JEE Main 31 January 2024 Shift 1 question paper with solutions

JEE Main 31 January 2024 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 · Complex Numbers and Quadratic Equations · Single correct

For $0 < c < b < a$, let $(a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements (I) If $\alpha \in (-1, 0)$, then $b$ cannot be the geometric mean of $a$ and $c$ (II) If $\alpha \in (0, 1)$, then $b$ may be the geometric mean of $a$ and $c$

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

Answer: (a)

Solution

Given $f(x) = (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b)$. $f(x) = a + b - 2c + b + c - 2a + c + a - 2b = 0$. $f(1) = 0$. Therefore, $\alpha \cdot 1 = \frac{c + a - 2b}{a + b - 2c}$. $\alpha = \frac{c + a - 2b}{a + b - 2c}$. If $-1 \frac{a + c}{2}$. Therefore, $b$ cannot be G.M. between $a$ and $c$. If $0 c$ and $b < \frac{a + c}{2}$. Therefore, $b$ may be the G.M. between $a$ and $c$.

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

Let a be the sum of all coefficients in the expansion of $(1 - 2x + 2x^2)^{2023} (3 - 4x^2 + 2x^3)^{2024}$ and $$b = \lim_{x \to 0} \left( \frac{\int_0^x \frac{\log(1+t)}{t^{2024}+1} \, dt}{x^2} \right).$$ If the equations $cx^2 + dx + e = 0$ and $2bx^2 + ax + 4 = 0$ have a common root, where $c, d, e \in \mathbb{R}$, then $d : c : e$ equals

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

Answer: (d)

Solution

Put $x = 1$. Therefore, $a = 1$. $$b = \lim_{x \to 0} \frac{\int_0^x \frac{\ln(1+t)}{1+t^{2024}} \, dt}{x^2}$$ Using L'HOPITAL Rule $$b = \lim_{x \to 0} \frac{\ln(1+x)}{(1+x^{2024})} \times \frac{1}{2x} = \frac{1}{2}$$ Now, $cx^2 + dx + e = 0$, $x^2 + x + 4 = 0$ $(D < 0)$ Therefore, $\frac{c}{1} = \frac{d}{1} = \frac{e}{4}$

Question 3

Maths · Conic Sections · Single correct

If the foci of a hyperbola are same as that of the ellipse $\frac{x^2}{9} + \frac{y^2}{25} = 1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point $\left( \sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}} \right)$ on the hyperbola, is equal to

  1. $7 \sqrt{\frac{2}{5}} - \frac{8}{3}$
  2. $14 \sqrt{\frac{2}{5}} - \frac{4}{3}$
  3. $14 \sqrt{\frac{2}{5}} - \frac{16}{3}$
  4. $7 \sqrt{\frac{2}{5}} + \frac{8}{3}$

Answer: (a)

Solution

$\frac{x^2}{9} + \frac{y^2}{25} = 1$ $a = 3, \quad b = 5$ $e = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} \therefore \text{ foci} = (0, \pm be) = (0, \pm 4)$ $\therefore e_H = \frac{4}{5} \times \frac{15}{8} = \frac{3}{2}$ Let equation of hyperbola: $\frac{x^2}{A^2} - \frac{y^2}{B^2} = -1$ $\therefore B \cdot e_H = 4 \quad \therefore B = \frac{8}{3}$ $\therefore A^2 = B^2(e_H^2 - 1) = \frac{64}{9}\left(\frac{9}{4} - 1\right) \therefore A^2 = \frac{80}{9}$ $\therefore \frac{x^2}{\dfrac{80}{9}} - \frac{y^2}{\dfrac{64}{9}} = -1$ $\text{Directrix : } y = \pm\frac{B}{e_H} = \pm\frac{16}{9}$ $PS = e \cdot PM = \frac{3}{2}\left|\frac{14}{3} \cdot \sqrt{\frac{2}{5}} - \frac{16}{9}\right|$ $= 7\sqrt{\frac{2}{5}} - \frac{8}{3}$

Question 4

Maths · Conic Sections · Single correct

If one of the diameters of the circle $x^2 + y^2 - 10x + 4y + 13 = 0$ is a chord of another circle $C$, whose center is the point of intersection of the lines $2x + 3y = 12$ and $3x - 2y = 5$, then the radius of the circle $C$ is

  1. $\sqrt{20}$
  2. 4
  3. 6
  4. 3$\sqrt{2}$

Answer: (c)

Solution

Given the equations: $$2x + 3y = 12$$ $$3x - 2y = 5$$ Solving these equations, we get: $$13x = 39$$ $$x = 3, \; y = 2$$ The center of the given circle is $(5, -2)$. The radius is calculated as: $$\sqrt{25 + 4 - 13} = 4$$ Therefore, $$CM = \sqrt{4 + 16} = 5\sqrt{2}$$ $$CP = \sqrt{16 + 20} = 6$$

Question 5

Maths · Applications of Integrals · Single correct

The area of the region $$\left\{(x, y) : y^2 \leq 4x,\ x 0,\ x \neq 3\right\}$$ is

  1. $\frac{16}{3}$
  2. $\frac{64}{3}$
  3. $\frac{8}{3}$
  4. $\frac{32}{3}$

Answer: (d)

Solution

Given $y^2 \leq 4x$, $x 0$$ Case-I: $y > 0$ $$\frac{x(x-1)(x-2)}{(x-3)(x-4)} > 0$$ $x \in (0, 1) \cup (2, 3)$ Case-II: $y < 0$ $$\frac{x(x-1)(x-2)}{(x-3)(x-4)} < 0$$ $x \in (1, 2) \cup (3, 4)$ Area $= 2 \int_{0}^{4} \sqrt{x} \, dx$ $$= 2 \cdot \frac{2}{3} \left[ x^{3/2} \right]_{0}^{4} = \frac{32}{3}$$

Question 6

Maths · Relations and Functions · Single correct

If $f(x) = \frac{4x+3}{6x-4}, \ x \neq \frac{2}{3}$ and $(f \circ f)(x) = g(x)$, where$g:\mathbb{R}-\left\{\frac{2}{3}\right\}\rightarrow\mathbb{R}-\left\{\frac{2}{3}\right\}$, then $(gogog)(4)$ is equal to

  1. $-\frac{19}{20}$
  2. $\frac{19}{20}$
  3. $-4$
  4. $4$

Answer: (d)

Solution

Given $f(x) = \frac{4x + 3}{6x - 4}$. $$g(x) = \frac{4 \left( \frac{4x + 3}{6x - 4} \right) + 3}{6 \left( \frac{4x + 3}{6x - 4} \right) - 4} = \frac{34x}{34} = x$$ Therefore, $g(x) = x$. Thus, $g(g(g(4))) = 4$.

Question 7

Maths · Limits and Derivatives · Single correct

$\lim_{x\to0}\frac{e^{4|\sin x|}-2|\sin x|-1}{x^2}$

  1. is equal to -1
  2. does not exist
  3. is equal to 1
  4. is equal to 2

Answer: (d)

Solution

Given $$\lim_{x \to 0} \frac{e^{\tan x} - 2|\sin x| - 1}{x^2}$$ Rewrite as $$\lim_{x \to 0} \frac{e^{4 \sin x} - 2|\sin x| - 1}{|\sin x|^2} \times \frac{\sin^2 x}{x^2}$$ Let $|\sin x| = t$. Then $$\lim_{t \to 0} \frac{e^{2t} - 2t - 1}{t^2} \times \lim_{x \to 0} \frac{\sin^2 x}{x^2}$$ Simplifies to $$= \lim_{t \to 0} \frac{2e^{2t} - 2}{2t} \times 1 = 2 \times 1 = 2$$

Question 8

Maths · Determinants · Single correct

If the system of linear equations $$x - 2y + z = -4$$ $$2x + \alpha y + 3z = 5$$ $$3x - y + \beta z = 3$$ has infinitely many solutions, then $12\alpha + 13\beta$ is equal to

  1. 60
  2. 64
  3. 54
  4. 58

Answer: (d)

Solution

Given $$D = \begin{vmatrix} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{vmatrix}$$ $$= 1(\alpha \beta + 3) + 2(2\beta - 9) + 1(-2 - 3\alpha)$$ $$= \alpha \beta + 3 + 4\beta - 18 - 2 - 3\alpha$$ For infinite solutions $D = 0$, $D_1 = 0$, $D_2 = 0$ and $$D_3 = 0$$ $$D = 0$$ $$\alpha \beta - 3\alpha + 4\beta = 17 \ldots$$ $$D_1 = \begin{vmatrix} -4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta \end{vmatrix} = 0$$ $$D_2 = \begin{vmatrix} 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta \end{vmatrix} = 0$$ $$\Rightarrow 1(5\beta - 9) + 4(2\beta - 9) + 1(6 - 15) = 0$$ $$13\beta - 9 - 36 - 9 = 0$$ $$13\beta = 54, \beta = \frac{54}{13} put in (1)$$ $$\frac{54}{13} \alpha - 3\alpha + 4\left(\frac{54}{13}\right) = 17$$ $$54\alpha - 39\alpha + 216 = 221$$ $$15\alpha = 5 \alpha = \frac{1}{3}$$ Now, $12\alpha + 13\beta = 12 \cdot \frac{1}{3} + 13 \cdot \frac{54}{13}$ $$= 4 + 54 = 58$$

Question 9

Maths · Differential Equations · Single correct

The solution curve of the differential equation $y \frac{dx}{dy} = x (\log_e x - \log_e y + 1)$, $x > 0$, $y > 0$ passing through the point $(e, 1)$ is

  1. $\left| \log_e \frac{y}{x} \right| = x$
  2. $\left| \log_e \frac{y}{x} \right| = y^2$
  3. $\left| \log_e \frac{x}{y} \right| = y$
  4. $2 \left| \log_e \frac{x}{y} \right| = y + 1$

Answer: (c)

Solution

Given $\($ $\frac{dx}{dy}$ = $\frac{x}{y}$ $\left$( $\ln$ $\left$( $\frac{x}{y}$ $\right$) + 1 $\right$) $\)$. Let $\($ $\frac{x}{y}$ = t $\Rightarrow$ x = ty $\)$. Then $\($ $\frac{dx}{dy}$ = t + y $\frac{dt}{dy}$ $\)$. Substitute to get $\($ t + y $\frac{dt}{dy}$ - t($\ln$(t) + 1) $\)$. This simplifies to $\($ y $\frac{dt}{dy}$ - t $\ln$(t) = $\frac{dt}{t \ln(t)}$ = $\frac{dy}{y}$ $\)$. Integrating both sides, we have $\($ $\int$ $\frac{dt}{t \cdot \ln(t)}$ - $\int$ $\frac{dy}{y}$ $\)$. This implies $\($ $\int$ $\frac{d}{p}$ - $\int$ $\frac{dy}{y}$ $\)$ where $\($ $\ln$ t = p $\)$. Thus, $\($ $\frac{1}{t}$ dt = dp $\)$. Integrating gives $\($ $\ln$ p = $\ln$ y + c $\)$. Therefore, $\($ $\ln$($\ln$ t) = $\ln$ y + c $\)$. Substituting back, $\($ $\ln$ $\left$( $\ln$ $\left$( $\frac{x}{y}$ $\right$) $\right$) - $\ln$ y + c $\)$. At $\($ x = e, y = 1 $\)$, $\($ $\ln$ $\left$( $\ln$ $\left$( $\frac{e}{1}$ $\right$) $\right$) - $\ln$(1) + c $\Rightarrow$ c = 0 $\)$. Thus, $\($ $\ln$ $\left$| $\ln$ $\left$( $\frac{x}{y}$ $\right$) $\right$| - $\ln$ y $\)$. Finally, $\($ $\ln$ $\left$( $\frac{x}{y}$ $\right$) $\bigg$| - e^{ky} $\)$ and $\($ $\ln$ $\left$( $\frac{x}{y}$ $\right$) $\bigg$| - y $\)$.

Question 10

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

Let $\alpha, \beta, \gamma, \delta \in \mathbb{Z}$ and let $A(\alpha, \beta)$, $B(1, 0)$, $C(\gamma, \delta)$ and $D(1, 2)$ be the vertices of a parallelogram $ABCD$. If $AB = \sqrt{10}$ and the points $A$ and $C$ lie on the line $3y = 2x + 1$, then $2(\alpha + \beta + \gamma + \delta)$ is equal to

  1. 10
  2. 5
  3. 12
  4. 8

Answer: (d)

Solution

Let $E$ is mid point of diagonals $\dfrac{\alpha + \gamma}{2} = \dfrac{1+1}{2}$ \quad \& \quad $\dfrac{\beta + \delta}{2} = \dfrac{2+0}{2}$ $\alpha + \gamma = 2$ \hfill $\beta + \delta = 2$ $2(\alpha + \beta + \gamma + \delta) = 2(2 + 2) = 8$

Question 11

Maths · Differential Equations · Single correct

Let $y = y(x)$ be the solution of the differential equation $$ \frac{dy}{dx} = \frac{(\tan x) + y}{\sin x (\sec x - \sin x \tan x)}, $$ $$ x \in \left(0, \frac{\pi}{2}\right) satisfying the condition y\left(\frac{\pi}{4}\right) = 2. $$ Then, $y\left(\frac{\pi}{3}\right)$ is

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

Answer: (a)

Solution

Given $\dfrac{dy}{dx}=\dfrac{\sin x+y\cos x}{\sin x\cos x\left(\dfrac{1}{\cos x}-\sin x\cdot\dfrac{\sin x}{\cos x}\right)}$ $=\dfrac{\sin x+y\cos x}{\sin x(1-\sin^2x)}$ $\dfrac{dy}{dx}=\sec^2x+y\cdot2\csc(2x)$ $\dfrac{dy}{dx}-2\csc(2x)\,y=\sec^2x$ $\dfrac{dy}{dx}+py=Q$ I.F. $=e^{\int p\,dx}$ $=e^{-\int 2\csc(2x)\,dx}$ Let $2x=t$. $dx=\dfrac{dt}{2}$ $\mathrm{I.F.}=e^{-\int \csc t\,dt}$ $=e^{-\ln\left|\tan\dfrac{t}{2}\right|}$ $=\dfrac{1}{|\tan x|}$ $y(\mathrm{I.F.})=\int Q(\mathrm{I.F.})\,dx+c$ $\Rightarrow y\cdot\dfrac{1}{|\tan x|}=\int \sec^2x\cdot\dfrac{1}{|\tan x|}\,dx+c$ For $\tan x=t$, $\Rightarrow y\cdot\dfrac{1}{|\tan x|}=\int\dfrac{dt}{|t|}+c$ $=\ln|t|+c$ $\Rightarrow y=|\tan x|(\ln|\tan x|+c)$ Put $x=\dfrac{\pi}{4},\ y=2$. $2=\ln1+c$ $\Rightarrow c=2$ $\therefore y=|\tan x|(\ln|\tan x|+2)$ $\therefore y\left(\dfrac{\pi}{3}\right)=\sqrt{3}\left(\ln\sqrt{3}+2\right)$

Question 12

Maths · Vector Algebra · Single correct

Let $\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}$, $\vec{b} = 4\hat{i} + \hat{j} + 7\hat{k}$ and $\vec{c} = \hat{i} - 3\hat{j} + 4\hat{k}$ be three vectors. If a vectors $\vec{p}$ satisfies $\vec{p} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{p} \cdot \vec{a} = 0$, then $\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k})$ is equal to

  1. 24
  2. 36
  3. 28
  4. 32

Answer: (d)

Solution

Given $\vec{p} \times \vec{b} - \vec{c} \times \vec{b} = \vec{0}$. $$(\vec{p} - \vec{c}) \times \vec{b} = \vec{0}$$ $$\vec{p} - \vec{c} = \lambda \vec{b} \implies \vec{p} = \vec{c} + \lambda \vec{b}$$ Now, $\vec{p} \cdot \vec{a} = 0$ (given). So, $\vec{c} \cdot \vec{a} + \lambda \vec{a} \cdot \vec{b} = 0$. $$(3 - 3 - 8) + \lambda (12 + 1 - 14) = 0$$ $$\lambda = -8$$ $$\vec{p} = \vec{c} - 8 \vec{b}$$ $$\vec{p} = -31\hat{i} - 11\hat{j} - 52\hat{k}$$ So, $\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k})$ $$= -31 + 11 + 52$$ $$= 32$$

Question 13

Maths · Sequences and Series · Single correct

The sum of the series $\frac{1}{1 - 3 \cdot 1^2 + 1^4} + \frac{2}{1 - 3 \cdot 2^2 + 2^4} + \frac{3}{1 - 3 \cdot 3^2 + 3^4} + \ldots$ up to 10 terms is

  1. $\frac{45}{109}$
  2. $-\frac{45}{109}$
  3. $\frac{55}{109}$
  4. $-\frac{55}{109}$

Answer: (d)

Solution

General term of the sequence, $$T_r = \frac{r}{1 - 3r^2 + r^4}$$ $$T_r = \frac{r}{r^4 - 2r^2 + 1 - r^2}$$ $$T_r = \frac{r}{(r^2 - 1)^2 - r^2}$$ $$T_r = \frac{r}{(r^2 - r - 1)(r^2 + r - 1)}$$ $$T_r = \frac{1}{2} \left[ \frac{(r^2 + r - 1) - (r^2 - r - 1)}{(r^2 - r - 1)(r^2 + r - 1)} \right]$$ $$= \frac{1}{2} \left[ \frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1} \right]$$ Sum of 10 terms, $$\sum_{r=1}^{10} T_r = \frac{1}{2} \left[ \frac{1}{1 - 1} - \frac{1}{109} \right] = \frac{-55}{109}$$

Question 14

Maths · Three Dimensional Geometry · Single correct

The distance of the point $Q(0, 2, -2)$ form the line passing through the point $P(5, -4, 3)$ and perpendicular to the lines $\vec{r} = (-3\hat{i} + 2\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 5\hat{k}), \lambda \in \mathbb{R}$ and $\vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \mu (-\hat{i} + 3\hat{j} + 2\hat{k}), \mu \in \mathbb{R}$ is

  1. $\sqrt{86}$
  2. $\sqrt{20}$
  3. $\sqrt{54}$
  4. $\sqrt{74}$

Answer: (d)

Solution

A vector in the direction of the required line can be obtained by cross product of $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix} = -9\hat{i} - 9\hat{j} + 9\hat{k}$$ Required line, $$\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda'(-9\hat{i} - 9\hat{j} + 9\hat{k})$$ $$\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} - \hat{k})$$ Now distance of $$(0, 2, -2)$$ $$P.V. of P = (5 + \lambda) \hat{i} + (\lambda - 4) \hat{j} + (3 - \lambda) \hat{k}$$ $$\overrightarrow{AP} = (5 + \lambda) \hat{i} + (\lambda - 6) \hat{j} + (5 - \lambda) \hat{k}$$ $$\overrightarrow{AP} \cdot (\hat{i} + \hat{j} - \hat{k}) = 0$$ $$5 + \lambda + \lambda - 6 - 5 + \lambda = 0$$ $$\lambda = 2$$ $$|\overrightarrow{AP}| = \sqrt{49 + 16 + 9}$$ $$|\overrightarrow{AP}| = \sqrt{74}$$

Question 15

Maths · Inverse Trigonometric Functions · Single correct

For $\alpha, \beta, \gamma \neq 0$. If $\sin^{-1} \alpha + \sin^{-1} \beta + \sin^{-1} \gamma = \pi$ and $(\alpha + \beta + \gamma)(\alpha - \gamma + \beta) = 3 \alpha \beta$, then $\gamma$ equal to

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

Answer: (a)

Solution

Let $\sin^{-1}\alpha = A$, $\sin^{-1}\beta = B$, $\sin^{-1}\gamma = C$. $A + B + C = \pi$. $(\alpha + \beta)^2 - \gamma^2 = 3\alpha\beta$. $\alpha^2 + \beta^2 - \gamma^2 = \alpha\beta$. $$\frac{\alpha^2 + \beta^2 - \gamma^2}{2\alpha\beta} = \frac{1}{2}$$ Therefore, $\cos C = \frac{1}{2}$. $\sin C = \gamma$. $\cos C = \sqrt{1 - \gamma^2} = \frac{1}{2}$. $\gamma = \frac{\sqrt{3}}{2}$.

Question 16

Maths · Probability · Single correct

Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is

  1. $\frac{2}{25}$
  2. $\frac{4}{25}$
  3. $\frac{2}{3}$
  4. $\frac{4}{75}$

Answer: (d)

Solution

Probability of drawing first red and then white $$= \frac{10}{75} \times \frac{30}{75} = \frac{4}{75}$$

Question 17

Maths · Continuity and Differentiability · Single correct

Let $g(x)$ be a linear function and $f(x) = \begin{cases} g(x), & x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}, & x > 0 \end{cases}$, is continuous at $x = 0$. If $f'(1) = f(-1)$, then the value of $g(3)$ is

  1. $\frac{1}{3} \log_e \left( \frac{4}{9e^{1/3}} \right)$
  2. $\frac{1}{3} \log_e \left( \frac{4}{9} \right) + 1$
  3. $\log_9 \left( \frac{4}{9} \right) - 1$
  4. $\log_6 \left( \frac{4}{9e^{1/3}} \right)$

Answer: (d)

Solution

Let $g(x) = ax + b$. Now function $f(x)$ is continuous at $x = 0$. Therefore, $\lim_{x \to 0^+} f(x) = f(0)$. $$\lim_{x \to 0} \left( \frac{1+x}{2+x} \right)^{\frac{1}{x}} = b$$ Thus, $0 = b$. Therefore, $g(x) = ax$. Now, for $x > 0$, $$f'(x) = \frac{1}{x} \cdot \left( \frac{1+x}{2+x} \right)^{\frac{1}{x} - 1} \cdot \frac{1}{(2+x)^2}$$ $$+ \left( \frac{1+x}{2+x} \right)^{\frac{1}{x}} \cdot \ln \left( \frac{1+x}{2+x} \right) \cdot \left( -\frac{1}{x^2} \right)$$ Therefore, $f'(1) = \frac{1}{9} - 2 \cdot \frac{3}{2} \cdot \ln \left( \frac{2}{3} \right)$. And $f(-1) = g(-1) = -a$. Therefore, $a = \frac{2}{3} \ln \left( \frac{2}{3} \right) - \frac{1}{9}$. Thus, $g(3) = 2 \ln \left( \frac{2}{3} \right) - \frac{1}{3}$. $$= \ln \left( \frac{4}{9 \cdot e^{1/3}} \right)$$

Question 18

Maths · Determinants · Single correct

If $f(x) = \begin{vmatrix} x^3 & 2x^2 + 1 & 1 + 3x \\ 3x^2 + 2 & 2x & x^3 + 6 \\ x^3 - x & 4 & x^2 - 2 \end{vmatrix}$ for all $x \in \mathbb{R}$, then $2f(0) + f'(0)$ is equal to

  1. 48
  2. 24
  3. 42
  4. 18

Answer: (c)

Solution

Given $$f(0) = \begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix} = 12$$ $$f'(x) = \begin{vmatrix} 3x^2 & 4x & 3 \\ 3x^2 + 2 & 2x & x^3 + 6 \\ x^3 - x & 4 & x^2 - 2 \end{vmatrix} + \begin{vmatrix} x^3 & 2x^2 + 1 & 1 + 3x \\ 6x & 2 & 3x^2 \\ x^3 - x & 4 & x^2 - 2 \end{vmatrix} + \begin{vmatrix} x^3 & 2x^2 + 1 & 1 + 3x \\ 3x^2 + 2 & 2x & x^3 + 6 \\ 3x^2 - 1 & 0 & 2x \end{vmatrix}$$ Therefore, $$f'(0) = \begin{vmatrix} 0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix} + \begin{vmatrix} 0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2 \end{vmatrix} - \begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0 \end{vmatrix}$$ $$= 24 - 6 = 18$$ Therefore, $$2f(0) + f'(0) = 42$$

Question 19

Maths · Probability · Single correct

Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable $x$ to be the number of rotten apples in a draw of two apples, the variance of $x$ is

  1. $\frac{37}{153}$
  2. $\frac{57}{153}$
  3. $\frac{47}{153}$
  4. $\frac{40}{153}$

Answer: (d)

Solution

3 bad apples, 15 good apples. Let X be no of bad apples. Then $P(X = 0) = \frac{{^{15}C_2}}{{^{18}C_2}} = \frac{105}{153}$. $P(X = 1) = \frac{{^3C_1 \times ^{15}C_1}}{{^{18}C_2}} = \frac{45}{153}$. $P(X = 2) = \frac{{^3C_2}}{{^{18}C_2}} = \frac{3}{153}$. $E(X) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + 2 \times \frac{3}{153} = \frac{51}{153} = \frac{1}{3}$. $Var(X) = E(X^2) - (E(X))^2$. $= 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + 4 \times \frac{3}{153} - \left(\frac{1}{3}\right)^2$. $= \frac{57}{153} - \frac{1}{9} = \frac{40}{153}$.

Question 20

Maths · Complex Numbers and Quadratic Equations · Single correct

Let S be the set of positive integral values of a for which $$\frac{ax^2 + 2(a+1)x + 9a + 4}{x^2 - 8x + 32} < 0, \forall x \in \mathbb{R}$$. Then, the number of elements in S is:

  1. 1
  2. 0
  3. $\infty$
  4. 3

Answer: (b)

Solution

Given $ax^2 + 2(a+1)x + 9a + 4 < 0 \forall x \in \mathbb{R}$. Therefore, $a < 0$.

Question 21

Maths · Integrals · Numerical

If the integral $525 \int_{0}^{\frac{\pi}{2}} \sin 2x \cos^{\frac{11}{2}} x \left(1 + \cos^{\frac{5}{2}} x \right)^{\frac{1}{2}} \, dx$ is equal to $(n \sqrt{2} - 64)$, then $n$ is equal to _______.

Answer: 176

Solution

Given $$I = \int_0^\pi \sin 2x \cdot (\cos x)^{\frac{11}{2}} \left(1 + (\cos x)^{\frac{5}{2}}\right)^{\frac{1}{2}} \, dx$$ Put $\cos x = t^2 \Rightarrow \sin x \, dx = -2t \, dt$ Therefore, $$I = 4 \int_0^1 t^2 \cdot t^{11} \sqrt{1 + t^5} (t) \, dt$$ $$I = 4 \int_0^1 t^{14} \sqrt{1 + t^5} \, dt$$ Put $1 + t^5 = k^2$ $$\Rightarrow 5t^4 \, dt = 2k \, dk$$ Therefore, $$I = 4 \cdot \int_1^{\sqrt{2}} (k^2 - 1)^2 \cdot k \cdot \frac{2k}{5} \, dk$$ $$I = \frac{8}{5} \int_1^{\sqrt{2}} k^6 - 2k^4 + k^2 \, dk$$ $$I = \frac{8}{5} \left[ \frac{k^7}{7} - \frac{2k^5}{5} + \frac{k^3}{3} \right]_1^{\sqrt{2}}$$ $$I = \frac{8}{5} \left[ \frac{8\sqrt{2}}{7} - \frac{8\sqrt{2}}{5} + \frac{2\sqrt{2}}{3} - \frac{1}{7} + \frac{2}{5} - \frac{1}{3} \right]$$ $$I = \frac{8}{5} \left[ \frac{22\sqrt{2}}{105} - \frac{8}{105} \right]$$ Therefore, $$525 \cdot I = 176\sqrt{2} - 64$$

Question 22

Maths · Applications of Derivatives · Numerical

Let $S=(-1,\infty)$ and $f:S\to\mathbb{R}$ be defined as \[ f(x)=\int_{-1}^{x}(e^t-1)^{11}(2t-1)^5(t-2)^7(t-3)^{12}(2t-10)^{61}\,dt \] Let $p=$ Sum of square of the values of $x$, where $f(x)$ attains local maxima on $S$. and $q=$ Sum of the values of $x$, where $f(x)$ attains local minima on $S$. Then, the value of $p^2+2q$ is ________.

Answer: 27

Solution

Given $$f'(x) = (e^x - 1)^{11} (2x - 1)^5 (x - 2)^7 (x - 3)^{12} (2x - 10)^{61}$$ Local minima at $x = \frac{1}{2}$, $x = 5$ Local maxima at $x = 0$, $x = 2$ Therefore, $p = 0 + 4 = 4$, $q = \frac{1}{2} + 5 = \frac{11}{2}$ Then $p^2 + 2q = 16 + 11 = 27$

Question 23

Maths · Permutations and Combinations · Numerical

The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to _______.

Answer: 3734

Solution

We have III, TT, D, S, R, B, U, O, N Number of words with selection (a, a, b) $$= \binom{8}{1} \times \frac{4!}{3!} = 32$$ Number of words with selection (a, a, b, b) $$= \frac{4!}{2!2!} = 6$$ Number of words with selection (a, a, b, c) $$= \binom{2}{1} \times \binom{8}{2} \times \frac{4!}{2!} = 672$$ Number of words with selection (a, b, c, d) $$= \binom{9}{4} \times 4! = 3024$$ Therefore, total = 3024 + 672 + 6 + 32 $$= 3734$$

Question 24

Maths · Three Dimensional Geometry · Numerical

Let $Q$ and $R$ be the feet of perpendiculars from the point $P(a, a, a)$ on the lines $x = y$, $z = 1$ and $x = -y$, $z = -1$ respectively. If $\angle QPR$ is a right angle, then $12a^2$ is equal to _____.

Answer: 12

Solution

Given $\($ $\frac{x}{1}$ = $\frac{y}{1}$ = $\frac{z-1}{0}$ = r $\rightarrow$ Q(r, r, 1) $\)$ and $\($ $\frac{x}{1}$ = $\frac{y}{-1}$ = $\frac{z+1}{0}$ = k $\rightarrow$ R(k, -k, -1) $\)$. $\($ $\overline{PQ}$ = (a-r)$\hat{i}$ + (a-r)$\hat{j}$ + (a-1)$\hat{k}$ $\)$ $\($ a = r + a - r = 0 $\)$ $\($ 2a = 2r $\rightarrow$ a = r $\)$ $\($ $\overline{PR}$ = (a-k)$\hat{i}$ + (a+k)$\hat{j}$ + (a+1)$\hat{k}$ $\)$ $\($ a-k-a-k = 0 $\Rightarrow$ k = 0 $\)$ As, $\($ PQ $\perp$ PR $\)$ $\($ (a-r)(a-k) + (a-r)(a+k) + (a-1)(a+1) = 0 $\)$ $\($ a = 1 or -1 $\)$ $\($ 12a^2 = 12 $\)$

Question 25

Maths · Binomial Theorem · Numerical

In the expansion of $(1+x)(1-x^2)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5$, $x \neq 0$, the sum of the coefficient of $x^3$ and $x^{-13}$ is equal to _______.

Answer: 118

Solution

Given $$(1+x)(1-x^2)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5$$ This simplifies to $$(1+x)(1-x^2)\left(\left(1+\frac{1}{x}\right)^3\right)^5$$ Which further simplifies to $$\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}}$$ This equals $$\frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}$$ The coefficient of $x^3$ in the expansion is approximately the coefficient of $x^{18}$ in $$(1+x)^{17} - x(1+x)^{17}$$ This equals $$0 - 1$$ Which simplifies to $$-1$$ The coefficient of $x^{-13}$ in the expansion is approximately the coefficient of $x^2$ in $$(1+x)^{17} - x(1+x)^{17}$$ This equals $$\binom{17}{2} - \binom{17}{1}$$ Which simplifies to $$17 \times 8 - 17$$ This equals $$17 \times 7$$ Which simplifies to $$119$$ Hence Answer = 119 - 1 = 118

Question 26

Maths · Complex Numbers and Quadratic Equations · Numerical

If $\alpha$ denotes the number of solutions of $|1 - i|^x = 2^x$ and $\beta = \left( \frac{|z|}{\arg(z)} \right)$, where $$z = \frac{\pi}{4} (1 + i)^4 \left( \frac{1 - \sqrt{\pi} i}{\sqrt{\pi} + i} + \frac{\sqrt{\pi} - i}{1 + \sqrt{\pi}} \right), i = \sqrt{-1},$$ then the distance of the point $(\alpha, \beta)$ from the line $4x - 3y = 7$ is _______.

Answer: 3

Solution

Given $\left(\sqrt{2}\right)^x = 2^x \Rightarrow x = 0 \Rightarrow \alpha = 1$. $$z = \frac{\pi}{4} (1+i)^4 \left[ \frac{\sqrt{\pi} - \pi i - i - \sqrt{\pi}}{\pi + 1} + \frac{\sqrt{\pi} - i - \pi i - \sqrt{\pi}}{1 + \pi} \right]$$ $$= -\frac{\pi i}{2} (1 + 4i + 6i^2 + 4i^3 + 1)$$ $$= 2\pi i$$ $$\beta = \frac{2\pi}{\frac{\pi}{2}} = 4$$ Distance from $(1, 4)$ to $4x - 3y = 7$ Will be $\frac{15}{5} = 3$

Question 27

Maths · Conic Sections · Numerical

Let the foci and length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ be $(\pm 5, 0)$ and $\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1$ equals

Answer: 51

Solution

Focii are $\equiv (\pm 5, 0)$; $\frac{2b^2}{a} = \sqrt{50}$. Let $a = 5$ and $b^2 = \frac{5\sqrt{2}a}{2}$. Then $b^2 = a^2 (1 - e^2) = \frac{5\sqrt{2}a}{2}$. This implies $a (1 - e^2) = \frac{5\sqrt{2}}{2}$. Thus, $\frac{5}{e} (1 - e^2) = \frac{5\sqrt{2}}{2}$. Solving $\sqrt{2} - \sqrt{2}e^2 = e$, we get $\sqrt{2}e^2 + e - \sqrt{2} = 0$. This simplifies to $\sqrt{2}e + 2e - e - \sqrt{2} = 0$. Further simplification gives $\sqrt{2}e(e + \sqrt{2}) - 1(1 + \sqrt{2}) = 0$. Therefore, $(e + \sqrt{2})(\sqrt{2}e - 1) = 0$. Thus, $e \neq -\sqrt{2}; e = \frac{1}{\sqrt{2}}$. The equation is $\frac{x^2}{b^2} - \frac{y^2}{a^2 b^2} = 1$ with $a = 5\sqrt{2}$ and $b = 5$. Then $a^2b^2 = b^2 (e_1^2 - 1) \Rightarrow e_1^2 = 51$.

Question 28

Maths · Vector Algebra · Numerical

Let $\vec{a}$ and $\vec{b}$ be two vectors such that |$\vec{a}$| = 1, |$\vec{b}$| = 4 and $\vec{a}$ $\cdot$ $\vec{b}$ = 2. If $\vec{c}$ = (2$\vec{a}$ $\times$ $\vec{b}$) - 3$\vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\alpha$, then 192 $\sin^2$ $\alpha$ is equal to .

Answer: 48

Solution

Given $\vec{b} \cdot \vec{c} = (2 \vec{a} \times \vec{b}) \cdot \vec{b} - 3 |\vec{b}|^2$. $|\vec{b}||\vec{c}| \cos \alpha = -3 |\vec{b}|^2$. $|\vec{c}| \cos \alpha = -12$, as $|\vec{b}| = 4$. $\vec{a} \cdot \vec{b} = 2$. $\cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}$. $|\vec{c}|^2 = |(2 \vec{a} \times \vec{b}) - 3 \vec{b}|^2$. $= 64 \times \frac{3}{4} + 144 = 192$. $|\vec{c}|^2 \cos^2 \alpha = 144$. $192 \cos^2 \alpha = 144$. $192 \sin^2 \alpha = 48$.

Question 29

Maths · Relations and Functions · Numerical

Let $\mathcal{A} = \{1, 2, 3, 4\}$ and $R = \{(1, 2), (2, 3), (1, 4)\}$ be a relation on $\mathcal{A}$. Let $S$ be the equivalence relation on $\mathcal{A}$ such that $R \subseteq S$ and the number of elements in $S$ is $n$. Then, the minimum value of $n$ is $\ldots$

Answer: 16

Solution

Question 30

Maths · Integrals · Numerical

Let f : $\mathbb{R} \to \mathbb{R}$ be a function defined by $f(x) = \frac{4^x}{4^x + 2}$ and $$M = \int_{f(a)}^{f(1-a)} x \sin^4(x(1-x)) \, dx$$ $$N = \int_{f(a)}^{f(1-a)} \sin^4(x(1-x)) \, dx; \ a \neq \frac{1}{2}.$$ If $\alpha M = \beta N, \alpha, \beta \in \mathbb{N}$, then the least value of $\alpha^2 + \beta^2$ is equal to _______.

Answer: 5

Solution

Given $f(a) + f(1-a) = 1$. $$M = \int_{f(a)}^{f(1-a)} (1-x) \cdot \sin^4 x (1-x) \, dx$$ M = N - M 2M = N $\alpha = 2$; $\beta = 1$; Ans. 5

Physics

Question 31

Physics · Kinetic Theory · Single correct

The parameter that remains the same for molecules of all gases at a given temperature is :

  1. kinetic energy
  2. momentum
  3. mass
  4. speed

Answer: (a)

Solution

Q11 $$\mathrm{KE} = \frac{f}{2} k T$$ Conceptual

Question 32

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

Identify the logic operation performed by the given circuit.

  1. NAND
  2. NOR
  3. OR
  4. AND

Answer: (c)

Solution

Using De-Morgan's law, we have: $$Y = \overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B.$$

Question 33

Physics · Motion in a Straight Line · Single correct

The relation between time ' t ' and distance ' x ' is $t = \alpha x^2 + \beta x$, where $\alpha$ and $\beta$ are constants. The relation between acceleration (a) and velocity (v) is:

  1. $a = -2\alpha v^3$
  2. $a = -5\alpha v^5$
  3. $a = -3\alpha v^2$
  4. $a = -4\alpha v^4$

Answer: (a)

Solution

Given $t = \alpha x^2 + \beta x$ (differentiating with respect to time) $$\frac{dt}{dx} = 2\alpha x + \beta$$ $$\frac{1}{v} = 2\alpha x + \beta$$ Differentiating with respect to time $$-\frac{1}{v^2} \frac{dv}{dt} = 2\alpha \frac{dx}{dt}$$ $$\frac{dv}{dt} = -2\alpha v^3$$

Question 34

Physics · Ray Optics and Optical Instruments · Single correct

The refractive index of a prism with apex angle $A$ is $\cot \frac{A}{2}$. The angle of minimum deviation is:

  1. $\delta_m = 180^\circ - A$
  2. $\delta_m = 180^\circ - 3A$
  3. $\delta_m = 180^\circ - 4A$
  4. $\delta_m = 180^\circ - 2A$

Answer: (d)

Solution

Given $$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\frac{A}{2}}$$ We have $$\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\frac{A}{2}}$$ This implies $$\sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)$$ Therefore, $$\frac{\pi}{2} - \frac{A}{2} = \frac{A}{2} + \frac{\delta_m}{2}$$ Solving for $\delta_m$ gives $$\delta_m = \pi - 2A$$

Question 35

Physics · Moving Charges and Magnetism · Single correct

A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immerged in a perpendicular magnetic field $B = B_0 \hat{j}$ as shown in figure. The magnetic force on the wire if it has a current $i$ is :

  1. $-iBR \hat{j}$
  2. $2iBR \hat{j}$
  3. $iBR \hat{j}$
  4. $-2iBR \hat{j}$

Answer: (d)

Solution

Note: Direction of magnetic field is in $+\hat{k}$. $$\vec{F} = i \vec{\ell} \times \vec{B}$$ $$\ell = 2R$$ $$\vec{F} = -2iRB \hat{j}$$

Question 36

Physics · Dual Nature of Radiation and Matter · Single correct

If the wavelength of the first member of Lyman series of hydrogen is $\lambda$. The wavelength of the second member will be

  1. $\frac{27}{32} \lambda$
  2. $\frac{32}{27} \lambda$
  3. $\frac{27}{5} \lambda$
  4. $\frac{5}{27} \lambda$

Answer: (a)

Solution

Given $$\frac{1}{\lambda} = \frac{13.6 z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] \cdots (i)$$ $$\frac{1}{\lambda'} = \frac{13.6 z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{3^2} \right] \cdots (ii)$$ On dividing (i) and (ii) $$\lambda' = \frac{27}{32} \lambda$$

Question 37

Physics · Gravitation · Single correct

Four identical particles of mass $m$ are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is $\left( \frac{2\sqrt{2} + 1}{32} \right) \frac{Gm^2}{L^2}$, the length of the sides of the square is

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

Answer: (b)

Solution

The net force $F_{net}$ is given by $F_{net} = \sqrt{2}F + F'$. The force $F$ is $F = \frac{Gm^2}{a^2}$ and $F'$ is $F' = \frac{Gm^2}{(\sqrt{2}a)^2}$. Therefore, $$F_{net} = \sqrt{2}\frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2}$$ Simplifying, $$\left(\frac{2\sqrt{2} + 1}{32}\right)\frac{Gm^2}{L^2} = \frac{G^2}{a^2}\left(\frac{2\sqrt{2} + 1}{2}\right)$$ Thus, $a = 4L$.

Question 38

Physics · Thermodynamics · Single correct

The given figure represents two isobaric processes for the same mass of an ideal gas, then

  1. $P_2 \geq P_1$
  2. $P_2 > P_1$
  3. $P_1 = P_2$
  4. $P_1 > P_2$

Answer: (d)

Solution

Given $PV = nRT$. $V = \left( \frac{nR}{P} \right) T$. Slope $= \frac{nR}{P}$. Slope $\propto \frac{1}{P}$. $(Slope)_2 > (Slope)_1$. $P_2 < P_1$.

Question 39

Physics · Mathematics in Physics · Single correct

If the percentage errors in measuring the length and the diameter of a wire are 0.1$\%$ each. The percentage error in measuring its resistance will be:

  1. 0.2$\%$
  2. 0.3$\%$
  3. 0.1$\%$
  4. 0.144$\%$

Answer: (b)

Solution

Given the formula for resistance, $$R = \frac{\rho L}{\pi \frac{d^2}{4}}$$ we have the relative change in resistance as $$\frac{\Delta R}{R} = \frac{\Delta L}{L} + \frac{2 \Delta d}{d}.$$ Given $$\frac{\Delta L}{L} = 0.1\%$$ and $$\frac{\Delta d}{d} = 0.1\%,$$ we find $$\frac{\Delta R}{R} = 0.3\%.$$

Question 40

Physics · Electromagnetic Waves · Single correct

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of $5 \times 10^{10} \, \mathrm{Hz}$ and an amplitude of $50 \, \mathrm{Vm}^{-1}$. The total average energy density of the electromagnetic field of the wave is: [Use $\varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{Nm}^2$]

  1. $1.106 \times 10^{-8} \, \mathrm{Jm}^{-3}$
  2. $4.425 \times 10^{-8} \, \mathrm{Jm}^{-3}$
  3. $2.212 \times 10^{-8} \, \mathrm{Jm}^{-3}$
  4. $2.212 \times 10^{-10} \, \mathrm{Jm}^{-3}$

Answer: (a)

Solution

Given $$U_E = \frac{1}{2} \varepsilon_0 E^2$$ Substituting the values, $$U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2$$ $$= 1.106 \times 10^{-8} \, \mathrm{J/m^3}$$

Question 41

Physics · Physical World, Units and Measurements · Single correct

\quad \text{A force is represented by } F=ax^2+bt^{1/2} \text{where }x=\text{distance and }t=\text{time. The dimensions of }\frac{b^2}{a}\text{ are:}

  1. $[ML^3T^{-1}]$
  2. $[MLT^{-2}]$
  3. $[ML^{-1}T^{-1}]$
  4. $[ML^2T^{-3}]$

Answer: (a)

Solution

Question 42

Physics · Electric Charges and Fields · Single correct

Two charges $q$ and $3q$ are separated by a distance $r$ in air. At a distance $x$ from charge $q$, the resultant electric field is zero. The value of $x$ is:

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

Answer: (c)

Solution

The net electric field at point P is zero. $$\left( \vec{E}_{net} \right)_P = 0$$ Therefore, $$\frac{kq}{x^2} = \frac{k \cdot 3q}{(r-x)^2}$$ Simplifying gives: $$(r-x)^2 = 3x^2$$ Solving for $r-x$: $$r-x = \sqrt{3}x$$ Solving for $x$: $$x = \frac{r}{\sqrt{3} + 1}$$

Question 43

Physics · Laws of Motion · Single correct

In the given arrangement of a doubly inclined plane two blocks of masses $M$ and $m$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is $0.25$. The value of $m$, for which $M = 10 \, \mathrm{kg}$ will move down with an acceleration of $2 \, \mathrm{m/s^2}$, is : ( take $g = 10 \, \mathrm{m/s^2}$ and $\tan 37^\circ = 3/4$)

  1. 9 kg
  2. 4.5 kg
  3. 6.5 kg
  4. 2.25 kg

Answer: (b)

Solution

For M block $$10 \, g \sin 53^\circ - \mu (10 \, g) \cos 53^\circ - T = 10 \times 2$$ $$T = 80 - 15 - 20$$ $$T = 45 \, \mathrm{N}$$ For m block $$T - mg \sin 37^\circ - \mu mg \cos 37^\circ = m \times 2$$ $$45 = 10 \, m$$ $$m = 4.5 \, \mathrm{kg}$$

Question 44

Physics · Moving Charges and Magnetism · Single correct

A coil is placed perpendicular to a magnetic field of 5000 $\mathrm{T}$. When the field is changed to 3000 $\mathrm{T}$ in 2 $\mathrm{s}$, an induced emf of 22 $\mathrm{V}$ is produced in the coil. If the diameter of the coil is 0.02 $\mathrm{m}$, then the number of turns in the coil is:

  1. 7
  2. 70
  3. 35
  4. 140

Answer: (b)

Solution

Given $\varepsilon = N \left( \frac{\Delta \phi}{t} \right)$. $\Delta \phi = (\Delta B)A$. $B_i = 5000 \, \mathrm{T}$, $B_f = 3000 \, \mathrm{T}$. $d = 0.02 \, \mathrm{m}$, $r = 0.01 \, \mathrm{m}$. $\Delta \phi = (\Delta B)A = (2000) \pi (0.01)^2 = 0.2 \pi$. $\varepsilon = N \left( \frac{\Delta \phi}{t} \right) \Rightarrow 22 = N \left( \frac{0.2 \pi}{2} \right)$. $N = 70$.

Question 45

Physics · Waves · Single correct

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe will be:

  1. 60 cm
  2. 45 cm
  3. 30 cm
  4. 15 cm

Answer: (d)

Solution

Given $\dfrac{\lambda}{4} = L_1$, $2\left(\dfrac{\lambda}{2}\right) = \lambda$. $v = f\lambda$, $\quad f_2 = \dfrac{2v}{2L_2}$ $v = f_1(4L_1)$, $\quad f_2 = \dfrac{v}{L_2}$ $f_1 = \dfrac{v}{4L_1}$ $f_1 = f_2$ $$\frac{v}{4L_1} = \frac{v}{L_2}$$ $$\Rightarrow L_2 = 4L_1$$ $60 = 4 \times L_1$ $L_1 = 15\,\mathrm{cm}$

Question 46

Physics · Motion in a Straight Line · Single correct

A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity time graph for the transit of the ball?

Answer: (b)

Solution

The equation of motion is given by $$mg - F_B - F_v = ma$$ Substituting the expressions for the forces, we have $$\left( \rho \frac{4}{3} \pi r^3 \right) g - \left( \rho_L \frac{4}{3} \pi r^3 \right) g - 6 \pi \eta r v = m \frac{dv}{dt}$$ Let $$\frac{4}{3} \pi R^3 g (\rho - \rho_L) = \mathrm{K}_1$$ and $$\frac{6 \pi \eta r}{m} = \mathrm{K}_2$$ Then the equation becomes $$\frac{dv}{dt} = \mathrm{K}_1 - \mathrm{K}_2 v$$ Integrating both sides, we get $$\int_0^v \frac{dv}{\mathrm{K}_1 - \mathrm{K}_2 v} = \int_0^t dt$$ This simplifies to $$-\frac{1}{\mathrm{K}_2} \ln[\mathrm{K}_1 - \mathrm{K}_2 v] \bigg|_0^v = t$$ Simplifying further, we have $$\ln \left( \frac{\mathrm{K}_1 - \mathrm{K}_2 v}{\mathrm{K}_1} \right) = -\mathrm{K}_2 t$$ Solving for $v$, we find $$\mathrm{K}_1 - \mathrm{K}_2 v = \mathrm{K}_1 e^{-\mathrm{K}_2 t}$$ Thus, $$v = \frac{\mathrm{K}_1}{\mathrm{K}_2} \left[ 1 - e^{-\mathrm{K}_2 t} \right]$$

Question 47

Physics · Laws of Motion · Single correct

A coin is placed on a disc. The coefficient of friction between the coin and the disc is $\mu$. If the distance of the coin from the center of the disc is $r$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is :

  1. $\frac{\mu g}{r}$
  2. $\sqrt{\frac{r}{\mu g}}$
  3. $\sqrt{\frac{\mu g}{r}}$
  4. $\frac{\mu}{\sqrt{rg}}$

Answer: (c)

Solution

The normal force is given by $N = mg$. The frictional force is $f = m \omega^2 r$. Also, $f = \mu N$. Equating the two expressions for $f$, we have $\mu mg = mr \omega^2$. Solving for $\omega$, we get $$\omega = \sqrt{\frac{\mu g}{r}}.$$

Question 48

Physics · Thermal Properties of Matter · Single correct

Two conductors have the same resistances at 0^$\circ$ $\mathrm{C}$ but their temperature coefficients of resistance are $\alpha_1$ and $\alpha_2$. The respective temperature coefficients for their series and parallel combinations are:

  1. $\alpha_1 + \alpha_2$, $\frac{\alpha_1 + \alpha_2}{2}$
  2. $\frac{\alpha_1 + \alpha_2}{2}$, $\frac{\alpha_1 + \alpha_2}{2}$
  3. $\alpha_1 + \alpha_2$, $\frac{\alpha_1 \alpha_2}{\alpha_1 + \alpha_2}$
  4. $\frac{\alpha_1 + \alpha_2}{2}$, $\alpha_1 + \alpha_2$

Answer: (b)

Solution

Series: $R_{eq} = R_1 + R_2$ $$2R \left(1 + \alpha_{eq} \Delta \theta \right) = R \left(1 + \alpha_1 \Delta \theta \right) + R \left(1 + \alpha_2 \Delta \theta \right)$$ $$2R \left(1 + \alpha_{eq} \Delta \theta \right) = 2R + (\alpha_1 + \alpha_2) R \Delta \theta$$ Parallel: $$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$ $$\frac{1}{\frac{R}{2} \left(1 + \alpha_{eq} \Delta \theta \right)} = \frac{1}{R \left(1 + \alpha_1 \Delta \theta \right)} + \frac{1}{R \left(1 + \alpha_2 \Delta \theta \right)}$$ $$2 \left[ (1 + \alpha_1 \Delta \theta) (1 + \alpha_2 \Delta \theta) \right]$$ $$= \left[ 2 + (\alpha_1 + \alpha_2) \Delta \theta \right] \left[ 1 + \alpha_{eq} \Delta \theta \right]$$ $$2 \left[ 1 + \alpha_1 \Delta \theta + \alpha_2 \Delta \theta + \alpha_1 \alpha_2 \Delta \theta \right]$$ $$= 2 + 2 \alpha_{eq} \Delta \theta + (\alpha_1 + \alpha_2) \Delta \theta + \alpha_{eq} (\alpha_1 + \alpha_2) \Delta \theta^2$$ Neglecting small terms $$2 + 2 (\alpha_1 + \alpha_2) \Delta \theta = 2 + 2 \alpha_{eq} \Delta \theta + (\alpha_1 + \alpha_2) \Delta \theta$$ $$(\alpha_1 + \alpha_2) \Delta \theta = 2 \alpha_{eq} \Delta \theta$$ $$\alpha_{eq} = \frac{\alpha_1 + \alpha_2}{2}$$

Question 49

Physics · Work, Energy and Power · Single correct

An artillery piece of mass $M_1$ fires a shell of mass $M_2$ horizontally. Instaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:

  1. $M_1 / (M_1 + M_2)$
  2. $M_2 / M_1$
  3. $M_2 / (M_1 + M_2)$
  4. $M_1 / M_2$

Answer: (b)

Solution

Given $|\vec{p}_1| = |\vec{p}_2|$. The kinetic energy is given by $\mathrm{KE} = \frac{p^2}{2M}$; $p$ is the same. Thus, $\mathrm{KE} \propto \frac{1}{m}$. Therefore, $$\frac{\mathrm{KE}_1}{\mathrm{KE}_2} = \frac{p^2/2M_1}{p^2/2M_2} = \frac{M_2}{M_1}.$$

Question 50

Physics · Dual Nature of Radiation and Matter · Single correct

When a metal surface is illuminated by light of wavelength $\lambda$, the stopping potential is $8 \, \mathrm{V}$. When the same surface is illuminated by light of wavelength $3\lambda$, stopping potential is $2 \, \mathrm{V}$. The threshold wavelength for this surface is:

  1. $5\lambda$
  2. $3\lambda$
  3. $9\lambda$
  4. $4.5\lambda$

Answer: (c)

Solution

Given $E = \phi + K_{max}$ and $\phi = \frac{hc}{\lambda_0}$. Also, $K_{max} = eV_0$. From equation (i): $$8e = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$$ From equation (ii): $$2e = \frac{hc}{3\lambda} - \frac{hc}{\lambda_0}$$ On solving (i) and (ii), we find $$\lambda_0 = 9\lambda$$

Question 51

Physics · Moving Charges and Magnetism · Numerical

An electron moves through a uniform magnetic field $\vec{B} = B_0 \hat{i} + 2B_0 \hat{j} \, \mathrm{T}$. At a particular instant of time, the velocity of electron is $\vec{u} = 3 \hat{i} + 5 \hat{j} \, \mathrm{m/s}$. If the magnetic force acting on electron is $\vec{F} = 5e \mathrm{kN}$, where $e$ is the charge of electron, then the value of $B_0$ is _____ T.

Answer: 5

Solution

Given $\vec{F} = q(\vec{v} \times \vec{B})$. $5e\hat{k} = e(3\hat{i} + 5\hat{j}) \times (B_0\hat{i} + 2B_0\hat{j})$. $5e\hat{k} = e \left(6B_0\hat{k} - 5B_0\hat{k}\right)$. Therefore, $B_0 = 5 \, \mathrm{T}$.

Question 52

Physics · Electrostatic Potential and Capacitance · Numerical

A parallel plate capacitor with plate separation 5 $\mathrm{\ mm}$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness 2 $\mathrm{\ mm}$, while keeping the battery connections intact, the capacitor draws 25$\%$ more charge from the battery than before. The dielectric constant of the sheet is

Answer: 2

Solution

Without dielectric $$Q = \frac{A \varepsilon_0}{d} V$$ With dielectric $$Q = \frac{A \varepsilon_0 V}{d - t + \frac{t}{K}}$$ Given $$\frac{A \varepsilon_0 V}{d - t + \frac{t}{K}} = (1.25) \frac{A \varepsilon_0 V}{d}$$ $$\Rightarrow 1.25 \left( 3 + \frac{2}{K} \right) = 5$$ $$\Rightarrow K = 2$$

Question 53

Physics · Current Electricity · Numerical

Equivalent resistance of the following network is _____ $\Omega$.

Answer: 1

Solution

The 6 $\Omega$ resistor is a short circuit. The equivalent circuit is simplified to three 3 $\Omega$ resistors in parallel. The equivalent resistance is calculated as follows: $$R_{eq} = 3 \times \frac{1}{3} = 1 \Omega$$

Question 54

Physics · System of Particles and Rotational Motion · Numerical

A solid circular disc of mass $50 \, \mathrm{kg}$ rolls along a horizontal floor so that its center of mass has a speed of $0.4 \, \mathrm{m/s}$. The absolute value of work done on the disc to stop it is_____ J.

Answer: 6

Solution

Using work energy theorem $$W = \Delta KE = 0 - \left( \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \right)$$ $$W = 0 - \frac{1}{2} mv^2 \left( 1 + \frac{K^2}{R^2} \right)$$ $$= -\frac{1}{2} \times 50 \times 0.4^2 \left( 1 + \frac{1}{2} \right) = -6 \, \mathrm{J}$$ Absolute work = +6 J $$W = -6 \, \mathrm{J} |W| = 6 \, \mathrm{J}$$

Question 55

Physics · Motion in a Plane · Numerical

A body starts falling freely from height $H$ hits an inclined plane in its path at height $h$. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of $\frac{H}{h}$ for which the body will take the maximum time to reach the ground is_____

Answer: 2

Solution

Total time of flight = T $$T = \sqrt{\frac{2h}{g}} + \sqrt{\frac{2(H-h)}{g}}$$ For max. time $\frac{dT}{dh} = 0$ $$\sqrt{\frac{2}{g}} \left( \frac{-1}{2\sqrt{H-h}} + \frac{1}{2\sqrt{h}} \right) = 0$$ $$\sqrt{H-h} = \sqrt{h}$$ $$h = \frac{H}{2} \Rightarrow \frac{H}{h} = 2$$

Question 56

Physics · Wave Optics · Numerical

Two waves of intensity ratio 1 : 9 cross each other at a point. The resultant intensities at the point, when (a) Waves are incoherent is $I_1$ (b) Waves are coherent is $I_2$ and differ in phase by $60^\circ$. If $\frac{l_1}{l_2} = \frac{10}{x}$ then $x=$ $\underline{\hspace{2cm}}$

Answer: 13

Solution

For incoherent wave $I_1 = I_A + I_B \Rightarrow I_1 = I_0 + 9I_0$ $I_1 = 10I_0$ For coherent wave $I_2 = I_A + I_B + 2\sqrt{I_A I_B} \cos 60^\circ$ $$I_2 = I_0 + 9I_0 + 2\sqrt{9I_0^2} \cdot \frac{1}{2} = 13I_0$$ $$\frac{I_1}{I_2} = \frac{10}{13}$$

Question 57

Physics · Electromagnetic Induction · Numerical

A small square loop of wire of side $\ell$ is placed inside a large square loop of wire of side $L$ $(L = \ell^2)$. The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is $\sqrt{x} \times 10^{-7} \, \mathrm{H}$, where x = ___________

Answer: 128

Solution

Flux linkage for inner loop. $$\phi = B_{center} \cdot \ell^2$$ $$= 4 \times \frac{\mu_0 i}{4 \pi \frac{L}{2}} (\sin 45 + \sin 45) \ell^2$$ $$\phi = 2 \sqrt{2} \frac{\mu_0 i}{\pi L} \ell^2$$ $$M = \frac{\phi}{i} = \frac{2 \sqrt{2} \mu_0 \ell^2}{\pi L} = 2 \sqrt{2} \frac{\mu_0}{\pi}$$ $$= 2 \sqrt{2} \frac{4 \pi}{\pi} \times 10^{-7}$$ $$= 8 \sqrt{2} \times 10^{-7} H$$ $$= \sqrt{128} \times 10^{-7} H$$ $$x = 128$$

Question 58

Physics · Mechanical Properties of Solids · Numerical

The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02$\%$ is_____ m. (Take density of sea water = $10^3 \, \mathrm{kgm^{-3}}$, Bulk modulus of rubber = $9 \times 10^8 \, \mathrm{Nm^{-2}}$, and $g = 10 \, \mathrm{ms^{-2}}$)

Answer: 18

Solution

Given $\($ $\beta$ = $\frac{-\Delta P}{\frac{\Delta V}{V}}$ $\)$. $\($ $\Delta$ P = -$\beta$ $\frac{\Delta V}{V}$ $\)$. $\($ $\rho$ g $\,$ h = -$\beta$ $\frac{\Delta V}{V}$ $\)$. $\($ 10^3 $\times$ 10 $\times$ h = -9 $\times$ 10^8 $\times$ $\left$( $\frac{-0.02}{100}$ $\right$) $\)$. Therefore, $\($ h = 18 $\,$ $\mathrm{m}$ $\)$.

Question 59

Physics · Oscillations · Numerical

A particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\frac{2A}{3}$. The new amplitude of motion is $\frac{nA}{3}$. The value of $n$ is

Answer: 7

Solution

Given $v = \omega \sqrt{A^2 - x^2}$. At $x = \frac{2A}{3}$, $$v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \frac{\sqrt{5}A\omega}{3}.$$ New amplitude $= A'$. $$v' = 3v = \sqrt{5}A\omega = \omega \sqrt{(A')^2 - \left(\frac{2A}{3}\right)^2}.$$ $$A' = \frac{7A}{3}.$$

Question 60

Physics · Nuclei · Numerical

The mass defect in a particular reaction is 0.4 g. The amount of energy liberated is $n \times 10^7$ kWh, where $n =$ $\underline{\hspace{2cm}}$ (speed of light $= 3 \times 10^8 \, \mathrm{m/s}$)

Answer: 1

Solution

E = $\Delta$ mc^2 = 0.4 $\times$ 10^{-3} $\times$ (3 $\times$ 10^8)^2 = 3600 $\times$ 10^7 $\mathrm{kWs}$ = $\frac{3600 \times 10^7}{3600}$ $\mathrm{kWh}$ = 1 $\times$ 10^7 $\mathrm{kWh}$

Chemistry

Question 61

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

Give below are two statements: Statement-I : Noble gases have very high boiling points. Statement-II: Noble gases are monoatomic gases. They are held together by strong dispersion forces. Because of this they are liquefied at very low temperature. Hence, they have very high boiling points. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (d)

Solution

Statement I and II are False. Noble gases have low boiling points. Noble gases are held together by weak dispersion forces.

Question 62

Chemistry · Equilibrium · Single correct

For the given reaction, choose the correct expression of $K_C$ from the following: $$\mathrm{Fe}^{3+}_{(aq)} + \mathrm{SCN}^-_{(aq)} \rightleftharpoons (\mathrm{FeSCN})^{2+}_{(aq)}$$

  1. $K_C = \dfrac{[\mathrm{FeSCN^{2+}}]}{[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]}$
  2. $K_C = \dfrac{[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]}{[\mathrm{FeSCN^{2+}}]}$
  3. $K_C = \dfrac{[\mathrm{FeSCN^{2+}}]}{[\mathrm{Fe^{3+}}]^2[\mathrm{SCN^-}]^2}$
  4. $K_C = \dfrac{[\mathrm{FeSCN^{2+}}]^2}{[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]}$

Answer: (a)

Solution

$K_C = \dfrac{\text{Products ion conc.}}{\text{Reactants ion conc.}}$ $K_C = \dfrac{[\mathrm{FeSCN^{2+}}]}{[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]}$

Question 63

Chemistry · Solutions · Single correct

Identify the mixture that shows positive deviations from Raoult's Law

  1. $(\mathrm{CH}_3)_2\mathrm{CO} + \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2$
  2. $\mathrm{CHCl}_3 + \mathrm{C}_6\mathrm{H}_6$
  3. $\mathrm{CHCl}_3 + (\mathrm{CH}_3)_2\mathrm{CO}$
  4. $(\mathrm{CH}_3)_2\mathrm{CO} + \mathrm{CS}_2$

Answer: (d)

Solution

($\mathrm{CH_3}$)_2$\mathrm{CO}$ + $\mathrm{CS_2}$ exhibits positive deviations from Raoult's Law.

Question 64

Chemistry · Analytical Chemistry · Single correct

The compound that is white in color is

  1. ammonium sulphide
  2. lead sulphate
  3. lead iodide
  4. ammonium arsinomolybdate

Answer: (b)

Solution

Lead sulphate is white. Ammonium sulphide is soluble. Lead iodide is bright yellow. Ammonium arsino molybdate is yellow.

Question 65

Chemistry · Electrochemistry · Single correct

The metals that are employed in the battery industries are A. Fe B. Mn C. Ni D. Cr E. Cd Choose the correct answer from the options given below:

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

Answer: (a)

Solution

Mn, Ni and Cd metals used in battery industries.

Question 66

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

A species having carbon with sextet of electrons and can act as electrophile is called

  1. carbon free radical
  2. carbanion
  3. carbocation
  4. pentavalent carbon

Answer: (c)

Solution

The given species is a carbocation with the formula $\mathrm{CH_3^+}$. It has three hydrogen atoms bonded to a central carbon atom, which carries a positive charge. This species has a total of six valence electrons: three from the hydrogen atoms and three from the carbon atom. Therefore, it is a six electron species.

Question 67

Chemistry · Electrochemistry · Single correct

Identify the factor from the following that does not affect electrolytic conductance of a solution.

  1. The nature of the electrolyte added.
  2. The nature of the electrode used.
  3. Concentration of the electrolyte.
  4. The nature of solvent used.

Answer: (b)

Solution

Conductivity of electrolytic cell is affected by concentration of electrolyte, nature of electrolyte and nature of solvent.

Question 68

Chemistry · Hydrocarbons · Single correct

The product (C) in the below mentioned reaction is:

  1. Propan-1-ol
  2. Propene
  3. Propyne
  4. Propan-2-ol

Answer: (d)

Solution

The reaction sequence starts with $\mathrm{CH_3CH_2Br}$ reacting with $\mathrm{KOH_{(alc)}}$ under heat $\Delta$ to form $\mathrm{CH_2=CH_2}$. This compound then reacts with $\mathrm{HBr}$ to form $\mathrm{CH_3CHBrCH_3}$. Finally, $\mathrm{CH_3CHBrCH_3}$ reacts with $\mathrm{KOH_{(aq)}}$ under heat $\Delta$ to form $\mathrm{CH_3CH(OH)CH_3}$.

Question 69

Chemistry · Alcohols, Phenols and Ethers · Single correct

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A: Alcohols react both as nucleophiles and electrophiles. Reason R: Alcohols react with active metals such as sodium, potassium and aluminum to yield corresponding alkoxides and liberate hydrogen. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (d)

Solution

As per NCERT, Assertion (A) and Reason (R) is correct but Reason (R) is not the correct explanation.

Question 70

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

The correct sequence of electron gain enthalpy of the elements listed below is A. Ar B. Br C. F D. S Choose the most appropriate from the options given below:

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

Answer: (b)

Solution

\begin{tabular}{|c|c|} \hline Element & $\Delta_{eg}H\ (\mathrm{kJ/mol})$ \\ \hline F & $-333$ \\ \hline S & $-200$ \\ \hline Br & $-325$ \\ \hline Ar & $+96$ \\ \hline \end{tabular}

Question 71

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

Identify correct statements from below: A. The chromate ion is square planar. B. Dichromates are generally prepared from chromates. C. The green manganate ion is diamagnetic. D. Dark green coloured $\mathrm{K_2MnO_4}$ disproportionates in a neutral or acidic medium to give permanganate. E. With increasing oxidation number of transition metal, ionic character of the oxides decreases. Choose the correct answer from the options given below:

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

Answer: (d)

Solution

A. $\mathrm{CrO_4^{2-}}$ is tetrahedral B. $2\mathrm{Na_2CrO_4} + 2\mathrm{H^+} \rightarrow \mathrm{Na_2Cr_2O_7} + 2\mathrm{Na^+} + \mathrm{H_2O}$ C. As per NCERT, green manganate is paramagnetic with 1 unpaired electron. D. Statement is correct E. Statement is correct

Question 72

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

'Adsorption' principle is used for which of the following purification method?

  1. Extraction
  2. Chromatography
  3. Distillation
  4. Sublimation

Answer: (b)

Solution

Principle used in chromatography is adsorption.

Question 73

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

Integrated rate law equation for a first order gas phase reaction is given by (where $P_i$ is initial pressure and $P_t$ is total pressure at time $t$)

  1. $k = \frac{2.303}{t} \times \log \frac{P_i}{(2P_i - P_t)}$
  2. $k = \frac{2.303}{t} \times \log \frac{2P_i}{(2P_i - P_t)}$
  3. $k = \frac{2.303}{t} \times \log \frac{(2P_i - P_t)}{P_i}$
  4. $k = \frac{2.303}{t} \times \log \frac{P_i}{(2P_i - P_t)}$

Answer: (a)

Solution

The reaction is given as $\mathrm{A} \rightarrow \mathrm{B} + \mathrm{C}$. Initially, the pressures are $P_i$ for A, and $0$ for B and C. After the reaction, the pressures are $P_i - x$ for A, $x$ for B, and $x$ for C. The total pressure $P_t$ is given by $P_t = P_i + x$. Solving for $x$, we have $P_i - x = P_i - P_t + P_i = 2P_i - P_t$. The rate constant $K$ is given by $$K = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}.$$

Question 74

Chemistry · Alcohols, Phenols and Ethers · Single correct

Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$: Assertion $A$: $\mathrm{p}K_a$ value of phenol is $10.0$ while that of ethanol is $15.9$. Reason $R$: Ethanol is stronger acid than phenol. In the light of the above statements, choose the correct 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 and $R$ is the correct explanation of $A$.
  4. Both $A$ and $R$ are true but $R$ is NOT the correct explanation of $A$.

Answer: (a)

Solution

Phenol is more acidic than ethanol because conjugate base of phenoxide is more stable than ethoxide.

Question 75

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

Given below are two statements: Statement I: IUPAC name of HO - $\mathrm{CH}_2$ - $(\mathrm{CH}_2)_3$ - $\mathrm{CH}_2$ - $\mathrm{COCH}_3$ is 7-hydroxyheptan-2-one. Statement II: 2-oxoheptan-7-ol is the correct IUPAC name for above compound. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (a)

Solution

7-Hydroxyheptan-2-one is correct IUPAC name

Question 76

Chemistry · Co-ordination Compounds · Single correct

The correct statements from following are: A. The strength of anionic ligands can be explained by crystal field theory. B. Valence bond theory does not give a quantitative interpretation of kinetic stability of coordination compounds. C. The hybridization involved in formation of $[Ni(CN)_4]^{2-}$ complex is $dsp^2$. D. The number of possible isomer(s) of cis- $[PtCl_2(en)_2]^{2+}$ is one Choose the correct answer from the options given below:

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

Answer: (d)

Solution

Q16 B. VBT does not explain stability of complex C. Hybridisation of $[\mathrm{Ni(CN)_4}]^{-2}$ is $\mathrm{dsp^2}$.

Question 77

Chemistry · Chemical Bonding and Molecular Structure · Single correct

The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals A. have the same energy B. have the minimum overlap C. have same symmetry about the molecular axis D. have different symmetry about the molecular axis Choose the most appropriate from the options given below:

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

Answer: (b)

Solution

Molecular orbital should have maximum overlap. Symmetry about the molecular axis should be similar.

Question 78

Chemistry · Biomolecules · Single correct

Match List I with List II \begin{tabular}{|c|l|c|l|} \hline & List - I & & List - II \\ \hline A. & Glucose / NaHCO$_3$ / $\Delta$ & I. & Gluconic acid \\ \hline B. & Glucose / HNO$_3$ & II. & No reaction \\ \hline C. & Glucose / HI / $\Delta$ & III. & n-hexane \\ \hline D. & Glucose / Bromine water & IV. & Saccharic acid \\ \hline \end{tabular} Choose the correct answer from the options given below:

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

Answer: (b)

Solution

Glucose $\xrightarrow{\mathrm{NaHCO_3}}$ no reaction Glucose $\xrightarrow{\mathrm{HNO_3}}$ saccharic acid Glucose $\xrightarrow[\Delta]{\mathrm{HI}}$ $n$-hexane Glucose $\xrightarrow[\Delta]{\mathrm{Br_2}}$ Gluconic acid

Question 79

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

Consider the oxides of group 14 elements $SiO_2, GeO_2, SnO_2, PbO_2, CO$ and $GeO$. The amphoteric oxides are

  1. GeO, GeO_2
  2. SiO_2, GeO_2
  3. SnO_2, PbO_2
  4. SnO_2, CO

Answer: (c)

Solution

SnO$_2$ and PbO$_2$ are amphoteric.

Question 80

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

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

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

Answer: (b)

Solution

Fact (NCERT)

Question 81

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

Molar mass of the salt from $NaBr$, $NaNO_3$, $KI$ and $CaF_2$ which does not evolve coloured vapours on heating with concentrated $H_2SO_4$ is $\mathrm{g \, mol^{-1}}$, (Molar mass in $gmol^{-1}$ : Na : 23, N : 14, K : 39, O : 16, Br : 80, I : 127, F : 19, Ca : 40)

Answer: 78

Solution

$CaF_2$ does not evolve any gas with concentrated $H_2SO_4$. NaBr $\rightarrow$ evolve $\mathrm{Br_2}$ $NaNO_3$ $\rightarrow$ evolve $\mathrm{NO_2}$ KI $\rightarrow$ evolve $\mathrm{I_2}$

Question 82

Chemistry · The d-and f-Block Elements · Numerical

The 'Spin only' Magnetic moment for $[\mathrm{Ni(NH_3)_6}]^{2+}$ is ______ $\times 10^{-1}$ BM. (given = Atomic number of Ni : 28)

Answer: 28

Solution

NH_3 acts as WFL with Ni^{2+}. Ni^{2+} = 3d^8. No. of unpaired electrons = 2. $$\mu = \sqrt{n(n+2)} = \sqrt{8} = 2.82 \, \mathrm{BM}$$ $$= 28.2 \times 10^{-1} \, \mathrm{BM}$$ x = 28

Question 83

Chemistry · Some Basic Concepts of Chemistry · Numerical

Number of moles of methane required to produce $22_g$ $CO_{2_(g)}$ after combustion is $x \times 10^{-2}$ moles. The value of $x$ is

Answer: 50

Solution

Given the reaction: $\($ $\mathrm{CH_4}$_{(g)} + 2$\mathrm{O_2}$_{(g)} $\rightarrow$ $\mathrm{CO_2}$_{(g)} + 2$\mathrm{H_2O}$_{($\ell$)} $\)$ $\($ n_{$\mathrm{CO_2}$} = $\frac{22}{44}$ = 0.5 $\)$ moles So moles of $\($ $\mathrm{CH_4}$ $\)$ required = 0.5 moles i.e. $\($ 50 $\times$ 10^{-2} $\)$ mole $\($ x = 50 $\)$

Question 84

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

The product of the following reaction is $P$. The number of hydroxyl groups present in the product $P$ is .

Answer: 0

Solution

The product benzene has zero hydroxyl group.

Question 85

Chemistry · Chemical Bonding and Molecular Structure · Numerical

The number of species from the following in which the central atom uses $sp^3$ hybrid orbitals in its bonding is

Answer: 4

Solution

Q18 $\mathrm{NH_3} \rightarrow \mathrm{sp^3}$ $\mathrm{SO_2} \rightarrow \mathrm{sp^2}$ $\mathrm{SiO_2} \rightarrow \mathrm{sp^3}$ $\mathrm{BeCl_2} \rightarrow \mathrm{sp}$ $\mathrm{CO_2} \rightarrow \mathrm{sp}$ $\mathrm{H_2O} \rightarrow \mathrm{sp^3}$ $\mathrm{CH_4} \rightarrow \mathrm{sp^3}$ $\mathrm{BF_3} \rightarrow \mathrm{sp^2}$

Question 86

Chemistry · Haloalkanes and Haloarenes · Numerical

The total number of hydrogen atoms in product A and product B is .

Answer: 10

Solution

The reaction given is: $\($ $\mathrm{CH_3CH_2Br + NaOH}$ $\rightarrow$ $\begin{cases}$ $\mathrm{C_2H_5OH}$ $\\$ $\mathrm{CH_2=CH_2}$ $\end{cases}$ $\)$ with $\($ $\mathrm{H_2O}$ $\)$. The total number of hydrogen atoms in A and B is 10.

Question 87

Chemistry · Hydrocarbons · Numerical

Number of alkanes obtained on electrolysis of a mixture of $CH_3COONa$ and $C_2H_5COONa$ is

Answer: 3

Solution

Q18 $\mathrm{CH_3COONa} \rightarrow \cdot \mathrm{CH_3}$ $\mathrm{C_2H_5COONa} \rightarrow \cdot \mathrm{C_2H_5}$ $2 \cdot \mathrm{C_2H_5} \rightarrow \mathrm{CH_3} - \mathrm{CH_2} - \mathrm{CH_2} - \mathrm{CH_3}$ $2 \cdot \mathrm{CH_3} \rightarrow \mathrm{CH_3} - \mathrm{CH_3}$ $\cdot \mathrm{CH_3} + \cdot \mathrm{C_2H_5} \rightarrow \mathrm{CH_3} - \mathrm{CH_2} - \mathrm{CH_3}$

Question 88

Chemistry · Thermodynamics · Numerical

Consider the following reaction at 298 K. $$\frac{3}{2} \mathrm{O}_2 (g) \rightleftharpoons \mathrm{O}_3 (g) \cdot K_P = 2.47 \times 10^{-29}.$$ $\Delta_r G^\oplus$ for the reaction is _______ kJ. (Given R = 8.314 J $K^{-1} mol^{-1}$)

Answer: 163

Solution

$\dfrac{3}{2}\,\mathrm{O_2(g)} \rightleftharpoons \mathrm{O_3(g)};\quad K_P = 2.47 \times 10^{-29}$ $\Delta_r G^\circ = -RT\ln K_P$ $= -8.314 \times 10^{-3} \times 298 \times \ln(2.47 \times 10^{-29})$ $= -8.314 \times 10^{-3} \times 298 \times (-65.87)$ $= 163.19\ \mathrm{kJ}$

Question 89

Chemistry · Structure of Atom · Numerical

The ionization energy of sodium in kJmol^{-1}. If electromagnetic radiation of wavelength 242 $\mathrm{nm}$ is just sufficient to ionize sodium atom is

Answer: 494

Solution

Given $$E = \frac{1240}{\lambda (\mathrm{nm})} \, \mathrm{eV}$$ Substituting the value of $\lambda$: $$E = \frac{1240}{242} \, \mathrm{eV}$$ $$= 5.12 \, \mathrm{eV}$$ Converting to joules per atom: $$= 5.12 \times 1.6 \times 10^{-19}$$ $$= 8.198 \times 10^{-19} \, \mathrm{J/atom}$$ Converting to kilojoules per mole: $$= 494 \, \mathrm{kJ/mol}$$

Question 90

Chemistry · Electrochemistry · Numerical

One Faraday of electricity liberates $x \times 10^{-1}$ gram atom of copper from copper sulphate, $x$ is

Answer: 5

Solution

The reaction is given by: $\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}$ 2 Faraday is required to deposit 1 mol of Cu. 1 Faraday will deposit 0.5 mol of Cu. 0.5 mol is equivalent to 0.5 g atom, which is equal to $5\times10^{-1}$. Therefore, $x=5$.