JEE Main 20 July 2021 Shift 2 question paper with solutions

JEE Main 20 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.

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Maths

Question 1

Maths · Binomial Theorem · Single correct

For the natural numbers $m$, $n$, if $(1-y)^m (1+y)^n = 1 + a_1 y + a_2 y^2 + \ldots + a_{m+n} y^{m+n}$ and $a_1 = a_2 = 10$, then the value of $(m+n)$ is equal to:

  1. 88
  2. 64
  3. 100
  4. 80

Answer: (d)

Solution

Given $(1-y)^m(1+y)^n$. Coefficient of $y$ $(a_1) = 1 \cdot \binom{n}{1} + \binom{m}{1}(-1)$. Therefore, $n - m = 10 \ldots (1)$. Coefficient of $y^2$ $(a_2) = 1 \cdot \binom{n}{2} - \binom{m}{1} \cdot \binom{n}{1} + 1 \cdot \binom{m}{2} = 10$. This simplifies to $$\frac{n(n-1)}{2} - m \cdot n + \frac{m(m-1)}{2} = 10$$ $$m^2 + n^2 - 2mn - (n + m) = 20$$ $$(n - m)^2 - (n + m) = 20$$ $$n + m = 80 \ldots (2)$$ By equation (1) and (2), $m = 35$, $n = 45$.

Question 2

Maths · Inverse Trigonometric Functions · Single correct

The value of $\tan \left( 2 \tan^{-1} \left( \frac{3}{5} \right) + \sin^{-1} \left( \frac{5}{13} \right) \right)$ is equal to:

  1. $\frac{-181}{69}$
  2. $\frac{220}{21}$
  3. $\frac{-291}{76}$
  4. $\frac{151}{63}$

Answer: (b)

Solution

Given $\tan^{-1} \frac{3}{5} + \tan^{-1} \frac{3}{5} + \tan^{-1} \frac{5}{12}$ with $x > 0, y > 0, xy 0, y > 0, xy < 1$. $$\tan^{-1} \frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8} \cdot \frac{5}{12}} = \tan^{-1} \frac{220}{21}$$ Therefore, $$\tan \left( \tan^{-1} \frac{220}{21} \right) = \frac{220}{21}$$

Question 3

Maths · Conic Sections · Single correct

Let $r_1$ and $r_2$ be the radii of the largest and smallest circles, respectively, which pass through the point $(-4, 1)$ and having their centres on the circumference of the circle $x^2 + y^2 + 2x + 4y - 4 = 0$ If $\frac{r_1}{r_2} = a + b\sqrt{2}$, then $a + b$ is equal to:

  1. 3
  2. 11
  3. 5
  4. 7

Answer: (c)

Solution

Centre of smallest circle is A. Centre of largest circle is B. $r_2 = |CP - CA| = 3\sqrt{2} - 3$. $r_3 = CP + CB = 3\sqrt{2} + 3$. $$\frac{r_1}{r_2} = \frac{3\sqrt{2} + 3}{3\sqrt{2} - 3} = \frac{(3\sqrt{2} + 3)^2}{9} = (\sqrt{2} + 1)^2 = 3 + 2\sqrt{2}$$ $a = 3$, $b = 2$.

Question 4

Maths · Mathematical Reasoning · Single correct

Consider the following three statements: $(A)$ If $3 + 3 = 7$ then $4 + 3 = 8$. $(B)$ If $5 + 3 = 8$ then earth is flat. $(C)$ If both $(A)$ and $(B)$ are true then $5 + 6 = 17$. Then, which of the following statements is correct?

  1. $(A)$ is false, but $(B)$ and $(C)$ are true
  2. $(A)$ and $(C)$ are true while $(B)$ is false
  3. $(A)$ is true while $(B)$ and $(C)$ are false
  4. $(A)$ and $(B)$ are false while $(C)$ is true

Answer: (b)

Solution

Truth Table

Question 5

Maths · Three Dimensional Geometry · Single correct

The lines $x = ay - 1 = z - 2$ and $x = 3y - 2 = bz - 2$, $(ab \neq 0)$ are coplanar, if:

  1. $b = 1, a \in \mathbb{R} - \{0\}$
  2. $a = 1, b \in \mathbb{R} - \{0\}$
  3. $a = 2, b = 2$
  4. $a = 2, b = 3$

Answer: (a)

Solution

Given the equations: $\n$ $\frac{x+1}{a}$ = y = $\frac{z-1}{a}$ $\n$ $\frac{x+2}{3}$ = y = $\frac{z}{3/b}$ Lines are coplanar $\begin{vmatrix} a & 1 & -a\\ 3 & 1 & 3\\ -1 & 0 & -1 \end{vmatrix}=0$ $\Rightarrow a\left(\frac{-3}{1}-0\right)-1(a-3)=0$ $\Rightarrow -3a-a+3=0$ $\Rightarrow a=\frac{3}{4}\in\mathbb{R}$ $\therefore\ b=\frac{1}{a}=\frac{4}{3}\in\mathbb{R}\setminus\{0\}$

Question 6

Maths · Integrals · Single correct

If [x] denotes the greatest integer less than or equal to x, then the value of the integral $$\int_{-\pi/2}^{\pi/2} [x] - \sin x \, dx$$ is equal to :

  1. -$\pi$
  2. $\pi$
  3. 0
  4. 1

Answer: (a)

Question 7

Maths · Complex Numbers and Quadratic Equations · Single correct

If the real part of the complex number $$\left(1 - \cos \theta + 2i \sin \theta \right)^{-1}$$ is $\frac{1}{5}$ for $\theta \in (0, \pi)$, then the value of the integral $$\int_{0}^{\theta} \sin x \, dx$$ is equal to:

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

Answer: (a)

Solution

Given $$z = \frac{1}{1 - \cos \theta + 2i \sin \theta}$$ $$= \frac{2 \sin^2 \frac{\theta}{2} - 2i \sin \theta}{(1 - \cos \theta)^2 + 4 \sin^2 \frac{\theta}{2}}$$ $$= \frac{\sin \frac{\theta}{2} - i \cos \frac{\theta}{2}}{4 \sin \frac{\theta}{2} \left( \sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2} \right)}$$ The real part of $z$ is $$\mathrm{Re}(z) = \frac{1}{2 \left( \sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2} \right)} = \frac{1}{5}$$ $$\sin \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2} = \frac{5}{2}$$ $$1 - \cos^2 \frac{\theta}{2} + 4 \cos \frac{\theta}{2} = \frac{5}{2}$$ $$3 \cos^2 \frac{\theta}{2} = \frac{3}{2}$$ $$\cos^2 \frac{\theta}{2} = \frac{1}{2}$$ $$\frac{\theta}{2} = n \pi \pm \frac{\pi}{4}$$ $$\theta = 2n \pi \pm \frac{\pi}{2}$$ $$\theta = 2n \pi \pm \frac{\pi}{2}$$ $$\theta \in (0, \pi)$$ $$\theta = \frac{\pi}{2}$$ $$\int_0^{\frac{\pi}{2}} \sin \theta d\theta - \left[ -\cos \theta \right]_0^{\frac{\pi}{2}}$$ $$= -(0 - 1)$$ $$= 1$$

Question 8

Maths · Relations and Functions · Single correct

Let $f : \mathbb{R} - \left\{ \frac{\alpha}{6} \right\} \to \mathbb{R}$ be defined by $f(x) = \frac{5x+3}{6x-\alpha}$. Then the value of $\alpha$ for which $(f \circ f)(x) = x$, for all $x \in \mathbb{R} - \left\{ \frac{\alpha}{6} \right\}$, is

  1. No such $\alpha$ exists
  2. 5
  3. 8
  4. 6

Answer: (b)

Solution

Given $$f(x) = \frac{5x+3}{6x-\alpha} = y ....(1)$$ We have $$5x + 3 = 6xy - \alpha y$$ Rearranging gives $$x(6y - 5) = \alpha y + 3$$ Solving for $x$, we get $$x = \frac{\alpha y + 3}{6y - 5}$$ The inverse function is $$f^{-1}(x) = \frac{\alpha x + 3}{6x - 5} ....(2)$$ Since $f \circ f(x) = x$, we have $$f(x) = f^{-1}(x)$$ From equations (i) and (ii), it is clear that $\alpha = 5$.

Question 9

Maths · Limits and Derivatives · Single correct

If $f : \mathbb{R} \to \mathbb{R}$ is given by $f(x) = x + 1$, then the value of $$\lim_{n \to \infty} \frac{1}{n} \left[ f(0) + f\left( \frac{5}{n} \right) + f\left( \frac{10}{n} \right) + \ldots + f\left( \frac{5(n-1)}{n} \right) \right]$$ is:

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

Answer: (d)

Solution

I = $\sum$_{r=0}^{n-1} f$\left$($\frac{5r}{n}$$\right$) $\frac{1}{n}$ I = $\int$_0^1 f(5x) $\,$ dx I = $\int$_0^1 (5x + 1) $\,$ dx I = $\left$[ $\frac{5x^2}{2}$ + x $\right$]_0^1 I = $\frac{5}{2}$ + 1 = $\frac{7}{2}$

Question 10

Maths · Probability · Single correct

Let A, B and C be three events such that the probability that exactly one of A and B occurs is $(1-k)$, the probability that exactly one of B and C occurs is $(1-2k)$, the probability that exactly one of C and A occurs is $(1-k)$ and the probability of all A, B and C occur simultaneously is $k^2$, where $0 < k < 1$. Then the probability that at least one of A, B and C occur is:

  1. greater than $\frac{1}{8}$ but less than $\frac{1}{4}$
  2. greater than $\frac{1}{2}$
  3. greater than $\frac{1}{4}$ but less than $\frac{1}{2}$
  4. exactly equal to $\frac{1}{2}$

Answer: (b)

Solution

Given $$\overline{P(A \cap B)} + P(A \cap \overline{B}) = 1 - k$$ $$\overline{P(A \cap C)} + P(A \cap \overline{C}) = 1 - 2k$$ $$\overline{P(B \cap C)} + P(B \cap \overline{C}) = 1 - k$$ $$P(A \cap B \cap C) = k^2$$ We have $$P(A) + P(B) - 2P(A \cap B) = 1 - k$$ $$P(B) + P(C) - 2P(B \cap C) = 1 - k$$ $$P(C) + P(A) - 2P(A \cap C) = 1 - 2k$$ Adding (1), (2), and (3): $$P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = \frac{-4k + 3}{2}$$ So $$P(A \cup B \cup C) = \frac{-4k + 3}{2} + k^2$$ $$P(A \cup B \cup C) = \frac{2k^2 - 4k + 3}{2}$$ $$= \frac{2(k - 1)^2 + 1}{2}$$ Therefore, $$P(A \cup B \cup C) > \frac{1}{2}$$

Question 11

Maths · Applications of Derivatives · Single correct

The sum of all the local minimum values of the twice differentiable function $f : \mathbb{R} \to \mathbb{R}$ defined by $$f(x) = x^3 - 3x^2 - \frac{3f''(2)}{2}x + f''(1)$$ is :

  1. -22
  2. 5
  3. -27
  4. 0

Answer: (c)

Solution

Given $f(x) = x^3 - 3x^2 - \frac{3}{2} f'(2)x + f''(1)$ as equation (i). $f(x) = 3x^2 - 6x - \frac{3}{2} f''$ as equation (ii). $f''(x) = 6x - 6$ as equation (iii). Now using the 3rd equation $f''(2) = 12 - 6 = 6$. $f''(11 = 0)$ Use (ii) $f'(x) = 3x^2 - 6x - \frac{3}{2} f''(2)$ $f(x) = 3x^2 - 6x - \frac{3}{2} \times 6$ $f(x) = 3x^2 - 6x - 9$ $f(x) = 0$ $3x^2 - 6x - 9 = 0$ $\Rightarrow x = -1 \& 3$ Use (iii) $f''(x) = 6x - 6$ $f''(-1) = -12 0$ minima. Use (i) $f(x) = x^3 - 3x^2 - \frac{3}{2} f''(2)x + f''(1)$ $f(x) = x^3 - 3x^2 - \frac{3}{2} \times 6 \times x + 0$ $f(x) = x^3 - 3x^2 - 9x$ $f(3) = 27 - 27 - 9 \times 3 = -27$

Question 12

Maths · Properties of Triangles · Single correct

Let in a right angled triangle, the smallest angle be $\theta$. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then $\sin \theta$ is equal to:

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

Answer: (b)

Solution

$\angle A=\theta$ $\angle B=90^\circ-\theta$ $a=$ smallest side $c^2=a^2+b^2$ $\dfrac{1}{b^2}=\dfrac{1}{c^2}+\dfrac{1}{a^2}$ $\Rightarrow \dfrac{bc^2}{a^2}=b^2+c^2$ Use $a=2R\sin A=2R\sin\theta$ $b=2R\sin B=2R\sin(90^\circ-\theta)=2R\cos\theta$ $c=2R\sin C=2R\sin90^\circ=2R$ $4R^2\cos^2\theta=4R^2\sin^2\theta+4R^2$ $\Rightarrow \cos^2\theta=\sin^2\theta\cos^2\theta+\sin^2\theta$ $\Rightarrow 1-\sin^2\theta=\sin^2\theta(1-\sin^2\theta)+\sin^2\theta$ $\Rightarrow \sin^4\theta-3\sin^2\theta+1=0$ $\Rightarrow \sin^2\theta=\dfrac{3-\sqrt{5}}{2}$ $\Rightarrow \sin\theta=\dfrac{\sqrt{5}-1}{2}$

Question 13

Maths · Differential Equations · Single correct

Let $y = y(x)$ satisfies the equation $\frac{dy}{dx} - \left| A \right| = 0$, for all $x > 0$, where $$A = \begin{bmatrix} y & \sin x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x} \end{bmatrix}$$. If $y(\pi) = \pi + 2$, then the value of $y\left(\frac{\pi}{2}\right)$ is:

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

Answer: (a)

Solution

Given $|A| = -\frac{y}{x} + 2 \sin x + 2$. $$\frac{dy}{dx} = |A|$$ $$\frac{dy}{dx} = -\frac{y}{x} + 2 \sin x + 2$$ $$\frac{dy}{dx} + \frac{y}{x} = 2 \sin x + 2$$ I.F. $= e^{\int \frac{1}{x} \, dx} = x$ $$\Rightarrow \, yx = \int x(2 \sin x + 2) \, dx$$ $$xy = x^2 - 2x \cos x + 2 \sin x + c$$ Now $x = \pi$, $y = \pi + 2$ Use in (i) $c = 0$ Now (i) becomes $xy = x^2 - 2x \cos x + 2 \sin x$ Put $x = \pi/2$ $$\frac{\pi}{2} y = \left(\frac{\pi}{2}\right)^2 - 2 \cdot \frac{\pi}{2} \cos \frac{\pi}{2} + 2 \sin \frac{\pi}{2}$$ $$\frac{\pi}{2} y = \frac{\pi^2}{4} + 2$$

Question 14

Maths · Three Dimensional Geometry · Single correct

Consider the line $L$ given by the equation $$\frac{x-3}{2} = \frac{y-1}{1} = \frac{z-2}{1}.$$ Let $Q$ be the mirror image of the point $(2, 3, -1)$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$, and the line $L$ is perpendicular to $P$. Then which of the following points is on the plane $P$?

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

Answer: (d)

Solution

Plane $p$ is perpendicular to line $$\frac{x-3}{2} = \frac{y-1}{1} = \frac{z-2}{1}$$ and passes through point $(2, 3)$. The equation of plane $p$ is $$2(x-2) + 1(y-3) + 1(z+1) = 0$$ $$2x + y + z - 6 = 0$$ Point $(1, 2, 2)$ satisfies the above equation.

Question 15

Maths · Statistics · Single correct

If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and $\frac{20}{3}$, respectively, then the value of |a - b| is equal to:

  1. 9
  2. 11
  3. 7
  4. 1

Answer: (d)

Solution

Given $$10 = \frac{7 + 10 + 11 + 15 + a + b}{6}$$ This implies $a + b = 17$. $$\frac{20}{3} = \frac{7^2 + 10^2 + 11^2 + 15^2 + a^2 + b^2}{6} - 10^2$$ $$a^2 + b^2 = 145$$ Solve (i) and (ii) $a = 9$, $b = 8$ or $a = 8$, $b = 9$ $$|a - b| = 1$$

Question 16

Maths · Integrals · Single correct

Let $g(t) = \displaystyle\int_{-\pi/2}^{\pi/2} \cos\left(\frac{\pi}{4}t + f(x)\right)dx$, where \[ f(x) = \log_e\left(x+\sqrt{x^2+1}\right), \quad x \in \mathbb{R}. \] Then which one of the following is correct?

  1. \[ \text{(1)}\quad g(1) = g(0) \]
  2. \[ \text{(2)}\quad \sqrt{2}\,g(1) = g(0) \]
  3. \[ \text{(3)}\quad g(1) = \sqrt{2}\,g(0) \]
  4. \[ \text{(4)}\quad g(1) + g(0) = 0 \]

Answer: (b)

Solution

Given $$g(t) = \int_{-\pi/2}^{\pi/2} \left( \cos \frac{\pi}{4} t + f(x) \right) dx$$ We have $$g(t) = \pi \cos \frac{\pi}{4} t + \int_{-\pi/2}^{\pi/2} f(x) dx$$ Thus, $$g(t) = \pi \cos \frac{\pi}{4} t$$ Finally, $$g(1) = \frac{\pi}{\sqrt{2}}, g(0) = \pi$$

Question 17

Maths · Conic Sections · Single correct

Let P be a variable point on the parabola $y = 4x^2 + 1$. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line $y = x$ is:

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

Answer: (b)

Solution

Given $\dfrac{K-C}{h-C}=-1$ $\Rightarrow C=\dfrac{h+K}{2}$ $R=\left(\dfrac{x+C}{2},\dfrac{y+C}{2}\right)$ $R=\left(\dfrac{x}{2}+\dfrac{h}{4}+\dfrac{K}{4},\dfrac{y}{2}+\dfrac{h}{4}+\dfrac{K}{4}\right)$ $h=\dfrac{x}{2}+\dfrac{h}{4}+\dfrac{K}{4}$ $K=\dfrac{y}{2}+\dfrac{h}{4}+\dfrac{K}{4}$ $\Rightarrow x=\dfrac{3h}{2}-\dfrac{K}{2},\qquad y=\dfrac{3K}{2}-\dfrac{h}{2}$ $y=4x^2+1$ $\left(\dfrac{3K-h}{2}\right)=4\left(\dfrac{3h-K}{2}\right)^2+1$

Question 18

Maths · Determinants · Single correct

The value of $k \in \mathbb{R}$, for which the following system of linear equations $$3x - y + 4z = 3$$ $$x + 2y - 3z = -2$$ $$6x + 5y + kz = -3$$ has infinitely many solutions, is:

  1. 3
  2. -5
  3. 5
  4. -3

Answer: (b)

Solution

Given the determinant equation: $$\begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & K \end{vmatrix} = 0$$ Expanding along the first row, we have: $$3(2K + 15) + K + 18 - 28 = 0$$ Simplifying gives: $$7K + 35 = 0$$ Solving for $K$, we find: $$K = -5$$

Question 19

Maths · Sequences and Series · Single correct

If sum of the first 21 terms of the series $\log_{9^{1/2}} x + \log_{9^{1/3}} x + \log_{9^{1/4}} x + \ldots$, where $x > 0$ is 504, then $x$ is equal to

  1. 243
  2. 9
  3. 7
  4. 81

Answer: (d)

Solution

Given $s = 2 \log_9 x + 3 \log_9 x + \ldots + 22 \log_9 x$. $s = \log_9 x (2 + 3 + \ldots + 22)$. $s = \log_9 x \left\{ \frac{21}{2} (2 + 22) \right\}$. Given $252 \log_9 x = 504$. $\Rightarrow \log_9 x = 2 \Rightarrow x = 81$.

Question 20

Maths · Vector Algebra · Single correct

In a triangle ABC, if $|\overrightarrow{BC}| = 3$, $|\overrightarrow{CA}| = 5$ and $|\overrightarrow{BA}| = 7$, then the projection of the vector $\overrightarrow{BA}$ on $\overrightarrow{BC}$ is equal to

  1. $\frac{19}{2}$
  2. $\frac{13}{2}$
  3. $\frac{11}{2}$
  4. $\frac{15}{2}$

Answer: (c)

Solution

Projection of $\overrightarrow{BA}$ on $\overrightarrow{BC}$ is equal to $$= |\overrightarrow{BA}| \cos \angle ABC$$ $$= 7 \left| \frac{7^2 + 3^2 - 5^2}{2 \times 7 \times 3} \right| = \frac{11}{2}$$

Question 21

Maths · Matrices · Numerical

Let $A = \{ a_{ij} \}$ be a $3 \times 3$ matrix, where $$a_{ij} = \begin{cases} (-1)^{j-i} & \text{if } i j \end{cases}$$ then $\det(3 \text{ Adj}(2A^{-1}))$ is equal to

Answer: 108

Solution

Given $$A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$$ The determinant is $$|A| = 4$$ We have $$|3 \operatorname{adj}(2A^{-1})| = |3 \cdot 2^2 \operatorname{adj}(A^{-1})|$$ This simplifies to $$= 12^3 \left| \operatorname{adj}(A^{-1}) \right| = 12^3 |A^{-1}|^2 = \frac{12^3}{|A|^2} = \frac{12^3}{16} = 108$$

Question 22

Maths · Relations and Functions · Numerical

The number of solutions of the equation $$\log_{(x+1)}\left(2x^2 + 7x + 5\right) + \log_{(2x+5)}\left(x+1\right)^2 - 4 = 0$$ $x > 0$, is

Answer: 1

Solution

Given the equation $$\log_{(x+1)}(2x^2 + 7x + 5) + \log_{(2x+5)}(x+1)^2 - 4 = 0$$ we simplify to $$\log_{(x+1)}(2x+5)(x+1) + 2\log_{(2x+5)}(x+1) = 4$$ which further simplifies to $$\log_{(x+1)}(2x+5) + 1 + 2\log_{(2x+5)}(x+1) = 4$$ Let $$\log_{(x+1)}(2x+5) = t$$ Then $$t + \frac{2}{t} = 3 \Rightarrow t^2 - 3t + 2 = 0$$ Solving gives $$t = 1, 2$$ For $$\log_{(x+1)}(2x+5) = 1$$ and $$\log_{(x+1)}(2x+5) = 2$$ we have $$x + 1 = 2x + 3$$ and $$2x + 5 = (x + 1)^2$$ Solving these gives $$x = -4 (rejected)$$ and $$x^2 = 4 \Rightarrow x = 2, -2 (rejected)$$ Therefore, $$x = 2$$ The number of solutions is 1.

Question 23

Maths · Differential Equations · Numerical

Let a curve $y = y(x)$ be given by the solution of the differential equation $$\cos\left(\frac{1}{2}\cos^{-1}(e^{-x})\right) dx = \sqrt{e^{2x} - 1} dy$$ If it intersects $y$-axis at $y = -1$, and the intersection point of the curve with $x$-axis is $(\alpha, 0)$, then $e^{\alpha}$ is equal to

Answer: 2

Solution

Given $\cos \left( \frac{1}{2} \cos^{-1}(e^{-x}) \right) dx = \sqrt{e^{2x} - 1} dy$. Put $\cos^{-1}(e^{-x}) = \theta$, $\theta \in [0, \pi]$. $\cos \theta = e^{-x} \Rightarrow 2 \cos^2 \frac{\theta}{2} - 1 = e^{-x}$. $\cos \frac{\theta}{2} = \sqrt{\frac{e^{-x} + 1}{2}} = \sqrt{\frac{e^{x} + 1}{2e^{x}}}$. $\sqrt{\frac{e^{x} + 1}{2e^{x}}} dx = \sqrt{e^{2x} - 1} dy$. $\frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{e^{x} - 1}} = \int dy$. Put $e^{x} = t$, $\frac{dt}{dx} = e^{x}$. $\frac{1}{\sqrt{2}} \int \frac{dt}{e^{x} \sqrt{e^{x} - 1}} = \int dy$. $\int \frac{dt}{t \sqrt{t - 1}} = \sqrt{2}y$. Put $t = \frac{1}{z}$, $dt = \frac{1}{z^2} dz$. $\int \frac{-dz}{z^2 \sqrt{1/z - 1}} = \sqrt{2}y$. $- \int \frac{dz}{\sqrt{1 - z}} = \sqrt{2}y$. $-2(1 - z)^{1/2} = \sqrt{2}y + c$. $2 \left( 1 - \frac{1}{t} \right)^{1/2} = \sqrt{2}y + c$. $2(1 - e^{-x})^{1/2} = \sqrt{2}y + c \xrightarrow{(0,-1)} \Rightarrow c = \sqrt{2}$. $2(1 - e^{-x})^{1/2} = \sqrt{2}(y + 1)$, passes through $(\alpha, 0)$. $2(1 - e^{-\alpha})^{1/2} = \sqrt{2}$. $\sqrt{1 - e^{-\alpha} = \frac{1}{\sqrt{2}} \Rightarrow 1 - e^{-\alpha} = \frac{1}{2}$. $e^{-\alpha} = \frac{1}{2} \Rightarrow e^{\alpha} = 2$.

Question 24

Maths · Vector Algebra · Numerical

For $p > 0$, a vector $\vec{v}_2 = 2\hat{i} + (p + 1)\hat{j}$ is obtained by rotating the vector $\vec{v}_1 = \sqrt{3}p\hat{i} + \hat{j}$ by an angle $\theta$ about origin in counter clockwise direction. If $\tan \theta = \frac{(\alpha \sqrt{3} - 2)}{(4\sqrt{3} + 3)}$, then the value of $\alpha$ is equal to

Answer: 6

Solution

Given $\vec{V}_2 = 2i + (P+1)j$ and $\vec{V}_1 = \sqrt{3}Pi + j$. The magnitudes are equal: $$|\vec{V}_1| = |\vec{V}_2|$$ This gives the equation: $$3P^2 + 1 = 4 + (P+1)^2$$ Simplifying, we have: $$2P^2 - 2P - 4 = 0 \Rightarrow P^2 - P - 2 = 0$$ Solving for $P$, we get $P = 2, -1$. The value $-1$ is rejected. Now, calculate $\cos \theta$: $$\cos \theta = \frac{\vec{V}_1 \cdot \vec{V}_2}{|\vec{V}_1||\vec{V}_2|} = \frac{2\sqrt{3}P + (P+1)}{(P+1)^2 + 4\sqrt{3}P^2 + 1}$$ Substituting $P = 2$, we find: $$\cos \theta = \frac{4\sqrt{3} + 3}{\sqrt{13 \sqrt{13}}} = \frac{4\sqrt{3} + 3}{13}$$ For $\tan \theta$: $$\tan \theta = \frac{\sqrt{112 - 24\sqrt{3}}}{4\sqrt{3} + 3} = \frac{6\sqrt{3} - 2}{4\sqrt{3} + 3} = \frac{\alpha \sqrt{3} - 2}{4\sqrt{3} + 3}$$ Thus, $\alpha = 6$.

Question 25

Maths · Straight Lines and Pair of Straight Lines · Numerical

Consider a triangle having vertices A(-2, 3), B(1, 9) and C(3, 8). If a line L passing through the circum-centre of triangle ABC, bisects line BC, and intersects y-axis at point $\left(0, \frac{\alpha}{2}\right)$, then the value of real number $\alpha$ is

Answer: 9

Solution

Given the triangle with vertices $A(-2, 3)$, $B(1, 9)$, and $C(3, 8)$, we have: $$(\sqrt{50})^2 = (\sqrt{45})^2 + (\sqrt{5})^2$$ This implies $\angle B = 90^\circ$. The circumcenter is given by: $$\left( \frac{1}{2}, \frac{11}{2} \right)$$ The midpoint of $BC$ is: $$\left( 2, \frac{17}{2} \right)$$ The equation of the line is: $$\left( y - \frac{11}{2} \right) = 2 \left( x - \frac{1}{2} \right) \Rightarrow y = 2x + \frac{9}{2}$$ This line passes through: $$\left( 0, \frac{\alpha}{2} \right)$$ Solving for $\alpha$: $$\frac{\alpha}{2} = \frac{9}{2} \Rightarrow \alpha = 9$$

Question 26

Maths · Applications of Derivatives · Numerical

If the point on the curve $y^2 = 6x$, nearest to the point $\left(3, \frac{3}{2}\right)$ is $(\alpha, \beta)$, then $2(\alpha + \beta)$ is equal to

Answer: 9

Solution

Given $P \equiv \left( \frac{3}{2} t^2, 3t \right)$. Normal at point $P$ is $tx + y = 3t + \frac{3}{2} t^3$. Passes through $\left( 3, \frac{3}{2} \right)$. Therefore, $$3t + \frac{3}{2} = 3t + \frac{3}{2} t^3$$ $$P \equiv \left( \frac{3}{2}, 3 \right) = (\alpha, \beta)$$ $$\Rightarrow t^3 = 1 \Rightarrow t = 1$$ $$2(\alpha + \beta) = 2 \left( \frac{3}{2} + 3 \right) = 9$$

Question 27

Maths · Continuity and Differentiability · Numerical

Let a function $g:[0,4]\to\mathbb{R}$ be defined as \[ g(x)= \begin{cases} \displaystyle \max_{0\leq t\leq x}\{t^3-6t^2+9t-3\}, & 0\leq x\leq 3,\\ 4-x, & 3<x\leq 4. \end{cases} \] Then the number of points in the interval $(0,4)$ where $g(x)$ is NOT differentiable is ________.

Answer: 1

Solution

Given $f(x) = x^3 - 6x^2 + 9x - 3$. The derivative is $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$. We have $f(1) = 1$ and $f(3) = -3$. For $g(x)$: $$g(x) = \begin{cases} f(x) & 0 \leq x \leq 1 \\ 0 & 1 < x \leq 3 \\ -1 & 3 < x \leq 4 \end{cases}$$ $g(x)$ is continuous. The derivative $g'(x)$ is: $$g'(x) = \begin{cases} 3(x-1)(x-3) & 0 \leq x \leq 1 \\ 0 & 1 < x \leq 3 \\ -1 & 3 < x \leq 4 \end{cases}$$ $g(x)$ is non-differentiable at $x = 3$.

Question 28

Maths · Sequences and Series · Numerical

For $k \in \mathbb{N}$, let $$\frac{1}{\alpha(\alpha+1)(\alpha+2)\ldots(\alpha+20)} = \sum_{K=0}^{20} \frac{A_k}{\alpha+k},$$ where $\alpha > 0$. Then the value of $100 \left( \frac{A_{14} + A_{15}}{A_{13}} \right)^2$ is equal to

Answer: 9

Solution

Given $\($ $\frac{1}{\alpha(\alpha+1)\ldots(\alpha+20)}$ = $\sum$_{k=0}^{20} $\frac{A_k}{\alpha+k}$ $\)$. $\($ A_{14} = $\frac{1}{(-14)(-13)\ldots(-1)(1)\ldots(6)}$ = $\frac{1}{14! \cdot 6!}$ $\)$ $\($ A_{15} = $\frac{1}{(-15)(-14)\ldots(-1)(1)\ldots(5)}$ = $\frac{1}{15! \cdot 5!}$ $\)$ $\($ A_{13} = $\frac{1}{(-13)\ldots(-1)(1)\ldots(7)}$ = $\frac{-1}{13! \cdot 7!}$ $\)$ $\($ $\frac{A_{14}}{A_{13}}$ = $\frac{1}{14! \cdot 6!}$ $\times$ $\frac{-13 \times 7!}{-1}$ = $\frac{-7}{14}$ = $\frac{-1}{2}$ $\)$ $\($ $\frac{A_{15}}{A_{13}}$ = $\frac{-1}{15! \cdot 5!}$ $\times$ $\frac{-13 \times 7!}{-1}$ = $\frac{42}{15 \times 14}$ = $\frac{1}{5}$ $\)$ $\($ 100 $\left$( $\frac{A_{14}}{A_{13}}$ + $\frac{A_{15}}{A_{13}}$ $\right$)^2 = 100 $\left$( $\frac{-1}{2}$ + $\frac{1}{5}$ $\right$)^2 = 9 $\)$

Question 29

Maths · Sequences and Series · Numerical

Let $\{a_n\}_{n=1}^{\infty}$ be a sequence such that $a_1 = 1$, $a_2 = 1$ and $a_{n+2} = 2a_{n+1} + a_n$ for all $n \geq 1$. Then the value of $47 \sum_{n=1}^{\infty} \frac{a_n}{2^{3n}}$ is equal to:

Answer: 7

Solution

Given $a_{n+2} = 2a_{n+1} + a_n$, let $\sum_{n=1}^{\infty} \frac{a_n}{8^n} = P$. Divide by $8^n$ we get $$\frac{a_{n+2}}{8^n} = \frac{2a_{n+1}}{8^n} + \frac{a_n}{8^n}$$ $$\Rightarrow 64 \frac{a_{n+2}}{8^{n+2}} = \frac{16a_{n+1}}{8^{n+1}} + \frac{a_n}{8^n}$$ $$64 \sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}} = 16 \sum_{n=1}^{\infty} \frac{a_{n+1}}{8^{n+1}} + \sum_{n=1}^{\infty} \frac{a_n}{8^n}$$ $$64 \left( P - \frac{a_1}{8} - \frac{a_2}{8^2} \right) = 16 \left( P - \frac{a_1}{8} \right) + P$$ $$\Rightarrow 64 \left( P - \frac{1}{8} - \frac{1}{64} \right) = 16 \left( P - \frac{1}{8} \right) + P$$ $$64P - 8 - 1 = 16P - 2 + P$$ $$47P = 7$$

Question 30

Maths · Limits and Derivatives · Numerical

If $\lim_{x \to 0} \frac{\alpha x e^{-x} - \beta \log_e (1+x) + \gamma x^2 e^{-x}}{x \sin^2 x} = 10$, $\alpha, \beta, \gamma \in \mathbb{R}$, then the value of $\alpha + \beta + \gamma$ is

Answer: 3

Solution

For limit to exist $$\alpha - \beta = 0, \alpha + \frac{\beta}{2} + \gamma = 0$$ $$\frac{\alpha}{2} - \frac{\beta}{3} - \gamma = 10$$ $$\beta = \alpha, \gamma = -\frac{3\alpha}{2}$$ Put in (i) $$\frac{\alpha}{2} - \frac{\alpha}{3} + \frac{3\alpha}{2} = 10$$ $$\frac{\alpha}{6} + \frac{3\alpha}{2} = 10 \Rightarrow \frac{\alpha + 9\alpha}{6} = 10$$ $$\Rightarrow \alpha = 6$$ $$\alpha = 6, \beta = 6, \gamma = -9$$ $$\alpha + \beta + \gamma = 3$$

Physics

Question 31

Physics · Work, Energy and Power · Single correct

If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :

  1. 100$\%$
  2. 200$\%$
  3. 300$\%$
  4. 400$\%$

Answer: (a)

Solution

Given $K_2 = 4K_1$. $$\frac{1}{2}mv_2^2 = 4 \times \frac{1}{2}mv_1^2$$ Thus, $v_2 = 2v_1$. Momentum $P = mv$. Therefore, $P_2 = mv_2 = 2mv_1$. And $P_1 = mv_1$. The percentage change is given by $$\frac{\Delta P}{P_1} \times 100 = \frac{2mv_1 - mv_1}{mv_1} \times 100 = 100\%$$

Question 32

Physics · Motion in a Plane · Single correct

A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time $t_1$. If he remains stationary on a moving escalator then the escalator takes him up in time $t_2$. The time taken by him to walk up on the moving escalator will be:

  1. $\frac{t_1 t_2}{t_2 - t_1}$
  2. $\frac{t_1 + t_2}{2}$
  3. $\frac{t_1 t_2}{t_2 + t_1}$
  4. $t_2 - t_1$

Answer: (c)

Solution

L = Length of escalator $V_{b/esc} = \frac{L}{t_1}$ When only escalator is moving. $V_{esc} = \frac{L}{t_2}$ when both are moving $$V_{b/g} = V_{b/esc} + V_{esc}$$ $$V_{b/g} = \frac{L}{t_1} + \frac{L}{t_2} \implies \left[ t = \frac{L}{V_{b/g}} = \frac{t_1 t_2}{t_1 + t_2} \right]$$

Question 33

Physics · Gravitation · Single correct

A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02R. The percentage difference in the time periods of the two satellites is:

  1. 1.5
  2. 2.0
  3. 0.7
  4. 3.0

Answer: (d)

Solution

Given $T^2 \propto R^3$. Therefore, $T = kR^{3/2}$. Differentiating, we have $$\frac{dT}{T} = \frac{3}{2} \frac{dR}{R}$$ $$= \frac{3}{2} \times 0.02 = 0.03$$ The percentage change is $3\%$.

Question 34

Physics · Waves · Single correct

With what speed should a galaxy move outward with respect to earth so that the sodium-D line at wavelength 5890$\AA$ is observed at 5896$\AA$?

  1. 306 km/sec
  2. 322 km/sec
  3. 296 km/sec
  4. 336 km/sec

Answer: (a)

Solution

Given $f = f_0 \sqrt{\frac{1+\beta}{1-\beta}}$ and $\beta = \frac{v}{c}$. $$\frac{f}{f_0} \sqrt{\frac{1+\beta}{1-\beta}}$$ $$\left(1 + \frac{\Delta f}{f_0}\right)^2 = (1 + \beta)(1 - \beta)^{-1}$$ $\beta$ is small compared to 1. $$\left(1 + \frac{2\Delta f}{f_0}\right) = (1 + 2\beta)$$ $$\beta = \frac{\Delta f}{f_0} = \frac{v}{c}$$ $$v = 6 \times \frac{c}{5890} = 305.6 \, \mathrm{km/s}$$

Question 35

Physics · Mechanical Properties of Solids · Single correct

The length of a metal wire is $\ell_1$, when the tension in it is $T_1$ and is $\ell_2$ when the tension is $T_2$. The natural length of the wire is:

  1. $\sqrt{\ell_1 \ell_2}$
  2. $\frac{\ell_1 \, T_2 - \ell_2 \, T_1}{T_2 - T_1}$
  3. $\frac{\ell_1 \, T_2 + \ell_2 \, T_1}{T_2 + T_1}$
  4. $\frac{\ell_1 + \ell_2}{2}$

Answer: (b)

Solution

Given $$T_1 = k (\ell_1 - \ell_0)$$ $$T_2 = k (\ell_2 - \ell_0)$$ The ratio is $$\frac{T_1}{T_2} = \frac{\ell_1 - \ell_0}{\ell_2 - \ell_0}$$ Solving for $\ell_0$, we have $$\frac{T_1 \ell_2 - T_2 \ell_1}{T_1 - T_2} = \ell_0$$

Question 36

Physics · Electromagnetic Waves · Single correct

In an electromagnetic wave the electric field vector and magnetic field vector are given as $\vec{E} = E_0 \hat{i}$ and $\vec{B} = B_0 \hat{k}$ respectively. The direction of propagation of electromagnetic wave is along:

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

Answer: (d)

Solution

Direction of propagation = $\vec{\mathbf{E}} \times \vec{\mathbf{B}} = \hat{\mathbf{i}} \times \hat{\mathbf{k}} = -\hat{\mathbf{j}}$

Question 37

Physics · Alternating Current · Single correct

For a series LCR circuit with $R = 100\Omega$, $L = 0.5\,\mathrm{mH}$ and $C = 0.1\,\mathrm{pF}$ connected across $220\,\mathrm{V} - 50\,\mathrm{Hz}$ AC supply, the phase angle between current and supplied voltage and the nature of the circuit is:

  1. $0^\circ$, resistive circuit
  2. $\approx 90^\circ$, predominantly inductive circuit
  3. $0^\circ$, resonance circuit
  4. $\approx 90^\circ$, predominantly capacitive circuit

Answer: (d)

Solution

Given $R = 100 \, \Omega$. $X_L = \omega L = 50 \pi \times 10^{-3}$. $X_C = \frac{1}{\omega C} = \frac{10^{11}}{100 \pi}$. $X_C \gg X_L$. $|X_C - X_L| \gg R$.

Question 38

Physics · Thermodynamics · Single correct

Which of the following graphs represent the behavior of an ideal gas ? Symbols have their usual meaning.

Answer: (c)

Solution

Given the equation $PV = nRT$, it follows that $PV \propto T$. This represents a straight line with a positive slope (nR).

Question 39

Physics · Oscillations · Single correct

A particle is making simple harmonic motion along the X-axis. If at a distances $x_1$ and $x_2$ from the mean position the velocities of the particle are $v_1$ and $v_2$ respectively. The time period of its oscillation is given as:

  1. $T = 2\pi \sqrt{\frac{x_2^2 + x_1^2}{v_1^2 - v_2^2}}$
  2. $T = 2\pi \sqrt{\frac{x_2^2 + x_1^2}{v_1^2 + v_2^2}}$
  3. $T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 + v_2^2}}$
  4. $T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}$

Answer: (d)

Solution

Given $v^2 = \omega^2 \left(A^2 - x^2\right)$. $A^2 = x_1^2 + \frac{v_1^2}{\omega^2} = x_2^2 + \frac{v_2^2}{\omega^2}$. $\omega^2 = \frac{v_2^2 - v_1^2}{x_1^2 - x_2^2}$. $T = 2\pi \sqrt{\frac{x_1^2 - x_2^2}{v_2^2 - v_1^2}}$.

Question 40

Physics · Dual Nature of Radiation and Matter · Single correct

An electron having de-Broglie wavelength $\lambda$ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is:

  1. 0
  2. $\frac{2 m^2 c^2 \lambda^2}{h^2}$
  3. $\frac{2mc\lambda^2}{h}$
  4. $\frac{hc}{mc}$

Answer: (c)

Solution

Given $\lambda = \frac{h}{mv}$. The kinetic energy is given by $$\frac{p^2}{2m} = \frac{h^2}{2m\lambda^2} = \frac{hc}{\lambda_c}.$$ Therefore, $$\lambda_c = \frac{2m\lambda^2 c}{h}.$$

Question 41

Physics · System of Particles and Rotational Motion · Single correct

A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50$\%$ of its translational kinetic energy. The body is :

  1. Solid sphere
  2. Solid cylinder
  3. Hollow cylinder
  4. Ring

Answer: (b)

Solution

Given $$\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{1}{2} mv^2$$ $$I = \frac{1}{2} mR^2$$ Body is solid cylinder

Question 42

Physics · Physical World, Units and Measurements · Single correct

If time $(t)$, velocity $(v)$, and angular momentum $(l)$ are taken as the fundamental units. Then the dimension of mass $(m)$ in terms of $t$, $v$ and $l$ is:

  1. $[t^{-1}v^{1}l^{-2}]$
  2. $[t^{1}v^{2}l^{-1}]$
  3. $[t^{-2}v^{-1}l^{1}]$
  4. $[t^{-1}v^{-2}l^{1}]$

Answer: (d)

Solution

Given $m \propto t^a v^b \ell^c$. $$m \propto [T]^a \left[ LT^{-1} \right]^b \left[ ML^2 T^{-1} \right]^c$$ $$\mathrm{M}^1 \mathrm{L}^0 \mathrm{T}^0 = \mathrm{M}^c \mathrm{L}^{b+2c} \mathrm{T}^{a-b-c}$$ Comparing powers $c = 1$, $b = -2$, $a = -1$. $$m \propto t^{-1} V^{-2} \ell^1$$

Question 43

Physics · Kinetic Theory · Single correct

The correct relation between the degrees of freedom $f$ and the ratio of specific heat $\gamma$ is:

  1. $f = \frac{2}{\gamma - 1}$
  2. $f = \frac{2}{\gamma + 1}$
  3. $f = \frac{\gamma + 1}{2}$
  4. $f = \frac{1}{\gamma + 1}$

Answer: (a)

Solution

Given $\($ $\gamma$ = 1 + $\frac{2}{f}$ $\)$. Solving for $\($ f $\)$, we have: $$ f = \frac{2}{\gamma - 1} $$

Question 44

Physics · Nuclei · Single correct

For a certain radioactive process the graph between $\ln R$ and $t(\mathrm{sec})$ is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately:

  1. 9.15$\mathrm{sec}$
  2. 6.93$\mathrm{sec}$
  3. 2.62$\mathrm{sec}$
  4. 4.62$\mathrm{sec}$

Answer: (d)

Solution

Given $R = R_0 e^{-\lambda t}$. Taking the natural logarithm, we have $\ln R = \ln R_0 - \lambda t$. The negative of $\lambda$ is the slope of the straight line. Given $\lambda = \frac{3}{20}$, we find $t_{1/2} = \frac{\ln 2}{\lambda} = 4.62$.

Question 45

Physics · Gravitation · Single correct

Consider a binary star system of star A and star B with masses $m_A$ and $m_B$ revolving in a circular orbit of radii $r_A$ and $r_B$, respectively. If $T_A$ and $T_B$ are the time period of star A and star B, respectively, then:

  1. $\frac{T_A}{T_B} = \left(\frac{r_A}{r_B}\right)^{\frac{3}{2}}$
  2. $T_A = T_B$
  3. $T_A > T_B$ (if $m_A > m_B$)
  4. $T_A > T_B$ (if $r_A > r_B$)

Answer: (b)

Solution

Given $T_\mathrm{A} = T_\mathrm{B}$ since $\omega_\mathrm{A} = \omega_\mathrm{B}$.

Question 46

Physics · Magnetism and Matter · Single correct

At an angle of $30^\circ$ to the magnetic meridian, the apparent dip is $45^\circ$. Find the true dip:

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

Answer: (d)

Solution

Given $A \tan \delta = \tan \delta' \cos \theta$. This equals $\tan 45^\circ \cos 30^\circ$. Therefore, $\tan \delta = 1 \times \frac{\sqrt{3}}{2}$. Thus, $\delta = \tan^{-1} \left( \frac{\sqrt{3}}{2} \right)$.

Question 47

Physics · Motion in a Straight Line · Single correct

A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time ' $t$ ' is proportional to:

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

Answer: (a)

Solution

P = constant $$\frac{1}{2}mv^2 = Pt$$ Therefore, $v \propto \sqrt{t}$ $$\frac{dx}{dt} = C\sqrt{t} C = constant$$ by integration. $$x \propto t^{3/2}$$

Question 48

Physics · Mathematics in Physics · Single correct

Two vectors $\vec{P}$ and $\vec{Q}$ have equal magnitudes. If the magnitude of $\vec{P} + \vec{Q}$ is $n$ times the magnitude of $\vec{P} - \vec{Q}$, then angle between $\vec{P}$ and $\vec{Q}$ is:

  1. $\sin^{-1}\left(\frac{n-1}{n+1}\right)$
  2. $\cos^{-1}\left(\frac{n-1}{n+1}\right)$
  3. $\sin^{-1}\left(\frac{n^2-1}{n^2+1}\right)$
  4. $\cos^{-1}\left(\frac{n^2-1}{n^2+1}\right)$

Answer: (d)

Solution

Given $|\vec{P}| = |\vec{Q}| = x$ ...(i) $|\vec{P} + \vec{Q}| = n|\vec{P} - \vec{Q}|$ $$P^2 + Q^2 + 2PQ \cos \theta = n^2 \left( P^2 + Q^2 - 2PQ \cos \theta \right)$$ Using (i) in above equation $$\cos \theta = \frac{n^2 - 1}{1 + n^2}$$ $$\theta = \cos^{-1} \left( \frac{n^2 - 1}{n^2 + 1} \right)$$

Question 49

Physics · Mechanical Properties of Fluids · Single correct

Two small drops of mercury each of radius R coalesce to form a single large drop. The ratio of total surface energy before and after the change is:

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

Answer: (a)

Solution

The equation for the volumes is given by: $$\frac{4}{3} \pi R^3 + \frac{4}{3} \pi R^3 = \frac{4}{3} \pi R'^3$$ Solving for $R'$ gives: $$R' = 2^{\frac{1}{3}} R \ldots (i)$$ The initial surface area $A_i$ is: $$A_i = 2 \left[ 4 \pi R^2 \right]$$ The final surface area $A_f$ is: $$A_f = 4 \pi R'^2$$ The ratio of initial to final surface area is: $$\frac{U_i}{U_f} = \frac{A_i}{A_f} = \frac{2R^2}{2^{2/3} R^2} = 2^{1/3}$$

Question 50

Physics · Magnetism and Matter · Single correct

The magnetic susceptibility of a material of a rod is 499. Permeability in vacuum is $4\pi \times 10^{-7} \, \mathrm{H/m}$. Absolute permeability of the material of the rod is:

  1. $4\pi \times 10^{-4} \, \mathrm{H/m}$
  2. $2\pi \times 10^{-4} \, \mathrm{H/m}$
  3. $3\pi \times 10^{-4} \, \mathrm{H/m}$
  4. $\pi \times 10^{-4} \, \mathrm{H/m}$

Answer: (b)

Solution

Given $\mu = \mu_0 (1 + x_m)$. $$= 4\pi \times 10^{-7} \times 500$$ $$= 2\pi \times 10^{-4} \, \mathrm{H/m}$$

Question 51

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

A zener diode having zener voltage 8 V and power dissipation rating of 0.5 W is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance $R_P$ is $\ldots$$\ldots$$\ldots$$\ldots$ $\Omega$.

Answer: 192

Solution

$P = Vi$ $0.5 = 8i$ $i = \dfrac{1}{16}\,\text{A}$ $E = 20 = 8 + iR_P$ $R_P = 12 \times 16 = 192\,\Omega$

Question 52

Physics · Laws of Motion · Numerical

A body of mass 'm' is launched up on a rough inclined plane making an angle of 30^$\circ$ with the horizontal. The coefficient of friction between the body and plane is $\frac{\sqrt{x}}{5}$ if the time of ascent is half of the time of descent. The value of x is

Answer: 3

Solution

Given $t_a = \frac{1}{2} t_d$ and $\sqrt{\frac{2s}{a_a}} = \frac{1}{2} \sqrt{\frac{2s}{a_d}} \ldots (i)$. $a_a = g \sin \theta + \mu g \cos \theta$ $= \frac{g}{2} + \frac{\sqrt{3}}{2} \mu g$ $a_d = g \sin \theta - \mu g \cos \theta$ $= \frac{g}{2} - \frac{\sqrt{3}}{2} \mu g$ Using the above values of $a_a$ and $a_d$ and putting in (i) we will get $\mu = \frac{\sqrt{3}}{5}$ equation.

Question 53

Physics · Current Electricity · Numerical

In the given figure switches $S_1$ and $S_2$ are in open condition. The resistance across $ab$ when the switches $S_1$ and $S_2$ are closed is _____ $\Omega$

Answer: 10

Solution

When switch $S_1$ and $S_2$ are closed, the equivalent resistance is calculated as follows: $$\frac{12 \times 6}{12 + 6} + 2 + \frac{6 \times 12}{6 + 12}$$ Simplifying, we have: $$\frac{72}{18} + 2 + \frac{72}{18} = 4 + 2 + 4 = 10 \, \Omega$$

Question 54

Physics · System of Particles and Rotational Motion · Numerical

Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is $\frac{\sqrt{x}}{2}$. Then, the value of $x$ is _____.

Answer: 3

Solution

In both cases, the moment of inertia $I$ is about the point of contact. Ring $$mgh = \frac{1}{2} I \omega^2$$ $$mgh = \frac{1}{2} \left(2mR^2\right) \frac{v_R^2}{R^2}$$ $$v_R = \sqrt{gh}$$ Solid cylinder $$mgh = \frac{1}{2} I \omega^2$$ $$mgh = \frac{1}{2} \left(\frac{3}{2} mR^2\right) \frac{v_C^2}{R^2}$$ $$v_C = \sqrt{\frac{4gh}{3}}$$ $$\frac{v_R}{v_C} = \frac{\sqrt{3}}{2}$$

Question 55

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

For the forward biased diode characteristics shown in the figure, the dynamic resistance at $I_D = 3 \, \mathrm{mA}$ will be _____ $\Omega$

Answer: 25

Solution

$R_d = \dfrac{dV}{di} = \dfrac{1}{\dfrac{di}{dv}} = \dfrac{1}{\dfrac{5-1\times10^{-3}}{0.75-0.65}}$ $\dfrac{100}{4} = 25\,\Omega$

Question 56

Physics · Alternating Current · Numerical

A series LCR circuit of $R = 5\, \Omega$, $L = 20\, mH$ and $C = 0.5\, \mu F$ is connected across an AC supply of $250\, V$, having variable frequency. The power dissipated at resonance condition is $\times 10^2\, W$

Answer: 125

Solution

Given $X_L = X_C$ (due to resonance). $Z = R$ so $i_{ms} = \frac{V}{Z} = \frac{V}{R}$. $$\frac{V^2}{R} = \frac{250 \times 250}{5} = 125 \times 10^2 \, W$$

Question 57

Physics · Thermodynamics · Numerical

One mole of an ideal gas at 27°C is taken from A to B as shown in the given PV indicator diagram. The work done by the system will be $\times 10^{-1}$ J. [ Given : $R = 8.3 \, \mathrm{J/moleK}$, $\ln 2 = 0.6931$] (Round off to the nearest integer)

Answer: 17258

Solution

Process of isothermal $$W = nRT \ln \left( \frac{V_2}{V_1} \right)$$ $$= 1 \times 8.3 \times 300 \times \ln 2$$ $$= 17258 \times 10^{-1} \, \mathrm{J}$$

Question 58

Physics · Dual Nature of Radiation and Matter · Numerical

A certain metallic surface is illuminated by monochromatic radiation of wavelength $\lambda$. The stopping potential for photoelectric current for this radiation is $3 \, V_0$. If the same surface is illuminated with a radiation of wavelength $2\lambda$, the stopping potential is $V_0$. The threshold wavelength of this surface for photoelectric effect is $\lambda$

Answer: 4

Solution

Given $$KE = \frac{hc}{\lambda} - \phi hc$$ $$e(3V_0) = \frac{hc}{\lambda_0} - \phi$$ $$eV_0 = \frac{hc}{2\lambda_0} - \phi$$ Using (i) and (ii) $$\phi = \frac{hc}{4\lambda_0} = \frac{hc}{\lambda_t}$$ $$\lambda_t = 4\lambda_0$$

Question 59

Physics · System of Particles and Rotational Motion · Numerical

A body rotating with an angular speed of 600 $\mathrm{rpm}$ is uniformly accelerated to 1800 $\mathrm{rpm}$ in 10 $\mathrm{sec}$. The number of rotations made in the process is

Answer: 200

Solution

Given $$\omega_f = \omega_0 + \alpha t$$ $$\alpha = 1200 \times 6$$ $$\theta = \omega_0 t + \frac{1}{2} \alpha t^2$$ $$= 600 \times \frac{10}{60} + \frac{1}{2} \times 1200 \times 6 \times \frac{1}{36}$$ $$\theta = 200$$

Question 60

Physics · Nuclei · Numerical

A radioactive substance decays to $\left( \frac{1}{16} \right)^{th}$ of its initial activity in 80 days. The half life of the radioactive substance expressed in days is _____.

Answer: 20

Solution

The decay process is shown as $N_0 \to \frac{N_0}{2} \to \frac{N_0}{4} \to \frac{N_0}{8} \to \frac{N_0}{16}$. Given $4 \times t_{1/2} = 80$, we solve for $t_{1/2}$. $$t_{1/2} = 20 days$$

Chemistry

Question 61

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

Which one of the following pairs of isomers is an example of metamerism?

  1. $\mathrm{CH_3CH_2CH_2CH_2CH_3}$ and

Answer: (d)

Question 62

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

In the above reactions, product $A$ and product $B$ respectively are:

Answer: (d)

Solution

The first reaction involves the Hofmann rearrangement using $\mathrm{KOBr}$, which converts the amide to an amine with one less carbon atom. The product is $\mathrm{C_6H_4BrNH_2}$, labeled as $[A]$. The second reaction involves the reduction of the amide using $\mathrm{LiAlH_4}$ followed by hydrolysis with $\mathrm{H_3O^+}$. This converts the amide to an amine without losing a carbon atom. The product is $\mathrm{C_6H_4BrCH_2NH_2}$, labeled as $(B)$.

Question 63

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The major product (P) in the following reaction is:

Answer: (b)

Solution

The reaction involves an intramolecular aldol condensation. The starting compound undergoes a reaction with (i) KOH (alc.) followed by (ii) acidification with $\mathrm{H^+}$ and heating ($\Delta$) to form the major product (P).

Question 64

Chemistry · Hydrogen · Single correct

The single largest industrial application of dihydrogen is:

  1. Manufacture of metal hydrides
  2. Rocket fuel in space research
  3. In the synthesis of ammonia
  4. In the synthesis of nitric acid

Answer: (c)

Solution

Informative, according to NCERT uses of dihydrogen. In fact $\mathrm{NH_3}$ largest production is used to manufacture nitrogenous fertilisers.

Question 65

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

Consider two chemical reactions (A) and (B) that take place during metallurgical process : (A) $\mathrm{ZnCO_3}_{(s)} \xrightarrow{\Delta} \mathrm{ZnO}_{(s)} + \mathrm{CO_2}_{(g)}$ (B) $\mathrm{2ZnS}_{(s)} + \mathrm{3O_2}_{(g)} \xrightarrow{\Delta} \mathrm{2ZnO}_{(s)} + \mathrm{2SO_2}_{(g)}$ The correct option of names given to them respectively is :

  1. is calcination and (B) is roasting
  2. Both (A) and (B) are producing same product so both are roasting
  3. Both (A) and (B) are producing same product so both are calcination
  4. is roasting and (B) is calcination

Answer: (a)

Solution

(A) $\mathrm{ZnCO_3 (s)} \xrightarrow{\Delta} \mathrm{ZnO(s)} + \mathrm{CO_2 (g)}$ Heating in absence of oxygen in calcination. (B) $2\mathrm{ZnS(s)} + 3\mathrm{O_2 (g)} \rightarrow 2\mathrm{ZnO(g)} + 2\mathrm{SO_2 (g)}$ Heating in presence of oxygen in roasting. Hence (A) is calcination while (B) in roasting.

Question 66

Chemistry · Equilibrium · Single correct

A solution is $0.1\,\mathrm{M}$ in $\mathrm{Cl^-}$ and $0.001\,\mathrm{M}$ in $\mathrm{CrO_4^{2-}}$. Solid $\mathrm{AgNO_3}$ is gradually added to it. Assuming that the addition does not change the volume and $K_{sp}(\mathrm{AgCl})=1.7\times10^{-10}\,\mathrm{M^2}$ and $K_{sp}(\mathrm{Ag_2CrO_4})=1.9\times10^{-12}\,\mathrm{M^3}$, select the correct statement from the following:

  1. AgCl precipitates first because its $K_{sp}$ is high.
  2. Ag_2CrO_4 precipitates first as its $K_{sp}$ is low.
  3. $Ag_2CrO_4$ precipitates first because the amount of $Ag^+$ needed is low.
  4. AgCl will precipitate first as the amount of $Ag^+$ needed to precipitate is low.

Answer: (d)

Solution

(i) $[\mathrm{Ag}^+]$ required to ppt $\mathrm{AgCl(s)}$ Ksp = IP = $[\mathrm{Ag}^+][\mathrm{Cl}^-] = 1.7 \times 10^{-10}$ $$[\mathrm{Ag}^+] = 1.7 \times 10^{-9}$$ (ii) $[\mathrm{Ag}^+]$ required to ppt $\mathrm{Ag_2CrO_4(s)}$ Ksp = IP = $[\mathrm{Ag}^+]^2 [\mathrm{CrO_4^{2-}}] = 1.9 \times 10^{-12}$ $$[\mathrm{Ag}^+] = 4.3 \times 10^{-5}$$ $[\mathrm{Ag}^+]$ required to ppt $\mathrm{AgCl}$ is low so $\mathrm{AgCl}$ will ppt $1^{st}$

Question 67

Chemistry · Structure of Atom · Single correct

Outermost electronic configuration of a group 13 element, E, is $4 \, s^2, 4p^1$. The electronic configuration of an element of p-block period-five placed diagonally to element, E is:

  1. $[\mathrm{Kr}]3 \, d^{10} 4 \, s^2 4p^2$
  2. $[\mathrm{Ar}]3 \, d^{10} 4 \, s^2 4p^2$
  3. $[\mathrm{Xe}]5 \, d^{10} 6 \, s^2 6p^2$
  4. $[\mathrm{Kr}]4 \, d^{10} 5 \, s^2 5p^2$

Answer: (d)

Solution

The element E is Ga and the diagonal element of 5th period is $^{50}\mathrm{Sn}$ having outer electronic configuration will be $[\mathrm{Kr}]5s^24d^{10}5p^2$.

Question 68

Chemistry · The s-Block Elements · Single correct

Metallic sodium does not react normally with:

  1. gaseous ammonia
  2. But-2-yne
  3. Ethyne
  4. tert-butyl alcohol

Answer: (b)

Solution

Metallic sodium does not react with 2-butyne because 2-butyne does not have acidic hydrogen.

Question 69

Chemistry · Co-ordination Compounds · Single correct

Spin only magnetic moment of an octahedral complex of $\mathrm{Fe}^{2+}$ in the presence of a strong field ligand in BM is :

  1. 4.89
  2. 2.82
  3. 0
  4. 3.46

Answer: (c)

Solution

In presence of SFL $\Delta_0 > P$ means pairing occurs therefore for $\mathrm{Fe^{+2} \rightarrow 3\, d^6}$. Therefore, the number of unpaired $e^-$ (s) $= 0$. Therefore, $\mu = \sqrt{n(n+2)}\, \mathrm{BM} = 0$ where $n =$ number of unpaired $e^-$ (s). In $\mathrm{NiCl_2}$, $\mathrm{Ni^{+2}}$ is having configuration $3\, d^8$. Therefore, the number of unpaired electrons $= 2$. After formation of oxidised product $[\mathrm{Ni(CN)_6}]^{-2}\mathrm{Ni^{+4}}$ is obtained. $\mathrm{Ni^{+4} \Rightarrow 3\, d^6}$ and CN$^-$ is a strong field ligand. Therefore, the number of unpaired electrons $= 0$. Therefore, the charge is $2 - 0 = 2$.

Question 70

Chemistry · Structure of Atom · Single correct

Which one of the following species doesn't have a magnetic moment of 1.73BM, (spin only value) ?

  1. $\mathrm{O_2^+}$
  2. $\mathrm{CuI}$
  3. $\mathrm{[Cu(NH_3)_4]Cl_2}$
  4. $\mathrm{O_2^-}$

Answer: (b)

Solution

Species must not contain single unpaired. (1) $\mathrm{O_2^+} \rightarrow \sigma_{1s}^2 < \sigma_{1s}^{*2} < \sigma_{2s}^2 < \sigma_{2s}^{*2} < \pi_{2px}^2 = \pi_{2py}^2 < \pi_{2px}^{*} = \pi_{2py}^{*}$ Unpaired $e^- = 1$. Therefore, $\mu = 1.73 \mathrm{BM}$. (1) $\mathrm{Cu^+ \Gamma Cu^+} \rightarrow [\mathrm{Ar}] 3d^{10}$. Therefore, unpaired $e^- = 0$. $\Gamma \rightarrow [\mathrm{Xe}]$. Therefore, unpaired $e^- = 0$. Therefore, $\mu = 0$. 3. $[\mathrm{Cu(NH_3)_4}] \mathrm{Cl_2}$ $\mathrm{Cu} \rightarrow [\mathrm{A}] 3d^3$. Unpaired $= 1$. Therefore, $\mu = 1.73 \mathrm{BM}$.

Question 71

Chemistry · Chemistry in Everyday Life · Single correct

Which one of the following statements is not true about enzymes?

  1. Enzymes are non-specific for a reaction and substrate.
  2. Almost all enzymes are proteins.
  3. Enzymes work as catalysts by lowering the activation energy of a biochemical reaction.
  4. The action of enzymes is temperature and pH specific.

Answer: (a)

Solution

Fact

Question 72

Chemistry · Chemical Bonding and Molecular Structure · Single correct

The hybridisations of the atomic orbitals of nitrogen in $\mathrm{NO}_2^-$, $\mathrm{NO}_2^+$ and $\mathrm{NH}_4^+$ respectively are.

  1. $\mathrm{sp}^3$, $\mathrm{sp}^2$ and $\mathrm{sp}$
  2. $\mathrm{sp}$, $\mathrm{sp}^2$ and $\mathrm{sp}^3$
  3. $\mathrm{sp}^3$, $\mathrm{sp}$ and $\mathrm{sp}^2$
  4. $\mathrm{sp}^2$, $\mathrm{sp}$ and $\mathrm{sp}^3$

Answer: (d)

Solution

The first structure has $2\sigma + 1 lp$, indicating $sp^2$ hybridization. The second structure has $2\sigma + 0 lp$, indicating $sp$ hybridization. The third structure has $4\sigma + 0$, indicating $sp^3$ hybridization.

Question 73

Chemistry · Polymers · Single correct

Bakelite is a cross-linked polymer of formaldehyde and:

  1. PHBV
  2. Buna-S
  3. Novolac
  4. Dacron

Answer: (c)

Solution

Novolac (phenol formaldehyde Resin) $\rightarrow$ Bakelite

Question 74

Chemistry · Hydrocarbons · Single correct

Benzene on nitration gives nitrobenzene in presence of $\mathrm{HNO}_3$ and $\mathrm{H}_2\mathrm{SO}_4$ mixture, where:

  1. both $\mathrm{H}_2\mathrm{SO}_4$ and $\mathrm{HNO}_3$ act as a bases
  2. $\mathrm{HNO}_3$ acts as an acid and $\mathrm{H}_2\mathrm{SO}_4$ acts as a base
  3. both $\mathrm{H}_2\mathrm{SO}_4$ and $\mathrm{HNO}_3$ act as an acids
  4. $\mathrm{HNO}_3$ acts as a base and $\mathrm{H}_2\mathrm{SO}_4$ acts as an acid

Answer: (d)

Solution

Reagent for nitration of Benzene. $$\mathrm{H_2SO_4 + HNO_3 \rightleftharpoons HSO_4^- + H_2NO_3^+}$$ (Acid) $\hspace{1cm}$ (Base) $$\mathrm{H_2NO_3^+ \rightleftharpoons H_2O + NO_2^+}$$ Benzene reacts with $\mathrm{NO_2^+}$ to form nitrobenzene.

Question 75

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Consider the above reaction, compound B is :

Answer: (c)

Question 76

Chemistry · Hydrocarbons · Single correct

Major product P of above reaction, is:

Answer: (d)

Solution

Question 77

Chemistry · Redox Reactions · Single correct

$Cu^{2+}$ salt reacts with potassium iodide to give

  1. $Cu_2I_2$
  2. $Cu_2I_3$
  3. $CuI$
  4. $Cu(I_3)_2$

Answer: (a)

Solution

The given reactions are: $$2\mathrm{Cu}^{+2} + 4\Gamma \rightarrow \mathrm{Cu}_2\mathrm{I}_2(\,s) + \mathrm{I}_2$$ $$2\mathrm{Cu}^{+2} + 3\Gamma \rightarrow 2\mathrm{CuI} + \mathrm{I}_2$$

Question 78

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

In Carius method, halogen containing organic compound is heated with fuming nitric acid in the presence of :

  1. HNO_3
  2. AgNO_3
  3. CuSO_4
  4. BaSO_4

Answer: (b)

Solution

Organic compound is heated with fuming nitric acid in the presence of silver nitrate in carius method. Lunar caustic ($\text{AgNO}_3$) is used as reagent hare to distinguish $\text{Cl}^-$, Br and $\text{I}^-$ respectively as follows. $\text{Cl}^-\text{(aq)} \xrightarrow{\text{AgNO}_3} \text{AgCl} \downarrow_{\text{ppt}} \text{ white}$ $\text{Br}^-\text{(aq)} \xrightarrow{\text{AgNO}_3} \text{AgBr} \downarrow_{\text{ppt}} \text{ pale yellow}$ $\text{I(aq)} \xrightarrow{\text{AgNO}_3} \text{AgI} \downarrow_{\text{ppt}} \text{ Dark yellow}$

Question 79

Chemistry · Environmental Chemistry · Single correct

Which one of the following gases is reported to retard photosynthesis?

  1. CO
  2. CFCs
  3. CO_2
  4. NO_2

Answer: (d)

Solution

According to NCERT only $\mathrm{NO_2}$ from the given options can retard the photosynthesis process in plants.

Question 80

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The correct order of their reactivity towards hydrolysis at room temperature is:

  1. $(A) > (B) > (C) > (D)$
  2. $(D) > (A) > (B) > (C)$
  3. $(D) > (B) > (A) > (C)$
  4. $(A) > (C) > (B) > (D)$

Answer: (a)

Solution

Reactivity Hydrolysis towards $A > B > C > D$

Question 81

Chemistry · Thermodynamics · Fill in the blank

For a given chemical reaction $\mathrm{A} \rightarrow \mathrm{B}$ at $300 \, \mathrm{K}$ the free energy change is $-49.4 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$ and the enthalpy of reaction is $51.4 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$. The entropy change of the reaction is $\mathrm{JK}^{-1} \mathrm{mol}^{-1}$

Answer: 360

Solution

Given chemical reaction: $$\mathrm{A \xrightarrow{T_{300\, \mathrm{K}}} B[\Delta G]_{P,T} = -49.4 \, \mathrm{kJ/mol}}$$ $$\Delta H_{\mathrm{rxn}} = 51.4 \, \mathrm{kJ/mol}$$ $$\Delta S_{\mathrm{rxn}} = ?$$ From the relation $[\Delta G]_{P,T} = \Delta H - T \Delta S$ $$\Rightarrow \Delta S_{\mathrm{rxn}} = \frac{\Delta H_{\mathrm{rxn}} - [\Delta G]_{P,T}}{T}$$ $$= \frac{[51.4 - (-49.4)] \times 1000}{300} \, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}$$ $$\Rightarrow \Delta S_{\mathrm{rxn}} = 336 \, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}$$

Question 82

Chemistry · Structure of Atom · Numerical

The wavelength of electrons accelerated from rest through a potential difference of 40 $\mathrm{kV}$ is $x \times 10^{-12}$ $\mathrm{m}$. The value of $x$ is ___ (Nearest integer) Given : Mass of electron = $9.1 \times 10^{-31}$ $\mathrm{kg}$ Charge on an electron = $1.6 \times 10^{-19}$ $\mathrm{C}$ Planck's constant = $6.63 \times 10^{-34}$ $\mathrm{Js}$

Answer: 6

Solution

De-broglie-wave length of electron: $$\lambda_e = \frac{h}{\sqrt{2 \, \mathrm{m(KE)}}}$$ Since $\cdot$ e$^-$ is accelerated from rest $\Rightarrow \mathrm{KE} = q \times V$ $$\lambda = \frac{h}{\sqrt{2 \mathrm{mqV}}}$$ $$= \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.6 \times 10^{-19} \times 9.1 \times 10^{-31} \times 40 \times 10^3}}$$ $$= 0.614 \times 10^{-11} \, \mathrm{m}$$ $$= 6.14 \times 10^{-12} \, \mathrm{m}$$ Nearest integer = 6 OR $$\lambda = \frac{12.3}{\sqrt{V}} \, \mathrm{\overset{\circ}{A}} \ldots$$ $$= \frac{12.3}{200} = 6.15 \times 10^{-12} \, \mathrm{m}$$ Ans. is 6

Question 83

Chemistry · Solutions · Numerical

The vapour pressures of A and B at $25^\circ C$ are $90 \, \mathrm{mmHg}$ and $15 \, \mathrm{mmHg}$ respectively. If A and B are mixed such that the mole fraction of A in the mixture is $0.6$, then the mole fraction of B in the vapour phase is $x \times 10^{-1}$. The value of $x$ is ___ (Nearest integer)

Answer: 1

Solution

Given $P_A^\circ=90\,\mathrm{mmHg}$ at $25^\circ\mathrm{C}$, $P_B^\circ=15\,\mathrm{mmHg}$ and $X_A=0.6$ $X_B=0.4$ $P_T=X_AP_A^\circ+X_BP_B^\circ$ $=(0.6\times90)+(0.4\times15)$ $=54+6=60\,\mathrm{mmHg}$ Now mol fraction of $B$ in the vapour phase i.e. $Y_B=\frac{P_B}{P_T}=\frac{X_BP_B^\circ}{60}=0.1=1\times10^{-1}$ Therefore, $x=1$

Question 84

Chemistry · Redox Reactions · Numerical

$4\,\mathrm{g}$ equimolar mixture of $\mathrm{NaOH}$ and $\mathrm{Na_2CO_3}$ contains $x\,\mathrm{g}$ of $\mathrm{NaOH}$ and $y\,\mathrm{g}$ of $\mathrm{Na_2CO_3}$. The value of $x$ is \_\_\_\_ $\mathrm{g}$ (Nearest integer)

Answer: 1

Solution

Total mass = $4\ \text{g}$ Now NaOH : $a$ mol Na$_2$CO$_3$ : $a$ mol $W_{\mathrm{NaOH}} + W_{\mathrm{Na_2CO_3}} = 4$ $\Rightarrow 40a + 106a = 4$ $\Rightarrow a = \frac{4}{146}\ \text{mol}$ $\Rightarrow$ Therefore, mass of NaOH is: $\frac{4}{146} \times 40\ \text{g}$ $= 1.095 \approx 1$

Question 85

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

When 0.15 g of an organic compound was analyzed using Carius method for estimation of bromine, 0.2397 g of AgBr was obtained. The percentage of bromine in the organic compound is ___ (Nearest integer) [Atomic mass: Silver = 108, Bromine = 80 ]

Answer: 68

Solution

Moles of Br = Moles of AgBr obtained. Therefore, the mass of Br is given by $$Mass of Br = \frac{0.2397}{188} \times 80 \, \mathrm{g}$$ Therefore, the percentage of Br in the organic compound is $$\frac{W_{\mathrm{Br}}}{W_T} \times 100$$ $$= \frac{0.2397 \times 80}{188 \times 0.15} \times 100 = 0.85 \times 80$$ $$= 68$$ Therefore, the nearest integer is '68'.

Question 86

Chemistry · Surface Chemistry · Numerical

100 ml of 0.0018$\%$ (w/v) solution of $\mathrm{Cl}^-$ ion was the minimum concentration of $\mathrm{Cl}^-$ required to precipitate a negative sol in one h. The coagulating value of $\mathrm{Cl}^-$ ion is ___ (Nearest integer)

Answer: 1

Question 87

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

$\mathrm{PCl}_5\,(\mathrm{g}) \rightarrow \mathrm{PCl}_3\,(\mathrm{g}) + \mathrm{Cl}_2\,(\mathrm{g})$ In the above first order reaction the concentration of $\mathrm{PCl}_5$ reduces from initial concentration $50 \, \mathrm{mol} \, \mathrm{L}^{-1}$ to $10 \, \mathrm{mol} \, \mathrm{L}^{-1}$ in $120 \, \mathrm{minutes}$ at $300 \, \mathrm{K}$. The rate constant for the reaction at $300 \, \mathrm{K}$ is $x \times 10^{-2} \, \mathrm{min}^{-1}$. The value of $x$ is [Given $\log 5 = 0.6989$]

Answer: 1

Solution

The reaction is given as: $$\mathrm{PCl_5(g)} \xrightarrow{1st order, 300K} \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$$ At time $t = 0$, the concentration is $50 \, \mathrm{M}$. At time $t = 120 \, \mathrm{min}$, the concentration is $10 \, \mathrm{M}$. The rate constant $K$ is calculated as follows: $$K = \frac{2.303}{t} \log \frac{[A_0]}{[A_t]}$$ Substituting the values, we get: $$K = \frac{2.303}{120} \log \frac{50}{10}$$ $$K = \frac{2.303}{120} \times 0.6989 = 0.013413 \, \mathrm{min^{-1}}$$ This simplifies to: = 1.3413 $\times$ $10^{-2}$ $\mathrm{min^{-1}}$ Rounding to the nearest integer gives 1.34 $\Rightarrow$ Nearest integer = 1

Question 88

Chemistry · The Solid State · Numerical

Diamond has a three dimensional structure of C atoms formed by covalent bonds. The structure of diamond has face centred cubic lattice where 50$\%$ of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is ___

Answer: 8

Solution

Carbon atoms occupy FCC lattice points as well as half of the tetrahedral voids. Therefore, number of carbon atoms per unit cell $= 8$.

Question 89

Chemistry · Co-ordination Compounds · Numerical

An aqueous solution of $\mathrm{NiCl_2}$ was heated with excess sodium cyanide in presence of strong oxidizing agent to form $[Ni(CN)_6]^{2-}$. The total change in number of unpaired electrons on metal centre is ___

Answer: 2

Solution

For $[\mathrm{Ni(CN)}_6]^{2-}$, $\mathrm{Ni}^{+4} \rightarrow d^6$ strong field ligand. Pairing will be there, zero unpaired electron. For $\mathrm{NiCl}_2 \rightarrow \mathrm{Ni}^{2+} \rightarrow d^8$, there are two unpaired electrons. Charge = 2.

Question 90

Chemistry · Electrochemistry · Numerical

Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation. $$6\mathrm{OH}^- + \mathrm{Cl}^- \rightarrow \mathrm{ClO}_3^- + 3\mathrm{H}_2\mathrm{O} + 6e^-$$ A current of $x\,\mathrm{A}$ has to be passed for 10 h to produce 10.0 g of potassium chlorate. the value of $x$ is ___ (Nearest integer) (Molar mass of $\mathrm{KClO}_3 = 122.6 \, \mathrm{g \, mol}^{-1}$) $F = 96500\,\mathrm{C}$

Answer: 1

Solution

Given balanced equation is $$60\;\mathrm{H}^+ + \mathrm{Cl}^- \rightarrow \mathrm{ClO}_3^- + 3\mathrm{H}_2\mathrm{O} + 6e^-$$ $$\rightarrow 10\;\mathrm{gKClO}_3 \Rightarrow \frac{10}{122.6}\;\mathrm{molKClO}_3\;is obtained$$ From the above reaction, it is concluded that by $6F$ charge $1\;\mathrm{molKClO}_3$ is obtained. By the passage of $6F$ charge $= 1\;\mathrm{molKClO}_3$ Thus, by the passage of $\frac{x \times 10 \times 60 \times 60}{96500}$ $F$ charge $$= \frac{1}{6} \times \frac{x \times 10 \times 60 \times 60}{96500}$$ Now $$\frac{x \times 10 \times 60 \times 60}{6 \times 96500} = \frac{10}{122.6}$$ $$\Rightarrow x = \frac{10 \times 965}{60 \times 122.6} = \frac{965}{735.6} = 1.311 \simeq 1$$ OR $$W = \frac{E}{F} \times I \times t$$ $$10 = \frac{122.6}{96500 \times 6} \times x \times 10 \times 3600$$ $$X = 1.311$$ Ans.(1)