JEE Main 25 June 2022 Shift 1 question paper with solutions

JEE Main 25 June 2022 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Conic Sections · Single correct

Let a circle C touch the lines $L_1 : 4x - 3y + K_1 = 0$ and $L_2 : 4x - 3y + K_2 = 0$, $K_1, K_2 \in \mathbb{R}$. If a line passing through the centre of the circle C intersects $L_1$ at $(-1, 2)$ and $L_2$ at $(3, -6)$, then the equation of the circle C is

  1. $(x - 1)^2 + (y - 2)^2 = 4$
  2. $(x + 1)^2 + (y - 2)^2 = 4$
  3. $(x - 1)^2 + (y + 2)^2 = 16$
  4. $(x - 1)^2 + (y - 2)^2 = 16$

Answer: (c)

Solution

Given the lines $L_1 : 4x - 3y + K_1 = 0$ and $L_2 : 4x - 3y + K_2 = 0$. Now, $$-4 - 6 + K_1 = 0 \implies K_1 = 10$$ $$12 + 18 + K_2 = 0 \implies K_2 = -30$$ Tangent to the circle are $$4x - 3y + 10 = 0$$ $$4x - 3y - 30 = 0$$ Length of diameter $2r = \left| \frac{10 + 30}{5} \right| = 8$ Therefore, $r = 4$. Now, the center is the midpoint of $A$ and $B$. $x = 1, \; y = -2$ Equation of the circle $$(x - 1)^2 + (y + 2)^2 = 16$$

Question 2

Maths · Integrals · Single correct

The value of $$\int_{0}^{\pi} \frac{e^{\cos x} \sin x}{(1 + \cos^2 x)(e^{\cos x} + e^{-\cos x})} \, dx$$ is equal to

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

Answer: (c)

Solution

Given $$I = \int_0^\pi \frac{e^{\cos x} \sin x}{(1 + \cos^2 x)(e^{\cos x} + e^{-\cos x})} \, dx \ldots (1)$$ Use King's property $$I = \int_0^\pi \frac{e^{-\cos x} \sin x}{(1 + \cos^2 x)(e^{-\cos x} + e^{\cos x})} \, dx \ldots (2)$$ On adding equation (1) and (2), we get $$2I = \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \, dx = 2 \int_0^{\pi/2} \frac{\sin x}{1 + \cos^2 x} \, dx$$ On putting $\cos x = t$, we get $$I = \int_0^1 \frac{dt}{1 + t^2} = \left( \tan^{-1} t \right)_0^1 = \frac{\pi}{4}$$

Question 3

Maths · Properties of Triangles · Single correct

Let a, b and c be the length of sides of a triangle ABC such that $\frac{a+b}{7}$ = $\frac{b+c}{8}$ = $\frac{c+a}{9}$ . If r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of $\($ $\frac{R}{r}$ $\)$ is equal to

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

Answer: (a)

Solution

$\frac{a+b}{7}=\lambda$ $\frac{b+c}{8}=\lambda$ $\frac{c+a}{9}=\lambda$ $a+b=7\lambda,\ b+c=8\lambda,\ a+c=9\lambda$ $\Rightarrow\ a+b+c=12\lambda$ Now $a=4\lambda,\ b=3\lambda,\ c=5\lambda$ $\therefore\ c^2=b^2+a^2$ $\angle C=90^\circ$ $\Delta=\frac12ab\sin C$ $=\frac12ab$ $\frac{R}{r}$ $=\frac{c}{2\sin C}\times\frac{s}{\Delta}$ $=\frac{c}{2}\times\frac{6\lambda}{\frac12ab}$ $=\frac{c}{ab}\times6\lambda$ $=\frac52$

Question 4

Maths · Relations and Functions · Single correct

Let f : $\mathbb{N}$ $\to$ $\mathbb{R}$ be a function such that f(x+y) = 2f(x)f(y) for natural numbers x and y. If f(1) = 2, then the value of $\alpha$ for which $$\sum_{k=1}^{10} f(\alpha + k) = \frac{512}{3} (2^{20} - 1)$$ holds, is

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

Answer: (c)

Solution

Given $f : \mathbb{N} \to \mathbb{R}$, $f(x + y) = 2 \cdot f(x) \cdot f(y)$ $\ldots$ (1) and $f(1) = 2$, we have $$\sum_{k=1}^{10} f(\alpha + k) = 2f(\alpha) \sum_{k=1}^{10} f(k)$$ $$= 2f(\alpha)(f(1) + f(2) + \ldots + f(10)) \ldots (2)$$ From (1), $f(2) = 2 \cdot f^2(1) = 2^3$, $f(3) = 2 \cdot f(2) \cdot f(1) = 2^5$, and continuing this pattern, $f(10) = 2^9 \cdot f^{10}(1) = 2^{19}$. Therefore, $f(\alpha) = 2^{2\alpha - 1}$ for $\alpha \in \mathbb{N}$. From (2), $$\sum_{k=1}^{10} f(\alpha + k) = 2 \left(2^{2\alpha - 1}\right)(2 + 2^3 + 2^5 + \ldots + 2^{19})$$ $$\frac{512}{3} (2^{20} - 1) = 2^{2\alpha} \left(\frac{2^{20} - 1}{3}\right)$$ Hence $\alpha = 4$.

Question 5

Maths · Matrices · Single correct

Let A be a 3 $\times$ 3 real matrix such that $$A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}; A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$$ and $$A \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}.$$ If $$X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}^T$$ and I is an identity matrix of order 3, then the system $$(A - 2I)X = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}$$ has

  1. no solution
  2. infinitely many solutions
  3. unique solution
  4. exactly two solutions

Answer: (b)

Solution

Given $$A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$$ We have $$A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$$ This implies $$c_1 = 1, \ c_2 = 1, \ c_3 = 2$$ Next, $$A \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} c_1 + a_1 \\ c_2 + a_2 \\ c_3 + a_3 \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$$ This implies $$a_1 = -2, \ a_2 = -1, \ a_3 = -1$$ Then, $$A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$$ This implies $$b_1 = 3, \ b_2 = 2, \ b_3 = 1$$ Thus, $$A = \begin{bmatrix} -2 & 3 & 1 \\ -1 & 2 & 1 \\ -1 & 1 & 2 \end{bmatrix}$$ Now, $$A - 2I = \begin{bmatrix} -4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix}$$ We have $$|A - 2I| = 0$$ Now, $$\begin{bmatrix} -4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \\ 1 \end{bmatrix}$$ This gives the equations: $$-4x_1 + 3x_2 + x_3 = 4 ....(1)$$ $$-x_1 + x_3 = 1 .....(2)$$ $$-x_1 + x_2 = 1 .....(3)$$ $(1)-[(2)+3(3)]$ $0=0$ $\Rightarrow\ \text{infinite solutions}$

Question 6

Maths · Continuity and Differentiability · Single correct

Let f : $\mathbb{R}$ $\to$ $\mathbb{R}$ be defined as f(x) = x^3 + x - 5. If g(x) is a function such that f(g(x)) = x, $\forall$ x $\in$ $\mathbb{R}$, then g'(63) is equal to .

  1. $\frac{1}{49}$
  2. $\frac{3}{49}$
  3. $\frac{43}{49}$
  4. $\frac{91}{49}$

Answer: (a)

Solution

Given $f(x) = x^3 + x - 5$. Differentiating, we have $f'(x) = 3x^2 + 1$, which is an increasing function. Therefore, $f(x)$ is invertible. Let $g(x)$ be the inverse of $f(x)$. Then $g(f(x)) = x$. Differentiating, we get $g'(f(x))f'(x) = 1$. Given $f(x) = 63$, we solve $x^3 + x - 5 = 63$, which gives $x = 4$. Substituting $x = 4$, we have $g'(f(4))f'(4) = 1$. Thus, $g'(63) \times 49 = 1$ since $f'(4) = 49$. Therefore, $g'(63) = \frac{1}{49}$.

Question 7

Maths · Mathematical Reasoning · Single correct

Consider the following two propositions: P1 : $\sim (p \rightarrow \sim q)$ P2 : $(p \land \sim q) \land ((\sim p) \lor q)$ If the proposition $p \rightarrow ((\sim p) \lor q)$ is evaluated as FALSE, then:

  1. P1 is TRUE and P2 is FALSE
  2. P1 is FALSE and P2 is TRUE
  3. Both P1 and P2 are FALSE
  4. Both P1 and P2 are TRUE

Answer: (c)

Solution

The expression $p \to (\sim p \lor q)$ is false when $p$ is true and $q$ is false. From the table, $P1$ and $P2$ both are false.

Question 8

Maths · Sequences and Series · Single correct

If $\frac{1}{2^3 \cdot 3^{10}} + \frac{1}{2^2 \cdot 3^9} + \ldots + \frac{1}{2^{10} \cdot 3} = \frac{K}{2^{10} \cdot 3^{10}}$, then the remainder when $K$ is divided by $6$ is

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

Answer: (d)

Solution

Given $\($ $\frac{1}{2^3 \cdot 3^{10}}$ + $\frac{1}{2^2 \cdot 3^9}$ + $\frac{1}{2^3 \cdot 3^8}$ + $\ldots$ + $\frac{1}{2^{10} \cdot 3}$ = $\frac{K}{2^{10} \cdot 3^{10}}$ $\)$. $\($ K = 2^9 + 2^8 $\cdot$ 3 + 2^7 $\cdot$ 3^2 + $\ldots$ + 3^9 $\)$ $\[$ = $\frac{2^9 \left( \frac{3}{2} \right)^{10} - 1}{\frac{3}{2} - 1}$ = 3^{10} - 2^{10} $\]$ Now, $\($ 3^{10} - 2^{10} = (3^5 - 2^5)(3^5 + 2^5) $\)$ $\($ = (211)(275) $\)$ $\($ = (35 $\times$ 6 + 1)(45 $\times$ 6 + 5) $\)$ $\($ = 6$\lambda$ + 5 $\)$ Remainder is 5.

Question 9

Maths · Applications of Derivatives · Single correct

Let $f(x)$ be a polynomial function such that $f(x) + f'(x) + f''(x) = x^5 + 64$. Then, the value of $$\lim_{x \to 1} \frac{f(x)}{x-1}$$

  1. -15
  2. -60
  3. 60
  4. 15

Answer: (a)

Solution

Given $\($ $\lim$_{{x $\to$ 1}} $\frac{f(x)}{x-1}$ = f'(1) $\)$ and $\($ f(1) = 0 $\)$. $\($ f(x) + f'(x) + f''(x) = x^5 + 64 $\)$ $\($ f'(x) + f''(x) + f'''(x) = 5x^4 $\)$ $\($ f''(x) + f'''(x) + f^{(iv)}(x) = 20x^3 $\)$ $\($ f'''(x) + f^{(iv)}(x) + f^{(v)}(x) = 60x^2 $\)$ Therefore, $\($ f^{(v)}(x) - f'(x) = 60x^2 - 20x^3 $\)$ $\($ $\Rightarrow$ 120 - f'(1) = 40 $\Rightarrow$ f'(1) = 80 $\)$ Also $\($ f(1) + f'(1) + f''(1) = 65 $\Rightarrow$ f''(1) = -15 $\)$. Ans.

Question 10

Maths · Probability · Single correct

Let $E_1$ and $E_2$ be two events such that the conditional probabilities $\mathrm{P}(E_1 | E_2) = \frac{1}{2}$, $\mathrm{P}(E_2 | E_1) = \frac{3}{4}$ and $\mathrm{P}(E_1 \cap E_2) = \frac{1}{8}$. Then:

  1. $\mathrm{P}(E_1 \cap E_2) = \mathrm{P}(E_1) \cdot \mathrm{P}(E_2)$
  2. $\mathrm{P}(E_1' \cap E_2') = \mathrm{P}(E_1') \cdot \mathrm{P}(E_2)$
  3. $\mathrm{P}(E_1 \cap E_2') = \mathrm{P}(E_1) \cdot \mathrm{P}(E_2)$
  4. $\mathrm{P}(E_1' \cap E_2) = \mathrm{P}(E_1) \cdot \mathrm{P}(E_2)$

Answer: (c)

Solution

(A) $\mathrm{P}(E_1) \cdot \mathrm{P}(E_2) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24} \neq \mathrm{P}(E_1 \cap E_2)$ (B) $\mathrm{P}(E_1' \cap E_2') = 1 - \mathrm{P}(E_1 \cup E_2)$ $$= 1 - (\mathrm{P}(E_1) + \mathrm{P}(E_2) - \mathrm{P}(E_1 \cap E_2))$$ $$= 1 - \left( \frac{1}{6} + \frac{1}{4} - \frac{1}{8} \right) = \frac{17}{24}$$ $\mathrm{P}(E_1')\mathrm{P}(E_2') = \frac{5}{6} \times \frac{1}{4} = \frac{5}{24}$ \text{(C)}$\mathrm{P}(E_1 \cap E_2') = \mathrm{P}(E_1) - \mathrm{P}(E_1 \cap E_2) = \frac{1}{6} - \frac{1}{8} = \frac{1}{24}$ (D) $\mathrm{P}(E_1' \cap E_2) = \mathrm{P}(E_2) - \mathrm{P}(E_1 \cap E_2) = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$

Question 11

Maths · Matrices · Single correct

Let $A= \begin{bmatrix} 0 & -2\\ 2 & 0 \end{bmatrix}$. If $M$ and $N$ are two matrices given by $M=\sum_{k=1}^{10}A^{2k}$ and $N=\sum_{k=1}^{10}A^{2k-1}$, then $MN^2$ is ______.

  1. a non-identity symmetric matrix
  2. a skew-symmetric matrix
  3. neither symmetric nor skew-symmetric matrix
  4. an identity matrix

Answer: (a)

Solution

Given $$A = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$$ We calculate $$A^2 = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix} = -4I$$ Then $$A^3 = -4A$$ $$A^4 = (-4I)(-4I) = (-4)^2 I$$ $$A^5 = (-4)^2 A, A^6 = (-4)^3 I$$ Now, $$M = \sum_{k=1}^{10} A^{2k} = A^2 + A^4 + \ldots + A^{20}$$ $$= [-4 + (-4)^2 + (-4)^3 + \ldots + (-4)^{20}]I$$ $$= -4\lambda I$$ Thus, $M$ is a symmetric matrix. For $N$, $$N = \sum_{k=1}^{10} A^{2k-1} = A + A^3 + \ldots + A^{19}$$ $$= A[1 + (-4) + (-4)^2 + \ldots + (-4)^9]$$ $$= \lambda A \implies skew symmetric$$ Therefore, $N^2$ is a symmetric matrix and $MN^2$ is a non-identity symmetric matrix.

Question 12

Maths · Applications of Integrals · Single correct

Let g : (0, $\infty$) $\to$ $\mathbb{R}$ be a differentiable function such that $\int$( $\frac{x(\cos x - \sin x)}{e^x + 1}$ + $\frac{g(x)(e^x + 1 - xe^x)}{(e^x + 1)^2}$) $\mathrm{d}$x = $\frac{xg(x)}{e^x + 1}$ + c, for all x > 0, where c is an arbitrary constant. Then.

  1. g is decreasing in $\left(0, \frac{\pi}{4}\right)$
  2. g' is increasing in $\left(0, \frac{\pi}{4}\right)$
  3. g + g' is increasing in $\left(0, \frac{\pi}{2}\right)$
  4. g - g' is increasing in $\left(0, \frac{\pi}{2}\right)$

Answer: (d)

Solution

On differentiating both sides with respect to $x$, we get $$\left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - xe^x)}{(e^x + 1)^2} \right) = \frac{(e^x + 1)(g(x) + xg'(x)) - e^x \cdot x \cdot g(x)}{(e^x + 1)^2}$$ $$(e^x + 1)x(\cos x - \sin x) + g(x)(e^x + 1 - xe^x)$$ $$= (e^x + 1)(g(x) + xg'(x)) - e^x \cdot x \cdot g(x)$$ $$\Rightarrow g'(x) = \cos x - \sin x$$ $$\Rightarrow g(x) = \sin x + \cos x + C$$ $g(x)$ is increasing in $(0, \pi/4)$ $$g''(x) = -\sin x - \cos x 0$$ $$\Rightarrow \phi$ is increasing Hence option D is correct.

Question 13

Maths · Applications of Derivatives · Single correct

Let f : $\mathbb{R} \to \mathbb{R}$ and g : $\mathbb{R} \to \mathbb{R}$ be two functions defined by $f(x) = \log_e(x^2 + 1) - e^{-x} + 1$ and $g(x) = \frac{1 - 2e^{2x}}{e^x}$. Then, for which of the following range of $\alpha$, the inequality $$f\left(g\left(\frac{(\alpha - 1)^2}{3}\right)\right) > f\left(g\left(\frac{\alpha - 5}{3}\right)\right)$$ holds?

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

Answer: (a)

Solution

Given $f(x) = \log_e(x^2 + 1) - e^{-x} + 1$. Differentiating, we have $$f'(x) = \frac{2x}{x^2 + 1} + e^{-x} > 0 \forall \; x \in \mathbb{R}$$ This implies $f$ is strictly increasing. Let $g(x) = \frac{1 - 2e^{2x}}{e^x} = e^{-x} - 2e^x$. Differentiating, we find $$g'(x) = -(2e^x + e^{-x}) f\left(g\left(\alpha - \frac{5}{3}\right)\right)$$ This implies $$g\left(\frac{(\alpha - 1)^2}{3}\right) > g\left(\alpha - \frac{5}{3}\right)$$ Thus, $$\frac{(\alpha - 1)^2}{3} < \alpha - \frac{5}{3}$$ Simplifying, we get $$\alpha^2 - 5\alpha + 6 < 0$$ Factoring, $$(\alpha - 2)(\alpha - 3) < 0$$ Therefore, $\alpha \in (2, 3)$.

Question 14

Maths · Vector Algebra · Single correct

Let \[ \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}, \] where $a_i>0,\ i=1,2,3$, be a vector which makes equal angles with the coordinate axes $\mathrm{OX}$, $\mathrm{OY}$ and $\mathrm{OZ}$. Also, let the projection of $\vec{a}$ on the vector $3\hat{i}+4\hat{j}$ be $7$. Let $\vec{b}$ be a vector obtained by rotating $\vec{a}$ through $90^\circ$. If $\vec{a}$, $\vec{b}$ and the $x$-axis are coplanar, then the projection of vector $\vec{b}$ on $3\hat{i}+4\hat{j}$ is equal to

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

Answer: (b)

Solution

Given $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$. Let $\vec{a} = \lambda \left( \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \right) = \frac{\lambda}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})$. Now projection of $\vec{a}$ on $\vec{b} = 7$. $$\Rightarrow \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 7$$ $$\frac{\lambda}{\sqrt{3}} \frac{(\hat{i} + \hat{j} + \hat{k}) \cdot (3\hat{i} + 4\hat{j})}{5} = 7$$ $$\lambda = 5\sqrt{3}$$ $$\vec{a} = 5(\hat{i} + \hat{j} + \hat{k})$$ Now $\vec{b} = 5\alpha (\hat{i} + \hat{j} + \hat{k}) + \beta (\hat{i})$ $$\vec{a} \cdot \vec{b} = 0$$ $$\Rightarrow 25\alpha (3) + 5\beta = 0$$ $$\Rightarrow 15\alpha + \beta = 0 \Rightarrow \beta = -15\alpha$$ $$\vec{b} = 5\alpha (-2\hat{i} + \hat{j} + \hat{k})$$ $$|\vec{b}| = 5\sqrt{3}$$ $$\Rightarrow \alpha = \pm \frac{1}{\sqrt{2}}$$ $$\vec{b} = \pm \frac{5}{\sqrt{2}} (-2\hat{i} + \hat{j} + \hat{k})$$ Projection of $\vec{b}$ on $3\hat{i} + 4\hat{j}$ is $$\frac{\vec{b} \cdot (3\hat{i} + 4\hat{j})}{5} = \pm \frac{5}{\sqrt{2}} \left( \frac{-6 + 4}{5} \right) = \pm \sqrt{2}$$

Question 15

Maths · Differential Equations · Single correct

Let y = y(x) be the solution of the differential equation (x + 1)y' - y = e^{3x}(x + 1)^2, with y(0) = $\frac{1}{3}$. Then, the point x = -$\frac{4}{3}$ for the curve y = y(x) is:

  1. not a critical point
  2. a point of local minima
  3. a point of local maxima
  4. a point of inflection

Answer: (b)

Solution

(x + 1)dy - y $\,$ dx = e^{3x}(x + 1)^2 $\frac{(x + 1)dy - y \, dx}{(x + 1)^2}$ = e^{3x} $\mathrm{d}$$\left$($\frac{y}{x+1}$$\right$) = e^{3x} $\implies$ $\frac{y}{x+1}$ = $\frac{e^{3x}}{3}$ + C $\left$(0, $\frac{1}{3}$$\right$) $\implies$ C = 0 $\implies$ y = $\frac{(x+1)e^{3x}}{3}$ $\frac{dy}{dx}$ = $\frac{1}{3}$((x+1)3e^{3x} + e^{3x}) = $\frac{3^{3x}}{3}$(3x + 4) Clearly, $\;$ x = $\frac{-4}{3}$ $\;$ is point of local minima

Question 16

Maths · Conic Sections · Single correct

If $y = m_1x + c_1$ and $y = m_2x + c_2$, $m_1 \neq m_2$ are two common tangents of circle $x^2 + y^2 = 2$ and parabola $y^2 = x$, then the value of $8|m_1m_2|$ is equal to

  1. 3 + 4$\sqrt{2}$
  2. -5 + 6$\sqrt{2}$
  3. -4 + 3$\sqrt{2}$
  4. 7 + 6$\sqrt{2}$

Answer: (c)

Solution

Given the equations $C_1: x^2 + y^2 = 2$ and $C_2: y^2 = x$. Let the tangent to the parabola be $y = mx + \frac{1}{4m}$. It is also a tangent of the circle, so the distance from the center of the circle $(0, 0)$ will be $\sqrt{2}$. $$\left| \frac{1}{\frac{4m}{\sqrt{1+m^2}}} \right| = \sqrt{2} \implies 1 = 32m^2 + 32m^4$$ By solving $$m^2 = \frac{3\sqrt{2} - 4}{8}, m^2 = \frac{-3\sqrt{2} - 4}{8} (rejected)$$ $$m = \pm \frac{\sqrt{3\sqrt{2} - 4}}{8}$$ So, $8 \left| m_1 m_2 \right| = 3\sqrt{2} - 4$

Question 17

Maths · Three Dimensional Geometry · Single correct

Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, -1, -1), parallel to the line PQ meets the plane S at R, then QR^2 is equal to:

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

Answer: (b)

Solution

Let parallel vector of L = $\vec{b}$. Mirror image of Q on given plane $x+y+z=5$: $$\frac{a-1}{1} = \frac{b-0}{1} = \frac{c-1}{1} = \frac{-2(2-5)}{3}$$ $a = 3$, $b = 2$, $c = 3$ $Q = (3, 2, 3)$ Therefore, $\vec{b} \parallel \overline{PQ}$ So, $\vec{b} = (1, 1, 1)$ Equation of line L: $$\frac{x-1}{1} = \frac{y+1}{1} = \frac{z+1}{1}$$ Let point R, $(\lambda + 1, \lambda - 1, \lambda - 1)$ lying on plane $x + y + z = 5$, so, $3\lambda - 1 = 5$ Therefore, $\lambda = 2$ Point R is $(3, 1, 1)$ $QR^2 = 5$ Ans.

Question 18

Maths · Differential Equations · Single correct

If the solution curve $y = y(x)$ of the differential equation $y^2 dx + (x^2 - xy + y^2) dy = 0$, which passes through the point $(1, 1)$ and intersects the line $y = \sqrt{3} \, x$ at the point $(\alpha, \sqrt{3} \, \alpha)$, then value of $\log_e(\sqrt{3} \, \alpha)$ is equal to

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

Answer: (c)

Solution

Given $$y^2 dx - xy \, dy = - (x^2 + y^2) dy$$ Rearranging, $$y(y \, dx - x \, dy) = - (x^2 + y^2) \, dy$$ $$- y(x \, dx - y \, dx) = - (x^2 + y^2) dy$$ This implies $$\frac{x \, dy - y \, dx}{x^2} = \left(1 + \frac{y^2}{x^2}\right) \frac{dy}{y}$$ Therefore, $$\Rightarrow \frac{d(y/x)}{1 + \frac{y^2}{x^2}} = \frac{dy}{y}$$ Integrating both sides, $$\Rightarrow \tan^{-1}\left(\frac{y}{x}\right) = \ln y + C$$ Given the points $$(\alpha, \sqrt{3} \alpha) \Rightarrow \frac{\pi}{3} = \ln(\sqrt{3} \alpha) + \frac{\pi}{4}$$ Thus, $$\therefore \ln(\sqrt{3} \alpha) = \frac{\pi}{12}$$

Question 19

Maths · Conic Sections · Single correct

Let x = 2t, y = $\frac{t}{3}$ be a conic. Let S be the focus and B be the point on the axis of the conic such that SA $\perp$ BA, where A is any point on the conic. If k is the ordinate of the centroid of $\Delta$ SAB, then $\lim$_{t $\to$ 1} k is equal to

  1. $\frac{17}{18}$
  2. $\frac{19}{18}$
  3. $\frac{11}{18}$
  4. $\frac{13}{18}$

Answer: (d)

Solution

Parabola $x^2=12y$ $SA\perp SB$ so, $m_{AS}\cdot m_{AB}=-1$ $\left( \frac{3-t^2}{-2t} \right) \left( \frac{\alpha-t^2}{0-2t} \right) =-1$ By solving, $3\alpha= \frac{27t^2+t^4}{t^2-9}$ Ordinate of centroid of $\triangle SAB$ $=K$ $=\frac{\alpha+\frac{t^2}{3}+3}{3}$ $K= \frac{9+3\alpha+t^2}{9}$ $\lim_{t\to1}K$ $=\lim_{t\to1} \frac1{9} \left( 9+t^2+ \frac{27t^2+t^4}{t^2-9} \right)$ $=\frac{13}{18}$

Question 20

Maths · Complex Numbers and Quadratic Equations · Single correct

Let a circle C in complex plane pass through the points $z_1 = 3 + 4i$, $z_2 = 4 + 3i$ and $z_3 = 5i$. If $z(\neq z_1)$ is a point on C such that the line through $z$ and $z_1$ is perpendicular to the line through $z_2$ and $z_3$, then $\arg(z)$ is equal to:

  1. $\tan^{-1}\left(\frac{2}{\sqrt{5}}\right) - \pi$
  2. $\tan^{-1}\left(\frac{24}{7}\right) - \pi$
  3. $\tan^{-1}(3) - \pi$
  4. $\tan^{-1}\left(\frac{3}{4}\right) - \pi$

Answer: (b)

Solution

Slope of BC = $\frac{3 - 5}{4 - 0}$ = -$\frac{1}{2}$. Slope of AP = 2. Equation of AP: y - 4 = 2(x - 3). Therefore, y = 2(x - 1). P lies on circle x^2 + y^2 = 25. Therefore, x^2 + (2(x - 1))^2 = 25. Therefore, x = -$\frac{7}{5}$ and y = -$\frac{24}{5}$. Therefore, arg (z) = $\tan$^{-1}$\left$($\frac{24}{7}$$\right$) - $\pi$.

Question 21

Maths · Binomial Theorem · Numerical

Let $C_r$ denote the binomial coefficient of $x^r$ in the expansion of $(1 + x)^{10}$. If $\alpha, \beta \in \mathbb{R}$. $C_1 + 3 \cdot 2C_2 + 5 \cdot 3C_3 + \ldots$ upto 10 terms $$= \frac{\alpha \times 2^{11}}{2^\beta - 1} \left( C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots upto 10 terms \right)$$ then the value of $\alpha + \beta$ is equal to

Answer: 286

Question 22

Maths · Permutations and Combinations · Numerical

The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is _____.

Answer: 63

Solution

Given $x$, $y$, $z$ are odd numbers. $z = 1, 3, 5, 7, 9$. $x + y + z = 7, 14, 21$ [sum of digit multiple of 7]. $\frac{x+y}{1+0+9} = 6, 4, 2, 13, 11, 9, 7, 5, 20, 18, 16, 14, 12$. $x + y = 6 \Rightarrow (1,5), (2,4), (3,3), (4,2), (5,1), (6,0)$ $\rightarrow \mathrm{T.N.} = 6$ $x + y = 4 \Rightarrow (1,3), (2,2), (3,1), (4,0)$ $\rightarrow \mathrm{T.N.} = 4$ $x + y = 2 \Rightarrow (1,1), (2,0)$ $\rightarrow \mathrm{T.N.} = 2$ $x + y = 13 \Rightarrow (4,9), (5,8), (6,7), (7,6), (8,5), (9,4)$ $\rightarrow \mathrm{T.N.} = 6$ $x + y = 11 \Rightarrow (2,9), (3,8), (4,7), (5,6), (6,5), (6,5), (7,4), (8,3), (9,2)$ $\rightarrow \mathrm{T.N.} = 8$ $x + y = 9 \Rightarrow (1,8), (2,7), (3,8), (4,5), (5,4), \ldots, (8,1), (9,0)$ $\rightarrow \mathrm{T.N.} = 9$ $x + y = 7 \Rightarrow (1,8), (2,5), (3,4), \ldots, (8,1), (7,0)$ $\rightarrow \mathrm{T.N.} = 7$ $x + y = 5 \Rightarrow (1,4), (2,3), (3,2), (4,1), (5,0)$ $\rightarrow \mathrm{T.N.} = 5$ $x + y = 20 \Rightarrow Not possible$ $x + y = 18 \Rightarrow (9,9)$ $\rightarrow \mathrm{T.N.} = 1$ $x + y = 16 \Rightarrow (7,9), (8,8), (9,7)$ $\rightarrow \mathrm{T.N.} = 3$ $x + y = 14 \Rightarrow (5,9), (6,8), (7,7), (8,6), (9,5)$ $\rightarrow \mathrm{T.N.} = 5$ $x + y = 12 \Rightarrow (3,9), (4,8), (5,7), (6,6), \ldots, (9,3)$ $\rightarrow \mathrm{T.N.} = 7$

Question 23

Maths · Vector Algebra · Fill in the blank

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$, where \[ |\vec{a}|=4,\qquad |\vec{b}|=3,\qquad \theta\in\left(\frac{\pi}{4},\frac{\pi}{3}\right). \] Then \[ \left|(\vec{a}-\vec{b})\times(\vec{a}+\vec{b})\right|^2 +4(\vec{a}\cdot\vec{b})^2 \] is equal to $\underline{\hspace{2cm}}$.

Answer: 576

Solution

Given $|\vec{a}| = 4$, $|\vec{b}| = 3$, $\theta \in \left( \frac{\pi}{4}, \frac{\pi}{3} \right)$. $$\left| (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) \right|^2 + 4 (\vec{a} \cdot \vec{b})^2$$ $$\left| \vec{a} \times \vec{b} - \vec{b} \times \vec{a} \right|^2 + 4a^2b^2 \cos^2 \theta$$ $$2 \left| \vec{a} \times \vec{b} \right|^2 + 4a^2b^2 \cos^2 \theta$$ $$4a^2b^2 \sin^2 \theta + 4a^2b^2 \cos^2 \theta$$ $$4a^2b^2 = 4 \times 16 \times 9 = 576$$

Question 24

Maths · Conic Sections · Numerical

Let the abscissae of the two points P and Q be the roots of $2x^2 - rx + p = 0$ and the ordinates of P and Q be the roots of $x^2 - sx - q = 0$. If the equation of the circle described on PQ as diameter is $2(x^2 + y^2) - 11x - 14y - 22 = 0$, then $2r + s - 2q + p$ is equal to

Answer: 7

Solution

Given the equations $2x^2 - rx + p = 0$ with roots $x_1, x_2$ and $y^2 - sy - q = 0$ with roots $y_1, y_2$. The equation of the circle with $PQ$ as diameter is $2(x^2 + y^2) - rx - 2sy + p - 2q = 0$. On comparing with the given equation, we have $r = 11$, $s = 7$, and $p - 2q = -22$. Therefore, $2r + s - 2q + p = 22 + 7 - 22 = 7$.

Question 25

Maths · Trigonometric Functions · Numerical

The number of values of $x$ in the interval $\left( \frac{\pi}{4}, \frac{7\pi}{4} \right)$ for which $14 \csc^2 x - 2 \sin^2 x = 21 - 4 \cos^2 x$ holds, is _______

Answer: 4

Solution

Given $x \in \left( \frac{\pi}{4}, \frac{7\pi}{4} \right)$. $14 \csc^2 x - 2 \sin^2 x = 21 - 4 \cos^2 x$ $= 21 - 4 (1 - \sin^2 x)$ $= 17 + 4 \sin^2 x$ $14 \csc^2 x - 6 \sin^2 x = 17$ Let $\sin^2 x = p$. $\frac{14}{p} - 6p = 17 \Rightarrow 14 - 6p^2 = 17p$ $6p^2 + 17p - 14 = 0$ $p = -3.5, \frac{2}{3} \Rightarrow \sin^2 x = \frac{2}{3}$ $\Rightarrow \sin x = \pm \sqrt{\frac{2}{3}}$

Question 26

Maths · Sequences and Series · Numerical

For a natural number n, let $a_n = 19^n - 12^n$. Then, the value of $$\frac{31 \alpha_9 - \alpha_{10}}{57 \alpha_8}$$ is

Answer: 4

Solution

Given $a_n = 19^n - 12^n$. $$\frac{31 \alpha_9 - \alpha_{10}}{57 \alpha_8} = \frac{31(19^9 - 12^9) - (19^{10} - 12^{10})}{57 \alpha_8}$$ $$= \frac{19^9 (31 - 19) - 12^9 (31 - 12)}{57 \alpha_8}$$ $$= \frac{19^9 \cdot 12 - 12^9 \cdot 19}{57 \alpha_8}$$ $$= \frac{12 \cdot 19 (19^8 - 12^8)}{57 \alpha_8} = 4$$

Question 27

Maths · Relations and Functions · Numerical

Let f : $\mathbb{R}$ $\to$ $\mathbb{R}$ be a function defined by $$f(x) = 2 \left( 1 - \frac{x^{25}}{2} \right) \left( 2 + x^{25} \right)^{\frac{1}{50}}.$$ If the function $$g(x) = f\left(f\left(f(x)\right)\right) + f\left(f(x)\right),$$ the greatest integer less than or equal to $g(1)$ is

Answer: 2

Solution

Given $$f(x) = \left[ 2 \left( 1 - \frac{x^{25}}{2} \right) (2 + x^{25}) \right]^{\frac{1}{50}}$$ Simplifying, we have $$f(x) = \left[ (2 - x^{25})(2 + x^{25}) \right]^{\frac{1}{50}}$$ $$= (4 - x^{50})^{\frac{1}{50}}$$ Now, $$f(f(x)) = \left( 4 - \left( (4 - x^{50})^{\frac{1}{50}} \right)^{50} \right)^{\frac{1}{50}} = x$$ Thus, $$g(x) = f(f(f(x))) + f(f(x))$$ $$= f(x) + x$$ For $$g(1) = f(1) + 1 = 3^{1/50} + 1$$ $$[g(1)] = [3^{1/50} + 1] = 2$$

Question 28

Maths · Three Dimensional Geometry · Numerical

Let the lines $$L_1 : \vec{r} = \lambda (\hat{i} + 2\hat{j} + 3\hat{k}), \lambda \in \mathbb{R}$$ $$L_2 : \vec{r} = (\hat{i} + 3\hat{j} + \hat{k}) + \mu (\hat{i} + \hat{j} + 5\hat{k}); \mu \in \mathbb{R}$$ intersect at the point $S$. If a plane $ax + by - z + d = 0$ passes through $S$ and is parallel to both the lines $L_1$ and $L_2$, then the value of $a + b + d$ is equal to .

Answer: 5

Solution

Both the lines lie in the same plane. Therefore, equation of the plane $$\begin{vmatrix} x & y & z \\ 1 & 2 & 3 \\ 1 & 1 & 5 \end{vmatrix} = 0$$ $$\Rightarrow 7x - 2y - z = 0$$ Therefore, $a + b + d = 5$

Question 29

Maths · Matrices · Numerical

Let A be a $3 \times 3$ matrix having entries from the set $\{-1, 0, 1\}$. The number of all such matrices A having sum of all the entries equal to 5, is

Answer: 414

Solution

Case-I: 1 occurs 7 times and -1 occurs 2 times. The number of possible matrices is $$\frac{9!}{7! \, 2!} = 36$$. Case-II: 1 occurs 6 times, -1 occurs 1 time, and 0 occurs 2 times. The number of possible matrices is $$\frac{9!}{6! \, 2!} = 252$$. Case-III: 1 occurs 5 times, and 0 occurs 4 times. The number of possible matrices is $$\frac{9!}{5! \, 4!} = 126$$. Hence the total number of all such matrices A is 414.

Question 30

Maths · Sequences and Series · Numerical

The greatest integer less than or equal to the sum of first 100 terms of the sequence $$\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, \ldots$$ is equal to

Answer: 98

Solution

The series is given by $$\frac{1}{3} + \frac{5}{9} + \frac{19}{27} + \frac{65}{81} + \ldots$$ This can be rewritten as $$\left(1 - \frac{2}{3}\right) + \left(1 - \frac{4}{9}\right) + \left(1 - \frac{8}{27}\right) + \left(1 - \frac{16}{81}\right) \ldots 100 terms$$ This simplifies to $$100 - \left[\frac{2}{3} + \left(\frac{2}{3}\right)^2 + \ldots \right]$$ Using the formula for the sum of a geometric series, we have $$100 - \frac{\frac{2}{3} \left(1 - \left(\frac{2}{3}\right)^{100}\right)}{1 - \frac{2}{3}}$$ This simplifies to $$100 - 2 \left(1 - \left(\frac{2}{3}\right)^{100}\right)$$ Therefore, $$S = 98 + 2 \left(\frac{2}{3}\right)^{100}$$ Thus, $$\Rightarrow [S] = 98$$

Physics

Question 31

Physics · Mathematics in Physics · Single correct

If $Z = \frac{A^2 B^3}{C^4}$, then the relative error in $Z$ will be:

  1. $\frac{\Delta A}{A} + \frac{\Delta B}{B} + \frac{\Delta C}{C}$
  2. $\frac{2 \Delta A}{A} + \frac{3 \Delta B}{B} - \frac{4 \Delta C}{C}$
  3. $\frac{2 \Delta A}{A} + \frac{3 \Delta B}{B} + \frac{4 \Delta C}{C}$
  4. $\frac{\Delta A}{A} + \frac{\Delta B}{B} - \frac{\Delta C}{C}$

Answer: (c)

Solution

Given $Z = \frac{A^2 B^3}{C^4}$. In case of error $$\frac{dZ}{Z} = \frac{2 \, dA}{A} + \frac{3 \, dB}{B} + \frac{4 \, dC}{C}$$ $$\frac{\Delta Z}{Z} = \frac{2 \, \Delta A}{A} + \frac{3 \, \Delta B}{B} + \frac{4 \, \Delta C}{C}$$

Question 32

Physics · Mathematics in Physics · Single correct

$\vec{A}$ is a vector quantity such that |$\vec{A}$| = non-zero constant. Which of the following expressions is true for $\vec{A}$?

  1. $\vec{A}$ $\cdot$ $\vec{A}$ = 0
  2. $\vec{A}$ $\times$ $\vec{A}$ < 0
  3. $\vec{A}$ $\times$ $\vec{A}$ = 0
  4. $\vec{A}$ $\times$ $\vec{A}$ > 0

Answer: (c)

Solution

Given $|\vec{A}| \neq 0$. $$\vec{A} \times \vec{A} = |\vec{A}| |\vec{A}| \sin 0^\circ \hat{n} = 0$$

Question 33

Physics · Mathematics in Physics · Single correct

Which of the following relations is true for two unit vectors $\hat{A}$ and $\hat{B}$ making an angle $\theta$ to each other?

  1. $|\hat{A} + \hat{B}| = |\hat{A} - \hat{B}| \tan \frac{\theta}{2}$
  2. $|\hat{A} - \hat{B}| = |\hat{A} + \hat{B}| \tan \frac{\theta}{2}$
  3. $|\hat{A} + \hat{B}| = |\hat{A} - \hat{B}| \cos \frac{\theta}{2}$
  4. $|\hat{A} - \hat{B}| = |\hat{A} + \hat{B}| \cos \frac{\theta}{2}$

Answer: (b)

Solution

Given $|\hat{A} + \hat{B}| = \sqrt{|\hat{A}|^2 + |\hat{B}|^2 + 2 |\hat{A}||\hat{B}| \cos \theta}$. $$= \sqrt{1 + 1 + 2 \cos \theta}$$ $$= \sqrt{2(1 + \cos \theta)}$$ $$= \sqrt{2 \times 2 \cos^2 \frac{\theta}{2}}$$ $$= 2 \cos \frac{\theta}{2}$$ For $|\hat{A} - \hat{B}| = \sqrt{|\hat{A}|^2 + |\hat{B}|^2 - 2 |\hat{A}||\hat{B}| \cos \theta}$. $$= \sqrt{2 - 2 \cos \theta}$$ $$= 2 \sin \frac{\theta}{2}$$ Thus, $$\frac{|\hat{A} + \hat{B}|}{|\hat{A} - \hat{B}|} = \cot \frac{\theta}{2}$$

Question 34

Physics · System of Particles and Rotational Motion · Single correct

If force $\mathbf{F} = 3\hat{i} + 4\hat{j} - 2\hat{k}$ acts on a particle having position vector $2\hat{i} + \hat{j} + 2\hat{k}$ then, the torque about the origin will be :-

  1. $3\hat{i} + 4\hat{j} - 2\hat{k}$
  2. $-10\hat{i} + 10\hat{j} + 5\hat{k}$
  3. $10\hat{i} + 5\hat{j} - 10\hat{k}$
  4. $10\hat{i} + \hat{j} - 5\hat{k}$

Answer: (b)

Solution

Given $\vec{\tau} = \vec{r} \times \vec{F}$. $$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 3 & 4 & -2 \end{vmatrix}$$ $$= \hat{i}(-2 - 8) - \hat{j}(-4 - 6) + \hat{k}(8 - 3)$$ $$= -10\hat{i} + 10\hat{j} + 5\hat{k}$$

Question 35

Physics · Gravitation · Single correct

The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given $g =$ acceleration due to gravity at the surface of earth)

  1. $g/2$
  2. $g/4$
  3. $g/3$
  4. $g/9$

Answer: (d)

Solution

Given $g = \frac{Gm}{r^2}$. For the new position, $g' = \frac{Gm}{(3r)^2}$. Simplifying, $g' = \frac{Gm}{9r^2}$. Therefore, $g' = \frac{g}{9}$.

Question 36

Physics · Mechanical Properties of Fluids · Single correct

The terminal velocity ($v_t$) of the spherical rain drop depends on the radius ($r$) of the spherical rain drop as:-

  1. $r^{1/2}$
  2. $r$
  3. $r^2$
  4. $r^3$

Answer: (c)

Solution

The terminal velocity $V_t$ is given by the equation $$V_t = \frac{2}{9} \frac{gr^2 (\rho_p - \rho_l)}{\eta};$$ where $V_t \propto r^2$.

Question 37

Physics · Kinetic Theory · Single correct

The relation between root mean square speed ($v_{rms}$) and most probable speed ($v_{p}$) for the molar mass $M$ of oxygen gas molecule at the temperature of 300 K will be :-

  1. $v_{rms} = \sqrt{\frac{2}{3}} v_{p}$
  2. $v_{rms} = \sqrt{\frac{3}{2}} v_{p}$
  3. $v_{rms} = v_{p}$
  4. $v_{rms} = \sqrt{\frac{1}{3}} v_{p}$

Answer: (b)

Solution

Given $v_{rms} = \sqrt{\frac{3RT}{M}}$ and $v_{mp} = \sqrt{\frac{2RT}{M}}$. Thus $v_{rms} = \sqrt{\frac{3}{2}} v_{mp}$.

Question 38

Physics · Electric Charges and Fields · Single correct

In the figure, a very large plane sheet of positive charge is shown. $P_1$ and $P_2$ are two points at distance $l$ and $2l$ from the charge distribution. If $\sigma$ is the surface charge density, then the magnitude of electric fields $E_1$ and $E_2$ at $P_1$ and $P_2$ respectively are:

  1. $E_1 = \sigma / \varepsilon_0$, $E_2 = \sigma / 2\varepsilon_0$
  2. $E_1 = 2\sigma / \varepsilon_0$, $E_2 = \sigma / \varepsilon_0$
  3. $E_1 = E_2 = \sigma / 2\varepsilon_0$
  4. $E_1 = E_2 = \sigma / \varepsilon_0$

Answer: (c)

Solution

As the sheet is very large, $\vec{E}$ is independent of distance from it. Thus $E_1 = E_2 = \frac{\sigma}{2 \varepsilon_0}$.

Question 39

Physics · Alternating Current · Single correct

Match List-I with List-II List-I (A) AC generator (B) Galvanometer (C) Transformer (D) Metal detector List-II (I) Detects the presence of current in the circuit (II) Converts mechanical energy into electrical energy (III) Works on the principle of resonance in AC circuit (IV) Changes an alternating voltage for smaller or greater value Choose the correct answer from the options given below :-

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

Answer: (a)

Solution

AC generator converts mechanical energy into electrical energy. Galvanometer shows deflection when current passes through it so it is used to show presence of current in any wire. Transformer is used to step up or step down the voltage. Metals detectors contain inductor coils and use principle of induction and resonance in AC circuit.

Question 40

Physics · Moving Charges and Magnetism · Single correct

A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r < R) from its centre will be :-

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

Answer: (b)

Solution

Question 41

Physics · Alternating Current · Single correct

If wattless current flows in the AC circuit, then the circuit is

  1. Purely Resistive circuit
  2. Purely Inductive circuit
  3. LCR series circuit
  4. RC series circuit only

Answer: (b)

Solution

Purely Inductive circuit $$\theta = \frac{\pi}{2}$$ $$\cos \frac{\pi}{2} = 0$$ Average power = 0

Question 42

Physics · Electromagnetic Waves · Single correct

The electric field in an electromagnetic wave is given by $\mathbf{E} = 56.5 \sin \omega (t - x/c) \, \mathrm{NC}^{-1}$. Find the intensity of the wave if it is propagating along $x$-axis in the free space. (Given $\varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C}^2 \mathrm{N}^{-1} \mathrm{m}^{-2}$)

  1. 5.65 $\mathrm{Wm}^{-2}$
  2. 4.24 $\mathrm{Wm}^{-2}$
  3. 1.9 $\times$ 10^{-7} $\mathrm{Wm}^{-2}$
  4. 56.5 $\mathrm{Wm}^{-2}$

Answer: (b)

Solution

Given $$I = \frac{1}{2} \varepsilon_0 E_0^2 c$$ $$I = \frac{1}{2} \times (8.85 \times 10^{-12})(56.5)^2 (3 \times 10^8)$$ $$= 4.24 \, \mathrm{Wm^{-2}}.$$

Question 43

Physics · Wave Optics · Single correct

The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point P and $\pi$ at point Q. Then the difference between the resultant intensities at P and Q will be:

  1. 2 I
  2. 6 I
  3. 5 I
  4. 7 I

Answer: (b)

Solution

Given $\($ I_P = I + 9I + 2 $\sqrt{I \times 9I}$ $\cos$ $\frac{\pi}{2}$ $\)$. $\($ I_P = 10I $\)$. $\($ I_Q = I + 9I + 2 $\sqrt{I \times 9I}$ $\cos$ $\pi$ $\)$. $\($ = 10I - 6I = 4I $\)$. Therefore, $\($ I_P - I_Q = 10I - 4I = 6I $\)$.

Question 44

Physics · Wave Optics · Single correct

A light wave travelling linearly in a medium of dielectric constant 4, incident on the horizontal interface separating medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be (Given : relative permeability of medium $\mu_r = 1$)

  1. $10^\circ$
  2. $20^\circ$
  3. $30^\circ$
  4. $60^\circ$

Answer: (d)

Solution

For total internal reflection, $i > \theta_C$. $$\Rightarrow \sin i > \sin \theta_C$$ $$\Rightarrow \sin i > \frac{\mu_R}{\mu_D} \ldots (1)$$ Also, $\mu = \sqrt{\mu_r \varepsilon_r}$. $$\frac{\mu_R}{\mu_D} = \frac{\sqrt{1 \times 1}}{\sqrt{4 \times 1}} = \frac{1}{2}$$ From (1), $\sin i > \frac{1}{2} \Rightarrow i > 30^\circ$, $i = 60^\circ$.

Question 45

Physics · Dual Nature of Radiation and Matter · Single correct

Given below are two statements :- Statement I : Davisson-Germer experiment establishes the wave nature of electrons. Statement II : If electrons have wave nature, they can interfere and show diffraction. In the light of the above statements choose the correct answer from the options given below:-

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

Answer: (a)

Solution

In Davisson-Germer experiment the electrons exhibit diffraction thereby proving that electrons have wave nature. Hence both statements are correct. Both the options are correct by concept.

Question 46

Physics · Atoms · Single correct

The ratio for the speed of the electron in the $3^{rd}$ orbit of $\mathrm{He}^{+}$ to the speed of the electron in the $3^{rd}$ orbit of hydrogen atom will be :-

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

Answer: (d)

Solution

Given $v \propto \frac{Z}{n} \propto Z$ (where $n$ is constant). Therefore, $$\frac{v_{\mathrm{He}^+}}{v_{\mathrm{H}}} = \frac{Z_{\mathrm{He}^+}}{Z_{\mathrm{H}}} = \frac{2}{1}.$$

Question 47

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

The photodiode is used to detect the optical signals. These diodes are preferably operated in reverse biased mode because.

  1. fractional change in majority carriers produce higher forward bias current
  2. fractional change in majority carriers produce higher reverse bias current
  3. fractional change in minority carriers produce higher forward bias current
  4. fractional change in minority carriers produce higher reverse bias current

Answer: (d)

Solution

Given $v \propto \frac{Z}{n} \propto Z$ (n = constant). Therefore, $$\frac{v_{\mathrm{He}^+}}{v_{\mathrm{H}}} = \frac{Z_{\mathrm{He}^+}}{Z_{\mathrm{H}}} = \frac{2}{1}.$$

Question 48

Physics · Communication Systems · Single correct

A signal of $100\,\mathrm{THz}$ frequency can be transmitted with maximum efficiency by:

  1. Coaxial cable
  2. Optical fibre
  3. Twisted pair of copper wires
  4. Water

Answer: (b)

Solution

Question 49

Physics · Ray Optics and Optical Instruments · Single correct

The difference of speed of light in the two media A and B ($v_A - v_B$) is $2.6 \times 10^7 \, \mathrm{m/s}$. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is: (Given: speed of light in vacuum $c = 3 \times 10^8 \, \mathrm{ms^{-1}}$)

  1. 1.303
  2. 1.318
  3. 1.13
  4. 0.12

Answer: (c)

Solution

Given $v = \frac{c}{\mu}$. Therefore, $$v_B = \frac{3 \times 10^8}{1.47} = 2.04 \times 10^8 = 20.4 \times 10^7 \, \mathrm{m/s}$$ Thus, $v_A - v_B = 2.6 \times 10^7 \, \mathrm{m/s}$. Therefore, $$v_A = (20.4 + 2.6) \times 10^7 = 23 \times 10^7 \, \mathrm{m/s}$$ Thus, $$\frac{\mu_B}{\mu_A} = \frac{v_A}{v_B} = \frac{23 \times 10^7}{20.4 \times 10^7} = 1.13$$

Question 50

Physics · Current Electricity · Single correct

A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for $\frac{1}{3}$ deflection in the galvanometer. Which of the below is true for measuring value of G?

  1. $\frac{1}{3}$ deflection method cannot be used for determining the resistance of the galvanometer.
  2. $\frac{1}{3}$ deflection method can be used and in this case the G equals to twice the value of shunt resistance(s).
  3. $\frac{1}{3}$ deflection method can be used and in this case, the G equals to three times the value of shunt resistance(s)
  4. $\frac{1}{3}$ deflection method can be used and in this case the G value equals to the shunt resistance(s).

Answer: (b)

Solution

In galvanometer, $( (I - I_g) S = I_g G )$. $$\frac{I_g}{I} = \frac{S}{S+G}$$ $$\Rightarrow \frac{1}{3} = \frac{S}{S+G} \Rightarrow S + G = 3S \Rightarrow G = 2S$$

Question 51

Physics · Laws of Motion · Numerical

A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is $\mathrm{m}$.

Answer: 2

Solution

Mass per unit length is $\lambda$. The normal force $N$ is given by $N = mg = \lambda (L-x) g$. The maximum static friction $f_{s_{max}}$ is $f_{s_{max}} = \mu_s N$. Therefore, $f_{s_{max}} = (0.5)(\lambda)(L-x)g$. Also, $f_{s_{max}} = m_x g$. Thus, $0.5 \lambda (L-x) g = \lambda x g$. Solving for $x$, we have $$\frac{L-x}{2} = x$$ which gives $$\frac{L}{2} = \frac{3x}{2} \implies x = \frac{L}{3} = \frac{6}{3} = 2 \, m.$$

Question 52

Physics · Work, Energy and Power · Numerical

A 0.5 kg block moving at a speed of 12 $ms^{-1}$ compresses a spring through a distance 30 $cm$ when its speed is halved. The spring constant of the spring will be ______ $Nm^{-1}$.

Answer: 600

Solution

Given $U_i + K_i = U_f + K_f$. $$\Rightarrow 0 + \frac{1}{2} m (12)^2 = \frac{1}{2} K (0.3)^2 + \frac{1}{2} m (6)^2$$ $$\Rightarrow 0.5 (12^2 - 6^2) = K (0.3)^2$$ $K = 600 \, \mathrm{N/m}$

Question 53

Physics · Mechanical Properties of Fluids · Numerical

The velocity of upper layer of water in a river is $36 \, \mathrm{kmh}^{-1}$. Shearing stress between horizontal layers of water is $10^{-3} \, \mathrm{Nm}^{-2}$. Depth of the river is _________ m. (Co-efficiency of viscosity of water is $10^{-2} \, \mathrm{Pa.s}$)

Answer: 100

Solution

Given the equation $F = \eta A \frac{\Delta v_x}{\Delta y}$. Dividing both sides by $A$, we have $$\frac{F}{A} = \eta \frac{\Delta v_x}{\Delta y}$$ which implies $$10^{-3} = 10^{-2} \times \frac{36 \times 1000}{h \times 3600}$$ Solving for $h$, we get $$h = 10^{-2} \times \frac{36 \times 1000}{10^{-3} \times 3600} = 100 \, \mathrm{m}$$

Question 54

Physics · Thermal Properties of Matter · Numerical

A steam engine intakes $50\,\mathrm{g}$ of steam at $100^\circ\mathrm{C}$ per minute and cools it down to $20^\circ\mathrm{C}$. If latent heat of vaporization of steam is $540\,\mathrm{cal\,g^{-1}}$, then the heat rejected by the steam engine per minute is $x \times 10^3\,\mathrm{cal}$.

Answer: 31

Solution

Heat rejected = mL_f + mS$\Delta$ T $$= (50 \times 540) + 50 \times 1 \times (100 - 20)$$ $$= 31000 \, Cal$$ $$= 31 \times 10^3 \, Cal$$

Question 55

Physics · Waves · Fill in the blank

The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20 cm. The length of the open organ pipe is _____ cm.

Answer: 80

Solution

$f_1 = \dfrac{2v}{2l_1}$ $f_2 = \dfrac{v}{4l_2}$ $f_1 = f_2$ $\Rightarrow \dfrac{2v}{2l_1} = \dfrac{v}{4l_2}$ $l_1 = 4l_2 = 80 \text{ cm}$

Question 56

Physics · Electrostatic Potential and Capacitance · Numerical

The equivalent capacitance between points A and B in below shown figure will be _____ $\mu$ $\mathrm{F}$.

Answer: 6

Solution

Two capacitors are short circuited. Finally, equivalent capacitance is calculated as follows: $$\frac{24 \times 8}{24 + 8} = \frac{24 \times 8}{32} = 6 \, \mu\mathrm{F}$$

Question 57

Physics · Current Electricity · Numerical

A resistor develops 300 J of thermal energy in 15s, when a current of 2A is passed through it. If the current increases to 3A, the energy developed in 10s is J.

Answer: 450

Solution

Given $H = i^2 R t$. $300 = 2^2 \times R \times 15$. $$\Rightarrow R = \frac{300}{60} = 5 \, \Omega$$ Now, for $i = 3 \, \mathrm{A}$, $t = 10 \, \mathrm{s}$, $R = 5 \, \Omega$. $H = 3^2 \times 5 \times 10 = 450 \, \mathrm{J}$

Question 58

Physics · Current Electricity · Numerical

The total current supplied to the circuit as shown in figure by the 5V battery is _________A

Answer: 2

Solution

Current supplied by 5V battery is given by $$\frac{5V}{2.5\, \Omega} = 2\, \mathrm{A}$$

Question 59

Physics · Electromagnetic Induction · Numerical

The current in a coil of self inductance 2.0 H is increasing according to $I = 2\sin(t^2)\,\mathrm{A}$. The amount of energy spent during the period when current changes from 0 to 2A is ______ J.

Answer: 4

Solution

Given $I = 2 \sin(t^2)$, we have $dI = 4t \sin(t^2) \, dt$. If $I = 0$, then $t = 0$. And if $I = 2$, then $2 = 2 \sin t^2$, which implies $t = \sqrt{\frac{\pi}{2}}$. The energy $E$ is given by the integral $$E = \int LI \, dI$$ which becomes $$= \int 2 \times 2 \sin(t^2) \times 4t \cos(t^2) \, dt$$ $$= 8 \int_0^{\sqrt{\pi/2}} t \sin(2t^2) \, dt$$ $$= 2 \left[ -\cos(2t^2) \right]_0^{\sqrt{\pi/2}}$$ $$= 2 \left[ -\cos \pi + \cos 0 \right] = 4.$$

Question 60

Physics · Laws of Motion · Numerical

A force on an object of mass 100g is $(10\hat{i} + 5\hat{j}) \, \mathrm{N}$. The position of that object at $t = 2 \, \mathrm{s}$ is $(a\hat{i} + b\hat{j}) \, \mathrm{m}$ after starting from rest. The value of $\frac{a}{b}$ will be

Answer: 2

Solution

Given $\vec{F} = 10\hat{i} + 5\hat{j}$. $m = 100 \, \mathrm{g} = 0.1 \, \mathrm{kg}$. $\vec{a} = \frac{\vec{F}}{m} = 100\hat{i} + 50\hat{j}$. $\vec{S} = \vec{u}t + \frac{1}{2} \vec{a} t^2 = \frac{1}{2} \vec{a} t^2 (as \vec{u} = 0)$. $$= \frac{1}{2} (100\hat{i} + 50\hat{j}) 2^2$$ $$= 200\hat{i} + 100\hat{j}$$ $$= a\hat{i} + b\hat{j}$$ $a = 200, \; b = 100$ Therefore, $\frac{a}{b} = 2$.

Chemistry

Question 61

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?

  1. NO
  2. N_2
  3. O_2
  4. C_2
  5. B_2

Answer: (c)

Solution

Bond strength is proportional to bond order. Removal of electron from antibonding MO increases B.O. $\mathrm{NO}$ and $\mathrm{O_2}$ have valence electrons in $\pi^*$ orbital.

Question 62

Chemistry · Surface Chemistry · Single correct

Incorrect statement for Tyndall effect is :-

  1. The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.
  2. The diameter of the dispersed particles is much smaller than the wavelength of the light used.
  3. During projection of movies in the cinemas hall, Tyndall effect is noticed.
  4. It is used to distinguish a true solution from a colloidal solution.

Answer: (b)

Solution

The diameter of dispersed particle should be somewhat below or near the wavelength of light.

Question 63

Chemistry · Structure of Atom · Single correct

The pair, in which ions are isoelectronic with $\mathrm{Al}^{3+}$ is :-

  1. $\mathrm{Br}^{-}$ and $\mathrm{Be}^{2+}$
  2. $\mathrm{Cl}^{-}$ and $\mathrm{Li}^{+}$
  3. $\mathrm{S}^{2-}$ and $\mathrm{K}^{+}$
  4. $\mathrm{O}^{2-}$ and $\mathrm{Mg}^{2+}$

Answer: (d)

Solution

Isoelectronic species have same number of electrons. $\mathrm{Al^{+3}}$, $\mathrm{O^{-2}}$, $\mathrm{Mg^{+2}}$ all have 10 electrons.

Question 64

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

Leaching of gold with dilute aqueous solution of NaCN in presence of oxygen gives complex [A], which on reaction with zinc forms the elemental gold and another complex [B]. [A] and [B], respectively are :-

  1. $[\mathrm{Au(CN)_4}]^{-}$ and $[\mathrm{Zn(CN)_2(OH)_2}]^{2-}$
  2. $[\mathrm{Au(CN)_2}]^{-}$ and $[\mathrm{Zn(OH)_4}]^{2-}$
  3. $[\mathrm{Au(CN)_2}]^{-}$ and $[\mathrm{Zn(CN)_4}]^{2-}$
  4. $[\mathrm{Au(CN)_4}]^{2-}$ and $[\mathrm{Zn(CN)_6}]^{4-}$

Answer: (c)

Solution

Au + $\mathrm{NaCN}$ $\rightarrow$ $\mathrm{Na[Au(CN)_2]}$ Zn + $\mathrm{Na[Au(CN)_2]}$ $\rightarrow$ $\mathrm{Na_2[Zn(CN)_4]}$ + $\mathrm{Au}$

Question 65

Chemistry · The p-Block Elements (Group-13 and 14) · Single correct

Number of electron deficient molecules among the following $\mathrm{PH_3}$, $\mathrm{B_2H_6}$, $\mathrm{CCl_4}$, $\mathrm{NH_3}$, $\mathrm{LiH}$ and $\mathrm{BCl_3}$ is

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

Answer: (c)

Solution

Electron deficient species have less than 8 electrons (or two electrons for H) in their valence (incomplete octet). $\mathrm{B_2H_6}$, $\mathrm{BCl_3}$ have incomplete octet.

Question 66

Chemistry · The s-Block Elements · Single correct

Which one of the following alkaline earth metal ions has the highest ionic mobility in its aqueous solution?

  1. $\mathrm{Be}^{2+}$
  2. $\mathrm{Mg}^{2+}$
  3. $\mathrm{Ca}^{2+}$
  4. $\mathrm{Sr}^{2+}$

Answer: (d)

Solution

Highest ionic mobility corresponds to lowest extent of hydration and highest size of gaseous ion. Hence $\mathrm{Sr^{2+}}$ has the highest ionic mobility in its aqueous solution.

Question 67

Chemistry · Co-ordination Compounds · Single correct

White precipitate of AgCl dissolves in aqueous ammonia solution due to formation of:

  1. [$\mathrm{Ag(NH_3)_4}$]$\mathrm{Cl_2}$
  2. [$\mathrm{Ag(Cl)_2(NH_3)_2}$]
  3. [$\mathrm{Ag(NH_3)_2}$]$\mathrm{Cl}$
  4. [$\mathrm{Ag(NH_3)Cl}$]$\mathrm{Cl}$

Answer: (c)

Solution

The reaction is given by: $$\mathrm{AgCl} + 2\mathrm{NH_3} \rightarrow [\mathrm{Ag(NH_3)_2}]^+\mathrm{Cl}^-$$ This compound is soluble.

Question 68

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

Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it?

  1. It will not prefer to undergo redox reactions.
  2. It will prefer to gain electron and act as an oxidizing agent
  3. It will prefer to give away an electron and behave as reducing agent
  4. It acts as both, oxidizing and reducing agent.

Answer: (b)

Solution

Cerium exists in two different oxidation states +3, +4. $$\mathrm{Ce^{+4} + e^- \rightarrow Ce^{3+}} E^0 = +1.61 \, \mathrm{V}$$ $$\mathrm{Ce^{+3} + 3e^- \rightarrow Ce} E^0 = -2.336 \, \mathrm{V}$$ It shows $\mathrm{Ce^{+4}}$ acts as a strong oxidising agent and accepts electrons.

Question 69

Chemistry · Redox Reactions · Single correct

Among the following, which is the strongest oxidizing agent?

  1. $Mn^{3+}$
  2. $Fe^{3+}$
  3. $Ti^{3+}$
  4. $Cr^{3+}$

Answer: (a)

Solution

Strongest oxidising agent have highest reduction potential value $$E^0_{\mathrm{Mn}^{+3}/\mathrm{Mn}^{+2}} = 1.51 \, \mathrm{V} (highest)$$

Question 70

Chemistry · Environmental Chemistry · Single correct

The eutrophication of water body results in :

  1. loss of Biodiversity
  2. breakdown of organic matter
  3. increase in biodiversity
  4. decrease in BOD.

Answer: (a)

Solution

Eutrophication of water body results in loss of Biodiversity.

Question 71

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

Phenol on reaction with dilute nitric acid, gives two products. Which method will be most effective for large scale separation?

  1. Chromatographic separation
  2. Fractional Crystallisation
  3. Steam distillation
  4. Sublimation

Answer: (c)

Solution

Para product has higher boiling point than ortho as intermolecular H-bond is possible in former, whereas intramolecular H-bond is possible in ortho product. Steam distillation can separate them as ortho product is steam volatile.

Question 72

Chemistry · Hydrocarbons · Single correct

In the following structures, which one is having staggered conformation with maximum dihedral angle?

Answer: (c)

Solution

Dihedral angle: It's the angle between 2 specified groups (–CH₃ here). Staggered form is given in option (C) and the angle is $180^\circ$.

Question 73

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The products formed in the following reaction.

Answer: (b)

Solution

Question 74

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

The IUPAC name of ethylidene chloride is :-

  1. 1-Chloroethene
  2. 1-Chloroethyne
  3. 1,2-Dichloroethane
  4. 1,1-Dichloroethane

Answer: (d)

Solution

1, 1-Dichloroethane is Ethylidene chloride.

Question 75

Chemistry · Haloalkanes and Haloarenes · Single correct

The major product in the reaction

  1. t-Butyl ethyl ether
  2. 2,2-Dimethyl butane
  3. 2-Methyl pent-1-ene
  4. 2-Methyl prop-1-ene

Answer: (d)

Solution

We have been given a bulky base, hence elimination will take place and not substitution.

Question 76

Chemistry · Alcohols, Phenols and Ethers · Single correct

The intermediate X, in the reaction is :

Answer: (c)

Solution

It's a classic Reimer-Tiemann reaction. Will be the intermediate formed.

Question 77

Chemistry · Alcohols, Phenols and Ethers · Single correct

In the following reaction: The compounds A and B respectively are :-

Answer: (c)

Solution

Given reaction is cumene-Peroxide method for the preparation of phenol. In this reaction, cumene hydroperoxide is converted to phenol and acetone in the presence of an acid catalyst.

Question 78

Chemistry · Amines · Single correct

The reaction of \begin{array}{c} R-C-NH_2\\ \| \ \\\ \ O \ \end{array} with bromine and $KOH$ gives $RNH_2$ as the end product. Which one of the following is the intermediate product formed in this reaction?

  1. R−NH−Br
  2. R−N=C=O

Answer: (c)

Solution

Question 79

Chemistry · Chemistry in Everyday Life · Single correct

Using very little soap while washing clothes, does not serve the purpose of cleaning of clothes because

  1. soap particles remain floating in water as ions
  2. the hydrophobic part of soap is not able to take away grease
  3. the micelles are not formed due to concentration of soap, below its CMC value
  4. colloidal structure of soap in water is completely disturbed.

Answer: (c)

Solution

Micelle formation only takes place above CMC.

Question 80

Chemistry · Chemistry in Everyday Life · Single correct

Which one of the following is an example of artificial sweetner?

  1. Bithional
  2. Alitame
  3. Salvarsan
  4. Lactose

Answer: (b)

Solution

Alitame is a second generation dipeptide sweetener that is 200 times sweeter than sucrose.

Question 81

Chemistry · Some Basic Concepts of Chemistry · Numerical

The number of N atoms is 681 g of $C_7H_5N_3O_6$ is $x \times 10^{21}$. The value of $x$ is _____ ($N_A = 6.02 \times 10^{23} \, \mathrm{mol}^{-1}$) (Nearest Integer)

Answer: 5418

Solution

M.M. of $\mathrm{C_7H_5N_3O_6}$ is $84 + 5 + 42 + 96 = 227$. $$n_{\mathrm{C_7H_5N_3O_6}} = \frac{681}{227} = 3$$ $$n_{\mathrm{N}} = \frac{681}{227} \times 3 = 9 \, \mathrm{mol}$$ The number of $\mathrm{N}$ atoms $= 9 \times 6.02 \times 10^{23}$. $$= 5418 \times 10^{21}$$ Therefore, the answer is 5418.

Question 82

Chemistry · The Solid State · Numerical

The distance between $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ ions in solid $\mathrm{NaCl}$ of density $43.1\,\mathrm{g\,cm^{-3}}$ is \_\_\_\_ $\times 10^{-10}\,\mathrm{m}$. (Nearest Integer) (Given: $N_A = 6.02 \times 10^{23}\,\mathrm{mol^{-1}}$)

Answer: 1

Solution

Unit cell formula – $\mathrm{Na_4Cl_4}$ Mass per unit cell = $\frac{Z \times \mathrm{M.M.}}{N_A}$ g $$= \frac{4 \times 58.5}{N_A} g$$ $$d_{unit cell} = \frac{m}{V} = \frac{m}{a^3}$$ $$\Rightarrow \frac{4 \times 58.5}{N_A \cdot a^3} = 43.1$$ $$\Rightarrow a^3 = 9.02 \times 10^{-24} cm^3$$ $$\Rightarrow a = 2.08 \times 10^{-8} cm$$ $$\Rightarrow a = 2.08 \times 10^{-10} m$$ Also $a = 2 \left( r_{\mathrm{Na^+}} + r_{\mathrm{Cl^-}} \right)$ $$\Rightarrow r_{\mathrm{Na^+}} + r_{\mathrm{Cl^-}} = 1.04 \times 10^{-10} m$$ Therefore, the answer is 1.

Question 83

Chemistry · Structure of Atom · Numerical

The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is $x \times 10^{-8}$ m. The value of $x$ is ___________. (Nearest Integer) (Given : Energy of the electron in the first shell of the hydrogen atom is $-2.2 \times 10^{-18}$ J; $h = 6.63 \times 10^{-34}$ Js and $c = 3 \times 10^{8}$ ms$^{-1}$)

Answer: 4

Solution

We can not calculate I.E. of lithium atom.

Question 84

Chemistry · Thermodynamics · Numerical

The standard entropy change for the reaction $$4\mathrm{Fe}(s) + 3\mathrm{O}_2(g) \rightarrow 2\mathrm{Fe}_2\mathrm{O}_3(s)$$ is $-550 \, \mathrm{JK}^{-1}$ at $298 \, \mathrm{K}$. [Given: The standard enthalpy change for the reaction is $-165 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$]. The temperature in K at which the reaction attains equilibrium is _________. (Nearest Integer)

Answer: 300

Solution

At equilibrium, $\Delta G = \Delta H - T \Delta S = 0$. $$\Rightarrow -165 \times 10^3 - T \times (-505) = 0$$ $$\Rightarrow T = 300 \, \mathrm{K}$$ The answer is 300.

Question 85

Chemistry · Solutions · Numerical

1 L aqueous solution of $\mathrm{H_2SO_4}$ contains $0.02$ m mol $\mathrm{H_2SO_4}$. 50$\%$ of this solution is diluted with deionized water to give 1 L solution (A). In solution (A), $0.01$ m mol of $\mathrm{H_2SO_4}$ are added. Total m mols of $\mathrm{H_2SO_4}$ in the final solution is $\times 10^3$ m mols.

Answer: 15

Solution

Given $n_{\mathrm{H_2SO_4}}$ in solution A is 50$\%$ of the original solution. $$= 0.01 \, \mathrm{mmol}.$$ $n_{\mathrm{H_2SO_4}}$ in the final solution is $0.01 + 0.01$. $$= 0.02 \, \mathrm{mmol}$$ $$= 0.00002 \times 10^3 \, \mathrm{mmol}$$ The answer is 0.

Question 86

Chemistry · Equilibrium · Numerical

The standard free energy change ($\Delta G^\circ$) for 50$\%$ dissociation of $\mathrm{N_2O_4}$ into $\mathrm{NO_2}$ at $27^\circ\mathrm{C}$ and $1 \, \mathrm{atm}$ pressure is $-x \, \mathrm{J \, mol^{-1}}$. The value of $x$ is _________. (Nearest Integer) [Given : $R = 8.31 \, \mathrm{J \, K}^{-1} \, \mathrm{mol}^{-1}$, $\log 1.33 = 0.1239$, $\ln 10 = 2.3$]

Answer: 710

Solution

$$\mathrm{N_2O_4} \rightleftharpoons 2\mathrm{NO_2}$$ \begin{tabular}{lll} $t = 0$ & $1\ \mathrm{mol}$ & \\[4pt] $t = t$ & $(1-0.5)\ \mathrm{mol}$ & $0.5\times 2\ \mathrm{mol}$ \\[4pt] & $= 0.5\ \mathrm{mol}$ & $= 1\ \mathrm{mol}$ \end{tabular} $$k_P = \frac{\left(\dfrac{1}{1.5}\times 1\right)^2}{\left(\dfrac{0.5}{1.5}\times 1\right)} = \frac{1}{0.75} = \frac{100}{75}$$ $= 1.33$ $\Delta G^0 = -RT\ln k_P$ $= -8.31 \times 300 \times \ln(1.33) = -710.45\ \mathrm{J/mol}$ $= -710\ \mathrm{J/mol}$

Question 87

Chemistry · Electrochemistry · Numerical

In a cell, the following reactions take place $$\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-} E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}} = 0.77 \, \mathrm{V}$$ $$\mathrm{2I^- \rightarrow I_2 + 2e^-} E^\circ_{\mathrm{I_2/I^-}} = 0.54 \, \mathrm{V}$$ The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \, \mathrm{V}$ at 298 K. The value of $x$ is _________ (Nearest Integer)

Answer: 23

Solution

The reaction is given by: $$\mathrm{Fe^{+3} + I^- \rightarrow I_2 + Fe^{+2}}$$ The cell potential is calculated as: $$E^0_{Cell} = E^0_{cathode} - E^0_{anode}$$ Substituting the values: $$= 0.77 - 0.54$$ $$= 0.23$$ $$= 23 \times 10^{-2} \, \mathrm{V}$$

Question 88

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

For a given chemical reaction $$\gamma_1\mathrm{A} + \gamma_2\mathrm{B} \rightarrow \gamma_3\mathrm{C} + \gamma_4\mathrm{D}$$ Concentration of C changes from $10\ \mathrm{mmol\ dm^{-3}}$ to $20\ \mathrm{mmol\ dm^{-3}}$ in $10$ seconds. Rate of appearance of D is $1.5$ times the rate of disappearance of B which is twice the rate of disappearance of A. The rate of appearance of D has been experimentally determined to be $9\ \mathrm{mmol\ dm^{-3}\ s^{-1}}$. Therefore the rate of reaction is \_\_\_\_\_ $\mathrm{mmol\ dm^{-3}\ s^{-1}}$. (Nearest Integer)

Answer: 1

Solution

Given: $$+\frac{d[D]}{dt} = \frac{-3 \, d[B]}{2 \, dt}$$ $$\Rightarrow \frac{-1 \, d[B]}{2 \, dt} = \frac{+1 \, d[D]}{3 \, dt}$$ $$-\frac{d[B]}{dt} = -2 \frac{d[A]}{dt} \Rightarrow \frac{-1 \, d[B]}{2 \, dt} = -\frac{d(A)}{dt}$$ $$+\frac{d[B]}{dt} = 9 \, \mathrm{mmol \, dm^{-3} \, s^{-1}}$$ $$+\frac{d[C]}{dt} = \frac{20 - 10}{10} = 1 \, \mathrm{mmol \, dm^{-3} \, s^{-1}}$$ $$+\frac{d[C]}{dt} = \frac{1}{9} \times +\frac{d[D]}{dt}$$ $$1A + 2B \xrightarrow{\frac{1}{3}} C + 3D$$ $$\Rightarrow 3A + 6B \longrightarrow C + 9D$$ Rate of reaction = $$+\frac{d[C]}{dt} = 1 \, \mathrm{mmol \, dm^{-3} \, s^{-1}}$$

Question 89

Chemistry · Co-ordination Compounds · Numerical

If $\mathrm{[Cu(H_2O)_4]^{2+}}$ absorbs a light of wavelength $600 \, \mathrm{nm}$ for d–d transition, then the value of octahedral crystal field splitting energy for $\mathrm{[Cu(H_2O)_6]^{2+}}$ will be $\times 10^{-21} \, \mathrm{J}$. (Nearest Integer) (Given: $h = 6.63 \times 10^{-34} \, \mathrm{Js}$ and $c = 3.08 \times 10^8 \, \mathrm{ms}^{-1}$)

Answer: 745

Solution

Given $\Delta_t = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.08 \times 10^8}{600 \times 10^{-9}}$. $$= \frac{6.63 \times 3.08 \times 10^{-17}}{600}$$ $$= 0.034034 \times 10^{-17}$$ $$= 340.34 \times 10^{-21} \, \mathrm{J}$$ Now, $\Delta_0 = \frac{9}{4} \Delta_t$. $$= \frac{9}{4} \times 340.34 \times 10^{-21}$$ $$= 765.765 \times 10^{-21} \, \mathrm{J}$$ $$\approx 766 \times 10^{-21} \, \mathrm{J}$$ Answer = 766

Question 90

Chemistry · Redox Reactions · Numerical

Number of grams of bromine that will completely react with $5.0 \, \mathrm{g}$ of pent-1-ene is _______ $\times 10^{-2} \, \mathrm{g}$. (Atomic mass of Br = 80 g/mol) [Nearest Integer]

Answer: 1136

Solution

Moles of $\mathrm{Br_2}$ $=$ moles of $\mathrm{C_5H_{10}}$. Therefore, $$\frac{w}{160} = \frac{5}{70}$$ Solving for $w$, we have $$w = \frac{5 \times 160}{70}\,\mathrm{g}$$ $$= 11.428\,\mathrm{g}$$ $$= 1142.8 \times 10^{-2}\,\mathrm{g} \approx 1143 \times 10^{-2}\,\mathrm{g}$$