_1 + e_2 = a$ and $e_1 e_2 = 2$, we have $e_2 = \frac{2}{e_1}$. Case 1: Both are eccentricities of hyperbola. Now, $e_1 > 1$ and $e_2 > 1$. $$\frac{2}{e_1} > 1 \implies e_1 2$$ Now $a = e_1 + \frac{2}{e_1}$. As $e_1 \to 1 \implies a \to 3$, $e_1 \to 0 \implies a \to \infty$. Therefore, $(3, \infty) \implies r = 3$. Now $\alpha^2 + \beta^2 + \gamma^2 = 8 + 9 + 9 = 26$.

Question 4

Maths · Complex Numbers and Quadratic Equations · Single correct

Let the set of all values of $k \in \mathbb{R}$ such that the equation $z(\overline{z} + 2 + i) + k(2 + 3i) = 0$, $z \in \mathbb{C}$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:

  1. -10
  2. -8
  3. 10$\sqrt{13}$
  4. 8$\sqrt{13}$

Answer: (a)

Solution

Put $z = x + iy$, $\overline{z} = x - iy$. $$\Rightarrow \overline{z} + 2 + i = x - iy + 2 + i$$ $$\Rightarrow z(\overline{z} + 2 + i) = (x + iy)(x + 2 + i(1-y))$$ Real part $= x(x + 2) - y(1 - y)$ Imaginary part $= x(1 - y) + y(x + 2)$ $$\Rightarrow x^2 + y^2 + 2x - y + 2k = 0 ..(1)$$ and $x + 2y + 3k = 0 ...(2)$ Eliminating $x$ from (1) and (2) $$5y^2 + (12k - 5)y + 9k^2 - 4k = 0$$ Since $y \in \mathbb{R}$ Use $D \geq 0$ $$(12k - 5)^2 - 4.5(9k^2 - 4k) \geq 0$$ $$\Rightarrow 36k^2 + 40k - 25 \leq 0$$ $$\Rightarrow \alpha + \beta = \frac{-10}{9}$$ $$\Rightarrow 9(\alpha + \beta) = -10$$

Question 5

Maths · Sequences and Series · Single correct

The value of $1^3 - 2^3 + 3^3 - \ldots + 15^3$ is:

  1. 1706
  2. 1856
  3. 1982
  4. 2403

Answer: (b)

Solution

Given $(1^3 - 2^3) + (3^3 - 4^3) + \ldots 13^3 - 14^3 + 15^3$. Use $n^3 - (n+1)^3 = -3n^2 - 3n - 1$. Put $n = 1, 3, 5, \ldots, 13$ (7 terms). Sum of pairs $= -1519$. Add last term $= 15^3$. $$\Rightarrow 3375 - 1519 = 1856$$

Question 6

Maths · Sequences and Series · Single correct

The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8. If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:

  1. $\frac{34}{9}$
  2. $\frac{34}{13}$
  3. $\frac{32}{9}$
  4. $\frac{32}{13}$

Answer: (a)

Solution

Sum of 10 terms of A.P, $S_{10} = 160$ $$5[2a + 9d] = 160$$ $$2a + 9d = 32$$ G.P. $\Rightarrow d, da, da^2 \ldots$ $\Rightarrow d + da = 8$ $$\Rightarrow \frac{d + d(32 - 9d)}{2} = 8$$ $$\Rightarrow 2d + 32d - 9d^2 = 8$$ $$\Rightarrow 9d^2 - 34d + 8 = 0$$ $$d_1 + d_2 = \frac{34}{9}$$

Question 7

Maths · Permutations and Combinations · Single correct

The number of 4-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word INCONSEQUENTIAL, without repeating any letter, is:

  1. 2670
  2. 2840
  3. 2920
  4. 3600

Answer: (d)

Solution

INCONSEQUENTIAL contains (I, E, O, A, U) and (N, C, S, Q, T, L). Total number of 4 letter words using two vowels and two consonants is $$\binom{5}{2} \times \binom{6}{2} \times 4!$$ $$= 3600$$

Question 8

Maths · Binomial Theorem · Single correct

If the coefficients of the middle terms in the binomial expansions of $(1 + \alpha x)^{26}$ and $(1 - \alpha x)^{28}$, $\alpha \neq 0$, are equal, then the value of $\alpha$ is:

  1. 1
  2. $\frac{14}{13}$
  3. $\frac{27}{7}$
  4. $\frac{7}{27}$

Answer: (d)

Solution

Given $(1 + \alpha x)^{26}$. Middle term, $T_{14} = \binom{26}{13} \alpha^{13}$. For $(1 - \alpha x)^{28}$, the middle term is $\binom{28}{14} \alpha^{14}$. Therefore, $$\binom{26}{13} \alpha^{13} = \binom{28}{14} \alpha^{14}$$ which implies $$\binom{26}{13} = \binom{28}{14} \alpha$$ leading to $$\binom{26}{13} = \frac{28}{14} \times \frac{27}{14} \cdot \binom{26}{13} \alpha$$ thus $$\alpha = \frac{7}{27}$$

Question 9

Maths · Statistics · Single correct

A data consists of 20 observations $x_1, x_2, \ldots, x_{20}$. If $$\sum_{i=1}^{20} (x_i + 5)^2 = 2500$$ and $$\sum_{i=1}^{20} (x_i - 5)^2 = 100$$, then the ratio of mean to standard deviation of this data is:

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

Answer: (b)

Solution

Given $$\sum_{i=1}^{20} (x_i + 5)^2 = 2500 (1)$$ $$\sum_{i=1}^{20} (x_i - 5)^2 = 100 (2)$$ On (1) - (2) we have $$20 \sum x_i = 2400 \implies \sum x_i = 120$$ The mean, $$\bar{x} = 6$$ Adding (1) and (2) gives $$2 \sum x_i^2 + 40(5^2) = 2600$$ Thus, $$\sum x_i^2 = 800$$ The variance is $$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{800}{20} - 36 = 4$$ The ratio $$\bar{x} : \sigma = 6 : 2 = 3 : 1$$ Answer: 3 : 1

Question 10

Maths · Probability · Single correct

A bag contains $(N + 1)$ coins — $N$ fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$, then $N$ is equal to:

  1. 5
  2. 7
  3. 8
  4. 9

Answer: (b)

Solution

Total $(N + 1)$ coins. $N$ implies fair coins. $1$ implies Head on both sides. $$P(1^+) = \left( \frac{N}{N+1} \right) \left( \frac{1}{2} \right) + \left( \frac{1}{N+1} \right) 1$$ $$= \frac{N+2}{2(N+1)} = \frac{9}{16}$$ $$8N + 16 = 9N + 9$$ $$\Rightarrow N = 7$$

Question 11

Maths · Conic Sections · Single correct

If the eccentricity e of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, passing through $\left(6, 4\sqrt{3}\right)$, satisfies $15(e^2 + 1) = 34e$, then the length of the latus rectum of the hyperbola $\frac{x^2}{b^2} - \frac{y^2}{2(a^2 + 1)} = 1$ is:

  1. 10
  2. 20
  3. 25
  4. 30

Answer: (a)

Solution

Given $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. It passes through $(6, 4\sqrt{3})$. $$\Rightarrow \frac{36}{a^2} - \frac{48}{b^2} = 1 ...(1)$$ Also $15e^2 - 34e + 15 = 0$. $$\Rightarrow 15e^2 - 25e - 9e + 15 = 0$$ $$e = \frac{5}{3} or \frac{3}{5} \Rightarrow e = \frac{5}{3}$$ $$1 + \frac{b^2}{a^2} = \frac{25}{9} \Rightarrow \frac{b^2}{a^2} = \frac{16}{9} ...(2)$$ Using (1) and (2) $$\Rightarrow \frac{36}{a^2} - \frac{48}{16a^2} \times 9 = 1 \Rightarrow a = 3, b = 4$$ Length of L.R. $$\frac{4(a^2 + 1)}{b} = \frac{4(10)}{4} = 10$$

Question 12

Maths · Conic Sections · Single correct

Let chord PQ of length $3\sqrt{13}$ of the parabola $y^2 = 12x$ be such that the ordinates of points P and Q are in the ratio 1 : 2. If the chord PQ subtends an angle $\alpha$ at the focus of the parabola, then $\sin \alpha$ is equal to:

  1. $\frac{3}{5}$
  2. $\frac{4}{5}$
  3. $\frac{5}{13}$
  4. $\frac{12}{13}$

Answer: (a)

Solution

Points are given as P(3t^2, 6t) and Q(12t^2, 12t). The distance PQ is calculated as $PQ = \sqrt{81t^4 + 36t^2}$. Solving the equation $81t^4 + 36t^2 = 117$ gives $9t^4 + 4t^2 - 13 = 0$. Factoring gives $(t^2 - 1)(9t^2 + 13) = 0$, leading to $t = 1$. Thus, the points are P(3, 6) and Q(12, 12). The slope of QS is $\frac{4}{3}$. The tangent of $\alpha$ is $\tan \alpha = \frac{3}{4}$, and the sine of $\alpha$ is $\sin \alpha = \frac{3}{5}$.

Question 13

Maths · Inverse Trigonometric Functions · Single correct

Let $0 < \alpha < 1$, $\beta = \frac{1}{3\alpha}$ and $\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \frac{\pi}{4}$. Then $6(\alpha + \beta)$ is equal to:

  1. 6
  2. 7
  3. 8
  4. 9

Answer: (b)

Solution

Given $\beta = \frac{1}{3\alpha}$, $\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \frac{\pi}{4}$. Therefore, $$\tan\left(\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta)\right) = 1$$ This implies $$\frac{1-\alpha + 1-\beta}{1-(1-\alpha)(1-\beta)} = 1$$ Simplifying, $$2 - \alpha - \beta = 1 - (1 - \alpha - \beta + \alpha \beta)$$ Thus, $$2 - \alpha - \beta = \alpha + \beta - \alpha \beta$$ Rearranging gives $$2(\alpha + \beta) = 2 + \alpha \beta$$ Therefore, $$2(\alpha + \beta) = 2 + \frac{1}{3} = \frac{7}{3}$$ So, $$\alpha + \beta = \frac{7}{6}$$ Finally, $$6(\alpha + \beta) = 7$$

Question 14

Maths · Trigonometric Functions · Single correct

Let $S = \left\{ \theta \in (-2\pi, 2\pi) : \cos \theta + 1 = \sqrt{3} \sin \theta \right\}$. Then $\sum_{\theta \in S} \theta$ is equal to:

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

Answer: (b)

Solution

Given $\cos \theta + 1 = \sqrt{3} \sin \theta$. $$\frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} + 1 = \sqrt{3} \left( \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} \right)$$ $$2 = 2 \sqrt{3} \tan \frac{\theta}{2}$$ This implies $\tan \frac{\theta}{2} = \frac{1}{\sqrt{3}}$. Given $\theta \in (-2\pi, 2\pi)$, $\frac{\theta}{2} \in (-\pi, \pi)$. $$\frac{\theta}{2} = -\frac{5\pi}{6}, \frac{\pi}{6}$$ $$\theta = -\frac{5\pi}{3}, \frac{\pi}{3}$$ Sum $= -\frac{5\pi}{3} + \frac{\pi}{3} = -\frac{4\pi}{3}$$

Question 15

Maths · Three Dimensional Geometry · Single correct

Let the image of the point P(1, 6, a) in the line L: $\frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$, $b > 0$, be $\left( \frac{a}{3}, 0, a+c \right)$. If $S(\alpha, \beta, \gamma)$, $\alpha > 0$, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $2\sqrt{14}$, then $\alpha + \beta + \gamma$ is equal to:

  1. 19
  2. 20
  3. 21
  4. 22

Answer: (c)

Solution

Let Q be the point perpendicular from P(1,6,a) to the line $$L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$$ The image of point P in line L is $P'\left(\frac{a}{3}, 0, a+c\right)$. Therefore, Q is the midpoint of PP': $$Q \equiv \left(\frac{1+\frac{a}{3}}{2}, \frac{6+0}{2}, \frac{a+a+c}{2}\right) = \left(\frac{3+a}{6}, 3, \frac{2a+c}{2}\right)$$ Comparing x and y, $\frac{3+a}{6} = 1 \implies a = 3$. Therefore, $Q \equiv \left(1, 3, \frac{6+c}{2}\right)$. Direction vector of L is $\vec{v} = (1, 2, b)$. Vector PQ is perpendicular to $\vec{v}$. Therefore, $\overrightarrow{PQ} = \left(1-1, 3-6, \frac{6+c}{2} - 3\right)$ $$= \left(0, -3, \frac{c}{2}\right)$$ $\overrightarrow{PQ} \cdot \vec{v} = 0$ $$(0)(1) + (-3)(2) + \left(\frac{c}{2}\right)(b) = 0$$ $$bc = 12$$ Now comparing z with $Q\left(1, 3, \frac{6+c}{2}\right)$ $$\frac{3-1}{2} = \frac{\frac{6+c}{2} - 3 + 1}{b}$$ $$2b - c = 2$$ Solving $bc = 12$ and $c = 2b - 2$ $$b = 3, -2$$ $$b = 3 (b > 0)$$ $$c = 4$$ Therefore, $Q(1, 3, 5)$ Now, any point S on the line L can be written as $(k, 2k+1, 3k+2)$ $$SQ = \sqrt{(k-1)^2 + (2k-2)^2 + (3k-3)^2} = 2\sqrt{14}$$ $$k = 3, -1$$ For $k = 3$, $S \equiv (3, 7, 11)$ Therefore, $\alpha + \beta + \gamma = 21$

Question 16

Maths · Three Dimensional Geometry · Single correct

Let a line L be perpendicular to both the lines $$L_1 : \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$$ and $$L_2 : \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}.$$ If $\theta$ is the acute angle between the lines L and $$L_3 : \frac{x-\frac{8}{7}}{2} = \frac{y-\frac{4}{7}}{1} = \frac{z}{2},$$ then $\tan \theta$ is equal to:

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

Answer: (b)

Solution

Given $\vec{r} = \vec{p} \times \vec{q}$ where $\vec{p} = (3, 5, 7)$ and $\vec{q} = (1, 4, 7)$. $$\vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix} = 7\hat{i} - 14\hat{j} + 7\hat{k}$$ This is equivalent to $7(\hat{i} - 2\hat{j} + \hat{k})$. Now, consider $\vec{r} = (1, -2, 1)$ and $\vec{s} = (2, 1, 2)$. Therefore, $$\cos \theta = \frac{2 - 2 + 2}{\sqrt{1 + 4 + 1} \sqrt{4 + 1 + 4}}$$ $$\Rightarrow \cos \theta = \frac{2}{3\sqrt{6}} \Rightarrow \tan \theta = \frac{5\sqrt{2}}{2}$$

Question 17

Maths · Limits and Derivatives · Single correct

The value of $\lim_{x \to 0} \left( \frac{x^2 \sin^2 x}{x^2 - \sin^2 x} \right)$ is:

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

Answer: (b)

Solution

Given $\lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x}$. Applying expansion, $$\lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \left( x - \frac{x^3}{3!} + \ldots \right)^2}$$ $$\lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - x^2 + \frac{2x^4}{3!} \ldots} = \frac{3!}{2} = 3$$

Question 18

Maths · Integrals · Single correct

The value of the integral $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{32 \cos^4 x}{1 + e^{\sin x}} \right) \, dx$$ is :

  1. $4\pi + 2$
  2. $3\pi + 8$
  3. $3\pi + 4$
  4. $4\pi + 3$

Answer: (b)

Solution

Apply (P-5) $$I = \int_{-\pi/4}^{\pi/4} \frac{32 \cos^4 \theta}{1 + e^{-\sin \theta}} \, d\theta$$ Add $2I = \int_{-\pi/4}^{\pi/4} 32 \cos^4 \theta \, d\theta = 2 \int_{0}^{\pi/4} 32 \cos^4 \theta \, d\theta$ $$I = 32 \int_{0}^{\pi/4} \cos^4 \theta \, d\theta$$ $$= 8 \int_{0}^{\pi/4} (2 \cos^2 \theta)^2 \, d\theta = 8 \int_{0}^{\pi/4} (1 + \cos 2\theta)^2 \, d\theta$$ $$= 8 \int_{0}^{\pi/4} (1 + 2 \cos 2\theta + \cos^2 2\theta) \, d\theta$$ $$= 8 \int_{0}^{\pi/4} \left(1 + 2 \cos 2\theta + \frac{1 + \cos 4\theta}{2}\right) \, d\theta$$ $$= 8 \left[ \frac{3\theta}{2} + \frac{2 \sin 2\theta}{2} + \frac{\sin 4\theta}{8} \right]_{0}^{\pi/4}$$ $$= 8 \left[ \frac{3\pi}{8} + 1 + 0 \right] = 3\pi + 8$$

Question 19

Maths · Applications of Integrals · Single correct

The area of the region $\{(x, y) : 0 \leq y \leq 6 - x, y^2 \geq 4x - 3, x \geq 0 \}$ is :

  1. 8
  2. 9
  3. 12
  4. 15

Answer: (b)

Solution

The area $A$ is given by the integral: $$A = \int_{0}^{3} \left(6 - y - \frac{y^2 + 3}{4}\right) \, dy = 9.$$

Question 20

Maths · Applications of Integrals · Single correct

Let $e$ be the base of natural logarithm and let $f : \{ 1, 2, 3, 4 \} \rightarrow \{ 1, e, e^2, e^3 \}$ and $g : \{ 1, e, e^2, e^3 \} \rightarrow \left\{ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \right\}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $\phi(x) = \left[ f^{-1} \left\{ g^{-1} \left( \frac{1}{2} \right) \right\} \right]^{x}$, then the area of the region $R = \{ (x, y) : x^2 \leq y \leq \phi(x), 0 \leq x \leq 1 \}$ is :

  1. $\frac{3 - \log_{e}(2)}{3 \log_{e}(2)}$
  2. $\frac{1}{3 \log_{e}(2)}$
  3. $3 + \log_{e}(2)$
  4. $\frac{3 + \log_{e}(2)}{2 + \log_{e}(3)}$

Answer: (a)

Solution

Given $f(x) = e^{x^{-1}}$. $g(x) = \frac{1}{\log_e x + 1}$ $f^{-1}(g^{-1}(1/2)) = f^{-1}(e) = 2$ $\phi(x) = 2^x$ Therefore, $$\int_0^1 (2^x - x^2) \, dx = \left( \frac{2^x}{\ln 2} - \frac{x^3}{3} \right)_0^1 = \frac{1}{\ln 2} - \frac{1}{3}$$ $$= \frac{3 - \ln 2}{3 \ln 2}$$

Question 21

Maths · Determinants · Numerical

Let $A = \begin{pmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ satisfy $$A^2 + \alpha (adj (adj (A))) + \beta (adj (A) (adj (adj (A)))) = \begin{pmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{pmatrix}$$ for Some $\alpha, \beta \in \mathbb{R}$. Then $(\alpha - \beta)^2$ is equal to $\ldots$

Answer: 4

Solution

Given $|A| = -1$. $adj (adj (A)) = |A|^{n-2} A = |A|^{3-2} \cdot A = |A| \cdot A$ $adj (A) \left( adj (adj (A)) \right) = adj (A) \left( |A| \cdot A \right)$ $= |A| \left( adj (A) \cdot A \right)$ Since $adj (A) A = |A| I$, then $adj (A) \left( adj (adj (A)) \right) = |A|^2 I$ or $adj (A) \left( adj (adj (A)) \right) = I$ Solving the matrix equation $$A^2 + \alpha (-A) + \beta (I) = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$$ $$A^2 - \alpha A + \beta I = M$$ $$\begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$$ $\alpha = 1$ and $\beta = -1$ Therefore, $(\alpha - \beta)^2 = 4$

Question 22

Maths · Conic Sections · Numerical

Let the centre of the circle $x^2 + y^2 + 2gx + 2fy + 25 = 0$ be in the first quadrant and lie on the line $2x - y = 4$. Let the area of an equilateral triangle inscribed in the circle be $27\sqrt{3}$. Then the square of the length of the chord of the circle on the line $x = 1$ is $\ldots$.

Answer: 80

Solution

Given $x^2 + y^2 + 2gx + 2fy + 25 = 0$. The point $C \equiv (-g, -f)$ lies on the line. Therefore, $-2g + f = 4$ (i). Also, $g^2 + f^2 - 25 = r^2$ (ii). The area of the equilateral triangle inside the circle is $$\frac{3\sqrt{3}}{4} r^2.$$ We have $$27\sqrt{3} = \frac{3\sqrt{3}}{4} \times r^2.$$ Solving for $r^2$, we get $$r^2 = 36.$$ Solving equations (i) and (ii), we find $g = -5$, $f = -6$. The diagram shows a circle with $r = 6$, centered at $(5, 6)$, and $d = 4$. The line $x = 1$ is tangent to the circle at point $A$. The length $\ell_{AB}$ is calculated as $$\ell_{AB} = \sqrt{36 - 16} = 2\sqrt{20}.$$ Therefore, $$(\ell_{AB})^2 = 80.$$

Question 23

Maths · Vector Algebra · Numerical

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$ and $\vec{c}$ be three vectors such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$, then $\vec{c} \cdot (\vec{a} - 2\vec{b})$ is equal to $\ldots$.

Answer: 3

Solution

Using triple product identity with $\vec{a} \times (\vec{a} \times \vec{c}) = \vec{a} \times \vec{b}$ $$(\vec{a} \cdot \vec{c}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{c} = \vec{a} \times \vec{b}$$ As $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} = -2\hat{i} + \hat{j} + \hat{k}$ Substituting given values $$3\vec{a} - 3\vec{c} = -2\hat{i} + \hat{j} + \hat{k}$$ $$3(\hat{i} + \hat{j} + \hat{k}) + 2\hat{i} + \hat{j} + \hat{k} = 3\vec{c}$$ $$\vec{c} = \frac{1}{3}(5\hat{i} + 2\hat{j} + 2\hat{k})$$ Now $\vec{c} \cdot (\vec{a} - 2\vec{b}) = \frac{1}{3}(5\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (\hat{i} - \hat{j} + 3\hat{k})$ $$= \frac{1}{3}[5 - 2 + 6] = 3$$

Question 24

Maths · Sequences and Series · Numerical

For the functions $f(\theta) = \alpha \tan^2 \theta + \beta \cot^2 \theta$, and $g(\theta) = \alpha \sin^2 \theta + \beta \cos^2 \theta$, $\alpha > \beta > 0$, let $\min_{0 < \theta < \frac{\pi}{2}} f(\theta) = \max_{0 < \theta < \pi} g(\theta)$. If the first term of a G.P. is $\left( \frac{\alpha}{2\beta} \right)$, its common ratio is $\left( \frac{2\beta}{\alpha} \right)$ and the sum of its first 10 terms is $\frac{m}{n}$, $\mathrm{gcd}(m, n) = 1$, then $m + n$ is equal to $\ldots$.

Answer: 1279

Solution

Given $f(\theta)_{\min} = 2\sqrt{\alpha \beta}$ (AM $\geq$ GM). $g(\theta) = \alpha \sin^2 \theta + \beta (1 - \sin^2 \theta) = (\alpha - \beta) \sin^2 \theta + \beta$. Therefore, $g(\theta)_{\max} = \alpha$. Equation $2\sqrt{\alpha \beta} = \alpha$. $4\alpha \beta = \alpha^2$. $\alpha = 4\beta$ ($\alpha \neq 0$). Now first term $\left( \frac{\alpha}{2\beta} \right) = 2$ and common ratio $$\left( \frac{2\beta}{\alpha} \right) = \frac{1}{2}$$ $$S_{10} = \frac{a \left( 1 - r^{10} \right)}{1 - r} = \frac{1023}{256} = \frac{m}{n}$$ $m + n = 1279$

Question 25

Maths · Differential Equations · Numerical

Let $y = y(x)$ be the solution of the differential equation $$ \left( x^2 - x \sqrt{x^2 - 1} \right) \mathrm{d}y + \left( y \left( x - \sqrt{x^2 - 1} \right) - x \right) \mathrm{d}x = 0, $$ $x \geq 1$. If $y(1) = 1$, then the greatest integer less than $y(\sqrt{5})$ is $\ldots$.

Answer: 3

Solution

Given $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x - \sqrt{x^2 - 1}}$. $\frac{dy}{dx} + \frac{y}{x} = x + \sqrt{x^2 - 1}$ Integrating factor is $e^{\ln x} = x$. $y \cdot x = \int \left( x + \sqrt{x^2 - 1} \right) x \, dx$ $$= \frac{x^2}{2} + \frac{(x^2 - 1)^{3/2}}{3} + C$$ Given $y(1) = 1$ $$1 = \frac{1}{2} + C \Rightarrow C = \frac{1}{2}$$ $$\left[ y(\sqrt{5}) \right] = \left[ \frac{\sqrt{5}}{2} + \frac{8}{3} + \frac{1}{2\sqrt{5}} \right] = 3$$

Physics

Question 26

Physics · Physical World, Units and Measurements · Single correct

The density $\rho$ of a uniform cylinder is determined by measuring its mass $m$, length $l$ and diameter $d$. The measured values of $m$, $l$ and $d$ are $97.42 \pm 0.02 \, \mathrm{g}$, $8.35 \pm 0.05 \, \mathrm{mm}$ and $20.20 \pm 0.02 \, \mathrm{mm}$, respectively. Calculated percentage fractional error in $\rho$ is $\ldots$

  1. 0.63$\%$
  2. 0.82$\%$
  3. 0.72$\%$
  4. 0.25$\%$

Answer: (b)

Solution

Given $m = 97.42 \pm 0.02 \, \mathrm{g}$, $l = 8.35 \pm 0.05 \, \mathrm{mm}$, $d = 20.20 \pm 0.02 \, \mathrm{mm}$. The density $\rho$ is given by $$\rho = \frac{m}{\frac{\pi}{4} d^2 \ell}$$ The relative error in $\rho$ is $$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{2 \Delta d}{d} + \frac{\Delta \ell}{\ell}$$ Substituting the values, $$= \frac{0.02}{97.42} + 2 \times \frac{0.02}{20.20} + \frac{0.05}{8.35} = 0.0082$$ The percentage error is $0.82\%$.

Question 27

Physics · Physical World, Units and Measurements · Single correct

The potential energy of a particle changes with distance $x$ from a fixed origin as $V = \frac{A \sqrt{x}}{x + B}$, where $A$ and $B$ are constant with appropriate dimensions. The dimension of $AB$ are $\ldots$.

  1. [M^1 L^{5/2} T^{-2}]
  2. [M^{3/2} L^{5/2} T^{-2}]
  3. [M^1 L^2 T^{-2}]
  4. [M^1 L^{7/2} T^{-2}]

Answer: (d)

Solution

[B] = [x] = L [U] = $\left[ \frac{A \sqrt{x}}{B + x} \right] \implies$ M L^2 T^{-2} = $\frac{[A] L^{1/2}}{L}$ [A] = M L^{5/2} T^{-2}

Question 28

Physics · Work, Energy and Power · Numerical

The rain drop of mass 1 g, starts with zero velocity from a height of 1 km. It hits the ground with a speed of 5 m/s. The work done by the unknown resistive force is $\ldots$ J. (take $g = 10 \, \mathrm{m/s^2}$)

  1. -8.75
  2. -8.35
  3. -9.55
  4. -9.98

Answer: (d)

Solution

Given $w_g + w_{res} = \Delta k$. $$mgh + w_{res} = \frac{1}{2}mv^2$$ Therefore, $$w_{res} = \frac{1}{2}mv^2 - mgh$$ $$= 10^{-3} \left( \frac{25}{2} - 10^4 \right)$$ $$= -9.98 \, \mathrm{J}$$

Question 29

Physics · Laws of Motion · Single correct

Two blocks (P and Q) with respectively masses $2\,\mathrm{kg}$ and $1.5\,\mathrm{kg}$ are joined by a massless thread. These blocks are mounted on a frictionless pully which is fixed on the edge of a cube (S), as shown in the figure below. Block P is positioned on the top surface which has no friction and block Q is in contact with side-surface, having coefficient friction $\mu$. The cube (S) moves towards the right with acceleration of $\frac{g}{2}$, where $g$ is gravitational acceleration. During this movement the block P and Q remain stationary. The value of $\mu$ is $\ldots$ (take $g = 10 \, \mathrm{m/s^2}$)

  1. 0.33
  2. 0.67
  3. 1
  4. 0.5

Answer: (b)

Solution

With respect to the bigger block: $$2 \times \frac{g}{2}$$ $$1.5 \times \frac{g}{2} N$$ $$1.5g$$ The friction force is given by $f = \mu N = \mu \times \frac{1.5g}{2}$. Balancing the forces, we have: $$1.5g = g + \mu + \frac{1.5g}{2}$$ This implies: $$0.5 = \mu \times \frac{1.5}{2}$$ Solving for $\mu$, we get: $$\left( \mu = \frac{2}{3} \right)$$

Question 30

Physics · Mechanical Properties of Solids · Single correct

A lift of mass $1600\,\mathrm{kg}$ is supported by thick iron wire. If the maximum stress which the wire can withstand is $4 \times 10^8 \, \mathrm{N/m^2}$ and its radius is $4 \, \mathrm{mm}$, then maximum acceleration the lift can take is $\ldots \, \mathrm{m/s^2}$. (take $g = 10 \, \mathrm{m/s^2}$ and $\pi = 3.14$)

  1. 2.56
  2. 3.89
  3. 4.32
  4. 5.16

Answer: (a)

Solution

T - mg = ma ($\sigma \pi$ r^2) - mg = ma $\Rightarrow$ 4 $\times$ 10^8 $\pi$ (4 $\times$ 10^{-2})^2 - 16000 = 1600 a $\Rightarrow$ a = $2.56\,\mathrm{m/s^2}$

Question 31

Physics · System of Particles and Rotational Motion · Single correct

A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm. It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are _____ and _____ respectively.

  1. 0.128 $\pi \mathrm{Nm}$, 100
  2. 0.0128 $\pi \mathrm{Nm}$, 50
  3. 0.128 $\pi \mathrm{Nm}$, 50
  4. 0.0128 $\pi \mathrm{Nm}$, 100

Answer: (d)

Solution

Given $\omega_i = 1200 \, rpm = 1200 \times \frac{2\pi}{60} = 40 \, \pi \, rad/sec$. $\rightarrow \omega_f = \omega_i + \alpha t \Rightarrow 0 = 40\pi + \alpha (10)$ $\alpha = -4\pi \, rad/sec^2$ $\rightarrow I = \frac{2}{5} MR^2 = \frac{2}{5} (5) (0.04)^2 = 2 \times 0.0016$ $I = 0.0032 \, Kg-m^2$ $\rightarrow \tau = I \alpha = 0.0032 \times 4\pi$ $\tau = 0.0128\pi \, N-m$ $\rightarrow \omega_f^2 = \omega_i^2 + 2\alpha \theta$ $0 = (40\pi)^2 + 2(4\pi)\theta$ $\theta = \frac{1600\pi^2}{8\pi} = 200\pi$ Number of revolutions $= \frac{\theta}{2\pi} = \frac{200\pi}{2\pi}$ Number of revolutions $= 100$

Question 32

Physics · Work, Energy and Power · Single correct

A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height $h$ as shown. If the body exerts a force of three times its weight on the plane at the highest point of circle then the height $h = \alpha R$. The value of $\alpha$ is

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

Answer: (b)

Solution

At the highest point, $$mg + N = \frac{mv^2}{R}$$ $$4mg = \frac{mv^2}{R}$$ $$v = \sqrt{4gR}$$ Now energy conservation $$mg(h - 2R) = \frac{1}{2}m \times 4gR$$ $$(h = 4R)$$

Question 33

Physics · System of Particles and Rotational Motion · Single correct

The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is $\ldots$ .

  1. $\left( \frac{\sqrt{3}}{4}, 1.25 \right)$
  2. $\left( \frac{\sqrt{3}}{4}, 1.0 \right)$
  3. $(0, 0)$
  4. $(1.25, 0)$

Answer: (a)

Solution

The coordinates of the masses are given as follows: For the 2 kg mass: $(-5\sqrt{3}, -\frac{5}{2})$ For the 3 kg mass: $(5\sqrt{3}, -\frac{5}{2})$ For the 15 kg mass: $(0, \frac{5}{2})$ The center of mass coordinates $(X_{cm}, Y_{cm})$ are calculated as: $$X_{cm} = \frac{2 \times (-5\sqrt{3}) + 3(5\sqrt{3}) + 15 \times 0}{20} = \frac{\sqrt{3}}{4} cm$$ $$Y_{cm} = \frac{2 \times (-5/2) + 3(5/2) + 15 \times (5/2)}{20}$$ $$Y_{cm} = 1.25 cm$$

Question 34

Physics · Mechanical Properties of Solids · Single correct

The two wires A and B of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire A and wire B is $20/11$. When the joined wire is kept under certain tension the elongations in the wires A and B are equal. If the length of wire A is $2.2 \, \mathrm{m}$, then the length of wire B is $\ldots$ m.

  1. 1.1
  2. 2.22
  3. 1.21
  4. 4.44

Answer: (c)

Solution

Given $$\Delta \ell = \frac{T \ell}{YA}$$ $$\ell \propto Y$$ $$\frac{\ell_1}{\ell_2} = \frac{Y_1}{Y_2} = \frac{20}{11}$$ Therefore, $$\ell_1 = 2.2 \, \mathrm{m}$$ Therefore, $$\ell_1 = 1.21 \, \mathrm{m}$$

Question 35

Physics · Thermodynamics · Single correct

Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure $90 \, \mathrm{kPa}$ and temperature $400 \, \mathrm{K}$. Keeping the temperature of one vessel constant at $400 \, \mathrm{K}$ the second vessel temperature is raised to $500 \, \mathrm{K}$. The final pressure in the vessels is $\ldots \mathrm{kPa}$.

  1. 100
  2. 120
  3. 90
  4. 105

Answer: (a)

Solution

Using mole conservation $$\frac{2 \times 9 \times 10^4 \times V}{R \times 400} = \frac{P \times V}{R \times 400} + \frac{P \times V}{R \times 500}$$ $P = 100 \, \mathrm{kPa}$

Question 36

Physics · Wave Optics · Single correct

In interference experiment the path difference between two interfering waves at a point A on the screen is $\lambda/3$, where $\lambda$ is the wavelength of these waves, and at another point B the path difference is $\lambda/6$. The ratio of intensities at points A and B is $\ldots$.

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

Answer: (c)

Solution

Given $\Delta \phi_A = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$. Therefore, $I_A = I + I + 2\sqrt{I} \cdot \sqrt{I} \cdot \cos\left(\frac{2\pi}{3}\right) = I$. For $\Delta \phi_B = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$. Therefore, $I_B = I + I + 2\sqrt{I} \cdot \sqrt{I} \cdot \cos\left(\frac{\pi}{3}\right) = 3I$. The ratio $\frac{I_A}{I_B} = \frac{I}{3I} = \frac{1}{3}$.

Question 37

Physics · Oscillations · Single correct

A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is $5 \, \mathrm{sec}$. the time required by it to move from $x = A$ to $x = \frac{A}{\sqrt{2}}$ is $\ldots$ sec.

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

Answer: (c)

Solution

Given $x = -A$, $x = 0$, $x = \frac{A}{\sqrt{2}}$, $x = +A$. $x = 0 \rightarrow \theta = 0^\circ$ $x = \frac{A}{\sqrt{2}} \rightarrow \theta = \frac{\pi}{4}$ Therefore, phase covered $= \frac{\pi}{4}$. $$time = \frac{phase}{w} = \frac{\pi}{4w} = \frac{\pi}{4} \times \frac{T}{2\pi} = \frac{T}{8}$$ Therefore, $t = \frac{5}{8}$ secs.

Question 38

Physics · Electric Charges and Fields · Numerical

A thin half ring of radius $35\,\mathrm{cm}$ is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100\,\mathrm{V/m}$, then the value of $Q$ is ______ nC. $(\epsilon_0=8.85\times10^{-12}\,\mathrm{C^2/Nm^2}$ and $\pi=3.14)$

  1. 2.14
  2. 2.44
  3. 3.25
  4. 0.7

Answer: (a)

Solution

The electric field $E$ is given by the equation $$E = \frac{2k\lambda}{R} \sin\left(\frac{\theta}{2}\right).$$ Solving for $Q$, we have $$100 = 2 \times \frac{9 \times 10^9 \times Q}{\pi R^2}.$$ Therefore, $$Q = 2.14 \times 10^{-9} \, \mathrm{C}.$$

Question 39

Physics · Current Electricity · Single correct

The maximum rated power of the LED is $2\,\mathrm{mW}$ and it is used in the circuit with input voltage of $5\,\mathrm{V}$ as shown in the figure below. The current through resistance $R_s$ is $0.5\,\mathrm{mA}$. The minimum value of the resistance of $R_s$, to ensure that the LED is not damaged is $\ldots \mathrm{k\Omega}$.

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

Answer: (b)

Solution

Diode in reverse bias so no current in it. $P_{bulb} = 2 \times 10^{-3} = V_{bulb}$ $V_{bulb} = 4\, \mathrm{V}$ So $5\, \mathrm{V} = V_{RS} + V_{bulb}$ $V_{RS} = 5 - 4 = 1$ $1 = 0.5 \times 10^{-3} R_{s} \Rightarrow R_{s} = 2\, \mathrm{k}\Omega$

Question 40

Physics · Electromagnetic Waves · Single correct

A point light source emits E.M. waves in free space. A detector, placed at a distance of L m, measures the intensity as $I_0$. The detector is now shifted to another location on the same spherical surface ensuring the angle between original location and new location as $45^\circ$. The measured intensity at new location will be $\ldots$.

  1. $\frac{I_0}{4}$
  2. $I_0$
  3. $\frac{I_0}{\sqrt{2}}$
  4. $\frac{I_0}{2}$

Answer: (b)

Solution

Distance is same. So intensity is also same. $$I = \frac{P}{4 \pi R^2}$$

Question 41

Physics · Ray Optics and Optical Instruments · Single correct

A spherical interface lens of radius $R$ separates two media of refractive indices 1 and 1.4 respectively as shown in the figure below. A point source is placed at a distance of $4R$ in front of spherical interface. The magnitude of the magnification of point source image is $\ldots$.

  1. 1.66
  2. 2.33
  3. 2.66
  4. 1.33

Answer: (a)

Solution

Given $m = \frac{n_1}{n_2} \frac{v}{u}$. $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ $$\frac{1.4}{v} + \frac{1}{4R} = \frac{1.4 - 1}{R}$$ $$\frac{1.4}{v} = \frac{4}{10R} - \frac{1}{4R}$$ $$v = \frac{56R}{6} = \frac{28}{3} R$$ $$m = \frac{1}{1.4} \cdot \frac{28R/3}{(-4R)} = -1.67$$

Question 42

Physics · Moving Charges and Magnetism · Single correct

A small cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying a current of 2 A. The magnetic energy stored inside the cube is $\alpha \times 10^{-14}$ J. The value of $\alpha$ is $\ldots$. ($\mu_0 = 4 \pi \times 10^{-7}$ Tm/A, $\pi = 3.14$)

  1. 6.28
  2. 6.28 $\times$ 10^{-6}
  3. 628
  4. 6.28 $\times$ 10^{-4}

Answer: (a)

Solution

Magnetic field at centre B = $\left( \frac{\mu_0}{4\pi} \right) \frac{2\pi i}{R}$ = $\frac{10^{-7} \times 2 \times \pi \times 2}{10 \times 10^{-2}}$ So energy stored U = $\frac{B^2}{2\mu_0} \times$ V = $\frac{10^{-14} \times 4 \times \pi^2 \times 4 \times 100}{2 \times 4 \pi \times 10^{-7}} \times \left( 1 \times 10^{-9} \right)$ = 2$\pi \times$ 10^{-14} $\mathrm{J}$ = 6.28 $\times$ 10^{-14} $\mathrm{J}$

Question 43

Physics · Electromagnetic Induction · Single correct

An inductor of inductance $10\,\mathrm{mH}$ having resistance of 100 $\Omega$ is connected to battery of E.M.F. $1.0\,\mathrm{V}$ through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is $2\,\mathrm{mA}$ and $4\,\mathrm{mA}$ is $\ldots$

  1. 4/3
  2. 3/4
  3. 5/3
  4. 3/5

Answer: (a)

Solution

Apply KVL $$E - iR - V_L = 0$$ $$V_L = E - iR = 1 - 100 \, i$$ at $i_1 = 2 \times 10^{-3} \, \mathrm{A}$ $$V_{L1} = 0.8 \, \mathrm{V}$$ at $i_2 = 4 \times 10^{-3} \, \mathrm{A}$ $$V_{L2} = 0.6 \, \mathrm{V}$$ Therefore, $$\frac{V_{L1}}{V_{L2}} = \frac{4}{3}$$

Question 44

Physics · Atoms · Single correct

The ratio of momentum of the photons of the $1^{st}$ and $2^{nd}$ line of Balmer series of Hydrogen atoms is $\alpha/\beta$. The possible values of $\alpha$ and $\beta$ are :-

  1. 27 and 20
  2. 3 and 16
  3. 5 and 36
  4. 20 and 27

Answer: (d)

Solution

Given $p = \frac{h}{\lambda}$. $$\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right)$$ $$\frac{1}{\lambda_1} = R \frac{5}{36}$$ $$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right)$$ $$\frac{1}{\lambda_2} = \frac{3R}{16}$$ $$\frac{p_1}{p_2} = \frac{\lambda_2}{\lambda_1} = \frac{16}{\frac{3R}{36}} = \frac{20}{5R} = \frac{20}{27}$$ $\alpha = 20$ and $\beta = 27$

Question 45

Physics · Alternating Current · Single correct

A LCR series circuit driven with $E_{rms} = 90 \, \mathrm{V}$ at frequency $f_d = 30 \, \mathrm{Hz}$ has resistance $R = 80 \, \Omega$, an inductance with inductive reactance $X_L = 20.0 \, \Omega$ and capacitance with capacitive reactance $X_C = 80.0 \, \Omega$. The power factor of the circuit is $\ldots$.

  1. 0.8
  2. 0.64
  3. 0.9
  4. 0.5

Answer: (a)

Solution

Power factor is $\cos \phi = \frac{R}{Z}$. $$Z = \sqrt{R^2 + (X_L - X_C)^2} = 100 \, \Omega$$ $$\cos \phi = \frac{80}{100} = 0.8$$

Question 46

Physics · Current Electricity · Numerical

Refer to the circuit diagram given below. The heat generated across the $6 \, \Omega$ resistance in $100$ second is $\frac{\alpha}{100} \, \mathrm{J}$. The value of $\alpha$ is $\ldots$. (Nearest integer)

Answer: 3477

Solution

Given $i_1 + i_2 + i_3 = 0$. $$\frac{x - 0 - 3}{6} + \frac{x - 2}{3} + \frac{x - 0}{4} = 0$$ Solving for $x$, we get $x = \frac{14}{9} \, \mathrm{V}$. Current across $6\Omega$ is given by: $$\frac{3 - (14/9)}{6} = \frac{13}{54} \, \mathrm{A}$$ Calculating $H$: $$H = \left(\frac{13}{54}\right)^2 \times 6 \times 100 = 34.77 = \frac{\alpha}{100}$$ Solving for $\alpha$, we find $\alpha = 3477$.

Question 47

Physics · Wave Optics · Numerical

An unpolarized light of intensity $I_0$ passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is $0^\circ$ and intensity of light emerged from analyser is $\frac{3}{8} I_0$, the angle of rotation of the light by the solution with respect to analyser is $\ldots$ degrees.

Answer: 30

Solution

The initial intensity $I_0$ is unpolarised. After passing through the polariser, the intensity becomes $\frac{I_0}{2}$, which is polarised. The intensity after the analyser is given by $I = I' \cos^2 \theta$. Substituting the given values: $$\frac{3I_0}{8} = \frac{I_0}{2} \cos^2 \theta$$ Solving for $\cos \theta$: $$\cos \theta = \frac{\sqrt{3}}{2}$$ Therefore, $\theta = 30^\circ$.

Question 48

Physics · Nuclei · Numerical

The energy released when $\frac{7}{17.13}$ kg of $^7_3\mathrm{Li}$ is converted into $^4_2\mathrm{He}$ by proton bombardment is $\alpha \times 10^{32}$ eV. The value of $\alpha$ is $\ldots$. (Nearest integer) (Mass of $^7_3\mathrm{Li} = 7.0183$ u, mass of $^4_2\mathrm{He} = 4.004$ u, mass of proton $= 1.008$ u and $1$ u $= 931$ MeV/$c^2$ and Avogadro number $= 6.0 \times 10^{23}$)

Answer: 6

Solution

The reaction is $^7_3\mathrm{Li} + ^1_1\mathrm{P} \rightarrow 2\,^4_2\mathrm{He}$. The mass defect $\Delta m = (7.018 + 1.008) - 2(4.004) = 0.0183 \, \mathrm{u}$. Energy released per reaction $E = \Delta mc^2$. Number of moles of Li $= \frac{7}{17.13} \times \frac{10^3}{7} = \frac{10^3}{17.13}$. Number of molecules $= \frac{10^3}{17.13} \times 6 \times 10^{23}$. Total energy released $$E_{Total} = 0.0183 \times 931 \times \frac{6}{17.13} \times 10^{32}$$ $$\approx 6 \times 10^{32}$$ $\alpha = 6$

Question 49

Physics · Electrostatic Potential and Capacitance · Numerical

A three coulomb charge moves from the point $(0, -2, -5)$ to the point $(5, 1, 2)$ in an electric field expressed as $\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k} \, \mathrm{N/C}$. The work done in moving the charge is $\ldots$ J.

Answer: 186

Solution

Given $\Delta U = -w_{electrostatic}$. $w_{electrostatic} = -q \Delta V$. $\Delta V = - \int 2x \, dx - \int 3y^2 \, dy - \int 4 \, dz$. $$= \left( -x^2 - y^3 - 4z \right)_{(0, -2, -5)}^{(5, 1, 2)}$$ $$= -62$$ $w_{electrostatic} = 3(62) = 186 \, J$.

Question 50

Physics · Thermodynamics · Numerical

A certain gas is isothermally compressed to $\left( \frac{1}{3} \right)^{rd}$ of its initial volume $(V_0 = 3 \, litre)$ by applying required pressure. If the bulk modulus of the gas is $3 \times 10^5 \, N/m^2$, the magnitude of work done on the gas is $\ldots$ J.

Answer: 989

Solution

For isothermal process initially $B = P_0$. $$W = nRT \ln \left( \frac{V_2}{V_1} \right)$$ $$= P_0 V_0 \ln \left( \frac{1}{3} \right)$$ $$W = BV_0 \ln \left( \frac{1}{3} \right)$$ $$W = -988.75 \, \mathrm{J} \approx 989 \, \mathrm{J}$$ Note: Bulk modulus of gas for isothermal process is not constant. In the given solution we have considered it as initial bulk modulus.

Chemistry

Question 51

Chemistry · Some Basic Concepts of Chemistry · Single correct

An oxide of iron contains 69.9$\%$ iron, its empirical formula, is : (Given : Molar mass of Fe and O are 56 and $16\,\mathrm{g \, mol^{-1}}$ respectively.)

  1. FeO
  2. Fe_2O_3
  3. Fe_3O_4
  4. FeO_3

Answer: (b)

Solution

Moles of Fe: $\frac{69.9}{56}$, Moles of Oxygen: $\frac{30.1}{16}$. Molar Ratio: $1 : 1.5$. Molar Ratio: $2 : 3$. Answer: $\mathrm{Fe_2O_3}$

Question 52

Chemistry · Structure of Atom · Single correct

If shortest wavelength of hydrogen atom in Lyman series is $x$, then longest wavelength in Balmer series of $\mathrm{He}^+$ is:

  1. $\frac{9x}{5}$
  2. $\frac{36x}{5}$
  3. $\frac{x}{4}$
  4. $\frac{5x}{9}$

Answer: (a)

Solution

Given $\frac{1}{x} = R \times 1^2 \times \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \ldots$(1) $\frac{1}{\lambda} = R \times 2^2 \times \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \ldots$(2) $\frac{x}{\lambda} = R \times 4 \times \frac{5}{4 \times 9}$ $\frac{1}{\lambda} = \frac{1}{x} \times \frac{5}{9}$ $\lambda = \frac{9x}{5}$

Question 53

Chemistry · Structure of Atom · Single correct

Match the List-I with List-II : Choose the correct answer from the options given below :

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

Answer: (d)

Solution

Number of Radial nodes is given by $n - \ell - 1$. Number of Angular nodes or nodal planes is $\ell$. For $2s \Rightarrow n = 2$, $\ell = 0$: 1 radial node + zero nodal plane. For $3s \Rightarrow n = 3$, $\ell = 0$: 2 radial nodes + zero nodal plane. For $3p \Rightarrow n = 3$, $\ell = 1$: 1 radial node + 1 nodal plane. For $4d \Rightarrow n = 4$, $\ell = 2$: 1 radial node + 2 nodal planes.

Question 54

Chemistry · Chemical Bonding and Molecular Structure · Single correct

The pairs among A = [$\mathrm{SO}_3^{2-}$, $\mathrm{CO}_3^{2-}$], B = [$\mathrm{O}_2^{2-}$, $\mathrm{F}_2$], C = [$\mathrm{CN}^-$, $\mathrm{CO}$], D = [$\mathrm{NH}_3$, $\mathrm{H}_3 \mathrm{O}^+$] and E = [$\mathrm{MnO}_4^{2-}$, $\mathrm{CrO}_4^{2-}$] that do not have similar Lewis dot structure are:

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

Answer: (b)

Solution

A and E do not have similar Lewis structure. $\mathrm{MnO_4^{2-}}$ have unpaired $e$ but $\mathrm{CrO_4^{2-}}$ does not have unpaired $e$. In B, C, D pair of species have same number of $e$. They have similar Lewis dot structure.

Question 55

Chemistry · Thermodynamics · Single correct

Arrange the following isothermal processes in order of the magnitude of the work (p - V) involved between states 1 and 2. A. Expansion in single stage $w_A$ B. Expansion in multi stages $w_B$ C. Compression in single stage $w_C$ D. Compression in multi stages $w_D$

  1. $|w_B| > |w_A| > |w_C| > |w_D|$
  2. $|w_C| > |w_D| > |w_A| > |w_B|$
  3. $|w_C| > |w_D| > |w_B| > |w_A|$
  4. $|w_B| > |w_A| > |w_D| > |w_C|$

Answer: (c)

Solution

In the expansion process, the work done in a single step is less than the work done in multiple steps. In the compression process, the work done in multiple steps is less than the work done in a single step. Therefore, $$|W_A| < |W_B| < |W_D| < |W_C|.$$

Question 56

Chemistry · Solutions · Single correct

When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was $x\%$ of vapor pressure of pure solvent. What is $x\%$ ?

  1. 50$\%$
  2. 60$\%$
  3. 70$\%$
  4. 80$\%$

Answer: (d)

Solution

Given $P_S = P^0 X_A$. $$\frac{P_S}{P^0} = X_A = \frac{1}{1 + 0.25} = 0.80$$ Percentage is $0.8 \times 100 = 80$.

Question 57

Chemistry · Equilibrium · Single correct

One mole each of He and A(g) are taken in a $10\,\mathrm{L}$ closed flask and heated to $400\,\mathrm{K}$ to establish the following equilibrium. A(g) $\rightleftharpoons$ B(g) K_c for this reaction at $400\,\mathrm{K}$ is 4.0. The partial pressures (in atm) of He and B(g) are respectively (at equilibrium). (Assume He, A(g) and B(g) behave as ideal gases) (Given : R = $0.082\,\mathrm{L} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1}$)

  1. 3.28, 2.624
  2. 2.624, 3.28
  3. 3.28, 0.656
  4. 0.656, 6.56

Answer: (a)

Solution

At time $t = 0$, the concentration of $A$ is $1 \, mol$ and $B$ is not present. At equilibrium, the concentration of $A$ is $1-x$ and $B$ is $x$. The equilibrium constant $K_C$ is given by $$K_C = \frac{x/10}{(1-x)/10} = 4.$$ Solving for $x$, we find $x = 0.8$. The partial pressure of helium $P_{He}$ is calculated as $$P_{He} = \frac{1 \times 0.082 \times 400}{10} = 3.28 \, atm.$$ The partial pressure of $B$, $P_B$, is $$P_B = \frac{0.8 \times 0.082 \times 400}{10} = 2.624 \, atm.$$

Question 58

Chemistry · Electrochemistry · Single correct

Consider the following data: $\mathrm{BaSO_4}$ is sparingly soluble in water. If the conductivity of the saturated $\mathrm{BaSO_4}$ solution is x $\mathrm{S\,cm^{-1}}$ then the solubility product of $\mathrm{BaSO_4}$ can be given as. (Here $\Lambda_m$ = $\Lambda_m^0$ )

  1. $\frac{10^6 x^2}{\alpha^2 (x_1 + x_2 - 2x_3)^2}$
  2. $\frac{x^2}{(x_1 + x_2 - 2x_3)^2}$
  3. $\frac{\alpha^2 (x_1 + x_2 - 2x_3)^2}{10^6 x^2}$
  4. $\frac{x^2}{(x_1 + x_2 + 2x_3)^2}$

Answer: (a)

Solution

The dissolution of $\mathrm{BaSO_4(s)}$ in water is represented by the equation: $$\mathrm{BaSO_4(s) \rightleftharpoons Ba^{2+}_{(aq)} + SO_4^{2-}_{(aq)}}$$ The solubility product is given by $K_{sp} = S^2$. The conductivity $k$ of $\mathrm{BaSO_4}$ is $x \, \mathrm{S \, cm^{-1}}$. The molar conductivity at infinite dilution $\Lambda_m^0$ of $\mathrm{BaSO_4}$ is $$(x_1 + x_2 - 2x_3)$$ The molar conductivity $\Lambda_m$ is given by $$\Lambda_m = \alpha \Lambda_m^0 = \frac{k \times 1000}{S}$$ Solving for $S$, we have: $$S = \frac{x \times 1000}{\alpha (x_1 + x_2 - 2x_3)}$$ Substituting back into the expression for $K_{sp}$, we get: $$k_{sp} = S^2 = \frac{1}{\alpha^2} \left[ \frac{x \times 1000}{x_1 + x_2 - 2x_3} \right]^2$$

Question 59

Chemistry · The p-Block Elements (Group-13 and 14) · Single correct

Given below are two statements: Statement-I: Aluminium is more electropositive than thallium as the standard electrode potential value of $E^0_{\mathrm{Al}^{3+}/\mathrm{Al}}$ is negative and $E^0_{\mathrm{Tl}^{3+}/\mathrm{Tl}}$ is positive. Statement-II: The sum of first three ionization enthalpies of boron is very high when compared to that of aluminium. Due to this reason boron forms covalent compounds only and aluminium forms $\mathrm{Al}^{3+}$ ion. 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

Given $E^0_{\mathrm{Al^{3+}/Al}} = -1.66 \, \mathrm{V}$ and $E^0_{\mathrm{Tl^{3+}/Tl}} = +1.26 \, \mathrm{V}$. For $\mathrm{B}$: $\mathrm{IE}_1 + \mathrm{IE}_2 + \mathrm{IE}_3 = 801 + 2427 + 3659 = 6887 \, \mathrm{kJ}$. Both statements are true.

Question 60

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

The correct statements among the following are: A. Basic vanadium oxide is used in the manufacture of $\mathrm{H_2SO_4}$. B. The spin-only magnetic moment value of the transition metal halide employed in Ziegler-Natta polymerization is $2.84 \, \mathrm{BM}$. C. The p-block metal compound employed in Ziegler-Natta polymerization has the metal in $+3$ oxidation state. D. The number of electrons present in the outermost 'd' orbital of metal halide employed in Wacker process is $8$. Choose the correct answer from the options given below:

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

Answer: (c)

Solution

A. $\mathrm{V_2O_5}$ is amphoteric. It is used in manufacture of $\mathrm{H_2SO_4}$. B. Ziegler natta catalyst is $\mathrm{TiCl_4} + \mathrm{AlMe_3}$ $$\mathrm{Ti^{4+}} : 3d^0 \, 4s^0 n = 0 \mu = 0$$ C. O.S. of Al is $+3$ D. $\mathrm{PdCl_2}$ is used in wacker process $$\mathrm{Pd^{2+}} : 4d^8 5s^0$$ C and D are correct.

Question 61

Chemistry · Co-ordination Compounds · Single correct

Match the List-I with List-II Choose the correct answer from the options given below:

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

Answer: (c)

Solution

In tetrahedral complex $$CFSE = \left[ -\frac{3}{5} n_1 + \frac{2}{5} n_2 \right] \Delta_t$$ For $d^2$: $e^{1,1} t_2^{0,0,0}$ with $n_1 = 2$, $n_2 = 0$; $$C.F.S.E = \left[ -\frac{3}{5} (2) \right] \Delta_t = -1.2 \Delta_t$$ For $d^4$: $e^{1,1} t_2^{1,1,0}$ with $n_1 = 2$, $n_2 = 2$; $$C.F.S.E = \left[ -\frac{3}{5} (2) + \frac{2}{5} (2) \right] \Delta_t = -0.4 \Delta_t$$ For $d^6$: $e^{2,1} t_2^{1,1,1}$ with $n_1 = 3$, $n_2 = 3$; $$C.F.S.E = \left[ -\frac{3}{5} (3) + \frac{2}{5} (3) \right] \Delta_t = -0.6 \Delta_t$$ For $d^8$: $e^{2,2} t_2^{1,1,1}$ with $n_1 = 4$, $n_2 = 4$; $$C.F.S.E = \left[ -\frac{3}{5} (4) + \frac{2}{5} (4) \right] \Delta_t = -0.8 \Delta_t$$ A - III, B - IV, C - I, D - II

Question 62

Chemistry · Co-ordination Compounds · Single correct

Which of the following are true about the energy of the given d-orbitals of a tetrahedral complex? (A) $d_{xy}=d_{xz}>d_{x^2-y^2}$ (B) $d_{xy}=d_{yz}>d_{z^2}$ (C) $d_{x^2-y^2}>d_{z^2}>d_{xz}$ (D) $d_{x^2-y^2}=d_{z^2}<d_{xz}$ Choose the correct answer from the options given below:

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

Answer: (a)

Solution

Order of energy in tetrahedral splitting $$(dxy = dyz = dzx) < d_{x^2-y^2} = d_{z^2}$$ A, B and D are correct.

Question 63

Chemistry · Hydrocarbons · Single correct

$R_f$ value for 2-methylpropene in a solvent system (Ethyl acetate + ether) is 0.42. 2-methylpropene is treated with dilute $\mathrm{H_2SO_4}$ to give major organic product (X). $R_f$ value for (X) in the same solvent system under identical condition will be:

  1. 0.42
  2. 0.82
  3. 0.62
  4. 0.12

Answer: (d)

Solution

Ter butanol (X) is more polar than 2-Methyl propene and $R_f \propto \frac{1}{polarity}$ so $R_f$ of ter.butanol is less than 2-Methylpropen. So, Ans is 0.12

Question 64

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Given below are two statements: Statement-I: 2,6-diethylcyclohexanone and 6-methyl-2-n-propylcyclohexanone are metamers. Statement-II: 2,2,6,6-tetramethylcyclohexanone exhibits keto-enol tautomerism. 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

Statement-I is correct and Statement-II is incorrect. The compounds shown are metamers. The compound with no $\alpha$-H does not exhibit keto-enol tautomerism.

Question 65

Chemistry · Hydrocarbons · Single correct

Given below are two statements : Statement-I : Methane can be prepared by decarboxylation of sodium ethanoate, Kolbe's electrolysis of sodium acetate and reaction of $\mathrm{CH_3MgBr}$ with water. Statement-II : Methane cannot be prepared from unsaturated hydrocarbons and by Wurtz 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: (d)

Solution

Methane can not be prepared by Kolbe's electrolysis process. So, statement-1 is false. $$\begin{array}{c} \mathrm{O} \\ || \\ 2\mathrm{CH_3-C-ONa} \xrightarrow{Kolbe electrolysis} \mathrm{CH_3-CH_3} + 2\mathrm{CO_2} + 2\mathrm{NaOH} + \mathrm{H_2} \\ Ethane \end{array}$$ Methane also can not be prepared by Wurtz reaction so statement-2 is true. $$\begin{array}{c} 2\mathrm{R-X} \xrightarrow{Na.oxyether} \mathrm{R-R} + 2\mathrm{NaX} \\ Wurtz reaction \end{array}$$ If $\mathrm{R} = \mathrm{-CH_3}$, then ethane is formed not methane.

Question 66

Chemistry · Haloalkanes and Haloarenes · Single correct

Given below are two statements: Statement-I: 3-phenylpropene reacts with HBr and gives secondary alkyl bromide having a chiral carbon atom as the major product. Statement-II: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann 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: (c)

Solution

Ph-$\mathrm{CH_2}$-$\mathrm{CH}$ = $\mathrm{CH_2} \xrightarrow{\mathrm{HBr}} \mathrm{Ph}$-$\mathrm{CH_2}$-$\overset{\oplus}{\mathrm{CH}}$-$\mathrm{CH_3} \rightarrow \mathrm{Ph}$-$\overset{\oplus}{\mathrm{CH}}$-$\mathrm{CH_2}$-$\mathrm{CH_3}$ (Hydride shift) $\xleftarrow{\mathrm{Br}^-} \mathrm{Ph}$-$\overset{*}{\mathrm{CH}}$-$\mathrm{CH_2}$-$\mathrm{CH_3}$ (Major product) $\newline$ 1 Chiral carbon $\newline$ Aryl cyanide is not formed by Gattermann reaction.

Question 67

Chemistry · Alcohols, Phenols and Ethers · Single correct

Consider the following sequence of reactions The major product P is:

Answer: (c)

Solution

The reaction starts with the dehydration of isopropanol using copper at 573 K to form propene. This is followed by an electrophilic aromatic substitution (EAR) with benzoic acid in the presence of $H^+$ to form isopropyl benzoate.

Question 68

Chemistry · Amines · Single correct

Arrange the following compounds according to increasing order of boiling points. n-$\mathrm{C}_4\mathrm{H}_9\mathrm{OH} (\mathrm{A})$, $n-\mathrm{C}_4\mathrm{H}_9\mathrm{NH}_2 (\mathrm{B})$, $n-\mathrm{C}_4\mathrm{H}_{10} (\mathrm{C})$ and $\mathrm{C}_2\mathrm{H}_5\mathrm{NH}\mathrm{C}_2\mathrm{H}_5 (\mathrm{D})$.

  1. $\mathrm{C} < \mathrm{B} < \mathrm{A} < \mathrm{D}$
  2. $\mathrm{D} < \mathrm{C} < \mathrm{B} < \mathrm{A}$
  3. $\mathrm{C} < \mathrm{D} < \mathrm{B} < \mathrm{A}$
  4. $\mathrm{D} < \mathrm{B} < \mathrm{A} < \mathrm{C}$

Answer: (c)

Solution

Boiling point $\propto$ Intermolecular attraction $\propto$ Number of H-bonding. n-butanol has strong H-bonding.

Question 69

Chemistry · Biomolecules · Single correct

Match the List-I with List-II : Choose the correct answer from the options given below :

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

Answer: (c)

Solution

Vitamin-A $\rightarrow$ Xerophthalmia Vitamin-C (Ascorbic acid) $\rightarrow$ Scurvy Vitamin-B_2 (Riboflavin) $\rightarrow$ Cheilosis Vitamin-B_6 (Pyridoxine) $\rightarrow$ Convulsions

Question 70

Chemistry · Biomolecules · Single correct

Match the List-I with List-II Choose the correct answer from the options given below:

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

Answer: (b)

Question 71

Chemistry · The s-Block Elements · Numerical

First and second ionization enthalpies of lithium are $520 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$ and $7297 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$ respectively. Energy required to convert $3.5 \, \mathrm{mg}$ lithium $(\mathrm{g})$ into $\mathrm{Li^{2+}}(\mathrm{g}) [\mathrm{Li(g)} \rightarrow \mathrm{Li^{2+}(g)}]$ is $\ldots \, \mathrm{kJ} \, \mathrm{mol}^{-1}$. (nearest integer) [Molar mass of $\mathrm{Li} = 7 \, \mathrm{g} \, \mathrm{mol}^{-1}$]

Answer: 4

Solution

Number of moles of Li is given by $$\frac{3.5 \times 10^{-3}}{7} = 5 \times 10^{-4}$$. Total energy needed to convert $$\mathrm{Li(g)} \rightarrow \mathrm{Li^{2+}(g)}$$ is $$5 \times 10^{-4} (520 + 7297) = 3.9085 \, \mathrm{kJ} = 4 \, \mathrm{kJ}$$.

Question 72

Chemistry · Amines · Numerical

Consider the following sequence of reactions. The percentage of nitrogen in the yellow product (X) formed is $\ldots \%$. (Nearest Integer) (Given Molar mass in $\mathrm{g \, mol^{-1}}$ H : 1, C : 12, N : 14)

Answer: 21

Solution

HCl catalysed hydrolysis. Aniline acts as a solvent and reactant. The mixture is gently heated to facilitate rearrangement. After the reaction is complete, the mixture is cooled and treated with acetic acid to neutralise excess acid and precipitate the final product. Mass percentage of nitrogen in final product $$= \frac{42}{197} \times 100 = 21.30\% \approx 21$$

Question 73

Chemistry · Alcohols, Phenols and Ethers · Numerical

$4.7\,\mathrm{g}$ of phenol is heated with Zn to give product $X$. If this reaction goes to $60\%$ completion, then the number of moles of compound $X$ formed will be_______$\times10^{-2}$ (Nearest Integer) (Given molar mass in $\mathrm{g\,mol^{-1}}$: H: $1$, C: $12$, O: $16$)

Answer: 3

Solution

Moles of phenol = $\frac{4.7}{94.0} = 0.05 mol$ Reaction is 1 : 1 so 0.05 mole of phenol will give 0.05 mole of benzene on 100$\%$ yield. For 60$\%$ yield = $0.05 \times 0.60$ = 0.03 mol. mol of 'X' = $3 \times 10^{-2} mol$ So answer is (3)

Question 74

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1/2} = 3$ hour. The percentage of sucrose remaining after 6 hours is $\ldots$. (Nearest integer) (Given : $\log$ 2 = 0.3010 and $\log$ 3 = 0.4771)

Answer: 25

Solution

Sucrose + $\mathrm{H_2O} \rightarrow$ glucose + fructose $t_{1/2} = 3 \, \mathrm{hr}$ For 1st order, $t_{75\%} = 2 \times t_{1/2}$ So $t = 6 \, \mathrm{hr}$, 75$\%$ of sucrose decomposed. So 25$\%$ of sucrose remains after 6 hr.

Question 75

Chemistry · Thermodynamics · Numerical

Consider the reaction $X \rightleftharpoons Y$ at $300 \, \mathrm{K}$. If $\Delta H^\theta$ and $K$ are $28.40 \, \mathrm{kJ \, mol^{-1}}$ and $1.8 \times 10^{-7}$ at the same temperature, then the magnitude of $\Delta S^\theta$ for the reaction in $\mathrm{J \, K^{-1} \, mol^{-1}}$ is $\ldots$ (Nearest Integer). (Given : $R = 8.3 \, \mathrm{J \, K^{-1} \, mol^{-1}}$, $\ln 10 = 2.3$, $\log 3 = 0.48$, $\log 2 = 0.30$).

Answer: 34

Solution

Given the reaction $\mathrm{X} \rightleftharpoons \mathrm{Y}$ with $K_{\mathrm{eq}} = 1.8 \times 10^{-7}$. The equation for the equilibrium constant is: $$\ln(K_{\mathrm{eq}}) = \frac{\Delta_r S^\circ}{R} - \frac{\Delta_r H^\circ}{R} \left( \frac{1}{T} \right)$$ Substituting the given values: $$2.3 \log(1.8 \times 10^{-7}) = \frac{\Delta_r S^\circ}{R} - \frac{28400}{R} \times \frac{1}{300}$$ Rearranging for $\Delta_r S^\circ$: $$\Delta_r S^\circ = 2.3 \times R \log(1.8 \times 10^{-7}) + \frac{284}{3}$$ Calculating the values: $$= 19.09 \left[-8 + 2 \times 0.48 + 0.3 \right] + \frac{284}{3}$$ Simplifying further: $$= -128.66 + \frac{284}{3}$$ Finally, we find: $$\Delta_r S^\circ = -33.99 \approx -34 \, \mathrm{J/K \cdot mol}$$