JEE Main 27 July 2021 Shift 1 question paper with solutions

JEE Main 27 July 2021 Shift 1: all 90 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 · Statistics · Single correct

If the mean and variance of the following data: 6, 10, 7, 13, a, 12, b, 12 are 9 and $\frac{37}{4}$ respectively, then $(a - b)^2$ is equal to:

  1. 24
  2. 12
  3. 32
  4. 16

Answer: (d)

Solution

Mean $\displaystyle = \frac{6+10+7+13+a+12+b+12}{8}=9$ $60+a+b=72$ $a+b=12 \qquad \ldots (1)$ variance $\displaystyle = \frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2=\frac{37}{4}$ $\sum x_i^2=6^2+10^2+7^2+13^2+a^2+b^2+12^2+12^2$ $=a^2+b^2+642$ $\displaystyle \frac{a^2+b^2+642}{8}-(9)^2=\frac{37}{4}$ $\displaystyle \frac{a^2+b^2}{8}+\frac{321}{4}-81=\frac{37}{4}$ $\displaystyle \frac{a^2+b^2}{8}=81+\frac{37}{4}-\frac{321}{4}$ $\displaystyle \frac{a^2+b^2}{8}=81-71$ $\therefore\ a^2+b^2=80 \qquad \ldots (2)$ From (1) $a^2+b^2+2ab=144$ $80+2ab=144$ $\therefore\ 2ab=64$ $(a-b)^2=a^2+b^2-2ab=80-64=16$

Question 2

Maths · Limits and Derivatives · Single correct

The value of $\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2j-1)+8n}{(2j-1)+4n}$ is equal to:

  1. 5 + $\\log_e$ ($\frac{3}{2}$)
  2. 2 - $\\log_e$ $(\frac{2}{3})$
  3. 3 + 2 $\\log_e$ $(\\frac{2}{3})$
  4. 1 + 2 $\\log_e$ $(\\frac{3}{2})$

Answer: (d)

Solution

Given $$\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{\left( \frac{2j}{n} - \frac{1}{n} + 8 \right)}{\left( \frac{2j}{n} - \frac{1}{n} + 4 \right)}$$ We have $$\int_{0}^{1} \frac{2x+8}{2x+4} \, dx = \int_{0}^{1} \, dx + \int_{0}^{1} \frac{4}{2x+4} \, dx$$ This simplifies to $$= 1 + 4 \frac{1}{2} \left( \ln |2x + 4| \right) \bigg|_{0}^{1}$$ Evaluating the integral, we get $$= 1 + 2 \ln \left( \frac{3}{2} \right)$$

Question 3

Maths · Vector Algebra · Single correct

Let $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=-\hat{i}+2\hat{j}+3\hat{k}$. Then the vector product $(\vec{a}+\vec{b})\times\left(\left(\vec{a}\times\left((\vec{a}-\vec{b})\times\vec{b}\right)\right)\times\vec{b}\right)$ is equal to $\underline{\hspace{2cm}}$.

  1. 5(34$\hat{i}$ - 5$\hat{j}$ + 3$\hat{k}$)
  2. 7(34$\hat{i}$ - 5$\hat{j}$ + 3$\hat{k}$)
  3. 7(30$\hat{i}$ - 5$\hat{j}$ + 7$\hat{k}$)
  4. 5(30$\hat{i}$ - 5$\hat{j}$ + 7$\hat{k}$)

Answer: (b)

Solution

Given $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j} + 3\hat{k}$. $\vec{a} + \vec{b} = 3\hat{j} + 5\hat{k}; \vec{a} \cdot \vec{b} = -1 + 2 + 6 = 7$ $((\vec{a} \times ((\vec{a} - \vec{b}) \times \vec{b})) \times \vec{b})$ $((\vec{a} \times (\vec{a} \times \vec{b} - \vec{b} \times \vec{b})) \times \vec{b})$ $(\vec{a} \times (\vec{a} \times \vec{b} - 0)) \times \vec{b}$ $(\vec{a} \times (\vec{a} \times \vec{b})) \times \vec{b}$ $((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) \times \vec{b}$ $((\vec{a} \cdot \vec{b})\vec{a} \times \vec{b} - (\vec{a} \cdot \vec{a})(\vec{b} \times \vec{b}))$ $(\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b})$ $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ -1 & 2 & 3 \end{vmatrix} = -\hat{i} - 5\hat{j} + 3\hat{k}$ $\therefore 7(-\hat{i} - 5\hat{j} + 3\hat{k})$ $(\vec{a} + \vec{b}) \times (7(-\hat{i} - 5\hat{j} + 3\hat{k}))$ $7(0\hat{i} + 3\hat{j} + 5\hat{k}) \times (-\hat{i} - 5\hat{j} + 3\hat{k})$ $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 5 \\ -1 & -5 & 3 \end{vmatrix}$ $\Rightarrow 34\hat{i} - (5)\hat{j} + (3)\hat{k}$ $\Rightarrow 34\hat{i} - 5\hat{j} + 3\hat{k}$ $\therefore 7(34\hat{i} - 5\hat{j} + 3\hat{k})$

Question 4

Maths · Integrals · Single correct

The value of the definite integral $$\int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{(1+e^x \cos x)(\sin^4 x + \cos^4 x)}$$ is equal to:

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

Answer: (b)

Solution

Given $$I = \int_{\pi/4}^{\pi/4} \frac{dx}{(1+e^{x \cos x})(\sin^4 x + \cos^4 x)} \ldots (1)$$ Using $$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$ $$I = \int_{-\pi/4}^{\pi/4} \frac{dx}{(1+e^{-x \cos x})(\sin^4 x + \cos^4 x)}$$ Add (1) and (2) $$2I = \int_{\pi/4}^{x/4} \frac{dx}{\sin^4 x + \cos^4 x}$$ $$2I = 2 \int_0^{\pi/4} \frac{dx}{\sin^4 x + \cos^4 x}$$ $$I = \int_0^{\pi/4} \frac{(1+\tan^4 x) \sec^2 x}{\tan^4 x + 1} dx$$ $$I = \int_0^{\pi/4} (1+\frac{1}{\tan^2 x}) \sec^2 x \left( \tan x - \frac{1}{\tan x} \right)^2 + 2 dx$$ Let $\tan x - \frac{1}{\tan x} = t$ $$\left(1 + \frac{1}{\tan^2 x}\right) \sec^2 x dx = dt$$ $$I = \int_{-\infty}^0 \frac{dt}{t^2 + 2} = \left[ \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) \right]_0^{-\infty}$$

Question 5

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $\mathbb{C}$ be the set of all complex numbers. Let $S_1=\{\,z\in\mathbb{C}\mid |z-3-2i|^2=8\,\}$ $S_2=\{\,z\in\mathbb{C}\mid \mathrm{Re}(z)\ge 5\,\}$ and $S_3=\{\,z\in\mathbb{C}\mid |z-\bar z|\ge 8\,\}$. Then the number of elements in $S_1\cap S_2\cap S_3$ is equal to:

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

Answer: (a)

Solution

Given $S_1 : |z - 3 - 2i|^2 = 8$. $|z - 3 - 2i| = 2\sqrt{2}$ $(x - 3)^2 + (y - 2)^2 = (2\sqrt{2})^2$ $S_2 : x \geq 5$ $S_3 : |z - \bar{z}| \geq 8$ $|2iy| \geq 8$ $2|y| \geq 8 \therefore y \geq 4, y \leq -4$ $n(S_1 \cap S_2 \cap S_3) = 1$

Question 6

Maths · Applications of Integrals · Single correct

If the area of the bounded region $$R = \left\{ (x, y) : \max \{0, \log_e x\} \leq y \leq 2^x, \frac{1}{2} \leq x \leq 2 \right\}$$ is, $\alpha (\log_e 2)^{-1} + \beta (\log_e 2) + \gamma$, then the value of $$(\alpha + \beta - 2\gamma)^2$$ is equal to :

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

Answer: (b)

Solution

Given the region $R = \{ (x, y) \cdot \max (0, \log_e x) \leq y \leq 2^x, \frac{1}{2} \leq x \leq 2 \}$. The integral is evaluated as follows: $$\int_{\frac{1}{2}}^2 2^x \, dx - \int_1^2 \ln x \, dx$$ This simplifies to: $$\left[ \frac{2^x}{\ln 2} \right]_{\frac{1}{2}}^2 - \left[ x \ln x - x \right]_1^2$$ Evaluating the first integral: $$\frac{(2^2) - 2^{1/2}}{\log_2 2} - (2 \ln 2 - 1)$$ Simplifying further: $$\frac{(2^2 - \sqrt{2})}{\log_2 2} - 2 \ln 2 + 1$$ Thus, $\alpha = 2^2 - \sqrt{2}$, $\beta = -2$, $\gamma = 1$. Now, evaluating: $$(\alpha + \beta + 2 \gamma)^2$$ This becomes: $$\left(2^2 - \sqrt{2} - 2 - 2 \right)^2$$ Finally: $$(\sqrt{2})^2 = 2$$

Question 7

Maths · Conic Sections · Single correct

A ray of light through $(2, 1)$ is reflected at a point $P$ on the $y$-axis and then passes through the point $(5, 3)$. If this reflected ray is the directrix of an ellipse with eccentricity $\frac{1}{3}$ and the distance of the nearer focus from this directrix is $\frac{8}{\sqrt{53}}$, then the equation of the other directrix can be:

  1. $11x + 7y + 8 = 0$ or $11x + 7y - 15 = 0$
  2. $11x - 7y - 8 = 0$ or $11x + 7y + 15 = 0$
  3. $2x - 7y + 29 = 0$ or $2x - 7y - 7 = 0$
  4. $2x - 7y - 39 = 0$ or $2x - 7y - 7 = 0$

Answer: (c)

Solution

Equation of reflected Ray $y - 1 = \frac{2}{7}(x + 2)$ $7y - 7 = 2x + 4$ $2x - 7y + 11 = 0$ Let the equation of other directrix is $2x - 7y + \lambda$ Distance of directrix from Focus $\frac{a}{e} - ae = \frac{8}{\sqrt{53}}$ $3a - \frac{a}{3} = \frac{8}{\sqrt{53}}$ or $a = \frac{3}{\sqrt{53}}$ Distance from other focus $\frac{a}{e} + ae$ $3a + \frac{a}{3} = \frac{10a}{3} = \frac{10}{3} \times \frac{3}{\sqrt{53}} = \frac{10}{\sqrt{53}}$ Distance between two directrix $= \frac{2a}{e}$ $= 2 \times 3 \times \frac{3}{\sqrt{53}} = \frac{18}{\sqrt{53}}$ $\frac{|\lambda - 11|}{\sqrt{53}} = \frac{18}{\sqrt{53}}$ $\lambda - 11 = 18$ or $-18$ $\lambda = 29$ or $-7$ $2x - 7y - 7 = 0$ or $2x - 7y + 29 = 0$

Question 8

Maths · Binomial Theorem · Single correct

If the coefficients of $x^7$ in $\left(x^2 + \frac{1}{bx}\right)^{11}$ and $x^{-7}$ in $\left(x - \frac{1}{bx^2}\right)^{11}$, $b \neq 0$, are equal, then the value of $b$ is equal to:

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

Answer: (c)

Solution

Coefficient of $x^7$ in $\left(x^2 + \frac{1}{bx}\right)^{11}$ is given by $$\binom{11}{r}(x^2)^{11-t} \cdot \left(\frac{1}{bx}\right)^t$$ $$= \binom{11}{r} x^{22-3t} \cdot \frac{1}{b^t}$$ Solving $22 - 3r = 7$ gives $r = 5$. Thus, $\binom{11}{5} \cdot \frac{1}{b^5} x^7$. Coefficient of $x^{-7}$ in $\left(x - \frac{b}{bx^2}\right)^{11}$ is given by $$\binom{11}{r}(x)^{11-t} \cdot \left(-\frac{1}{bx^2}\right)^t$$ $$= \binom{11}{r} x^{11-3t} \cdot \frac{(-1)^t}{b^t}$$ Solving $11 - 3r = -7$ gives $r = 6$. Thus, $\binom{11}{6} \cdot \frac{1}{b^6} x^{-7}$. Equating $\binom{11}{5} \cdot \frac{1}{b^5} = \binom{11}{6} \cdot \frac{1}{b^6}$ gives $b \neq 0$, therefore $b = 1$.

Question 9

Maths · Mathematical Reasoning · Single correct

The compound statement $(P \lor Q) \land (\sim P) \Rightarrow Q$ is equivalent to:

  1. $P \lor Q$
  2. $P \land \sim Q$
  3. $\sim (P \Rightarrow Q)$
  4. $\sim (P \Rightarrow Q) \iff P \land \sim Q$

Answer: (d)

Solution

Using Truth Table For the first table:

Question 10

Maths · Trigonometric Functions · Single correct

If $\sin \theta + \cos \theta = \frac{1}{2}$, then $16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta))$ is equal to:

  1. 23
  2. -27
  3. -23
  4. 27

Answer: (c)

Solution

Given $\sin \theta + \cos \theta = \frac{1}{2}$. $\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = \frac{1}{4}$ $\sin 2\theta = -\frac{3}{4}$ Now: $$\cos 4\theta = 1 - 2 \sin^2 2\theta$$ $$= 1 - 2 \left( -\frac{3}{4} \right)^2$$ $$= 1 - 2 \times \frac{9}{16} = -\frac{1}{8}$$ $$\sin 6\theta = 3 \sin 2\theta - 4 \sin^3 2\theta$$ $$= \left( 3 - 4 \sin^2 2\theta \right) \cdot \sin 2\theta$$ $$= \left[ 3 - 4 \left( \frac{9}{16} \right) \right] \cdot \left( -\frac{3}{4} \right)$$ $$\Rightarrow \left[ \frac{3}{4} \right] \times \left( -\frac{3}{4} \right) = -\frac{9}{16}$$ $$16[\sin 2\theta + \cos 4\theta + \sin 6\theta]$$ $$16 \left( -\frac{3}{4} - \frac{1}{8} - \frac{9}{16} \right) = -23$$

Question 11

Maths · Matrices · Single correct

Let $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$. If $A^{-1} = \alpha I + \beta A$, $\alpha, \beta \in \mathbb{R}$, $I$ is a $2 \times 2$ identity matrix, then $4(\alpha - \beta)$ is equal to:

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

Answer: (d)

Solution

Given the matrix $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$ with $|A| = 6$. The inverse of $A$ is given by $$A^{-1} = \frac{adj A}{|A|} = \frac{1}{6} \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix}.$$ We equate this to $$\begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix} = \begin{bmatrix} \alpha & 0 \\ 0 & \alpha \end{bmatrix} + \begin{bmatrix} \beta & 2\beta \\ -\beta & 4\beta \end{bmatrix}.$$ Solving the equations, we have $$\alpha + \beta = \frac{2}{3}$$ $$\beta = -\frac{1}{6}$$ which implies $$\alpha = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}.$$ Therefore, $$4(\alpha - \beta) = 4(1) = 4.$$

Question 12

Maths · Continuity and Differentiability · Single correct

Let $f : \left( -\frac{\pi}{4}, \frac{\pi}{4} \right) \to \mathbb{R}$ be defined as $$f(x) = \begin{cases} (1 + |\sin x|)^{\frac{3a}{|\sin x|}}, & -\frac{\pi}{4} < x < 0 \\ b, & x = 0 \\ e^{\cot 4x / \cot 2x}, & 0 < x < \frac{\pi}{4} \end{cases}$$ If $f$ is continuous at $x = 0$, then the value of $6a + b^2$ is equal to:

  1. 1 - e
  2. e - 1
  3. 1 + e
  4. e

Answer: (c)

Solution

Given $\lim_{x \to 0} f(x) = b$. $\lim_{x \to 0^+} x e^{\frac{\cot 4x}{\cot 2x}} = e^{\frac{1}{2}} = b$. $\lim_{x \to 0} \left(1 + |\sin x|\right)^{\frac{3a}{\sin x}} = e^{3a} = e^{\frac{1}{2}}$. $\lim_{x \to 0} \left(1 + |\sin x|\right)^{\sin x} = e^{3a} = e^{\frac{1}{2}}$. $a = \frac{1}{6} \Rightarrow 6a = 1$. $(6a + b^2) = (1 + e)$.

Question 13

Maths · Differential Equations · Single correct

Let y = y(x) be solution of the differential equation $$ \log_e \left( \frac{dy}{dx} \right) = 3x + 4y, with y(0) = 0 $$ If $$ y \left( -\frac{2}{3} \log_e 2 \right) = \alpha \log_e 2, $$ then the value of $\($ $\alpha$ $\)$ is equal to:

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

Answer: (a)

Solution

\[ \frac{dy}{dx} = e^{3x}\cdot e^{4y} \Rightarrow \int e^{-4y}\,dy = \int e^{3x}\,dx \] \[ \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C \Rightarrow -\frac{1}{4} - \frac{1}{3} = C \Rightarrow C = -\frac{7}{12} \] \[ \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} - \frac{7}{12} \Rightarrow e^{-4y} = \frac{4e^{3x}-7}{-3} \] \[ e^{4y} = \frac{3}{7-4e^{3x}} \Rightarrow 4y = \ln\left(\frac{3}{7-4e^{3x}}\right) \] \[ 4y = \ln\left(\frac{3}{6}\right) \text{ when } x = -\frac{2}{3}\ln 2 \] \[ y = \frac{1}{4}\ln\left(\frac{1}{2}\right) = -\frac{1}{4}\ln 2 \]

Question 14

Maths · Three Dimensional Geometry · Single correct

Let the plane passing through the point $(-1, 0, -2)$ and perpendicular to each of the planes $2x + y - z = 2$ and $x - y - z = 3$ be $ax + by + cz + 8 = 0$. Then the value of $a + b + c$ is equal to:

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

Answer: (d)

Solution

Normal of required plane $(2\hat{i}+\hat{j}-\hat{k})\times(\hat{i}-\hat{j}-\hat{k})$ $=-2\hat{i}+\hat{j}-3\hat{k}$ Equation of plane $-2(x+1)+1(y-0)-3(z+2)=0$ $-2x+y-3z-8=0$ $\Rightarrow 2x-y+3z+8=0$ $\therefore\ a+b+c=4$

Question 15

Maths · Conic Sections · Single correct

Two tangents are drawn from the point $P(-1, 1)$ to the circle $x^2 + y^2 - 2x - 6y + 6 = 0$. If these tangents touch the circle at points $A$ and $B$, and if $D$ is a point on the circle such that length of the segments $AB$ and $AD$ are equal, then the area of the triangle $ABD$ is equal to:

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

Answer: (c)

Solution

The area of triangle $\Delta ABD$ is calculated as follows: $$\Delta ABD = \frac{1}{2} \times 2 \times 4$$ which simplifies to $$= 4$$

Question 16

Maths · Limits and Derivatives · Single correct

Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that $f(2) = 4$ and $f'(2) = 1$. Then, the value of $\lim_{x \to 2} \frac{x^2 f(2) - 4f(x)}{x - 2}$ is equal to:

  1. 4
  2. 8
  3. 16
  4. 12

Answer: (d)

Solution

Apply L'Hopital Rule $\lim_{x \to 2} \left( \frac{2x f(2) - 4f'(x)}{1} \right)$ $$= \frac{4(4) - 4}{1} = 12$$

Question 17

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

Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set $\{$P, Q$\}$ is equal to

  1. $\{$(4, 0), (0, 6)$\}$
  2. $\{$(2 + 2$\sqrt{2}$, 3 - $\sqrt{5}$), (2 - 2$\sqrt{2}$, 3 + $\sqrt{5}$)$\}$
  3. $\{$(2 + 2$\sqrt{2}$, 3 + $\sqrt{5}$), (2 - 2$\sqrt{2}$, 3 - $\sqrt{5}$)$\}$
  4. $\{$(-1, 5), (5, 1)$\}$

Answer: (d)

Solution

Given $\tan \theta = -\frac{2}{3}$. Using symmetric form of line $P, Q$: $(2 \pm \sqrt{13} \cos \theta, 3 \pm \sqrt{13} \sin \theta)$, we have: $$\left(2 \pm \sqrt{13} \cdot \left(-\frac{3}{\sqrt{13}}\right), 3 \pm \sqrt{13} \left(\frac{2}{\sqrt{13}}\right)\right)$$ $(-1, 5)$ and $(5, 1)$.

Question 18

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $\alpha, \beta$ be two roots of the equation $x^2 + (20)^{1/4}x + (5)^{1/2} = 0$. Then $\alpha^8 + \beta^8$ is equal to

  1. 10
  2. 100
  3. 50
  4. 160

Answer: (c)

Solution

Given the equation $$\left( x^2 + \sqrt{5} \right)^2 = \sqrt{20} x^2$$. We have $$x^4 = -5 \Rightarrow x^8 = 25$$. Also, $$\alpha^8 + \beta^8 = 50$$.

Question 19

Maths · Probability · Single correct

The probability that a randomly selected 2-digit number belongs to the set $\{ n \in N : (2^n - 2) \}$ is a multiple of 3 is equal to

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

Answer: (c)

Solution

Total number of cases $={}^{90}C_{1}=90$ Now, $n^2-2=(3-1)^n$ $={}^{n}C_{0}3^n-{}^{n}C_{1}3^{\,n-1}+\cdots+(-1)^{n-1}\,{}^{n}C_{n-1}3+(-1)^n\,{}^{n}C_{n}$ $=3\left(3^{\,n-1}-{}^{n}C_{1}3^{\,n-2}+\cdots+(-1)^{n-1}n\right)+(-1)^n-2$ $(2^n-2)$ is divisible by $3$ only when $n$ is odd. Required Probability $=\dfrac{45}{90}=\dfrac{1}{2}$

Question 20

Maths · Conic Sections · Single correct

Let \[ A=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}\mid 2x^2+2y^2-2x-2y=1\right\}, \] \[ B=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}\mid 4x^2+4y^2-16y+7=0\right\}, \] and \[ C=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}\mid x^2+y^2-4x-2y+5\le r^2\right\}. \] Then the minimum value of $|r|$ such that \[ A\cup B\subseteq C \] is equal to

  1. $\frac{3 + \sqrt{10}}{2}$
  2. $\frac{2 + \sqrt{10}}{2}$
  3. $\frac{3 + 2\sqrt{5}}{2}$
  4. 1 + $\sqrt{5}$

Answer: (c)

Solution

Given the equations: $S_1: x^2 + y^2 - x - y - \frac{1}{2} = 0; \ C_1 \left( \frac{1}{2}, \frac{1}{2} \right)$ $r_1 = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{2}} = 1$ $S_2: x^2 + y^2 - 4y + \frac{7}{4} = 0; \ C_2: (0, 2)$ $r_2 = \sqrt{4 - \frac{7}{4}} = \frac{3}{2}$ $S_3: x^2 + y^2 - 4x - 2y + 5 - r^2 = 0$ $C_3: (2, 1)$ $r_3 = \sqrt{4 + 1 - 5 + r^2} = |r|$ For $C_1C_3 = \sqrt{\frac{5}{2}}$ $\sqrt{\frac{5}{2}} \leq |r - 1| \Rightarrow \begin{cases} r \leq 1 + \sqrt{\frac{5}{2}} \\ r \geq \frac{3}{2} + \sqrt{5} \end{cases}$ For $C_2C_3 = \sqrt{5} \leq |r - \frac{3}{2}|$ $r - \frac{3}{2} \geq \sqrt{5}$ $r - \frac{3}{2} \leq -\sqrt{5}$

Question 21

Maths · Determinants · Numerical

For real numbers $\alpha$ and $\beta$, consider the following system of linear equations: $$x + y - z = 2, x + 2y + \alpha z = 1, 2x - y + z = \beta$$ If the system has infinite solutions, then $\alpha + \beta$ is equal to

Answer: 5

Solution

For infinite solutions $$\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$$ $$\Delta = \begin{vmatrix} 1 & 1 & -1 \\ 1 & 2 & \alpha \\ 2 & -1 & 1 \end{vmatrix} = 0$$ $$\Delta = \begin{vmatrix} 1 & 2 & \alpha \\ 2 & -1 & 1 \\ 3 & 0 & 0 \end{vmatrix}$$ $$\Delta = 3(2 + \alpha) = 0$$ Thus, $\alpha = -2$ $$\Delta_2 = \begin{vmatrix} 1 & 2 & -1 \\ 1 & 1 & -2 \\ 2 & \beta & 1 \end{vmatrix} = 0$$ $$1(1 + 2\beta) - 2(1 + 4) - (\beta - 2) = 0$$ $$\beta - 7 = 0$$ $$\beta = 7$$ Therefore, $\alpha + \beta = 5$ Ans.

Question 22

Maths · Vector Algebra · Numerical

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b}$ and $\vec{c} = \hat{j} - \hat{k}$ be three vectors such that $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 1$. If the length of projection vector of the vector $\vec{b}$ on the vector $\vec{a} \times \vec{c}$ is $l$, then the value of $3l^2$ is equal to

Answer: 2

Solution

Given $\vec{a} \times \vec{b} = \vec{c}$. Take dot with $\vec{c}$. $$(\vec{a} \times \vec{b}) \cdot \vec{c} = |\vec{c}|^2 = 2$$ Projection of $\vec{b}$ or $\vec{a} \times \vec{c} = \ell$. $$\frac{|\vec{b} \cdot (\vec{a} \times \vec{c})|}{|\vec{a} \times \vec{c}|} = \ell$$ Therefore, $$\ell = \frac{2}{\sqrt{6}} \implies \ell^2 = \frac{4}{6}$$ $$3\ell^2 = 2$$

Question 23

Maths · Sequences and Series · Numerical

If $\log_3 2$, $\log_3 \left(2^x - 5\right)$, $\log_3 \left(2^x - \frac{7}{2}\right)$ are in an arithmetic progression, then the value of $x$ is equal to

Answer: 3

Solution

Given $$2 \log_3 (2^x - 5) = \log_3 2 + \log_3 \left(2^x - \frac{7}{2}\right)$$ Let $2^x = t$. $$\log_3 (t - 5)^2 = \log_3 2 \left(t - \frac{7}{2}\right)$$ $$(t - 5)^2 = 2t - 7$$ $$t^2 - 12t + 32 = 0$$ $$(t - 4)(t - 8) = 0$$ Thus, $2^x = 4$ or $2^x = 8$. $X = 2$ (Rejected) Or $x = 3$

Question 24

Maths · Relations and Functions · Numerical

Let the domain of the function $$f(x) = \log_4 \left( \log_5 \left( \log_3 \left( 18x - x^2 - 77 \right) \right) \right)$$ be (a, b) Then the value of the integral $$\int_a^b \frac{\sin^3 x}{\sin^3 x + \sin^3 (a + b - x)} \, dx$$ is equal to

Answer: 1

Solution

For domain $$\log_5 \left( \log_3 (18x - x^2 - 77) \right) > 0$$ $$\log_3 (18x - x^2 - 77) > 1$$ $$18x - x^2 - 77 > 3$$ $$x^2 - 18x + 80 < 0$$ $$x \in (8, 10)$$ Thus, $a = 8$ and $b = 10$. $$I = \int_a^b \frac{\sin^3 x}{\sin^3 x + \sin^3 (a+b-x)} \, dx$$ $$I = \int_a^b \frac{\sin^3 (a+b-x)}{\sin^3 x + \sin^3 (a+b-x)} \, dx$$ $$2I = (b-a) \implies I = \frac{b-a}{2} (\because a = 8 and b = 10)$$ $$I = \frac{10-8}{2} = 1$$

Question 25

Maths · Determinants · Numerical

Let $f(x) = \begin{vmatrix} \sin^2 x & -2 + \cos^2 x & \cos 2x \\ 2 + \sin^2 x & \cos^2 x & \cos 2x \\ \sin^2 x & \cos^2 x & 1 + \cos 2x \end{vmatrix}, x \in [0, \pi]$ Then the maximum value of $f(x)$ is equal to

Answer: 6

Solution

Given the matrix: $$\begin{vmatrix} -2 & -2 & 0 \\ 2 & 0 & -1 \\ \sin^2 x & \cos^2 x & 1 + \cos 2x \end{vmatrix}$$ Perform the row operations: $$(R_1 \rightarrow R_1 - R_2)$$ $$(R_2 \rightarrow R_2 - R_3)$$ Calculate: $$-2 (\cos^2 x) + 2 \left(2 + 2 \cos 2x + \sin^2 x\right)$$ Simplify to: $$4 + 4 \cos 2x - 2 \left(\cos^2 x - \sin^2 x\right)$$ Thus, $$f(x) = 4 + \underbrace{2 \cos 2x}_{max = 1}$$ The maximum value of $f(x)$ is: $$f(x)_{max} = 4 + 2 = 6$$

Question 26

Maths · Integrals · Numerical

Let $F: [3,5] \to \mathbb{R}$ be a twice differentiable function on $(3,5)$ such that \[ F(x) = e^{-x}\int_3^x \left(3t^2 + 2t + 4F'(t)\right)dt \] If $F'(4) = \dfrac{\alpha e^{\beta} - 224}{\left(e^{\beta}-4\right)^2}$, then $\alpha + \beta$ is equal to

Answer: 16

Solution

Given $F(3) = 0$. $e^x F(x) = \int_3^x (3t^2 + 2t + 4 \, F'(t)) \, dt$ $e^x F(x) + e^x F'(x) = 3x^2 + 2x + 4 \, F'(x)$ $(e^x - 4) \frac{dy}{dx} + e^x y = (3x^2 + 2x)$ $\frac{dy}{dx} + \frac{e^x}{(e^x - 4)} y = \frac{(3x^2 + 2x)}{(e^x - 4)}$ $y e^{\int \frac{e^x}{(e^x - 4)} \, dx} = \int \frac{(3x^2 + 2x)}{(e^x - 4)} e^{\int \frac{e^x}{e^x - 4} \, dx} \, dx$ $\cdot \,(e^x - 4) = \int (3x^2 + 2x) \, dx + c$ $y \,(e^x - 4) = x^3 + x^2 + c$ Put $x = 3 \Rightarrow c = -36$ $F(x) = \frac{(x^3 + x^2 - 36)}{(e^x - 4)}$ $F'(x) = \frac{(3x^2 + 2x)(e^x - 4) - (x^3 + x^2 - 36)e^x}{(e^x - 4)^2}$ $F'(4) = \frac{56(e^4 - 4) - 4ne^n}{(e^4 - 4)^2}$ $= \frac{12e^4 - 22y}{(e^y - 4)^2} \Rightarrow \alpha = 12$ $\beta = 4$ $\alpha + \beta = 16$

Question 27

Maths · Three Dimensional Geometry · Numerical

Let a plane P pass through the point $(3, 7, -7)$ and contain the line, $\frac{x-2}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$. If distance of the plane P from the origin is $d$, then $d^2$ is equal to

Answer: 3

Solution

Given $\overrightarrow{BA} = (\hat{i} + 4\hat{j} - 5\hat{k})$. $\overrightarrow{BA} = (\hat{i} + 4\hat{j} - 5\hat{k})$. $\overrightarrow{BA} \times \vec{\ell} = \overrightarrow{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 1 \\ 1 & 4 & -5 \end{vmatrix}$. $\hat{a} + b\hat{j} + c\hat{k} = -14\hat{i} - \hat{j}(m) + \hat{k}(-14)$. $a = 1$, $b = 1$, $c = 1$. Plane is $(x - 2) + (y - 3) + (z + z) = 0$. $x + y + z - 3 = 0$. $d = \sqrt{3} \Rightarrow d^2 = 3$.

Question 28

Maths · Relations and Functions · Numerical

Let $S = \{1, 2, 3, 4, 5, 6, 7\}$. Then the number of possible functions $f : S \to S$ such that $f(m \cdot n) = f(m) \cdot f(n)$ for every $m, n \in S$ and $m \cdot n \in S$ is equal to

Answer: 490

Solution

Given $\(F(mn)=f(m)\cdot f(n)\). Put \(m=1\)$. \[ f(n)=f(1)\cdot f(n)\Rightarrow f(1)=1. \] Put \(m=n=2\). \[ f(4)=f(2)\cdot f(2) \] \[ f(2)=1 \Rightarrow f(4)=1 \qquad \text{or} \qquad f(2)=2 \Rightarrow f(4)=4 \] Put \(m=2,\; n=3\). \[ f(6)=f(2)\cdot f(3) \] When \(f(2)=1\), \[ f(3)=1 \text{ to } 7 \] When \(f(2)=2\), \[ f(3)=1 \text{ or } 2 \text{ or } 3 \] \(f(5)\) and \(f(7)\) can take any value. \[ |x|\times 7\times 1\times 7\times 1\times 7 \] \[ +\,|x|\times 3\times 1\times 7\times 1\times 7 \] \[ =490 \]

Question 29

Maths · Differential Equations · Numerical

If $y = y(x)$, $y \in \left[0, \frac{\pi}{2}\right]$ is the solution of the differential equation $$\sec y \frac{dy}{dx} - \sin(x + y) - \sin(x - y) = 0, with y(0) = 0$$ then $5y'\left(\frac{\pi}{2}\right)$ is equal to

Answer: 2

Solution

Given $\sec y \frac{dy}{dx} = 2 \sin x \cos y$. Multiply both sides by $\sec y$ to get $\sec^2 y \, dy = 2 \sin x \, dx$. Integrating both sides, we have $\tan y = -2 \cos x + c$. Given $c = 2$, we substitute to get $\tan y = -2 \cos x + 2$. At $x = \frac{\pi}{2}$, $\tan y = 2$. Differentiating again, $\sec^2 y \frac{dy}{dx} = 2 \sin x$. Substitute $\tan y = 2$ to get $5 \frac{dy}{dx} = 2$.

Question 30

Maths · Continuity and Differentiability · Numerical

Let $f : [0, 3] \to \mathbb{R}$ be defined by $$f(x) = \min\{x - \lfloor x \rfloor, 1 + \lfloor x \rfloor - x\}$$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. Let $P$ denote the set containing all $x \in [0, 3]$ where $f$ is discontinuous, and $Q$ denote the set containing all $x \in (0, 3)$ where $f$ is not differentiable. Then the sum of number of elements in $P$ and $Q$ is equal to

Answer: 5

Solution

The function is given by $1 - \{x\} = 1 - x$ for $0 \leq x < 1$. The function is non-differentiable at $x = \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}$.

Physics

Question 31

Physics · Current Electricity · Single correct

In the given figure, a battery of emf $E$ is connected across a conductor $PQ$ of length $l$ and different area of cross-sections having radii $r_1$ and $r_2$ ($r_2 < r_1$). Choose the correct option as one moves from $P$ to $Q$:

  1. Drift velocity of electron increases.
  2. Electric field decreases.
  3. Electron current decreases.
  4. All of these

Answer: (a)

Solution

Current is constant in conductor, $i = constant$. Resistance of element $dR = \frac{\rho \, dx}{\pi r^2}$. $$dV = i \, dR = \frac{i \rho \, dx}{\pi r^2}$$ $$E = \frac{dV}{dx} = \frac{i \rho}{\pi r^2}$$ $$V_d = \frac{e E \tau}{m}$$ Thus, $V_d \propto E$. Therefore, $E \propto \frac{1}{r^2}$. If $r$ decreases, $E$ will increase. Therefore, $V_d$ will increase.

Question 32

Physics · Kinetic Theory · Single correct

The number of molecules in one litre of an ideal gas at 300 K and 2 atmospheric pressure with mean kinetic energy $2 \times 10^{-9}$ J per molecules is :

  1. 0.75 $\times$ 10^{11}
  2. 3 $\times$ 10^{11}
  3. 1.5 $\times$ 10^{11}
  4. 6 $\times$ 10^{11}

Answer: (c)

Solution

KE = $\frac{3}{2}$ k T PV = $\frac{N}{N_A}$ RT N = $\frac{PV}{k T}$ = N = 1.5 $\times$ 10^{11}

Question 33

Physics · Electric Charges and Fields · Single correct

The relative permittivity of distilled water is 81. The velocity of light in it will be: ( Given $\mu_r = 1$)

  1. $4.33 \times 10^7 \, \mathrm{m/s}$
  2. $2.33 \times 10^7 \, \mathrm{m/s}$
  3. $3.33 \times 10^7 \, \mathrm{m/s}$
  4. $5.33 \times 10^7 \, \mathrm{m/s}$

Answer: (c)

Solution

Given $$V = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$$ The velocity is $$= 3.33 \times 10^7 \, \mathrm{m/sec}$$

Question 34

Physics · Ray Optics and Optical Instruments · Single correct

Choose the correct answer from the options given below :

  1. (\text{a}) - (\text{ii}), \quad (\text{b}) - (\text{iii}), \quad (\text{c}) - (\text{i}), \quad (\text{d}) - (\text{iv})
  2. (\text{a}) - (\text{ii}), \quad (\text{b}) - (\text{i}), \quad (\text{c}) - (\text{iii}), \quad (\text{d}) - (\text{iv})
  3. (\text{a}) - (\text{iii}), \quad (\text{b}) - (\text{iv}), \quad (\text{c}) - (\text{ii}), \quad (\text{d}) - (\text{i})
  4. (\text{a}) - (\text{iii}), \quad (\text{b}) - (\text{iv}), \quad (\text{c}) - (\text{i}), \quad (\text{d}) - (\text{ii})

Answer: (c)

Solution

(a) $m, L$ $$I = \frac{mL^2}{12}$$ (b) $2m, L$ $$I = \frac{(2m)(L^2)}{3}$$ (c) $m, 2L$ $$I = \frac{m(2L)^2}{12} = \frac{mL^2}{3}$$ (d) $2m, 2L$ $$I = \frac{2m(2L)^2}{3} = \frac{8}{3}mL^2$$

Question 35

Physics · System of Particles and Rotational Motion · Single correct

Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. The masses of A, B and C are $m$, $2m$ and $2m$ respectively. A moves towards B with a speed of $9 \, \mathrm{m/s}$ and makes an elastic collision with it. Thereafter B makes a completely inelastic collision with C. All motions occur along same straight line. The final speed of C is:

  1. $6 \, \mathrm{m/s}$
  2. $9 \, \mathrm{m/s}$
  3. $4 \, \mathrm{m/s}$
  4. $3 \, \mathrm{m/s}$

Answer: (d)

Solution

Collision between A and B $$m \times 9 = mv_1 + 2 \, mv_2 (from momentum conservation)$$ $$e = 1 = \frac{v_2 - v_1}{9}$$ $$\Rightarrow v_2 = 6 \, \mathrm{m/s}, \ v_1 = -3 \, \mathrm{m/s}$$ Collision between B and C $$2 \, m \times 6 = 4 \, mv (from momentum conservation)$$ $$v = 3 \, \mathrm{m/s}$$

Question 36

Physics · Electrostatic Potential and Capacitance · Single correct

A capacitor of capacitance $C = 1 \, \mu \mathrm{F}$ is suddenly connected to a battery of $100 \, \mathrm{volt}$ through a resistance $R = 100 \, \Omega$. The time taken for the capacitor to be charged to get $50 \, \mathrm{V}$ is: [ Take $\ln 2 = 0.69$ ]

  1. $1.44 \times 10^{-4} \, \mathrm{s}$
  2. $3.33 \times 10^{-4} \, \mathrm{s}$
  3. $0.69 \times 10^{-4} \, \mathrm{s}$
  4. $0.30 \times 10^{-4} \, \mathrm{s}$

Answer: (c)

Solution

Given the equation for voltage: $$V = V_0 \left(1 - e^{-\frac{t}{RC}}\right)$$ Substitute the values: $$50 = 100 \left(1 - e^{-\frac{t}{RC}}\right)$$ Solving for $t$: $$t = 0.69 \times 10^{-4} sec.$$

Question 37

Physics · Electrostatic Potential and Capacitance · Single correct

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be: (Given area of plate = A)

  1. $\frac{15}{34} \frac{K \varepsilon_0 A}{d}$
  2. $\frac{15}{6} \frac{K \varepsilon_0 A}{d}$
  3. $\frac{25}{6} \frac{K \varepsilon_0 A}{d}$
  4. $\frac{9}{6} \frac{K \varepsilon_0 A}{d}$

Answer: (a)

Solution

Given the equation for the effective capacitance: $$\frac{1}{C_{eff}} = \frac{d}{K \epsilon_0 A} + \frac{2d}{3K \epsilon_0 A} + \frac{3d}{5K \epsilon_0 A}$$ Solving for $C_{eff}$, we have: $$C_{eff} = \frac{15K \epsilon_0 A}{34d}$$

Question 38

Physics · System of Particles and Rotational Motion · Single correct

The figure shows two solid discs with radii $R$ and $r$, respectively. If mass per unit area is same for both, what is the ratio of MI of bigger disc about axis $AB$ (which is $\perp$ to the plane of the disc and passing through its centre) of MI of smaller disc around one of its diameters lying on its plane? Given '$M$' is the mass of the larger disc. (MI stands for moment of inertia.)

  1. $R^2 : r^2$
  2. $2r^4 : R^4$
  3. $2R^2 : r^2$
  4. $2R^4 : r^4$

Answer: (d)

Solution

Ratio of moment of inertia = $\($ $\frac{\frac{1}{2} MR^2}{\frac{1}{4} mr^2}$ $\)$ \[ = \frac{2 \sigma \pi R^2 R^2}{\sigma \pi r^2 r^2} = \frac{2R^4}{r^4} \]

Question 39

Physics · Wave Optics · Single correct

In Young's double slit experiment, if the source of light changes from orange to blue then:

  1. the central bright fringe will become a dark fringe.
  2. the distance between consecutive fringes will decrease.
  3. the distance between consecutive fringes will increase.
  4. the intensity of the minima will increase.

Answer: (b)

Solution

Fringe width = $\frac{\lambda D}{d}$ as $\lambda$ decreases, fringe width also decreases

Question 40

Physics · Thermodynamics · Single correct

In the reported figure, there is a cyclic process ABCDA on a sample of $1 \, \mathrm{mol}$ of a diatomic gas. The temperature of the gas during the process $A \rightarrow B$ and $C \rightarrow D$ are $T_1$ and $T_2$ $(T_1 > T_2)$ respectively. Choose the correct option out of the following for work done if processes BC and DA are adiabatic.

  1. $W_{AB} = W_{DC}$
  2. $W_{AD} = W_{BC}$
  3. $W_{BC} + W_{DA} > 0$
  4. $W_{AB} < W_{CD}$

Answer: (b)

Solution

Work done in adiabatic process = $$\frac{-nR}{\gamma - 1} \left( T_f - T_i \right)$$ Therefore, $$W_{AD} = \frac{-nR}{\gamma - 1} \left( T_2 - T_1 \right)$$ and $$W_{BC} = \frac{-nR}{\gamma - 1} \left( T_2 - T_1 \right)$$ Therefore, $$W_{AD} = W_{BC}$$

Question 41

Physics · Mathematics in Physics · Single correct

Assertion $A$: If $A, B, C, D$ are four points on a semi-circular arc with centre at 'O' such that $|\overrightarrow{AB}| = |\overrightarrow{BC}| = |\overrightarrow{CD}|$, then $\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} = 4\overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{OC}$ Reason $R$: Polygon law of vector addition yields $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{AD} = 2\overrightarrow{AO}$ In the light of the above statements, choose the most appropriate answer from the options given below:

  1. $A$ is correct but $R$ is not correct.
  2. $A$ is not correct but $R$ is correct.
  3. Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$.
  4. Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$.

Answer: (d)

Solution

Polygon law is applicable in both but the equation given in the reason is not useful in explaining the assertion.

Question 42

Physics · Mechanical Properties of Fluids · Single correct

A light cylindrical vessel is kept on a horizontal surface. Area of base is $A$. A hole of cross-sectional area $a$ is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is (a<<A)

  1. $\frac{A}{2a}$
  2. None of these
  3. $\frac{2a}{A}$
  4. $\frac{a}{A}$

Answer: (c)

Solution

For no sliding $$f \geq \rho a v^2$$ $$\mu mg \geq \rho a v^2$$ $$\mu \rho A h g \geq \rho a 2 g h$$ $$\mu \geq \frac{2a}{A}$$ Option (3)

Question 43

Physics · Oscillations · Single correct

A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is $\frac{3E}{4}$ then its displacement 'y' is given by:

  1. $y = a$
  2. $y = \frac{a}{\sqrt{2}}$
  3. $y = \frac{a\sqrt{3}}{2}$
  4. $y = \frac{a}{2}$

Answer: (d)

Solution

Given $E = \frac{1}{2} K a^2$. $$\frac{3E}{4} = \frac{1}{2} K \left(a^2 - y^2\right)$$ $$\frac{3}{4} \times \frac{1}{2} K a^2 = \frac{1}{2} K \left(a^2 - y^2\right)$$ $$y^2 = a^2 - \frac{3a^2}{4}$$ $$y = \frac{a}{2}$$

Question 44

Physics · Nuclei · Single correct

If 'f denotes the ratio of the number of nuclei decayed ($N_d$) to the number of nuclei at $t = 0$ ($N_0$) then for a collection of radioactive nuclei, the rate of change of 'f with respect to time is given as: $[\lambda$ is the radioactive decay constant $]$

  1. $-\lambda (1 - e^{-\lambda t})$
  2. $\lambda (1 - e^{-\lambda t})$
  3. $\lambda e^{-\lambda t}$
  4. $-\lambda e^{-\lambda d}$

Answer: (c)

Solution

Given $N = N_0 e^{-\lambda t}$. $N_d = N_0 - N$. $N_d = N_0 \left(1 - e^{-\lambda t}\right)$. $\($ $\frac{N_d}{N_0}$ = f = 1 - e^{-$\lambda$ t} $\)$. $\($ $\frac{df}{dt}$ = $\lambda$ e^{-$\lambda$ t} $\)$.

Question 45

Physics · Electrostatic Potential and Capacitance · Single correct

Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be:

  1. $\frac{V}{K+2}$
  2. $\frac{V}{K}$
  3. $\frac{3V}{K+2}$
  4. $\frac{3V}{K}$

Answer: (c)

Solution

Now, $$V_C = \frac{2CV + CV}{KC + 2C}$$ $$= \frac{3V}{K + 2}$$

Question 46

Physics · Motion in a Straight Line · Single correct

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching $\frac{h}{3}$ in both the directions.

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

Answer: (c)

Solution

Given $u = \sqrt{2gh}$. Now, $S = \frac{h}{3}$ and $a = -g$. The equation for $S$ is $S = ut + \frac{1}{2}at^2$. Substituting the values, we have $$\frac{h}{3} = \sqrt{2gh}t + \frac{1}{2}(-g)t^2$$ which simplifies to $$t^2 \left( \frac{g}{2} \right) - \sqrt{2gh}t + \frac{h}{3} = 0.$$ From the quadratic equation, $$t_1, t_2 = \frac{\sqrt{2gh} \pm \sqrt{2gh - \frac{4g}{2} \frac{h}{3}}}{g}.$$ Simplifying further, $$\frac{t_1}{t_2} = \frac{\sqrt{2gh} - \sqrt{\frac{4gh}{3}}}{\sqrt{2gh} + \sqrt{\frac{4gh}{3}}} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}.$$

Question 47

Physics · Alternating Current · Single correct

A 0.07$\mathrm{H}$ inductor and a 12$\Omega$ resistor are connected in series to a 220$\mathrm{V}$, 50$\mathrm{Hz}$ ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $\pi$ as $\frac{22}{7}$]

  1. 8.8$\mathrm{A}$ and $\tan^{-1}\left(\frac{11}{6}\right)$
  2. 88$\mathrm{A}$ and $\tan^{-1}\left(\frac{11}{6}\right)$
  3. 0.88$\mathrm{A}$ and $\\tan^{-1}\left(\frac{11}{6}\right)$
  4. 8.8$\mathrm{A}$ and $\tan^{-1}\left(\frac{6}{11}\right)$

Answer: (a)

Solution

Given $\phi = \tan^{-1}\left(\frac{X_L}{R}\right)$. $X_L = \omega L$. $X_L = 2 \times \frac{22}{7} \times 50 \times 0.07 = 22\,\Omega$. $\phi = \tan^{-1}\left(\frac{22}{12}\right)$. $R = 12\,\Omega$. $\phi = \tan^{-1}\left(\frac{11}{6}\right)$. $Z = \sqrt{X_L^2 + R^2} = 25.059$. $I = \frac{V}{Z} = \frac{220}{25.059} = 8.77\,\mathrm{A}$.

Question 48

Physics · Oscillations · Single correct

Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle 'θ' with the vertical?

  1. $x = \left( \frac{q^2 l}{2 \pi \varepsilon_0 m g} \right)^{\frac{1}{2}}$
  2. $x = \left( \frac{q^2 l}{2 \pi \varepsilon_0 m g} \right)^{\frac{1}{3}}$
  3. $x = \left( \frac{q^2 l^2}{2 \pi \varepsilon_0 m^2 g} \right)^{\frac{1}{3}}$
  4. $x = \left( \frac{q^2 l^2}{2 \pi \varepsilon_0 m^2 g^2} \right)^{\frac{1}{3}}$

Answer: (b)

Solution

Given the forces, we have: $$T \cos \theta = mg$$ $$T \sin \theta = \frac{kq^2}{x^2}$$ Thus, $$\tan \theta = \frac{kq^2}{x^2 mg}$$ As $\tan \theta \approx \sin \theta \approx \frac{x}{2L}$, we have: $$\frac{x}{2L} = \frac{kq^2}{x^2 mg}$$ Solving for $x$, we get: $$x = \left( \frac{q^2 L}{2 \pi \varepsilon_0 mg} \right)^{1/3}$$

Question 49

Physics · Experimental Physics · Single correct

Assertion A: If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5 \, \mathrm{mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001 \, \mathrm{cm}$. Reason R: Least Count $= \frac{\text{Pitch}}{\text{Total divisions on circular scale}}$ In the light of the above statements, choose the most appropriate answer from the options given below:

  1. A is not correct but R is correct.
  2. Both A and R are correct and R is the correct explanation of A.
  3. A is correct but R is not correct.
  4. Both A and R are correct and R is NOT the correct explanation of A.

Answer: (a)

Solution

Least count = $\($ $\frac{Pitch}{total division on circular scale}$ $\)$ In 5 revolutions, distance traveled is 5 mm. In 1 revolution, it will travel 1 mm. So least count = $\($ $\frac{1}{50}$ = 0.02 $\)$

Question 50

Physics · Thermal Properties of Matter · Single correct

A body takes 4 min. to cool from $61^\circ \mathrm{C}$ to $59^\circ \mathrm{C}$. If the temperature of the surroundings is $30^\circ \mathrm{C}$, the time taken by the body to cool from $51^\circ \mathrm{C}$ to $49^\circ \mathrm{C}$ is:

  1. 4 min.
  2. 3 min.
  3. 8 min.
  4. 6 min.

Answer: (d)

Solution

$\dfrac{\Delta T}{\Delta t} = K(T_t - T_s) \quad T_t = \text{average temp.}$ $T_S = \text{surrounding temp}$ $\dfrac{61-59}{4} = K\left(\dfrac{61+59}{2} - 30\right) \ldots(1)$ $\dfrac{51-49}{t} = K\left(\dfrac{51+49}{2} - 30\right) \ldots(2)$ $\text{Divide (1) \& (2) } \dfrac{t}{4} = \dfrac{60-30}{50-30} = \dfrac{30}{20}$ $\text{so, } t = 6 \text{ minutes}$

Question 51

Physics · Electromagnetic Induction · Numerical

Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then $T_1$ is connected to $T_2$. As the current in $R = 6\Omega$ attains a maximum value of steady state level, $T_1$ is disconnected from $T_2$ and immediately connected to $T_3$. Potential drop across $r = 3\Omega$ resistor immediately after $T_1$ is connected to $T_3$ is ___ V. (Round off to the Nearest Integer)

Answer: 3

Solution

When $T_1$ and $T_2$ are connected, then the steady state current in the inductor $I = \frac{6}{6} = 1 \, \mathrm{A}$. When $T_1$ and $T_3$ are connected then current through inductor remains same. So potential difference across $3 \, \Omega$ $$V = Ir = 1 \times 3 = 3 \, volt$$

Question 52

Physics · Gravitation · Numerical

Suppose two planets (spherical in shape) of radii $R$ and $2R$, but mass $M$ and $9M$ respectively have a centre to centre separation $8R$ as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is $\sqrt{\frac{a}{7} \frac{GM}{R}}$ then the value of 'a' is ____ [Given : The two planets are fixed in their position]

Answer: 4

Solution

Acceleration due to gravity will be zero at P therefore, $$\frac{GM}{x^2} = \frac{G9M}{(8R-x)^2}$$ $$8R - x = 3x$$ $$x = 2R$$ Apply conservation of energy and consider velocity at P is zero. $$\frac{1}{2} mv^2 - \frac{GMm}{R} - \frac{G9Mm}{7R} = 0 - \frac{GMm}{2R} - \frac{G9Mm}{6R}$$ $$\therefore \; V = \sqrt{\frac{4}{7} \frac{GM}{R}}$$

Question 53

Physics · Atoms · Numerical

In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5$\AA$. If the speed of electron is $2.2 \times 10^6 \, \mathrm{m/s}$, then the current associated with the electron will be ___ $\times 10^{-2}$ mA. [ Take $\pi$ as $\frac{22}{7}$ ]

Answer: 112

Solution

Given $$I = \frac{e}{T} = \frac{e \omega}{2 \pi} = \frac{eV}{2 \pi}$$ Calculate $$I = \frac{1.6 \times 10^{-19} \times 2.2 \times 10^6 \times 7}{2 \times 22 \times 0.5 \times 10^{-10}}$$ This simplifies to $$= 1.12 \, \mathrm{mA}$$ Which is equivalent to $$112 \times 10^{-2} \, \mathrm{mA}$$

Question 54

Physics · Nuclei · Numerical

A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 $\mu$F is first charged and later connected with resistor 'R'. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of 'R' should be ___ $\Omega$.

Answer: 150

Solution

Given $T_m = 30 \, \mathrm{ms}$ and $C = 200 \, \mu \mathrm{F}$. The equation is $$\frac{q}{N} = \frac{Q_0 e^{-t/RC}}{N_0 e^{-\lambda t}} = \frac{Q_0}{N_0} e^{\left( \lambda - \frac{1}{RC} \right)}.$$ Since $q/N$ is constant, hence $$\lambda = \frac{1}{RC}.$$ Therefore, $$R = \frac{1}{\lambda C} = \frac{T_m}{C} = \frac{30 \times 10^{-3}}{200 \times 10^{-6}} = 150 \, \Omega.$$

Question 55

Physics · Dual Nature of Radiation and Matter · Numerical

A particle of mass $9.1 \times 10^{-31} \, \mathrm{kg}$ travels in a medium with a speed of $10^6 \, \mathrm{m/s}$ and a photon of a radiation of linear momentum $10^{-27} \, \mathrm{kg \, m/s}$ travels in vacuum. The wavelength of photon is ___ times the wavelength of the particle.

Answer: 910

Solution

For photon $\lambda_1 = \frac{h}{P} = \frac{6.6 \times 10^{-34}}{10^{-27}}$. For particle $\lambda_2 = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^6}$. Therefore, $\frac{\lambda_1}{\lambda_2} = 910$.

Question 56

Physics · Ray Optics and Optical Instruments · Numerical

A prism of refractive index $n_1$ and another prism of refractive index $n_2$ are stuck together (as shown in the figure). $n_1$ and $n_2$ depend on $\lambda$, the wavelength of light, according to the relation $$n_1 = 1.2 + \frac{10.8 \times 10^{-14}}{\lambda^2} and n_2 = 1.45 + \frac{1.8 \times 10^{-14}}{\lambda^2}$$ The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be ___ nm

Answer: 600

Solution

For no bending, $n_1 = n_2$ $$1.2 + \frac{10.8 \times 10^{-14}}{\lambda^2} = 1.45 + \frac{1.8 \times 10^{-4}}{\lambda^2}$$ On solving, $$9 \times 10^{-14} = 25 \lambda^2$$ $$\lambda = 6 \times 10^{-7}$$ $$\lambda = 600 \, \mathrm{nm}$$

Question 57

Physics · Mechanical Properties of Solids · Numerical

A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section $10^{-6} \, \mathrm{m}^2$ stretched by an amount 0.04 m. The velocity of the projected stone is ___ m/s (Young's modulus of rubber = $0.5 \times 10^9 \, \mathrm{N/m}^2$)

Answer: 20

Solution

By energy conservation $$\frac{1}{2} \cdot \frac{YA}{L} \cdot x^2 = \frac{1}{2} mv^2$$ $$\frac{0.5 \times 10^9 \times 10^{-6} \times (0.04)^2}{0.1} = \frac{20}{1000} v^2$$ Therefore, $$v^2 = 400$$ $$v = 20 \, \mathrm{m/s}$$

Question 58

Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Numerical

A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 $\Omega$ in the collector circuit is 0.6 V. If the current gain factor ($\beta$) is 24, then the base current is ___ $\mu$A. (Round off to the Nearest Integer)

Answer: 25

Solution

$\beta = \dfrac{I_C}{I_B} = 24; \quad R_C = 1000$ $\Delta V = 0.6$ $I_C = \dfrac{0.6}{1000}$ $I_C = 6 \times 10^{-4}$ $I_B = \dfrac{I_C}{\beta} = \dfrac{6 \times 10^{-4}}{24} = 25\,\mu\text{A}$

Question 59

Physics · Communication Systems · Numerical

The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5 V amplitude are $\frac{a}{10}$ V and $\frac{b}{10}$ V respectively. Then the value of $\frac{a}{b}$ is ____

Answer: 1

Solution

Given the amplitude diagram, we have the following equation: $$\frac{a}{10} = \frac{b}{10} = \frac{\mu A_c}{2}$$ which implies $$\frac{a}{b} = 1$$

Question 60

Physics · Magnetism and Matter · Numerical

In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85 \times 10^{-2} \, \mathrm{A/m^2}$ and moment of inertia $5 \times 10^{-6} \, \mathrm{kg \, m^2}$. If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is ___ mT [Take $\pi^2$ as 9.85]

Answer: 8

Solution

Given the formula for the period, $$T = 2\pi \sqrt{\frac{I}{MB}}$$. The magnetic field is calculated as $$B = 80 \times 10^{-4} = 8 \, \mathrm{mT}$$.

Chemistry

Question 61

Chemistry · Alcohols, Phenols and Ethers · Single correct

Which one of the following compounds will give orange precipitate when treated with 2,4-dinitrophenyl hydrazine?

Answer: (d)

Solution

Explanation: 2,4-D.N.P test is used for carbonyl compound (aldehyde and ketone).

Question 62

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

The product obtained from the electrolytic oxidation of acidified sulphate solutions, is :

  1. $\mathrm{HSO}_4^-$
  2. $\mathrm{HO}_3\mathrm{SOOSO}_3\mathrm{H}$
  3. $\mathrm{HO}_2\mathrm{SOSO}_2\mathrm{H}$
  4. $\mathrm{HO}_3\mathrm{SOSO}_3\mathrm{H}$

Answer: (b)

Solution

Electrolysis of concentrated solution of acidified sulphate solution yields $\mathrm{H_2S_2O_8}$.

Question 63

Chemistry · The Solid State · Single correct

The parameters of the unit cell of a substance are $a = 2.5, \ b = 3.0, \ c = 4.0, \ \alpha = 90^\circ, \ \beta = 120^\circ, \ \gamma = 90^\circ$ The crystal system of the substance is :

  1. Hexagonal
  2. Orthorhombic
  3. Monoclinic
  4. Triclinic

Answer: (c)

Solution

Given $a \neq b \neq c$ and $\alpha = \gamma = 90^\circ \neq \beta$ are parameters of monoclinic unit cell.

Question 64

Chemistry · Redox Reactions · Single correct

The oxidation states of 'P' in $\mathrm{H_4P_2O_7}$, $\mathrm{H_4P_2O_5}$ and $\mathrm{H_4P_2O_6}$, respectively, are:

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

Answer: (c)

Solution

Oxidation state of P in $\mathrm{H_4P_2O_7}$, $\mathrm{H_4P_2O_5}$ and $\mathrm{H_4P_2O_6}$ is 5, 3, and 4 respectively. $\mathrm{H_4P_2O_7}$ $$2x + 4(+1) + 7(-2) = 0$$ $$x = +5$$ $\mathrm{H_4P_2O_5}$ $$2x + 4(+1) + 5(-2) = 0$$ $$x = +3$$ $\mathrm{H_4P_2O_6}$ $$2x + 4(+1) + 6(-2) = 0$$ $$x = +4$$

Question 65

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

For a reaction of order $n$, the unit of the rate constant is:

  1. $\mathrm{mol}^{1-n} \mathrm{L}^{1-n} \mathrm{s}^{-1}$
  2. $\mathrm{mol}^{1-n} \mathrm{L}^{2n} \mathrm{s}^{-1}$
  3. $\mathrm{mol}^{1-n} \mathrm{L}^{n-1} \mathrm{s}^{-1}$
  4. $\mathrm{mol}^{1-n} \mathrm{L}^{1-n} \mathrm{s}^{-1}$

Answer: (c)

Solution

Rate = k[A]^n Comparing units $$\frac{\mathrm{mol}/\ell}{\mathrm{sec}} = \mathrm{k} \left( \frac{\mathrm{mol}}{\ell} \right)^n$$ $$\Rightarrow \mathrm{k} = \mathrm{mol}^{(1-n)} \ell^{(n-1)} \mathrm{s}^{-1}$$

Question 66

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

Given below are two statements: Statement I: Aniline is less basic than acetamide. Statement II: In aniline, the lone pair of electrons on nitrogen atom is delocalised over benzene ring due to resonance and hence less available to a proton. Choose the most appropriate option:

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

Answer: (b)

Solution

Explanation: aniline is more basic than acetamide because in acetamide, lone pair of nitrogen is delocalised to more electronegative element oxygen. In aniline lone pair of nitrogen delocalised over benzene ring.

Question 67

Chemistry · Co-ordination Compounds · Single correct

The type of hybridisation and magnetic property of the complex $[MnCl_6]^{3-}$, respectively, are:

  1. $sp^3 d^2$ and diamagnetic
  2. $d^2 sp^3$ and diamagnetic
  3. $d^2 sp^3$ and paramagnetic
  4. $sp^3 d^2$ and paramagnetic

Answer: (d)

Solution

The complex $[\mathrm{MnCl}_6]^{3-}$ has the electron configuration for $\mathrm{Mn}^{3+}$ as $[\mathrm{Ar}]\,3d^4$. The $3d$ orbitals have 4 unpaired electrons. This makes the complex paramagnetic and having 4 unpaired electrons.

Question 68

Chemistry · Co-ordination Compounds · Single correct

The number of geometrical isomers found in the metal complexes $[PtCl_2(NH_3)_2]$, $[Ni(CO)_4]$, $[Ru(H_2O)_3Cl_3]$ and $[CoCl_2(NH_3)_4]^+$ respectively, are:

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

Answer: (b)

Solution

[$\mathrm{PtCl_2(NH_3)_2}$] 2 Geometrical isomers [$\mathrm{Ni(CO)_4}$] $\rightarrow$ All ligands are same Zero Geometrical isomers [$\mathrm{Ru(H_2O)_3Cl_3}$] 2 Geometrical isomers [$\mathrm{CoCl_3(NH_3)_3}$]^+ 2 Geometrical isomers

Question 69

Chemistry · Environmental Chemistry · Single correct

Which one of the following statements is NOT correct?

  1. Eutrophication indicates that water body is polluted?
  2. The dissolved oxygen concentration below 6 ppm inhibits fish growth
  3. Eutrophication leads to increase in the oxygen level in water
  4. Eutrophication leads to anaerobic conditions

Answer: (c)

Solution

Eutrophication leads to decrease in oxygen level of water. 3rd statement is incorrect.

Question 70

Chemistry · Structure of Atom · Single correct

Given below are two statements: Statement I : Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom. Statement II : Bohr's model of hydrogen atom contradicts Heisenberg's uncertainty principle. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (d)

Solution

Rutherford's gold foil experiment only proved that electrons are held towards the nucleus by electrostatic forces of attraction and move in circular orbits with very high speeds. Bohr's model gave an exact formula for simultaneous calculation of speed and distance of electron from the nucleus, something which was deemed impossible according to Heisenberg.

Question 71

Chemistry · Hydrocarbons · Single correct

Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction:

  1. $HOCl$
  2. dilute $HNO_2$
  3. Liquid $NH_3$
  4. Concentrated $HIO_3$

Answer: (d)

Solution

Iodination of alkane is a reversible reaction. It can be irreversible in the presence of a strong oxidising agent like conc. $\mathrm{HNO_3}$ or conc. $\mathrm{HIO_3}$.

Question 72

Chemistry · Biomolecules · Single correct

Which one among the following chemical tests is used to distinguish monosaccharide from disaccharide?

  1. Seliwanoff's test
  2. Iodine test
  3. Barfoed test
  4. Tollen's test

Answer: (c)

Solution

Barford test is used for distinguish monosaccharide from disaccharide.

Question 73

Chemistry · Chemistry in Everyday Life · Single correct

Match List-I with List-II : \begin{tabular}{|l|l|} \hline List-I & List-II \\ (Drug) & (Class of Drug) \\ \hline (a) Furacin & (i) Antibiotic \\ \hline (b) Arsphenamine & (ii) Tranquilizers \\ \hline (c) Dimetone & (iii) Antiseptic \\ \hline (d) Valium & (iv) Synthetic antihistamines \\ \hline \end{tabular} Choose the most appropriate match:

  1. (a) - (i), (b) - (iii), $(c)$ - (iv), (d) - (ii)
  2. (a) -(iii), (b)-(iv), $(c)$-(ii), (d)-(i)
  3. (a) -(ii), (b)-(i), $(c)$-(iii), (d)-(iv)
  4. (a) -(iii), (b)-(i), $(c)$-(iv), (d)-(ii)

Answer: (d)

Solution

Furacine acts as antiseptic. Arsphenamine, also known as salvarsan, acts as antibiotic. Dimetone is synthetic histamine. Valium is a tranquilizer.

Question 74

Chemistry · General Principles and Processes of Isolation of Elements · Single correct

The statement that is INCORRECT about Ellingham diagram is

  1. provides idea about the reaction rate.
  2. provides idea about free energy change.
  3. provides idea about changes in the phases during the reaction.
  4. provides idea about reduction of metal oxide.

Answer: (a)

Solution

Ellingham diagram is a plot between $\Delta G^\circ$ and $T$ and does not give any information regarding rate of reaction.

Question 75

Chemistry · Alcohols, Phenols and Ethers · Single correct

Consider the above reaction and identify the Product P:

Answer: (d)

Solution

Dehydration of alcohol by $E^1$ reaction. H-attached at more hindered site while OH attached at less hindered site in HBO reaction.

Question 76

Chemistry · Biomolecules · Single correct

The compound 'A' is a complementary base of in DNA stands.

  1. Uracil
  2. Guanine
  3. Adenine
  4. Cytosine

Answer: (c)

Solution

Given structure is Thymine and Thymine being paired with adenine

Question 77

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

Staggered and eclipsed conformers of ethane are:

  1. Polymers
  2. Rotamers
  3. Enantiomers
  4. Mirror images

Answer: (b)

Solution

Staggered and eclipsed conformers of ethane are also known as rotamers.

Question 78

Chemistry · The s-Block Elements · Single correct

\text{Match List-I with List-II :} \begin{tabular}{|l|l|} \hline $\textbf{List-I}$ & $\textbf{List-II}$ \\[4pt] \hline \text{(a) } $\mathrm{NaOH}$ & \text{(i) Acidic} \\ \hline \text{(b) } $\mathrm{Be(OH)_2}$ & \text{(ii) Basic} \\ \hline \text{(c) } $\mathrm{Ca(OH)_2}$ & \text{(iii) Amphoteric} \\ \hline \text{(d) } $\mathrm{B(OH)_3}$ & \\ \hline \text{(e) } $\mathrm{Al(OH)_3}$ & \\ \hline \end{tabular} Choose the most appropriate answer from the options given below

  1. (a)-(ii), (b)-(ii), ($c$)-(iii), (d)-(ii), (e)-(iii)
  2. (a)-(ii), (b)-(iii), ($c$)-(ii), (d)-(i), (e)-(iii)
  3. (a)-(ii), (b)-(ii), ($c$)-(iii), (d)-(i), (e)-(iii)
  4. (a)-(ii), (b)-(i), ($c$)-(ii), (d)-(iii), (e)-(iii)

Answer: (b)

Solution

$\mathrm{NaOH}\rightarrow$ Basic $\mathrm{Be(OH)_2}\rightarrow$ Amphoteric $\mathrm{Ca(OH)_2}\rightarrow$ Basic $\mathrm{B(OH)_3}\rightarrow$ Acidic $\mathrm{Al(OH)_3}\rightarrow$ Amphoteric

Question 79

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

The correct order of stability of given carbocation is:

  1. A > C > B > D
  2. D > B > C > A
  3. D > B > A > C
  4. C > A > D > B

Answer: (a)

Solution

The stability order is given as follows: $Ph-CH_2^+ > CH_3-CH_2^+ > CH_2=CH^+ > CH \equiv C^+$. The phenyl group is stable due to resonance. The positive charge is more stable on less electronegative elements.

Question 80

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Given below are two statements: One is labelled as Assertion $A$ and the other labelled as Reason $R$. Assertion $A$: Lithium halides are somewhat covalent in nature. Reason $R$: Lithium possess high polarisation capability. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. $A$ is true but $R$ is false
  2. $A$ is false but $R$ is true
  3. Both $A$ and $R$ are true but $R$ is NOT the correct explanation of $A$
  4. Both $A$ and $R$ are true and $R$ is the correct explanation of $A$

Answer: (d)

Solution

Lithium due to small size has very high polarization capability and thus increases covalent nature in Halides.

Question 81

Chemistry · Solutions · Numerical

The density of NaOH solution is $1.2 \, \mathrm{g \, cm^{-3}}$. The molality of this solution is ___ $\mathrm{m}$. (Round off to the Nearest Integer) [Use : Atomic masses: Na : $23.0\mathrm{u}$ O : $16.0\mathrm{u}$ H : $1.0\mathrm{u}$ Density of $\mathrm{H_2O}$ : $1.0 \, \mathrm{g \, cm^{-3}}$]

Answer: 5

Solution

Consider $1 \, \ell$ solution. Mass of solution $= (1.2 \times 1000) \, \mathrm{g}$ $= 1200 \, \mathrm{gm}$ Neglecting volume of NaOH, mass of water $= 1000 \, \mathrm{gm}$ $\Rightarrow$ Mass of NaOH $= (1200 - 1000) \, \mathrm{gm}$ $= 200 \, \mathrm{gm}$ $\Rightarrow$ Moles of NaOH $= \frac{200 \, \mathrm{g}}{50 \, \mathrm{g/mol}} = 5 \, \mathrm{mol}$ $\Rightarrow$ molality $= \frac{5 \, \mathrm{mol}}{1 \, \mathrm{kg}} = 5 \, \mathrm{m}$

Question 82

Chemistry · Surface Chemistry · Numerical

$\mathrm{CO_2}$ gas adsorbs on charcoal following Freundlich adsorption isotherm. For a given amount of charcoal, the mass of $\mathrm{CO_2}$ adsorbed becomes $64$ times when the pressure of $\mathrm{CO_2}$ is doubled. The value of $n$ in the Freundlich isotherm equation is ____ $\times 10^{-2}$. (Round off to the Nearest Integer)

Answer: 17

Question 83

Chemistry · Electrochemistry · Numerical

The conductivity of a weak acid HA of concentration $0.001 \, \mathrm{mol} \, \mathrm{L}^{-1}$ is $2.0 \times 10^{-5} \, \mathrm{S} \, \mathrm{cm}^{-1}$. If $\Lambda_m^\circ(\mathrm{HA}) = 190 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$, the ionization constant $(K_a)$ of HA is equal to $\times 10^{-6}$ (Round off to the Nearest Integer)

Answer: 12

Solution

Given $\Lambda_m = 1000 \times \frac{K}{M}$. $$= 1000 \times \frac{2 \times 10^{-5}}{0.001} = 20 \, \mathrm{S \, cm^2 \, mol^{-1}}$$ Therefore, $$\alpha = \frac{\Lambda_m}{\Lambda_m^\infty} = \frac{20}{190} = \left( \frac{2}{19} \right)$$ The reaction is $$\mathrm{HA \rightleftharpoons H^+ + A^-}$$ The concentrations are $0.001(1 - \alpha)$, $0.001\alpha$, $0.001\alpha$. Thus, $$k_a = 0.001 \left( \frac{\alpha^2}{1 - \alpha} \right) = \frac{0.001 \times \left( \frac{2}{19} \right)^2}{1 - \left( \frac{2}{19} \right)}$$ This simplifies to $$= 12.3 \times 10^{-6}$$

Question 84

Chemistry · Solutions · Numerical

1.46 $\mathrm{g}$ of a biopolymer dissolved in a 100 $\mathrm{mL}$ water at 300 $\mathrm{K}$ exerted an osmotic pressure of 2.42 $\times 10^{-3}$ $\mathrm{bar}$. The molar mass of the biopolymer is $-10^4$ $\mathrm{g \, mol^{-1}}$. (Round off to the Nearest Integer) [Use: R = 0.083 $\mathrm{L}$ \, $\mathrm{bar}$ \, $\mathrm{mol^{-1}}$ \, $\mathrm{K^{-1}}$]

Answer: 15

Solution

Given $\pi = CRT$; $\pi =$ osmotic pressure. $C =$ molarity. $T =$ Temperature of solution. Let the molar mass be $M \mathrm{gm/mol}$. $$2.42 \times 10^{-3} \, \mathrm{bar} = \left( \frac{1.46 \, \mathrm{g}}{M \mathrm{gm/mol}} \right) \times \left( \frac{0.083 \, \mathrm{L} - \mathrm{bar}}{\mathrm{mol} \cdot \mathrm{K}} \right) \times (300 \, \mathrm{K})$$ $$\Rightarrow M = 15.02 \times 10^4 \, \mathrm{g/mol}$$

Question 85

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical

An organic compound is subjected to chlorination to get compound A using 5.0 $\mathrm{g}$ of chlorine. When 0.5 $\mathrm{g}$ of compound A is reacted with $\mathrm{AgNO_3}$ [Carius Method], the percentage of chlorine in compound A is when it forms 0.3849 $\mathrm{g}$ of $\mathrm{AgCl}$. (Round off to the Nearest Integer) (Atomic masses of Ag and Cl are 107.87 and 35.5 respectively)

Answer: 19

Solution

Mass of organic compound = 0.5 $\,$ $\mathrm{gm}$. mass of formed $\mathrm{AgCl}$ = 0.3849 $\,$ $\mathrm{gm}$. $\%$ of Cl = $\frac{atomic mass of Cl \times mass formed \mathrm{AgCl}}{molecular mass of \mathrm{AgCl} \times mass of organic compound}$ $\times$ 100 = $\frac{35.5 \times 0.3849}{143.37 \times 0.5}$ $\times$ 100 = 19.06 $\approx$ 19

Question 86

Chemistry · Co-ordination Compounds · Numerical

The number of geometrical isomers possible in triamminetrinitrocobalt (III) is $X$ and in trioxalatochromate (III) is $Y$. Then the value of $X + Y$ is

Answer: 2

Solution

Triamminetrinitrocobalt(III) goes to $[\mathrm{Co(NO_2)_3(NH_3)_3}]$. Ion goes to $[\mathrm{Cr(C_2O_4)_3}]^{3-}$. $[\mathrm{Co(NO_2)_3(NH_3)_3}]$ has two geometrical isomers (X). $[\mathrm{Cr(C_2O_4)_3}]^{3-}$ has zero geometrical isomer (Y). $X + Y = 2 + 0 = 2.0$

Question 87

Chemistry · Chemical Bonding and Molecular Structure · Numerical

In gaseous triethyl amine the "-C-N-C-" bond angle is ______ degree.

Answer: 108

Solution

In gaseous triethyl amine the "-C-N-C-" bond angle is 108 degree.

Question 88

Chemistry · Thermodynamics · Numerical

For water at $100^\circ\mathrm{C}$ and $1\,\mathrm{bar}$, $\Delta_{\mathrm{vap}}H-\Delta_{\mathrm{vap}}U=$ _______ $\times10^2\,\mathrm{J\,mol^{-1}}$ (Round off to the Nearest Integer) $\left[K_b=0.52\,\mathrm{K\,kg\,mol^{-1}}\right]$ [Use: $R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$] Assume volume of $\mathrm{H_2O(l)}$ is much smaller than volume of $\mathrm{H_2O(g)}$. Assume $\mathrm{H_2O(g)}$ is treated as an ideal gas.

Answer: 31

Solution

Question 89

Chemistry · Equilibrium · Numerical

$\mathrm{PCl_5 \rightleftharpoons PCl_3+Cl_2}$, $K_c=1.844$. $3.0$ moles of $\mathrm{PCl_5}$ are introduced into a $1\,\mathrm{L}$ closed reaction vessel at $380\,\mathrm{K}$. The number of moles of $\mathrm{PCl_5}$ at equilibrium is $\underline{\hspace{1cm}}\times10^{-3}$. (Round off to the nearest integer.)

Answer: 1400

Solution

Given the reaction $\mathrm{PCl_5}_{(g)} \rightleftharpoons \mathrm{PCl_3}_{(g)} + \mathrm{Cl_2}_{(g)}$, with $K_2 = 1.844$. At $t = 0$, there are 3 moles. At $t = \infty$, let the change be $x$ for both products. $$\frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]} = \frac{x^2}{3-x} = 1.844$$ This leads to the equation: $$x^2 + 1.844x - 5.532 = 0$$ Solving for $x$ using the quadratic formula: $$x = \frac{-1.844 + \sqrt{(1.844)^2 + 4 \times 5.532}}{2}$$ This gives $x \approx 1.604$. Therefore, the moles of $\mathrm{PCl_5}$ remaining are $3 - 1.604 \approx 1.396$.

Question 90

Chemistry · Chemical Bonding and Molecular Structure · Fill in the blank

The difference between the bond orders of $\mathrm{CO}$ and $\mathrm{NO^+}$ is $\frac{x}{2}$, where $x=\underline{\hspace{1cm}}$. (Round off to the nearest integer.)

Answer: 0

Solution

Bond order of CO = 3. Bond order of NO^+ = 3. Difference = 0 = $\frac{x}{2}$. x = 0.