JEE Main 27 July 2021 Shift 2 question paper with solutions

JEE Main 27 July 2021 Shift 2: 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.

Register free to attempt this shift as a timed 180-minute test with instant scoring and chapter-wise analysis.

Maths

Question 1

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

The point P(a, b) undergoes the following three transformations successively: (a) reflection about the line $y = x$. (b) translation through 2 units along the positive direction of $x$-axis. ($c$) rotation through angle $\frac{\pi}{4}$ about the origin in the anti-clockwise direction. If the co-ordinates of the final position of the point P are $\left( -\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right)$, then the value of $2a + b$ is equal to:

  1. 13
  2. 9
  3. 5
  4. 7

Answer: (b)

Solution

Image of $A(a, b)$ along $y = x$ is $B(b, a)$. Translating it 2 units it becomes $C(b + 2, a)$. Now, applying rotation theorem $$-\frac{1}{2} + \frac{7}{\sqrt{2}} i = ((b + 2) + ai) \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$$ $$-\frac{1}{\sqrt{2}} + \frac{7}{\sqrt{2}} i = \left( \frac{b+2}{\sqrt{2}} - \frac{a}{\sqrt{2}} \right) + i \left( \frac{b+2}{\sqrt{2}} + \frac{a}{\sqrt{2}} \right)$$ $\Rightarrow b - a + 2 = -1$ and $b + 2 + a = 7$ $\Rightarrow a = 4; b = 1$ $\Rightarrow 2a + b = 9$

Question 2

Maths · Binomial Theorem · Single correct

A possible value of '$x$', for which the ninth term in the expansion of $$\left\{ 3^{\log_3 \sqrt{25^{n-1} + 7}} + 3\left(-\frac{1}{8}\right)^{\log_3 (5^{x-1} + 1)} \right\}^{10}$$ in the increasing powers of $3\left(-\frac{1}{8}\right)^{\log_3 (5^{x-1} + 1)}$ is equal to 180, is:

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

Answer: (d)

Solution

$\frac{{}^{10}C_{8}}{25^{x-1}+7}\times\left(5^{x-1}+1\right)^{-1}=180$ $\Rightarrow \frac{25^{x-1}+7}{5^{x-1}+1}=4$ Let $t=5^{x-1}$ $\Rightarrow \frac{t^{2}+7}{t+1}=4$ $\Rightarrow t^{2}-4t+3=0$ $\Rightarrow t=1,3$ $\Rightarrow 5^{x-1}=1$ (one of the possible values) $\Rightarrow x-1=0$ $\Rightarrow x=1$

Question 3

Maths · Three Dimensional Geometry · Single correct

For real numbers $\alpha$ and $\beta \neq 0$, if the point of intersection of the straight lines \[ \frac{x-\alpha}{1} = \frac{y-1}{2} = \frac{z-1}{3} and \frac{x-4}{\beta} = \frac{y-6}{3} = \frac{z-7}{3}, \] lies on the plane $x + 2y - z = 8$, then $\alpha - \beta$ is equal to:

  1. 5
  2. 9
  3. 3
  4. 7

Answer: (d)

Solution

First line is $(\phi+\alpha,\;2\phi+1,\;3\phi+1)$ and second line is $(q\beta+4,\;3q+6,\;3q+7)$ For intersection $\phi+\alpha=q\beta+4 \qquad \ldots (i)$ $2\phi+1=3q+6 \qquad \ldots (ii)$ $3\phi+1=3q+7 \qquad \ldots (iii)$ For (ii) and (iii), $\phi=1,\;q=-1$ So, from (i) $\alpha+\beta=3$ Now, point of intersection is $(\alpha+1,\;3,\;4)$ It lies on the plane. Hence, $\alpha=5 \ \&\ \beta=-2$

Question 4

Maths · Trigonometric Functions · Single correct

Let $f : \mathbb{R} \to \mathbb{R}$ be defined as $$f(x+y) + f(x-y) = 2f(x)f(y), f\left(\frac{1}{2}\right) = -1.$$ Then the value of $$\sum_{k=1}^{20} \frac{1}{\sin(k) \sin(k+f(k))}$$ is equal to:

  1. $\text{cosec}^2(21) \cos(20) \cos(2)$
  2. $\sec^2(1) \sec(21) \cos(20)$
  3. $\text{cosec}^2(1) \text{cosec}(21) \sin(20)$
  4. $\sec^2(21) \sin(20) \sin(2)$

Answer: (c)

Solution

Given $f(x) = \cos \lambda x$. Therefore, $f\left(\frac{1}{2}\right) = -1$. So, $-1 = \cos \frac{\lambda}{2}$. This implies $\lambda = 2\pi$. Thus $f(x) = \cos 2\pi x$. Now $k$ is a natural number. Thus $f(k) = 1$. $$\sum_{k=1}^{20} \frac{1}{\sin k \sin(k+1)} = \frac{1}{\sin 1} \sum_{k=1}^{20} \left[ \frac{\sin((k+1)-k)}{\sin k \cdot \sin(k+1)} \right]$$ $$= \frac{1}{\sin 1} \sum_{k=1}^{20} (\cot k - \cot(k+1))$$ $$= \frac{\cot 1 - \cot 21}{\sin 1} = \cosec^2 1 \cdot \cosec(21) \cdot \sin 20$$

Question 5

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $\mathbb{C}$ be the set of all complex numbers. Let $$S_1 = \{ z \in \mathbb{C} : |z - 2| \leq 1 \}$$ and $$S_2 = \{ z \in \mathbb{C} : z(1+i) + \overline{z}(1-i) \geq 4 \}.$$ Then, the maximum value of $$\left| z - \frac{5}{2} \right|^2$$ for $$z \in S_1 \cap S_2$$ is equal to:

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

Answer: (d)

Solution

Given $|t - 2| \leq 1$. Put $t = x + iy$. $$(x - 2)^2 + y^2 \leq 1$$ Also, $t(1 + i) + \bar{t}(1 - i) \geq 4$ gives $x - y \geq 2$. Let point on circle be $A(2 + \cos \theta, \sin \theta)$. $$\theta \in \left[ -\frac{3\pi}{4}, \frac{\pi}{4} \right]$$ $$(AP)^2 = \left(2 + \cos \theta - \frac{5}{2}\right)^2 + \sin^2 \theta$$ $$= \cos^2 \theta - \cos \theta + \frac{1}{4} + \sin^2 \theta$$ $$= \frac{5}{4} - \cos \theta$$ For $(AP)^2$ maximum $\theta = -\frac{3\pi}{4}$. $$(AP)^2 = \frac{5}{4} + \frac{1}{\sqrt{2}} = \frac{5\sqrt{2} + 4}{4\sqrt{2}}$$

Question 6

Maths · Probability · Single correct

A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability. The smallest value of n, so that the probability of guessing at least 'n' correct answers is less than $\frac{1}{2}$, is

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

Answer: (a)

Solution

$P(E) 128$ $\Rightarrow n-1\geq4$ $\Rightarrow n\geq5$

Question 7

Maths · Trigonometric Functions · Single correct

If $\tan\left(\frac{\pi}{9}\right)$, $x$, $\tan\left(\frac{7\pi}{18}\right)$ are in arithmetic progression and $\tan\left(\frac{\pi}{9}\right)$, $y$, $\tan\left(\frac{5\pi}{18}\right)$ are also in arithmetic progression, then $|x - 2y|$ is equal to :

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

Answer: (c)

Solution

Given $x = \frac{1}{2} \left( \tan \frac{\pi}{9} + \tan \frac{7\pi}{18} \right)$ and $2y = \tan \frac{\pi}{9} + \tan \frac{5\pi}{18}$. So, $x - 2y = \frac{1}{2} \left( \tan \frac{\pi}{9} + \tan \frac{7\pi}{18} \right) - \left( \tan \frac{\pi}{9} + \tan \frac{5\pi}{18} \right)$. Therefore, $$|x - 2y| = \left| \frac{\cot \frac{\pi}{9} - \tan \frac{\pi}{9}}{2} - \tan \frac{5\pi}{18} \right|$$ $$= \left| \cot \frac{2\pi}{9} - \cot \frac{2\pi}{9} \right| = 0$$ $$\left( as \tan \frac{5\pi}{18} = \cot \frac{2\pi}{9} ; \tan \frac{7\pi}{18} = \cot \frac{\pi}{9} \right)$$

Question 8

Maths · Statistics · Single correct

Let the mean and variance of the frequency distribution \begin{tabular}{|l|l|l|l|l|} \hline x : & $x_1$ = 2 & $x_2$ = 6 & $x_3$ = 8 & $x_4$ = 9 \\ \hline f : & 4 & 4 & $\alpha$ & $\beta$ \\ \hline \end{tabular} be 6 and 6.8 respectively. If $x_3$ is changed from 8 to 7, then the mean for the new data will be:

  1. 4
  2. 5
  3. $\frac{17}{3}$
  4. $\frac{16}{3}$

Answer: (c)

Solution

Given $32 + 8\alpha + 9\beta = (8 + \alpha + \beta) \times 6$ which implies $2\alpha + 3\beta = 16$. Also, $4 \times 16 + 4 \times \alpha + 9\beta = (8 + \alpha + \beta) \times 6.8$ which implies $640 + 40\alpha + 90\beta = 544 + 68\alpha + 68\beta$. This simplifies to $28\alpha - 22\beta = 96$ and further to $14\alpha - 11\beta = 48$. From (i) and (ii), $\alpha = 5$ and $\beta = 2$. So, the new mean is $\frac{32 + 35 + 18}{15} = \frac{85}{15} = \frac{17}{3}$.

Question 9

Maths · Applications of Integrals · Single correct

The area of the region bounded by $y - x = 2$ and $x^2 = y$ is equal to :-

  1. $\frac{16}{3}$
  2. $\frac{2}{3}$
  3. $\frac{9}{2}$
  4. $\frac{4}{3}$

Answer: (c)

Solution

Given $y - x = 2$, $x^2 = y$. Now, $x^2 = 2 + x$. This implies $x^2 - x - 2 = 0$. Therefore, $(x + 1)(x - 2) = 0$. The area is given by $$\int_{-1}^{2} (2 + x - x^2) \, dx$$ which equals $$\left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^{2}$$. This simplifies to $$\left( 4 + 2 - \frac{8}{3} \right) - \left( -2 + \frac{1}{2} + \frac{1}{3} \right)$$ which equals $$6 - 3 + 2 - \frac{1}{2} = \frac{9}{2}$$.

Question 10

Maths · Differential Equations · Single correct

Let $y = y(x)$ be the solution of the differential equation $(x - x^3) \, dy = (y + yx^2 - 3x^4) \, dx, \, x > 2$ If $y(3) = 3$, then $y(4)$ is equal to:

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

Answer: (b)

Solution

(x - x^3) $\,$ dy = (y + yx^2 - 3x^4) $\,$ dx $\Rightarrow$ x $\,$ dy - y $\,$ dx = (yx^2 - 3x^4) $\,$ dx + x^3 $\,$ dy $\Rightarrow$ $\frac{x \, dy - y \, dx}{x^2}$ = (y $\,$ dx + x $\,$ dy) - 3x^2 $\,$ dx $\Rightarrow$ $\mathrm{d}$ $\left$( $\frac{y}{x}$ $\right$) = $\mathrm{d}$(xy) - $\mathrm{d}$(x^3) Integrate $\Rightarrow$ $\frac{y}{x}$ = xy - x^3 + c given f(3) = 3 $\Rightarrow$ $\frac{3}{3}$ = 3 $\times$ 3 - 3^3 + c $\Rightarrow$ c = 19 $\therefore$ $\frac{y}{x}$ = xy - x^3 + 19 at x = 4, $\frac{y}{4}$ = 4y - 64 + 19 15y = 4 $\times$ 45 $\Rightarrow$ y = 12

Question 11

Maths · Limits and Derivatives · Single correct

The value of $\lim_{x \to 0} \left( \frac{x}{\sqrt[3]{1-\sin x} - \sqrt[3]{1+\sin x}} \right)$ is equal to:

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

Answer: (c)

Solution

\[ \begin{aligned} \lim_{x\to 0} \left( \frac{x} {\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}} \right) &= \lim_{x\to 0} \left( \frac{x} {\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}} \right) \\[6pt] &\times \left( \frac{\sqrt[8]{1-\sin x}+\sqrt[8]{1+\sin x}} {\sqrt[4]{1-\sin x}-\sqrt[4]{1+\sin x}} \right) \\[6pt] &\times \left( \frac{\sqrt[4]{1-\sin x}+\sqrt[4]{1+\sin x}} {\sqrt{1-\sin x}-\sqrt{1+\sin x}} \right) \\[6pt] &\times \left( \frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}} {(1-\sin x)-(1+\sin x)} \right) \\[10pt] &= \lim_{x\to 0} \frac{x} {-2\sin x} \\[6pt] &\times \Bigl(\sqrt[8]{1-\sin x}+\sqrt[8]{1+\sin x}\Bigr) \\[6pt] &\times \Bigl(\sqrt[4]{1-\sin x}+\sqrt[4]{1+\sin x}\Bigr) \\[6pt] &\times \Bigl(\sqrt{1-\sin x}+\sqrt{1+\sin x}\Bigr) \\[10pt] &= \left(-\frac12\right)(2)(2)(2) \left\{ \lim_{x\to 0}\frac{\sin x}{x} \right\} \\[6pt] &= -4. \end{aligned} \]

Question 12

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

Two sides of a parallelogram are along the lines $4x + 5y = 0$ and $7x + 2y = 0$. If the equation of one of the diagonals of the parallelogram is $11x + 7y = 9$, then other diagonal passes through the point:

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

Answer: (b)

Solution

Both the lines pass through origin. Point D is equal to the intersection of $4x + 5y = 0$ and $11x + 7y = 9$. So, coordinates of point D are $\left( \frac{5}{3}, -\frac{4}{3} \right)$. Also, point B is the point of intersection of $7x + 2y = 0$ and $11x + 7y = 9$. So, coordinates of point B are $\left( -\frac{2}{3}, \frac{7}{3} \right)$. Diagonals of the parallelogram intersect at the middle. Let the middle point of B, D be: $$\left( \frac{\frac{5}{3} - \frac{2}{3}}{2}, \frac{-\frac{4}{3} + \frac{7}{3}}{2} \right) = \left( \frac{1}{2}, \frac{1}{2} \right)$$ Equation of diagonal AC: $$\Rightarrow (y - 0) = \frac{\frac{1}{a} - 0}{\frac{1}{a} - 0}(\pi - 0)$$ $$y = x$$ Diagonal AC passes through $(2, 2)$.

Question 13

Maths · Relations and Functions · Single correct

Let $\alpha = \max_{x \in \mathbb{R}} \left\{ 8^{2 \sin 3x} \cdot 4^{4 \cos 3x} \right\}$ and $\beta = \min_{x \in \mathbb{R}} \left\{ 8^{2 \sin 3x} \cdot 4^{4 \cos 3x} \right\}$. If $8x^2 + bx + c = 0$ is a quadratic equation whose roots are $\alpha^{1/5}$ and $\beta^{1/5}$, then the value of $c - b$ is equal to:

  1. 42
  2. 47
  3. 43
  4. 50

Answer: (a)

Solution

Given ($\alpha=\max\left\{8^{2\sin3x}\cdot4^{4\cos3x}\right\}$\) \[ =\max\left\{2^{6\sin3x}\cdot2^{8\cos3x}\right\} \] \[ =\max\left\{2^{6\sin3x+8\cos3x}\right\} \] and \[ \beta=\min\left\{8^{2\sin3x}\cdot4^{4\cos3x}\right\} =\min\left\{2^{6\sin3x+8\cos3x}\right\} \] Now range of $\(6\sin3x+8\cos3x\)$ \[ \left[-\sqrt{6^2+8^2},\sqrt{6^2+8^2}\right] =[-10,10] \] \[ \alpha=2^{10},\qquad \beta=2^{-10} \] So, \[ \alpha^{1/5}=2^2=4 \] \[ \Rightarrow \beta^{1/5}=2^{-2}=\frac14 \] Quadratic $\(8x^2+bx+c=0\)$, \[ c-b=8\Big[(\text{product of roots})+(\text{sum of roots})\Big] \] \[ =8\left[4\cdot\frac14+4+\frac14\right] =8\left[\frac{21}{4}\right] =42 \]

Question 14

Maths · Continuity and Differentiability · Single correct

Let $f : [0, \infty) \to [0, 3]$ be a function defined by $f(x) = \begin{cases} \max\{\sin t : 0 \leq t \leq x\}, & 0 \leq x \leq \pi \\ 2 + \cos x, & x > \pi \end{cases}$ Then which of the following is true?

  1. $f$ is continuous everywhere but not differentiable exactly at one point in $(0, \infty)$
  2. $f$ is differentiable everywhere in $(0, \infty)$
  3. $f$ is not continuous exactly at two points in $(0, \infty)$
  4. $f$ is continuous everywhere but not differentiable exactly at two points in $(0, \infty)$

Answer: (b)

Solution

Graph of $\max \{ \sin t : 0 \leq t \leq x \}$ in $x \in [0, \pi]$. & graph of $\cos x$ for $x \in [\pi, \infty)$. So graph of $$f(x) = \begin{cases} \max \{ \sin t : 0 \leq t \leq x \}, & 0 \leq x \leq \pi \\ 2 + \cos x, & x > h \end{cases}$$ $f(x)$ is differentiable everywhere in $(0, \infty)$.

Question 15

Maths · Sets · Single correct

Let $\mathbb{N}$ be the set of natural numbers and a relation R on $\mathbb{N}$ be defined by $$R = \{(x,y) \in \mathbb{N} \times \mathbb{N} : x^3 - 3x^2y - xy^2 + 3y^3 = 0\}$$ Then the relation R is :

  1. symmetric but neither reflexive nor transitive
  2. reflexive but neither symmetric nor transitive
  3. reflexive and symmetric, but not transitive
  4. an equivalence relation

Answer: (b)

Solution

Given the equation $$x^3 - 3x^2y - xy^2 + 3y^3 = 0$$ we have: $$\Rightarrow x \left(x^2 - y^2\right) - 3y \left(x^2 - y^2\right) = 0$$ $$\Rightarrow (x - 3y)(x - y)(x + y) = 0$$ Now, $x = y$ for all $(x, y) \in \mathbb{N} \times \mathbb{N}$ so reflexive. But not symmetric and transitive. See, $(3, 1)$ satisfies but $(1, 3)$ does not. Also $(3, 1)$ and $(1, -1)$ satisfies but $(3, -1)$ does not.

Question 16

Maths · Mathematical Reasoning · Single correct

Which of the following is the negation of the statement "for all $M > 0$, there exists $x \in S$ such that $x \geq M$"?

  1. there exists $M > 0$, such that $x < M$ for all $x \in S$
  2. there exists $M > 0$, there exists $x \in S$ such that $x \geq M$
  3. there exists $M > 0$, there exists $x \in S$ such that $x < M$
  4. there exists $M > 0$, such that $x \geq M$ for all $x \in S$

Answer: (a)

Solution

P: for all $M > 0$, there exists $x \in S$ such that $x \geq M$. Not P: there exists $M > 0$, for all $x \in S$ such that $x < m$. Negation of 'there exists' is 'for all'.

Question 17

Maths · Conic Sections · Single correct

Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept $6\sqrt{5}$ on the x-axis. Then the radius of the circle C is equal to:

  1. $\sqrt{53}$
  2. 9
  3. 8
  4. $\sqrt{82}$

Answer: (b)

Solution

The radius $r$ is calculated using the Pythagorean theorem. $$r = \sqrt{6^2 + (3\sqrt{5})^2}$$ $$= \sqrt{36 + 45} = 9$$

Question 18

Maths · Vector Algebra · Single correct

Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a} = \vec{b} \times (\vec{b} \times \vec{c})$. If magnitudes of the vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ are $\sqrt{2}$, $1$ and $2$ respectively and the angle between $\vec{b}$ and $\vec{c}$ is $\theta \left(0 < \theta < \frac{\pi}{2}\right)$, then the value of $1 + \tan \theta$ is equal to:

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

Answer: (b)

Solution

Given $\vec{a} = (\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{c}$. This simplifies to $1.2 \cos \theta \vec{b} - \vec{c}$. Therefore, $\vec{a} = 2 \cos \theta \vec{b} - \vec{c}$. The magnitude $|\vec{a}|^2 = (2 \cos \theta)^2 + 2^2 - 2 \cdot 2 \cos \theta \cdot \vec{b} \cdot \vec{c}$. This implies $2 = 4 \cos^2 \theta + 4 - 4 \cos \theta \cdot 2 \cos \theta$. Thus, $-2 = -4 \cos^2 \theta$. Solving gives $\cos^2 \theta = \frac{1}{2}$. Therefore, $\sec^2 \theta = 2$. This implies $\tan^2 \theta = 1$. Thus, $\theta = \frac{\pi}{4}$. Finally, $1 + \tan \theta = 2$.

Question 19

Maths · Matrices · Single correct

Let A and B be two $3 \times 3$ real matrices such that $$\left( A^2 - B^2 \right)$$ is invertible matrix. If $A^5 = B^5$ and $A^3 B^2 = A^2 B^3$, then the value of the determinant of the matrix $A^3 + B^3$ is equal to:

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

Answer: (d)

Solution

Given $C = A^2 - B^2$; $|C| \neq 0$. $A^5 = B^5$ and $A^3 B^2 = A^2 B^2$. Now, $A^5 - A^3 B^2 = B^5 - A^2 B^3$. $$\Rightarrow A^3 \left( A^2 - B^2 \right) + B^3 \left( A^2 - B^2 \right) = 0$$ $$\Rightarrow \left( A^3 + B^3 \right) \left( A^2 - B^2 \right) = 0$$ Post multiplying inverse of $A^2 - B^2$: $$A^3 + B^3 = 0$$

Question 20

Maths · Applications of Derivatives · Single correct

Let $f : (a, b) \to \mathbb{R}$ be twice differentiable function such that $f(x) = \int_a^x g(t)\,dt$ for a differentiable function $g(x)$. If $f(x) = 0$ has exactly five distinct roots in $(a, b)$, then $g(x)g'(x) = 0$ has at least :

  1. twelve roots in (a, b)
  2. five roots in (a, b)
  3. seven roots in (a, b)
  4. three roots in (a, b)

Answer: (c)

Solution

Given $f(x) = \int_a^x g(t) \, dt$. As $f(x) \to 5$, $f'(x) \to 4$, $g(x) \to 4$, and $g'(x) \to 3$.

Question 21

Maths · Vector Algebra · Numerical

Let $\vec{a}=\hat{i}-\alpha\hat{j}+\beta\hat{k}$, $\vec{b}=3\hat{i}+\beta\hat{j}-\alpha\hat{k}$, and $\vec{c}=-\alpha\hat{i}-2\hat{j}+\hat{k}$, where $\alpha$ and $\beta$ are integers. If $\vec{a}\cdot\vec{b}=-1$ and $\vec{b}\cdot\vec{c}=10$, then $(\vec{a}\times\vec{b})\cdot\vec{c}$ is equal to $\underline{\hspace{2cm}}$.

Answer: 9

Solution

Given $\vec{a} = (1, -\alpha, \beta)$, $\vec{b} = (3, \beta, -\alpha)$, $\vec{c} = (-\alpha, -2, 1)$; $\alpha, \beta \in I$. $\vec{a} \cdot \vec{b} = -1 \implies 3 - \alpha \beta - \alpha \beta = -1$ $\implies \alpha \beta = 2$ $\vec{b} \cdot \vec{c} = 10$ $\implies -3\alpha - 2\beta - \alpha = 10$ $\implies 2\alpha + \beta + 5 = 0$ Therefore, $\alpha = -2; \beta = -1$. $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & -2 & 1 \end{vmatrix}$ $= 1(-1 + 4) - 2(3 - 4) - 1(-6 + 2)$ $= 3 + 2 + 4 = 9$

Question 22

Maths · Three Dimensional Geometry · Numerical

The distance of the point $P(3, 4, 4)$ from the point of intersection of the line joining the points $Q(3, -4, -5)$ and $R(2, -3, 1)$ and the plane $2x + y + z = 7$, is equal to

Answer: 7

Solution

$\overrightarrow{QR}:\ \frac{x-3}{1}=\frac{y+4}{-1}=\frac{z+5}{-6}=r$ $\Rightarrow (x,y,z)=(r+3,\,-r-4,\,-6r-5)$ Now, satisfying it in the given plane. We get $r=-2$. So, required point of intersection is $T(1,\,-2,\,7)$. Hence, $PT=7$.

Question 23

Maths · Complex Numbers and Quadratic Equations · Numerical

If the real part of the complex number $z = \frac{3 + 2i \cos \theta}{1 - 3i \cos \theta}, \theta \in \left(0, \frac{\pi}{2}\right)$ is zero, then the value of $\sin^2 3\theta + \cos^2 \theta$ is equal to

Answer: 1

Solution

Given $\mathrm{Re}(z)=\frac{3-6\cos^2\theta}{1+9\cos^2\theta}=0$ $\Rightarrow 3-6\cos^2\theta=0$ $\Rightarrow \cos^2\theta=\frac{1}{2}$ $\Rightarrow \theta=\frac{\pi}{4}$ Hence, $\sin^2(3\theta)+\cos^2\theta=1$

Question 24

Maths · Conic Sections · Numerical

Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3, -4)$, one focus at $(4, -4)$ and one vertex at $(5, -4)$. If $mx - y = 4$, $m > 0$ is a tangent to the ellipse E, then the value of $5 \, m^2$ is equal to

Answer: 3

Solution

Given $C(3, -4)$, $S(4, -4)$ and $A(5, -4)$. Hence, $a = 2$ and $ae = 1$. Therefore, $e = \frac{1}{2}$ which implies $b^2 = 3$. So, $E : \frac{(x-3)^2}{4} + \frac{(y+4)^2}{3} = 1$. Intersecting with given tangent. $$\frac{x^2 - 6x + 9}{4} + \frac{m^2 x^2}{3} = 1$$ Now, $D = 0$ (as it is tangent). So, $5m^2 = 3$.

Question 25

Maths · Integrals · Numerical

If $\int_{0}^{\pi} (\sin^3 x) e^{-\sin^2 x} dx = \alpha - \frac{\beta}{e} \int_{0}^{1} \sqrt{t e^t} dt$, then $\alpha + \beta$ is equal to

Answer: 5

Solution

I = 2 $\int$_0^{$\pi$/2} $\sin$^3 x e^{-$\sin$^2 x} dx = 2 $\int$_0^{$\pi$/2} $\sin$ x e^{-$\sin$^2 x} dx + $\int$_0^{$\pi$/2} $\cos$ x $\,$ e^{-$\sin$^2 x} (-$\sin$ 2x) dx = 2 $\int$_0^{$\pi$/2} $\sin$ x e^{-$\sin$^2 x} dx + $\left$[ $\cos$ x e^{-$\sin$^2 x} $\right$]_0^{$\pi$/2} + $\int$_0^{$\pi$/2} $\sin$ x e^{-$\sin$^2 x} dx = 3 $\int$_0^{$\pi$/2} $\sin$ x e^{-$\sin$^2 x} dx - 1 = $\frac{3}{2}$ $\int$_{-1}^{0} $\frac{e^{\alpha} d\alpha}{\sqrt{1+\alpha}}$ - 1 (Put -$\sin$^2 x = t) = $\frac{3}{2e}$ $\int$_0^1 $\frac{e^x}{\sqrt{x}}$ dx - 1 (put 1 + $\alpha$ = x) = $\frac{3}{2e}$ $\int$_0^1 e^u $\frac{1}{\sqrt{u}}$ dx - 1 = 2 - $\frac{3}{e}$ $\int$_0^1 e^x $\sqrt{x}$ dx Hence, $\alpha$ + $\beta$ = 5

Question 26

Maths · Complex Numbers and Quadratic Equations · Numerical

The number of real roots of the equation $$e^{4x} - e^{3x} - 4e^{2x} - e^{x} + 1 = 0$$ is equal to

Answer: 2

Solution

Given $t^4 - t^3 - 4t^2 - t + 1 = 0$, $e^x = t > 0$. $$\Rightarrow t^2 - t - 4 - \frac{1}{t} + \frac{1}{t^2} = 0$$ $$\Rightarrow \alpha^2 - \alpha - 6 = 0, \alpha = t + \frac{1}{t} \geq 2$$ $$\Rightarrow \alpha = 3, -2 (reject)$$ $$\Rightarrow t + \frac{1}{t} = 3$$ Therefore, the number of real roots $= 2$.

Question 27

Maths · Differential Equations · Numerical

Let y = y(x) be the solution of the differential equation dy = e^{(x+y)} dx; $\alpha\in\mathbb{N}$. If y($\log_e$ 2) = $\log_e$ 2 and y(0) = $\log_e$ $( \frac{1}{2})$, then the value of $\alpha$ is equal to

Answer: 2

Solution

Given $$\int e^{-y} \, dy = \int e^{ax} \, dx$$ $$\Rightarrow e^{-y} = \frac{e^{ax}}{\alpha} + c$$ Put $$(x, y) = (\ln 2, \ln 2)$$ $$-\frac{1}{2} = \frac{2^\alpha}{\alpha} + C$$ Put $$(x, y) \equiv (0, -\ln 2)$$ in (i) $$-2 = \frac{1}{\alpha} + C$$ (ii) - (iii) $$\frac{2^\alpha - 1}{\alpha} = \frac{3}{2}$$ $$\Rightarrow \alpha = 2 (as \alpha \in \mathbb{N})$$

Question 28

Maths · Permutations and Combinations · Numerical

Let n be a non-negative integer. Then the number of divisors of the form "4n + 1" of the number $$(10)^{10} \cdot (11)^{11} \cdot (13)^{13}$$ is equal to

Answer: 924

Solution

Given $N = 2^{10} \times 5^{10} \times 11^{11} \times 13^{13}$. Now, power of 2 must be zero, power of 5 can be anything, power of 13 can be anything. But, power of 11 should be even. So, required number of divisors is $$1 \times 11 \times 14 \times 6 = 924$$

Question 29

Maths · Sets · Numerical

Let $A = \{ n \in \mathbb{N} \mid n^2 \leq n + 10,000 \}$, $B = \{ 3k + 1 \mid k \in \mathbb{N} \}$ and $C = \{ 2k \mid k \in \mathbb{N} \}$, then the sum of all the elements of the set $A \cap (B - C)$ is equal to

Answer: 832

Solution

B - C $\equiv$ $\{$7, 13, 19, $\ldots$, 97, $\ldots$$\}$ Now, $n^2 - n \leq 100 \times 100$ $\($$\Rightarrow$ n(n - 1) $\leq$ 100 $\times$ 100$\)$ $\($$\Rightarrow$ A = $\{$1, 2, $\ldots$, 100$\}$$\)$ So, $\($A $\cap$ (B - C) = $\{$7, 13, 19, $\ldots$, 97$\}$$\)$ Hence, sum = $\($$\frac{16}{2}$(7 + 97) = 832$\)$

Question 30

Maths · Matrices · Numerical

If $A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ and $M = A + A^2 + A^3 + \ldots + A^{20}$ then the sum of all the elements of the matrix $M$ is equal to

Answer: 2020

Solution

Given $$A^n = \begin{bmatrix} 1 & n & \frac{n^2+n}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{bmatrix}$$ So, required sum $$= 20 \times 3 + 2 \times \left( \frac{20 \times 21}{2} \right) + \sum_{r=1}^{20} \left( \frac{r^2+r}{2} \right)$$ $$= 60 + 420 + 105 + 35 \times 41 = 2020$$

Physics

Question 31

Physics · Dual Nature of Radiation and Matter · Single correct

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy $3 \, \mathrm{eV}$. The proton captures the electrons and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength $4000 \, \mathrm{\mathring{A}}$. What is the maximum kinetic energy of the emitted photoelectron?

  1. $7.61 \, \mathrm{eV}$
  2. $1.41 \, \mathrm{eV}$
  3. $3.3 \, \mathrm{eV}$
  4. No photoelectron would be emitted

Answer: (b)

Solution

Initially, energy of electron $= +3 \, \mathrm{eV}$ finally, in $2^{nd}$ excited state, energy of electron $= -\frac{(13.6 \, \mathrm{eV})}{3^2}$ $= -1.51 \, \mathrm{eV}$ Loss in energy is emitted as photon. So, photon energy $\frac{hc}{\lambda} = 4.51 \, \mathrm{eV}$ $\mathrm{KE_{max}} = \frac{hc}{\lambda} - \phi = 4.51 - \left( \frac{hc}{\lambda_{th}} \right)$ $= 4.51 \, \mathrm{eV} - \frac{12400 \, \mathrm{eV} \cdot \AA}{4000 \AA}$ $= 1.41 \, \mathrm{eV}$

Question 32

Physics · Ray Optics and Optical Instruments · Single correct

The expected graphical representation of the variation of angle of deviation ' $\delta$ ' with angle of incidence ' $i$ ' in a prism is :

Answer: (a)

Solution

Standard graph between angle of deviation and incident angle.

Question 33

Physics · Mechanical Properties of Fluids · Single correct

A raindrop with radius $R = 0.2 \, \mathrm{mm}$ falls from a cloud at a height $h = 2000 \, \mathrm{m}$ above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is: [Density of water $f_w = 1000 \, \mathrm{kg} \, \mathrm{m}^{-3}$ and Density of air $f_a = 1.2 \, \mathrm{kg} \, \mathrm{m}^{-3}$, $g = 10 \, \mathrm{m/s}^2$ Coefficient of viscosity of air $= 1.8 \times 10^{-5} \, \mathrm{Nsm}^{-2}$]

  1. $250.6 \, \mathrm{ms}^{-1}$
  2. $43.56 \, \mathrm{ms}^{-1}$
  3. $4.94 \, \mathrm{ms}^{-1}$
  4. $14.4 \, \mathrm{ms}^{-1}$

Answer: (c)

Solution

At terminal speed $a = 0$ $F_{net} = 0$ $mg = F_v = 6 \pi \eta R v$ $$v = \frac{mg}{6 \pi R v}$$ $$v = \frac{\rho_w \frac{4 \pi}{3} R^3 g}{6 \pi \eta R}$$ $$= \frac{2 \rho_w R^2 g}{9 \eta}$$ $$= \frac{400}{81} \, \mathrm{m/s}$$ $$= 4.94 \, \mathrm{m/s}$$

Question 34

Physics · Thermodynamics · Single correct

One mole of an ideal gas is taken through an adiabatic process where the temperature rises from $27^\circ \mathrm{C}$ to $37^\circ \mathrm{C}$. If the ideal gas is composed of polyatomic molecule that has $4$ vibrational modes, which of the following is true? \[ \mathrm{R} = 8.314 \, \mathrm{J} \, \mathrm{mol}^{-1} \mathrm{K}^{-1} \]

  1. The work done by the gas is close to \(332\,\mathrm{J}\).
  2. The work done on the gas is close to \(582\,\mathrm{J}\).
  3. The work done by the gas is close to \(582\,\mathrm{J}\).
  4. The work done on the gas is close to \(332\,\mathrm{J}\).

Answer: (b)

Solution

Since, each vibrational mode, corresponds to two degrees of freedom, hence, $f = 3 (trans.) + 3 (rot.) + 8 (vib.) = 14$. $\gamma = 1 + \frac{2}{f}$ $$\gamma = 1 + \frac{2}{14} = \frac{8}{7}$$ $$W = \frac{nR\Delta T}{\gamma - 1} = -582$$ As $W < 0$, work is done on the gas.

Question 35

Physics · Oscillations · Single correct

An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $t = \frac{T}{4}$ s starting from mean position. Assume that the initial phase of the oscillation is zero.

  1. 0.62 J
  2. 6.2 $\times$ 10^{-3} J
  3. 1.2 $\times$ 10^{3} J
  4. 6.2 $\times$ 10^{3} J

Answer: (a)

Solution

Given $T = 2\pi \sqrt{\frac{m}{k}}$. $0.2 = 2\pi \sqrt{\frac{0.5}{k}}$. $k = 50\pi^2$. $\approx 500$. $x = A \sin(\omega t + \phi)$. $= 5 \, cm \sin\left(\frac{\omega T}{4} + 0\right)$. $= 5 \, cm \sin\left(\frac{\pi}{2}\right)$. $= 5 \, cm$. $PE = \frac{1}{2} k x^2$. $= \frac{1}{2} (500) \left(\frac{5}{100}\right)^2$. $= 0.6255$.

Question 36

Physics · Physical World, Units and Measurements · Single correct

Match List I with List II. $$ \begin{array}{ll} \text{List-I} & \text{List-II} \\ \hline (a) \text{ Capacitance, } C & (i) \, M^1 L^1 T^{-3} A^{-1} \\ (b) \text{ Permittivity of free space, } \varepsilon_0 & (ii) \, M^{-1} L^{-3} T^4 A^2 \\ (c) \text{ Permeability of free space, } \mu_0 & (iii) \, M^{-1} L^{-2} T^4 A^2 \\ (d) \text{ Electric field, } E & (iv) \, M^1 L^1 T^{-2} A^{-2} \end{array} $$ Choose the correct answer from the options given below

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

Answer: (a)

Solution

Given $q = CV$. $$[C] = \left[ \frac{q}{V} \right] = \frac{(A \times T)^2}{ML^2 \, T^{-2}}$$ $$= M^{-1} \, L^{-2} \, T^4 \, A^2$$ $$[E] = \left[ \frac{F}{q} \right] = \frac{MLT^{-2}}{AT}$$ $$= MLT^{-3} \, A^{-1}$$ $$F = \frac{q_1 \, q_2}{4 \pi \varepsilon_o \, r^2}$$ $$[\varepsilon_o] = M^{-1} \, L^{-3} \, T^4 \, A^2$$ Speed of light $c = \frac{1}{\sqrt{\mu_o \varepsilon_o}}$ $$\mu_0 = \frac{1}{\varepsilon_o \, c^2}$$ $$[\mu_0] = \frac{1}{\left[M^{-1} \, L^{-3} \, T^4 \, A^2 \right] \left[L \, T^{-1} \right]^2}$$ $$= \left[M^1 \, L^1 \, T^{-2} \, A^{-2} \right]$$

Question 37

Physics · Work, Energy and Power · Single correct

Given below is the plot of a potential energy function $U(x)$ for a system, in which a particle is in one dimensional motion, while a conservative force $F(x)$ acts on it. Suppose that $E_{mech} = 8 \, \mathrm{J}$, the incorrect statement for this system is :

  1. at $x > x_4$, K. E. is constant throughout the region.
  2. at $x < x_1$, K. E. is smallest and the particle is moving at the slowest speed.
  3. at $x = x_2$, KE. is greatest and the particle is moving at the fastest speed.
  4. at $x = x_3$, K. E. = 4 J.

Answer: (b)

Solution

Given $E_{mech.} = 8 \, J$. (A) At $x > x_4$, $U = constant = 6 \, J$. $K = E_{mech.} - U = 2 \, J = constant$. (B) At $x < x_1$, $U = constant = 8 \, J$. $K = E_{mech.} - U = 8 - 8 = 0 \, J$. Particle is at rest. (C) At $x = x_2$, $U = 0 \implies E_{mech.} = K = 8 \, J$. KE is greatest, and particle is moving at fastest speed. (D) At $x = x_3$, $U = 4 \, J$. $U + K = 8 \, J$. $K = 4 \, J$.

Question 38

Physics · Alternating Current · Single correct

A 100 $\Omega$ resistance, a 0.1 $\mu\mathrm{F}$ capacitor and an inductor are connected in series across a 250 $\mathrm{V}$ supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 $\mathrm{Hz}$.

  1. 0.70 $\mathrm{H}$
  2. 70.3 $\mathrm{mH}$
  3. 7.03 $\times 10^{-5} \mathrm{H}$
  4. 70.3 $\mathrm{H}$

Answer: (d)

Solution

Given $C = 0.1 \, \mu \mathrm{F} = 10^{-7} \, \mathrm{F}$. Resonant frequency $= 60 \, \mathrm{Hz}$, $\omega_0 = \frac{1}{\sqrt{LC}}$. $$2 \pi f_0 = \frac{1}{\sqrt{LC}} \Rightarrow L = \frac{1}{4 \pi^2 f_0^2 C}$$ by putting values $L = 70.3 \, \mathrm{Hz}$.

Question 39

Physics · Electric Charges and Fields · Single correct

A simple pendulum of mass ' m ', length ' l ' and charge '+q' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be

  1. $\tan^{-1}$[ $\frac{q}{mg}$ $\times$ $\frac{C_1(V_2-V_1)}{(C_1+C_2)(d-t)}$ ]
  2. $\tan^{-1}$ [ $\frac{q}{mg}$ $\times$ $\frac{C_2(V_2-V_1)}{(C_1+C_2)(d-t)}$ ]
  3. $\tan^{-1}$ [ $\frac{q}{mg}$ $\times$ $\frac{C_2(V_1+V_2)}{(C_1+C_2)(d-t)}$ ]
  4. $\tan^{-1}$ [ $\frac{q}{mg}$ $\times$ $\frac{C_1(V_1+V_2)}{(C_1+C_2)(d-t)}$ ]

Answer: (c)

Solution

Let $E$ be the electric field in air. $T \sin \theta = qE$. $T \cos \theta = mg$. $$\tan \theta = \frac{qE}{mg}$$ $$Q = \left[ \frac{C_1 C_2}{C_1 + C_2} \right] [V_1 + V_2]$$ $$E = \frac{Q}{A \varepsilon_0} = \left[ \frac{C_1 C_2}{C_1 + C_2} \right] \frac{[V_1 + V_2]}{A \varepsilon_0}$$ $$C_1 = \frac{\varepsilon_0 A}{d-t} \implies E = \frac{C_2 [V_1 + V_2]}{(C_1 + C_2)(d-t)}$$ Now $$\theta = \tan^{-1} \left[ \frac{qE}{mg} \right]$$ $$\theta = \tan^{-1} \left[ \frac{q}{mg} \times \frac{C_2 (V_1 + V_2)}{(C_1 + C_2)(d-t)} \right]$$

Question 40

Physics · Thermodynamics · Single correct

Two Carnot engines A and B operate in series such that engine A absorbs heat at $T_1$ and rejects heat to a sink at temperature $T$. Engine B absorbs half of the heat rejected by Engine A and rejects heat to the sink at $T_3$. When workdone in both the cases is equal, to value of $T$ is :

  1. $\frac{2}{3} T_1 + \frac{3}{2} T_3$
  2. $\frac{1}{3} T_1 + \frac{2}{3} T_3$
  3. $\frac{3}{2} T_1 + \frac{1}{3} T_3$
  4. $\frac{2}{3} T_1 + \frac{1}{3} T_3$

Answer: (d)

Solution

Given the system, we have: $$W_A = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T}{T_1} \Rightarrow \frac{Q_2}{Q_1} = \frac{T}{T_1}$$ $$W_B = 1 - \frac{Q_3}{(Q_2/2)} = 1 - \frac{T_3}{T} \Rightarrow \frac{2Q_3}{Q_2} = \frac{T_3}{T}$$ Now, $W_A = W_B$. $$Q_1 - Q_2 = \frac{Q_2}{2} - Q_3$$ $$\Rightarrow \frac{2Q_1}{Q_2} + \frac{2Q_3}{Q_2} = 3$$ $$\Rightarrow \frac{2 \ T_1}{T} + \frac{T_3}{T} = 3$$ $$\frac{2 \ T_1}{3} + \frac{T_3}{3} = T$$

Question 41

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

Find the truth table for the function Y of A and B represented in the following figure.

Answer: (b)

Solution

Question 42

Physics · Moving Charges and Magnetism · Single correct

Figure A and B shown two long straight wires of circular cross-section (a and b with a < b), carrying current I which is uniformly distributed across the cross-section. The magnitude of magnetic field B varies with radius r and can be represented as:

Answer: (c)

Question 43

Physics · Gravitation · Single correct

Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :

  1. $\sqrt{\frac{G}{2R^3}}$
  2. $\frac{1}{2} \sqrt{\frac{G}{R^3}}$
  3. $\frac{1}{2R} \sqrt{\frac{1}{G}}$
  4. $\sqrt{\frac{2G}{R^3}}$

Answer: (b)

Solution

The force $F$ is given by $$F = \frac{Gm^2}{(2R)^2} = mR\omega^2.$$ Solving for $\omega$, we have $$\omega = \frac{1}{2} \sqrt{\frac{G}{R^3}}.$$

Question 44

Physics · Nuclei · Single correct

Consider the following statements: A. Atoms of each element emit characteristics spectrum. B. According to Bohr's Postulate, an electron in a hydrogen atom, revolves in a certain stationary orbit. C. The density of nuclear matter depends on the size of the nucleus. D. A free neutron is stable but a free proton decay is possible. E. Radioactivity is an indication of the instability of nuclei. Choose the correct answer from the options given below :

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

Answer: (b)

Solution

(A) True, atom of each element emits characteristic spectrum. (B) True, according to Bohr's postulates $$mvr = \frac{nh}{2\pi}$$ and hence electron resides into orbits of specific radius called stationary orbits. ($C$) False, density of nucleus is constant (D) False, A free neutron is unstable decays into proton and electron and antineutrino. (E) True unstable nucleus show radioactivity.

Question 45

Physics · Electric Charges and Fields · Single correct

What will be the magnitude of electric field at point O as shown in figure? Each side of the figure is $l$ and perpendicular to each other?

  1. $\frac{1}{4\pi\varepsilon_0} \frac{q}{l^2}$
  2. $\frac{1}{4\pi\varepsilon_0} \frac{q}{(2l)^2} (2\sqrt{2} - 1)$
  3. $\frac{q}{4\pi\varepsilon_0 (2l)^2}$
  4. $\frac{1}{4\pi\varepsilon_0} \frac{2q}{2l^2} (\sqrt{2})$

Answer: (b)

Solution

Given $$E_1 = \frac{kq}{\ell^2} = E_2$$ $$E_3 = \frac{kq}{(\sqrt{2} \ell)^2} = \frac{kq}{2 \ell^2}$$ $$E = \frac{\sqrt{2} kq}{\ell^2} - \frac{kq}{2 \ell^2} = \frac{kq}{2 \ell^2} (2 \sqrt{2} - 1)$$

Question 46

Physics · Mathematics in Physics · Single correct

A physical quantity 'y' is represented by the formula $y = m^2 r^{-4} g^{-x} l^{-\frac{3}{2}}$ If the percentage errors found in $y, m, l$ and $g$ are $18, 1, 0.5, 4$ and $p$ respectively, then find the value of $x$ and $p$.

  1. 5 and $\pm 2$
  2. 4 and $\pm 3$
  3. $\frac{16}{3}$ and $\pm \frac{3}{2}$
  4. 8 and $\pm 2$

Answer: (c)

Solution

Given $$\frac{\Delta y}{y} = \frac{2 \Delta m}{m} + \frac{4 \Delta r}{r} + \frac{x \Delta g}{g} + \frac{3}{2} \frac{\Delta \ell}{\ell}$$ $$18 = 2(1) + 4(0.5) + xp + \frac{3}{2}(4)$$ $$8 = xp$$ By checking from options, $$x = \frac{16}{3}, p = \pm \frac{3}{2}$$

Question 47

Physics · Work, Energy and Power · Single correct

An automobile of mass ' m ' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by:

  1. $\left( \frac{9P}{8m} \right)^{\frac{1}{2}} t^{\frac{3}{2}}$
  2. $\left( \frac{8P}{9m} \right)^{\frac{1}{2}} t^{\frac{2}{3}}$
  3. $\left( \frac{9m}{8P} \right)^{\frac{1}{2}} t^{\frac{3}{2}}$
  4. $\left( \frac{8P}{9m} \right)^{\frac{1}{2}} t^{\frac{3}{2}}$

Answer: (d)

Solution

Given $P = const.$ $P = Fv = \frac{mv^2 dv}{dx}$ $$\int_0^x \frac{P}{m} \, dx = \int_0^v v^2 \, dv$$ $$\frac{Px}{m} = \frac{v^3}{3}$$ $$\left( \frac{3Px}{m} \right)^{1/3} = v = \frac{dx}{dt}$$ $$\left( \frac{3P}{m} \right)^{1/3} \int_0^t dt = \int_0^x x^{-1/3} \, dx$$ Thus, $$x = \left( \frac{8P}{9m} \right)^{1/2} t^{3/2}$$

Question 48

Physics · Gravitation · Single correct

The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of $9.0 \times 10^3$ km. Find the mass of Mars. $\{$ Given $\frac{4\pi^2}{G}$ = 6 $\times$ $10^{11} N^{-1} m^{-2} kg^2$ $\}$

  1. $5.96 \times 10^{19}$ kg
  2. $3.25 \times 10^{21}$ kg
  3. $7.02 \times 10^{25}$ kg
  4. $6.00 \times 10^{23}$ kg

Answer: (d)

Solution

Option D is correct $$T^2 = \frac{4\pi^2}{GM} \cdot r^3$$ $$M = \frac{4\pi^2}{G} \cdot \frac{r^3}{T^2}$$ by putting values $$M = 6 \times 10^{23}$$

Question 49

Physics · Motion in a Straight Line · Single correct

A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation $$ F = F_0 \left[ 1 - \left( \frac{t - T}{T} \right)^2 \right] $$ Where $F_0$ and $T$ are constants. The force acts only for the time interval $2T$. The velocity $v$ of the particle after time $2T$ is :

  1. $2 F_0 T/M$
  2. $F_0 T/2M$
  3. $4 F_0 T/3M$
  4. $F_0 T/3M$

Answer: (c)

Solution

Given $t = 0$, $u = 0$. $$a = \frac{F_0}{M} - \frac{F_0}{MT^2}(t - T)^2 = \frac{dv}{dt}$$ $$\int_0^v dv = \int_{t=0}^{2T} \left( \frac{F_0}{M} - \frac{F_0}{MT^2}(t - T)^2 \right) dt$$ $$V = \left[ \frac{F_0}{M} t \right]_0^{2T} - \frac{F_0}{MT^2} \left[ \frac{t^3}{3} - t^2 T + T^2 t \right]_0^{2T}$$ $$V = \frac{4 F_0 T}{3M}$$

Question 50

Physics · Current Electricity · Single correct

The resistance of a conductor at $15^{\circ}C$ is $16\,\Omega$ and at $100^{\circ}C$ is $20\,\Omega$. What will be the temperature coefficient of resistance of the conductor?

  1. 0.010^{\circ}C^{-1}$
  2. 0.033^{\circ}C^{-1}$
  3. 0.003^{\circ}C^{-1}$
  4. 0.042^{\circ}C^{-1}$

Answer: (c)

Solution

Given $$16 = R_o \left[ 1 + \alpha (15 - T_o) \right]$$ $$20 = R_o \left[ 1 + \alpha (100 - T_o) \right]$$ Assuming $T_o = 0^\circ \mathrm{C}$, as a general convention. $\Rightarrow$ $\frac{16}{20}$ = $\frac{1 + \alpha \times 15}{1 + \alpha \times 100}$ $\Rightarrow$ $\alpha$ = $0.003^{\circ}C^{-1}$

Question 51

Physics · System of Particles and Rotational Motion · Numerical

In the given figure, two wheels $P$ and $Q$ are connected by a belt $B$. The radius of $P$ is three times as that of $Q$. In case of same rotational kinetic energy, the ratio of rotational inertias $\left( \frac{I_1}{I_2} \right)$ will be $x : 1$. The value of $x$ will be

Answer: 9

Solution

The equation for rotational kinetic energy is given by $$\frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} I_2 \omega_2^2.$$ By substituting the expressions for angular velocity, we have $$I_1 \left( \frac{v}{3R} \right)^2 = I_2 \left( \frac{v}{R} \right)^2.$$ Solving for the ratio of moments of inertia, we find $$\frac{I_1}{I_2} = \left( \frac{3R}{R} \right)^2 = \frac{9}{1}.$$

Question 52

Physics · Wave Optics · Numerical

The difference in the number of waves when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is mm. [Refractive index of air = 1.0003, wavelength of yellow light in vacuum = 6000$\AA$...]

Answer: 2

Solution

Thickness $t = n \lambda$ So, $n \lambda_{vac} = (n+1) \lambda_{air}$ $$n \lambda = (n+1) \frac{\lambda}{\mu_{air}}$$ $$n = \frac{1}{\mu_{air} - 1} = \frac{10^4}{3}$$ $$t = n \lambda$$ $$= \frac{10^4}{3} \times 6000 \AA \ldots$$ $$= 2 \, mm$$

Question 53

Physics · Communication Systems · Numerical

The maximum amplitude for an amplitude modulated wave is found to be $12 \, \mathrm{V}$ while the minimum amplitude is found to be $3 \, \mathrm{V}$. The modulation index is $0.6x$ where $x$ is

Answer: 1

Solution

Given $A_{max} = A_c + A_m = 12$ and $A_{min} = A_c - A_m = 3$. Therefore, $A_c = \frac{15}{2}$ and $A_m = \frac{9}{2}$. The modulation index is given by $\frac{A_m}{A_c} = \frac{9/2}{15/2} = 0.6$. Thus, $x = 1$.

Question 54

Physics · Electromagnetic Induction · Numerical

In the given figure the magnetic flux through the loop increases according to the relation $\phi_B(t) = 10t^2 + 20t$, where $\phi_B$ is in milliwebers and $t$ is in seconds. The magnitude of current through $R = 2\Omega$ resistor at $t = 5 \, \mathrm{s}$ is_____ mA.

Answer: 60

Solution

Given $|\epsilon| = \frac{d\phi}{dt} = 20t + 20 \, \mathrm{mV}$. $|i| = \frac{|\epsilon|}{R} = 10t + 10 \, \mathrm{mA}$. At $t = 5$, $|i| = 60 \, \mathrm{mA}$.

Question 55

Physics · Oscillations · Numerical

A particle executes simple harmonic motion represented by displacement function as $$x(t) = A \sin(\omega t + \phi)$$ If the position and velocity of the particle at $t = 0 \, \mathrm{s}$ are $2 \, \mathrm{cm}$ and $2\omega \, \mathrm{cm/s}^{-1}$ respectively, then its amplitude is $x\sqrt{2} \, \mathrm{cm}$ where the value of $x$ is

Answer: 2

Solution

Given $$x(t) = A \sin(\omega t + \phi)$$ $$v(t) = A \omega \cos(\omega t + \phi)$$ From equation (1): $$2 = A \sin \phi$$ From equation (2): $$2\omega = A \omega \cos \phi$$ From (1) and (2), we have $\($ $\tan$ $\phi$ = 1 $\)$. Thus, $\($ $\phi$ = 45^$\circ$ $\)$. Putting the value of $\($ $\phi$ $\)$ in equation (1): $$2 = A \left\{ \frac{1}{\sqrt{2}} \right\}$$ Therefore, $$A = 2\sqrt{2}$$ Finally, $$x = 2$$

Question 56

Physics · Motion in a Plane · Numerical

A swimmer wants to cross a river from point $A$ to point $B$. Line $AB$ makes an angle of $30^\circ$ with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle $\theta$ with the line $AB$ should be $\underline{\hspace{1cm}}$, so that the swimmer reaches point $B$.

Answer: 30

Solution

Both velocity vectors are of same magnitude, therefore the resultant would pass exactly midway through them. \[ \theta = 30^\circ \]

Question 57

Physics · Current Electricity · Numerical

For the circuit shown, the value of current at time $t=3.2\ \mathrm{s}$ will be $\underline{\qquad}\ \mathrm{A}.$ [Voltage distribution $V(t)$ is shown by Fig.\,(1) and the circuit is shown in Fig.\,(2)]

Answer: A

Solution

From graph voltage at t $= 3.2sec$ is $6 \, volt$. $$i = \frac{6 - 5}{1}$$ $i = 1 \, A$

Question 58

Physics · Motion in a Plane · Numerical

A small block slides down from the top of hemisphere of radius $R = 3 \, \mathrm{m}$ as shown in the figure. The height 'h' at which the block will lose contact with the surface of the sphere is m. (Assume there is no friction between the block and the hemisphere)

Answer: 2

Solution

Given $mg \cos \theta = \frac{mv^2}{R}$ and $\cos \theta = \frac{h}{R}$. Using energy conservation, $mg\{R - h\} = \frac{1}{2} mv^2$. From (1) and (2), $mg \left\{ \frac{h}{R} \right\} = \frac{2mg\{R-h\}}{R}$. Solving for $h$, we get $h = \frac{2R}{3} = 2 \, \mathrm{m}$.

Question 59

Physics · Atoms · Numerical

The $K_\alpha$ X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms with a $K$ electron knocked out is 27.5 keV, the energy of this atom when an $L$ electron is knocked out will be keV. (Round off to the nearest integer) $$[h = 4.14 \times 10^{-15} \, \mathrm{eVs}, c = 3 \times 10^8 \, \mathrm{ms^{-1}}]$$

Answer: 10

Solution

Given $E_{K_a} = E_k - E_L$. $$\frac{hc}{\lambda_{K_a}} = E_k - E_L$$ Therefore, $$E_L = E_k - \frac{hc}{\lambda_{K_a}}$$ Substituting the values, $$E_L = 27.5 \, \mathrm{keV} - \frac{12.42 \times 10^{-7} \, \mathrm{eVm}}{0.071 \times 10^{-9} \, \mathrm{m}}$$ $$E_L = (27.5 - 17.5) \, \mathrm{keV}$$ $$= 10 \, \mathrm{keV}$$

Question 60

Physics · Motion in a Plane · Numerical

The water is filled upto height of 12 m in a tank having vertical sidewalls. A hole is made in one of the walls at a depth 'h' below the water level. The value of 'h' for which the emerging stream of water strikes the ground at the maximum range is m.

Answer: 6

Solution

Given the problem, we have: $$R = \sqrt{2gh} \times \sqrt{\frac{(12-h) \times 2}{g}}$$ Simplifying, we get: $$\sqrt{4h(12-h)} = R$$ For maximum $R$, we set the derivative to zero: $$\frac{dR}{dh} = 0$$ This implies $h = 6 \, \mathrm{m}$.

Chemistry

Question 61

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

Which one of the following set of elements can be detected using sodium fusion extract?

  1. Sulfur, Nitrogen, Phosphorous, Halogens
  2. Phosphorous, Oxygen, Nitrogen, Halogens
  3. Nitrogen, Phosphorous, Carbon, Sulfur
  4. Halogens, Nitrogen, Oxygen, Sulfur

Answer: (a)

Solution

By sodium fusion extract we can detect sulphur, nitrogen, Phosphorous and halogens, because they are converted into their ionic form with sodium metal.

Question 62

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Consider the above reaction, the major product $"P"$ formed is

Answer: (b)

Solution

Question 63

Chemistry · Hydrogen · Single correct

The number of neutrons and electrons, respectively, present in the radioactive isotope of hydrogen is :-

  1. 1 and 1
  2. 3 and 1
  3. 2 and 1
  4. 2 and 2

Answer: (c)

Solution

Radioactive isotope of hydrogen is Tritium $\left( ^3_1 \mathrm{T} \right)$. No. of neutrons $(A - Z) = 3 - 1 = 2$ No. of electrons $= 1$

Question 64

Chemistry · Co-ordination Compounds · Single correct

Match List - I with List II :

  1. A. (a)- (v), (b) - (i), (c ) - (ii), (d) - (iv)
  2. B. (a)- (v), (b) - (ii), (c ) - (iv), (d) - (i)
  3. C. (a)- (iv), (b) - (iii), (c ) - (i), (d) - (ii)
  4. D. (a)- (v), (b) - (iii), (c ) - (ii), (d) - (i)

Answer: (d)

Solution

Li makes alloy with Lead to make white metal bearings for motor engines. Liquid Na metal is used as coolant in fast breeder nuclear reactor. K is a very absorbent of $\mathrm{CO_2}$. Cs is used in making photoelectric cell.

Question 65

Chemistry · Surface Chemistry · Single correct

Given below are two statement : one is labelled as Assertion $\textbf{A}$ and the other is labelled as Reason $\textbf{R}$. Assertion $\textbf{A}$ : $\mathrm{SO_2 (g)}$ is adsorbed to a large extent than $\mathrm{H_2 (g)}$ on activated charcoal. Reason $\textbf{R}$ : $\mathrm{SO_2 (g)}$ has a higher critical temperature than $\mathrm{H_2 (g)}$. In the light of the above statements, choose the most appropriate answer from the options given below.

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

Answer: (b)

Question 66

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

The CORRECT order of first ionisation enthalpy is :

  1. Mg < S < Al < P
  2. Mg < Al < S < P
  3. Al < Mg < S < P
  4. Mg < Al < P < S

Answer: (a)

Solution

Given the elements Mg, Al, P, and S, the ionization energy order is Al < Mg < S < P. The valence configuration is $[\mathrm{Ne}] : 3s^2 3p^1 3s^2 3p^3 3s^2 3p^4$. Properties include full and filled stable half-filled shells.

Question 67

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

Given below are two statements: Statement I : Hyperconjugation is a permanent effect. Statement II : Hyperconjugation in ethyl cation ($\text{CH}_3 - \text{CH}_2^+$) involves the overlapping of $\text{C}_{sp^2} - \text{H}_{1s}$ bond with empty 2p orbital of other carbon. Choose the correct option: (1) Both statement I and statement II are false (2) Statement I is incorrect but statement II is true (3) Statement I is correct but statement II is false (4) Both Statement I and statement II are true.

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

Answer: (c)

Solution

Statement I: It is correct statement. Statement II: $\mathrm{CH_3 - CH_2}$ involve $\mathrm{C_{sp^3} - H_{1s}}$ bond with empty $2p$ orbital hence given statement is false.

Question 68

Chemistry · Co-ordination Compounds · Single correct

Given below are two statements: Statement I: $[\mathrm{Mn(CN)}_6]^{3-}$, $[\mathrm{Fe(CN)}_6]^{3-}$ and $[\mathrm{Co(C_2O_4)}_3]^{3-}$ are $d^2sp^3$ hybridised. Statement II: $[\mathrm{MnCl}_6]^{3-}$ and $[\mathrm{FeF}_6]^{3-}$ are paramagnetic and have 4 and 5 unpaired electrons, respectively. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (d)

Solution

$[\mathrm{Mn(CN)_6}]^{3-}$ $[\mathrm{Fe(CN)_6}]^{3-}$ $[\mathrm{Co(C_2O_4)_3}]^{3-}$ $\downarrow$ $\mathrm{Mn}^{3+},\ \mathrm{CN^-}$ $\mathrm{Fe}^{3+},\ \mathrm{CN^-}$ $\mathrm{Co}^{3+},\ \mathrm{C_2O_4^{2-}}$ $d^4$ configuration, strong field ligand (SFL) $d^5$ configuration, strong field ligand (SFL) $d^6$ configuration, chelating ligand $\Rightarrow$ All have large crystal field splitting; hence all undergo $d^2sp^3$ hybridisation. $[\mathrm{MnCl_6}]^{3-}$ and $[\mathrm{FeF_6}]^{3-}$ $d^4$ configuration, $\mathrm{Cl^-}$ (WFL) $d^5$ configuration, $\mathrm{F^-}$ (WFL) $\Rightarrow\ [\mathrm{MnCl_6}]^{3-}$ has $4$ unpaired electrons. $\Rightarrow\ [\mathrm{FeF_6}]^{3-}$ has $5$ unpaired electrons.

Question 69

Chemistry · Analytical Chemistry · Single correct

To an aqueous solution containing ions such as $\mathrm{Al}^{3+}$, $\mathrm{Zn}^{2+}$, $\mathrm{Ca}^{2+}$, $\mathrm{Fe}^{3+}$, $\mathrm{Ni}^{2+}$, $\mathrm{Ba}^{2+}$ and $\mathrm{Cu}^{2+}$ was added conc. HCl, followed by $\mathrm{H}_2 \mathrm{S}$. The total number of cations precipitated during this reaction is/are:

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

Answer: (a)

Solution

$Al^{3+}$ and $Fe^{3+}$ sulphides hydrolyse in water. $Ni^{2+}$ and $Zn^{2+}$ require basic medium with $H_2 S$ to form ppt. $Ca^{2+}$ and $Ba^{2+}$ sulphides are soluble hence we will receive only $CuS$ ppt.

Question 70

Chemistry · Chemistry in Everyday Life · Single correct

Given below are two statements: Statement I: Penicillin is a bacteriostatic type antibiotic. Statement II: The general structure of Penicillin is:

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

Answer: (b)

Solution

Statement I: Penicillin is bactericidal not bacteriostatic hence given statement is false. Statement II: Structure of penicillin given is correct.

Question 71

Chemistry · Biomolecules · Single correct

Compound A gives D-Galactose and D-Glucose on hydrolysis. The compound A is:

  1. Amylose
  2. Sucrose
  3. Maltose
  4. Lactose

Answer: (d)

Solution

Lactose: It is a disaccharide of $\beta - \mathrm{D}$-Galactose and $\beta - \mathrm{D}$-Glucose with $C_1$ of galactose and $C_4$ of glucose link. Lactose: $\beta - \mathrm{D}$-Galactose $+ \beta - \mathrm{D}$-Glucose

Question 72

Chemistry · Amines · Single correct

R - CN $\xrightarrow{\begin{array}{c} (i) \ DIBAL-H \\ (ii) \ H_2O \end{array}}$ R - Y Consider the above reaction and identify "Y"

  1. -CH_2NH_2
  2. -CONH_2
  3. -CH
  4. -COO

Answer: (c)

Solution

The reaction given is: $$\mathrm{R{-}C \equiv N \xrightarrow{(1)DiBAL-H}{(2)H_2O} R{-}C{-}H}$$ Here Y is an aldehyde.

Question 73

Chemistry · Hydrocarbons · Single correct

consider the above reaction, and choose the correct statement:

  1. The reaction is not possible in acidic medium
  2. Both compounds A and B are formed equally
  3. Compound A will be the major product
  4. Compound B will be the major product

Answer: (c)

Solution

The reaction begins with the protonation of the alcohol group by $\mathrm{H^+}$ from $\mathrm{H_2SO_4}$, forming a water molecule attached to the benzene ring. This intermediate loses water ($\mathrm{-H_2O}$) to form a carbocation. The carbocation undergoes rearrangement to form a more stable structure. The elimination of $\mathrm{H_2SO_4}$ leads to the formation of an alkene. The reaction shows geometric isomerism (GI) resulting in two products: (A) Trans (more stable product, Saytzeff's alkene, major) and (B) Cis.

Question 74

Chemistry · Co-ordination Compounds · Single correct

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

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

Answer: (a)

Question 75

Chemistry · Structure of Atom · Single correct

If the Thompson model of the atom was correct, then the result of Rutherford's gold foil experiment would have been:

  1. All of the $\alpha$-particles pass through the gold foil without decrease in speed.
  2. $\alpha$-Particles are deflected over a wide range of angles.
  3. All $\alpha$-particles get bounced back by $180^\circ$
  4. $\alpha$-Particles pass through the gold foil deflected by small angles and with reduced speed.

Answer: (d)

Solution

As in Thomson model, protons are diffused (charge is not centred) $\alpha$-particles deviate by small angles and due to repulsion from protons, their speed decreases.

Question 76

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

Number of Cl = O bonds in chlorous acid, chloric acid and perchloric acid respectively are:

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

Answer: (c)

Solution

Question 77

Chemistry · The Solid State · Multiple correct

"Select the correct statements. (A) Crystalline solids have long range order. (B) Crystalline solids are isotropic. $(C)$ Amorphous solid are sometimes called pseudo solids. (D) Amorphous solids soften over a range of temperatures. (E) Amorphous solids have a definite heat of fusion. Choose the most appropriate answer from the options given below."

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

Answer: (c)

Solution

Crystalline solids have definite arrangement of constituent particles and have long range order. Different constituent particles of an amorphous solid have different bond strengths and soften over a range of temperatures.

Question 78

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

What is A in the following reaction?

Answer: (d)

Solution

Question 79

Chemistry · Hydrocarbons · Single correct

The correct sequence of correct reagents for the following transformation is :-

  1. $(i)Fe, HCl$ $(ii) Cl_2, HCl,$ $(iii) NaNO_2, HCl$, $0^\circ$ C $(iv) H_2O/H^+$
  2. $(i)Fe, HCl$ (ii) $NaNO_2$, HCl, $0^\circ$ C $(iii) H_2O/H^+$ $(iv) Cl_2, FeCl_3$
  3. (i) $\mathrm{Cl_2,\ FeCl_3}$, (ii) $\mathrm{Fe,\ HCl}$, (iii) $\mathrm{NaNO_2,\ HCl,\ 0^\circ C}$, (iv) $\mathrm{H_2O/H^+}$
  4. (i) $\mathrm{Cl_2,\ FeCl_3}$, (ii) $\mathrm{NaNO_2,\ HCl,\ 0^\circ C}$, (iii) $\mathrm{Fe,\ HCl}$, (iv) $\mathrm{H_2O/H^+}$

Answer: (c)

Solution

The reaction sequence starts with a nitrobenzene compound where $\mathrm{NO_2}$ is meta directing. The first step involves chlorination using $\mathrm{Cl_2/FeCl_3}$, resulting in the addition of a chlorine atom to the benzene ring. Next, reduction is carried out using $\mathrm{Fe/HCl}$, converting the $\mathrm{NO_2}$ group to an $\mathrm{NH_2}$ group. Following this, diazotisation is performed using $\mathrm{NaNO_2 + HCl}$ at $0^\circ \mathrm{C}$, forming a diazonium salt. Finally, the diazonium group is replaced by an $\mathrm{OH}$ group using $\mathrm{H_2O/H^+}$, resulting in the final product.

Question 80

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

The addition of silica during the extraction of copper from its sulphide ore :-

  1. converts copper sulphide into copper silicate
  2. converts iron oxide into iron silicate
  3. reduces copper sulphide into metallic copper
  4. reduces the melting point of the reaction mixture

Answer: (b)

Solution

Silica is used to remove FeO impurity from the ore of copper. $$\mathrm{FeO + SiO_2 \rightarrow FeSiO_3}$$ iron silicate (Slag)

Question 81

Chemistry · Equilibrium · Numerical

The equilibrium constant for the reaction $$\mathrm{A(s) \rightleftharpoons M(s) + \frac{1}{2}O_2(g)}$$ is $K_p = 4$. At equilibrium, the partial pressure of $O_2$ is $\_$$\_$$\_$ atm. (Round off to the nearest integer)

Answer: 16

Solution

Given $k_p = P_{\mathrm{O}_2}^{1/2} = 4$. Therefore, $P_{\mathrm{O}_2} = 16 bar = 16 atm$.

Question 82

Chemistry · Thermodynamics · Numerical

When $400 \, \mathrm{mL}$ of $0.2 \, \mathrm{M}$ $\mathrm{H_2SO_4}$ solution is mixed with $600 \, \mathrm{mL}$ of $0.1 \, \mathrm{M}$ $\mathrm{NaOH}$ solution, the increase in temperature of the final solution is _______ $\times 10^{-2} \, \mathrm{K}$. (Round off to the nearest integer). $\left[ \text{Use: } \mathrm{H^+} (\mathrm{aq}) + \mathrm{OH^-} (\mathrm{aq}) \rightarrow \mathrm{H_2O} : \Delta_r H = -57.1 \, \mathrm{kJ \, mol^{-1}} \right]$ Specific heat of $\mathrm{H_2O} = 4.18 \, \mathrm{J \, K^{-1} \, g^{-1}}$ density of $\mathrm{H_2O} = 1.0 \, \mathrm{g \, cm^{-3}}$ Assume no change in volume of solution on mixing.

Answer: 2

Solution

Given $n_{\mathrm{H^+}} = \frac{400 \times 0.2}{1000} \times 2 = 0.16$ and $n_{\mathrm{OH^-}} = \frac{600 \times 0.1}{1000} = 0.06$. Now, heat liberated from reaction equals heat gained by solutions. Therefore, $0.06 \times 57.1 \times 10^3 = (1000 \times 1.0) \times 4.18 \times \Delta T$ Thus, $\Delta T = 0.8196 \, \mathrm{K}$ Therefore, $81.96 \times 10^{-2} \, \mathrm{K} \approx 82 \times 10^{-2} \, \mathrm{K}$

Question 83

Chemistry · Some Basic Concepts of Chemistry · Numerical

$2\mathrm{SO_2}(g)+\mathrm{O_2}(g)\rightarrow2\mathrm{SO_3}(g)$ The above reaction is carried out in a vessel starting with partial pressure $P_{\mathrm{SO_2}}=250\ \mathrm{mbar}$, $P_{\mathrm{O_2}}=750\ \mathrm{mbar}$ and $P_{\mathrm{SO_3}}=0\ \mathrm{bar}$. When the reaction is complete, the total pressure in the reaction vessel is $\mathrm{mbar}$. (Round off to the nearest integer).

Answer: 375

Solution

The reaction is given by: $$2\mathrm{SO_2} (\mathrm{g}) + \mathrm{O_2} (\mathrm{g}) \rightarrow 2\mathrm{SO_3} (\mathrm{g})$$ Initial pressures are 250 $\mathrm{\, mbar}$ for $\mathrm{SO_2}$ and 750 $\mathrm{\, mbar}$ for $\mathrm{O_2}$. Since $\mathrm{SO_2}$ is the limiting reactant (L. R.), the final pressures are: Final pressure of $\mathrm{SO_2}$ = -250 $\mathrm{\, mbar}$ Final pressure of $\mathrm{O_2}$ = -125 $\mathrm{\, mbar}$ Final pressure of $\mathrm{SO_3}$ = 250 $\mathrm{\, mbar}$ Therefore, the final total pressure is: $$625 + 250 = 875 \mathrm{\, mbar}$$

Question 84

Chemistry · Redox Reactions · Numerical

$10.0\,\mathrm{mL}$ of $0.05\,\mathrm{M}$ $\mathrm{KMnO_4}$ solution was consumed in a titration with $10.0\,\mathrm{mL}$ of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is \_\_\_\_ $\times 10^{-2}\,\mathrm{g/L}$. (Round off to the nearest integer)

Answer: 1575

Solution

Given $n_{eq} KMnO_4 = n_{eq} H_2C_2O_4 \cdot 2H_2O$. Or, $$\frac{10 \times 0.05}{1000} \times 5 = \frac{10 \times M}{1000} \times 2$$ Therefore, the concentration of oxalic acid solution is $0.125 M$. $$= 0.125 \times 126 \, g/L = 15.75 \, g/L$$ $$= 1575 \times 10^{-2} \, g/L$$

Question 85

Chemistry · Chemical Bonding and Molecular Structure · Numerical

The total number of electrons in all bonding molecular orbitals of $\mathrm{O}_2^{2-}$ is ........ (Round off to the nearest integer)

Answer: 10

Solution

M. O. Configuration of $\mathrm{O_2^{2-}}$ (18$\bar{e}$) $$\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2$$ $$\pi 2p_x^2 = \pi 2p_y^2$$ Total B.M.O electrons = 10

Question 86

Chemistry · Co-ordination Compounds · Numerical

Three moles of a metal complex with formula $\mathrm{Co(en)_2Cl_3}$ give $3$ moles of Silver Chloride on treatment with excess Silver Nitrate. The secondary valency of Co in the complex is $\underline{\qquad}$. (Round off to the nearest integer)

Answer: 6

Solution

The reaction is given by: $$3 \ [\mathrm{Co(en)_2Cl_2}] \ Cl + \mathrm{AgNO_3} \xrightarrow{excess} 3 \ \mathrm{AgCl} \ (white ppt)$$ The secondary valency of Co is 6. (C. N.)

Question 87

Chemistry · Solutions · Numerical

In a solvent 50$\%$ of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is $\times 10^{-2}$ (Round off to the nearest integer)

Answer: 125

Solution

The reaction is given by: $$2\mathrm{HA} \rightleftharpoons \mathrm{H_2} \mathrm{A_2HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A}$$ Initial moles are $a \times \frac{50}{100}$ for $\mathrm{HA}$, and $0$ for $\mathrm{H^+}$ and $\mathrm{A}$. Final moles are $0$ for $\mathrm{HA}$, $0.25a$ for $\mathrm{H^+}$, and $0.5a$ for $\mathrm{A}$. Now, $$i = \frac{final moles}{initial moles} = \frac{0.25a + 0.5a + 0.5a}{0.5a + 0.5a}$$ $$= 1.25 = 125 \times 10^{-2}$$

Question 88

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

The dihedral angle in staggered form of Newman projection of 1, 1, 1 -Trichloro ethane is ........ degree. (Round off to the nearest integer)

Answer: 60

Solution

1, 1, 1– Trichloro ethane $[\mathrm{CCl_3} - \mathrm{CH_3}]$ Dihedral angle $(\phi) = 60^\circ$ (Newmanns staggered form)

Question 89

Chemistry · Chemical Kinetics and Nuclear Chemistry · Fill in the blank

For the first order reaction $\mathrm{A\rightarrow 2B}$, $1$ mole of reactant $\mathrm{A}$ gives $0.2$ moles of $\mathrm{B}$ after $100$ minutes. The half-life of the reaction is $\underline{\hspace{1cm}}\,\mathrm{min}$. (Round off to the nearest integer). [Use: $\ln 2=0.69,\ \ln 10=2.3$. Properties of logarithms: $\ln x^y=y\ln x$, $\ln\left(\frac{x}{y}\right)=\ln x-\ln y$]

Answer: 300

Solution

A $\rightarrow 2 \, \mathrm{B}$ At $t = 0$, $1 \, mole$ of A and $0$ of B. At $t = 100 \, min$, $1 - x$ moles of A and $2x$ moles of B. $= 0.9 \, mol$ of A and $= 0.2 \, mol$ of B. Now, $t = \frac{t_{1/2}}{\ln 2} \times \frac{[A_0]}{[A_t]}$ $$100 = \frac{t_{1/2}}{\ln 2} \times \ln \frac{1}{0.9} \implies t_{1/2} = 690 \, min.$$ (taking $\ln 3 = 1.11$) Ans. 600 to 700

Question 90

Chemistry · Electrochemistry · Numerical

For the cell $\text{Cu(s)} \mid \text{Cu}^{2+}\text{(aq)(0.1M)} \| \text{Ag}^+\text{(aq) (0.01M)} \mid \text{Ag(s)}$ the cell potential $E_1 = 0.3095$ V For the cell $\text{Cu(s)} \mid \text{Cu}^{2+}\text{(aq) (0.01M)} \| \text{Ag}^+\text{(aq) (0.001M)} \mid \text{Ag(s)}$ the cell potential $= x \times 10^{-2}$ V. (Round off the Nearest Integer). $\left[\text{Use : } \dfrac{2.303RT}{F} = 0.059\right]$

Answer: 28

Solution

Cell reaction is: $$\mathrm{Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)}$$ Now, $E_{cell} = E^\circ_{Cell} - \frac{0.059}{2} \log \left( \frac{[\mathrm{Cu^{2+}}]}{[\mathrm{Ag^+}]^2} \right) \ldots (1)$ Therefore, $E_1 = 0.3095 = E^\circ_{Cell} - \frac{0.059}{2} \cdot \log \left( \frac{0.01}{(0.001)^2} \right) \ldots (2)$ From (1) and (2), $E_2 = 0.28 \, \mathrm{V} = 28 \times 10^{-2} \, \mathrm{V}$