JEE Main 1 February 2024 Shift 1 question paper with solutions

JEE Main 1 February 2024 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Probability · Single correct

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

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

Answer: (b)

Solution

The probability $P(4W4B/2W2B)$ is given by \[ \frac{ P(4W4B)\times P(2W2B/4W4B) }{ \begin{matrix} P(2W6B)\times P(2W2B/2W6B)+P(3W5B)\times P(2W2B/3W5B)\\ +\cdots+P(6W2B)\times P(2W2B/6W2B) \end{matrix} } \] This simplifies to $$= \frac{\frac{1}{5} \times \frac{4 \binom{C_2}{4} \binom{C_2}{4}}{8 \binom{C_4}{4}}}{\frac{1}{5} \times \frac{2 \binom{C_2}{6} \binom{C_2}{2}}{8 \binom{C_4}{4}} + \frac{1}{5} \times \frac{3 \binom{C_2}{5} \binom{C_2}{2}}{8 \binom{C_4}{4}} + \ldots + \frac{1}{5} \times \frac{6 \binom{C_2}{2} \binom{C_2}{2}}{8 \binom{C_4}{4}}}$$ Finally, we have $$= \frac{2}{7}$$

Question 2

Maths · Integrals · Single correct

The value of the integral $$\int_{0}^{\frac{\pi}{4}} \frac{xdx}{\sin^4(2x) + \cos^4(2x)}$$ equals :

  1. $\($ $\frac{\sqrt{2}\pi^2}{8}$ $\)$
  2. $\($ $\frac{\sqrt{2}\pi^2}{16}$ $\)$
  3. $\($ $\frac{\sqrt{2}\pi^2}{32}$ $\)$
  4. $\($ $\frac{\sqrt{2}\pi^2}{64}$ $\)$

Answer: (c)

Solution

Let $2x = t$ then $dx = \frac{1}{2} dt$. $$I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{tdt}{\sin^4 t + \cos^4 t}$$ $$I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \left( \frac{\pi}{2} - t \right) \frac{dt}{\sin^4 \left( \frac{\pi}{2} - t \right) + \cos^4 \left( \frac{\pi}{2} - t \right)}$$ $$I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{\sin^4 t + \cos^4 t}{\sin^4 t + \cos^4 t} - I$$ $$2I = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t}$$ $$2I = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{\sec^4 t dt}{\tan^4 t + 1}$$ Let $\tan t = y$ then $\sec^2 t dt = dy$. $$2I = \frac{\pi}{8} \int_0^{\infty} \frac{(1 + y^2) \, dy}{1 + y^4}$$ $$= \frac{\pi}{16} \int_0^{\infty} \frac{1 + \frac{1}{y^2}}{y^2 + \frac{1}{y^2}} \, dy$$ Put $y - \frac{1}{y} = p$. $$I = \frac{\pi}{16} \int_{-\infty}^{\infty} \frac{dp}{p^2 + (\sqrt{2})^2}$$ $$= \frac{\pi}{16\sqrt{2}} \left[ \tan^{-1} \left( \frac{p}{\sqrt{2}} \right) \right]_{-\infty}^{\infty}$$ $$I = \frac{\pi^2}{16\sqrt{2}}$$

Question 3

Maths · Matrices · Single correct

If $A=\begin{bmatrix} \sqrt{2} & 1\\ -1 & \sqrt{2} \end{bmatrix}$, $B=\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}$, $C=ABA^{T}$ and $X=A^{T}C^{2}A$, then $\det X$ is equal to:

  1. 243
  2. 729
  3. 27
  4. 891

Answer: (b)

Solution

Given $$A = \begin{bmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{bmatrix} \implies \det(A) = 3$$ $$B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \implies \det(B) = 1$$ Now $C = ABA^T \implies \det(C) = (\det(A))^2 \times \det(B)$ $$|C| = 9$$ Now $$|X| = |A^T C^2 A|$$ $$= |A^T| |C|^2 |A|$$ $$= |A|^2 |C|^2$$ $$= 9 \times 81$$ $$= 729$$

Question 4

Maths · Trigonometric Functions · Single correct

If $\tan A = \frac{1}{\sqrt{x(x^2+x+1)}}$, $\tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan C = \left(x^{-3} + x^{-2} + x^{-1}\right)^{\frac{1}{2}}$, $0 < A, B, C < \frac{\pi}{2}$, then $A + B$ is equal to :

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

Answer: (a)

Solution

Finding $\tan(A + B)$ we get $$\Rightarrow \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{1}{\sqrt{x(x^2 + x + 1)}} + \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}}{1 - \frac{1}{x^2 + x + 1}}$$ $$\Rightarrow \tan(A + B) = \frac{(1 + x) \left( \sqrt{x^2 + x + 1} \right)}{(x^2 + x)(\sqrt{x})}$$ $$\frac{(1 + x) \left( \sqrt{x^2 + x + 1} \right)}{(x^2 + x)(\sqrt{x})}$$ $$\tan(A + B) = \frac{\sqrt{x^2 + x + 1}}{x \sqrt{x}} = \tan C$$ Therefore, $A + B = C$.

Question 5

Maths · Permutations and Combinations · Single correct

If n is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then n is equal to:

  1. 47
  2. 53
  3. 51
  4. 43

Answer: (c)

Solution

Total ways to partition 5 into 4 parts are: $5, 0, 0, 0 \Rightarrow 1$ way $4, 1, 0, 0 \Rightarrow \frac{5!}{4!} = 5$ ways $3, 2, 0, 0 \Rightarrow \frac{5!}{3!2!} = 10$ ways $2, 2, 0, 1 \Rightarrow \frac{5!}{2!2!} = 15$ ways $2, 1, 1, 1 \Rightarrow \frac{5!}{2!(1!)^3} = 10$ ways $3, 1, 1, 0 \Rightarrow \frac{5!}{3!2!} = 10$ ways Total $\Rightarrow 1 + 5 + 10 + 15 + 10 + 10 = 51$ ways

Question 6

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $S = \{ z \in \mathbb{C} : |z - 1| = 1 and (\sqrt{2} - 1)(z + \bar{z}) - i(z - \bar{z}) = 2\sqrt{2} \}$. Let $z_1, \ z_2 \in S$ be such that $|z_1| = \max_{z \in S} |z|$ and $|z_2| = \min_{z \in S} |z|$. Then $\left| \sqrt{2} z_1 - z_2 \right|^2$ equals:

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

Answer: (d)

Solution

Let $Z = x + iy$. Then $(x - 1)^2 + y^2 = 1 \rightarrow [OPTION1]$ and $\, (\sqrt{2} - 1)(2x) - i(2iy) = 2\sqrt{2}$ implies $(\sqrt{2} - 1)x + y = \sqrt{2} \rightarrow (2)$. Solving (1) and (2) we get either $x = 1$ or $x = \frac{1}{2 - \sqrt{2}} \rightarrow (3)$. On solving (3) with (2) we get for $x = 1 \Rightarrow y = 1 \Rightarrow Z_2 = 1 + i$ and for $x = \frac{1}{2 - \sqrt{2}} \Rightarrow y = \sqrt{2} - \frac{1}{\sqrt{2}} \Rightarrow Z_1 = \left(1 + \frac{1}{\sqrt{2}}\right) + \frac{i}{\sqrt{2}}$. Now $$\left|\sqrt{2}z_1 - z_2\right|^2$$ $$= \left|\left(\frac{1}{\sqrt{2}} + 1\right)\sqrt{2} + i - (1 + i)\right|^2$$ $$= (\sqrt{2})^2$$ $$= 2$$

Question 7

Maths · Statistics · Single correct

Let the median and the mean deviation about the median of 7 observation 170, 125, 230, 190, 210, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is :

  1. 31
  2. 28
  3. 30
  4. 32

Answer: (c)

Solution

Median $= 170 \Rightarrow 125, a, b, 170, 190, 210, 230$ Mean deviation about Median $= \frac{0 + 45 + 60 + 20 + 40 + 170 - a + 170 - b}{7} = \frac{205}{7}$ $\Rightarrow a + b = 300$ Mean $= \frac{170 + 125 + 230 + 190 + 210 + a + b}{7} = 175$ Mean deviation About mean $= \frac{50 + 175 - a + 175 - b + 5 + 15 + 35 + 55}{7} = 30$

Question 8

Maths · Vector Algebra · Single correct

Let $\vec{a}$ = -5$\hat{i}$ + $\hat{j}$ - 3$\hat{k}$, $\vec{b}$ = $\hat{i}$ + 2$\hat{j}$ - 4$\hat{k}$ and $\vec{c}$ = ((($\vec{a}$ $\times$ $\vec{b}$) $\times$ $\hat{i}$) $\times$ $\hat{i}$) $\times$ $\hat{i}$ . Then $\vec{c}$ $\cdot$ (-$\hat{i}$ + $\hat{j}$ + $\hat{k}$) is equal to

  1. -12
  2. -10
  3. -13
  4. -15

Answer: (a)

Solution

Given $\vec{a} = -5\hat{i} + \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 4\hat{k}$. The expression $(\vec{a} \times \vec{b}) \times \hat{i}$ is calculated as follows: $$(\vec{a} \cdot \hat{i}) \vec{b} - (\vec{b} \cdot \hat{i}) \vec{a}$$ This simplifies to: $$-5\vec{b} - \vec{a}$$ Further simplification gives: $$((( -5\vec{b} - \vec{a}) \times \hat{i}) \times \hat{i})$$ This becomes: $$((-11\hat{j} + 23\hat{k}) \times \hat{i}) \times \hat{i}$$ Which simplifies to: $$(11\hat{k} + 23\hat{j}) \times \hat{i}$$ Finally, this results in: $$(11\hat{j} - 23\hat{k})$$ The dot product $\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$ is calculated as: $$11 - 23 = -12$$

Question 9

Maths · Sequences and Series · Single correct

Let $S = \left\{x \in \mathbb{R} : \left(\sqrt{3} + \sqrt{2}\right)^x + \left(\sqrt{3} - \sqrt{2}\right)^x = 10\right\}$. Then the number of elements in $S$ is:

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

Answer: (c)

Solution

Given $\left( \sqrt{3} + \sqrt{2} \right)^x + \left( \sqrt{3} - \sqrt{2} \right)^x = 10$. Let $\left( \sqrt{3} + \sqrt{2} \right)^x = t$. Then $t + \frac{1}{t} = 10$. Solving the quadratic equation $t^2 - 10t + 1 = 0$, we find $$t = \frac{10 \pm \sqrt{100 - 4}}{2} = 5 \pm 2\sqrt{6}.$$ Therefore, $\left( \sqrt{3} + \sqrt{2} \right)^x = \left( \sqrt{3} \pm \sqrt{2} \right)^2$. Thus, $x = 2$ or $x = -2$. Number of solutions = 2.

Question 10

Maths · Applications of Integrals · Single correct

The area enclosed by the curves $xy + 4y = 16$ and $x + y = 6$ is equal to:

  1. 28 - 30 $\log$_2 2
  2. 30 - 28 $\log$_2 2
  3. 30 - 32 $\log$_2 2
  4. 32 - 30 $\log$_2 2

Answer: (c)

Solution

Given the equations $xy + 4y = 16$ and $x + y = 6$. Rewriting the first equation as $y(x + 4) = 16$ and the second as $x + y = 6$. On solving equations (1) and (2), we get $x = 4$ and $x = -2$. The area is given by the integral: $$Area = \int_{-2}^{4} \left( (6 - x) - \left( \frac{16}{x + 4} \right) \right) \, dx$$ which evaluates to $30 - 32 \ln 2$.

Question 11

Maths · Relations and Functions · Single correct

Let f : $\mathbb{R}$ $\to$ $\mathbb{R}$ and g : $\mathbb{R}$ $\to$ $\mathbb{R}$ be defined as $$f(x) = \begin{cases} \log_e x , & x > 0 \\ e^{-x} , & x \leq 0 \end{cases}$$ and $$g(x) = \begin{cases} x , & x \geq 0 \\ e^x , & x < 0 \end{cases}.$$ Then, gof : $\mathbb{R}$ $\to$ $\mathbb{R}$ is :

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

Answer: (b)

Solution

Given the function $g(f(x))$ defined as follows: $$g(f(x)) = \begin{cases} f(x), & f(x) \geq 0 \\ e^{f(x)}, & f(x) < 0 \end{cases}$$ And another definition: $$g(f(x)) = \begin{cases} e^{-x}, & (-\infty, 0] \\ e^{\ln x}, & (0, 1) \\ \ln x, & [1, \infty) \end{cases}$$ The graph of $g(f(x))$ is shown. $g(f(x))$ is many one into.

Question 12

Maths · Determinants · Single correct

If the system of equations $$2x + 3y - z = 5$$ $$x + \alpha y + 3z = -4$$ $$3x - y + \beta z = 7$$ has infinitely many solutions, then $13\alpha\beta$ is equal to

  1. 1110
  2. 1120
  3. 1210
  4. 1220

Answer: (b)

Solution

Using family of planes $$2x + 3y - z - 5 = k_1(x + \alpha y + 3z + 4) + k_2(3x - y + \beta z - 7)$$ $$2 = k_1 + 3k_2, 3 = k_1 \alpha - k_2, -1 = 3k_1 + \beta k_2, -5 = 4k_1 - 7k_2$$ On solving we get $$k_2 = \frac{13}{19}, k_1 = \frac{-1}{19}, \alpha = -70, \beta = \frac{-16}{13}$$ $$13 \alpha \beta = 13(-70) \left(\frac{-16}{13}\right)$$ $$= 1120$$

Question 13

Maths · Conic Sections · Single correct

For $0 < \theta < \pi/2$, if the eccentricity of the hyperbola $x^2 - y^2 \operatorname{cosec}^2\theta = 5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^2 \operatorname{cosec}^2\theta + y^2 = 5$, then the value of $\theta$ is :

  1. $\frac{\pi}{6}$
  2. $\frac{5\pi}{12}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{4}$

Answer: (c)

Solution

Given $$e_h = \sqrt{1 + \sin^2 \theta}$$ $$e_c = \sqrt{1 - \sin^2 \theta}$$ $$e_h = \sqrt{7} e_c$$ We have $$1 + \sin^2 \theta = 7 \left(1 - \sin^2 \theta\right)$$ Solving for $\sin^2 \theta$: $$\sin^2 \theta = \frac{6}{8} = \frac{3}{4}$$ Thus, $$\sin \theta = \frac{\sqrt{3}}{2}$$ Therefore, $$\theta = \frac{\pi}{3}$$

Question 14

Maths · Differential Equations · Single correct

Let y = y(x) be the solution of the differential equation $\frac{dy}{dx}$ = 2x(x+y)^3 - x(x+y) - 1, y(0) = 1. Then, ( $\frac{1}{\sqrt{2}}$ + y( $\frac{1}{\sqrt{2}}$ ) )^2 equals:

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

Answer: (d)

Solution

Given $\($ $\frac{dy}{dx}$ = 2x(x+y)^3 - x(x+y) - 1 $\)$. Let $\($ x+y = t $\)$. Then $\($ $\frac{dt}{dx}$ - 1 = 2xt^3 - xt - 1 $\)$. Rearranging gives $\($ $\frac{dt}{2t^3 - t}$ = xdx $\)$. This implies $\($ $\frac{tdt}{2t^4 - t^2}$ = xdx $\)$. Let $\($ t^2 = z $\)$. Then $\($ $\int$ $\frac{dz}{2(2z^2 - z)}$ = $\int$ xdx $\)$. This simplifies to $\($ $\int$ $\frac{dz}{4z\left(z - \frac{1}{2}\right)}$ = $\int$ xdx $\)$. Integrating gives $\($ $\ln$ $\left$| $\frac{z - \frac{1}{2}}{z}$ $\right$| = x^2 + k $\)$. Solving for $\($ z $\)$ gives $\($ z = $\frac{1}{2 - \sqrt{e}}$ $\)$.

Question 15

Maths · Continuity and Differentiability · Single correct

Let f : $\mathbb{R} \to \mathbb{R}$ be defined as $$f(x) = \begin{cases} \frac{a - b \cos 2x}{x^2} & ; x 1 \end{cases}$$ If $f$ is continuous everywhere in $\mathbb{R}$ and $m$ is the number of points where $f$ is NOT differential then $m + a + b + c$ equals :

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

Answer: (d)

Solution

At $x = 1$, $f(x)$ is continuous therefore, $f(1^-) = f(1) = f(1^+)$. $f(1) = 3 + c$. $$f(1^+) = \lim_{h \to 0} 2(1 + h) + 1$$ $$f(1^+) = \lim_{h \to 0} 3 + 2h = 3$$ from (1) and (2), $c = 0$. At $x = 0$, $f(x)$ is continuous therefore, $f(0^-) = f(0) = f(0^+)$. $f(0) = f(0^+) = 2$. $f(0^-)$ has to be equal to 2. $$\lim_{h \to 0} \frac{a - b \cos(2h)}{h^2}$$ $$\lim_{h \to 0} \frac{a - b \left\{ 1 - \frac{4h^2}{2!} + \frac{16h^4}{4!} + \ldots \right\}}{h^2}$$ $$\lim_{h \to 0} \frac{a - b + b \left\{ 2h^2 - \frac{3}{2} h^4 \ldots \right\}}{h^2}$$ for limit to exist $a - b = 0$ and limit is $2b$. From (3), (4) and (5), $a = b = 1$. Checking differentiability at $x = 0$. LHD: $$\lim_{h \to 0} \frac{\frac{1 - \cos 2h}{h^2} - 2}{-h}$$ $$1 - \left( 1 - \frac{4h^2}{2!} + \frac{16h^4}{4!} \cdots \right) - 2h^2$$ $$\lim_{h \to 0} \frac{-h^3}{-h} = 0$$ RHD: $$\lim_{h \to 0} \frac{(0 + h)^2 + 2 - 2}{h} = 0$$ Function is differentiable at every point in its domain. Therefore, $m = 0$. $m + a + b + c = 0 + 1 + 1 + 0 = 2$

Question 16

Maths · Conic Sections · Single correct

Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a > b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is:

  1. 3
  2. 7/2
  3. 3/2
  4. 5/2

Answer: (c)

Solution

Given $e = \frac{1}{\sqrt{2}} = \sqrt{1 - \frac{b^2}{a^2}} \Rightarrow \frac{1}{2} = 1 - \frac{b^2}{a^2}$. Then, $\frac{2b^2}{a} = 14$. Now, $e_H = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}$. Thus, $(e_H)^2 = \frac{3}{2}$.

Question 17

Maths · Sequences and Series · Single correct

Let 3, a, b, c be in A.P. and 3, a - 1, b + 1, c + 9 be in G.P. Then, the arithmetic mean of a, b and c is :

  1. -4
  2. -1
  3. 13
  4. 11

Answer: (d)

Solution

Given the sequence $3, a, b, c$ is an arithmetic progression (A.P), we have $3, 3 + d, 3 - 2d, 3 + 3d$. The sequence $3, a - 1, b + 1, c + 9$ is a geometric progression (G.P), giving $3, 2 + d, 4 + 2d, 12 + 3d$. Solving for $a$, $b$, and $c$, we find $a = 3 + d$, $b = 3 + 2d$, and $c = 3 + 3d$. The possible values for $d$ are $4$ and $-2$. If $d = 4$, the G.P becomes $3, 6, 12, 24$. Then $a = 7$, $b = 11$, and $c = 15$. The average is given by $$\frac{a + b + c}{3} = 11.$$

Question 18

Maths · Conic Sections · Single correct

Let $C : x^2 + y^2 = 4$ and $C' : x^2 + y^2 - 4\lambda x + 9 = 0$ be two circles. If the set of all values of $\lambda$ so that the circles $C$ and $C'$ intersect at two distinct points, is $\mathbb{R} - [a, b]$, then the point $(8a + 12, 16b - 20)$ lies on the curve :

  1. $x^2 + 2y^2 - 5x + 6y = 3$
  2. $5x^2 - y = -11$
  3. $x^2 - 4y^2 = 7$
  4. $6x^2 + y^2 = 42$

Answer: (d)

Solution

Given $x^2 + y^2 = 4$. $C(0,0)$, $r_1 = 2$. $C'(2\lambda, 0)$, $r_2 = \sqrt{4\lambda^2 - 9}$. $|r_1 - r_2| < CC' < |r_1 + r_2|$. $|2 - \sqrt{4\lambda^2 - 9}| < |2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$. $4 + 4\lambda^2 - 9 - 4\sqrt{4\lambda^2 - 9} < 4\lambda^2$. True $\lambda \in \mathbb{R}$.... (1) $4\lambda^2 < 4 + 4\lambda^2 - 9 + 4\sqrt{4\lambda^2 - 9}$. $5 < 4\sqrt{4\lambda^2 - 9}$ and $\lambda^2 \geq \frac{9}{4}$. $$\frac{25}{16} < 4\lambda^2 - 9 \lambda \in \left(-\infty, -\frac{3}{2}\right) \cup \left[\frac{3}{2}, \infty\right)$$ $$\frac{169}{64} < \lambda^2$$ $$\lambda \in \left(-\infty, -\frac{13}{8}\right) \cup \left(\frac{13}{8}, \infty\right)$$ From (1) and (2) $\lambda \in$ $$\lambda \in \left(-\infty, -\frac{13}{8}\right) \cup \left(\frac{13}{8}, \infty\right) \Rightarrow \mathbb{R} - \left[-\frac{13}{8}, \frac{13}{8}\right]$$ As per question $a = -\frac{13}{8}$ and $b = \frac{13}{8}$. Therefore, the required point is $(-1, 6)$ which satisfies option (4).

Question 19

Maths · Applications of Derivatives · Single correct

If $5f(x) + 4f\left(\frac{1}{x}\right) = x^2 - 2, \forall x \neq 0$ and $y = 9x^2 f(x)$, then $y$ is strictly increasing in :

  1. $\left(0, \frac{1}{\sqrt{5}}\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)$
  2. $\left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)$
  3. $\left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(0, \frac{1}{\sqrt{5}}\right)$
  4. $(-\infty, \frac{1}{\sqrt{5}}) \cup \left(0, \frac{1}{\sqrt{5}}\right)$

Answer: (b)

Solution

Given $5f(x) + 4f\left(\frac{1}{x}\right) = x^2 - 2$, $\forall x \neq 0$. Substitute $x \to \frac{1}{x}$. $$5f\left(\frac{1}{x}\right) + 4f(x) = \frac{1}{x^2} - 2$$ On solving (1) and (2) $$f(x) = \frac{5x^4 - 2x^2 - 4}{9x^2}$$ $y = 9x^2 f(x)$ $$y = 5x^4 - 2x^2 - 4 \ldots (3)$$ $$\frac{dy}{dx} = 20x^3 - 4x$$ For strictly increasing $$\frac{dy}{dx} > 0$$ $$4x \left(5x^2 - 1\right) > 0$$ $$x \in \left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)$$

Question 20

Maths · Three Dimensional Geometry · Single correct

If the shortest distance between the lines $\frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}$ is 1, then the sum of all possible values of $\lambda$ is:

  1. 0
  2. 2$\sqrt{3}$
  3. 3$\sqrt{3}$
  4. -2$\sqrt{3}$

Answer: (b)

Solution

Passing points of lines $\mathbf{L_1}$ and $\mathbf{L_2}$ are $(\lambda, 2, 1)$ and $(\sqrt{3}, 1, 2)$. $$S.D = \frac{\begin{vmatrix} \sqrt{3} - \lambda & -1 & 1 \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix}}{\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix}}$$ $$1 = \left| \frac{\sqrt{3} - \lambda}{\sqrt{3}} \right|$$ $$\lambda = 0, \lambda = 2\sqrt{3}$$

Question 21

Maths · Differential Equations · Numerical

If $x = x(t)$ is the solution of the differential equation $(t + 1) dx = \left(2x + (t + 1)^4\right) dt, x(0) = 2$, then, $x(1)$ equals

Answer: 14

Solution

(t+1)dx = $\left$(2x + (t+1)^4$\right$) dt $\frac{dx}{dt}$ = $\frac{2x + (t+1)^4}{t+1}$ $\frac{dx}{dt}$ - $\frac{2x}{t+1}$ = (t+1)^3 I $\cdot$ F = e^{-$\int$ $\frac{2}{t+1}$ dt} = e^{-2 $\ln$(t+1)} = $\frac{1}{(t+1)^2}$ $\frac{x}{(t+1)^2}$ = $\int$ $\frac{1}{(t+1)^2}$ (t+1)^3 dt + c $\frac{x}{(t+1)^2}$ = $\frac{(t+1)^2}{2}$ + c $\Rightarrow$ c = $\frac{3}{2}$ x = $\frac{(t+1)^4}{2}$ + $\frac{3}{2}$ (t+1)^2 put, t = 1 x = 2^3 + 6 = 14

Question 22

Maths · Sets · Numerical

The number of elements in the set $$S = \{(x, y, z) : x, y, z \in \mathbb{Z}, x + 2y + 3z = 42, x, y, z \geq 0\}$$ equals $\ldots$.

Answer: 169

Solution

Question 23

Maths · Binomial Theorem · Numerical

If the Coefficient of $x^{30}$ in the expansion of $$\left(1 + \frac{1}{x}\right)^6 (1 + x^2)^7 (1 - x^3)^8$$ ; $x \neq 0$ is $\alpha$, then $|\alpha|$ equals

Answer: 678

Solution

coeff. of $x^{30}$ in $\dfrac{(x+1)^6(1+x^2)^7(1-x^3)^8}{x^6} =\text{ coeff. of }x^{36}\text{ in }(1+x)^6(1+x^2)^7(1-x^3)^8$ General term ${}^{6}C_{r_1}\,{}^{7}C_{r_2}\,{}^{8}C_{r_3}(-1)^{r_3}x^{r_1+2r_2+3r_3}$ $r_1+2r_2+3r_3=36$ Case-I : \[ \begin{array}{|c|c|c|} \hline r_1 & r_2 & r_3\\ \hline 0 & 6 & 8\\ 2 & 5 & 8\\ 4 & 4 & 8\\ 6 & 3 & 8\\ \hline \end{array} \] $r_1+2r_2=12\;(\text{Taking }r_3=8)$ Case-II : \[ \begin{array}{|c|c|c|} \hline r_1 & r_2 & r_3\\ \hline 1 & 7 & 7\\ 3 & 6 & 7\\ 5 & 5 & 7\\ \hline \end{array} \] $r_1+2r_2=15\;(\text{Taking }r_3=7)$ Case-III : \[ \begin{array}{|c|c|c|} \hline r_1 & r_2 & r_3\\ \hline 4 & 7 & 6\\ 6 & 6 & 6\\ \hline \end{array} \] $r_1+2r_2=18\;(\text{Taking }r_3=6)$ Coeff. $=7+(15\times21)+(15\times35)+(35)$ $-(6\times8)-(20\times7\times8)-(6\times21\times8)+(15\times28)$ $-(7\times28)-678$ $\therefore\ |a|=678$

Question 24

Maths · Sequences and Series · Numerical

Let 3, 7, 11, 15, $\ldots$, 403 and 2, 5, 8, 11, $\ldots$, 404 be two arithmetic progressions. Then the sum, of the common terms in them, is equal to

Answer: 6699

Solution

The sequences are 3, 7, 11, 15, $\ldots$, 403 and 2, 5, 8, 11, $\ldots$, 404. The least common multiple of 4 and 3 is 12. The sequence 11, 23, 35, $\ldots$ is considered, ending at 403. The equation is given by: $$403 = 11 + (n - 1) \times 12$$ Solving for $n$: $$\frac{392}{12} = n - 1$$ $$33 = n$$ The sum is calculated as: $$Sum = \frac{33}{2} (22 + 32 \times 12)$$ $$= 6699$$

Question 25

Maths · Limits and Derivatives · Numerical

Let $\{$x$\}$ denote the fractional part of $x$ and $f(x) = \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}$, $x \neq 0$. If $L$ and $R$ respectively denotes the left hand limit and the right hand limit of $f(x)$ at $x = 0$, then $\frac{32}{\pi^2} \left(L^2 + R^2\right)$ is equal to

Answer: 18

Solution

Finding right hand limit $$\lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h)$$ $$= \lim_{h \to 0} f(h)$$ $$= \lim_{h \to 0} \cos^{-1}(1 - h^2) \sin^{-1}(1 - h)$$ $$= \lim_{h \to 0} \frac{\cos^{-1}(1 - h^2)}{h} \frac{\sin^{-1}(1 - h)}{1 - h^2}$$ $$= \lim_{h \to 0} \frac{\cos^{-1}(1 - h^2)}{h} \left( \frac{\sin^{-1} 1}{1} \right)$$ Let $\($ $\cos$^{-1}(1 - h^2) = $\theta$ $\Rightarrow$ $\cos$ $\theta$ = 1 - h^2 $\)$ $$= \frac{\pi}{2} \lim_{\theta \to 0} \frac{\theta}{\sqrt{1 - \cos \theta}}$$ $$= \frac{\pi}{2} \lim_{\theta \to 0} \frac{1}{\sqrt{\frac{1 - \cos \theta}{\theta^2}}}$$ $$= \frac{\pi}{2} \frac{1}{\sqrt{1/1/2}}$$ $$R = \frac{\pi}{\sqrt{2}}$$ Now finding left hand limit $$L = \lim_{x \to 0^-} f(x)$$ $$= \lim_{h \to 0} f(-h)$$ $$= \lim_{h \to 0} \frac{\cos^{-1}(1 - \{-h\}^2) \sin^{-1}(1 - \{-h\}h)}{\{-h\} - \{-h\}^3}$$ $$= \lim_{h \to 0} \frac{\cos^{-1}(1 - (-h + 1)^2) \sin^{-1}(1 - (-h + 1))}{(-h + 1) - (-h + 1)^3}$$ $$= \lim_{h \to 0} \frac{\cos^{-1}(-h^2 + 2h) \sin^{-1} h}{(1 - h)(1 - (1 - h)^2)}$$ $$= \lim_{h \to 0} \left( \frac{\pi}{2} \right) \frac{\sin^{-1} h}{(1 - (1 - h)^2)}$$ $$= \frac{\pi}{2} \lim_{h \to 0} \frac{\sin^{-1} h}{\sin^{-1} h} \frac{1}{-h + 2}$$ $$= \frac{\pi}{2} \lim_{h \to 0} \frac{\sin^{-1} h}{h} \frac{1}{-h + 2}$$ $$L = \frac{\pi}{4}$$ $$\frac{32}{\pi^2} \left( L^2 + R^2 \right) = \frac{32}{\pi^2} \left( \frac{\pi^2}{4} + \frac{\pi^2}{16} \right)$$ $$= 16 + 2$$ $$= 18$$

Question 26

Maths · Conic Sections · Numerical

Let the line $L : \sqrt{2}x + y = \alpha$ pass through the point of the intersection $P$ (in the first quadrant) of the circle $x^2 + y^2 = 3$ and the parabola $x^2 = 2y$. Let the line $L$ touch two circles $C_1$ and $C_2$ of equal radius $2\sqrt{3}$. If the centres $Q_1$ and $Q_2$ of the circles $C_1$ and $C_2$ lie on the y-axis, then the square of the area of the triangle $PQ_1Q_2$ is equal to

Answer: 72

Solution

Given $x^2 + y^2 = 3$ and $x^2 = 2y$. $y^2 + 2y - 3 = 0 \Rightarrow (y + 3)(y - 1) = 0$ $y = -3$ or $y = 1$ $y = 1x = \sqrt{2} \Rightarrow P(\sqrt{2}, 1)$ $p$ lies on the line $\sqrt{2}x + y = \alpha$ $\sqrt{2}(\sqrt{2}) + 1 = \alpha$ $\alpha = 3$ For circle $C_1$ $Q_1$ lies on $y$ axis Let $Q_1(0, \alpha)$ coordinates $R_1 = 2\sqrt{3}$ (Given) Line $L$ act as tangent Apply $P = r$ (condition of tangency) $$\Rightarrow \left| \frac{\alpha - 3}{\sqrt{3}} \right| = 2\sqrt{3}$$ $$\Rightarrow |\alpha - 3| = 6$$ $\alpha - 3 = 6$ or $\alpha - 3 = -6$ $$\Rightarrow \alpha = 9 \alpha = -3$$ $$\triangle PQ_1Q_2 = \frac{1}{2} \begin{vmatrix} \sqrt{2} & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1 \end{vmatrix}$$ $$= \frac{1}{2} (\sqrt{2}(12)) = 6\sqrt{2}$$ $$(\Delta PQ_1Q_2)^2 = 72$$

Question 27

Maths · Complex Numbers and Quadratic Equations · Numerical

Let $\mathbb{P} = \{ z \in \mathbb{C} : |z + 2 - 3i| \leq 1 \}$ and $\mathbb{Q} = \{ z \in \mathbb{C} : z(1+i) + \bar{z}(1-i) \leq -8 \}$. Let in $\mathbb{P} \cap \mathbb{Q}$, $|z - 3 + 2i|$ be maximum and minimum at $z_1$ and $z_2$ respectively. If $|z_1|^2 + 2|z_2|^2 = \alpha + \beta\sqrt{2}$, where $\alpha$, $\beta$ are integers, then $\alpha + \beta$ equals

Answer: 36

Solution

Clearly for the shaded region $z_1$ is the intersection of the circle and the line passing through $P$ ($L_1$) and $z_2$ is the intersection of line $L_1$ and $L_2$. Circle: $(x + 2)^2 + (y - 3)^2 = 1$. $L_1$: $x + y - 1 = 0$. $L_2$: $x - y + 4 = 0$. On solving circle and $L_1$ we get $$z_1 : \left(-2 - \frac{1}{\sqrt{2}}, 3 + \frac{1}{\sqrt{2}}\right)$$ On solving $L_1$ and $z_2$ is intersection of line $L_1$ and $L_2$ we get $$z_2 : \left(-\frac{3}{2}, \frac{5}{2}\right)$$ $$|z_1|^2 + 2|z_2|^2 = 14 + 5\sqrt{2} + 17$$ $$= 31 + 5\sqrt{2}$$ $$\alpha = 31$$ So $\beta = 5$. $\alpha + \beta = 36$

Question 28

Maths · Integrals · Numerical

If $$\int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2} \cos x \, dx}{(1+e^{\sin x})(1+\sin^4 x)} = \alpha \pi + \beta \log_e(3 + 2\sqrt{2})$$, where $\alpha$, $\beta$ are integers, then $\alpha^2 + \beta^2$ equals

Answer: 8

Solution

Given $$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{8\sqrt{2} \cos x}{(1+\sin x^2)(1+\sin^4 x)} \, dx$$ Apply king $$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{8\sqrt{2} \cos z (\sin z)}{(1+\sin z)(1+\sin^4 x)} \, dx$$ Adding (1) & (2) $$2I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{8\sqrt{2} \cos x}{1+\sin^4 x} \, dx$$ $$I = \int_{0}^{\frac{\pi}{2}} \frac{8\sqrt{2} \cos x}{1+\sin^4 x} \, dx$$ Let $\sin x = t$ $$I = \int_{0}^{1} \frac{8\sqrt{2}}{1+t^4} \, dx$$ $$I = 4\sqrt{2} \int_{0}^{1} \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} - \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}} \, dt$$ $$I = 4\sqrt{2} \int_{0}^{1} \left(1+\frac{1}{t^2}\right) \left(\frac{1}{t}-\frac{1}{t}\right)^2 + 2 - \frac{1}{\left(t+\frac{1}{t}\right)^2} - 2 \, dt$$ Let $t - \frac{1}{t} = z$ and $dt + \frac{1}{t} = k$ $$= 4\sqrt{2} \left[ \int_{0}^{\infty} \frac{dz}{z^2+2} - \int_{\infty}^{0} \frac{dk}{k^2-2} \right]$$ $$= 4\sqrt{2} \left[ \frac{1}{\sqrt{2}} \tan^{-1} \frac{z}{\sqrt{2}} \right]_{0}^{\infty} - \left[ \frac{1}{\sqrt{2}} \ln \left(\frac{k-\sqrt{2}}{k+\sqrt{2}}\right) \right]_{\infty}^{0}$$ $$= 4\sqrt{2} \left[ \frac{\pi}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \left[ \ln \frac{2-\sqrt{2}}{2+\sqrt{2}} \right] \right]$$ $$= 2\pi + 2 \ln(3+2\sqrt{2})$$ $$\alpha = 2$$ $$\beta = 2$$

Question 29

Maths · Three Dimensional Geometry · Numerical

Let the line of the shortest distance between the lines $$L_1 : \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k})$$ and $$L_2 : \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu (\hat{i} + \hat{j} - \hat{k})$$ intersect $L_1$ and $L_2$ at $P$ and $Q$ respectively. If $(\alpha, \beta, \gamma)$ is the midpoint of the line segment $PQ$, then $$2(\alpha + \beta + \gamma)$$ is equal to

Answer: 21

Solution

Given $\($ $\vec{b}$ = $\hat{i}$ - $\hat{j}$ + $\hat{k}$ $\)$ (DR's of $\($ L_1 $\)$) and $\($ $\vec{d}$ = $\hat{i}$ + $\hat{j}$ - $\hat{k}$ $\)$ (DR's of $\($ L_2 $\)$). $\($ $\vec{b}$ $\times$ $\vec{d}$ = $\begin{vmatrix}$ $\hat{i}$ & $\hat{j}$ & $\hat{k}$ $\\$ 1 & -1 & 1 $\\$ 1 & 1 & -1 $\end{vmatrix}$ $\)$ $\($ = 0 $\hat{i}$ + 2 $\hat{j}$ + 2 $\hat{k}$ $\)$ (DR's of Line perpendicular to $\($ L_1 $\)$ and $\($ L_2 $\)$) DR of AB line $\($ = (0, 2, 2) = $\left$( $\frac{3 + \mu - \lambda}{0}$, $\frac{3 + \mu + \lambda}{2}$, $\frac{3 - \mu - \lambda}{2}$ $\right$) $\)$ Solving above equation we get $\($ $\mu$ = -$\frac{3}{2}$ $\)$ and $\($ $\lambda$ = $\frac{3}{2}$ $\)$ Point A = $\($ $\left$( $\frac{5}{2}$, $\frac{1}{2}$, $\frac{9}{2}$ $\right$) $\)$ Point B = $\($ $\left$( $\frac{5}{2}$, $\frac{7}{2}$, $\frac{15}{2}$ $\right$) $\)$ Point of AB = $\($ $\left$( $\frac{5}{2}$, 2, 6 $\right$) = ($\alpha$, $\beta$, $\gamma$) $\)$ $\($ 2($\alpha$ + $\beta$ + $\gamma$) = 5 + 4 + 12 = 21 $\)$

Question 30

Maths · Relations and Functions · Numerical

Let A = {1, 2, 3, $\ldots$, 20}. Let $R_1$ and $R_2$ two relation on A such that $R_1$ = $\{$(a, b) : b is divisible by a$\}$ $R_2$ = $\{$(a, b) : a is an integral multiple of b$\}$. Then, number of elements in $R_1 - R_2$ is equal to $\ldots$.

Answer: 46

Solution

Physics

Question 31

Physics · Mechanical Properties of Solids · Single correct

With rise in temperature, the Young's modulus of elasticity

  1. changes erratically
  2. decreases
  3. increases
  4. remains unchanged

Answer: (b)

Solution

Q1 Conceptual questions

Question 32

Physics · Gravitation · Single correct

If $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g = \pi^2 \, \mathrm{m/s^2}$, then the length of the second's pendulum at a height $h = 2R$ from the surface of earth will be,:

  1. $\frac{2}{9} \, \mathrm{m}$
  2. $\frac{1}{9} \, \mathrm{m}$
  3. $\frac{4}{9} \, \mathrm{m}$
  4. $\frac{8}{9} \, \mathrm{m}$

Answer: (b)

Solution

Given $g' = \frac{GMe}{(3R)^2} = \frac{1}{9} g$. The time period $T = 2\pi \sqrt{\frac{\ell}{g'}}$. Since the time period of the second pendulum is 2 sec, $T = 2 sec$. $2 = 2\pi \sqrt{\frac{\ell}{9g}}$. Therefore, $\ell = \frac{1}{9} m$.

Question 33

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

In the given circuit if the power rating of Zener diode is 10 $\mathrm{mW}$, the value of series resistance $R_s$ to regulate the input unregulated supply is:

  1. 5k$\Omega$
  2. 10$\Omega$
  3. 1k$\Omega$
  4. None of these

Answer: (d)

Solution

Pd across $R_s$ $V_1 = 8 - 5 = 3 \, \mathrm{V}$ Current through the load resistor $$I = \frac{5}{1 \times 10^3} = 5 \, \mathrm{mA}$$ Maximum current through Zener diode $$I_{z max.} = \frac{10}{5} = 2 \, \mathrm{mA}$$ And minimum current through Zener diode $I_{z min.} = 0$ Therefore, $I_s max. = 5 + 2 = 7 \, \mathrm{mA}$ And $R_s min. = \frac{V_1}{I_s max.} = \frac{3}{7} \, \mathrm{k}\Omega$ Similarly $I_s min. = 5 \, \mathrm{mA}$ And $R_s max. = \frac{V_1}{I_s min.} = \frac{3}{5} \, \mathrm{k}\Omega$ Therefore, $\frac{3}{7} \, \mathrm{k}\Omega < R_s < \frac{3}{5} \, \mathrm{k}\Omega$

Question 34

Physics · Current Electricity · Single correct

The reading in the ideal voltmeter (V) shown in the given circuit diagram is :

  1. 5 \, $\mathrm{V}$
  2. 10 \, $\mathrm{V}$
  3. 0 \, $\mathrm{V}$
  4. 3 \, $\mathrm{V}$

Answer: (c)

Solution

Given $\($ i = $\frac{E_{eq}}{r_{eq}}$ = $\frac{8 \times 5}{8 \times 0.2}$ $\)$. $\($ I = 25 \, A $\)$ $\($ V = E - ir $\)$ $\($ = 5 - 0.2 $\times$ 25 $\)$ $\($ = 0 $\)$

Question 35

Physics · Electrostatic Potential and Capacitance · Single correct

Two identical capacitors have same capacitance $C$. One of them is charged to the potential $V$ and other to the potential $2 \, V$. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :

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

Answer: (a)

Solution

Given $V_C = \frac{q_{net}}{C_{net}} = \frac{CV + 2CV}{2C}$. $V_C = \frac{3V}{2}$. Loss of energy $$= \frac{1}{2}CV^2 + \frac{1}{2}C(2V)^2 - \frac{1}{2}2C\left(\frac{3V}{2}\right)^2$$ $$= \left(\frac{CV^2}{4}\right)$$

Question 36

Physics · Kinetic Theory · Single correct

Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :

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

Answer: (a)

Solution

Given \[ C_V=\frac{n_1C_{v_1}+n_2C_{v_2}}{n_1+n_2} \] \[ =\frac{2\times\frac{3}{2}R+6\times\frac{5}{2}R}{2+6} \] \[ =\frac{9}{4}R \]

Question 37

Physics · System of Particles and Rotational Motion · Single correct

A ball of mass 0.5 kg is attached to a string of length 50 cm. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is 400 N. The maximum possible value of angular velocity of the ball in rad/s is,;

  1. 1600
  2. 40
  3. 1000
  4. 20

Answer: (b)

Solution

Given $T = m \omega^2 \ell$. $$400 = 0.5 \omega^2 \times 0.5$$ $$\omega = 40 \, \mathrm{rad/s}.$$

Question 38

Physics · Alternating Current · Single correct

A parallel plate capacitor has a capacitance $C = 200 \, \mathrm{pF}$. It is connected to $230 \, \mathrm{V}$ ac supply with an angular frequency $300 \, \mathrm{rad/s}$. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :

  1. $1.38 \, \mu \mathrm{A}$ and $1.38 \, \mu \mathrm{A}$
  2. $14.3 \, \mu \mathrm{A}$ and $143 \, \mu \mathrm{A}$
  3. $13.8 \, \mu \mathrm{A}$ and $138 \, \mu \mathrm{A}$
  4. $13.8 \, \mu \mathrm{A}$ and $13.8 \, \mu \mathrm{A}$

Answer: (d)

Solution

Given the formula for current, we have: $$I = \frac{V}{X_C} = 230 \times 300 \times 200 \times 10^{-12} = 13.8 \, \mu A$$

Question 39

Physics · Thermodynamics · Single correct

The pressure and volume of an ideal gas are related as $PV^{3/2} = K$ (Constant). The work done by gas when the gas is taken from state $A (P_1, V_1, T_1)$ to state $B (P_2, V_2, T_2)$ is :

  1. $2 (P_1 V_1 - P_2 V_2)$
  2. $2 (P_2 V_2 - P_1 V_1)$
  3. $2 (\sqrt{P_1} V_1 - \sqrt{P_2} V_2)$
  4. $2 (P_2 \sqrt{V_2} - P_1 \sqrt{V_1})$

Answer: (a)

Solution

For $PV^x = constant$ If work done by gas is asked then $$W = \frac{nR \Delta T}{1-x}$$ Here $x = \frac{3}{2}$ Therefore, $$W = \frac{P_2 V_2 - P_1 V_1}{-\frac{1}{2}}$$ $$= 2 (P_1 V_1 - P_2 V_2) \ldots \ldots Option (1) is correct$$ If work done by external is asked then $W = -2 (P_1 V_1 - P_2 V_2) \ldots \ldots Option (2) is correct$

Question 40

Physics · Current Electricity · Single correct

A galvanometer has a resistance of $50\,\Omega$ and it allows maximum current of $5\,\mathrm{mA}$. It can be converted into voltmeter to measure upto $100\,\mathrm{V}$ by connecting in series a resistor of resistance

  1. $5975\,\Omega$
  2. $20050\,\Omega$
  3. $19950\,\Omega$
  4. $19500\,\Omega$

Answer: (c)

Solution

The resistance is calculated as follows: $$R = \frac{V}{I_g} - R_g = \frac{100}{5 \times 10^{-3}} - 50$$ $$= 20000 - 50$$ $$= 19950 \, \Omega$$

Question 41

Physics · Dual Nature of Radiation and Matter · Single correct

The de Broglie wavelengths of a proton and an $\alpha$ particle are $\lambda$ and $2\lambda$ respectively. The ratio of the velocities of proton and $\alpha$ particle will be:

  1. 1 : 8
  2. 1 : 2
  3. 4 : 1
  4. 8 : 1

Answer: (d)

Solution

Given $\lambda = \frac{h}{p} = \frac{h}{mv}$, we have $v = \frac{h}{m\lambda}$. The ratio of velocities is given by $$\frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} \times \frac{\lambda_\alpha}{\lambda_p}$$ Calculating this gives $$= 4 \times 2 = 8$$

Question 42

Physics · Experimental Physics · Single correct

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is:

  1. $\frac{1}{2}$
  2. $\frac{10}{11}$
  3. $\frac{50}{11}$
  4. $\frac{5}{11}$

Answer: (d)

Solution

Given $10 \mathrm{MSD} = 11 \mathrm{VSD}$. $1 \mathrm{VSD} = \frac{10}{11} \mathrm{MSD}$. $\mathrm{LC} = 1 \mathrm{MSD} - 1 \mathrm{VSD}$. $= 1 \mathrm{MSD} - \frac{10}{11} \mathrm{MSD}$. $= \frac{1 \mathrm{MSD}}{11}$. $= \frac{5}{11}$ units.

Question 43

Physics · Alternating Current · Single correct

In series LCR circuit, the capacitance is changed from C to 4C. To keep the resonance frequency unchanged, the new inductance should be:

  1. reduced by $\frac{1}{4} L$
  2. increased by $2 L$
  3. reduced by $\frac{3}{4} L$
  4. increased to $4 L$

Answer: (c)

Solution

Given $\omega' = \omega$. $$\frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{LC}}$$ Therefore, $L'C' = LC$. $$L'(4C) = LC$$ $$L' = \frac{L}{4}$$ Thus, inductance must be decreased by $\frac{3L}{4}$.

Question 44

Physics · Mathematics in Physics · Single correct

The radius $(r)$, length $(l)$ and resistance $(R)$ of a metal wire was measured in the laboratory as $r = (0.35 \pm 0.05)\, \mathrm{cm}$ $R = (100 \pm 10)\, \mathrm{ohm}$ $l = (15 \pm 0.2)\, \mathrm{cm}$ The percentage error in resistivity of the material of the wire is:

  1. 25.6%
  2. 39.9%
  3. 37.3%
  4. 35.6%

Answer: (b)

Solution

Given $\rho = R \frac{\rho}{\ell}$. The relative change in $\rho$ is given by $$\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta r}{r} + \frac{\Delta \ell}{\ell}$$ Substituting the given values, $$= \frac{10}{100} + 2 \times \frac{0.05}{0.35} + \frac{0.2}{15}$$ Simplifying, $$= \frac{1}{10} + \frac{2}{7} + \frac{1}{75}$$ Therefore, $$\frac{\Delta \rho}{\rho} = 39.9\%$$

Question 45

Physics · Physical World, Units and Measurements · Single correct

The dimensional formula of angular impulse is:

  1. [M L^{-2} T^{-1}]
  2. [M L^2 T^{-2}]
  3. [M L T^{-1}]
  4. [M L^2 T^{-1}]

Answer: (d)

Solution

Question 46

Physics · System of Particles and Rotational Motion · Single correct

A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass $10^{-2} \, \mathrm{kg}$ moving with a speed of $2 \times 10^2 \, \mathrm{ms^{-1}}$. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use $g = 10 \, \mathrm{m/s^2}$)

  1. 0.30 m
  2. 0.20 m
  3. 0.35 m
  4. 0.40 m

Answer: (b)

Solution

Given the equation for momentum conservation: $$mu = (M + m)V$$ Substituting the given values: $$10^{-2} \times 2 \times 10^2 \cong 1 \times V$$ Solving for $V$: $$V \cong 2 \, \mathrm{m/s}$$ The height $h$ is given by: $$h = \frac{V^2}{2g} = 0.2 \, \mathrm{m}$$

Question 47

Physics · Motion in a Plane · Single correct

A particle moving in a circle of radius $R$ with uniform speed takes time $T$ to complete one revolution. If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to $4R$. The angle of projection $\theta$ is then given by:

  1. $\sin^{-1} \left[ \frac{2gT^2}{\pi^2 R} \right]^{\frac{1}{2}}$
  2. $\sin^{-1} \left[ \frac{\pi^2 R}{2gT^2} \right]^{\frac{1}{2}}$
  3. $\cos^{-1} \left[ \frac{2gT^2}{\pi^2 R} \right]^{\frac{1}{2}}$
  4. $\cos^{-1} \left[ \frac{\pi R}{2gT^2} \right]^{\frac{1}{2}}$

Answer: (a)

Solution

Given $\frac{2\pi R}{T} = V$. Maximum height $H = \frac{v^2 \sin^2 \theta}{2g}$. $$4R = \frac{4\pi^2 R^2}{T^2 2g} \sin^2 \theta$$ $$\sin \theta = \sqrt{\frac{2gT^2}{\pi^2 R}}$$ $$\theta = \sin^{-1} \left( \frac{2gT^2}{\pi^2 R} \right)^{\frac{1}{2}}$$

Question 48

Physics · Laws of Motion · Single correct

Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in $\mathrm{ms^{-2}}$ is: (Consider that the string is massless and unstretchable and the pulley is also massless and frictionless):

  1. 3
  2. 4
  3. 2
  4. 1.2

Answer: (c)

Solution

Given $f_k = \mu N = 0.04 \times 20 \, g = 8 \, Newton$. $$a = \frac{60 - 8}{26} = 2 \, \mathrm{m/s^2}$$

Question 49

Physics · Atoms · Single correct

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly:

  1. 1.5 $\mathrm{eV}$
  2. 13.6 $\mathrm{eV}$
  3. 1.9 $\mathrm{eV}$
  4. 12.1 $\mathrm{eV}$

Answer: (d)

Solution

Transition from $n = 1$ to $n = 3$ with $\Delta E = 12.1 \, \mathrm{eV}$.

Question 50

Physics · Wave Optics · Single correct

A monochromatic light of wavelength $6000\,\mathrm{\AA}$ is incident on the single slit of width $0.01\,\mathrm{mm}$. If the diffraction pattern is formed at the focus of the convex lens of focal length $20\,\mathrm{cm}$, the linear width of the central maximum is:

  1. 60 mm
  2. 24 mm
  3. 120 mm
  4. 12 mm

Answer: (b)

Solution

Linear width $$W = \frac{2 \lambda d}{a} = \frac{2 \times 6 \times 10^{-7} \times 0.2}{1 \times 10^{-5}}$$ $$= 2.4 \times 10^{-2} = 24 \, \mathrm{mm}$$

Question 51

Physics · Moving Charges and Magnetism · Numerical

A regular polygon of 6 sides is formed by bending a wire of length $4\pi$ meter. If an electric current of $4\pi\sqrt{3}$ A is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be $x \times 10^{-7}$ T. The value of $x$ is_____

Answer: 72

Solution

The magnetic field B is calculated as follows: $$B = 6 \left( \frac{\mu_0 I}{4 \pi r} \right) (\sin 30^\circ + \sin 30^\circ)$$ Substituting the values: $$= 6 \times \frac{10^{-7} \times 4 \pi \sqrt{3}}{\left( \frac{\sqrt{3} \times 4 \pi}{2 \times 6} \right)}$$ Simplifying gives: $$= 72 \times 10^{-7} \, \mathrm{T}$$

Question 52

Physics · Electromagnetic Induction · Numerical

A rectangular loop of sides $12\,\mathrm{cm}$ and $5\,\mathrm{cm}$, with its sides parallel to the x-axis and y-axis respectively moves with a velocity of $5\,\mathrm{cm/s}$ in the positive x axis direction, in a space containing a variable magnetic field in the positive z direction. The field has a gradient of $10^{-3}\,\mathrm{T/cm}$ along the negative x direction and it is decreasing with time at the rate of $10^{-3}\,\mathrm{T/s}$. If the resistance of the loop is $6\,\mathrm{m\Omega}$, the power dissipated by the loop as heat is $\underline{\hspace{1cm}}\times10^{-9}\,\mathrm{W}$.

Answer: 216

Solution

Given $B_0$ is the magnetic field at origin. $$\frac{dB}{dx} = \frac{-10^{-3}}{10^{-2}}$$ $$\int_{B_0}^{B} dB = -\int_{0}^{x} 10^{-1} dx$$ $$B - B_0 = -10^{-1} x$$ $$B = \left( B_0 - \frac{x}{10} \right)$$ Motional emf in $AB = 0$. Motional emf in $CD = 0$. Motional emf in $AD = \varepsilon_1 = B_0 \ell v$. Magnetic field on rod $BC$ is $$\left( B_0 - \frac{(-12 \times 10^{-2})}{10} \right)$$ Motional emf in $BC = \varepsilon_2 = \left( B_0 + \frac{12 \times 10^{-2}}{10} \right) \ell \times v$. $$\varepsilon_{eq} = \varepsilon_2 - \varepsilon_1 = 300 \times 10^{-7} \, \mathrm{V}$$ For time variation: $$(\varepsilon_{eq})' = A \frac{dB}{dt} = 60 \times 10^{-7} \, \mathrm{V}$$ $$(\varepsilon_{eq})_{net} = \varepsilon_{eq} + (\varepsilon_{eq})' = 360 \times 10^{-7} \, \mathrm{V}$$ Power $= \frac{(\varepsilon_{eq})_{net}^2}{R} = 216 \times 10^{-9} \, \mathrm{W}$$

Question 53

Physics · Ray Optics and Optical Instruments · Fill in the blank

The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20cm. The focal length of the lens used is ____ cm

Answer: 15

Solution

Given $v = 3u$. $v - u = 20 \, \mathrm{cm}$ $2u = 20 \, \mathrm{cm}$ $u = 10 \, \mathrm{cm}$ $$\frac{1}{(-30)} - \frac{1}{(-10)} = \frac{1}{f}$$ $f = 15 \, \mathrm{cm}$

Question 54

Physics · Electric Charges and Fields · Numerical

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is $1.5 \, \mathrm{g/cc}$, the dielectric constant of water will be _____ (Take density of water $= 1 \, \mathrm{g/cc}$)

Answer: 3

Solution

In air, $\tan \frac{\theta}{2} = \frac{F}{mg} = \frac{q^2}{4\pi \varepsilon_0 r^2 mg}$. In water, $\tan \frac{\theta}{2} = \frac{F'}{mg'} = \frac{q^2}{4\pi \varepsilon_0 \varepsilon_r r^2 mg_{eff}}$. Equate both equations: $$\varepsilon_0 \, g = \varepsilon_0 \varepsilon_r g \left[ 1 - \frac{1}{1.5} \right]$$ $$\varepsilon_r = 3$$

Question 55

Physics · Nuclei · Numerical

The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is $\frac{1000}{x}$, where $x$ is ____.

Answer: 27

Solution

Given $R = R_0 A^{1/3}$. $R^3 \propto A$. $$\left( \frac{4.8}{4} \right)^3 = \frac{64}{A}$$ $$\frac{64}{A} = 1.44 \times 1.2$$ $$A = \frac{64}{1.44 \times 1.2} = \frac{1000}{x}$$ $$x = \frac{144 \times 12}{64} = 27$$

Question 56

Physics · System of Particles and Rotational Motion · Numerical

The identical spheres each of mass $2M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $4 \, \mathrm{m}$ each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is $\frac{4\sqrt{2}}{x}$, where the value of $x$ is _____

Answer: 3

Solution

Position vector $\vec{r}_{COM} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2 + m_3 \vec{r}_3}{m_1 + m_2 + m_3}$. $$\vec{r}_{COM} = \frac{2M \times 0 + 2M \times 4 \hat{i} + 2M \times 4 \hat{j}}{6M}$$ $$\vec{r} = \frac{4}{3} \hat{i} + \frac{4}{3} \hat{j}$$ $$|\vec{r}| = \frac{4 \sqrt{2}}{3}$$ $x = 3$

Question 57

Physics · Waves · Numerical

A tuning fork resonates with a sonometer wire of length $1\,\mathrm{m}$ stretched with a tension of $6\,\mathrm{N}$. When the tension in the wire is changed to $54\,\mathrm{N}$, the same tuning fork produces $12$ beats per second with it. The frequency of the tuning fork is ___ Hz.

Answer: 6

Solution

The frequency is given by the formula $$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}.$$ For the first frequency, $$f_1 = \frac{1}{2} \sqrt{\frac{6}{\mu}}.$$ For the second frequency, $$f_2 = \frac{1}{2} \sqrt{\frac{54}{\mu}}.$$ The ratio of the frequencies is $$\frac{f_1}{f_2} = \frac{1}{3}.$$ The difference between the frequencies is $$f_2 - f_1 = 12.$$ Therefore, $$f_1 = 6 \, \mathrm{Hz}.$$

Question 58

Physics · Mechanical Properties of Fluids · Numerical

A plane is in level flight at constant speed and each of its two wings has an area of $40 \, \mathrm{m}^2$. If the speed of the air is $180 \, \mathrm{km/h}$ over the lower wing surface and $252 \, \mathrm{km/h}$ over the upper wing surface, the mass of the plane is _____ kg. (Take air density to be $1 \, \mathrm{kg} \, \mathrm{m}^{-3}$ and $g = 10 \, \mathrm{ms}^{-2}$)

Answer: 9600

Solution

Given $A = 80 \, \mathrm{m^2}$. Using Bernoulli equation $$A (P_2 - P_1) = \frac{1}{2} \rho \left( V_1^2 - V_2^2 \right) A$$ $$mg = \frac{1}{2} \times 1 \left( 70^2 - 50^2 \right) \times 80$$ $$mg = 40 \times 2400$$ $$m = 9600 \, \mathrm{kg}$$

Question 59

Physics · Current Electricity · Numerical

The current in a conductor is expressed as $I = 3t^2 + 4t^3$, where $I$ is in Ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor during $t = 1 \, \mathrm{s}$ to $t = 2 \, \mathrm{s}$ is_____ C.

Answer: 22

Solution

Given $$q = \int_1^2 i \, dt = \int_1^2 (3t^2 + 4t^3) \, dt$$ Evaluate the integral: $$q = \left( t^3 + t^4 \right) \bigg|_1^2$$ Substitute the limits: $$q = 22C$$

Question 60

Physics · Motion in a Straight Line · Numerical

A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position (x) with time (t) is given as $x = -3t^3 + 18t^2 + 16t$, where $x$ is in $m$ and $t$ is in $s$. The velocity of the particle when its acceleration becomes zero is_____ m/s.

Answer: 52

Solution

Given $$x = 3t^3 + 18t^2 + 16t$$ Velocity is given by $$v = -9t^2 + 36 + 16$$ Acceleration is given by $$a = -18t + 36$$ Setting acceleration to zero at $t = 2 \, \mathrm{s}$: $$a = 0 at t = 2 \, \mathrm{s}$$ Substitute $t = 2$ into the velocity equation: $$v = -9(2)^2 + 36 \times 2 + 16$$ Calculate the velocity: $$v = 52 \, \mathrm{m/s}$$

Chemistry

Question 61

Chemistry · Biomolecules · Single correct

If one strand of a DNA has the sequence ATGCTTCA, sequence of the bases in complementary strand is:

  1. CATTAGCT
  2. TACGAAGT
  3. GTACTTAC
  4. ATGCGACT

Answer: (b)

Solution

Adenine base pairs with thymine with 2 hydrogen bonds and cytosine base pairs with guanine with 3 hydrogen bonds.

Question 62

Chemistry · Haloalkanes and Haloarenes · Single correct

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason $(R)$. Assertion (A): Haloalkanes react with KCN to form alkyl cyanides as a main product while with AgCN form isocyanide as the main product. Reason $(R)$: KCN and AgCN both are highly ionic compounds. In the light of the above statement, choose the most appropriate answer from the options given below:

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

Answer: (a)

Solution

When $\mathrm{KCN}$ reacts with $\mathrm{R-X}$, the major product is $\mathrm{R-CN}$ due to the ionic nature of $\mathrm{KCN}$. When $\mathrm{AgCN}$ reacts with $\mathrm{R-X}$, the major product is $\mathrm{R-NC}$ due to the covalent nature of $\mathrm{AgCN}$. AgCN is mainly covalent in nature and nitrogen is available for attack, so alkyl isocyanide is formed as the main product.

Question 63

Chemistry · Redox Reactions · Single correct

In acidic medium, $K_2Cr_2O_7$ shows oxidising action as represented in the half reaction $$Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$ $X, Y, Z$ and $A$ are respectively are:

  1. $8, 6, 4$ $and$ $Cr_2O_3$
  2. $14, 7, 6$ $and$ $Cr^{3+}$
  3. $8, 4, 6$ $and$ $Cr_2O_3$
  4. $14, 6, 7$ $and$ $Cr^{3+}$

Answer: (d)

Solution

The balanced reaction is, $\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6e^- \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O}$ X = 14 Y = 6 A = 7

Question 64

Chemistry · Redox Reactions · Single correct

Which of the following reactions are disproportionation reactions? (A) $\mathrm{Cu}^+ \rightarrow \mathrm{Cu}^{2+} + \mathrm{Cu}$ (B) $3\mathrm{MnO}_4^{2-} + 4\mathrm{H}^+ \rightarrow 2\mathrm{MnO}_4^- + \mathrm{MnO}_2 + 2\mathrm{H}_2\mathrm{O}$ (C) $2\mathrm{KMnO}_4 \rightarrow \mathrm{K}_2\mathrm{MnO}_4 + \mathrm{MnO}_2 + \mathrm{O}_2$ (D) $2\mathrm{MnO}_4^- + 3\mathrm{Mn}^{2+} + 2\mathrm{H}_2\mathrm{O} \rightarrow 5\mathrm{MnO}_2 + 4\mathrm{H}^+$ Choose the correct answer from the options given below:

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

Answer: (a)

Solution

When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation. $$\mathrm{Cu^+ \rightarrow Cu^{2+}}$$ $+$ $Cu$ $$3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O}$$

Question 65

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

In case of isoelectronic species the size of $\mathrm{F}^-$, Ne and $\mathrm{Na}^+$ is affected by:

  1. Principal quantum number (n)
  2. None of the factors because their size is the same
  3. Electron-electron interaction in the outer orbitals
  4. Nuclear charge (z)

Answer: (d)

Solution

In $\mathrm{F}^-$, Ne, $\mathrm{Na}^+$ all have $1s^2, 2s^2, 2p^6$ configuration. They have different size due to the difference in nuclear charge.

Question 66

Chemistry · Structure of Atom · Single correct

According to the wave-particle duality of matter by de-Broglie, which of the following graph plot presents most appropriate relationship between wavelength of electron ($\lambda$) and momentum of electron (p) ?

Answer: (a)

Solution

Given $\lambda = \frac{h}{p} \left[ \lambda \propto \frac{1}{p} \right]$. Therefore, $\lambda p = h$ (constant). So, the plot is a rectangular hyperbola.

Question 67

Chemistry · Co-ordination Compounds · Single correct

Given below are two statements: Statement (I): A solution of $[\mathrm{Ni(H_2O)_6}]^{2+}$ is green in colour. Statement (II): A solution of $[\mathrm{Ni(CN)_4}]^{2-}$ is colourless. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (b)

Solution

The complex $[\mathrm{Ni(H_2O)_6}]^{+2}$ forms a green colour solution due to $d-d$ transition. The complex $[\mathrm{Ni(CN)_4}]^{-2}$ is diamagnetic and it is colourless.

Question 68

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason $(R)$. Assertion (A) : PH$_3$ has lower boiling point than NH$_3$. Reason $(R)$ : In liquid state NH$_3$ molecules are associated through vander waal's forces, but PH$_3$ molecules are associated through hydrogen bonding. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (d)

Solution

Unlike $\mathrm{NH_3}$, $\mathrm{PH_3}$ molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of $\mathrm{PH_3}$ is lower than $\mathrm{NH_3}$.

Question 69

Chemistry · Haloalkanes and Haloarenes · Single correct

Identify $A$ and $B$ in the following sequence of reaction

Answer: (b)

Solution

The reaction sequence starts with toluene, which is converted to benzal chloride by the action of $\mathrm{Cl_2}$ in the presence of light ($\mathrm{hv}$). Benzal chloride is then converted to benzaldehyde by hydrolysis with water at $373 \, \mathrm{K}$.

Question 70

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

Given below are two statements: Statement (I) : Aminobenzene and aniline are same organic compounds. Statement (II) : Aminobenzene and aniline are different organic compounds. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (b)

Solution

Aniline is also known as amino benzene.

Question 71

Chemistry · Co-ordination Compounds · Single correct

Which of the following complex is homoleptic?

  1. $[\mathrm{Ni(CN)_4}]^{2-}$
  2. $[\mathrm{Ni(NH_3)_2Cl_2}]$
  3. $[\mathrm{Fe(NH_3)_4Cl_2}]^{+}$
  4. $[\mathrm{Co(NH_3)_4Cl_2}]^{+}$

Answer: (a)

Solution

In Homoleptic complex all the ligand attached with the central atom should be the same. Hence $[\mathrm{Ni(CN)_4}]^{2-}$ is a homoleptic complex.

Question 72

Chemistry · Hydrocarbons · Single correct

Which of the following compound will most easily be attacked by an electrophile?

Answer: (d)

Solution

Higher the electron density in the benzene ring more easily it will be attacked by an electrophile. Phenol has the highest electron density amongst all the given compounds.

Question 73

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

Ionic reactions with organic compounds proceed through: (A) Homolytic bond cleavage (B) Heterolytic bond cleavage (C) Free radical formation (D) Primary free radical (E) Secondary free radical Choose the correct answer from the options given below:

  1. only
  2. $(C)$ only
  3. only
  4. and (E) only

Answer: (c)

Solution

Heterolytic cleavage of a bond leads to the formation of ions.

Question 74

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Arrange the bonds in order of increasing ionic character in the molecules. $LiF, K_2O, N_2, SO_2$ and $ClF_3.$

  1. ClF_3 < N_2 < SO_2 < K_2O < LiF
  2. LiF < K_2O < ClF_3 < SO_2 < N_2
  3. N_2 < SO_2 < ClF_3 < K_2O < LiF
  4. N_2 < ClF_3 < SO_2 < K_2O < LiF

Answer: (c)

Solution

Increasing order of ionic character $\mathrm{N_2} < \mathrm{SO_2} < \mathrm{ClF_3} < \mathrm{K_2O} < \mathrm{LiF}$. Ionic character depends upon difference of electronegativity (bond polarity).

Question 75

Chemistry · Solutions · Single correct

We have three aqueous solutions of $\mathrm{NaCl}$ labelled as 'A', 'B' and 'C' with concentration $0.1\,\mathrm{M}$, $0.01\,\mathrm{M}$ $\&$ $0.001\,\mathrm{M}$, respectively. The value of van t' Hoff factor $(i)$ for these solutions will be in the order.

  1. $i_A < i_B < i_C$
  2. $i_A < i_C < i_B$
  3. $i_A = i_B = i_C$
  4. $i_A > i_B > i_C$

Answer: (a)

Solution

The table shows the values of $i$ for NaCl at different concentrations. As the solution becomes very dilute, the value of $i$ approaches 2.

Question 76

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

In Kjeldahl's method for estimation of nitrogen, $CuSO_4$ acts as:

  1. Reducing agent
  2. Catalytic agent
  3. Hydrolysis agent
  4. Oxidising agent

Answer: (b)

Solution

Kjeldahl's method is used for estimation of Nitrogen where $\mathrm{CuSO_4}$ acts as a catalyst.

Question 77

Chemistry · Redox Reactions · Single correct

Given below are two statements: Statement (I): Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution. Statement (II): In this titration phenolphthalein can be used as indicator. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (a)

Solution

Statement (I): Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution as it is economical and its concentration does not change with time. Phenolphthalein can act as an indicator in acid-base titration as it shows colour in pH range 8.3 to 10.1.

Question 78

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

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

  1. (A)-(III), (B)-(IV), \text(C)-(I), (D)-(II)
  2. (A)-(IV), (B)-(II), \text(C)-(I), (D)-(III)
  3. (A)-(IV), (B)-(II), \text(C)-(III), (D)-(I)
  4. (A)-(III), (B)-(IV), \text(C)-(II), (D)-(I)

Answer: (b)

Solution

The sequence of reactions is as follows: 1. $\mathrm{CH_3(CH_2)_5COOC_2H_5}$ is treated with DIBAL-H and $\mathrm{H_2O}$ to form $\mathrm{CH_3(CH_2)_5CHO}$. 2. $\mathrm{C_6H_5COC_6H_5}$ is treated with $\mathrm{Zn(Hg)}$ and conc. $\mathrm{HCl}$ to form $\mathrm{C_6H_5CH_2C_6H_5}$. 3. $\mathrm{C_6H_5CHO}$ is treated with $\mathrm{CH_3MgBr}$ followed by $\mathrm{H_2O}$ to form $\mathrm{C_6H_5CH(OH)CH_3}$. 4. $\mathrm{CH_3COCH_2COOC_2H_5}$ is treated with $\mathrm{NaBH_4}$ and $\mathrm{H^+}$ to form $\mathrm{CH_3CH(OH)CH_2COOC_2H_5}$.

Question 79

Chemistry · Thermodynamics · Single correct

Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following:

  1. $q = 0, \Delta T \neq 0, w = 0$
  2. $q = 0, \Delta T < 0, w \neq 0$
  3. $q \neq 0, \Delta T = 0, w = 0$
  4. $q = 0, \Delta T = 0, w = 0$

Answer: (d)

Solution

During free expansion of an ideal gas under adiabatic condition $q = 0$, $\Delta T = 0$, $w = 0$.

Question 80

Chemistry · Hydrocarbons · Single correct

Given below are two statements: Statement (I) : The $\mathrm{NH}_2$ group in Aniline is ortho and para directing and a powerful activating group. Statement (II) : Aniline does not undergo FriedelCraft's reaction (alkylation and acylation). In the light of the above statements, choose the most appropriate answer from the options given below :

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

Answer: (a)

Solution

The $\mathrm{NH_2}$ group in Aniline is ortho and para directing and a powerful activating group as $\mathrm{NH_2}$ has strong $+M$ effect. Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation) as Aniline will form complex with $\mathrm{AlCl_3}$ which will deactivate the benzene ring.

Question 81

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

Number of optical isomers possible for 2-chlorobutane

Answer: 2

Solution

There is one chiral centre present in the given compound. So, total optical isomers = 2.

Question 82

Chemistry · Electrochemistry · Fill in the blank

The potential for the given half cell at 298 K is $(-) \ldots \times 10^{-2} \mathrm{V}$. $2\mathrm{H}^+_{(aq)} + 2\mathrm{e}^- \rightarrow \mathrm{H}_2(g)$ $[\mathrm{H}^+] = 1\mathrm{M}, \;P_{\mathrm{H}_2} = 2 \mathrm{atm}$ (Given: $2.303RT/F = 0.06 \mathrm{V}, \log 2 = 0.3$)

Answer: 1

Solution

Given $$E = E^\circ_{\mathrm{H^+/H_2}} - \frac{0.06}{2} \log \frac{P_{\mathrm{H_2}}}{[\mathrm{H^+}]^2}$$ Substituting the values, we have $$E = 0.00 - \frac{0.06}{2} \log \frac{2}{[1]^2}$$ Simplifying, $$E = -0.03 \times 0.3 = -0.9 \times 10^{-2} \, \mathrm{V}$$

Question 83

Chemistry · Analytical Chemistry · Single correct

The number of white coloured salts among the following is \begin{enumerate} \item[(A)] $SrSO_4$ \item[(B)] $Mg(NH_4)PO_4$ \item[(C)] $BaCrO_4$ \item[(D)] $Mn(OH)_2$ \item[(E)] $PbSO_4$ \item[(F)] $PbCrO_4$ \item[(G)] $AgBr$ \item[(H)] $PbI_2$ \item[(I)] $CaC_2O_4$ \item[(J)] $[Fe(OH)_2(CH_3COO)]$ \end{enumerate}

  1. SrSO_4
  2. $\mathrm{Mg(NH_4)PO_4}$
  3. BaCrO_4
  4. Mn(OH)_2

Answer: e

Solution

Q1 $\mathrm{SrSO_4}$ - white $\mathrm{Mg(NH_4)PO_4}$ - white $\mathrm{BaCrO_4}$ - yellow $\mathrm{Mn(OH)_2}$ - white $\mathrm{PbSO_4}$ - white $\mathrm{PbCrO_4}$ - yellow $\mathrm{AgBr}$ - pale yellow $\mathrm{PbI_2}$ - yellow $\mathrm{CaC_2O_4}$ - white $[\mathrm{Fe(OH)_2(CH_3COO)}]$ - Brown Red

Question 84

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

The ratio of $\frac{^{14}C}{^{12}C}$ in a piece of wood is $\frac{1}{8}$ part that of atmosphere. If half life of $^{14}C$ is 5730 years, the age of wood sample is _____ years.

Answer: 17190

Solution

Given $$ \lambda t = \ln \frac{(^{14}\mathrm{C}/^{12}\mathrm{C})_{atmosphere}}{(^{14}\mathrm{C}/^{12}\mathrm{C})_{wood sample}} $$ As per the question, $$ \frac{(^{14}\mathrm{C}/^{12}\mathrm{C})_{wood}}{(^{14}\mathrm{C}/^{12}\mathrm{C})_{atmosphere}} = \frac{1}{8} $$ So, $$ \lambda t = \ln 8 $$ $$ \frac{\ln 2}{t_{1/2}} t = \ln 8 $$ $$ t = 3 \times t_{1/2} = 17190 years $$

Question 85

Chemistry · Chemical Bonding and Molecular Structure · Numerical

The number of molecules/ion/s having trigonal bipyramidal shape is ........ . $\mathrm{PF}_5$, $\mathrm{BrF}_5$, $\mathrm{PCl}_5$, $[\mathrm{PtCl}_4]^{2-}$, $\mathrm{BF}_3$, $\mathrm{Fe(CO)}_5$

Answer: 3

Solution

$PF_5, PCl_5, Fe(CO)_5$; Trigonal bipyramidal BrF_5; square pyramidal [$\mathrm{PtCl}_4]^{-2}$; square planar $BF_3$; Trigonal planar

Question 86

Chemistry · Hydrocarbons · Numerical

Total number of deactivating groups in aromatic electrophilic substitution reaction among the following is

Answer: 2

Solution

The groups are classified based on their mesomeric effect. The first group $-\mathrm{OCH_3}$ is a $-M$ group. The second group $-\mathrm{NH}$ is a $+M$ group. The third group $-\mathrm{CH_3}$ is a $+M$ group. The fourth group $-\mathrm{C \equiv N}$ is a $-M$ group. The fifth group $-\mathrm{OCH_3}$ is a $+M$ group.

Question 87

Chemistry · Classification of Elements and Periodicity in Properties · Numerical

Lowest Oxidation number of an atom in a compound $\mathrm{A_2 B}$ is $-2$. The number of an electron in its valence shell is

Answer: 6

Solution

Given $\mathrm{A_2B \rightarrow 2A^+ + B^{2-}}$, $\mathrm{B^{2-}}$ has complete octet in its dianionic form, thus in its atomic state it has 6 electrons in its valence shell. As it has negative charge, it has acquired two electrons to complete its octet.

Question 88

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

Among the following oxide of p-block elements, number of oxides having amphoteric nature is $\mathrm{Cl_2O_7}$, CO, $\mathrm{PbO_2}$, $\mathrm{N_2O}$, NO, $\mathrm{Al_2O_3}$, $\mathrm{SiO_2}$, $\mathrm{N_2O_5}$, $\mathrm{SnO_2}$

Answer: 3

Solution

Acidic oxide: $\mathrm{Cl_2O_7}$, $\mathrm{SiO_2}$, $\mathrm{N_2O_5}$ Neutral oxide: $\mathrm{CO}$, $\mathrm{NO}$, $\mathrm{N_2O}$ Amphoteric oxide: $\mathrm{Al_2O_3}$, $\mathrm{SnO_2}$, $\mathrm{PbO_2}$

Question 89

Chemistry · Some Basic Concepts of Chemistry · Numerical

Consider the following reaction: $$3\mathrm{PbCl}_2 + 2(\mathrm{NH}_4)_3\mathrm{PO}_4 \rightarrow \mathrm{Pb}_3(\mathrm{PO}_4)_2 + 6\mathrm{NH}_4\mathrm{Cl}$$ If $72\,\mathrm{mmol}$ of $\mathrm{PbCl}_2$ is mixed with $50\,\mathrm{mmol}$ of $(\mathrm{NH}_4)_3\mathrm{PO}_4$, then amount of $\mathrm{Pb}_3(\mathrm{PO}_4)_2$ formed is ___ mmol. (nearest integer)

Answer: 24

Solution

Limiting Reagent is $\mathrm{PbCl_2}$. mmol of $\mathrm{Pb_3(PO_4)_2}$ formed is given by $$\frac{mmol of \mathrm{PbCl_2} reacted}{3}$$ $$= 24 \, \mathrm{mmol}$$

Question 90

Chemistry · Equilibrium · Numerical

$K_a$ for $\mathrm{CH_3COOH}$ is $1.8 \times 10^{-5}$ and $K_b$ for $\mathrm{NH_4OH}$ is $1.8 \times 10^{-5}$. The pH of ammonium acetate solution will be

Answer: 7

Solution

Given $\($ pH = $\frac{pK_w + pK_a - pK_b}{2}$ $\)$ and $\($ pK_a = pK_b $\)$. Therefore, $\($ pH = $\frac{pK_w}{2}$ = 7 $\)$.