JEE Main 28 June 2022 Shift 1 question paper with solutions

JEE Main 28 June 2022 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 · Binomial Theorem · Single correct

If $$\sum_{k=1}^{31} \binom{31}{k} \binom{31}{k-1} - \sum_{k=1}^{30} \binom{30}{k} \binom{30}{k-1} = \frac{\alpha (60!)}{(30!)(31!)},$$ Where $\alpha \in \mathbb{R}$, then the value of $16\alpha$ is equal to

  1. 1411
  2. 1320
  3. 1615
  4. 1855

Answer: (a)

Solution

Given $\($ $\sum$_{R=1}^{31} $\binom{31}{R}$ $\cdot$ $\binom{31}{R-1}$ $\)$ $\[$ = $\binom{31}{1}$ $\cdot$ $\binom{31}{0}$ + $\binom{31}{2}$ $\cdot$ $\binom{31}{1}$ + $\ldots$ + $\binom{31}{31}$ $\cdot$ $\binom{31}{30}$ $\]$ $\[$ = $\binom{31}{0}$ $\cdot$ $\binom{31}{30}$ + $\binom{31}{1}$ $\cdot$ $\binom{31}{29}$ + $\ldots$ + $\binom{31}{30}$ $\cdot$ $\binom{31}{0}$ $\]$ $\[$ = $\binom{62}{30}$ $\]$ Similarly, $\($ $\sum$_{R=1}^{30} $\left$( $\binom{30}{R}$ $\cdot$ $\binom{30}{R-1}$ $\right$) = $\binom{60}{29}$ $\)$ $\[$ $\binom{62}{30}$ - $\binom{60}{29}$ = $\frac{62!}{30!32!}$ - $\frac{60!}{29!31!}$ $\]$ $\[$ = $\frac{60!}{29! \cdot 31!}$ $\left$$\{$ $\frac{62 \cdot 61}{30 \cdot 32}$ - 1 $\right$$\}$ $\]$ $\[$ = $\frac{60!}{30!31!}$ $\left$( $\frac{2822}{32}$ $\right$) $\]$ Therefore, $\($ 16 $\alpha$ = 16 $\times$ $\frac{2822}{32}$ = 1411 $\)$

Question 2

Maths · Relations and Functions · Single correct

Let a function $f : \mathbb{N} \to \mathbb{N}$ be defined by $$f(n) = \begin{cases} 2n, & n = 2, 4, 6, 8, \ldots \\ n - 1, & n = 3, 7, 11, 15, \ldots \\ \frac{n+1}{2}, & n = 1, 5, 9, 13, \ldots \end{cases}$$ then, $f$ is

  1. one-one but not onto
  2. onto but not one-one
  3. neither one-one nor onto
  4. one-one and onto

Answer: (d)

Solution

Given $$f(x) = \begin{cases} 4R & ; \ n = 2R \\ 4R - 2 & ; \ n = 4R - 1 \\ 2R - 1 & ; \ n = 4R - 3 \end{cases}$$ where $R \in \mathbb{N}$. Note that for any element, it will fall into exactly one of these sets. $$\{ y : y = 4R; \ y \in \mathbb{N} \}$$ $$\{ y : y = 4R - 2; \ y \in \mathbb{N} \}$$ $$\{ y : y = 2R - 1; \ y \in \mathbb{N} \}$$ Corresponding to that $y$, we will get exactly one value of $n$. Thus, $f$ is one-to-one and onto.

Question 3

Maths · Determinants · Single correct

If the system of linear equations $$2x + 3y - z = -2$$ $$x + y + z = 4$$ $$x - y + |\lambda| z = 4\lambda - 4$$ where $\lambda \in \mathbb{R}$, has no solution, then

  1. $\lambda = 7$
  2. $\lambda = -7$
  3. $\lambda = 8$
  4. $\lambda^2 = 1$

Answer: (b)

Solution

Given the determinant: $$\begin{vmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & \lambda \end{vmatrix} = 0$$ This implies $|\lambda| = 7 \Rightarrow \lambda = \pm 7$ $\hspace{0.5cm}$ ...(1) System: $$2x + 3y - z = -2 \hspace{0.5cm} ...(2)$$ $$x + y + z = 4 \hspace{0.5cm} ...(3)$$ $$x - y + |\lambda| z = 4\lambda - 4 \hspace{0.5cm} ...(4)$$ Eliminating $y$ from equations (2) and (3) we get: $$x + 4z = 14 \hspace{0.5cm} ...(5)$$ Adding equations (3) and (4) gives: $$x + \left( \frac{|\lambda| + 1}{2} \right) z = 2\lambda \hspace{0.5cm} ...(6)$$ Clearly for $\lambda = -7$, the system is inconsistent.

Question 4

Maths · Determinants · Single correct

Let A be a matrix of order 3 $\times$ 3 and $\det$(A) = 2. Then $\det$($\det$(A) $\operatorname{adj}$(5 $\operatorname{adj}(A^3)))$ is equal to

  1. 512 $\times$ 10^6
  2. 256 $\times$ 10^6
  3. 1024 $\times$ 10^6
  4. 256 $\times$ 10^{11}

Answer: (a)

Solution

Given $|(\det(A)) \ \mathrm{adj}(5 \ \mathrm{adj}(A))|$. $$= |2 \mathrm{adj}(5 \mathrm{adj}(A^3))|$$ $$= 2^3 \ |\mathrm{adj}(5 \ \mathrm{adj}(A^3))|$$ $$= 2^3 \ |5 \mathrm{adj}(A^3)|^2$$ $$= 2^3 \ (5^3 \cdot |\mathrm{adj}(A^3)|)^2$$ $$= 2^3 \cdot 5^6 \cdot |\mathrm{adj} A^3|^2$$ $$= 2^3 \cdot 5^6 \cdot (|A|^3)^2$$ $$= 2^3 \cdot 5^6 \cdot 2^{12} = 2^{15} \times 5^6$$ $$= 2^9 \times 10^6$$ $$= 512 \times 10^6.$$

Question 5

Maths · Permutations and Combinations · Single correct

The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is

  1. 36
  2. 48
  3. 60
  4. 72

Answer: (d)

Solution

To make a number divisible by 3 we can use the digits 1, 2, 5, 6, 7 or 1, 2, 3, 5, 7. Using 1, 2, 5, 6, 7, number of even numbers is $$= 4 \times 3 \times 2 \times 1 \times 2 = 48$$ Using 1, 2, 3, 5, 7, number of even numbers is $$= 4 \times 3 \times 2 \times 1 \times 1 = 24$$ Required answer is 72.

Question 6

Maths · Sequences and Series · Single correct

Let $A_1, A_2, A_3, \ldots$ be an increasing geometric progression of positive real numbers. If $A_1 A_3 A_5 A_7 = \frac{1}{1296}$ and $A_2 + A_4 = \frac{7}{36}$, then, the value of $A_6 + A_8 + A_{10}$ is equal to

  1. 33
  2. 37
  3. 43
  4. 47

Answer: (c)

Solution

Given $A_1 \cdot A_3 \cdot A_5 \cdot A_7 = \frac{1}{1296}$. $$(A_4)^4 = \frac{1}{1296}$$ Therefore, $A_4 = \frac{1}{6}$ $\hspace{0.5cm}$ ...(1) $A_2 + A_4 = \frac{7}{36}$ Thus, $A_2 = \frac{1}{36}$ $\hspace{0.5cm}$ ...(2) $A_6 = 1$ $A_8 = 6$ $A_{10} = 36$ $A_6 + A_8 + A_{10} = 43$

Question 7

Maths · Integrals · Single correct

Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $$\int_{0}^{1} [-8x^2 + 6x - 1] \, dx$$ is equal to

  1. -1
  2. -$\frac{5}{4}$
  3. $\frac{\sqrt{17} - 13}{8}$
  4. $\frac{\sqrt{17} - 16}{8}$

Answer: (c)

Solution

Evaluate the integral from 0 to 1 of $[-8x^2 + 6x - 1] \, dx$. This can be split into three integrals: $$\int_0^{1/4} -1 \, dx + \int_{1/4}^{1/2} 0 \, dx + \int_{1/2}^{3/4} -1 \, dx$$ Using the graph, we add: $$\int_{3/4}^{\frac{3+\sqrt{17}}{8}} -2 \, dx + \int_{\frac{3+\sqrt{17}}{8}}^1 -3 \, dx$$ This simplifies to: $$= -\left[ x \right]_0^{1/4} + 0 - \left[ x \right]_{1/2}^{3/4} - 2 \left[ x \right]_{3/4}^{\frac{3+\sqrt{17}}{8}} - 3 \left[ x \right]_{\frac{3+\sqrt{17}}{8}}^1$$ Evaluating the integrals: $$= -\left( \frac{1}{4} - 0 \right) - \left( \frac{3}{4} - \frac{1}{2} \right) - 2 \left( \frac{3+\sqrt{17}}{8} - \frac{3}{4} \right) - 3 \left( 1 - \frac{3+\sqrt{17}}{8} \right)$$ Simplifying further: $$= -\frac{1}{4} - \frac{1}{4} + \frac{-6 - 2\sqrt{17}}{8} + \frac{3}{2} - 3 + \frac{9 + 3\sqrt{17}}{8}$$ Finally, $$= \frac{\sqrt{17} - 13}{8}$$

Question 8

Maths · Continuity and Differentiability · Single correct

Let $f : \mathbb{R} \to \mathbb{R}$ be defined as $$f(x) = \begin{cases} e^x, & x < 0 \\ ae^x + [x-1], & 0 \leq x < 1 \\ b + [\sin(\pi x)], & 1 \leq x < 2 \\ [e^{-x}] - c, & x \geq 2 \end{cases}$$ where $a, b, c \in \mathbb{R}$ and $[t]$ denotes greatest integer less than or equal to $t$. Then, which of the following statements is true?

  1. There exists $a, b, c \in \mathbb{R}$ such that $f$ is continuous on $\mathbb{R}$.
  2. If $f$ is discontinuous at exactly one point, then $a + b + c = 1$.
  3. If $f$ is discontinuous at exactly one point, then $a + b + c \neq 1$.
  4. $f$ is discontinuous at at least two points, for any values of $a, b$ and $c$.

Answer: (c)

Solution

Given $f(x)$ is discontinuous at $x = 1$. For continuous at $x = 0$; $a = 1$. For continuous at $x = 2$; $b + c = 1$. $a + b + c = 2$.

Question 9

Maths · Applications of Integrals · Single correct

The area of the region S = $\{$(x,y) : $y^2$ $\leq$ 8x, y $\geq$ $\sqrt{2}$x, x $\geq$ 1$\}$ is

  1. $\frac{13\sqrt{2}}{6}$
  2. $\frac{11\sqrt{2}}{6}$
  3. $\frac{5\sqrt{2}}{6}$
  4. $\frac{19\sqrt{2}}{6}$

Answer: (b)

Solution

Given $y^2 = 8x$ (1) $y = \sqrt{2x}$ (2) $y^2 = 2x^2$ Solving $8x = 2x^2$ gives $x = 0$ and $x = 4$. The area is given by: $$ Area := \int_{1}^{4} \left( 2\sqrt{2}\sqrt{x} - \sqrt{2}x \right) \, dx $$ Evaluating the integral: $$ = 2\sqrt{2} \left( \frac{x^{3/2}}{3/2} \right) \bigg|_1^4 - \sqrt{2} \left( \frac{x^2}{2} \right) \bigg|_1^4 $$ $$ = \frac{4\sqrt{2}}{3} (8 - 1) - \frac{\sqrt{2}}{3} (16 - 1) $$ $$ = \frac{28\sqrt{2}}{3} - \frac{15\sqrt{2}}{2} = \frac{11\sqrt{2}}{6} $$

Question 10

Maths · Differential Equations · Single correct

Let the solution curve $y = y(x)$ of the differential equation, $$\left[ \frac{x}{\sqrt{x^2 - y^2}} + e^x \right] x \frac{dy}{dx} = x + \left[ \frac{x}{\sqrt{x^2 - y^2}} + e^x \right] y$$ pass through the points $(1, 0)$ and $(2\alpha, \alpha),\ \alpha > 0$. Then $\alpha$ is equal to

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

Answer: (a)

Solution

Given $\frac{x}{\sqrt{x^2-y^2}} + e^x - \frac{y}{x} \frac{dy}{dx} = x + \left( \frac{x}{\sqrt{x^2-y^2}} + e^x - \frac{y}{x} \right) y$ $$\Rightarrow \frac{y}{x} e^x (x\,dy - y\,dx) + \frac{y}{\sqrt{x^2-y^2}} (x\,dy - y\,dx) = x\,dx$$ Dividing both sides by $x^2$ $$\Rightarrow \frac{y}{x} e^x \left( \frac{x\,dy - y\,dx}{x^2} \right) + \frac{1}{\sqrt{1-\left(\frac{y}{x}\right)^2}} \left( \frac{x\,dy - y\,dx}{x^2} \right) = \frac{dx}{x}$$ $$\Rightarrow \frac{y}{x} e^x \left[ d\left(\frac{y}{x}\right) \right] + \frac{1}{\sqrt{1-\left(\frac{y}{x}\right)^2}}\, d\left(\frac{y}{x}\right) = \frac{dx}{x}$$ Integrate both sides. $$\int \frac{y}{x} e^x\, d\left(\frac{y}{x}\right) + \int \frac{1}{\sqrt{1-\left(\frac{y}{x}\right)^2}}\, d\left(\frac{y}{x}\right) = \int \frac{dx}{x}$$ $$\Rightarrow \frac{y}{x} e^x + \sin^{-1}\left(\frac{y}{x}\right) = \ln x + c$$ It passes through $(1, 0)$: $1 + 0 = 0 + c \Rightarrow c = 1$. It passes through $(2\alpha, \alpha)$: $$\frac{1}{e^2} + \sin^{-1}\frac{1}{2} = \ln 2\alpha + 1$$ $$\Rightarrow \ln 2\alpha = \frac{1}{e^2} + \frac{\pi}{6} - 1$$ $$\Rightarrow 2\alpha = e^{\left(\frac{1}{e^2} + \frac{\pi}{6} - 1\right)}$$ $$\Rightarrow \alpha = \frac{1}{2} e^{\left(\frac{\pi}{6} - 1 + e^{-2}\right)}$$

Question 11

Maths · Differential Equations · Single correct

Let $y = y(x)$ be the solution of the differential equation $x(1-x^2)\frac{dy}{dx} + (3x^2y - y - 4x^3) = 0, x > 1,$ with $y(2) = -2$. Then $y(3)$ is equal to

  1. -18
  2. -12
  3. -6
  4. -3

Answer: (a)

Solution

Given the equation $x(1-x^2) \frac{dy}{dx} + (3x^2y - y - 4x^3) = 0$. Simplifying, we have $x(1-x^2) \frac{dy}{dx} + (3x^2 - 1)y = 4x^3$. Using the integrating factor method, we find the integrating factor (IF) as follows: $$IF = e^{\int P \, dx} = e^{\int \frac{3x^2 - 1}{x - x^3} \, dx}$$ Let $x - x^3 = t$, then $IF = e^{\int \frac{-dt}{t}} = e^{-\ln t} = \frac{1}{t}$. Therefore, $$IF = \frac{1}{x - x^3}$$ The solution is given by $$y \times IF = \int Q \times IF \, dx$$ $$y \left( \frac{1}{x - x^3} \right) = \int \frac{4x^3}{x - x^3} \times \frac{1}{x - x^3} \, dx$$ $$= \int \frac{4x^3}{(x - x^3)^2} \, dx$$ $$= \int \frac{4x}{(1-x^2)^2} \, dx$$ Let $1-x^2 = K$, then $$-2x \, dx = dK$$ $$= 2 \int \frac{-dK}{K^2}$$ $$= -2 \left( -\frac{1}{K} \right) + c$$ $$\frac{y}{x - x^3} = \frac{2}{K} + c$$ $$\frac{y}{x - x^3} = \frac{2}{1-x^2} + c$$ At $x = 2$, $y = -2$, $$\frac{-2}{2-8} = \frac{2}{1-4} + c$$ $$\frac{1}{3} = \frac{-2}{3} + c$$ Solving, $$\frac{y}{3-27} = \frac{2}{1-9} + 1$$ $$\frac{y}{-24} = \frac{1}{4}$$ $$\frac{y}{-24} = \frac{3}{4}$$ $$y = \frac{3}{4}(-24) = -18$$

Question 12

Maths · Applications of Derivatives · Single correct

The number of real solutions of $x^7 + 5x^3 + 3x + 1 = 0$ is equal to _______.

  1. 0
  2. 1
  3. 3
  4. 5

Answer: (b)

Solution

Given $f(x) = x^7 + 5x^3 + 3x + 1$. The derivative is $f'(x) = 7x^6 + 15x^2 + 3 > 0$. Therefore, $f(x)$ is a strictly increasing function. As $x \to -\infty$, $y \to -\infty$. As $x \to \infty$, $y \to \infty$. Therefore, the number of real solutions is 1.

Question 13

Maths · Conic Sections · Single correct

Let the eccentricity of the hyperbola $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $\sqrt{\frac{5}{2}}$ and length of its latus rectum be $6\sqrt{2}$. If $y = 2x + c$ is a tangent to the hyperbola $H$, then the value of $c^2$ is equal to

  1. 18
  2. 20
  3. 24
  4. 32

Answer: (b)

Solution

Given the equation $y = mx \pm \sqrt{a^2 m^2 - b^2}$. Let $m = 2$, $c^2 = a^2 m^2 - b^2$. Then $c^2 = 4a^2 - b^2$. We have $e^2 = 1 + \frac{b^2}{a^2}$. Therefore, $\($ $\frac{5}{2}$ = 1 + $\frac{b^2}{a^2}$ $\)$. Solving, $\($ $\frac{3}{2}$ = $\frac{b^2}{a^2}$ $\Rightarrow$ b^2 = $\frac{3a^2}{2}$ $\)$. Then, $\($ $\frac{2b^2}{a}$ = 6$\sqrt{2}$ $\)$. Substituting, $\($ $\frac{2}{a}$ $\times$ $\frac{3a^2}{2}$ = 6$\sqrt{2}$ $\)$. Simplifying gives $3a = 6\sqrt{2}$. Thus, $a = 2\sqrt{2}$. Therefore, $b = 2\sqrt{3}$. Finally, $c^2 = 4 \times 8 - 12$. Hence, $c^2 = 20$.

Question 14

Maths · Conic Sections · Single correct

If the tangents drawn at the points $O(0,0)$ and $P(1+\sqrt{5},2)$ on the circle \[ x^2+y^2-2x-4y=0 \] intersect at the point $Q$, then the area of the triangle $OPQ$ is equal to

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

Answer: (c)

Solution

Tangent at $O$ $$-(x + 0) - 2(y + 0) = 0$$ $$\Rightarrow x + 2y = 0$$ Tangent at $P$ $$x(1 + \sqrt{5}) + y \cdot 2 - (x + 1 + \sqrt{5}) - 2(y + 2 = 0)$$ Put $x = -2y$ $$-2y(1 + \sqrt{5}) + 2y + 2y - 1 - \sqrt{5} - 2y - 4 = 0$$ $$-2\sqrt{5}y = 5 + \sqrt{5} \Rightarrow y = \frac{\sqrt{5} + 1}{2}$$ $Q \left( \sqrt{5} + 1, -\frac{\sqrt{5} + 1}{2} \right)$ Length of tangent $OQ = \frac{5 + \sqrt{5}}{2}$ Area $= \frac{RL^3}{R^2 + L^2}$ $R = \sqrt{5}$ $$= \sqrt{5} \times \left( \frac{5 + \sqrt{5}}{2} \right)^3$$ $$= \frac{5 + \left( \frac{5 + \sqrt{5}}{2} \right)^2}{\sqrt{5} \times \frac{4 \times (125 + 75 + 75\sqrt{5} + 5\sqrt{5})}{(20 + 25 + 10\sqrt{5} + 5)}}$$ $$= \frac{5 + 3\sqrt{5}}{2}$$

Question 15

Maths · Three Dimensional Geometry · Single correct

If two distinct point Q, R lie on the line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$ and $PQ = PR = \sqrt{18}$ where the point P is $(1, -2, 3)$, then the area of the triangle PQR is equal to

  1. $\frac{2}{3} \sqrt{38}$
  2. $\frac{4}{3} \sqrt{38}$
  3. $\frac{8}{3} \sqrt{38}$
  4. $\frac{\sqrt{152}}{3}$

Answer: (b)

Solution

Given the equations of the planes: $$-x + 2y - z = 0$$ $$3x - 5y + 2z = 0$$ The normal vector $\mathbf{n}$ is given by: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & -1 \\ 3 & -5 & 2 \end{vmatrix} = \hat{i}(-1) - \hat{j}(1) + \hat{k}(-1)$$ Hints and Solutions: Direction ratios (DR) of $PT \rightarrow \alpha - 1, \alpha + 2, \alpha - 3$ DR of $QR \rightarrow 1, 1, 1$ $$\Rightarrow (\alpha - 1) \times 1 + (\alpha + 2) \times 1 + (\alpha - 3) \times 1 = 0$$ $$3\alpha = 2$$ $$\alpha = \frac{2}{3}$$ $$PT^2 = \frac{1}{9} + \frac{64}{9} + \frac{49}{9}$$ $$PT^2 = \frac{114}{9}$$ $$PT = \frac{\sqrt{114}}{3}$$ $$\cos \theta = \frac{\sqrt{114}}{3} \times \frac{1}{3\sqrt{2}} = \frac{\sqrt{57}}{9} = \frac{\sqrt{19 \times 3}}{3 \times 3}$$ $$= \frac{\sqrt{19}}{3\sqrt{3}}$$ $$\cos 2\theta = \frac{2 \times 19}{27} - 1 = \frac{11}{27}$$ $$\sin 2\theta = \sqrt{1 - \left(\frac{11}{27}\right)^2} = \frac{\sqrt{38 \sqrt{16}}}{27}$$ $$= \frac{4}{27} \sqrt{38}$$ Area: $$= \frac{1}{2} \times \sqrt{18} \times \sqrt{18} \times \frac{4}{27} \sqrt{38}$$ $$= \frac{18}{2} \times \frac{4}{27} \sqrt{38} = \frac{36}{27} \sqrt{38} = \frac{4}{3} \sqrt{38}$$

Question 16

Maths · Three Dimensional Geometry · Single correct

The acute angle between the planes $P_1$ and $P_2$, when $P_1$ and $P_2$ are the planes passing through the intersection of the planes $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ and the points $(2, 1, 3)$ and $(0, 1, 2)$, respectively, is

  1. $\frac{\pi}{3}$
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{6}$
  4. $\frac{\pi}{12}$

Answer: (a)

Solution

Equation of plane passing through the intersection of planes $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ is $$5x + 8y + 3z - 29 + \lambda (8x - 7y + z - 20) = 0$$ and if it is passing through $(2, 1, 3)$ then $\lambda = \frac{7}{2}$. $P_1$: Equation of plane through intersection of $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ and the point $(2, 1, 3)$ is $$5x + 8y + 3z - 29 + \frac{7}{2}(8x - 7y + z - 20) = 0$$ $$\Rightarrow 2x - y + z = 6$$ Similarly $P_2$: Equation of plane through intersection of $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ and the point $(0, 1, 2)$ is $$\Rightarrow x + y + 2z = 5$$ Angle between planes $\theta = \cos^{-1} \left( \frac{3}{\sqrt{6} \sqrt{6}} \right) = \frac{\pi}{3}$

Question 17

Maths · Three Dimensional Geometry · Single correct

Let the plane $P: \vec{r} \cdot \vec{a} = d$ contain the line of intersection of two planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 6$ and $\vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) = 7$. If the plane $P$ passes through the point $\left(2, 3, \frac{1}{2}\right)$, then the value of $\frac{|13\vec{a}|^2}{d^2}$ is equal to

  1. 90
  2. 93
  3. 95
  4. 97

Answer: (b)

Solution

Equation of plane passing through line of intersection of planes $P_1 : \vec{r} \cdot \left( \hat{i} + 3 \hat{j} - \hat{k} \right) = 6$ and $P_2 : \vec{r} \cdot \left( -6 \hat{i} + 5 \hat{j} - \hat{k} \right) = 7$ is $P_1 + \lambda P_2 = 0$ $$\left( \vec{r} \cdot \left( \hat{i} + 3 \hat{j} - \hat{k} \right) - 6 \right) + \lambda \left( \vec{r} \cdot \left( -6 \hat{i} + 5 \hat{j} - \hat{k} \right) - 7 \right) = 0$$ and it passes through point $\left( 2, 3, \frac{1}{2} \right)$ $$\Rightarrow \left( 2 + 9 - \frac{1}{2} - 6 \right) + \lambda \left( -12 + 15 - \frac{1}{2} - 7 \right) = 0$$ $$\Rightarrow \lambda = 1$$ Equation of plane is $\vec{r} \cdot \left( -5 \hat{i} + 8 \hat{j} - 2 \hat{k} \right) = 13$ $$|\vec{a}|^2 = 25 + 64 + 4 = 93 ; \ d = 13$$ Value of $\frac{|13 \vec{a}|^2}{d^2} = 93$

Question 18

Maths · Probability · Single correct

The probability, that in a randomly selected 3-digit number at least two digits are odd, is

  1. $\frac{19}{36}$
  2. $\frac{15}{36}$
  3. $\frac{13}{36}$
  4. $\frac{23}{36}$

Answer: (d)

Solution

At least two digits are odd. This is equal to exactly two digits are odd plus exactly three digits are odd. For exactly three digits are odd: $$5 \times 5 \times 5 = 125$$ For exactly two digits odd: If 0 is used then: $$2 \times 5 \times 5 = 50$$ If 0 is not used then: $$^3C_1 \times 4 \times 5 \times 5 = 300$$ Required Probability: $$\frac{475}{900} = \frac{19}{36}$$

Question 19

Maths · Heights and Distances · Single correct

Let AB and PQ be two vertical poles, 160 $\,$ $\mathrm{m}$ apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let $\frac{\pi}{8}$ and $\theta$ be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then $\tan^2$ $\theta$ is equal to

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

Answer: (c)

Solution

Let $BC = CQ = x$ and $AB = h$ and $PQ = 2h$. $\tan \theta = \frac{h}{x}$, $\tan \frac{\pi}{8} = \frac{2h}{x}$. $$\frac{\tan \theta}{\tan \left( \frac{\pi}{8} \right)} = \frac{1}{2}$$ $$\tan \theta = \frac{1}{2} \tan \left( \frac{\pi}{8} \right) = \frac{1}{2} \left( \sqrt{2} - 1 \right)$$ $$\tan^2 \theta = \frac{1}{4} \left( 3 - 2 \sqrt{2} \right)$$

Question 20

Maths · Mathematical Reasoning · Single correct

Let $p,q,r$ be three logical statements. Consider the compound statements $S_1:\ ((\sim p)\vee q)\vee((\sim p)\vee r)$ and $S_2:\ p\rightarrow(q\vee r)$ Then, which of the following is NOT true?

  1. \text{If } S_2 \text{ is True, then } S_1 \text{ is True}
  2. \text{If } S_2 \text{ is False, then } S_1 \text{ is False}
  3. \text{If } S_2 \text{ is False, then } S_1 \text{ is True}
  4. \text{If } S_1 \text{ is False, then } S_2 \text{ is False}

Answer: (c)

Solution

Given $$s_1: (\sim p \lor q) \lor (\sim p \lor r)$$ $$\equiv \sim p \lor (q \lor r)$$ $$s_2: p \rightarrow (q \lor r)$$ $$\equiv \sim p \lor (q \lor r) \rightarrow$$ By conditional law $$s_1 \equiv s_2$$

Question 21

Maths · Relations and Functions · Fill in the blank

Let $R_1$ and $R_2$ be relations on the set $\{1, 2, \ldots, 50\}$ such that $R_1 = \{(p, p^n) : p$ is a prime and $n \geq 0$ is an integer$\}$ and $R_2 = \{(p, p^n) : p$ is a prime and $n = 0$ or $1\}$. Then, the number of elements in $R_1 - R_2$ is .

Answer: 8

Solution

Here, $p, p^n \in \{1, 2, \ldots, 50\}$. Now $p$ can take values $2, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37, 41, 43$ and $47$. Therefore, we can calculate the number of elements in $R_1$ as $(2, 2^0), (2, 2^1), \ldots, (2, 2^5)$; $(3, 3^0), \ldots, (3, 3^3)$; $(5, 5^0), \ldots, (5, 5^2)$; $(7, 7^0), \ldots, (7, 7^2)$; $(11, 11^0), \ldots, (11, 11^1)$. And rest for all other two elements each. Therefore, $$n(R_1) = 6 + 4 + 3 + 3 + (2 \times 10) = 36$$ Similarly for $R_2$, $(2, 2^0), (2, 2^1)$; $(47, 47^0), (47, 47^1)$. Therefore, $$n(R_2) = 2 \times 14 = 28$$ Therefore, $$n(R_1) - n(R_2) = 36 - 28 = 8$$

Question 22

Maths · Complex Numbers and Quadratic Equations · Numerical

The number of real solutions of the equation $$e^{4x} + 4e^{3x} - 58e^{2x} + 4e^{x} + 1 = 0$$ is .

Answer: 2

Solution

Given $e^{4x} + 4e^{3x} - 58e^{2x} + 4e^x + 1 = 0$. Let $f(x) = e^{2x} \left( e^{2x} + \frac{1}{e^{2x}} + 4 \left( e^x + \frac{1}{e^x} \right) - 58 \right)$. $e^x + \frac{1}{e^x}$. Let $h(t) = t^2 + 4t - 58 = 0$. $$t = \frac{-4 \pm \sqrt{16 + 4 \cdot 58}}{2}$$ $$\frac{-4 \pm 2 \sqrt{62}}{2}$$ $t_1 = -2 + 2 \sqrt{62}$ $t_2 = -2 - 2 \sqrt{62}$ (not possible) $t \geq 2$ $e^x + \frac{1}{e^x} = -2 + 2 \sqrt{62}$ $e^{2x} - (-2 + 2 \sqrt{62}) e^x + 1 = 0$ $(-2 + 2 \sqrt{62}) - 4$ $4 + 4.62 - 8 \sqrt{62} - 4$ $248 - 8 \sqrt{62} > 0$ $$\frac{-b}{2a} > 0$$ Both roots are positive. 2 real roots.

Question 23

Maths · Statistics · Numerical

The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to .

Answer: 17

Solution

We have $$Variance = \frac{\sum_{r=1}^{15} x_r^2}{15} - \left( \frac{\sum_{r=1}^{15} x_r}{15} \right)^2$$ Now, as per information given in equation $$\frac{\sum x_r^2}{15} - 8^2 = 3^2 \implies \sum x_r^2 = \log 5$$ Now, the new $$\sum x_r^2 = \log 5 - 5^2 + 20^2 = 1470$$ And, new $$\sum x_r = (15 \times 8) - 5 + (20) = 135$$ Therefore, $$Variance = \frac{1470}{15} - \left( \frac{135}{15} \right)^2 = 98 - 81 = 17$$

Question 24

Maths · Vector Algebra · Numerical

If $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$, $\vec{b} = 3\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$ are coplanar vectors and $\vec{a} \cdot \vec{c} = 5$, $\vec{b} \perp \vec{c}$, then $122 (c_1 + c_2 + c_3)$ is equal to

Answer: 150

Solution

Given $\overline{\mathbf{a}} \cdot \overline{\mathbf{c}} = 5 \implies 2c_1 + c_2 + 3c_3 = 5$ ...(1) $\overline{\mathbf{b}} \cdot \overline{\mathbf{c}} = 0 \implies 3c_1 + 3c_2 + c_3 = 0$ ...(2) And $\left[ \overline{\mathbf{a}} \ \overline{\mathbf{b}} \ \overline{\mathbf{c}} \right] = 0 \implies \begin{vmatrix} 2 & 1 & 3 \\ 3 & 3 & 1 \\ 3 & 1 & 1 \end{vmatrix} = 0$ $$\implies 8c_1 - 7c_2 - 3c_3 = 0$$ ...(3) By solving (1), (2), (3) we get $$c_1 = \frac{10}{122}, \ c_2 = \frac{-85}{122}, \ c_3 = \frac{225}{122}$$ Therefore, $122(c_1 + c_2 + c_3) = 150$

Question 25

Maths · Straight Lines and Pair of Straight Lines · Numerical

A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ratio 2 : 1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be $(\alpha, \beta)$. Then, the value of $7\alpha + 3\beta$ is equal to _______.

Answer: 31

Solution

By observation we see that $A(\alpha, 0)$. And $\beta = y-coordinate of R$ $$= \frac{2 \times 4 + 1 \times 0}{2 + 1} = \frac{8}{3} \ldots (1)$$ Now $P'$ is image of $P$ in $y = 0$ which will be $P'(2, -3)$. Therefore, equation of $P'Q$ is $(y + 3) = \frac{4 + 3}{5 - 2}(x - 2)$ i.e. $3y + 9 = 7x - 14$. $A \equiv \left(\frac{23}{7}, 0\right)$ by solving with $y = 0$. Therefore, $\alpha = \frac{23}{7} \ldots (2)$ By (1), (2) $$7\alpha + 3\beta = 23 + 8 = 31$$

Question 26

Maths · Applications of Derivatives · Numerical

Let $\ell$ be a line which is normal to the curve $y = 2x^2 + x + 2$ at a point $P$ on the curve. If the point $Q(6, 4)$ lies on the line $\ell$ and $O$ is origin, then the area of the triangle $OPQ$ is equal to .

Answer: 13

Solution

Given $y = 2x^2 + x + 2$. The derivative is $\frac{dy}{dx} = 4x + 1$. Let $P$ be $(h, k)$, then the normal at $P$ is $$y - k = -\frac{1}{4h + 1}(x - h)$$ This passes through $Q(6, 4)$. Therefore, $$4 - k = -\frac{1}{4h + 1}(6 - h)$$ $$\Rightarrow (4h + 1)(4 - k) + 6 - h = 0$$ Also $k = 2h^2 + h + 2$. Thus, $$(4h + 1)(4 - 2h^2 - h - 2) + 6 + h = 0$$ $$\Rightarrow 4h^3 - 3h^2 + 3h - 8 = 0$$ $$\Rightarrow h = 1, k = 5$$ Now the area of $\Delta OPQ$ will be $$\frac{1}{2}\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 5 \\ 1 & 6 & 4 \end{vmatrix} = 13$$

Question 27

Maths · Permutations and Combinations · Numerical

Let $A=\{1,a_1,a_2,\ldots,a_{18},77\}$ be a set of integers with $1<a_1<a_2<\cdots<a_{18}<77$. Let the set $A+A=\{x+y:x,y\in A\}$ contain exactly $39$ elements. Then, the value of $a_1+a_2+\cdots+a_{18}$ is equal to ______.

Answer: 702

Solution

$t_1+t_2+t_3+t_4=30$ Coefficient of $x^{30}$ in $(1+x+\cdots+x^{30})^2$ $=(x^4+x^5+x^6+x^7)$ $(x^2+x^3+x^4+x^5+x^6)$ $\left(\frac{1-x^{31}}{1-x}\right)^2$ $=x^6 \left(\frac{1-x^{31}}{1-x}\right)^2$ $(1+x+x^2+x^3) (1+x+x^2+x^3+x^4)$ $=x^6(1-x^3)(1-x^4)(1-x)^{-2}$ $=x^6(1-x^4-x^5+x^9)(1-x)^{-2}$ Coefficient of $x^n$ in $(1-x)^{-r}$ is ${}^{\,n+r-1}C_{r-1}$ $\Rightarrow {}^{27}C_3 -{}^{23}C_3 -{}^{22}C_3 +{}^{18}C_3$ $=2925-1771-1540+816$ $=430$ OR $x_2\in[4,7]$ $x_3\in[2,6]$ $\Rightarrow t_1+t_2+t_3+t_4=24$ total ways $= {}^{24+4-1}C_{4-1} -{}^{20+4-1}C_{4-1} -{}^{19+4-1}C_{4-1} +{}^{15+4-1}C_{4-1}$ $= {}^{27}C_3 -{}^{23}C_3 -{}^{22}C_3 +{}^{18}C_3$ $=430$

Question 28

Maths · Binomial Theorem · Numerical

The number of positive integers $k$ such that the constant term in the binomial expansion of $$\left(2x^3 + \frac{3}{x^k}\right)^{12}$$, $x \neq 0$ is $2^8 \cdot \ell$, where $\ell$ is an odd integer, is _______.

Answer: 3,6

Solution

Given $\($ $\left$( 2x^3 + $\frac{3}{x^k}$ $\right$)^{12} $\)$ $\($ t_{r+1} = $\binom{12}{r}$ $\left$( 2x^3 $\right$)^r $\left$( $\frac{3}{x^k}$ $\right$)^{12-r} $\)$ $\($ x^{3r - (12-r)k} $\rightarrow$ constant $\)$ Therefore, $\($ 3r - 12k + rk = 0 $\)$ $\($ $\Rightarrow$ k = $\frac{3r}{12 - r}$ $\)$ Therefore, possible values of $\($ r $\)$ are 3, 6, 8, 9, 10 and corresponding values of $\($ k $\)$ are 1, 3, 6, 9, 15. Now $\($ $\binom{12}{r}$ = 220, 924, 495, 220, 66 $\)$ Therefore, possible values of $\($ k $\)$ for which we will get $\($ 2^8 $\)$ are 3, 6.

Question 29

Maths · Complex Numbers and Quadratic Equations · Numerical

The number of elements in the set $\{$z = a + ib $\in$ $\mathbb{C}$ : a, b $\in$ $\mathbb{Z}$ and 1 < |z - 3 + 2i| < 4$\}$ is

Answer: 40

Solution

Given $1 < |Z - 3 + 2i| < 4$. This represents an annular region centered at $(3, -2)$ with inner radius $1$ and outer radius $4$. The inequality $1 < (a - 3)^2 + (b + 2)^2 < 16$ describes the same region in terms of $a$ and $b$. The possible integer solutions for $(a, b)$ are: $(0, \pm 2)$, $(\pm 2, 0)$, $(\pm 1, \pm 2)$, $(\pm 2, \pm 1)$, $(\pm 2, \pm 3)$, $(3 \pm 2, \pm 2)$, $(\pm 1, \pm 1)$, $(2 \pm, \pm 2)$, $(\pm 3, 0)$, $(0, \pm 3)$, $(\pm 3, \pm 1)$, $(\pm 1, \pm 3)$. Total 40 points.

Question 30

Maths · Conic Sections · Numerical

Let the lines $y + 2x = \sqrt{11} + 7\sqrt{7}$ and $2y + x = 2\sqrt{11} + 6\sqrt{7}$ be normal to a circle $C: (x-h)^2 + (y-k)^2 = r^2$. If the line $\sqrt{11}y - 3x = \frac{5\sqrt{77}}{3} + 11$ is tangent to the circle $C$, then the value of $(5h - 8k)^2 + 5r^2$ is equal to ____.

Answer: 816

Solution

Normal are $$y + 2x = \sqrt{11} + 7\sqrt{7},$$ $$2y + x = 2\sqrt{11} + 6\sqrt{7}$$ Center of the circle is point of intersection of normals i.e. $$\left( \frac{8\sqrt{7}}{3}, \sqrt{11} + \frac{5\sqrt{7}}{3} \right)$$ Tangent is $\sqrt{11}y - 3x = \frac{5\sqrt{77}}{3} + 11$ Radius will be perpendicular distance of tangent from center i.e. $4\sqrt{\frac{7}{5}}$ Now $(5h - 8k)^2 + 5r^2 = 816$

Physics

Question 31

Physics · Mechanical Properties of Fluids · Single correct

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity. Reason R: Coefficient of viscosity = $\frac{Force}{Velocity gradient}$ Question: Choose the correct answer from the options given below :

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

Answer: (c)

Solution

Pressure and time $$P : \frac{N}{\mathrm{m}^2}, Time : Sec$$ $$Pt = \frac{N \cdot \mathrm{sec}}{\mathrm{m}^2}$$ $$\eta = \frac{F}{6 \pi r v} : \frac{N}{\mathrm{m} \cdot \mathrm{m/sec}} : \frac{N \cdot \mathrm{sec}}{\mathrm{m}^2}$$

Question 32

Physics · Work, Energy and Power · Single correct

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as $a = k^2 r t^2$. where k is a constant. The power delivered to the particle by the force acting on it is given as

  1. zero
  2. mk^2 r^2 t^2
  3. mk^2 r t
  4. mk^2 rt

Answer: (c)

Solution

Given $a = k^2 r t^2 = \frac{V^2}{r}$. $V = k r t$. $a_t = \frac{dv}{dt} = k r$. $F_t = m a_t = m k r$. $P = \mathbf{F} \cdot \mathbf{V}$. $= F \cos \theta V = F_t V = m k r (k r t)$. $P = m k^2 r^2 t$.

Question 33

Physics · Motion in a Plane · Single correct

Motion of a particle in x-y plane is described by a set of following equations $x = 4 \sin \left( \frac{\pi}{2} - \omega t \right)$ m and $y = 4 \sin (\omega t)$ m. The path of particle will be –

  1. circular
  2. helical
  3. parabolic
  4. elliptical

Answer: (a)

Solution

Given $$x = 4 \sin \left( \frac{\pi}{2} - \omega t \right)$$ $$y = 4 \cos(\omega t)$$ or $$x = 4 \cos(\omega t)$$ $$y = 4 \sin(\omega t)$$ Eliminate 't' to find relation between $x$ and $y$ $$x^2 + y^2 = y^2 \cos^2 \omega t + y^2 \sin^2 \omega t = 4^2$$ Thus, $$x^2 + y^2 = 4^2$$

Question 34

Physics · System of Particles and Rotational Motion · Single correct

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

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

Answer: (a)

Solution

Solid sphere $$I_0 = I_{com} + MR^2$$ (Parallel Axis theorem) $$I_0 = \frac{2}{5} MR^2 + MR^2$$ $$I_0 = \frac{7}{5} MR^2$$ Hollow sphere $$I_0 = I_{com} + MR^2$$ $$= \frac{2}{3} MR^2 + MR^2 = \frac{5}{3} MR^2$$ $$I_1 + I_2 + I_3$$ (Perpendicular axis theorem) By symmetry MOI About 1'' and 2'' Axis are same i.e. $$I_1 = I_2$$ $$\therefore 2I_1 = I_3 = MR^2 \left(I_{com} = MR^2\right)$$ $$I_1 = \frac{MR^2}{2}$$ Similarly in disc $$2I_1 = \frac{MR^2}{2} \left\{ I_{com} = \frac{MR^2}{2} \right\}$$ $$I_1 = \frac{MR^2}{4}$$

Question 35

Physics · Gravitation · Single correct

Two planets A and B of equal mass are having their period of revolutions $T_A$ and $T_B$ such that $T_A = 2T_B$. These planets are revolving in the circular orbits of radii $r_A$ and $r_B$ respectively. Which out of the following would be the correct relationship of their orbits?

  1. $2r_A^2 = r_B^2$
  2. $r_A^3 = 2r_B^3$
  3. $r_A^3 = 3r_B^3$
  4. $T_A^2 - T_B^2 = \frac{\pi^2}{\mathrm{GM}} \left( r_B^3 - 4r_A^3 \right)$

Answer: (c)

Solution

Given the formula for the period $T$: $$T = \frac{2\pi}{\sqrt{Gm_a}} r^{\frac{3}{2}}$$ We have $T^2 \propto r^3$. Using the ratio of periods and radii: $$\left( \frac{T_A}{T_B} \right)^2 = \left( \frac{r_A}{r_B} \right)^3$$ Substituting the given values: $$\left( \frac{2}{1} \right)^2 = \frac{r_A}{r_B} \Rightarrow r_A^3 = 4r_B^3$$

Question 36

Physics · Mechanical Properties of Fluids · Single correct

A water drop of diameter $\mathrm{cm}$ is broken into 64 equal droplets. The surface tension of water is $0.075\,\mathrm{N\,m^{-1}}$. In this process, the gain in surface energy will be:

  1. $2.8 \times 10^{-4}\,\mathrm{J}$
  2. $1.5 \times 10^{-3}\,\mathrm{J}$
  3. $1.9 \times 10^{-4}\,\mathrm{J}$
  4. $9.4 \times 10^{-5}\,\mathrm{J}$

Answer: (a)

Solution

Given $d = 2 \, \mathrm{cm}$; $r = 1 \, \mathrm{cm}$; $T = 0.075$. $\Delta SE = T \, \Delta A$ $$= 0.075 (A_f - A_i)$$ $$A_i = 4 \pi r^2$$ $$A_f = 4 \pi r_0^2 \times 64$$ By volume conservation $$\frac{4}{3} \pi r^3 = 64 \cdot \frac{4}{3} \pi r_0^3$$ $$r_0 = \frac{r}{4}$$ $$A_f = 4 \pi \left( \frac{r}{4} \right)^2 \cdot 64 = 16 \pi r^2$$ $$\Delta SE = 0.075 \left( 16 \pi r^2 - 4 \pi r^2 \right)$$ $$= 0.075 \left( 12 \pi (0.01)^2 \right)$$ $$= 2.8 \times 10^{-4} \, \mathrm{J}$$

Question 37

Physics · Kinetic Theory · Single correct

Given below are two statement : Statement – I : What $\mu$ amount of an ideal gas undergoes adiabatic change from state $(P_1, V_1, T_1)$ to state $(P_2, V_2, T_2)$, the work done is $W = \frac{1R(T_2 - T_1)}{1 - \gamma}$, where $\gamma = \frac{C_P}{C_V}$ and $R$ = universal gas constant, Statement — II: In the above case, when work is done on the gas, the temperature of the gas would rise. Choose the correct answer from the options given below:

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

Answer: (a)

Solution

Given $$W_{adiabatic} = \frac{NR(T_f - T_i)}{1 - \gamma}$$ statement 1. $$Q = W + \Delta U$$ $$0 = W + \Delta U$$ $$\Delta U = -W$$ If work is done on the gas, i.e. work is negative, $$\therefore \Delta U$$ is positive. $$\therefore$$ Temperature will increase.

Question 38

Physics · Electric Charges and Fields · Single correct

Given below are two statements : Statement-I : A point charge is brought in an electric field. The value of electric field at a point near to the charge may increase if the charge is positive. Statement-II : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero. Choose the correct answer from the options given below :

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

Answer: (a)

Solution

If the electric field is in the positive direction and the positive charge is to the left of that point then the electric field will increase. But to the left of the positive charge the electric field would decrease. If the dipole is kept at the point where the electric field is maximum then the force on it will be zero.

Question 39

Physics · Electric Charges and Fields · Single correct

The three charges $q/2$, $q$ and $q/2$ are placed at the corners $A$, $B$ and $C$ of a square of side $'a'$ as shown in figure. The magnitude of electric field $(E)$ at the corner $D$ of the square, is:

  1. $\frac{q}{4 \pi \varepsilon_0 \, a^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{2} \right)$
  2. $\frac{q}{4 \pi \varepsilon_0 \, a^2} \left( 1 + \frac{1}{\sqrt{2}} \right)$
  3. $\frac{q}{4 \pi \varepsilon_0 \, a^2} \left( 1 - \frac{1}{\sqrt{2}} \right)$
  4. $\frac{q}{4 \pi \varepsilon_0 \, a^2} \left( \frac{1}{\sqrt{2}} - \frac{1}{2} \right)$

Answer: (a)

Solution

The electric field at point D due to the charges is calculated as follows. The net electric field at D is given by: $$ (E_{net})_D = \frac{kq}{2a^2} + \frac{\sqrt{2}kq}{2a^2} $$ This can be simplified to: $$ (E_{net})_D = \frac{kq}{a^2} \left( \frac{1}{2} + \frac{1}{\sqrt{2}} \right) $$ Further simplifying, we have: $$ (E_{net})_D = \frac{q}{4\pi \varepsilon_0 a^2} \left( \frac{1}{2} + \frac{1}{\sqrt{2}} \right) $$

Question 40

Physics · Moving Charges and Magnetism · Single correct

An infinitely long hollow conducting cylinder with radius $R$ carries a uniform current along its surface. Choose the correct representation of magnetic field $(B)$ as a function of radial distance $(r)$ from the axis of cylinder.

Answer: (d)

Solution

1) For $r < R$, $B_p = 0$. 2) For $r \geq R$, $B_p = \frac{\mu_0 I}{2 \pi r}$. $B_p \propto \frac{1}{r}$.

Question 41

Physics · Electromagnetic Waves · Single correct

A radar sends an electromagnetic signal of electric field $(E_0) = 2.25 \, \mathrm{V/m}$ and magnetic field $(B_0) = 1.5 \times 10^{-8} \, \mathrm{T}$ which strikes a target on line of sight at a distance of $3 \, \mathrm{km}$ in a medium. After that, a pail of signal (echo) reflects back towards the radar with same velocity and by same path. If the signal was transmitted at time $t_0$ from radar, then after how much time echo will reach to the radar?

  1. $2.0 \times 10^{-5} \, \mathrm{s}$
  2. $4.0 \times 10^{-5} \, \mathrm{s}$
  3. $1.0 \times 10^{-5} \, \mathrm{s}$
  4. $8.0 \times 10^{-5} \, \mathrm{s}$

Answer: (b)

Solution

Given $C = \frac{E_0}{B_0} = \frac{2.25}{1.5 \times 10^{-8}} = 1.5 \times 10^8 \, \mathrm{ms^{-1}}$. Then, $t = \frac{6 \times 10^3}{1.5 \times 10^8} = 4 \times 10^{-5} \, \mathrm{s}$.

Question 42

Physics · Ray Optics and Optical Instruments · Single correct

The refracting angle of a prism is $A$ and refractive index of the material of the prism is $\cot (A/2)$. Then the angle of minimum deviation will be -

  1. 180 - 2A
  2. 90 - A
  3. 180 + 2A
  4. 180 - 3A

Answer: (a)

Solution

Given $\($ $\mu$ = $\frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \frac{A}{2}}$ $\)$. $\($ $\mu$ = $\cot$ $\frac{A}{2}$ $\)$ implies $\($ $\sin$ $\left$( $\frac{A + \delta_m}{2}$ $\right$) = $\cos$ $\frac{A}{2}$ $\)$. Therefore, $\($ $\delta$_m = 180 - 2A $\)$.

Question 43

Physics · Wave Optics · Single correct

The aperture of the objective is 24.4 $\,$ $\mathrm{cm}$. The resolving power of this telescope. If a light of wavelength 2440 $\,$ $\mathrm{\AA}$ is used to see the object will be

  1. 8.1 $\times$ 10^6
  2. 10.0 $\times$ 10^7
  3. 8.2 $\times$ 10^5
  4. 1.0 $\times$ 10^{-8}

Answer: (c)

Solution

R.P = $\frac{d}{1.22 \lambda}$ = $\frac{24.4 \times 10^{-2}}{1.22 \times 2440 \times 10^{-10}}$ = 8.2 $\times$ 10^{5}

Question 44

Physics · Dual Nature of Radiation and Matter · Single correct

The de Brogue wavelengths for an electron and a photon are $\lambda_e$ and $\lambda_p$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Brogue wavelengths of two?

  1. $\lambda_p \propto \lambda_e^2$
  2. $\lambda_p \propto \lambda_e$
  3. $\lambda_p \propto \sqrt{\lambda_e}$
  4. $\lambda_p \propto \sqrt{\frac{1}{\lambda_e}}$

Answer: (a)

Solution

Given $$\lambda_e = \frac{h}{\sqrt{2mk}}$$. Also for photon, $$k = \frac{hc}{\lambda_p}$$. Therefore, $$\lambda_e = \frac{h \sqrt{\lambda_p}}{\sqrt{2m hc}}$$. Hence, $$\lambda_p \propto \lambda e^2$$.

Question 45

Physics · Nuclei · Single correct

The Q-value of a nuclear reaction and kinetic energy of the projectile particle, $K_p$ are related as:

  1. $Q = K_p$
  2. $(K_p + Q) < 0$
  3. $Q < K_p$
  4. $(K_p + Q) > 0$

Answer: (d)

Solution

The reaction is given by $x + p \rightarrow \gamma + b$. The equation for $Q$ is $Q = k_\gamma + k_b - k_p$. Rearranging gives $Q + k_p = k_\gamma + k_b$. Therefore, $Q + k_p > 0$.

Question 46

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

In the following circuit, the correct relation between output (Y) and inputs A and B will be:

  1. Y = AB
  2. Y = A + B
  3. Y = $\overline{AB}$
  4. Y = $\overline{A + B}$

Answer: (c)

Solution

This is a NAND gate.

Question 47

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

For using a multimeter to identify diode from electrical components, choose the correct statement out of the following about the diode:

  1. It is two terminal device which conducts current in both directions.
  2. It is two terminal device which conducts current in one direction only.
  3. It does not conduct current gives an initial deflection which decays to zero.
  4. It is three terminal device which conducts current in one direction only between central terminal and either of the remaining two terminals.

Answer: (b)

Solution

In forward bias diode conducts. In reverse bias it does not conduct.

Question 48

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

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: n-p-n transistor permits more current than a p-n-p transistor. Reason R: Electrons have greater mobility as a charge carrier. Choose the correct answer from the options given below:

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

Answer: (a)

Solution

Theory

Question 49

Physics · Communication Systems · Single correct

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

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

Answer: (c)

Solution

Theory

Question 50

Physics · Waves · Single correct

The velocity of sound in a gas, in which two wavelengths 4.08 m and 4.16 m produce 40 beats in 12 s, will be:

  1. $2.8\ \text{m s}^{-1}$
  2. $175.5\ \text{m s}^{-1}$
  3. $353.6\ \text{m s}^{-1}$
  4. $707.2\ \text{m s}^{-1}$

Answer: (d)

Solution

Given $f_b = f_1 - f_2$. $$\frac{v}{4.08} - \frac{v}{4.16} = \frac{40}{12}$$ Therefore, $v = 707.2$.

Question 51

Physics · Oscillations · Numerical

A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at $60^\circ$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ______ ms$^{-1}$. (if $g = 10$ m/s$^2$)

Answer: 5

Solution

The maximum speed is given by $V_{max} = \sqrt{2gh}$. The speed will be highest at the lowest position. $$h = (\ell - \ell \cos 60^\circ) = \frac{\ell}{2}$$ $$V_{max} = \sqrt{2 \times g \times \frac{\ell}{2}} = \sqrt{10 \times 2.5} = 5 \, \mathrm{m/s}$$

Question 52

Physics · Current Electricity · Numerical

A meter bridge setup is shown in the figure. It is used to determine an unknown resistance $R$ using a given resistor of $15 \, \Omega$. The galvanometer (G) shows null deflection when tapping key is at $43 \, \mathrm{cm}$ mark from end A. If the end correction for end A is $2 \, \mathrm{cm}$, then the determined value of $R$ will be _______ $\Omega$.

Answer: 19

Solution

Using the conditions of a balanced wheatstone bridge and adding the end correction. $$\frac{15}{(43+2)} = \frac{R}{(102-45)} \Rightarrow R = \frac{57}{45} \times 15$$ $$R = 19 \, \Omega$$

Question 53

Physics · Current Electricity · Numerical

Current measured by the ammeter \text{Ⓐ} in the reported circuit when no current flows through 10 $\Omega$ resistance, will be A.

Answer: 10

Solution

Using the condition of a balanced wheat stone bridge, $$\Rightarrow \frac{R}{3} = \frac{4}{6} \Rightarrow R = 2\, \Omega$$ So the effective resistance of the circuit is $$R_{eq} = \frac{6 \times 9}{6 + 9} = \frac{18}{5} \, \Omega$$ $$i = \frac{36}{R_{eq}} = 10\, A$$

Question 54

Physics · Alternating Current · Numerical

An AC source is connected to an inductance of $100\,\mathrm{mH}$, a capacitance of $100\,\mu\mathrm{F}$ and a resistance of $120\,\Omega$ as shown in figure. The time in which the resistance having a thermal capacity $2\,\mathrm{J/°C}$ will get heated by $16°\mathrm{C}$ is ________ s.

Answer: 15

Solution

Given $|X_L - X_C| = |10 - 10^2| = 90 \, \Omega$. $Z =$ Impedance $$Z = \sqrt{(X_L - X_C)^2 + R^2} = \sqrt{(90)^2 + (20)^2} = 150 \, \Omega$$ $$i_{rms} = \frac{V_{rms}}{Z} = \left( \frac{2}{15} \right) \, A$$ Now $i_{rms}^2 R \Delta t = ms(\Delta T)$ $$\Rightarrow \Delta t = 15 \, sec$$

Question 55

Physics · System of Particles and Rotational Motion · Numerical

The position vector of 1 kg object is $\vec{r} = (3\hat{i} - \hat{j}) \, \mathrm{m}$ and its velocity $\vec{v} = (3\hat{j} + \hat{k}) \, \mathrm{ms^{-1}}$. The magnitude of its angular momentum is $\sqrt{x} \, \mathrm{Nm}$ where $x$ is _______.

Answer: 91

Solution

Using $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m \vec{v}$, $m = 1 \mathrm{kg}$ $$\vec{L} = (3\hat{i} - \hat{j}) \times (3\hat{j} + \hat{k}) = (9\hat{k} - 3\hat{i}) \mathrm{N\!\cdot\!s}$$ $$\Rightarrow |\vec{L}| = \sqrt{91} \mathrm{N\!\cdot\!s}$$

Question 56

Physics · Work, Energy and Power · Numerical

A man of 60 $\mathrm{\, kg}$ is running on the road and suddenly jumps into a stationary trolly car of mass 120 kg. Then, the trolly car starts moving with velocity 2 $ms^{-1}$. The velocity of the running man was $ms^{-1}$ when he jumps into the car.

Answer: 6

Solution

Taking the system as man and trolley and using conservation of linear momentum. $$60 \times v = (60 + 120) \times 2$$ $$\Rightarrow v = 6 \, \mathrm{m/s}$$

Question 57

Physics · Laws of Motion · Fill in the blank

A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement as shown in the figure. The bigger mass is placed on a horizontal ice slab and is being pulled by a force of 2Mg. In this situation, the tension in the string is $\frac{x}{5}Mg$ for $x = \_\_\_\_$. Neglect the mass of the string and the friction between the bigger mass and the ice slab. (Given $g$ = acceleration due to gravity)

Answer: 6

Solution

Using $\vec{F}_{net} = \mu \vec{a}$, $$2Mg - T = 4Ma$$ $$T - Mg = Ma$$ $$\Rightarrow a = \frac{g}{5}$$ $$T = Mg + Ma = Mg + \frac{Mg}{5} = \frac{6}{5}Mg$$

Question 58

Physics · Thermodynamics · Numerical

The total internal energy of two mole monoatomic ideal gas at temperature $T = 300 \, \mathrm{K}$ will be J. (Given $R = 8.31 \, \mathrm{J/mol.K}$)

Answer: 7479

Solution

Question 59

Physics · Moving Charges and Magnetism · Numerical

A singly ionized magnesium atom (A24) ion is accelerated to kinetic energy $5 \, \mathrm{keV}$ and is projected perpendicularly into a magnetic field $B$ of the magnitude $0.5 \, \mathrm{T}$. The radius of path formed will be __________ cm.

Answer: 10

Solution

The radius of the circular path is given by the equation $$R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}.$$

Question 60

Physics · Alternating Current · Numerical

A telegraph line of length 100 km has a capacity of $0.01 \, \mu\mathrm{F/km}$ and it carries an alternating current at $0.5$ kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is ________ mH. (if $\pi = \sqrt{10}$)

Answer: 100

Solution

For minimum impedance $X_L = X_C$ $$\Rightarrow \omega L = \frac{1}{\omega C} \Rightarrow L = \frac{1}{\omega^2 C} = 10^{-1} \, \mathrm{H} = 100 \, \mathrm{mH}$$

Chemistry

Question 61

Chemistry · The Solid State · Single correct

The incorrect statement about the imperfections in solids is :

  1. Schottky defect decreases the density of the substance.
  2. Interstitial defect increases the density of the substance.
  3. Frenkel defect does not alter the density of the substance.
  4. Vacancy defect increases the density of the substance.

Answer: (d)

Solution

Due to vacancy defect density of the substance will decrease.

Question 62

Chemistry · Surface Chemistry · Single correct

The Zeta potential is related to which property of colloids?

  1. Colour
  2. Tyndall effect
  3. Charge on the surface of colloidal particles
  4. Brownian movement

Answer: (c)

Solution

The potential difference between the fixed and diffused layer of charges in a colloidal particle is called zeta potential.

Question 63

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

Element "E" belongs to the period 4 and group 16 of the periodic table. The valence shell electron configuration of the element, which is just above ‘E’ in the group is

  1. $3s^2$, $3p^4$
  2. $3d^{10}$, $4s^2$, $4p^4$
  3. $4d^{10}$, $5s^2$, $5p^4$
  4. $2s^2$, $p^4$

Answer: (a)

Solution

E $\Rightarrow [\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^4$ Element above E $\Rightarrow [\mathrm{Ne}]\,3s^2\,3p^4$

Question 64

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

Given are two statements one is labelled as Assertion A and other is labelled as Reason R. Assertion A : Magnesium can reduce Al$_2$O$_3$ at a temperature below 1350$^\circ$C, while above 1350$^\circ$C aluminium can reduce MgO. Reason R : The melting and boiling points of magnesium are lower than those of aluminium. In light of the above statements, choose most appropriate answer from the options given below:

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

Answer: (b)

Solution

From the Ellingham diagram given in NCERT, it can be seen that the $\mathrm{Mg,MgO}$ line crosses the $\mathrm{Al,Al_2O_3}$ line after $1350^\circ\mathrm{C}$. Hence, the Assertion is true. Yes, Mg has a lower melting point (MP) and boiling point (BP) than aluminium, but this does not explain the above fact.

Question 65

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

Dihydrogen reacts with CuO to give

  1. CuH_2
  2. Cu
  3. Cu_2O
  4. Cu(OH)_2

Answer: (b)

Solution

The reaction is: $$\mathrm{CuO + H_2 \rightarrow Cu + H_2O}$$ under hot conditions.

Question 66

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

Nitrogen gas is obtained by thermal decomposition of

  1. Ba(NO$_3$)$_2$
  2. Ba(N$_3$)$_2$
  3. NaNO$_2$
  4. NaNO$_3$

Answer: (b)

Solution

The reaction is given by the equation: $$\mathrm{Ba(N_3)_2} \rightarrow \mathrm{Ba} + 3\mathrm{N_2}$$

Question 67

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

Given below are two statements : Statement -I :The pentavalent oxide of group- 15 element. $\mathrm{E}_2\mathrm{O}_5$. is less acidic than trivalent oxide. $\mathrm{E}_2\mathrm{O}_3$. of the same element. Statement -II :The acidic character of trivalent oxide of group 15 elements. $\mathrm{E}_2\mathrm{O}_3$. decreases down the group. In light of the above statements, choose most appropriate answer from the options given below:

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

Answer: (d)

Solution

As positive oxidation state increases, electronegativity of element increases hence acidic character increases. Down the group, non-metallic character decreases, acidic character decreases. Acidic character: $E_2O_5 > E_2O_3$. Down the group, acidic character of $E_2O_3$ decreases.

Question 68

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

Which one of the lanthanoids given below is the most stable in divalent form?

  1. Ce (Atomic Number 58)
  2. Sm (Atomic Number 62)
  3. Eu (Atomic Number 63)
  4. Yb (Atomic Number 70)

Answer: (c)

Solution

Given $E^\circ_{\mathrm{M^{3+}/M^{2+}}} \Rightarrow \begin{array}{cc} \mathrm{Eu} & \mathrm{Yb} \\ -0.35 & -1.05 \end{array}$. Hence, due to more reduction potential in Eu as compared to Yb, it can be concluded that $\mathrm{Eu^{2+}}$ is more stable than $\mathrm{Yb^{2+}}$.

Question 69

Chemistry · Co-ordination Compounds · Single correct

Given below are two statements: Statement I: $[\mathrm{Ni(CN)}_4]^{2-}$ is square planar and diamagnetic complex, with $dsp^2$ hybridization for Ni but $[\mathrm{Ni(CO)}_4]$ is tetrahedral, paramagnetic and with $sp^3$-hybridization for Ni. Statement II: $[\mathrm{NiCl}_4]^{2-}$ and $[\mathrm{Ni(CO)}_4]$ both have same d-electron configuration, have same geometry and are paramagnetic. In light of the above statements, choose the correct answer from the options given below:

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

Answer: (b)

Solution

[$\mathrm{Ni(CN)_4}$]^{2-}: $\;$ d^8 configuration, SFL, sq. planar splitting (dsp^2), diamagnetic. [$\mathrm{Ni(CO)_4}$]: $\;$ d^{10} config (after excitation), SFL, tetrahedral splitting (sp^3), diamagnetic. [$\mathrm{NiCl_4}$]^{2-}: $\;$ d^8 config, WFL, tetrahedral splitting (sp^3), paramagnetic (2 unpaired e^-).

Question 70

Chemistry · Environmental Chemistry · Single correct

Which amongst the following is not a pesticide?

  1. DDT
  2. Organophosphates
  3. Dieldrin
  4. Sodium arsenite

Answer: (d)

Solution

Sodium arsenite is a herbicide.

Question 71

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

Which one of the following techniques is not used to spot components of a mixture separated on thin layer chromatographic plate?

  1. $\mathrm{I}_2$ (Solid)
  2. U.V. Light
  3. Visualisation agent as a component of mobile phase
  4. Spraying of an appropriate reagent

Answer: (c)

Solution

The function of mobile phase is to carry the components present on TLC.

Question 72

Chemistry · Hydrocarbons · Single correct

Which of the following structures are aromatic in nature?

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

Answer: (b)

Solution

A, B aromatic. C, D is nonaromatic.

Question 73

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The major product (P ) in the reaction is

Answer: (c)

Solution

The reaction involves the addition of HBr to the alkene. The bromine ion attacks the more substituted carbon, leading to the formation of the product shown.

Question 74

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The correct structure of product ‘A’ formed in the following reaction.

Answer: (a)

Solution

PhCH = O + PhCH $\xrightarrow{OD/D_2O}$ $PhCH_2OD + PhCO_2^-$

Question 75

Chemistry · Haloalkanes and Haloarenes · Single correct

Which one of the following compounds is inactive towards $\mathrm{S_N1}$ reaction?

Answer: (c)

Solution

Question 76

Chemistry · Amines · Single correct

Identify the major product formed in the following sequence of reactions:

Answer: (c)

Solution

The reaction starts with aniline ($\mathrm{C_6H_5NH_2}$) which undergoes bromination in the presence of $\mathrm{Br_2}$ and $\mathrm{H_2O}$ to form 2,4,6-tribromoaniline. This compound is then treated with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ to form the diazonium salt. Finally, the diazonium salt is reduced using $\mathrm{H_3PO_2}$ to yield 1,3,5-tribromobenzene.

Question 77

Chemistry · Amines · Single correct

A primary aliphatic amine on reaction with nitrous acid in cold (273 $\mathrm{K}$) and there after raising temperature of reaction mixture to room temperature (298 $\mathrm{K}$). Gives a/an

  1. nitrile
  2. alcohol
  3. diazonium salt
  4. secondary amine

Answer: (b)

Solution

$R-NH_2 \xrightarrow[\mathrm{+HCl}]{\mathrm{NaNO_2}} R-N_2^+ \rightarrow R^+ \xrightarrow{\mathrm{H_2O}} R-OH$

Question 78

Chemistry · Polymers · Single correct

Which one of the following is NOT a copolymer ?

  1. Buna-S
  2. Neoprene
  3. PHBV
  4. Butadiene-styrene

Answer: (b)

Solution

Buna-S, PHBr and Butadiene-styrene are copolymer. Only neoprene is namopolymer.

Question 79

Chemistry · Biomolecules · Single correct

Stability of $\alpha$ - Helix structure of proteins depends upon

  1. dipolar interaction
  2. H-bonding interaction
  3. van der Waals forces
  4. $\pi$ -stacking interaction

Answer: (b)

Solution

Mostly H-bonding is responsible for the stability of $\alpha$-helix form.

Question 80

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

The formula of the purple colour formed in Laissagne’s test for sulphur using sodium nitroprusside is

  1. NaFe[Fe(CN)_6]
  2. Na[Cr(NH_3)_2(NCS)_4]
  3. Na_2[Fe(CN)_5(NO)]
  4. Na_4[Fe(CN)_5(NOS)]

Answer: (d)

Solution

The reaction is given by: $$\mathrm{Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NO_5]}$$

Question 81

Chemistry · Redox Reactions · Numerical

A $2.0\,\mathrm{g}$ sample containing $\mathrm{MnO_2}$ is treated with $\mathrm{HCl}$ liberating $\mathrm{Cl_2}$. The $\mathrm{Cl_2}$ gas is passed into a solution of $\mathrm{KI}$ and $60.0\,\mathrm{mL}$ of $0.1\,\mathrm{M}$ $\mathrm{Na_2S_2O_3}$ is required to titrate the liberated iodine. The percentage of $\mathrm{MnO_2}$ in the sample is \_\_\_\_. (Nearest integer) [Atomic masses (in u): $\mathrm{Mn} = 55$; $\mathrm{Cl} = 35.5$; $\mathrm{O} = 16$; $\mathrm{I} = 127$; $\mathrm{Na} = 23$; $\mathrm{K} = 39$; $\mathrm{S} = 32$]

Answer: 13

Solution

The reaction is given by: $$\mathrm{MnO_2 + HCl \rightarrow Cl_2 + Mn^{+2}}$$ with 6 meq of $\mathrm{Cl_2}$ produced, which equals 3 mmol. Next reaction: $$\mathrm{Cl_2 + KI \rightarrow Cl^- + I_2}$$ with 6 meq of $\mathrm{I_2}$ produced. Then: $$\mathrm{I_2 + Na_2S_2O_3 \rightarrow I^- + Na_2S_4O_6}$$ with 6 meq of $\mathrm{I_2}$ reacting, which equals 6 mmol. The percentage of $\mathrm{MnO_2}$ is calculated as: $$\%\mathrm{MnO_2} = \frac{3 \times 10^{-3} \times 87}{2} \times 100$$ which equals 13.05%. Answer: 13

Question 82

Chemistry · Structure of Atom · Numerical

If the work function of a metal is $6.63 \times 10^{-19} \, \mathrm{J}$, the maximum wavelength of the photon required to remove a photoelectron from the metal is ______ nm. (Nearest integer) [Given : $h = 6.63 \times 10^{-34} \, \mathrm{J \, s}$, and $c = 3 \times 10^{8} \, \mathrm{m \, s^{-1}}$]

Answer: 300

Solution

Given $\phi = 6.63 \times 10^{-19} \, \mathrm{J} = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{\lambda}$. Therefore, $\lambda = 3 \times 10^{-7} \, \mathrm{m} = 300 \, \mathrm{nm}$.

Question 83

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

The hybridization of P exhibited in PF$_5$ is sp$^x$d$^y$. The value of $y$ is .

Answer: 1

Solution

PF$_5$ implies $sp^3d$ hybridisation. (5 sigma bonds, zero lone pair on central atom) Value of $y = 1$

Question 84

Chemistry · Thermodynamics · Numerical

4.0 $\mathrm{L}$ of an ideal gas is allowed to expand isothermally into vacuum until the total volume is 20 $\mathrm{L}$. The amount of heat absorbed in this expansion is ____ $\mathrm{L}$ \, $\mathrm{atm}$.

Answer: 0

Solution

For free expansion, $\($ P_{ext} = 0 $\)$, $\($ w = P_{ext} $\cdot$ 0 = 0 $\)$, $\($ w = 0 $\)$. Therefore, $\($ q = 0 $\)$, $\($ $\Delta$ U = q + w = 0 $\)$, $\($ $\Delta$ U = 0 $\)$. Ans. 0

Question 85

Chemistry · Solutions · Numerical

The vapour pressures of two volatile liquids A and B at 25°C are 50 Torr and 100 Torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is $\frac{x}{17}$. The value of x is _______.

Answer: 14

Solution

Given $\($ $\frac{y_B}{1-y_B}$ = $\frac{P_B^\circ}{P_A^\circ}$ $\left$[ $\frac{X_B}{1-X_B}$ $\right$] $\)$. Therefore, $\($ $\frac{y_B}{1-y_B}$ = $\frac{100}{50}$ $\left$[ $\frac{0.7}{0.3}$ $\right$] = $\frac{14}{3}$ $\)$. Thus, $\($ y_B = $\frac{14}{17}$ $\)$.

Question 86

Chemistry · Electrochemistry · Numerical

The solubility product of a sparingly soluble salt $\mathrm{A}_2\mathrm{X}_3$ is $1.1 \times 10^{-23}$. If specific conductance of the solution is $3 \times 10^{-5} \, \mathrm{S} \, \mathrm{m}^{-1}$, the limiting molar conductivity of the solution is $x \times 10^{-3} \, \mathrm{S} \, \mathrm{m}^2 \, \mathrm{mol}^{-1}$. The value of $x$ is .

Answer: 3

Solution

The reaction is given by $$\mathrm{A_2X_3(s) \rightleftharpoons 2A^{+3}_{(aq)} + 3X^{-2}_{(aq)}}$$ Solubility is $s \mathrm{M}$. Therefore, $2s$ and $3s$. $$(2s)^2 (3s)^3 = 1.1 \times 10^{-23}$$ $$108 \, s^5 = 1.1 \times 10^{-23}$$ $$s \approx 10^{-5} \, \mathrm{M} = 10^{-5} \, \mathrm{mol/L} = 0.01 \, \mathrm{mol/m^3}$$ Now, $\wedge_m \approx \wedge_m^\infty = \frac{k}{m} = \frac{k}{s}$. $$\Rightarrow \wedge_m^\infty = \frac{3 \times 10^{-5}}{0.01} = 3 \times 10^{-3} \, \mathrm{S \cdot m^2/mol}$$ Ans. 3

Question 87

Chemistry · Electrochemistry · Numerical

The quantity of electricity in Faraday needed to reduce 1 mol of $\mathrm{Cr_2O_7^{2-}}$ to $\mathrm{Cr^{3+}}$ is ________.

Answer: 6

Solution

Given the reaction: $$\mathrm{Cr_2O_7^{2-} + 6e^- \rightarrow 2Cr^{3+}}$$ 1 mol of $$\mathrm{Cr_2O_7^{2-}}$$ reacts with 6 mol of electrons. Therefore, the number of faradays equals the moles of electrons, which is 6.

Question 88

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

For a first order reaction $A \rightarrow B$, the rate constant, $k = 5.5 \times 10^{-14} \, \mathrm{s}^{-1}$. The time required for 67$\%$ completion of reaction is $x \times 10^{-1}$ times the half life of reaction. The value of $x$ is _______ (Nearest integer) (Given : $\log 3 = 0.4771$)

Answer: 16

Solution

Given $$t_{67\%} = \frac{1}{k} \ln \left( \frac{1}{1 - 0.67} \right) = \frac{t_{1/2}}{\ln 2} \times \ln \left( \frac{1}{1 - \frac{2}{3}} \right)$$ We have $$t_{67\%} = \frac{t_{1/2}}{\log 2} \times \log 3 = \frac{t_{1/2} \times 0.4771}{0.301}$$ Thus, $$\Rightarrow t_{67\%} = 1.585 \times t_{1/2}$$ Given $$X \times 10^{-1} = 1.585$$ Therefore, $$\Rightarrow X = 15.85$$ The answer is 16.

Question 89

Chemistry · Co-ordination Compounds · Numerical

Number of complexes which will exhibit synergic bonding amongst, $[Cr(CO)_6]$, $[Mn(CO)_5]$ and $[Mn_2(CO)_{10}]$ is .

Answer: 3

Solution

Carbonyl complex compounds have tendency to show synergic bonding.

Question 90

Chemistry · Analytical Chemistry · Numerical

In the estimation of bromine, 0.5 $\mathrm{g}$ of an organic compound gave 0.40 $\mathrm{g}$ of silver bromide. The percentage of bromine in the given compound is ________$\%$ (nearest integer) (Relative atomic masses of Ag and Br are 108u and 80u, respectively).

Answer: 34

Solution

The reaction is given as: $$\mathrm{O.C} \rightarrow \mathrm{AgBr}$$ with masses 0.5 g and 0.4 g respectively. The moles of Br are equal to the moles of AgBr: $$mol of Br = mol of AgBr = \frac{0.4}{188}$$ The percentage of Br is calculated as: $$\% Br = \frac{\frac{0.4}{188} \times 80}{0.5} \times 100$$ This results in: $$= 34.04\%$$