JEE Main 8 April 2026 Shift 2 question paper with solutions

JEE Main 8 April 2026 Shift 2: all 75 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Relations and Functions · Single correct

Consider the relation R on the set $\{$-2, -1, 0, 1, 2$\}$ defined by (a, b) $\in$ R if and only if 1 + ab > 0. Then among the statements: I. The number of elements in R is 17 II. R is an equivalence relation

  1. Only I is true
  2. Only II is true
  3. Both I and II are true
  4. Neither I nor II is true

Answer: (a)

Solution

Reflexive If $(a, a) \in R$ $$\Rightarrow 1 + a^2 > 0$$ $$\Rightarrow$$ It is reflexive relation Symmetric If $1 + ab > 0 \Rightarrow 1 + ba > 0$ i.e. If $(a, b) \in R$ and $(b, a) \in R$ Therefore it is symmetric relation Transitive $\therefore (-2, 0) \in R$ and $(0, 2) \in R$ But $(-2, 2) \notin R$ Hence it is not transitive $$\Rightarrow$$ It is not an equivalence relation. Check Number of elements in $R$ $\therefore 1 + ab > 0$ a and b each can be filled by 5 ways $\therefore$ total elements $= 25$ But out of these $(-2, -1), (-2, 2), (-1, 2), (1, -2), (2, -2), (2, -1)$ $(-1, 1)$ and $(1, -1)$ are not in relation.

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

The number of values of $z \in \mathbb{C}$, satisfying the equations $|z - (4 + 8i)| = \sqrt{10}$ and $|z - (3 + 5i)| + |z - (5 + 11i)| = 4\sqrt{5}$, is :

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

Answer: (b)

Solution

Given $|z - (3 + 5i)| + |z - (5 + 11i)| = 4 \sqrt{5}$. So, $S_1(3, 5)$ and $S_2(5, 11)$ are two foci of the ellipse, where $2ae = S_1S_2 = 2 \sqrt{10}$ and $2a = 4 \sqrt{5}$. Therefore, $a = 2 \sqrt{5}$, $e = \frac{1}{\sqrt{2}}$ and $b^2 = a^2 (1 - e^2)$ gives $b^2 = 10$. The center of an ellipse is the midpoint of both foci, thus the center is $(4, 8)$. The equation of the circle is $|Z - (4 + 8i)| = \sqrt{10}$. Therefore, the ellipse and the given circle are concentric, and the radius of the given circle is equal to the length of the minor axis of the given ellipse. Two points are common to both.

Question 3

Maths · Determinants · Single correct

If the system of linear equations : x + y + z = 6, x + 2y + 5z = 10, 2x + 3y + $\lambda$ z = $\mu$. has infinitely many solutions, then the value of $\lambda$ + $\mu$ equals.

  1. 12
  2. 16
  3. 22
  4. 28

Answer: (c)

Solution

Given $$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda \end{vmatrix} = \lambda - 6$$ $$D_1 = \begin{vmatrix} 6 & 1 & 1 \\ 10 & 2 & 5 \\ \mu & 3 & \lambda \end{vmatrix} = 2\lambda + 3\mu - 60$$ $$D_2 = \begin{vmatrix} 1 & 6 & 1 \\ 1 & 10 & 5 \\ 2 & \mu & \lambda \end{vmatrix} = 4(\lambda - \mu + 10)$$ $$D_3 = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 2 & 3 & \mu \end{vmatrix} = \mu - 16$$ For infinite many solution $$D = D_1 = D_2 = D_3 = 0$$ $$\Rightarrow \lambda = 6 \& \mu = 16$$ $$\Rightarrow \lambda + \mu = 22$$

Question 4

Maths · Determinants · Single correct

Let $A=\begin{pmatrix}\alpha&1&2\\2&3&0\\0&4&5\end{pmatrix}$ and $B=\begin{pmatrix}1&0&0\\0&-5\alpha&0\\0&4\alpha&-2\alpha\end{pmatrix}+\operatorname{adj}(A)$. If $\det(B)=66$, then $\det(\operatorname{adj}(A))$ equals:

  1. 289
  2. 361
  3. 441
  4. 529

Answer: (c)

Solution

Given $A = \begin{bmatrix} \alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix}$ and $|A| = 5\alpha + 6$. Now adj $A = \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix}$. Therefore, $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + adj(A)$. $B = \begin{bmatrix} 16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & \alpha - 2 \end{bmatrix}$. And $|B| = 6(5\alpha + 6) = 66$ (given in question). Therefore, $\alpha = 1$. Thus, $|A| = 21$. Therefore, $\det(adj A) = |A|^{(n-1)^2}$. Here $n = 3$. $= 21^2 = 441$.

Question 5

Maths · Sequences and Series · Single correct

Let $\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \ldots$ upto 40 terms. If $\left( \tan \beta \right)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2 + x - 2 = 0$, $\beta \in \left( 0, \frac{\pi}{2} \right)$, then $\sin^2 \beta + 3 \cos^2 \beta$ is equal to:

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

Answer: (a)

Solution

Given $$a = (3 + 8 + 13 \ldots up to 20 terms) + (4 + 9 + 14 + \ldots up to 20 terms)$$ $$= \frac{20}{2} [6 + 19 \times 5] + \frac{20}{2} (8 + 19 \times 5)$$ $$= 2040$$ $$\left( \tan \beta \right)^{\frac{2040}{1020}} = \tan^2 \beta$$ $$x^2 + x - 2 = 0 gives roots 1, -2$$ Therefore, $$\tan^2 \beta = 1 ; \sin^2 \beta = \frac{1}{2} \& \cos^2 \beta = \frac{1}{2}$$ $$\sin^2 \beta + 3 \cos^2 \beta = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$$

Question 6

Maths · Probability · Single correct

A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are $\frac{2}{5}$, $\frac{1}{5}$ and $\frac{2}{5}$. The probabilities that the candidate reaches late at the examination centre are $\frac{1}{5}$, $\frac{1}{3}$ and $\frac{1}{4}$ if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is :

  1. $\frac{11}{37}$
  2. $\frac{12}{37}$
  3. $\frac{13}{37}$
  4. $\frac{14}{37}$

Answer: (b)

Solution

The probability of taking the bus is given by $\mathrm{P(Bus)} = \mathrm{P(B)} = \frac{2}{5}$. The probability of being late given that the bus is taken is $\mathrm{P\left(\frac{late}{B}\right)} = \frac{1}{5}$. The probability of taking the scooter is $\mathrm{P(Scooter)} = \mathrm{P(S)} = \frac{1}{5}$. The probability of being late given that the scooter is taken is $\mathrm{P\left(\frac{late}{S}\right)} = \frac{1}{3}$. The probability of taking the car is $\mathrm{P(Car)} = \mathrm{P(C)} = \frac{2}{5}$. The probability of being late given that the car is taken is $\mathrm{P\left(\frac{late}{C}\right)} = \frac{1}{4}$. The probability of taking the bus given that the person is late is calculated as follows: $$\mathrm{P\left(\frac{B}{late}\right)} = \frac{\frac{2}{5} \times \frac{1}{5}}{\frac{2}{5} \times \frac{1}{5} + \frac{1}{5} \times \frac{1}{3} + \frac{2}{5} \times \frac{1}{4}} = \frac{12}{37}$$

Question 7

Maths · Statistics · Single correct

A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Then, the variance of all these 10 observations is equal to:

  1. 5.96
  2. 6.14
  3. 6.04
  4. 6.24

Answer: (c)

Solution

Given $\bar{x} = 1$, $\sigma_1^2 = 13$ and $\bar{y} = 2$, $\sigma_2^2 = 1$. Combined variance is given by: $$\frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2} + \frac{n_1 n_2}{(n_1 + n_2)^2} (\bar{x} - \bar{y})^2$$ $$= \frac{4 \times 13 + 6 \times 1}{10} + \frac{4 \times 6}{10^2} (2 - 1)^2$$ $$= \frac{58}{10} + \frac{24}{100} = \frac{580 + 24}{100} = \frac{604}{100} = 6.04$$

Question 8

Maths · Binomial Theorem · Single correct

If $$ 26\left(\frac{2^3}{3}{}^{13}C_2+\frac{2^5}{5}{}^{12}C_4+\frac{2^7}{7}{}^{12}C_6+\ldots+\frac{2^{13}}{13}{}^{12}C_{12}\right)=3^{13}-\alpha, $$ then $\alpha$ is equal to:

  1. 45
  2. 48
  3. 51
  4. 54

Answer: (c)

Question 9

Maths · Permutations and Combinations · Single correct

A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is

  1. 18
  2. 36
  3. 39
  4. 72

Answer: (b)

Solution

Given $B_1$, $B_2$, $B_3$ with values 1, 1, 2. The number of ways is calculated as follows: $$\frac{4!}{2!2!} \times 3! = 6 \times 6 = 36.$$

Question 10

Maths · Straight Lines and Pair of Straight Lines · Single correct

If a straight line drawn through the point of intersection of the lines $4x + 3y - 1 = 0$ and $3x + 4y - 1 = 0$, meet the co-ordinate axes at the points $P$ and $Q$, then the locus of the mid point of $PQ$ is:

  1. $x + y - 7 = 0$
  2. $x + y - 14xy = 0$
  3. $2x + y + 14xy = 0$
  4. $x + 2y - 14xy = 0$

Answer: (b)

Solution

Point of intersection of the lines $3x + 4y = 1$ and $4x + 3y = 1$ is $\left( \frac{1}{7}, \frac{1}{7} \right)$. Let the line be $\frac{x}{2h} + \frac{y}{2k} = 1$. Satisfy $\left( \frac{1}{7}, \frac{1}{7} \right)$. $$\frac{1}{14h} + \frac{1}{14k} = 1$$ $$\frac{1}{x} + \frac{1}{y} = 14$$

Question 11

Maths · Conic Sections · Single correct

Let O be the vertex of the parabola $y^2 = 4x$ and its chords OP and OQ are perpendicular to each other. If the locus of the mid-point of the line segment PQ is a conic C, then the length of its latus rectum is:

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

Answer: (b)

Solution

Given $t_1 t_2 = -4$. $2h = t_1^2 + t_2^2$ $2k = 2(t_1 + t_2)$ $2h = k^2 - 2t_1 t_2$ $2h = k^2 + 8$ $k^2 = 2h - 8$ $= 2(h - 4)$ L.R. = 2

Question 12

Maths · Inverse Trigonometric Functions · Single correct

Let $\alpha = 3 \sin^{-1} \left( \frac{6}{11} \right)$ and $\beta = 3 \cos^{-1} \left( \frac{4}{9} \right)$, where inverse trigonometric functions take only the principal values. Given below are two statements: Statement I : $\cos (\alpha + \beta) > 0$ Statement II : $\cos (\alpha) < 0$ 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 $\frac{1}{2} 0$$

Question 13

Maths · Continuity and Differentiability · Single correct

For the function $f(x) = e^{\sin |x|} - |x|$, $x \in \mathbb{R}$, consider the following statements: Statement I: $f$ is differentiable for all $x \in \mathbb{R}$. Statement II: $f$ is increasing in $\left(-\pi, -\frac{\pi}{2}\right)$. 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

Given $f(x) = e^{|\sin x|} - |x|$, it is non-differentiable at $x = \pi$. For $x \in \left(-\pi, -\frac{\pi}{2}\right)$, we have $f(x) = e^{-\sin x} + x$. The derivative is $f'(x) = -e^{-\sin x} \cos x + 1$. For $x \in \left(-\pi, -\frac{\pi}{2}\right)$, since $\cos x 0$. Therefore, $f(x)$ is increasing.

Question 14

Maths · Vector Algebra · Single correct

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$ and a vector $\vec{c}$ be such that $2 (\vec{a} \times \vec{b}) + 3 (\vec{b} \times \vec{c}) = \vec{0}$. If $\vec{a} \cdot \vec{c} = 15$, then $\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})$ is equal to :

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

Answer: (b)

Solution

Given $2(\vec{a} \times \vec{b}) - 3(\vec{c} \times \vec{b}) = \vec{0}$. $$(2\vec{a} - 3\vec{c}) \times \vec{b} = \vec{0}$$ $$\vec{b} \parallel (2\vec{a} - 3\vec{c})$$ $$2\vec{a} - 3\vec{c} = \lambda \vec{b}$$ $$\vec{c} = \frac{2\vec{a} - \lambda \vec{b}}{3} \ldots (1)$$ $\vec{a} \cdot \vec{c} = 15$ $$\left( \frac{2\vec{a} - \lambda \vec{b}}{3} \right) \cdot \vec{a} = 15$$ $$\frac{2|\vec{a}|^2 - \lambda \vec{b} \cdot \vec{a}}{3} = 15$$ $$2(26) - \lambda (40 - 2 - 3) = 45$$ $$\lambda = \frac{1}{5}$$ Therefore, $$\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k}) = \frac{\left( 2(4\hat{i} - \hat{j} + 3\hat{k}) - \frac{1}{5}(10\hat{i} + 2\hat{j} - \hat{k}) \right) \cdot (\hat{i} + \hat{j} - 3\hat{k})}{3}$$ $$= \frac{2(4 - 1 - 9) - \frac{1}{5}(10 + 2 + 3)}{3}$$ $$= -5$$

Question 15

Maths · Three Dimensional Geometry · Single correct

Let the foot of perpendicular from the point $(\lambda, 2, 3)$ on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}$ is equal to

  1. $\frac{12}{7}$
  2. $\frac{\sqrt{145}}{7}$
  3. $\frac{\sqrt{146}}{7}$
  4. $\frac{\sqrt{143}}{7}$

Answer: (c)

Solution

Given points $P(\lambda, 2, 3)$ and $Q(1, \mu, 2)$, line $L$ is given by the equation $$\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}.$$ Point $Q(1, \mu, 2)$ satisfies the line $L$. Therefore, $$\frac{1-4}{1} = \frac{\mu-9}{2} = \frac{2-5}{1}.$$ This implies $$-3 = \frac{\mu-9}{2} = -3 \Rightarrow \mu = 3.$$ The direction ratios of $PQ$ are $\lambda - 1$, $2 - \mu$, $1$. Since $PQ \perp L$, we have $$(\lambda - 1) \cdot 1 + 2(2 - \mu) + 1 \cdot 1 = 0.$$ This simplifies to $$\lambda - 2\mu + 4 = 0 \Rightarrow \lambda = 2.$$ Now the lines are $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6} and \frac{x-2}{2} = \frac{y-3}{3} = \frac{z+5}{6}.$$ In vector form, $$\mathbf{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + t_1 (2\hat{i} + 3\hat{j} + 6\hat{k})$$ and $$\mathbf{r} = (2\hat{i} + 3\hat{j} - 5\hat{k}) + t_2 (2\hat{i} + 3\hat{j} + 6\hat{k}).$$ The distance is given by $$\frac{|((\hat{i} + 2\hat{j} - 4\hat{k}) - (2\hat{i} + 3\hat{j} - 5\hat{k})) \times (2\hat{i} + 3\hat{j} + 6\hat{k})|}{|2\hat{i} + 3\hat{j} + 6\hat{k}|} = \frac{| -9\hat{i} + 8\hat{j} - \hat{k} |}{7} = \frac{\sqrt{146}}{7}.$$

Question 16

Maths · Integrals · Single correct

The value of the integral $$\int_{0}^{2} \frac{\sqrt{x\left(x^2 + x + 1\right)}}{\left(\sqrt{x+1}\right)\left(\sqrt{x^4 + x^2 + 1}\right)} \, dx$$ is equal to:

  1. $\frac{1}{3} \log_e \left(3 - 2\sqrt{2}\right)$
  2. $\frac{2}{3} \log_e \left(4 + \sqrt{2}\right)$
  3. $\frac{2}{3} \log_e \left(3 + 2\sqrt{2}\right)$
  4. $\frac{1}{3} \log_e \left(1 + 6\sqrt{2}\right)$

Answer: (c)

Solution

The integral is given by $$I = \int_0^2 \frac{x(x^2 + x + 1)}{\sqrt{(x+1)(x^2-x+1)(x^2+x+1)}} \, dx$$ which simplifies to $$= \int_0^2 \sqrt{\frac{x}{x^3 + 1}} \, dx$$ Let $x^{3/2} = t$, then $\frac{3}{2} x^{1/2} \, dx = dt$. The integral becomes $$= \frac{2}{3} \int_0^{2^{3/2}} \frac{dt}{\sqrt{t^2 + 1}}$$ Evaluating the integral, we have $$= \frac{2}{3} \left( \ln(t + \sqrt{t^2 + 1}) \right)_0^{2^{3/2}}$$ which results in $$= \frac{2}{3} \ln(2^{3/2} + 3)$$

Question 17

Maths · Differential Equations · Single correct

Let $y = y(x)$ be the solution of the differential equation $$x\sqrt{1-x^2} \, dy + \left( y\sqrt{1-x^2} - x\cos^{-1}x \right) \, dx = 0,$$ $x \in (0, 1)$, $\lim_{x \to 1^-} y(x) = 1$. Then $y\left( \frac{1}{2} \right)$ equals:

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

Answer: (a)

Solution

Given $x \sqrt{1-x^2} \, dy + \left( y \sqrt{1-x^2} - x \cos^{-1} x \right) dx = 0$. $$\Rightarrow \frac{dy}{dx} + \frac{y}{x} = \frac{\cos^{-1} x}{\sqrt{1-x^2}}$$ It is L.D.E. Therefore, I.F. $= e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x$. Thus, the solution is $y.x = \int \frac{x \cos^{-1} x}{\sqrt{1-x^2}} \, dx + C$. $y.x = I_1 + C$ ....(1) $I_1 = \int \frac{x \cos^{-1} x}{\sqrt{1-x^2}} \, dx$. Let $\cos^{-1} x = t \Rightarrow x = \cos t$. $$\frac{-1}{\sqrt{1-x^2}} \, dx = dt$$ $I_1 = \int -t \cdot \cot t \, dt$. $$= -(t \sin t + \cos t)$$ $I_1 = -\left( \sqrt{1-x^2} \cdot \cos^{-1} x + x \right)$ ....(2) Therefore, $y.x = -\left( \sqrt{1-x^2} \cdot \cos^{-1} x + x \right) + C$. Thus, $\lim_{x \to 1^-} y(x) = 1$. Therefore, $1 = -(0 + 1) + C$ $$\Rightarrow C = 2$$ Thus, $y.x = -\sqrt{1-x^2} \cdot \cos^{-1} x - x + 2$. $$\left( put x = \frac{1}{2} \right)$$ $$y \left( \frac{1}{2} \right) \times \frac{1}{2} = -\frac{\sqrt{3}}{2} \times \frac{\pi}{3} - \frac{1}{2} + 2$$ $$\Rightarrow y \left( \frac{1}{2} \right) = 3 - \frac{\pi}{\sqrt{3}}$$

Question 18

Maths · Applications of Integrals · Single correct

Let $f : (1, \infty) \to \mathbb{R}$ be function defined as $f(x) = \frac{x-1}{x+1}$. Let $f^{i+1}(x) = f\left(f^i(x)\right)$, $i = 1, 2, \ldots, 25$, where $f^1(x) = f(x)$. If $g(x) + f^{26}(x) = 0$, $x \in (1, \infty)$, then the area of the region bounded by the curves $y = g(x)$, $2y = 2x - 3$, $y = 0$ and $x = 4$ is:

  1. $\frac{1}{8} + \log_e 2$
  2. $\frac{1}{4} + \log_e 2$
  3. $\frac{5}{6} + 3 \log_e 2$
  4. $\frac{5}{6} + \log_e 2$

Answer: (a)

Solution

Given $f^2(x) = f(f(x)) = \frac{x-1}{x+1} - 1 = \frac{x-1}{x+1} + 1 = \frac{-1}{x}$. $f^3(x) = f(f^2(x)) = f\left(\frac{-1}{x}\right) = \frac{1+x}{1-x}$. $f^4(x) = f(f^3(x)) = f\left(f\left(\frac{-1}{x}\right)\right) = x$. $f^5(x) = f(f^4(x)) = f(x)$. $f^6(x) = f(f^5(x)) = f(f(x)) = \frac{-1}{x}$. $$\cdots$$ $f^{26}(x) = \frac{-1}{x}$. $g(x) = \frac{1}{x}$. From the graph, $x - 3/2 = 1/x \Rightarrow 2x^2 - 3x - 2 = 0 \Rightarrow x = -1/2, 2$. Required area = $\int_2^4 \frac{1}{x} \, dx + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$. = $\frac{1}{8} + (\ln x)_2^4 = \frac{1}{8} + \ln 2$.

Question 19

Maths · Continuity and Differentiability · Single correct

Let $f(x) = \begin{cases} \frac{1}{3}, & x \leq \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x)^2}, & x > \frac{\pi}{2} \end{cases}$, If $f$ is continuous at $x = \frac{\pi}{2}$, then the value of $$\int_{0}^{3b-6} \left| x^2 + 2x - 3 \right| \, dx$$ is:

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

Answer: (d)

Solution

Given the limit $$\lim_{x \to \frac{\pi}{2}} \frac{b(1 - \cos(\frac{\pi}{2} - x))}{4\left(\frac{\pi}{2} - x\right)^2} = \frac{1}{3} \implies b = \frac{8}{3}$$ Evaluate the integral $$\int_0^2 |x^2 + 2x - 3| \, dx = \int_0^2 |(x + 3)(x - 1)| \, dx$$ Let $$x - 1 = t$$ Then, $$\int_{-1}^1 |(t + 4)t| \, dt = \int_{-1}^0 (-(t^2 - 4t)) \, dt + \int_0^1 (t^2 + 4t) \, dt$$ Evaluating the integrals, $$\left(\frac{-t^3}{3} - 2t^2\right)\bigg|_{-1}^0 + \left(\frac{t^3}{3} + 2t^2\right)\bigg|_0^1 = 4$$

Question 20

Maths · Conic Sections · Single correct

Let $\frac{x^2}{f(a^2 + 7a + 3)}$ + $\frac{y^2}{f(3a + 15)}$ = 1 represent an ellipse with major axis along y-axis, where f is a strictly decreasing positive function on $\mathbb{R}$. If the set of all possible values of a is $\mathbb{R}$ - [$\alpha$, $\beta$], then $\alpha^2$ + $\beta^2$ is equal to:

  1. 28
  2. 40
  3. 61
  4. 24

Answer: (b)

Solution

Given ellipse is vertical. Therefore, $f(3a + 15) > f(a^2 + 7a + 3)$. Since $f(x)$ is decreasing for all $x \in \mathbb{R}$, we have: $$3a + 15 0$$ $$(a + 2)^2 > 16$$ $$|(a + 2)| > 4$$ Thus, $a > 2$ or $a < -6$. Therefore, $a \in \mathbb{R} - [-6, 2]$. Hence, $\alpha = -6$, $\beta = 2$. Therefore, $\alpha^2 + \beta^2 = 36 + 4 = 40$.

Question 21

Maths · Complex Numbers and Quadratic Equations · Numerical

The sum of squares of all real solution of the equation $$\log_{(x+1)}(2x^2 + 5x + 3) = 4 - \log_{(2x+3)}(x^2 + 2x + 1)$$ is equal to $\ldots$.

Answer: 2

Solution

Given $\log_{(x+1)}(2x+3)(x+1) + \log_{(2x+3)}(x+1)^2 = 4$. $1 + \log_{(x+1)}(2x+3) + 2\log_{(2x+3)}(x+1) = 4$. Let $\log_{x+1}(2x+3) = t$. Then, $$t + \frac{2}{t} = 3$$ $$t^2 - 3t + 2 = 0 \implies t = 1, 2$$ For $t = 1 \implies \log_{(x+1)}(2x+3) = 1 \implies 2x+3 = x+1$ $$\implies x = -2 (rejected)$$ For $t = 2 \implies 2x+3 = x^2 + 2x + 1 \implies x^2 = 2$ $$\implies x = \pm \sqrt{2}$$ Rejecting $x = -\sqrt{2}$, we get $x = \sqrt{2}$. Therefore, the sum of squares of all the roots = 2.

Question 22

Maths · Integrals · Numerical

If $$\int_{\pi/6}^{\pi/4} \left( \cot\left(x - \frac{\pi}{3}\right) \cot\left(x + \frac{\pi}{3}\right) + 1 \right) \, dx = \alpha \log_e\left(\sqrt{3} - 1\right),$$ then $9\alpha^2$ is equal to $\ldots$.

Answer: 12

Solution

Given $$\int_{\pi/6}^{\pi/4} \left( \cot(x - \frac{\pi}{3}) \cot(x + \frac{\pi}{3}) + 1 \right) \, dx$$ using $\cot(A - B) = \frac{\cot B \cot A + 1}{\cot B - \cot A}$ $$\Rightarrow \int_{\pi/6}^{\pi/4} -\frac{1}{\sqrt{3}} \left[ \cot(x - \frac{\pi}{3}) - \cot(x + \frac{\pi}{3}) \right] \, dx$$ $$\Rightarrow -\frac{1}{\sqrt{3}} \left[ \ln \left| \frac{\sin(x - \frac{\pi}{3})}{\sin(x + \frac{\pi}{3})} \right| \right]_{\pi/6}^{\pi/4}$$ $$= -\frac{1}{\sqrt{3}} \left[ \ln \left| \frac{\sin 15^\circ}{\sin 105^\circ} \right| - \ln \left| \frac{\sin 30^\circ}{\sin 90^\circ} \right| \right]$$ $$= -\frac{1}{\sqrt{3}} \left[ \ln(\tan 15^\circ) - \ln \left( \frac{1}{2} \right) \right]$$ $$= -\frac{1}{\sqrt{3}} \left[ \ln(2 - \sqrt{3}) + \ln 2 \right]$$ $$\Rightarrow -\frac{1}{\sqrt{3}} \ln(4 - 2\sqrt{3}) = -\frac{2}{\sqrt{3}} \ln(\sqrt{3} - 1)$$ Therefore, $$\alpha = -\frac{2}{\sqrt{3}} \therefore 9\alpha^2 = 12$$

Question 23

Maths · Three Dimensional Geometry · Numerical

Let a line $L_1$ pass through the origin and be perpendicular to the lines $$L_2 : \vec{r} = (3 + t) \hat{i} + (2t - 1) \hat{j} + (2t + 4) \hat{k}$$ and $$L_3 : \vec{r} = (3 + 2s) \hat{i} + (3 + 2s) \hat{j} + (2 + s) \hat{k},$$ $t, s \in \mathbb{R}$. If $(a, b, c), a \in \mathbb{Z},$ is the point on $L_3$ at a distance of $\sqrt{17}$ from the point of intersection of $L_1$ and $L_2$, then $(a + b + c)^2$ is equal to $\ldots$.

Answer: 4

Solution

Given the lines $L_2 : \frac{x-3}{1} = \frac{y+1}{2} = \frac{z-4}{2} = t$ and $L_5 : \frac{x-3}{2} = \frac{y-3}{2} = \frac{z-2}{1} = s$. The vector perpendicular to $L_2$ and $L_3$ is given by the determinant: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(-2) - \hat{j}(-3) + \hat{k}(-2)$$ which simplifies to $$-2\hat{i} + 3\hat{j} - 2\hat{k}$$ and further to $$-(2\hat{i} - 3\hat{j} + 2\hat{k}).$$ Now, consider $L_1 : \frac{x}{2} = \frac{y}{-3} = \frac{z}{2} = \ell$. The intersection of $L_1$ and $L_2$ gives: $$t + 3 = 2\ell, \; 2t - 1 = -3\ell, \; 2t + 4 = 2\ell.$$ Solving $t + 3 = 2t + 4$ gives $t = -1$. Therefore, $P = (2, -3, 2)$. A point on $L_3$ is $Q(2s + 3, 2s + 3, s + 2)$. The distance $PQ^2 = (2s + 1)^2 + (2s + 6)^2 + s^2 = 17$. Solving gives: $$9s^2 + 28s + 20 = 0$$ which factors to $$(9s + 10)(s + 2) = 0$$ giving $s = -2$. Thus, $Q = (-1, -1, 0) \equiv (a, b, c)$ and $(a + b + c)^2 = 4$.

Question 24

Maths · Conic Sections · Numerical

Consider the circle $C : x^2 + y^2 - 6x - 8y - 11 = 0$. Let a variable chord $AB$ of the circle $C$ subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord $AB$ is the circle $x^2 + y^2 - \alpha x - \beta y - \gamma = 0$, then $\alpha + \beta + 2\gamma$ is equal to $\ldots$.

Answer: 18

Solution

The equation of the circle is given by $C : x^2 + y^2 - 6x - 8y - 11 = 0$. The equation of chord $AB$ is given by $$y - k = -\frac{h}{k}(x - h)$$ which simplifies to $$hx + ky = h^2 + k^2 ....(2)$$ Homogenising the equation of the circle, we have $$x^2 + y^2 - 6x \left( \frac{hx + ky}{h^2 + k^2} \right) - 8y \left( \frac{hx + ky}{h^2 + k^2} \right) - 11 \left( \frac{hx + ky}{h^2 + k^2} \right)^2 = 0$$ The coefficient of $x^2$ plus the coefficient of $y^2$ equals zero: $$1 - \frac{6h}{h^2 + k^2} - 11 \cdot \frac{h^2}{(h^2 + k^2)^2} + 1 - \frac{8k}{h^2 + k^2} - 11 \cdot \frac{k^2}{(h^2 + k^2)^2} = 0$$ Simplifying, we get $$2(h^2 + k^2) - 6h - 8k - 11 = 0$$ which further simplifies to $$x^2 + y^2 - 3x - 4y - \frac{11}{2} = 0$$ Therefore, $\alpha + \beta + 2\gamma = 3 + 4 + 11 = 18$.

Question 25

Maths · Sequences and Series · Numerical

Let $f$ be a polynomial function such that $$ \log_2(f(x))=\log_2\left(2+\frac{2}{3}+\frac{2}{9}+\ldots\right). $$ Then $$ \log_3\left(1+\frac{f(x)}{f\left(\frac{1}{x}\right)}\right),\qquad x>0 $$ and $f(6)=37$. Then $$ \sum_{n=1}^{10}f(n) $$ is equal to $\ldots$.

Answer: 395

Solution

Given $\log_2 f(x) = \log_2 \left( \frac{2}{1 - \frac{1}{3}} \right) \cdot \log_3 \left( 1 + \frac{f(x)}{f\left( \frac{1}{x} \right)} \right)$. This simplifies to $\log_2 f(x) = \log_2 3 \cdot \log_3 \left( 1 + \frac{f(x)}{f\left( \frac{1}{x} \right)} \right)$. Therefore, $f(x) = 1 + \frac{f(x)}{f\left( \frac{1}{x} \right)}$. This implies $f(x) f\left( \frac{1}{x} \right) = f(x) + f\left( \frac{1}{x} \right)$. Thus, $f(x) = 1 \pm x^n$. Given $f(6) = 37$, we have $1 \pm 6^n = 37$. This gives $6^n = 36$, so $n = 2$. Therefore, $f(x) = 1 + x^2$. Finally, $\sum_{n=1}^{10} (1 + n^2) = 10 + \frac{10 \cdot 11 \cdot 21}{6} = 395$.

Physics

Question 26

Physics · Physical World, Units and Measurements · Single correct

A new unit ($\alpha$) of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of $\alpha$ units if light takes 6 min. 40 s to cover this distance?

  1. 200 $\alpha$
  2. 400 $\alpha$
  3. 300 $\alpha$
  4. 500 $\alpha$

Answer: (b)

Solution

Given $\alpha = 3 \times 10^8 \, \mathrm{m}$. Distance $= (400) \times 3 \times 10^8 \, \mathrm{m} = 400 \alpha$.

Question 27

Physics · Physical World, Units and Measurements · Single correct

Consider the equation $H = \frac{x^p \epsilon^q E^r}{t^s}$ Where $H =$ magnetic field; $E =$ electric field; $\epsilon =$ permittivity, $x =$ distance, $t =$ time. The values of $p$, $q$, $r$ and $s$ respectively are:

  1. 1, 1, 1, 1
  2. -1, 1, 2, 1
  3. 1, -1, -2, 1
  4. -1, -2, -2, 1

Answer: (a)

Solution

As per the question, considering H as magnetic field $$[H] = \frac{MLT^{-2}}{LT^{-1}} = MI^{-1}T^{-2}$$ $$[E] = \frac{MLT^{-2}}{I} = MLI^{-1}T^{-3}$$ $$[\epsilon] = \frac{I^2T^2}{MLT^{-2}L^2} = M^{-1}L^{-3}I^2T^4$$ $$MI^{-1}T^{-2} = M^{-q+r}L^{p-3q+r}I^{2q-r}T^{4q-3r-S}$$ $$-q + r = 1$$ $$p - 3q + r = 0$$ $$2q - r = -1$$ $$4q - 3r - S = -2$$ Solving we get $$q = 0, \ r = 1$$ $$p = -1, \ S = -1$$ No option matching Considering H as magnetic field intensity $$H = \frac{B}{\mu_0} = Ni \Rightarrow current per unit length \Rightarrow IL^{-1}$$ $$-q + r = 0 q = 1$$ $$p - 3q + r = -1 r = 1$$ $$2q - r = 1 p = +1$$ $$4q - 3r - s = 0 s = +1$$ Option (1) is matching Note: As per the question if we consider H as magnetic field, no option is matching but if we consider H as magnetic field intensity then option (1) is matching.

Question 28

Physics · Laws of Motion · Single correct

A car moving with a speed of $54\,\mathrm{km/h}$ takes a turn of radius $20\,\mathrm{m}$. A simple pendulum is suspended from the ceiling of the car. Determine the angle made by the string of the pendulum with the vertical during the turning. (Take g = $10\,\mathrm{m/s^2}$)

  1. $\tan^{-1}$(0.5)
  2. $\tan^{-1}$(0.75)
  3. $\tan^{-1}$(1.125)
  4. $\tan^{-1}$(0.25)

Answer: (c)

Solution

Given $\tan \theta = \frac{v^2}{rg}$. Calculating $v$: $$v = 54 \times \frac{5}{18} = 15 \, \mathrm{m/s}$$ Substitute the values: $$\tan \theta = \frac{225}{20 \times 10}$$ Simplify: $$\tan \theta = \frac{9}{8}$$ Therefore, $$\theta = \tan^{-1}(1.125)$$

Question 29

Physics · Motion in a Straight Line · Single correct

A gas balloon is going up with a constant velocity of $10\,\mathrm{m/s}$. When this balloon reached a height of $75\,\mathrm{m}$, a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is $\ldots \mathrm{m}$. (Take g = $10\,\mathrm{m/s^2}$)

  1. 85
  2. 150
  3. 129
  4. 125

Answer: (d)

Solution

Given $U = 10 \, \mathrm{m/s}$ and $H = 75 \, \mathrm{m}$. The height $h$ is given by $$h = \frac{u^2}{2g} \Rightarrow \frac{100}{2 \times 10} = 5 \, \mathrm{m}.$$ Using the equation of motion: $$-75 = 10 \times t - \frac{1}{2} \times 10 \times t^2$$ which simplifies to $$t^2 - 2t - 15 = 0.$$ Factoring gives $$(t - 5)(t + 3) = 0 \Rightarrow t = 5.$$ The height is calculated as $$Height = 75 + 10 \times 5 = 125 \, \mathrm{m}.$$

Question 30

Physics · Ray Optics and Optical Instruments · Single correct

A thin biconvex lens is prepared from the glass ($\mu = 1.5$) both curved surfaces of which have equal radii of 20 cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed, from the lens at a distance of $\ldots$ cm.

  1. 10
  2. 12.5
  3. 13
  4. 13.5

Answer: (a)

Solution

Combination of lens + mirror $f_m = \frac{R}{2} = -10 \, \mathrm{cm}$; $f_L = \frac{R}{2(\mu - 1)} = 20 \, \mathrm{cm}$ $$\frac{1}{f} = \frac{1}{f_m} - \frac{2}{f_L}$$ $$\frac{1}{f} = \frac{-1}{10} - \frac{2}{20} = -\frac{1}{5}$$ $f = -5 \, \mathrm{cm}$ To form image at the position of object, it should be placed at center of curvature of mirror + lens combination. $\therefore \; u = 2f = 2 \times 5 = 10 \, \mathrm{cm}$

Question 31

Physics · Motion in a Plane · Single correct

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range $R$. If the time of flight of these bodies are $5 \, \mathrm{s}$ and $10 \, \mathrm{s}$, respectively, then the value of $R$ is $\ldots$ m. (Take $g = 10 \, \mathrm{m/s^2}$)

  1. 250
  2. 25
  3. 500
  4. 125

Answer: (a)

Solution

For range to be same, angle of projection must be complementary angles. $$T_1 = \frac{2u \sin \theta}{g} = 10, \ T_2 = \frac{2u \cos \theta}{g} = 5$$ Range $$= \frac{2(u \sin \theta)(u \cos \theta)}{g}$$ $$= \frac{2(50)(25)}{10} \, \mathrm{m} = 250 \, \mathrm{m}$$

Question 32

Physics · System of Particles and Rotational Motion · Single correct

A solid cylinder having radius $R$ and length $L$ is slipping on a rough horizontal plane. At time $t = 0$ the cylinder has a translational velocity $v_0 = 49 \, \mathrm{m/s}$, perpendicular to its axis and a rotational velocity $v_0 / 4R$ about the centre. The time taken by the cylinder to start rolling is $\ldots$ seconds. (coefficient of kinetic friction $\mu_k = 0.25$ and $g = 9.8 \, \mathrm{m/s^2}$)

  1. 15
  2. 5
  3. 10
  4. 7.5

Answer: (b)

Solution

Given $a = \frac{\mu_k mg}{m} = \mu_k g$. $\tau = I \alpha \Rightarrow \mu_k mg R = \frac{MR^2}{2} \times \alpha$ $$\Rightarrow \alpha = \frac{2 \mu_k g}{R}$$ Therefore, $v = v_0 - \mu_k gt$ (i) Also, $\omega = \omega_0 + \frac{2 \mu_k gt}{R}$ (ii) $$\frac{v}{R} = \frac{v_0}{4R} + \frac{2 \mu_k gt}{R}$$ Solving Eq. (i) and (ii) $$\frac{v_0}{4} + 2 \mu_k gt = v_0 - \mu_k gt$$ $$3 \mu_k gt = v_0 - \frac{v_0}{4}$$ $$3 \mu_k gt = \frac{3v_0}{4}$$ $$\Rightarrow t = \frac{v_0}{4 \mu_k g} = \frac{49}{4 \times \frac{1}{4} \times 9.8} = 5 sec.$$

Question 33

Physics · Mechanical Properties of Fluids · Single correct

A liquid of density $600 \, \mathrm{kg/m^3}$ flowing steadily in a tube of varying cross-section. The cross-section at a point A is $1.0 \, \mathrm{cm^2}$ and that at B is $20 \, \mathrm{mm^2}$. Both the points A and B are in same horizontal plane, the speed of the liquid at A is $10 \, \mathrm{cm/s}$. The difference is pressures at A and B points is $\ldots$ Pa.

  1. 18
  2. 144
  3. 36
  4. 72

Answer: (d)

Solution

Using continuity equation, $$A_1 V_1 = A_2 V_2$$ $$1 \times 10 = \frac{1}{5} \times V_2$$ $$\Rightarrow V_2 = 50 \, \mathrm{cm/s}$$ Using Bernoulli's theorem, $$P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2$$ $$P_1 - P_2 = \frac{1}{2} \rho \left[ V_2^2 - V_1^2 \right]$$ $$= \frac{1}{2} \times 600 \left[ (50)^2 - (10)^2 \right] \times 10^{-4} = 72 \, \mathrm{Pa}.$$

Question 34

Physics · Mechanical Properties of Fluids · Single correct

A spherical liquid drop of radius $R$ acquires the terminal velocity $v_1$ when falls through a gas of viscosity $\eta$. Now the drop is broken into 64 identical droplets and each droplet acquires terminal velocity $v_2$ falling through the same gas. The ratio of terminal velocities $v_1/v_2$ is $\ldots$.

  1. 4
  2. 0.25
  3. 32
  4. 16

Answer: (d)

Solution

Using volume conservation, $$\frac{4}{3} \pi R_1^3 = 64 \times \frac{4}{3} \pi R_2^3 \Rightarrow R_2 = \frac{R_1}{4}$$ Terminal velocity $v \propto R^2$ $$\frac{v_1}{v_2} \Rightarrow \frac{R_1^2}{R_2^2} \Rightarrow (4)^2 = 16$$

Question 35

Physics · Thermodynamics · Single correct

One mole of diatomic gas having rotational modes only is kept is a cylinder with a piston system. The cross-section area of the cylinder is $4 \, \mathrm{cm}^2$. The gas is heated slowly to raise the temperature by $1.2 \, ^\circ \mathrm{C}$ during which the piston moves by $25 \, \mathrm{mm}$. The amount of heat supplied to the gas is $\ldots$ J. (Atmospheric pressure $= 100 \, \mathrm{kPa}$, $R = 8.3 \, \mathrm{J/mol.K}$) (Neglect mass of the piston)

  1. 24.8
  2. 25
  3. 15.04
  4. 29.98

Answer: (b)

Solution

1st Method Q = $\Delta$ U + W $$Q = \frac{f}{2} nR \Delta T + nR \Delta T = \left( \frac{f}{2} + 1 \right) nR \Delta T$$ If $f = 2$ then $Q = 2nR \Delta T \Rightarrow 2 \times 1 \times 8.3 \times 1.2 = 19.92$ (No option is matching) If $f = 5$ then $Q = 3.5nR \Delta T$ $$\Rightarrow 3.5 \times 1 \times 8.3 \times 1.2 = 34.86$$ (No option is matching) Method-2 Q = $\Delta$ U + W $$Q = \frac{f}{2} nR \Delta T + P \Delta V$$ If $f = 2$ $$\Rightarrow Q = nR \Delta T + P \Delta V$$ $$= 1 \times 8.3 \times 1.2 + 10^5 \times 25 \times 10^{-3} \times 4 \times 10^{-4}$$ $$= 10.96$$ (No option is matching) If $f = 5$ $$\Rightarrow Q = 2.5nR \Delta T + P \Delta V$$ $$2.5 \times 1 \times 8.3 \times 1.2 + 10^5 \times 25 \times 10^{-3} \times 4 \times 10^{-4}$$ $$= 25.9$$ (closest is 25) Note: As $P \Delta V \neq nR \Delta T$, given state is inconsistent. Hence we can opt the most suitable option as (2).

Question 36

Physics · Thermodynamics · Single correct

Initial pressure and volume of a monoatomic ideal gas are $P$ and $V$. The change in internal energy of this gas in adiabatic expansion to volume $V_{final} = 27 \, V$ is $\ldots$ J.

  1. $-2PV\left(3\sqrt{3}-1\right)$
  2. $\frac{4}{3}PV$
  3. $-\frac{4}{3}PV$
  4. $\frac{3}{4}PV$

Answer: (c)

Solution

Given $P_1 V_1^\gamma = P_2 V_2^\gamma$. $\gamma = \frac{5}{3}$ $PV^{5/3} = P_2 \times (27V)^{5/3}$ $P_2 = \frac{P}{(27)^{5/3}} = \frac{P}{3^5}$ $\Delta U \Rightarrow nC_V \Delta T = \frac{nR \Delta T}{(\gamma - 1)}$ $= \frac{P_2 V_2 - P_1 V_1}{(\gamma - 1)}$ $= \frac{\frac{P}{3^5} \times 3^3 \cdot V - PV}{\frac{5}{3} - 1} \Rightarrow \frac{-PV \left\{ 1 - \frac{1}{9} \right\}}{\frac{2}{3}}$ $= -PV \times \frac{8}{9} \times \frac{3}{2}$ $\Rightarrow -\frac{4}{3} PV$

Question 37

Physics · Oscillations · Single correct

The frequency of oscillation of a mass $m$ suspended by spring is $\nu_1$. If the length of the spring is cut to half, the same mass oscillates with frequency $\nu_2$. The value of $\frac{\nu_2}{\nu_1}$ is $\ldots$.

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

Answer: (c)

Solution

For a spring, $k \ell = constant$. $$k_1 \ell = k_2 \frac{\ell}{2} \Rightarrow k_2 = 2k_1$$ $$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$$ $$\Rightarrow \frac{f_1}{f_2} = \sqrt{\frac{k_1}{k_2}}$$ $$\Rightarrow \frac{f_1}{f_2} = \sqrt{\frac{1}{2}} \Rightarrow \frac{f_2}{f_1} = \sqrt{2}$$

Question 38

Physics · Electromagnetic Waves · Single correct

A monochromatic source of light operating at $15\,\mathrm{kW}$ emits 2.5 $\times$ 10^{22} photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to $\ldots$ (Take h = 6.6 $\times$ 10^{-34} $\mathrm{J.s}$ and c = 3 $\times$ 10^{8} $\mathrm{m/s}$).

  1. Microwave
  2. Infrared
  3. Visible
  4. Ultraviolet

Answer: (d)

Solution

Given $\frac{hC}{\lambda} = \frac{15 \times 10^3}{2.5 \times 10^{22}}$. Therefore, $\lambda = \frac{6.6 \times 10^{-34} \times 3 \times 10^8 \times 2.5 \times 10^{22}}{15 \times 10^3}$. $\lambda = 3.3 \times 10^{-7}$.

Question 39

Physics · Moving Charges and Magnetism · Single correct

A current carrying circular loop of radius 2 cm with unit normal $\hat{n} = \frac{\hat{k} + \hat{i}}{\sqrt{2}}$ is placed in a magnetic field, $\mathbf{B} = B_0 (3\hat{i} + 2\hat{k})$. If $B_0 = 4 \times 10^{-3} \, \mathrm{T}$ and current $I = 100\sqrt{2} \, \mathrm{A}$, the torque experienced by the loop is $\ldots$ Wb.A. $(\pi = 3.14)$

  1. $16 \times 10^{-5} \hat{k}$
  2. $5024 \times 10^{-7} \hat{k}$
  3. $5024 \times 10^{-7} \hat{i}$
  4. $5024 \times 10^{-7} \hat{j}$

Answer: (d)

Solution

The torque $\vec{\tau}$ is given by $\vec{\tau} = \vec{M} \times \vec{B}$. This implies: $$\Rightarrow 100 \sqrt{2} \times 3.14 \times 4 \times 10^{-4} \left[ \left( \frac{\hat{k} + \hat{i}}{\sqrt{2}} \right) \times (3 \hat{i} + 2 \hat{k}) \right] B_0$$ Simplifying further: $$\Rightarrow 16 \times 3.14 \times [3 \hat{j} - 2 \hat{j}] \times 10^{-5}$$ Finally, we get: $$\Rightarrow 50.24 \times 10^{-5} \hat{j}$$

Question 40

Physics · Electromagnetic Induction · Single correct

A $30\,\mathrm{cm}$ long solenoid has 10 turns per cm and area of $5\,\mathrm{cm}^2$. The current through the solenoid coil varies from $2\,\mathrm{A}$ to $4\,\mathrm{A}$ is $3.14\,\mathrm{s}$. The e.m.f. induced in the coil is $\alpha \times$ 10^{-5} $\mathrm{V}$. The value $\alpha$ is $\ldots$

  1. 60
  2. 12
  3. 120
  4. 34

Answer: (b)

Solution

The electromotive force $\varepsilon$ is given by the equation $$\varepsilon = - \frac{L \, di}{dt} = - \mu_0 n^2 A \ell \frac{di}{dt}.$$ Substituting the given values, we have $$= - 4 \pi \times 10^{-7} \times 10^6 \times 5 \times 10^{-4} \times 0.3 \times \frac{2}{\pi}.$$ This simplifies to $$\Rightarrow -12 \times 10^{-5}.$$

Question 41

Physics · Electric Charges and Fields · Single correct

Two point charges $q_1 = 3 \, \mu \mathrm{C}$ and $q_2 = -4 \, \mu \mathrm{C}$ are placed at points $(2\hat{i} + 3\hat{j} + 3\hat{k})$ and $(\hat{i} + \hat{j} + \hat{k})$ respectively. Force on charge $q_2$ is $\ldots$ N. $\left( Take \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 SI Units \right)$

  1. $(12\hat{i} + 24\hat{j} + 24\hat{k}) \times 10^{-3}$
  2. $(4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3}$
  3. $(3\hat{i} + 6\hat{j} + 6\hat{k}) \times 10^{-3}$
  4. $(-4\hat{i} - 8\hat{j} - 8\hat{k}) \times 10^{-3}$

Answer: (b)

Solution

The vector $\vec{r} = \vec{r}_2 - \vec{r}_1 \Rightarrow -\hat{i} - 2\hat{j} - 2\hat{k}$. The force is given by $$Force \Rightarrow \frac{k q_1 q_2}{r^3} \vec{r}$$ $$\Rightarrow \frac{9 \times 10^9 \times (3)(+4) \times 10^{-12}}{27} \left( \hat{i} + 2\hat{j} + 2\hat{k} \right)$$ $$\Rightarrow 4 \times 10^{-3} \left[ \hat{i} + 2\hat{j} + 2\hat{k} \right]$$

Question 42

Physics · Ray Optics and Optical Instruments · Single correct

Light ray incident along a vector $\overrightarrow{AO} \left( \overrightarrow{AO} = 2\hat{i} - 3\hat{j} \right)$ emerges out along vector $\overrightarrow{OB} \left( \overrightarrow{OB} = C\hat{i} - 4\hat{j} \right)$ as shown in the figure below. The value of C is $\ldots$.

  1. 1.6
  2. 0.16
  3. 11.6
  4. 16

Answer: (a)

Solution

Given $1 (\mathbf{AO} \times \hat{\mathbf{j}}) \Rightarrow (\mathbf{OB} \times \hat{\mathbf{j}}) \times 1.5$. $$\frac{2 \hat{\mathbf{k}}}{\sqrt{13}} = \frac{3}{2} \left[ \frac{C \hat{\mathbf{k}}}{\sqrt{C^2 + 16}} \right]$$ Therefore, $$\frac{4}{13} = \frac{9C^2}{4(C^2 + 16)} \Rightarrow 16C^2 + 256 = 117C^2$$ $$256 = 101 C^2$$ $$C = 1.59 \approx 1.6$$

Question 43

Physics · Dual Nature of Radiation and Matter · Single correct

$K_1$ and $K_2$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1 = 2 \lambda_2$ then the work function of material is given by:

  1. $K_2 + 2K_1$
  2. $2K_2 - K_1$
  3. $K_1 - 2K_2$
  4. $K_2 - 2K_1$

Answer: (d)

Solution

Given $$2K_1 = \frac{2hC}{\lambda_1} - 2\phi (i)$$ $$K_2 = \frac{2hC}{\lambda_1} - \phi (ii)$$ Solving (i) and (ii), we get $$K_2 - 2K_1 = \phi$$

Question 44

Physics · Nuclei · Single correct

Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous $\alpha$-decay with same Q value of $1\,\mathrm{MeV}$. The ratio of energies of $\alpha$-rays produced by A and B is $\ldots$.

  1. $\frac{2548}{2650}$
  2. $\frac{2706}{2646}$
  3. $\frac{2597}{2600}$
  4. $\frac{2862}{2499}$

Answer: (c)

Solution

Given $k_\alpha = \left( \frac{A - 4}{A} \right) Q$. Therefore, $$\frac{k_{\alpha_1}}{k_{\alpha_2}} = \frac{196}{200} \times \frac{212}{208} = \frac{2597}{2600}.$$

Question 45

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

The output Y for the given inputs A and B to the circuit is:

Answer: (b)

Solution

Given the circuit, the expression is $A \cdot B + \overline{A} \cdot B$. This simplifies to $ (A + \overline{A}) \cdot B = B$. Note: In the lower gate, input is taken from $A$ as $\overline{A}$ due to the node given in the figure.

Question 46

Physics · Electrostatic Potential and Capacitance · Numerical

A parallel plate capacitor is having separation between plates 0.885 mm. It has a capacitance of 1 $\mu$ F when the space between the plates is filled with an insulating material of resistivity 1 $\times$ 10^{13} $\Omega$ m and resistance 17.7 $\times$ 10^{14} $\Omega$. Relative permittivity of the insulating material is $\alpha \times 10^7$. The value of $\alpha$ is $\ldots$. (Take permittivity of free space = 8.85 $\times$ 10^{-12} $\mathrm{\ F/m}$)

Answer: 2

Solution

Given $\mathrm{RC} = \rho \mathrm{K} \varepsilon_0$. $$\mathrm{K} = \frac{\mathrm{RC}}{\rho \varepsilon_0} = \frac{17.7 \times 10^{14} \times 10^{-6}}{10^{13} \times 8.85 \times 10^{-12}} = 2 \times 10^7$$

Question 47

Physics · Ray Optics and Optical Instruments · Numerical

Some distance star is to be observed by some telescope of diameter of objective lens $a$, at an angular resolution of $3.0 \times 10^{-7}$ radian. If the wavelength of light from the star reaching the telescope is $500 \, \mathrm{nm}$, the minimum diameter of the objective lens of the telescope is $\ldots$ cm. (Nearest integer)

Answer: 203

Solution

Given $\theta = \frac{1.22 \lambda}{b}$. $$3 \times 10^{-7} = \frac{1.22 \times 5 \times 10^{-7}}{b} \implies b = 203.33 \, \mathrm{cm}$$

Question 48

Physics · Moving Charges and Magnetism · Numerical

A $5\,\mathrm{mg}$ particle carrying a charge of $5 \pi \times 10^{-6} \, \mathrm{C}$ is moving with velocity of $(3 \hat{i} + 2 \hat{k}) \times 10^{-2} \, \mathrm{m/s}$ in a region having magnetic field $\mathbf{B} = 0.1 \hat{k} \, \mathrm{Wb/m^2}$. It moves a distance of $\alpha$ meter along $\hat{k}$ when it completes 5 revolutions. The value of $\alpha$ is $\ldots$

Answer: 2

Solution

Given $5 \left( \frac{2 \pi m}{qB} \right)$, we have $$\Rightarrow \frac{5 \times 2 \times \pi \times 5 \times 10^{-6} \times 2 \times 10^{-2}}{5 \pi \times 10^{-6} \times 0.1} = 2 \, \mathrm{m}$$

Question 49

Physics · Electrostatic Potential and Capacitance · Numerical

The stored charge in the capacitor in steady state of the following circuit is $\ldots \mu \mathrm{C}$.

Answer: 200

Solution

At steady state, $$I = \frac{E}{R_{eq}} = \frac{12}{6} = 2 \, A$$ $$I_1 = I_2 = 1 \, A$$ $$I_3 = I_4 = \frac{1}{2} \, A$$ $$V_{AB} = \frac{1}{2} \times 4 = 2 \, V$$ Charge on capacitor $$Q = CV_{AB} = 100 \times 2 = 200 \, \mu C$$

Question 50

Physics · Motion in a Straight Line · Numerical

Two masses of 3.4 kg and 2.5 kg are accelerated from an initial speed of 5 m/s and 12 m/s, respectively. The distances traversed by the masses in the 5th second are 104 m and 129 m, respectively. The ratio of their momenta after 10 s is $\frac{x}{8}$. The value of $x$ is $\ldots$

Answer: 9

Solution

Given $S = u + \frac{a}{2}(2n-1)$, we have $104 = 5 + \frac{a}{2}(9)$ which implies $a = 22 \, \mathrm{m/s^2}$. Next, $129 = 12 + \frac{a}{2} \times 9$ which gives $a = 26 \, \mathrm{m/s^2}$. The ratio $\frac{p_1}{p_2} = \frac{(5 + 22 \times 10) \, 3.4}{(12 + 26 \times 10)(2.5)} = \frac{225 \times (3.4)}{272 \times 2.5} = \frac{9}{8}$.

Chemistry

Question 51

Chemistry · Some Basic Concepts of Chemistry · Single correct

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

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

Answer: (c)

Solution

(A) No. of atoms in 1.8 mg water $$= \frac{1.8 \times 10^{-3}}{18} \times N_A \times 3 = 3 \times 10^{-4} N_A$$ (B) No. of atoms in 9.8 mg $\mathrm{H_2SO_4}$ $$= \frac{9.8 \times 10^{-3}}{98} \times N_A \times 7 = 7 \times 10^{-4} \times N_A$$ (C) No. of atoms in 1.8 mg carbon $$= \frac{1.8 \times 10^{-3}}{12} \times N_A = 1.5 \times 10^{-4} \times N_A$$ (D) No. of atoms in 5.85 mg NaCl $$= \frac{5.85 \times 10^{-3}}{58.5} \times N_A \times 2 = 2 \times 10^{-4} N_A$$ Correct match: A-III, B-IV, C-II, D-I

Question 52

Chemistry · Solutions · Single correct

Given below are two statements : Given : Molar mass of C, H, O, Cl are 12, 1, 16 and $35.5\,\mathrm{g \, mol^{-1}}$, respectively. Statement I : In 30% (w/w) solution of methanol in $\mathrm{CCl_4}$ (at T K), the mole fraction of $\mathrm{CCl_4}$ is equal to 0.33. Statement II : Mixture of methanol and $\mathrm{CCl_4}$ shows positive deviation deviation from Raoult’s law. In the light of the above statements, choose the correct answer from the option given below :

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

Answer: (a)

Solution

(i) Let wt of solution = $100\,\mathrm{g} \newline$ wt of methanol = $30\,\mathrm{g} \newline$ wt of $\mathrm{CCl_4}$ = 100 - 30 = $70\,\mathrm{g} \newline \newline$ mole fraction of $\mathrm{CCl_4}$ = $\frac{n_{\mathrm{CCl_4}}}{n_{\mathrm{CCl_4}} + n_{\mathrm{CH_3OH}}}$ = $\frac{\frac{70}{154}}{\frac{70}{154} + \frac{30}{32}} \newline$ = 0.3267 $\approx$ 0.33 $\newline$ (ii) $\mathrm{CCl_4} \&$ methanol is mixture of positive deviation from Raoult’s law. $\newline$ So, statement (I) $\&$ statement (II) are correct.

Question 53

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

Bromine trifluoride autoionizes to form $\mathrm{BrF_2^+}$ and $\mathrm{BrF_4^-}$. The shapes of the cation and anion are respectively $\ldots$, and $\ldots$.

  1. bent, square planar
  2. linear, square planar
  3. bent, see-saw
  4. linear, tetrahedral

Answer: (a)

Solution

BrF_2^+ $\Rightarrow$ sp^3 [Bond pair = 2, lone pair = 2] $\Rightarrow$ bent BrF_2^{$\oplus$} $\Rightarrow$ sp^3 d^2 [Bond pair = 4, lone pair = 2] $\Rightarrow$ square planar

Question 54

Chemistry · Solutions · Single correct

Which of the following are not correct? A. For water, magnitude of $K_b$ is more than the magnitude of $K_f$. B. The elevation in boiling point of water when a non-volatile solute is added to it is larger in magnitude than its depression in freezing point. C. Osmotic pressure measurement is preferred over any other colligative property to determine molar mass of proteins and polymers. D. The dimerised form of benzoic acid in benzene Choose the correct answer from the options given below:

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

Answer: (c)

Solution

(A) $K_b$ for water $= 0.512 \, ^\circ \mathrm{C} - \mathrm{Kg/mol}$ $K_f$ for water $= 1.86 \, ^\circ \mathrm{C} - \mathrm{Kg/mol}$ So, $K_b < K_f$ for water (B) Since, $K_b < K_f$ so $\Delta T_b < \Delta T_f$ [same equimolal solution] (C) Osmotic pressure is used to determine molar mass of proteins & polymers (D) Dimerised form of benzoic acid in benzene $\mathrm{C_6H_5-C \overset{O}{\underset{O}{---H---O}} \underset{C_6H_5}{C}}$ Option (A), (B) & (D) are incorrect

Question 55

Chemistry · Equilibrium · Single correct

Consider the following reactions in which all the reactants and products are present in gaseous state $$2xy \rightleftharpoons x_2 + y_2 K_1 = 2.5 \times 10^5$$ $$xy + \frac{1}{2} z_2 \rightleftharpoons xyz K_2 = 5 \times 10^{-3}$$ The value of $K_3$ for the equilibrium $$\frac{1}{2} x_2 + \frac{1}{2} y_2 + \frac{1}{2} z_2 \rightleftharpoons xyz$$ is:

  1. $2.5 \times 10^{-3}$
  2. $2.5 \times 10^3$
  3. $1.0 \times 10^{-5}$
  4. $5 \times 10^{-3}$

Answer: (c)

Solution

Given $x_2 + y_2 \rightleftharpoons 2xy$, $K' = \frac{1}{K_1} = \frac{1}{2.5 \times 10^5} = \frac{1}{25 \times 10^4}$. $$\frac{1}{2} x_2 + \frac{1}{2} y_2 \rightleftharpoons xy, K'' = \left( \frac{1}{25 \times 10^4} \right)^{1/2} = \frac{1}{5 \times 10^2} \ldots (1)$$ $xy + \frac{1}{2} z_2 \rightleftharpoons xyz, K_2 = 5 \times 10^{-3} \ldots (2)$ On adding equation (1) and (2) $$\frac{1}{2} x_2 + \frac{1}{2} y_2 + \frac{1}{2} z_2 \rightleftharpoons xyz,$$ $$K_3 = \frac{1}{5 \times 10^2} \times 5 \times 10^{-3} = 1 \times 10^{-5}$$

Question 56

Chemistry · Electrochemistry · Single correct

Given at 298 K : $E^\circ_{\mathrm{Fe}^{2+}/\mathrm{Fe}} = X$ Volt $E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}} = Y$ Volt The $E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}$ in Volt at 298 K is given by :

  1. 2X - 3Y
  2. 3Y - 2X
  3. 3Y + 2X
  4. Y + X

Answer: (b)

Solution

Sol. (i) $\mathrm{Fe^{+2}(aq) + 2e^- \rightarrow Fe(s)} \ldots \Delta$ G_1^$\circ$ (ii) $\mathrm{Fe^{+3}(aq) + 3e^- \rightarrow Fe(s)} \ldots \Delta$ G_2^$\circ$ (iii) $\mathrm{Fe^{+3}(aq) + e^- \rightarrow Fe^{+2}(aq)} \ldots \Delta$ G_3^$\circ$ $\mathrm{iii = ii - i}$ -1 $\times$ F $\times$ E^$\circ_{\mathrm{Fe^{3+}/Fe^{2+}}}$ = -3 $\times$ F $\times$ y - (-2 $\times$ F $\times$ x) E^$\circ_{\mathrm{Fe^{3+}/Fe^{2+}}}$ = 3y - 2x Option (2) is correct

Question 57

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

Given below are two statements : R = $8.314\,\mathrm{J \, K^{-1} \, mol^{-1}}$ and $1\,\mathrm{cal}$ = $4.2\,\mathrm{J}$ $\textbf{Statement I :}$ When $E_a$ = $12.6\,\mathrm{kcal/mol}$, the room temperature rate constant is doubled by a $10^\circ \mathrm{C}$ increase in temperature (298 K to 308 K) $\textbf{Statement II :}$ For a first order reaction $\mathrm{A \rightarrow B}$, Here $[\mathrm{A}]_0$ is the initial concentration of A and $t_{1/2}$ is half-life of reaction. In the light of the above statements, choose the correct answer from the option 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 are false
  4. Statement I is false but Statement II are true

Answer: (c)

Solution

S-I: $\ln \left( \frac{K_2}{K_1} \right) = \frac{12.6 \times 10^3}{\left( \frac{8.314}{4.2} \right)} \left[ \frac{10}{298 \times 308} \right]$ $\Rightarrow \ln \left( \frac{K_2}{K_1} \right) = 0.693$ $\Rightarrow \frac{K_2}{K_1} = 2$ S-II: For 1$^{st}$ order $t_{1/2} = \frac{\ln 2}{K} ; \ t_{1/2} \propto [A]^0 = constant.$ Statement-I is true but statement-II is false.

Question 58

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

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

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

Answer: (a)

Solution

(A) $2s^2 \Rightarrow \mathrm{Be}$ (B) $2s^2 2p^1 \Rightarrow \mathrm{B}$ (C) $2s^2 2p^3 \Rightarrow \mathrm{N}$ (D) $2s^2 2p^6 \Rightarrow \mathrm{Ne}$ IE order: $\mathrm{B} < \mathrm{Be} < \mathrm{N} < \mathrm{Ne}$ A-II, B-III, C-IV, D-I

Question 59

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

Find the correct statements related to group 15 hydrides. A. Reducing nature increases from $\mathrm{NH}_3$ to $\mathrm{BiH}_3$ B. Tendency to donate lone pair of electrons decreases from $\mathrm{NH}_3$ to $\mathrm{BiH}_3$ C. The stability of hydrides decreases from $\mathrm{NH}_3$ to $\mathrm{BiH}_3$ D. HEH bond angle decreases from $\mathrm{NH}_3$ to $\mathrm{SbH}_3$ (E = Elements of group 15) Choose the correct answer from the options given below:

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

Answer: (c)

Solution

(1) Reducing nature: $\mathrm{NH_3} \mathrm{PH_3} > \mathrm{AsH_3} > \mathrm{SbH_3} > \mathrm{BiH_3}$$ (3) Stability of Hydrides: $$\mathrm{NH_3} > \mathrm{PH_3} > \mathrm{AsH_3} > \mathrm{SbH_3} > \mathrm{BiH_3}$$ (4) HEH bond angle: $\mathrm{NH_3} > \mathrm{PH_3} > \mathrm{AsH_3} > \mathrm{SbH_3} > \mathrm{BiH_3}$

Question 60

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

Given below are two statements : Statement I : The number of pairs among $[\mathrm{Ti}^{4+}, \mathrm{V}^{2+}]$, $[\mathrm{V}^{2+}, \mathrm{Mn}^{2+}]$, $[\mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}]$ and $[\mathrm{V}^{2+}, \mathrm{Cr}^{2+}]$ in which both ions are coloured is 3. Statement II : The number pairs among $[\mathrm{La}^{3+}, \mathrm{Yb}^{2+}]$, $[\mathrm{Lu}^{3+}, \mathrm{Ce}^{4+}]$ and $[\mathrm{Ac}^{3+}, \mathrm{Lr}^{3+}]$ ions in which both are diamagnetic is 3. In the light of the above statements, choose the correct answer from the option given below :

  1. Both Statement I and Statement II are correct
  2. Both Statement I and Statement II are incorrect
  3. Statement I is correct but Statement II are incorrect
  4. Statement I is incorrect but Statement II are correct

Answer: (a)

Solution

Ti^{+4} is colourless due to d^0 configuration. Therefore, statement 1 is correct. La^{+3}, Yb^{+2}, Lu^{+3}, Ce^{+4}, Al^{+3} and Lr^{+3} all are diamagnetic. Therefore, statement 2 is correct.

Question 61

Chemistry · Surface Chemistry · Single correct

Given below are two statements for catalytic properties of transition metals: Statement I: First row transition metals which act as catalyst utilise their 3d electrons only for formation of bonds between reactant molecules and atoms on the surface of catalyst. Statement II: There is increase in the concentration of reactants on the surface of catalyst which strengthens the bonds in reacting molecules. In the light of the above statements, choose the correct answer from the option given below:

  1. Both Statement I and Statement II are correct
  2. Both Statement I and Statement II are incorrect
  3. Statement I is correct but Statement II are incorrect
  4. Statement I is incorrect but Statement II are correct

Answer: (b)

Solution

First row transition metals utilize 3d and 4s electrons for bonding on the catalyst surface. Therefore, Statement 1 is incorrect. Adsorption increases reactant concentration on the catalyst surface and weakens bonds facilitating formation of activated complex and lowering activation energy. Therefore, Statement 2 is incorrect.

Question 62

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

Given below are two statements: Statement I: Vapours of the liquid with higher boiling point condense before vapours of the liquid with lower boiling points in fractional distillation. Statement II: The vapours rising up in the fractionating column become richer in high boiling component of the mixture. In the light of the above statements, choose the correct answer from the option 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 are false
  4. Statement I is false but Statement II are true

Answer: (c)

Solution

Statement I: True Statement II: False Reason - As vapours rise, they become richer in the lower-boiling component, not high-boiling component.

Question 63

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

The major product of which of the following reaction is not obtained by rearrangement reaction?

Answer: (b)

Solution

The reaction proceeds via an $E_1$ mechanism. The hydroxyl group is protonated by $\mathrm{H^+}$ to form $\mathrm{OH_2^+}$. This is followed by the rate-determining step (RDS) where the leaving group departs, forming a carbocation. The product is obtained without rearrangement.

Question 64

Chemistry · Hydrocarbons · Single correct

The total number of aromatic compounds/species from the following is

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

Answer: (b)

Solution

Aromatic compounds are

Question 65

Chemistry · Haloalkanes and Haloarenes · Single correct

n-Butane on monochlorination under photochemical conditions gives an optically active compound "P". "P" on further chlorination gives dichloro compounds. The number of dichloro compounds obtained (ignore stereoisomers) is :

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

Answer: (b)

Solution

The reaction of n-butane with $\mathrm{Cl_2/h\nu}$ produces a chlorinated product that is optically active. Further chlorination with $\mathrm{Cl_2/h\nu}$ leads to the formation of dichloro products. The total number of dichloro products, excluding stereoisomers, is 4.

Question 66

Chemistry · Haloalkanes and Haloarenes · Single correct

Given below are two statements: Statement-I: Due to increase in van der Waals forces, the order of boiling points is $\mathrm{CH_3CH_2CH_2I > CH_3CH_2I > CH_3I}$. In the light of the above statement, choose the correct answer from the options given below:

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

Answer: (a)

Solution

Statement I: True Order of boiling point: $\mathrm{CH_3CH_2CH_2I > CH_3CH_2I > CH_3I}$ Vanderwaals forces increase due to increase in molecular weight. Statement 2: True Order of melting point: $$Cl Cl Cl$$ $$ $$ $$Cl Cl$$ ($\mu \neq 0$) ($\mu = 0$)

Question 67

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Consider the following reaction. The major product (P) formed is :

Answer: (c)

Solution

The reaction begins with the reduction of the amide using $\mathrm{NaBH_4/MeOH}$, resulting in the formation of an alcohol. The next step involves treatment with $\mathrm{NaOH(aq.)/\Delta}$, leading to the formation of a carboxylate ion. Finally, the carboxylate ion is protonated by $\mathrm{H_3O^+}$ to yield the final product.

Question 68

Chemistry · Amines · Single correct

Which statements are True? A. In Hoffmann bromamide degradation, $4$ moles of $\mathrm{NaOH}$ and $2$ moles of $\mathrm{Br_2}$ are consumed per mole of an amide. B. Hoffmann bromamide reaction is not given by alkyl amides. C. Primary amines can be synthesized by Hoffmann bromamide degradation. D. Secondary amide on reaction with $\mathrm{Br_2}$ and $\mathrm{NaOH}$ will give secondary amine. E. The by-products of Hoffmann degradation are $\mathrm{Na_2CO_3}$, $\mathrm{NaBr}$ and $\mathrm{H_2O}$. Choose the correct answer from the options given below:

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

Answer: (c)

Solution

Hoffmann bromamide degradation reaction: $$\mathrm{R{-}C{-}NH_2 + 4NaOH + Br_2 \rightarrow R{-}NH_2 + Na_2CO_3 + 2NaBr + 2H_2O}$$ 1° Amide to 1° Amine Statement (A) – False Statement (B) – False Statement (C) – True Statement (D) – False Statement (E) – True

Question 69

Chemistry · Biomolecules · Single correct

The $\textbf{incorrect}$ statement from the following with respect to carbohydrates is:

  1. All monosaccharides are reducing sugars.
  2. The monosaccharide units obtained from hydrolysis of oligosaccharides are always the same.
  3. Starch and cellulose are typical examples of polysaccharides, which are very high molecular weight compounds of more than ten monosaccharide units.
  4. Open chain and cyclic structures co-exist at equilibrium that are responsible for certain properties as in the case of D-(+)-glucose.

Answer: (b)

Solution

The monosaccharide units obtained from hydrolysis of oligosaccharides may be or may not be same. Example: Lactose $\xrightarrow{\mathrm{H_3O^+}} \underline{Glucose + Galactose}$ different monosaccharide units Maltose $\xrightarrow{\mathrm{H_3O^+}} \underline{Glucose + Glucose}$ same monosaccharide units

Question 70

Chemistry · Biomolecules · Single correct

Which of the following amino acid will give violet coloured complex with neutral ferric chloride solution?

  1. Threonine
  2. Serine
  3. Tyrosine
  4. Cysteine

Answer: (c)

Solution

Tyrosine amino acid will give violet coloured complex with neutral $\mathrm{FeCl_3}$ solution because it contains phenolic group.

Question 71

Chemistry · Co-ordination Compounds · Numerical

Number of paramagnetic complexes among the following is $\ldots$. $[\mathrm{MnBr}_4]^{2-}$, $[\mathrm{NiCl}_4]^{2-}$, $[\mathrm{Ni(CN)}_4]^{2-}$, $[\mathrm{Ni(CO)}_4]$, $[\mathrm{CoF}_6]^{3-}$, $[\mathrm{Fe(CN)}_6]^{4-}$, $[\mathrm{Mn(CN)}_6]^{3-}$, $[\mathrm{Ti(CN)}_6]^{3-}$, $[\mathrm{Cu(H_2O)}_6]^{2+}$, $[\mathrm{Co(C_2O_4)}_3]^{3-}$

Answer: 6

Solution

[$\mathrm{MnBr_4}$]^{2-}, [$\mathrm{NiCl_4}$]^{2-}, [$\mathrm{CoF_6}$]^{3-}, [$\mathrm{Mn(CN)_6}$]^{3-}, [$\mathrm{Ti(CN)_6}$]^{3-}, [$\mathrm{Cu(H_2O)_6}$]^{2+} are paramagnetic.

Question 72

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

'x' is the product which is obtained from benzene by reacting it with carbon monoxide and hydrogen chloride in the presence of cuprous chloride. 'y' is the major product obtained from the benzene by reacting it with ethanoyl chloride in the presence of anhydrous AlCl_3. Product (major) obtained by heating x and y in the presence of alkali is z. Total number of $\pi$ (pi) electrons in z is $\ldots$.

Answer: 16

Solution

The reaction starts with benzene reacting with CO/HCl in the presence of CuCl to form benzaldehyde (x) via the Gattermann Koch reaction. Next, benzaldehyde (x) undergoes Friedel-Crafts acylation with acetyl chloride in the presence of anhydrous AlCl₃ to form acetophenone (y). Acetophenone (y) then reacts with benzaldehyde (x) in the presence of aqueous NaOH to form the crossed aldol condensation product (z), which is chalcone. The total number of pi ($\pi$) electrons in 'z' is 16 $\pi$ electrons.

Question 73

Chemistry · Structure of Atom · Numerical

Consider two radiations of wavelengths: 1. $\lambda_1 = 2000 \, \AA$ 2. $\lambda_2 = 6000 \, \AA$ The ratio of the energies of these two radiations $$\left( \frac{E_1}{E_2} \right)$$ is $\ldots$. (Nearest integer)

Answer: 3

Solution

The energy of a photon is given by $E_{photon} = \frac{hc}{\lambda}$. Therefore, $$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$$ which implies $$\frac{E_1}{E_2} = \frac{6000}{2000}$$ leading to $$\frac{E_1}{E_2} = 3$$

Question 74

Chemistry · Thermodynamics · Numerical

Consider the reaction: $2\,\mathrm{H_2S}(g)+3\,\mathrm{O_2}(g)\rightarrow2\,\mathrm{H_2O}(l)+2\,\mathrm{SO_2}(g)$ The magnitude of enthalpy change for the reaction in $\mathrm{kJ\,mol^{-1}}$ is $\ldots$ (Nearest integer) Given: $\Delta_f H^\circ(\mathrm{H_2S})=-20.1\,\mathrm{kJ\,mol^{-1}}$ $\Delta_f H^\circ(\mathrm{H_2O})=-286.0\,\mathrm{kJ\,mol^{-1}}$ $\Delta_f H^\circ(\mathrm{SO_2})=-297.0\,\mathrm{kJ\,mol^{-1}}$

Answer: 1126

Solution

Given $\Delta_r H^\circ = \Sigma \Delta_f H^\circ(product) - \Sigma \Delta_f H^\circ(Reactant)$. $\Delta_r H^\circ = 2 \times \Delta_f H^\circ(SO_2, g) + 2 \times \Delta_f H^\circ(H_2O, \ell) - 2 \times \Delta_f H^\circ(H_2S, g)$. $\Delta_r H^\circ = 2 \times (-297) + 2 \times (-286) - 2 \times (-20.1)$. $= -594 - 572 + 40.2$. $\Delta_r H^\circ = -1125.8 \, kJ/mol$. $|\Delta_r H^\circ| = 1125.8 \, kJ/mol$.

Question 75

Chemistry · Equilibrium · Numerical

Solid carbon, $\mathrm{CaO}$ and $\mathrm{CaCO_3}$ are mixed and allowed to attain equilibrium at $T\,\mathrm{K}$. $\mathrm{CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)}$ $K_{p_1}=0.08\,\mathrm{atm}$ $\mathrm{C(s)+CO_2(g)\rightleftharpoons2CO(g)}$ $K_{p_2}=2\,\mathrm{atm}$ The partial pressure of $\mathrm{CO}$ is $\ldots\times10^{-1}\,\mathrm{atm}$.

Answer: 4

Solution

Given the reaction $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$, with $K_{p_1} = 0.08 \, \mathrm{atm}$. $K_{p_1} = P_{\mathrm{CO_2}} = 0.08 \, \mathrm{atm}$. For the reaction $\mathrm{C(s) + CO_2(g) \rightleftharpoons 2CO(g)}$, $K_{p_2} = 2$. $$K_{p_2} = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO_2}}} = 2$$ $$\frac{P_{\mathrm{CO}}^2}{0.08} = 2$$ $$P_{\mathrm{CO}}^2 = 16 \times 10^{-2}$$ $$P_{\mathrm{CO}} = 4 \times 10^{-1}$$