JEE Main 26 August 2021 Shift 1 question paper with solutions

JEE Main 26 August 2021 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Trigonometric Functions · Single correct

The sum of solutions of the equation $\frac{\cos x}{1 + \sin x} = |\tan 2x|$, $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) - \left\{ \frac{\pi}{4}, -\frac{\pi}{4} \right\}$ is:

  1. $-\frac{11\pi}{30}$
  2. $\frac{\pi}{10}$
  3. $-\frac{7\pi}{30}$
  4. $-\frac{\pi}{15}$

Answer: (a)

Solution

$\dfrac{\cos x}{1+\sin x}=|\tan 2x|$ $\Rightarrow \dfrac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}} {\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2} =|\tan 2x|$ $\Rightarrow \tan^2\left(\frac{\pi}{4}-\frac{x}{2}\right)=\tan^2 2x$ $\Rightarrow 2x=n\pi\pm\left(\frac{\pi}{4}-\frac{x}{2}\right)$ $\Rightarrow x=-\frac{3\pi}{10},\ -\frac{\pi}{6},\ \frac{\pi}{10}$ or sum $=-\frac{11\pi}{6}$

Question 2

Maths · Statistics · Single correct

The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If $\alpha$ and $\sqrt{\beta}$ are the mean and standard deviation respectively for correct data, then $(\alpha, \beta)$ is:

  1. (11, 26)
  2. (10.5, 25)
  3. (11, 25)
  4. (10.5, 26)

Answer: (d)

Solution

Given: Mean $\left( \bar{x} \right) = \frac{\Sigma x_i}{20} = 10$ or $\Sigma x_i = 200$ (incorrect) or $200 - 25 + 35 = 210 = \Sigma x_i$ (Correct) Now correct $\bar{x} = \frac{210}{20} = 10.5$ Again given S.D = 2.5 ($\sigma$) $$\sigma^2 = \frac{\Sigma x_i^2}{20} - (10)^2 = (2.5)^2$$ or $\Sigma x_i^2 = 2125$ (incorrect) or $\Sigma x_i^2 = 2125 - 25^2 + 35^2$ $$= 2725 (Correct)$$ Therefore, correct $\sigma^2 = \frac{2725}{20} - (10.5)^2$ $$\sigma^2 = 26$$ or $\sigma = 26$ Therefore, $\alpha = 10.5, \beta = 26$

Question 3

Maths · Conic Sections · Single correct

On the ellipse $\frac{x^2}{8} + \frac{y^2}{4} = 1$ let $P$ be a point in the second quadrant such that the tangent at $P$ to the ellipse is perpendicular to the line $x + 2y = 0$. Let $S$ and $S'$ be the foci of the ellipse and $e$ be its eccentricity. If $A$ is the area of the triangle $SPS'$ then, the value of $(5 - e^2) \cdot A$ is:

  1. 6
  2. 12
  3. 14
  4. 24

Answer: (a)

Solution

Equation of tangent: $y = 2x + 6$ at $P$. Therefore, $P(-8/3, 2/3)$. $e = \frac{1}{\sqrt{2}}$. $S$ and $S'$ are $(-2, 0)$ and $(2, 0)$. Area of $\Delta SPS' = \frac{1}{2} \times 4 \times \frac{2}{3}$. $A = \frac{4}{3}$. Therefore, $(5 - e^2) A = \left(5 - \frac{1}{2}\right) \frac{4}{3} = 6$.

Question 4

Maths · Differential Equations · Single correct

Let $y = y(x)$ be a solution curve of the differential equation $(y + 1) \tan^2 x \, dx + \tan x \, dy + y \, dx = 0$, $x \in \left(0, \frac{\pi}{2}\right)$. If $\lim_{x \to 0^+} xy(x) = 1$, then the value of $y\left(\frac{\pi}{4}\right)$ is:

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

Answer: (d)

Solution

(y + 1) $\tan$^2 x $\,$ dx + $\tan$ x $\,$ dy + y $\,$ dx = 0 or $\frac{dy}{dx}$ + $\frac{\sec^2 x}{\tan x}$ $\cdot$ y = -$\tan$ x IF = e^{$\int$ $\frac{\sec^2 x}{\tan x}$ $\,$ dx} = e^{$\ln$ $\tan$ x} = $\tan$ x $\therefore$ $\,$ y $\tan$ x = - $\int$ $\tan$^2 x $\,$ dx or y $\tan$ x = -$\tan$ x + x + C or $\lim$_{x $\to$ 0} xy = -x + $\frac{x^2}{\tan x}$ + $\frac{Cx}{\tan x}$ = 1 or $\lim$_{x $\to$ 0} xy = -x + $\frac{x^2}{\tan x}$ + $\frac{Cx}{\tan x}$ = 1 or C = 1 y(x) = $\cot$ x + x $\cot$ x - 1 y $\left$( $\frac{\pi}{4}$ $\right$) = $\frac{\pi}{4}$

Question 5

Maths · Probability · Single correct

Let A and B be independent events such that P(A) = p, P(B) = 2p. The largest value of p, for which P (exactly one of A, B occurs ) = $\frac{5}{9}$, is :

  1. $\frac{1}{3}$
  2. $\frac{2}{9}$
  3. $\frac{4}{9}$
  4. $\frac{5}{12}$

Answer: (d)

Solution

P(Exactly one of $A$ or $B$) = P(A $\cap$ $\overline{B}$) + P($\overline{A}$ $\cap$ B) = $\frac{5}{9}$ = P(A)P($\overline{B}$) + P($\overline{A}$)P(B) = $\frac{5}{9}$ $\Rightarrow$ P(A)(1 - P(B)) + (1 - P(A))P(B) = $\frac{5}{9}$ $\Rightarrow$ p(1 - 2p) + (1 - p)2p = $\frac{5}{9}$ $\Rightarrow$ 36p^2 - 27p + 5 = 0 $\Rightarrow$ p = $\frac{1}{3}$ or $\frac{5}{12}$ $P_{max}$ = $\frac{5}{12}$

Question 6

Maths · Determinants · Single correct

Let $\theta \in \left(0, \frac{\pi}{2}\right)$. If the system of linear equations $$(1 + \cos^2 \theta) x + \sin^2 \theta y + 4 \sin 3\theta z = 0$$ $$\cos^2 \theta x + (1 + \sin^2 \theta) y + 4 \sin 3\theta z = 0$$ $$\cos^2 \theta x + \sin^2 \theta y + (1 + 4 \sin 3\theta) z = 0$$ has a non-trivial solution, then the value of $\theta$ is :

  1. $\frac{4\pi}{9}$
  2. $\frac{7\pi}{18}$
  3. $\frac{\pi}{18}$
  4. $\frac{5\pi}{18}$

Answer: (b)

Solution

Case-I $$\begin{vmatrix} 1 + \cos^2 \theta & \sin^2 \theta & 4 \sin 3\theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4 \sin 3\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \sin 3\theta \end{vmatrix} = 0$$ $C_1 \rightarrow C_1 + C_2$ $$\begin{vmatrix} 2 & \sin^2 \theta & 4 \sin 3\theta \\ 2 & 1 + \sin^2 \theta & 4 \sin 3\theta \\ 1 & \sin^2 \theta & 1 + 4 \sin 3\theta \end{vmatrix} = 0$$ $R_1 \rightarrow R_1 - R_2, R_2 \rightarrow R_2 - R_3$ $$\begin{vmatrix} 0 & -1 & 0 \\ 1 & 1 & -1 \\ 1 & \sin^2 \theta & 1 + 4 \sin 3\theta \end{vmatrix} = 0$$ or $4 \sin 3\theta = -2$ $$\sin 3\theta = -\frac{1}{2}$$ $$\theta = \frac{7\pi}{18}$$

Question 7

Maths · Inverse Trigonometric Functions · Single correct

Let $f(x) = \cos \left( 2 \tan^{-1} \sin \left( \cot^{-1} \sqrt{\frac{1-x}{x}} \right) \right)$, $0 < x < 1$. Then :

  1. $(1-x)^2 f'(x) - 2(f(x))^2 = 0$
  2. $(1+x)^2 f'(x) + 2(f(x))^2 = 0$
  3. $(1-x)^2 f'(x) + 2(f(x))^2 = 0$
  4. $(1+x)^2 f'(x) - 2(f(x))^2 = 0$

Answer: (c)

Solution

Given $$f(x) = \cos \left( 2 \tan^{-1} \sin \left( \cot^{-1} \sqrt{\frac{1-x}{x}} \right) \right)$$ $$\cot^{-1} \sqrt{\frac{1-x}{x}} = \sin^{-1} \sqrt{x}$$ or $$f(x) = \cos \left( 2 \tan^{-1} \sqrt{x} \right)$$ $$= \cos \tan^{-1} \left( \frac{2\sqrt{x}}{1-x} \right)$$ $$f(x) = \frac{1-x}{1+x}$$ Now $$f'(x) = \frac{-2}{(1+x)^2}$$ or $$f'(x)(1-x)^2 = -2 \left( \frac{1-x}{1+x} \right)^2$$ or $$(1-x)^2 f'(x) + 2(f(x))^2 = 0.$$

Question 8

Maths · Sequences and Series · Single correct

The sum of the series $$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{2^2}{x^4+1} + \cdots + \frac{2^{100}}{x^{2^{100}}+1}$$ when $x = 2$ is:

  1. $1 + \frac{2^{101}}{4^{101} - 1}$
  2. $1 + \frac{2^{100}}{4^{101} - 1}$
  3. $1 - \frac{2^{100}}{4^{100} - 1}$
  4. $1 - \frac{2^{101}}{4^{101} - 1}$

Answer: (d)

Solution

Given $$S = \frac{1}{x+1} + \frac{2}{x^2+1} + \frac{2^2}{x^4+1} + \ldots + \frac{2^{100}}{x^{2^{100}}+1}$$ We have $$S + \frac{1}{1-x} = \frac{1}{1-x} + \frac{1}{x+1} + \ldots = \frac{2}{1-x^2} + \frac{2}{1+x^2} + \ldots$$ Thus, $$S + \frac{1}{1-x} = \frac{2^{101}}{1-x^{2^{101}}}$$ Put $x = 2$ Then $$S = 1 - \frac{2^{101}}{2^{2^{101}}-1}$$ Not in option (BONUS)

Question 9

Maths · Binomial Theorem · Single correct

If ${}^{20}\mathrm{C}_r\,x^r$ is the coefficient of $x^r$ in the expansion of $(1+x)^{20}$, then the value of \[ \sum_{r=0}^{20} r\,{}^{20}\mathrm{C}_r \] is equal to:

  1. 420 $\times$ 2^{19}
  2. 380 $\times$ 2^{19}
  3. 380 $\times$ 2^{18}
  4. 420 $\times$ 2^{18}

Answer: (d)

Solution

Given $$\sum_{r=0}^{20} r^2 \cdot 20 \binom{20}{r}$$ $$\sum (4(r-1) + r) \cdot 20 \binom{20}{r}$$ $$\sum \left( r(r-1) \cdot \frac{20 \cdot 19}{r(r-1)} \cdot 1.8 \binom{20}{r} + r \cdot \frac{20}{r} \cdot 20 \cdot \sum_{r=1}^{19} \binom{20}{r-1} \right)$$ $$\Rightarrow 20 \times 19 \cdot 2^{18} + 20 \cdot 2^{19}$$ $$\Rightarrow 420 \times 2^{18}$$

Question 10

Maths · Sets · Single correct

Out of all the patients in a hospital 89$\%$ are found to be suffering from heart ailment and 98$\%$ are suffering from lungs infection. If K$\%$ of them are suffering from both ailments, then K can not belong to the set:

  1. {80, 83, 86, 89}
  2. {84, 86, 88, 90}
  3. {79, 81, 83, 85}
  4. {84, 87, 90, 93}

Answer: (c)

Solution

Given $n(A \cup B) \geq n(A) + n(B) - n(A \cap B)$. $$100 \geq 89 + 98 - n(A \cup B)$$ $$n(A \cup B) \geq 87$$ $$87 \leq n(A \cup B) \leq 89$$ Option (3)

Question 11

Maths · Complex Numbers and Quadratic Equations · Single correct

The equation $\arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{4}$ represents a circle with:

  1. centre at $(0, -1)$ and radius $\sqrt{2}$
  2. centre at $(0, 1)$ and radius $\sqrt{2}$
  3. centre at $(0, 0)$ and radius $\sqrt{2}$
  4. centre at $(0, 1)$ and radius $2$

Answer: (b)

Solution

In $\triangle OAC$, $\sin\left(\frac{\pi}{4}\right)=\frac{1}{AC}$ $\Rightarrow AC=\sqrt{2}$ Also, $\tan\left(\frac{\pi}{4}\right)=\frac{OA}{OC}=\frac{1}{OC}$ $\Rightarrow OC=1$ $\therefore$ Centre $=(0,1)$ Radius $=\sqrt{2}$

Question 12

Maths · Vector Algebra · Single correct

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k}$. If $\vec{c}$ is a vector such that $\vec{a}\times\vec{c}=\vec{b}$ and $\vec{a}\cdot\vec{c}=3$, then $\vec{a}\cdot(\vec{b}\times\vec{c})$ is equal to $\underline{\hspace{2cm}}$.

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

Answer: (a)

Solution

Given $|\vec{a}| = \sqrt{3}$, $\vec{a} \cdot \vec{c} = 3$, $\vec{a} \times \vec{b} = -2\hat{i} + \hat{j} + \hat{k}$, $\vec{a} \times \vec{c} = \vec{b}$. Cross with $\vec{a}$. $$\vec{a} \times (\vec{a} \times \vec{c}) = \vec{a} \times \vec{b}$$ $$\Rightarrow (\vec{a} \cdot \vec{c}) \vec{a} - a^2 \vec{c} = \vec{a} \times \vec{b}$$ $$\Rightarrow 3 \vec{a} - 3 \vec{c} = -2\hat{i} + \hat{j} + \hat{k}$$ $$\Rightarrow 3\hat{i} + 3\hat{j} + 3\hat{k} - 3\vec{c} = -2\hat{i} + \hat{j} + \hat{k}$$ $$\Rightarrow \vec{c} = \frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}$$ Therefore, $\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c} = \frac{-10}{3} + \frac{2}{3} + \frac{2}{3} = -2$

Question 13

Maths · Conic Sections · Single correct

If a line along a chord of the circle $4x^2 + 4y^2 + 120x + 675 = 0$, passes through the point $(-30, 0)$ and is tangent to the parabola $y^2 = 30x$, then the length of this chord is :

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

Answer: (d)

Solution

Equation of tangent to $y^2 = 30x$ $y = mx + \frac{30}{4m}$ Pass through $(-30, 0)$: $a = -30m + \frac{30}{4m} \implies m^2 = \frac{1}{4}$ $\implies m = \frac{1}{2}$ or $m = -\frac{1}{2}$ At $m = \frac{1}{2}$: $y = \frac{x}{2} + 15 \implies x - 2y + 30 = 0$ $P = \frac{15}{\sqrt{5}}$ $\ell_{AB} = 2\sqrt{R^2 - P^2} = 2\sqrt{\frac{225}{4} - \frac{225}{5}}$ $\implies \ell_{AB} = 30 \cdot \sqrt{\frac{1}{20}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$

Question 14

Maths · Integrals · Single correct

The value of $$\int_{-1/\sqrt{2}}^{1/\sqrt{2}} \left( \left( \frac{x+1}{x-1} \right)^2 + \left( \frac{x-1}{x+1} \right)^2 - 2 \right)^{1/2} \, dx$$ is:

  1. $\log$_e 4
  2. $\log$_e 16
  3. 2 $\log$_e 16
  4. 4 $\log$_e (3 + 2$\sqrt{2}$)

Answer: (b)

Solution

Given $$I = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \left( \left( \frac{x+1}{x-1} - \frac{x-1}{x+1} \right)^2 \right)^{1/2} \, dx$$ We have $$I = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \frac{4x}{x^2-1} \, dx \Rightarrow I = 2.4 \int_0^{1/\sqrt{2}} \left| \frac{x}{x^2-1} \right| \, dx$$ This implies $$\Rightarrow I = -4 \int_0^{1/\sqrt{2}} \frac{2x}{x^2-1} \, dx \Rightarrow I = -4 \ln \left| x^2 - 1 \right|_0^{1/\sqrt{2}}$$ Therefore, $$\Rightarrow I = 4 \ln 2 \Rightarrow I = \ln 16$$

Question 15

Maths · Three Dimensional Geometry · Single correct

A plane P contains the line $$x + 2y + 3z + 1 = 0 = x - y - z - 6,$$ and is perpendicular to the plane $$-2x + y + z + 8 = 0.$$ Then which of the following points lies on P?

  1. (-1, 1, 2)
  2. (0, 1, 1)
  3. (1, 0, 1)
  4. (2, -1, 1)

Answer: (b)

Solution

Equation of plane P can be assumed as $$P: x + 2y + 3z + 1 + \lambda (x - y - z - 6) = 0$$ $$\Rightarrow P: (1 + \lambda)x + (2 - \lambda)y + (3 - \lambda)z + 1 - 6\lambda = 0$$ $$\Rightarrow \vec{n}_1 = (1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (3 - \lambda)\hat{k}$$ Therefore, $$\vec{n}_1 \cdot \vec{n}_2 = 0$$ $$\Rightarrow 2(1 + \lambda) - (2 - \lambda) - (3 - \lambda) = 0$$ $$\Rightarrow 2 + 2\lambda - 2 + \lambda - 3 + \lambda = 0 \Rightarrow \lambda = \frac{3}{4}$$ $$\Rightarrow P: \frac{7}{4}x + \frac{5}{4}y + \frac{9}{4}z - \frac{14}{4} = 0$$ $$\Rightarrow 7x + 5y + 9z = 14$$ $$(0, 1, 1)$$ lies on P

Question 16

Maths · Matrices · Single correct

If $A = \begin{pmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ -\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ i & 1 \end{pmatrix}$, $i = \sqrt{-1}$, and $Q = A^T BA$, then the inverse of the matrix $A Q^{2021} A^T$ is equal to :

  1. $\begin{pmatrix} \frac{1}{\sqrt{5}} & -2021 \\ 2021 & \frac{1}{\sqrt{5}} \end{pmatrix}$
  2. $\begin{pmatrix} 1 & 0 \\ -2021i & 1 \end{pmatrix}$
  3. $\begin{pmatrix} 1 & 0 \\ 2021i & 1 \end{pmatrix}$
  4. $\begin{pmatrix} 1 & -2021i \\ 0 & 1 \end{pmatrix}$

Answer: (b)

Solution

Given $\mathrm{AA}^{\mathrm{T}} = \begin{pmatrix} \frac{1}{5} & \frac{2}{\sqrt{5}} \\ -\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{pmatrix}$. $\mathrm{AA}^{\mathrm{T}} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$. $Q^2 = \mathrm{A}^{\mathrm{T}} \mathrm{BAA}^{\mathrm{T}} \mathrm{BA} = \mathrm{A}^{\mathrm{T}} \mathrm{BIBA}$. Therefore, $Q^2 = \mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{A}$. $Q^3 = \mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{A} \mathrm{A}^{\mathrm{T}} \mathrm{BA} \Rightarrow Q^3 = \mathrm{A}^{\mathrm{T}} \mathrm{B}^3 \mathrm{A}$. Similarly: $Q^{2021} = \mathrm{A}^{\mathrm{T}} \mathrm{B}^{2021} \mathrm{A} \ldots (1)$. Now $\mathrm{B}^2 = \begin{pmatrix} 1 & 0 \\ i & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ i & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2i & 1 \end{pmatrix}$. $\mathrm{B}^3 = \begin{pmatrix} 1 & 0 \\ 2i & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ i & 1 \end{pmatrix} \Rightarrow \mathrm{B}^3 = \begin{pmatrix} 1 & 0 \\ 3i & 1 \end{pmatrix}$. Similarly $\mathrm{B}^{2021} = \begin{pmatrix} 1 & 0 \\ 2021i & 1 \end{pmatrix}$. Therefore, $\mathrm{AQ}^{2021} \mathrm{A}^{\mathrm{T}} = \mathrm{AA}^{\mathrm{T}} \mathrm{B}^{2021} \mathrm{AA}^{\mathrm{T}} = \mathrm{IB}^{2021} \mathrm{I}$. Thus, $\Rightarrow \mathrm{AQ}^{2021} \mathrm{A}^{\mathrm{T}} = \mathrm{B}^{2021} = \begin{pmatrix} 1 & 0 \\ 2021i & 1 \end{pmatrix}$. Therefore, $\left(\mathrm{AQ}^{2021} \mathrm{A}^{\mathrm{T}}\right)^{-1} = \begin{pmatrix} 1 & 0 \\ 2021i & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ -2021i & 1 \end{pmatrix}$.

Question 17

Maths · Sequences and Series · Single correct

If the sum of an infinite GP $a, ar, ar^2, ar^3, \ldots$ is 15 and the sum of the squares of its each term is 150, then the sum of $ar^2, ar^4, ar^6, \ldots$ is:

  1. $\frac{5}{2}$
  2. $\frac{1}{2}$
  3. $\frac{25}{2}$
  4. $\frac{9}{2}$

Answer: (b)

Solution

Sum of infinite terms: $$\frac{a}{1-r} = 15 \ldots (i)$$ Series formed by square of terms: $a^2, a^2r^2, a^2r^4, a^2r^6, \ldots$ Sum $$= \frac{a^2}{1-r^2} = 150$$ $$\Rightarrow \frac{a}{1-r} \cdot \frac{a}{1+r} = 150 \Rightarrow 15 \cdot \frac{a}{1+r} = 150$$ $$\Rightarrow \frac{a}{1+r} = 10 \ldots (ii)$$ By (i) and (ii) $a = 12; r = \frac{1}{5}$ Now series: $ar^2, ar^4, ar^6$ Sum $$= \frac{ar^2}{1-r^2} = \frac{12(1/25)}{1-1/25} = \frac{1}{2}$$

Question 18

Maths · Sequences and Series · Single correct

The value of $\lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{2n-1} \frac{n^2}{n^2 + 4r^2}$ is:

  1. $\frac{1}{2} \tan^{-1}(2)$
  2. $\frac{1}{2} \tan^{-1}(4)$
  3. $\tan^{-1}(4)$
  4. $\frac{1}{4} \tan^{-1}(4)$

Answer: (b)

Solution

Given $$L = \lim_{n \to \infty} \frac{1}{n} \cdot \sum_{r=0}^{2n-1} \frac{1}{1+4\left(\frac{r}{n}\right)^2}$$ This implies $$L = \int_0^2 \frac{1}{1+4x^2} \, dx$$ Therefore, $$L = \frac{1}{2} \tan^{-1}(2x) \bigg|_0^2 \Rightarrow L = \frac{1}{2} \tan^{-1} 4$$

Question 19

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

Let ABC be a triangle with A(-3, 1) and $\angle$ ACB = $\theta$, 0 < $\theta$ < $\frac{\pi}{2}$. If the equation of the median through B is 2x + y - 3 = 0 and the equation of angle bisector of C is 7x - 4y - 1 = 0, then $\tan$ $\theta$ is equal to:

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

Answer: (c)

Solution

Therefore, $M \left( \frac{a-3}{2}, \frac{b+1}{2} \right)$ lies on $2x + y - 3 = 0$. Thus, $2a + b = 11 \ldots (i)$. Therefore, $C$ lies on $7x - 4y = 1$. Thus, $7a - 4b = 1 \ldots (ii)$. Therefore, by (i) and (ii): $a = 3$, $b = 5$. Thus, $C(3, 5)$. Therefore, $m_{AC} = 2/3$. Also, $m_{CD} = 7/4$. Thus, $\tan \frac{\theta}{2} = \frac{\frac{2}{3} - \frac{4}{7}}{1 + \frac{2}{3} \cdot \frac{4}{7}} \Rightarrow \tan \frac{\theta}{2} = \frac{1}{2}$. Thus, $\tan \theta = \frac{2 \cdot \frac{1}{2}}{1 - \frac{1}{4}} = \frac{4}{3}$.

Question 20

Maths · Mathematical Reasoning · Single correct

If the truth value of the Boolean expression $((p \lor q) \land (q \rightarrow r) \land (\sim r)) \rightarrow (p \land q)$ is false then the truth values of the statements p, q, r respectively can be:

  1. TFT
  2. FFT
  3. TFF
  4. FTF

Answer: (c)

Solution

Question 21

Maths · Complex Numbers and Quadratic Equations · Numerical

Let $z = \frac{1-i\sqrt{3}}{2}$, $i = \sqrt{-1}$. Then the value of $$21 + \left( z + \frac{1}{z} \right)^3 + \left( z^2 + \frac{1}{z^2} \right)^3 + \left( z^3 + \frac{1}{z^3} \right)^3 + \ldots + \left( z^{21} + \frac{1}{z^{21}} \right)^3$$ is.

Answer: 13

Solution

Given $Z = \frac{1 - \sqrt{3}i}{2} = e^{-i \frac{\pi}{3}}$. $z^r + \frac{1}{z^r} = 2 \cos \left( -\frac{\pi}{3} \right) r = 2 \cos \frac{r \pi}{3}$. Therefore, $$21 + \sum_{r=1}^{21} \left( z^r + \frac{1}{z^r} \right)^3 = 8 \left( \cos^3 \frac{r \pi}{3} \right) = 2 \left( \cos r \pi + 3 \cos \frac{r \pi}{3} \right)$$ Thus, $$21 + \left( z + \frac{1}{z} \right)^3 + \left( z^2 + \frac{1}{z^2} \right)^3 + \ldots + \left( z^{21} + \frac{1}{z^{21}} \right)^3$$ This simplifies to $$21 + \sum_{r=1}^{21} \left( z^r + \frac{1}{z^r} \right)^3$$ Which equals $$21 + \sum_{r=1}^{21} \left( 2 \cos r \pi + 6 \cos \frac{r \pi}{3} \right)$$ Finally, $$21 - 2 - 6$$ The result is $$= 13$$

Question 22

Maths · Complex Numbers and Quadratic Equations · Numerical

The sum of all integral values of $k(k \neq 0)$ for which the equation $\frac{2}{x-1} - \frac{1}{x-2} = \frac{2}{k}$ in $x$ has no real roots, is .

Answer: 66

Solution

$\dfrac{2}{x-1}-\dfrac{1}{x-2}=\dfrac{2}{k}$ $x\in\mathbb{R}-\{1,2\}$ $\Rightarrow k(2x-4-x+1)=2(x^2-3x+2)$ $\Rightarrow k(x-3)=2(x^2-3x+2)$ For $x\neq3$, $k=2\left(x-3+\dfrac{2}{x-3}+3\right)$ $x-3+\dfrac{2}{x-3}\ge2\sqrt{2},\ \forall\,x>3$ $x-3+\dfrac{2}{x-3}\le-2\sqrt{2},\ \forall\,x<3$ $\Rightarrow 2\left(x-3+\dfrac{2}{x-3}+3\right)\in(-\infty,\;6-4\sqrt{2}]\cup[6+4\sqrt{2},\;\infty)$ For no real roots, $k\in(6-4\sqrt{2},\;6+4\sqrt{2})-\{0\}$ Integral $k\in\{1,2,\ldots,11\}$ Sum of $k=66$

Question 23

Maths · Three Dimensional Geometry · Numerical

Let the line L be the projection of the line \[ \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-4}{2} \] in the plane $x - 2y - z = 3$. If $d$ is the distance of the point $(0, 0, 6)$ from $L$, then $d^2$ is equal to.

Answer: 26

Solution

Given $L_1: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-4}{2}$. For the foot of the perpendicular from $(1, 3, 4)$ on $x - 2y - z - 3 = 0$, we have: $$(1 + t) - 2(3 - 2t) - (4 - t) - 3 = 0$$ which implies $$t = 2.$$ So the foot of the perpendicular is $(3, -1, 2)$. The point of intersection of $L_1$ with the plane is $(-11, -3, -8)$. The direction ratios of $L$ are $ $ which is approximately $ $. The distance $d = AB \sin \theta$ is given by: $$d = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & -4 \\ 7 & 1 & 5 \end{array} \right| \bigg/ \sqrt{7^2 + 1^2 + 5^2}$$ which simplifies to $$\Rightarrow d^2 = \frac{12^2 + (43)^2 + (10)^2}{49 + 1 + 25} = 26.$$

Question 24

Maths · Permutations and Combinations · Numerical

If $^1P_1 + 2 \cdot ^2P_2 + 3 \cdot ^3P_3 + \cdots + 15 \cdot ^{15}P_{15} = ^4P_{r-s}$, $0 \leq s \leq 1$ then $q+sC_{r-s}$ is equal to .

Answer: 136

Solution

Given $^1P_1 + 2 \cdot \, ^2P_2 + 3 \cdot \, ^3P_3 + \ldots + 15 \cdot \, ^{15}P_{15}$. This equals $1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + 15 \times 15!$. This can be expressed as $$\sum_{r=1}^{15} (r+1-1)r!$$ which simplifies to $$\sum_{r=1}^{15} (r+1)! - (r)!$$ This equals $16! - 1$. Therefore, it is equal to $^{16}P_{16} - 1$. Thus, $q = r = 16$, $s = 1$. Finally, $q+sC_{r-s} = ^{17}C_{15} = 136$.

Question 25

Maths · Applications of Derivatives · Numerical

A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then ( $\frac{4}{\pi}$ + 1 ) k is equal to .

Answer: 36

Solution

Let $x + y = 36$. $x$ is the perimeter of the square and $y$ is the perimeter of the circle. The side of the square is $x/4$. The radius of the circle is $\frac{y}{2\pi}$. Sum of Areas is given by: $$\left(\frac{x}{4}\right)^2 + \pi \left(\frac{y}{2\pi}\right)^2$$ This simplifies to: $$\frac{x^2}{16} + \frac{(36-x)^2}{4\pi}$$ For minimum Area: $$x = \frac{144}{\pi + 4}$$ Thus, the Radius $y = 36 - \frac{144}{\pi + 4}$. Therefore, $k = \frac{36\pi}{\pi + 4}$. Finally, $\left(\frac{4}{\pi} + 1\right) k = 36$.

Question 26

Maths · Applications of Integrals · Numerical

The area of the region $$S = \{(x, y) : 3x^2 \leq 4y \leq 6x + 24\}$$ is .

Answer: 27

Solution

For A and B $$3x^2 = 6x + 24 \Rightarrow x^2 - 2x - 8 = 0$$ $$\Rightarrow x = -2, 4$$ Area = $$\int_{-2}^{4} \left( \frac{3}{2}x + 6 - \frac{3}{4}x^2 \right) \, dx$$ $$= \left[ \frac{3x^2}{4} + 6x - \frac{x^3}{4} \right]_{-2}^{4} = 27$$

Question 27

Maths · Conic Sections · Numerical

The locus of a point, which moves such that the sum of squares of its distances from the points $(0,0), (1,0), (0,1), (1,1)$ is $18$ units, is a circle of diameter $d$. Then $d^2$ is equal to.

Answer: 16

Solution

Let $P(x, y)$ $$x^2 + y^2 + x^2 + (y - 1)^2 + y^2 + (x - 1)^2 + (y - 1)^2;$$ $$\Rightarrow 4 \left( x^2 + y^2 \right) - 4y - 4x = 14$$ $$\Rightarrow x^2 + y^2 - x - y - \frac{7}{2} = 0$$ $$d = 2 \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{7}{2}}$$ $$\Rightarrow d^2 = 16$$

Question 28

Maths · Continuity and Differentiability · Numerical

If $y = y(x)$ is an implicit function of $x$ such that $\log_e(x + y) = 4xy$, then $\frac{d^2y}{dx^2}$ at $x = 0$ is equal to .

Answer: 40

Solution

Given $\ln(x+y) = 4xy$ at $x = 0, y = 1$. Then $x + y = e^{4xy}$. This implies $$1 + \frac{dy}{dx} = e^{4xy} \left(4x \frac{dy}{dx} + 4y \right).$$ At $x = 0$, $$\frac{dy}{dx} = 3.$$ Now, $$\frac{d^2y}{dx^2} = e^{4xy} \left(4x \frac{dy}{dx} + 4y \right)^2 + e^{4xy} \left(4x \frac{d^2y}{dx^2} + 4y \right).$$ At $x = 0$, $$\frac{d^2y}{dx^2} = e^0 (4)^2 + e^0 (24).$$ Thus, $$\frac{d^2y}{dx^2} = 40.$$

Question 29

Maths · Permutations and Combinations · Numerical

The number of three-digit even numbers, formed by the digits 0, 1, 3, 4, 6, 7 if the repetition of digits is not allowed, is .

Answer: 52

Solution

(i) When '0' is at unit place Number of numbers = 20 (ii) When 4 or 6 are at unit place Number of numbers = 32 So number of numbers = 52

Question 30

Maths · Continuity and Differentiability · Numerical

Let a, b $\in$ $\mathbb{R}$, b $\neq$ 0, Define a function $$f(x) = \begin{cases} a \sin \frac{\pi}{2}(x-1), & for x \leq 0 \\ \frac{\tan 2x - \sin 2x}{bx^3}, & for x > 0 \end{cases}$$ If f is continuous at $x = 0$, then $10 - ab$ is equal to .

Answer: 14

Solution

Given $$f(x) = \begin{cases} \asin \frac{\pi}{2}(x-1), & x \leq 0 \\ \frac{\tan 2x - \sin 2x}{bx^3}, & x > 0 \end{cases}$$ For continuity at '0' $$\lim_{x \to 0^+} f(x) = f(0)$$ $$\Rightarrow \lim_{x \to 0^+} \frac{\tan 2x - \sin 2x}{bx^3} = -a$$ $$\Rightarrow \lim_{x \to 0^+} \frac{8x^3}{3} + \frac{8x^3}{3!}}{bx^3} = -a$$ $$\Rightarrow 8 \left( \frac{1}{3} + \frac{1}{3!} \right) = -ab$$ $$\Rightarrow 4 = -ab$$ $$\Rightarrow 10 - ab = 14$$

Physics

Question 31

Physics · Moving Charges and Magnetism · Single correct

The fractional change in the magnetic field intensity at a distance 'r' from centre on the axis of current carrying coil of radius 'a' to the magnetic field intensity at the centre of the same coil is: (Take $r < a$)

  1. $\frac{3}{2} \frac{a^2}{r^2}$
  2. $\frac{2}{3} \frac{a^2}{r^2}$
  3. $\frac{2}{3} \frac{r^2}{a^2}$
  4. $\frac{3}{2} \frac{r^2}{a^2}$

Answer: (d)

Solution

Given $B_{axis} = \frac{\mu_0 i R^2}{2 \left(R^2 + x^2\right)^{3/2}}$ and $B_{centre} = \frac{\mu_0 i}{2R}$. Therefore, $B_{centre} = \frac{\mu_0 i}{2a}$. Thus, $B_{axis} = \frac{\mu_0 i a^2}{2(a^2 + r^2)^{3/2}}$. Therefore, fractional change in magnetic field is $$\frac{\mu_0 i}{2a} - \frac{\mu_0 i a^2}{2(a^2 + r^2)^{3/2}} = 1 - \frac{1}{\left[1 + \left(\frac{r^2}{a^2}\right)\right]^{3/2}}$$ $$2\right] - \left[1 - \frac{3}{2} \frac{r^2}{a^2}\right] = \frac{3}{2} \frac{r^2}{a^2}$$ Note: $\left(1 + \frac{r^2}{a^2}\right)^{-3/2} \approx \left(1 - \frac{3}{2} \frac{r^2}{a^2}\right)$ [True only if $r << a$] Hence option (4) is the most suitable option.

Question 32

Physics · Mathematics in Physics · Single correct

The magnitude of vectors $\overrightarrow{OA}$, $\overrightarrow{OB}$ and $\overrightarrow{OC}$ in the given figure are equal. The direction of $\overrightarrow{OA} + \overrightarrow{OB} - \overrightarrow{OC}$ with x-axis will be :-

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

Answer: (a)

Solution

Let magnitude be equal to $\lambda$ $$\overrightarrow{OA} = \lambda \left[ \cos 30^\circ \hat{i} + \sin 30^\circ \hat{j} \right] = \lambda \left[ \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right]$$ $$\overrightarrow{OB} = \lambda \left[ \cos 60^\circ \hat{i} - \sin 60^\circ \hat{j} \right] = \lambda \left[ \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right]$$ $$\overrightarrow{OC} = \lambda \left[ \cos 45^\circ (-\hat{i}) + \sin 45^\circ \hat{j} \right] = \lambda \left[ -\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right]$$ Therefore, $\overrightarrow{OA} + \overrightarrow{OB} - \overrightarrow{OC}$ $$= \lambda \left[ \left( \frac{\sqrt{3} + 1}{2} + \frac{1}{\sqrt{2}} \right) \hat{i} + \left( \frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \right) \hat{j} \right]$$ Therefore, angle with $x$-axis $$\tan^{-1} \left[ \frac{\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}}{\frac{\sqrt{3} + 1}{2} + \frac{1}{\sqrt{2}}} \right] = \tan^{-1} \left[ \frac{\sqrt{2} - \sqrt{6} - 2}{\sqrt{6} + \sqrt{2} + 2} \right]$$ $$= \tan^{-1} \left[ \frac{1 - \sqrt{3} - \sqrt{2}}{\sqrt{3} + 1 + \sqrt{2}} \right]$$ Hence option (1)

Question 33

Physics · Ray Optics and Optical Instruments · Single correct

Car B overtakes another car A at a relative speed of $40 \, \mathrm{ms}^{-1}$. How fast will the image of car B appear to move in the mirror of focal length $10 \, \mathrm{cm}$ fitted in car A, when the car B is $1.9 \, \mathrm{m}$ away from the car A?

  1. $4 \, \mathrm{ms}^{-1}$
  2. $0.2 \, \mathrm{ms}^{-1}$
  3. $40 \, \mathrm{ms}^{-1}$
  4. $0.1 \, \mathrm{ms}^{-1}$

Answer: (d)

Solution

Mirror used is convex mirror (rear-view mirror). Therefore, $V_{I/m} = -m^2 V_{O/m}$. Given, $$V_{O/m} = 40 \, \mathrm{m/s}$$ $$m = \frac{f}{f-u} = \frac{10}{10+190} = \frac{10}{200}$$ Therefore, $$V_{I/m} = -\frac{1}{400} \times 40 = -0.1 \, \mathrm{m/s}$$ Therefore, the car will appear to move with speed $0.1 \, \mathrm{m/s}$. Hence option (4)

Question 34

Physics · Gravitation · Single correct

Inside a uniform spherical shell: (a) the gravitational field is zero (b) the gravitational potential is zero $(c)$ the gravitational field is same everywhere (d) the gravitation potential is same everywhere (e) all of the above Choose the most appropriate answer from the options given below:

  1. (a), $(c)$ and (d) only
  2. (e) only
  3. (a), (b) and $(c)$ only
  4. (b), $(c)$ and (d) only

Answer: (a)

Solution

Inside a spherical shell, gravitational field is zero and hence potential remains same everywhere. Hence option (1)

Question 35

Physics · Mechanical Properties of Fluids · Single correct

Two narrow bores of diameter 5.0 mm and 8.0 mm are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube. [Take surface tension of water $T = 7.3 \times 10^{-2} \, \mathrm{Nm}^{-1}$, angle of contact $= 0$, $g = 10 \, \mathrm{ms}^{-2}$ and density of water $= 1.0 \times 10^{3} \, \mathrm{kg} \, \mathrm{m}^{-3}$]

  1. 3.62 mm
  2. 2.19 mm
  3. 5.34 mm
  4. 4.97 mm

Answer: (b)

Solution

We have $P_A = P_B$. [Points A $\&$ B at same horizontal level] Therefore, $P_{atm} - \frac{2 \, T}{r_1} + \rho g (x + \Delta h) = P_{atm} - \frac{2 \, T}{r_2} + \rho g x$ Thus, $\rho g \Delta h = 2 \, T \left[ \frac{1}{r_1} - \frac{1}{r_2} \right]$ $$= 2 \times 7.3 \times 10^{-2} \left[ \frac{1}{2.5 \times 10^{-3}} - \frac{1}{4 \times 10^{-3}} \right]$$ Therefore, $\Delta h = \frac{2 \times 7.3 \times 10^{-2} \times 10^3}{10^3 \times 10} \left[ \frac{1}{2.5} - \frac{1}{4} \right]$ $$= 2.19 \times 10^{-3} \, \mathrm{m} = 2.19 \, \mathrm{mm}$$ Hence option (2)

Question 36

Physics · Thermodynamics · Single correct

An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90 W. How long it would take to increase the internal energy by $2.5 \times 10^3$ J?

  1. $2.5 \times 10^2$ s
  2. $4.1 \times 10^1$ s
  3. $2.4 \times 10^3$ s
  4. $2.5 \times 10^1$ s

Answer: (a)

Solution

Question 37

Physics · Alternating Current · Single correct

An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8 A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds:

  1. 0.4
  2. 0.8
  3. 0.125
  4. 0.2

Answer: (d)

Solution

Given $U = \frac{1}{2}Li^2 = 64$, therefore $L = 2$. $i^2 R = 640 R = \frac{640}{(8)^2} = 10$. $\tau = \frac{L}{R} = \frac{1}{5} = 0.2$. Option (4)

Question 38

Physics · Alternating Current · Single correct

A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance $R = 3 \, \mathrm{k\Omega}$ an inductor of inductive reactance $X_L = 250 \pi \Omega$ and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take $\pi^2 = 10$ )

  1. $4 \mu \mathrm{F}$
  2. $25 \mu \mathrm{F}$
  3. $400 \mu \mathrm{F}$
  4. $40 \mu \mathrm{F}$

Answer: (a)

Solution

For maximum average power, $X_L = X_C$. $$250\pi = \frac{1}{2\pi (50) C}$$ $$C = 4 \times 10^{-6}$$

Question 39

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

Identify the logic operation carried out by the given circuit :-

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

Answer: (c)

Solution

The table represents a truth table with inputs A, B and outputs X, Y, Z. Each row corresponds to a combination of inputs and their respective outputs.

Question 40

Physics · Atoms · Single correct

A particular hydrogen like ion emits radiation of frequency $2.92 \times 10^{15} \, \mathrm{Hz}$ when it makes transition from $n = 3$ to $n = 1$. The frequency in $\mathrm{Hz}$ of radiation emitted in transition from $n = 2$ to $n = 1$ will be:

  1. $0.44 \times 10^{15}$
  2. $6.57 \times 10^{15}$
  3. $4.38 \times 10^{15}$
  4. $2.46 \times 10^{15}$

Answer: (d)

Solution

Given $\($ nf_1 = k $\left$( $\frac{1}{1}$ - $\frac{1}{3^2}$ $\right$) $\)$ and $\($ nf_2 = k $\left$( 1 - $\frac{1}{2^2}$ $\right$) $\)$. The ratio $\($ $\frac{f_1}{f_2}$ = $\frac{8/9}{3/4}$ $\)$ implies $\($ f_2 = 2.46 $\times$ 10^{15} $\)$.

Question 41

Physics · Dual Nature of Radiation and Matter · Single correct

In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function $\phi = 2.5 \, \mathrm{eV}$. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. $(h = 6.63 \times 10^{-34} \, \mathrm{Js}, \, c = 3 \times 10^{8} \, \mathrm{ms^{-1}})$

  1. 1.3V
  2. 1.1V
  3. 1.9V
  4. 0.6V

Answer: (a)

Solution

Given $\($ $\mathrm{KE_{max}}$ = $eV_S$ = $\frac{hc}{\lambda}$ - $\phi$ $\)$ $\($ $\Rightarrow$ $eV_S$ = $\frac{1240}{280}$ - 2.5 = 1.93 $\mathrm{eV}$ $\)$ $\($ $\rightarrow$ $V_{S_1}$ = 1.93 $\mathrm{V}$ $\ldots$ (i) $\)$ $\($ $\rightarrow$ $eV_{S_2}$ = $\frac{1240}{400}$ - 2.5 = 0.6 $\mathrm{eV}$ $\)$ $\($ $\Rightarrow$ $V_{S_2}$ = 0.6 $\mathrm{V}$ $\ldots$ (ii) $\)$ $\($ $\Delta$ V = $V_{S_1} - V_{S_2}$ = 1.93 - 0.6 = 1.33 $\mathrm{V}$ $\)$

Question 42

Physics · Current Electricity · Single correct

In the given figure, the emf of the cell is 2.2 V and if internal resistance is 0.6 $\Omega$. Calculate the power dissipated in the whole circuit:

  1. 1.32 W
  2. 0.65 W
  3. 2.2 W
  4. 4.4 W

Answer: (c)

Solution

The equivalent resistance is calculated as follows: $$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{6} = \frac{6 + 3 + 2 + 4}{24} = \frac{15}{24}$$ Therefore, $$R_{eq} = \frac{24}{15} = 1.6 \Rightarrow R_T = 1.6 + 0.6 = 2.2 \, \Omega$$ The power is given by $$P = \frac{V^2}{R_T} = \frac{(2.2)^2}{2.2} = 2.2 \, W$$ Option (3)

Question 43

Physics · Electric Charges and Fields · Single correct

A solid metal sphere of radius $R$ having charge $q$ is enclosed inside the concentric spherical shell of inner radius $a$ and outer radius $b$ as shown in figure. The approximate variation electric field $\vec{E}$ as a function of distance $r$ from centre $O$ is given by:

Answer: (a)

Solution

Considering outer spherical shell is nonconducting. Electric field inside a metal sphere is zero. $r R \Rightarrow E = \dfrac{kQ}{r^2}$

Question 44

Physics · Kinetic Theory · Single correct

The rms speeds of the molecules of Hydrogen, Oxygen and Carbondioxide at the same temperature are $V_H$, $V_O$ and $V_C$ respectively then:

  1. $V_H > V_O > V_C$
  2. $V_C > V_O > V_H$
  3. $V_H = V_O > V_C$
  4. $V_H = V_O = V_C$

Answer: (a)

Solution

Given $$V_{RMS} = \sqrt{\frac{3RT}{M_w}}$$. At the same temperature, $$V_{RMS} \propto \frac{1}{\sqrt{M_w}}$$ which implies $$V_H > V_O > V_C$$.

Question 45

Physics · Experimental Physics · Single correct

In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the $20^{\text{th}}$ division of the circular scale coincides with reference line. Calculate the true reading.

  1. 5.00\,$\mathrm{mm}$
  2. 5.25\,$\mathrm{mm}$
  3. 5.15\,$\mathrm{mm}$
  4. 5.20\,$\mathrm{mm}$

Answer: (c)

Solution

Least count (L. C) = $\frac{0.5}{50}$ True reading = 5 + $\frac{0.5}{50}$ $\times$ 20 - $\frac{0.5}{50}$ $\times$ 5 = 5 + $\frac{0.5}{50}$ (15) = 5.15 \, $\mathrm{mm}$ Option (3)

Question 46

Physics · Current Electricity · Single correct

What equal length of an iron wire and a copper-nickel alloy wire, each of 2 $\mathrm{mm}$ diameter connected parallel to give an equivalent resistance of $3\Omega$? (Given resistivities of iron and copper-nickel alloy wire are $12\mu\Omega \mathrm{cm}$ and $51\mu\Omega \mathrm{cm}$ respectively)

  1. 82m
  2. 97m
  3. 110m
  4. 90m

Answer: (b)

Solution

$\dfrac{R_1R_2}{R_1+R_2}=3$ $\dfrac{ \left(\dfrac{(12\times10^{-6})(10^{-2})\,\ell\times4} {\pi(2)^2\times10^{-6}}\right) \left(\dfrac{(51\times10^{-6})(10^{-2})\,\ell\times4} {\pi(2)^2\times10^{-6}}\right) } { \dfrac{63\times10^{-6}\times10^{-2}\times\ell\times4} {\pi(2)^2\times10^{-6}} } =3$ $\Rightarrow \ell=97\ \mathrm{m}$

Question 47

Physics · Laws of Motion · Single correct

The initial mass of a rocket is $1000\ \mathrm{kg}$. Calculate at what rate the fuel should be burnt so that the rocket is given an acceleration of $20\ \mathrm{m\,s^{-2}}$. The gases come out at a relative speed of $500\ \mathrm{m\,s^{-1}}$ with respect to the rocket. [Use $g=10\ \mathrm{m/s^{2}}$]

  1. $6.0$ $\times$ $10^2$ $kg$ $s^{-1}$
  2. $500$ $kg$ $s^{-1}$
  3. $10$ $kg$ $s^{-1}$
  4. $60$ $kg$ $s^{-1}$

Answer: (d)

Solution

The thrust force is given by the equation: $$F_{thrust} = \left( \frac{dm}{dt} \cdot V_{rel} \right)$$ The equation of motion is: $$\left( \frac{dm}{dt} V_{rel} - mg \right) = ma$$ Substituting the given values: $$\Rightarrow \left( \frac{dm}{dt} \right) \times 500 - 10^3 \times 10 = 10^3 \times 20$$ Solving for $\frac{dm}{dt}$ gives: $$\frac{dm}{dt} = (60 \, \mathrm{kg/s})$$

Question 48

Physics · Physical World, Units and Measurements · Single correct

If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula $P = EL^2M^{-5}G^{-2}$ are :-

  1. $M^0 \, L^1 \, T^0$
  2. $M^{-1} \, L^{-1} \, T^2$
  3. $M^1 \, L^1 \, T^{-2}$
  4. $M^0 \, L^0 \, T^0$

Answer: (d)

Solution

$E = ML^2 T^{-2}$ $L = ML^2 T^{-1}$ $m = M$ $G = M^{-1} L^{+3} T^{-2}$ $P = \frac{EL^2}{M^5G^2}$ \[ [P] = \frac{(ML^2 \, T^{-2})(M^2 \, L^4 \, T^{-2})}{M^5(M^{-2} \, L^6 \, T^{-4})} = M^0 \, L^0 \, T^0 \]

Question 49

Physics · Electrostatic Potential and Capacitance · Single correct

The material filled between the plates of a parallel plate capacitor has resistivity $200 \, \Omega \mathrm{m}$. The value of capacitance of the capacitor is $2 \, \mathrm{pF}$. If a potential difference of $40 \, \mathrm{V}$ is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is: (given the value of relative permittivity of material is $50$ )

  1. $9.0 \, \mu \mathrm{A}$
  2. $9.0 \, \mathrm{mA}$
  3. $0.9 \, \mathrm{mA}$
  4. $0.9 \, \mu \mathrm{A}$

Answer: (c)

Solution

Given $\rho = 200 \, \Omega \mathrm{m}$, $C = 2 \times 10^{-12} \, \mathrm{F}$, $V = 40 \, \mathrm{V}$, $K = 56$. The current $i$ is given by $$i = \frac{q}{\rho k \varepsilon_0} = \frac{q_0}{\rho k \varepsilon_0} e^{-\frac{t}{\rho k \varepsilon_0}}.$$ The maximum current $i_{\max}$ is $$i_{\max} = \frac{2 \times 10^{-12} \times 40}{200 \times 50 \times 8.85 \times 10^{-12}}$$ $$= \frac{80}{10^4 \times 8.85} = 903 \, \mu \mathrm{A} = 0.9 \, \mathrm{mA}.$$

Question 50

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

Statement-I : By doping silicon semiconductor with pentavalent material, the electrons density increases. Statement-II : The n-type semiconductor has net negative charge. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (a)

Solution

Pentavalent activities have excess free $e^-$. So $e^-$ density increases but overall semiconductor is neutral. Option (1)

Question 51

Physics · Work, Energy and Power · Numerical

A uniform chain of length 3 meter and mass 3 kg overhangs a smooth table with 2 meter laying on the table. If k is the kinetic energy of the chain in joule as it completely slips off the table, then the value of k is .......... (Take $g = 10 \, \mathrm{m/s^2}$)

Answer: 40

Solution

From energy conservation $$K_i + U_i = k_f + U_f$$ $$0 + \left(-1 \times 10 \times \frac{1}{2}\right) = k_f + \left(-3 \times 10 \times \frac{3}{2}\right)$$ $$-5 = k_f - 45$$ $$k_f = 40 \, \mathrm{J}$$ Ans. 40.00

Question 52

Physics · Electromagnetic Waves · Numerical

The electric field in a plane electromagnetic wave is given by $$\vec{E} = 200 \cos \left[ \left( \frac{0.5 \times 10^3}{\mathrm{m}} \right) x - \left( 1.5 \times 10^{11} \frac{\mathrm{rad}}{\mathrm{s}} \right) t \right] \frac{\mathrm{V}}{\mathrm{m}} \hat{\jmath}$$ If this wave falls normally on a perfectly reflecting surface having an area of $100 \, \mathrm{cm}^2$. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is $\frac{x}{10^9} \, \frac{\mathrm{N}}{\mathrm{m}^2}$. Find the value of $x$.

Answer: 354

Solution

Given $E_0 = 200$. $I = \frac{1}{2} \varepsilon_0 E_0^2 \cdot C$ Radiation pressure $$P = \frac{2I}{C}$$ $$= \left( \frac{2}{C} \right) \left( \frac{1}{2} \varepsilon_0 E_0^2 C \right)$$ $$= \varepsilon_0 E_0^2$$ $$= 8.85 \times 10^{-12} \times 200^2$$ $$= 8.85 \times 10^{-8} \times 4$$ $$= \frac{354}{10^9}$$ Ans. 354.0

Question 53

Physics · Waves · Numerical

A source and a detector move away from each other in absence of wind with a speed of 20 $\mathrm{m/s}$ with respect to the ground. If the detector detects a frequency of 1800 $\mathrm{Hz}$ of the sound coming from the source, then the original frequency of source considering speed of sound in air 340 $\mathrm{m/s}$ will be ........ $\mathrm{Hz}$

Answer: 2025

Solution

Given $V_S = 20 \, \mathrm{m/s}$ and $V_O = 20 \, \mathrm{m/s}$. The formula for the observed frequency $f'$ is given by: $$f' = f \left( \frac{C - V_O}{C + V_S} \right)$$ Substituting the values: $$1800 = f \left( \frac{340 - 20}{340 + 20} \right)$$ Solving for $f$ gives: $$f = 2025 \, \mathrm{Hz}$$

Question 54

Physics · Motion in a Straight Line · Numerical

Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3 s with the same initial velocity of 35 m/s, then these balls collide at a height of .......... m. (Take $g = 10 \, \mathrm{m/s^2}$)

Answer: 50

Solution

When both balls will collide $y_1 = y_2$ $$35t - \frac{1}{2} \times 10 \times t^2 = 35(t - 3) - \frac{1}{2} \times 10 \times (t - 3)^2$$ $$35t - \frac{1}{2} \times 10 \times t^2 = 35t - 105 - \frac{1}{2} \times 10 \times t^2$$ $$-\frac{1}{2} \times 10 \times 3^2 + \frac{1}{2} \times 10 \times 6t$$ $$0 = 150 - 30t$$ $t = 5 sec$ Therefore, height at which both balls will collide $$h = 35t - \frac{1}{2} \times 10 \times t^2$$ $$= 35 \times 5 - \frac{1}{2} \times 10 \times 5^2$$ $h = 50 \, m$ Ans. 50.00

Question 55

Physics · Mechanical Properties of Fluids · Numerical

A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is $\ldots$ $\ldots$ $\ldots$ cm.

Answer: 2

Solution

Excess pressure inside the smaller soap bubble $$\Delta P = \frac{4S}{r_1} + \frac{4S}{r_2}$$ The excess pressure inside the equivalent soap bubble $$\Delta P = \frac{4S}{R_{eq}}$$ From (i) $\&$ (ii) $$\frac{4S}{R_{eq}} = \frac{4S}{r_1} + \frac{4S}{r_2}$$ $$\frac{1}{R_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}$$ $$= \frac{1}{6} + \frac{1}{3}$$ $$R_{eq} = 2 \, cm$$ Ans. 2.00

Question 56

Physics · Communication Systems · Numerical

An amplitude modulated wave is represented by $C_m(t) = 10(1 + 0.2 \cos 12560t) \sin(111 \times 10^4 t)$ volts. The modulating frequency in kHz will be

Answer: 2

Solution

Given $W_m = 12560 = 2 \pi f_m$. $$f_m = \frac{12560}{2\pi}$$ $$= 2000 \, \mathrm{Hz}$$

Question 57

Physics · Moving Charges and Magnetism · Numerical

Two short magnetic dipoles $m_1$ and $m_2$ each having magnetic moment of $1 \, \mathrm{Am}^2$ are placed at point $O$ and $P$ respectively. The distance between $OP$ is $1 \, \mathrm{meter}$. The torque experienced by the magnetic dipole $m_2$ due to the presence of $m_1$ is ...... $\times 10^{-7} \, \mathrm{Nm}$

Answer: 1

Solution

The torque $\vec{\tau}$ is given by the cross product of $\mathrm{M_2}$ and $\mathrm{B_1}$. Therefore, $\tau = \mathrm{M_2} \cdot \mathrm{B_1} \sin 90^\circ$. This simplifies to $$\tau = 1 \times \frac{\mu_0}{4\pi} \frac{\mathrm{M_1}}{(1)^3} \times 1$$ which equals $$10^{-7} \, \mathrm{N \cdot m}.$$

Question 58

Physics · Waves · Numerical

Two travelling waves produces a standing wave represented by equation, $$y = 1.0 \, \mathrm{mm} \cos(1.57 \, \mathrm{cm}^{-1}) \times \sin(78.5 \, \mathrm{s}^{-1}) t$$ The node closest to the origin in the region $x > 0$ will be at $x = \ldots \ldots \ldots \ldots$ cm.

Answer: 1

Solution

For node $$\cos(1.57 \, \mathrm{cm}^{-1}) x = 0$$ $$(1.57 \, \mathrm{cm}^{-1}) x = \frac{\pi}{2}$$ $$x = \frac{\pi}{2(1.57)} \, \mathrm{cm} = 1 \, \mathrm{cm}$$

Question 59

Physics · Wave Optics · Numerical

White light is passed through a double slit and interference is observed on a screen $1.5 \, \mathrm{m}$ away. The separation between the slits is $0.3 \, \mathrm{mm}$. The first violet and red fringes are formed $2.0 \, \mathrm{mm}$ and $3.5 \, \mathrm{mm}$ away from the central white fringes. The difference in wavelengths of red and violet light is ....... nm

Answer: 300

Solution

Position of bright fringe $y = n \frac{D \lambda}{d}$. $y_1$ of red $= \frac{D \lambda_r}{d} = 3.5 \, \mathrm{mm}$. $\lambda_r = 3.5 \times 10^{-3} \frac{d}{D}$. Similarly $\lambda_v = 2 \times 10^{-3} \frac{d}{D}$. $\lambda_r - \lambda_v = \left(1.5 \times 10^{-3}\right) \left(\frac{0.3 \times 10^{-3}}{1.5}\right)$ $= 3 \times 10^{-7} = 300 \, \mathrm{nm}$. Ans. 300.0

Question 60

Physics · System of Particles and Rotational Motion · Numerical

Consider a badminton racket with length scales as shown in the figure. If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $\frac{r}{2}$ distance from the end A of the handle will be $\ldots$$\ldots$$\ldots$ $Mr^2$

Answer: 52

Solution

The moment of inertia is given by $$I = \left[ I_1 + M \left( \frac{5}{2} r \right)^2 \right] + \left[ I_2 + M \left( \frac{13}{2} r \right)^2 \right]$$ This simplifies to $$= \left[ \frac{M(36r^2)}{12} + \frac{M(25r^2)}{4} \right] + \left[ \frac{Mr^2}{2} + \frac{169Mr^2}{4} \right]$$ Finally, we have $$= 52Mr^2$$ The answer is 52.00.

Chemistry

Question 61

Chemistry · Co-ordination Compounds · Single correct

Which one of the following complexes is violet in colour?

  1. $[\mathrm{Fe(CN)}_6]^{4-}$
  2. $[\mathrm{Fe(SCN)}_6]^{4-}$
  3. $\mathrm{Fe}_4[\mathrm{Fe(CN)}_6]_3 \cdot \mathrm{H}_2\mathrm{O}$
  4. $[\mathrm{Fe(CN)}_5\mathrm{NOS}]^{4}$

Answer: (d)

Solution

(1) $[\mathrm{Fe(CN)_6}]^{4-} \rightarrow$ Pale yellow solution (2) $[\mathrm{Fe(SCN)_6}]^{4-} \rightarrow$ Blood red colour (3) $\mathrm{Fe_4[Fe(CN)_6]_3 \cdot H_2O \rightarrow}$ Prussian blue (4) $[\mathrm{Fe(CN)_5NOS}]^{4-} \rightarrow$ Violet colour

Question 62

Chemistry · Surface Chemistry · Single correct

Which one of the following is correct for the adsorption of a gas at a given temperature on a solid surface?

  1. $\Delta H > 0, \Delta S > 0$
  2. $\Delta H > 0, \Delta S < 0$
  3. $\Delta H < 0, \Delta S < 0$
  4. $\Delta H 0$

Answer: (c)

Solution

(i) Adsorption of gas at metal surface is an exothermic process so $\Delta H < 0$. (ii) As the adsorption of gas on metal surface reduces the free movement of gas molecules thus restricting its randomness hence $\Delta S < 0$.

Question 63

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

Which one of the following when dissolved in water gives coloured solution in nitrogen atmosphere?

  1. CuCl_2
  2. AgCl
  3. ZnCl_2
  4. Cu_2Cl_2

Answer: (a)

Solution

(1) $\mathrm{CuCl_2 + nH_2O \rightarrow Cu^{+2}_{(aq.)}}$ blue colour (2) $\mathrm{AgCl + nH_2O \rightarrow}$ Insoluble (3) $\mathrm{ZnCl_2 + nH_2O \rightarrow Zn^{+2}_{(aq)}}$ Colourless (4) $\mathrm{Cu_2Cl_2 + nH_2O \rightarrow}$ Insoluble

Question 64

Chemistry · Haloalkanes and Haloarenes · Single correct

The major products formed in the following reaction sequence $A$ and $B$ are:

Answer: (a)

Solution

Question 65

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The major product formed in the following reaction is:

Answer: (c)

Solution

Question 66

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The major product formed in the following reaction is:

Answer: (a)

Solution

Question 67

Chemistry · Polymers · Single correct

The polymer formed on heating Novolac with formaldehyde is:

  1. Bakelite
  2. Polyester
  3. Melamine
  4. Nylon6, 6

Answer: (a)

Solution

Novolac plus formaldehyde produces Bakelite.

Question 68

Chemistry · Electrochemistry · Single correct

Given below are two statements : Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH$_3$COOH (weak electrolyte). Statement II : Molar conductivity decreases with decrease in concentration of electrolyte. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Answer: (d)

Solution

So $$\Lambda_m^{\infty \mathrm{CH_3COOH}} = \Lambda_m^{\infty(\mathrm{H^+})} + \Lambda_m^{\infty(\mathrm{CH_3COO^-})}$$ $$= 349.8 + 40.9$$ $$= 390.7 \mathrm{Scm^2/mole}$$ $$\Lambda_m^{\infty \mathrm{KCl}} = \Lambda_m^{\infty(\mathrm{K^+})} + \Lambda_m^{\infty(\mathrm{Cl^-})}$$ $$= 73.5 + 76.3$$ $$= 149.3 \mathrm{Scm^2/mole}$$ So Statement $-I$ is wrong or False. As the concentration decreases, the dilution increases which increases the degree of dissociation, thus increasing the number of ions, which increases the molar conductance. So Statement $-II$ is False.

Question 69

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The correct options for the products A and B of the following reactions are:

Answer: (b)

Solution

Question 70

Chemistry · Environmental Chemistry · Single correct

The conversion of hydroxyapatite occurs due to presence of $\mathrm{F}^{-}$ ions in water. The correct formula of hydroxyapatite is:

  1. $[3\mathrm{Ca}_3(\mathrm{PO}_4)_2 \cdot \mathrm{Ca(OH)}_2]$
  2. $[3\mathrm{Ca(OH)}_2 \cdot \mathrm{CaF}_2]$
  3. $[\mathrm{Ca}_3(\mathrm{PO}_4)_2 \cdot \mathrm{CaF}_2]$
  4. $[3\mathrm{Ca}_3(\mathrm{PO}_4)_2 \cdot \mathrm{CaF}_2]$

Answer: (a)

Solution

The $\mathrm{F}^{\theta}$ ions make the enamel on teeth much harder by converting hydroxyapatite, $[3(\mathrm{Ca}_3(\mathrm{PO}_4)) \cdot \mathrm{Ca(OH)}_2]$, the enamel on the surface of the teeth into much harder fluroappatite. $[3\mathrm{Ca}_3(\mathrm{PO}_4)_2 \cdot \mathrm{CaF}_2]$

Question 71

Chemistry · Equilibrium · Single correct

Given below are two statements. Statement I: In the titration between strong acid and weak base methyl orange is suitable as an indicator. Statement II: For titration of acetic acid with NaOH phenolphthalein is not a suitable indicator. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (b)

Solution

Question 72

Chemistry · Hydrocarbons · Single correct

Among the following compounds I - IV, which one forms a yellow precipitate on reacting sequentially with (i) $NaOH$ (ii) dil. $HNO_3$ (iii) $AgNO_3$ ?

  1. II
  2. IV
  3. I
  4. III

Answer: (b)

Solution

Other compounds' halide can't be removed because the corresponding $C^+$ is highly unstable.

Question 73

Chemistry · Hydrogen · Single correct

Which one of the following methods is most suitable for preparing deionized water?

  1. Synthetic resin method
  2. Clark's method
  3. Calgon's method
  4. Permutit method

Answer: (a)

Solution

Pure demineralised (de-ionized) water free from all soluble mineral salts is obtained by passing water successively through a cation exchange (in the H form) and an anion exchange (in the OH^- form) resins.

Question 74

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

Given below are two statements. Statement I: The choice of reducing agents for metals extraction can be made by using Ellingham diagram, a plot of $\Delta G$ vs temperature. Statement II: The value of $\Delta S$ increases from left to right in Ellingham diagram. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (d)

Solution

Given Statement-I is true as in a number of processes, one element is used to reduce the oxide of another metal. Any element will reduce the oxide of other metal which lie above it in the Ellingham diagram because the free energy change will become more negative. Given Statement-II is false as the value of $\Delta S$ is decreases from left to right in Ellingham diagram.

Question 75

Chemistry · The s-Block Elements · Single correct

What are the products formed in sequence when excess of $CO_2$ is passed in slaked lime?

  1. Ca(HCO_3)_2, CaCO_3
  2. CaCO_3, Ca(HCO_3)_2
  3. CaO, Ca(HCO_3)_2
  4. CaO, CaCO_3

Answer: (b)

Solution

The reactions are as follows: $$\mathrm{Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow + H_2O}$$ $$\mathrm{CaCO_3 \downarrow + CO_2 + H_2O \rightarrow Ca(HCO_3)_2}$$

Question 76

Chemistry · Structure of Atom · Single correct

Given below are two statements. Statement I: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus. Statement II: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (c)

Solution

Velocity of electron in Bohr's atom is given by $$V \propto \frac{Z}{n}$$ $Z$ is the atomic number of the atom, corresponds to positive charge so as $Z$ increases, velocity increases so Statement-I is wrong. And as $n$ decreases, velocity increases so Statement-II is correct.

Question 77

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The correct sequential addition of reagents in the preparation of 3-nitrobenzoic acid from benzene is:

  1. Br_2/AlBr_3, HNO_3/H_2SO_4, Mg/ether, CO_2, H_3O^+
  2. Br_2/AlBr_3, NaCN, H_3O^+, HNO_3/H_2SO_4
  3. Br_2/AlBr_3, HNO_3/H_2SO_4, NaCN, H_3O^+
  4. HNO_3/H_2SO_4, Br_2/AlBr_3, Mg/ether, CO_2, H_3O^+

Answer: (d)

Solution

The reaction sequence begins with benzene reacting with $\mathrm{HNO_3}$ and $\mathrm{H_2SO_4}$ to form nitrobenzene. Next, nitrobenzene reacts with $\mathrm{Br_2}$ and $\mathrm{AlBr_3}$ to form bromonitrobenzene. This compound then reacts with $\mathrm{Mg}$ in ether to form a Grignard reagent, which is $\mathrm{MgBr}$. Finally, the Grignard reagent reacts with $\mathrm{CO_2}$ and $\mathrm{H^+}$ to form 3-nitrobenzoic acid.

Question 78

Chemistry · The Solid State · Single correct

Given below are two statements. Statement I: Frenkel defects are vacancy as well as interstitial defects. Statement II: Frenkel defect leads to colour in ionic solids due to presence of F-centres. Choose the most appropriate answer for the statements 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: (c)

Solution

Theory based.

Question 79

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

The incorrect statement is:

  1. $Cl_2$ is more reactive than $ClF$.
  2. $F_2$ is more reactive than $ClF$.
  3. On hydrolysis $ClF$ forms $HOCl$ and $HF$.
  4. $F_2$ is a stronger oxidizing agent than $Cl_2$ in aqueous solution

Answer: (a)

Solution

(i) Reactivity order: $$\mathrm{F_2 > ClF \ (inter \ halogen) > Cl_2}$$ (ii) $$\mathrm{ClF + H_2O \rightarrow HOCl + HF}$$ (iii) Oxidizing power in aqueous solution $$\mathrm{F_2 > Cl_2 > Br_2 > I_2}$$

Question 80

Chemistry · Hydrocarbons · Single correct

Excess of isobutane on reaction with $\mathrm{Br}_2$ in presence of light at $125^\circ \mathrm{C}$ gives which one of the following, as the major product?

Answer: (d)

Solution

The reaction involves the bromination of isobutane in the presence of $\mathrm{Br_2}$ and light or heat ($hv/\Delta$). The major product is formed by the substitution of a hydrogen atom by a bromine atom at the tertiary carbon, resulting in the formation of 2-bromo-2-methylpropane.

Question 81

Chemistry · Chemical Bonding and Molecular Structure · Numerical

$\mathrm{AB_3}$ is an interhalogen T-shaped molecule. The number of lone pairs of electrons on A is _____. (Integer answer)

Answer: 2

Solution

T-shaped molecule means 3 sigma bonds and 2 lone pairs of electrons on the central atom.

Question 82

Chemistry · Electrochemistry · Numerical

These are physical properties of an element The total number of above properties that affect the reduction potential is ____ (Integer answer)

  1. Sublimation enthalpy
  2. Ionisation enthalpy
  3. Hydration enthalpy
  4. Electron gain enthalpy

Answer: (c)

Solution

Sublimation enthalpy, ionisation enthalpy and hydration enthalpy affect the reduction potential.

Question 83

Chemistry · Solutions · Numerical

Of the following four aqueous solutions, total number of those solutions whose freezing point is lower than that of 0.10 M $C_2H_5OH$ is ____ (Integer answer)

  1. 0.10 M $Ba_3(PO_4)_2$
  2. 0.10 M $Na_2SO_4$
  3. 0.10 M $KCl$
  4. 0.10 M $Li_3PO_4$

Answer: (d)

Solution

As $0.1 \, \mathrm{M} \, \mathrm{C_2H_5OH}$ is non-dissociative and rest all salt given are electrolyte so in each case effective molarity $> 0.1$ so each will have lower freezing point.

Question 84

Chemistry · Equilibrium · Numerical

The $\mathrm{OH^-}$ concentration in a mixture of $5.0\,\mathrm{mL}$ of $0.0504\,\mathrm{M}$ $\mathrm{NH_4Cl}$ and $2.0\,\mathrm{mL}$ of $0.0210\,\mathrm{M}$ $\mathrm{NH_3}$ solution is $x\times10^{-6}\,\mathrm{M}$. The value of $x$ is $\underline{\hspace{1cm}}$. (Nearest integer) Given: $K_w=1.0\times10^{-14}$, $K_b=1.8\times10^{-5}$.

Answer: 3

Solution

Given $[\mathrm{NH_4^+}] = 0.0504$ and $[\mathrm{NH_3}] = 0.0210$. So $K_b = \frac{[\mathrm{NH_4^+}][\mathrm{HO^-}]}{[\mathrm{NH_3}]}$. $$[\mathrm{HO^-}] = \frac{K_b \times [\mathrm{NH_3}]}{[\mathrm{NH_4^+}]} = 1.8 \times 10^{-5} \times \frac{2}{5} \times \frac{210}{504}$$ $$= 3 \times 10^{-6}$$

Question 85

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

The number of $4f$ electrons in the ground state electronic configuration of $\mathrm{Gd}^{+}$ is _____. [Atomic number of $\mathrm{Gd} = 64$]

Answer: 7

Solution

The electronic configuration of $^{64}\mathrm{Gd} : [\mathrm{Xe}]4f^7 5d^1 6s^2$. So the electronic configuration of $^{64}\mathrm{Gd}^{2+} : [\mathrm{Xe}]4f^5 5d^1 6s^0$. i.e. the number of 4f electrons in the ground state electronic configuration of $\mathrm{Gd}^{2+}$ is 7.

Question 86

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

The ratio of number of water molecules in Mohr's salt and potash alum is ____ $\times$ $10^{-1}$. (Integer answer)

Answer: 5

Solution

Mohr's salt: $\mathrm{(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O}$ The number of water molecules in Mohr's salt $= 6$ Potash alum: $\mathrm{KAl(SO_4)_2 \cdot 12H_2O}$ The number of water molecules in potash alum $= 12$ So ratio of number of water molecules in Mohr's salt and potash alum $= \frac{6}{12}$ $$= \frac{1}{2}$$ $$= 0.5$$ $$= 5 \times 10^{-1}$$

Question 87

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

The following data was obtained for the chemical reaction given below at $975\,\mathrm{K}$: $\mathrm{2NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(g)}$ The order of the reaction with respect to $\mathrm{NO}$ is \_\_\_\_ . [Integer answer]

Answer: 1

Solution

$7 \times 10^{-9} = K \times (8 \times 10^{-5})^{x}(8 \times 10^{-5})^{y} \ldots (1)$ $2.1 \times 10^{-8} = K \times (24 \times 10^{-5})^{x}(8 \times 10^{-5})^{y} \ldots (2)$ $\dfrac{1}{3} = \left(\dfrac{1}{3}\right)^{x} \Rightarrow x = 1$

Question 88

Chemistry · Thermodynamics · Numerical

The Born-Haber cycle for KCl is evaluated with the following data: $$\Delta_f H^\circ \text{ for KCl} = -436.7 \, \mathrm{kJ \, mol^{-1}}$$ $$\Delta_{sub} H^\circ \text{ for K} = 89.2 \, \mathrm{kJ \, mol^{-1}}$$ $$\Delta_{ion} H^\circ \text{ for K} = 419.0 \, \mathrm{kJ \, mol^{-1}}; \, \Delta_{eg} H^\circ \text{ for } \mathrm{Cl_2} = -348.6 \, \mathrm{kJ \, mol^{-1}}; \, \Delta_{diss} H^\circ \text{ for } \mathrm{Cl_2} = 243.0 \, \mathrm{kJ \, mol^{-1}}$$ The magnitude of lattice enthalpy of KCl in $\mathrm{kJ \, mol^{-1}}$ is ____. (Nearest integer)

Answer: 718

Solution

The formation enthalpy of KCl is given by the equation: $$\Delta_f H^\Theta_{KCl} = \Delta_{sub} H^\Theta_{(K)} + \Delta_{ionization} H^\Theta_{(K)} + \frac{1}{2} \Delta_{bond} H^\Theta_{(Cl_2)}$$ $$+ \Delta_{electron gain} H^\Theta_{(Cl)} + \Delta_{lattice} H^\Theta_{(KCl)}$$ Substituting the given values: $$\Rightarrow -436.7 = 89.2 + 419.0 + \frac{1}{2} (243.0) + \{-348.6\}$$ Solving for the lattice enthalpy: $$\Rightarrow \Delta_{lattice} H^\Theta_{(KCl)} = -717.8 \, kJ mol^{-1}$$ The magnitude of lattice enthalpy of KCl in kJmol$^{-1}$ is 718 (Nearest integer).

Question 89

Chemistry · Biomolecules · Numerical

The total number of negative charge in the tetrapeptide, Gly-Glu-Asp-Tyr, at pH 12.5 will be ____. (Integer answer)

Answer: 4

Solution

At pH = 12, the carboxylic acid groups and the phenolic group are deprotonated, resulting in negative charges. The total negative charge produced is 4.

Question 90

Chemistry · Solutions · Numerical

An aqueous KCl solution of density $1.20 \, \mathrm{g} \, \mathrm{mL}^{-1}$ has a molality of $3.30 \, \mathrm{mol} \, \mathrm{kg}^{-1}$. The molarity of the solution in $\mathrm{mol} \, \mathrm{L}^{-1}$ is ____. (Nearest integer) [Molar mass of KCl = $74.5$]

Answer: 3

Solution

1000 kg solvent has 3.3 moles of KCl. 1000 kg solvent $\rightarrow 3.3 \times 74.5 gm KCl$ $\rightarrow 245.85$ Weight of solution $= 1245.85 gm$ Volume of solution $= \frac{1245.85}{1.2} ml$ So molarity $= \frac{3.3 \times 1.2}{1245.85} \times 1000 = 3.17$