JEE Main 27 June 2022 Shift 2 question paper with solutions
JEE Main 27 June 2022 Shift 2: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Complex Numbers and Quadratic Equations · Single correct
The number of points of intersection of $|z - (4 + 3i)| = 2$ and $|z| + |z - 4| = 6$, $z \in \mathbb{C}$ is:
0
1
2
3
Answer: (c)
Solution
Given the circle $C: (x-4)^2 + (y-3)^2 = 4$ and the ellipse $E: \frac{(x-2)^2}{9} + \frac{y^2}{5} = 1$. The lower extremity of the vertical diameter of the circle is $(4, 1)$. Put in ellipse: $$\frac{(4-2)^2}{9} + \frac{1}{5} - 1$$ $$= \frac{4}{9} + \frac{1}{5} - 1$$ $$= \frac{29}{45} - 1 < 0$$ Two solutions. Answer (c)
Question 2
Maths · Determinants · Single correct
Let \[ f(x) = \begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{vmatrix}, \quad a \in \mathbb{R}. \] Then the sum of the squares of all the values of $a$ for which $2f'(10) - f'(5) + 100 = 0$ is:
Let for some real numbers $\alpha$ and $\beta$, $a = \alpha - i \beta$. If the system of equations $4ix + (1+i)y = 0$ and $$8 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) x + \overline{a} y = 0$$ has more than one solution then $\frac{\alpha}{\beta}$ is equal to:
$-2 + \sqrt{3}$
$2 - \sqrt{3}$
$2 + \sqrt{3}$
$-2 - \sqrt{3}$
Answer: (b)
Solution
Given $a = \alpha - i \beta$; $\alpha \in \mathbb{R}$; $\beta \in \mathbb{R}$. $4ix + (1+i) y = 0$ and $$8 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) x + \overline{a} y = 0$$ $$\begin{vmatrix} 4i & 1+i \\ 8e^{i2\pi/3} & \frac{1}{\overline{a}} \end{vmatrix} = 0$$ $$\Rightarrow 4i \overline{a} - (1+i) 8e^{i2\pi/3} = 0$$ $$\Rightarrow 4i (\alpha + i \beta) - 8 (1+i) \left( \frac{-1+i\sqrt{3}}{2} \right) = 0$$ $$\Rightarrow i \alpha - \beta + 1 + \sqrt{3} + i (1 - \sqrt{3}) = 0$$ $$\Rightarrow \beta = \sqrt{3} + 1$$ $$\alpha = \sqrt{3} - 1$$ So, $\($ $\frac{\alpha}{\beta}$ = $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ = 2 - $\sqrt{3}$ $\)$
Question 4
Maths · Matrices · Single correct
Let A and B be two 3 $\times$ 3 matrices such that AB = I and $|A| = \frac{1}{8}$ then $|adj(B adj(2A))|$ is equal to
16
32
64
128
Answer: (c)
Solution
$AB=I$ $adj(B)\,adj(2A)$ $=|B\,adj(2A)|^2$ $=|B|^2\,|adj(2A)|^2$ $=|B|^2\,(|2A|^2)^2$ $=|B|^2\,(2^6|A|^2)^2$ $|A|=\frac18$ and $|AB|=1$ $\Rightarrow |A||B|=1$ $\Rightarrow \frac18|B|=1$ $\Rightarrow |B|=8$ required value $=64$
Question 5
Maths · Sequences and Series · Single correct
Let $S = 2 + \dfrac{6}{7} + \dfrac{12}{7^2} + \dfrac{20}{7^3} + \dfrac{30}{7^4} + \ldots$ then $4S$ is equal to
If $a_1, a_2, a_3, ....$ and $b_1, b_2, b_3, ....$ are A.P. and $a_1 = 2$, $a_{10} = 3$, $a_1 b_1 = 1 = a_{10} b_{10}$ then $a_4 b_4$ is equal to
$\frac{35}{27}$
1
$\frac{27}{28}$
$\frac{28}{27}$
Answer: (d)
Solution
Given the sequences $a_1, a_2, a_3, \ldots$ in arithmetic progression (A.P.) with $a_1 = 2$, $a_{10} = 3$, and $d_1 = \frac{1}{9}$, and $b_1, b_2, b_3, \ldots$ in A.P. with $b_1 = \frac{1}{2}$, $b_{10} = \frac{1}{3}$, and $d_2 = -\frac{1}{54}$. Using $a_1 b_1 = 1 = a_{10} b_{10}$, $d_1$ and $d_2$ are common differences respectively. The expression for $a_4 \cdot b_4$ is given by: $$a_4 \cdot b_4 = \left(2 + 3d_1\right) \left(\frac{1}{2} + 3d_2\right)$$ Substituting the values, we have: $$= \left(2 + \frac{1}{3}\right) \left(\frac{1}{2} - \frac{1}{18}\right)$$ Simplifying further: $$= \left(\frac{7}{3}\right) \left(\frac{8}{18}\right) = \frac{28}{27}$$
Question 7
Maths · Applications of Derivatives · Single correct
If m and n respectively are the number of local maximum and local minimum points of the function $f(x) = \int_{0}^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \, dt$, then the ordered pair (m, n) is equal to
(3, 2)
(2, 3)
(2, 2)
(3, 4)
Answer: (b)
Solution
Given $m = L \cdot max$ and $N = L \cdot min$. $$f(x) = \int_{0}^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \, dt$$ $$f'(x) = \frac{(x^4 - 5x^2 + 4)2x}{2 + e^{x^2}} = \frac{2x(x^2 - 1)(x^2 - 4)}{2 + e^{x^2}}$$ $$= \frac{2x(x-1)(x+1)(x-2)(x+2)}{2 + e^{x^2}}$$ From the number line, we have $m = 2$ and $n = 3$.
Question 8
Maths · Integrals · Single correct
Let f be a differentiable function in $\left( 0, \frac{\pi}{2} \right)$. If $\int_{\cos x}^{1} t^2 f(t)\,dt=\sin^3 x+\cos x$, then $\frac{1}{\sqrt{3}}\,f'\!\left(\frac{1}{\sqrt{3}}\right)$ is equal to :
6 - 9$\sqrt{2}$
6 - $\frac{9}{\sqrt{2}}$
$\frac{9}{2}$ - 6$\sqrt{2}$
$\frac{9}{\sqrt{2}}$ - 6
Answer: (b)
Solution
At right hand vicinity of $x = 0$ given equation does not satisfy $$\therefore LHS = \int_1^t t^2 f(t) \, dt = 0, RHS = \lim_{x \to 0^+} (\sin^3 x + \cos x) = 1$$ LHS $\neq$ RHS hence data given in question is wrong hence BONUS. Correct data should have been $$\int_{\cos x}^1 t^2 f(t) \, dt = \sin^3 x + \cos x - 1$$ Calculation for option differentiating both sides $$-\cos^2 x \, f(\cos x) \cdot (-\sin x) = 3 \sin^2 x \cdot \cos x - \sin x$$ $$\Rightarrow f(\cos x) = 3 \tan x - \sec^2 x$$ $$\Rightarrow f'(\cos x)(-\sin x) = 3 \sec^2 x - 2 \sec^2 x \tan x$$ $$\Rightarrow f'(\cos x) \cos x = \frac{2}{\cos^2 x} - \frac{3}{\sin x \cdot \cos x}$$ When $\cos x = \frac{1}{\sqrt{3}}$ ; $\sin x = \frac{\sqrt{2}}{\sqrt{3}}$ $$\therefore f'\left(\frac{1}{\sqrt{3}}\right) \frac{1}{\sqrt{3}} = 6 - \frac{9}{\sqrt{2}}.$$
Question 9
Maths · Integrals · Single correct
The integral $\int_{0}^{1} \frac{1}{7 \left[ \frac{1}{x} \right]} \, dx$, where $[\cdot]$ denotes the greatest integer function is equal to
$1 + 6 \log_e \left( \frac{6}{7} \right)$
$1 - 6 \log_e \left( \frac{6}{7} \right)$
$\log_e \left( \frac{7}{6} \right)$
$1 - 7 \log_e \left( \frac{6}{7} \right)$
Answer: (a)
Solution
The integral from 0 to 1 of $\($ $\frac{1}{7^{\left\lfloor \frac{1}{x} \right\rfloor}}$ $\,$ dx $\)$ is equal to the negative of the integral from 1 to 0 of $\($ $\frac{1}{7^{\left\lfloor \frac{1}{x} \right\rfloor}}$ $\,$ dx $\)$. This can be expressed as: $$ = (-1) \left[ \int_{1}^{1/2} \frac{1}{7} \, dx + \int_{1/2}^{1/3} \frac{1}{7^2} \, dx + \int_{1/3}^{1/4} \frac{1}{7^3} \, dx + \ldots \right] $$ This simplifies to: $$ = \left( \frac{1}{7} + \frac{1}{2 \cdot 7^2} + \frac{1}{3 \cdot 7^3} + \ldots \right) - \left( \frac{1}{7} + \frac{1}{7^2 \cdot 3} + \frac{1}{7^3 \cdot 4} + \ldots \right) $$ This further simplifies to: $$ = -\ln \left( 1 - \frac{1}{7} \right) - 7 \left( \frac{1}{7^2 \cdot 2} + \frac{1}{7^3 \cdot 3} + \frac{1}{7^4 \cdot 4} + \ldots \right) $$ Using the expansion for $\($ $\ln$(1 + x) = x - $\frac{x^2}{2}$ + $\frac{x^3}{3}$ - $\frac{x^4}{4}$ + $\ldots$ $\)$ and $\($ $\ln$(1 - x) = - $\left$( x + $\frac{x^2}{2}$ + $\frac{x^3}{3}$ + $\frac{x^4}{4}$ + $\ldots$ $\right$) $\)$, we have: $$ = -\ln \frac{6}{7} - 7 \left( -\ln \left( 1 - \frac{1}{7} \right) - \frac{1}{7} \right) $$ Finally, this simplifies to: $$ = 6 \ln \frac{6}{7} + 1 $$
Question 10
Maths · Differential Equations · Single correct
If the solution curve of the differential equation $(\tan^{-1} y - x)\,dy = (1 + y^2)\,dx$ passes through the point $(1, 0)$ then the abscissa of the point on the curve whose ordinate is $\tan(1)$ is:
2e
$\frac{2}{e}$
2
$\frac{1}{e}$
Answer: (b)
Solution
Given $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1} y}{1+y^2}$. The integrating factor is $\mathrm{I.F.} = e^{\int \frac{1}{1+y^2}\,dy} = e^{\tan^{-1} y}$. Therefore, $x e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y}\,dy$. This simplifies to $x \cdot e^{\tan^{-1} y} = \left(\tan^{-1} y - 1\right) e^{\tan^{-1} y} + c$. Since $(1, 0)$ lies on the curve, $c = 2$. For $y = \tan 1$, it follows that $x = \frac{2}{e}$.
Question 11
Maths · Conic Sections · Single correct
If the equation of the parabola, whose vertex is at (5, 4) and the directrix is $3x + y - 29 = 0$, is $x^2 + ay^2 + bxy + cx + dy + k = 0$ then a + b + c + d + k is equal to
575
-575
576
-576
Answer: (d)
Solution
Vertex $(5,4)$ Directrix: $3x + y - 29 = 0$ Co-ordinates of $B$ (foot of directrix) $$\frac{x-5}{3} = \frac{y-4}{1} = -\left(\frac{15+4-29}{10}\right) = 1$$ $x = 8, y = 5$ $S = (2, 3)$ (focus) Equation of parabola $PS = PM$ so equation is $$x^2 + 9y^2 - 6xy + 134x - 2y - 711 = 0$$ $a + b + c + d + k = 9 - 6 + 134 - 2 - 711 = -576$
Question 12
Maths · Conic Sections · Single correct
The set of values of $k$ for which the circle $C : 4x^2 + 4y^2 - 12x + 8y + k = 0$ lies inside the fourth quadrant and the point $\left(1, -\frac{1}{3}\right)$ lies on or inside the circle $C$ is:
An empty set
$\left(6, \frac{95}{9}\right]$
$\left[\frac{80}{9}, 10\right)$
$\left(9, \frac{92}{9}\right]$
Answer: (d)
Solution
Given the circle equation $C: 4x^2 + 4y^2 - 12x + 8y + k = 0$. This simplifies to $x^2 + y^2 - 3x + 2y + \frac{k}{4} = 0$. The center is $\left( \frac{3}{2}, -1 \right)$ and the radius is $r = \frac{\sqrt{13 - k}}{2}$, which implies $k \leq 13$ (1). (i) The point $\left( 1, -\frac{1}{3} \right)$ lies on or inside circle $C$. This gives $S_1 \leq 0 \Rightarrow k \leq \frac{92}{9}$ (2). (ii) Circle $C$ lies in the 4th quadrant. The condition $r < 1$ implies $\frac{\sqrt{13 - k}}{2} < 1$, leading to $k < 9$ (3). Hence, combining (1), (2), and (3), we have $k \in \left( 9, \frac{92}{9} \right]$.
Question 13
Maths · Three Dimensional Geometry · Single correct
Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3}$ be $P$. Then the distance of $P$ from the plane $3x + 4y + 12z + 23 = 0$
Maths · Three Dimensional Geometry · Single correct
The shortest distance between the lines $\frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{-1}$ and $\frac{x+3}{2} = \frac{y-6}{1} = \frac{z-5}{3}$ is:
$\frac{18}{\sqrt{5}}$
$\frac{22}{3\sqrt{5}}$
$\frac{46}{3\sqrt{5}}$
$6\sqrt{3}$
Answer: (a)
Solution
Given the equations: $$\frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{-1}$$ $$\frac{x+3}{2} = \frac{y-6}{1} = \frac{z-5}{3}$$ Points are given as $A = (3, 2, 1)$ and $B = (-3, 6, 5)$. The direction vectors are: $$\vec{n_1} = 2\hat{i} + 3\hat{j} - \hat{k}$$ $$\vec{n_2} = 2\hat{i} + \hat{j} - 3\hat{k}$$ The vector $\overrightarrow{BA}$ is: $$\overrightarrow{BA} = 6\hat{i} - 4\hat{j} - 4\hat{k}$$ The shortest distance is given by: $$SHORTEST DISTANCE = \frac{\left[ \overrightarrow{BA} \ \vec{n_1} \ \vec{n_2} \right]}{\left| \vec{n_1} \times \vec{n_2} \right|}$$ Calculating the cross product: $$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{vmatrix}$$ This results in: $$= 10\hat{i} - 8\hat{j} - 4\hat{k}$$ The numerator is: $$\left[ \overrightarrow{BA} \ \vec{n_1} \ \vec{n_2} \right] = 60 + 32 + 16 = 108$$ The magnitude of the cross product is: $$\left| \vec{n_1} \times \vec{n_2} \right| = \sqrt{100 + 64 + 16} = \sqrt{180}$$ Thus, the shortest distance (S.D) is: $$S.D = \frac{108}{\sqrt{180}} = \frac{108}{6\sqrt{5}} = \frac{18}{\sqrt{5}}$$
Question 15
Maths · Vector Algebra · Single correct
Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonal of a parallelogram having area 2$\sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute. |$\vec{a}$| = 1 and | $\vec{a}$ $\cdot$ $\vec{b}$| = |$\vec{a}$ $\times$ $\vec{b}$|. If $\vec{c}$ = 2$\sqrt{2}$ ($\vec{a}$ $\times$ $\vec{b}$) - 2$\vec{b}$, then an angle between $\vec{b}$ and $\vec{c}$ is:
Let $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Define $f : S \to S$ as $$f(n) = \begin{cases} 2n, & \text{if } n = 1, 2, 3, 4, 5 \\ 2n - 11, & \text{if } n = 6, 7, 8, 9, 10 \end{cases}$$ Let $g : S \to S$ be a function such that $$f \circ g(n) = \begin{cases} n+1, & \text{if } n \text{ is odd} \\ n-1, & \text{if } n \text{ is even} \end{cases}$$ then $g(10)\big(g(1) + g(2) + g(3) + g(4) + g(5)\big)$ is equal to:
Answer: 190
Solution
Given $$f^{-1}(n) = \begin{cases} \frac{n}{2} & ; \; n = 2, 4, 6, 8, 10 \\ \frac{n+11}{2} & ; \; n = 1, 3, 5, 7, 9 \end{cases}$$ $$f(g(n)) = \begin{cases} n+1 & ; \; n \in odd \\ n-1 & ; \; n \in even \end{cases}$$ Therefore, $$g(n) = \begin{cases} f^{-1}(n+1) & ; \; n \in odd \\ f^{-1}(n-1) & ; \; n \in even \end{cases}$$ Thus, $$g(n) = \begin{cases} \frac{n+1}{2} & ; \; n \in odd \\ \frac{n+10}{2} & ; \; n \in even \end{cases}$$ Calculating, $$g(10) \cdot [g(1) + g(2) + g(3) + g(4) + g(5)]$$ $$= 10 \cdot [1 + 6 + 2 + 7 + 3] = 190$$
Question 22
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $\alpha$, $\beta$ be the roots of the equation $x^2 - 4\lambda x + 5 = 0$ and $\alpha$, $\gamma$ be the roots of the equation $x^2 - \left(3\sqrt{2} + 2\sqrt{3}\right)x + 7 + 3\lambda \sqrt{3} = 0$. If $\beta + \gamma = 3\sqrt{2}$, then $(\alpha + 2\beta + \gamma)^2$ is equal to :
Answer: 98
Solution
Given the equation $x^2 - 4\lambda x + 5 = 0$ with roots $\langle \alpha, \beta \rangle$. Another equation is $x^2 - (3\sqrt{2} + 2\sqrt{3})x + (7 + 3\lambda \sqrt{3}) = 0$ with roots $\langle \alpha, \gamma \rangle$. We have $\alpha + \beta = 4\lambda$ and $\alpha + \gamma = 3\sqrt{2} + 2\sqrt{3}$. Also, $\beta + \lambda = 3\sqrt{2}$ and $\alpha \gamma = 7 + 3\lambda \sqrt{3}$. Therefore, $\alpha = 2\lambda + \sqrt{3}$ and $\beta = 2\lambda - \sqrt{3}$. We know $\alpha \beta = 5$ and $4\lambda^2 = 8 \Rightarrow \lambda = \sqrt{2}$. Thus, $(\alpha + 2\beta + \lambda)^2 = (4\alpha + 3\sqrt{2})^2 = (7\sqrt{2})^2 = 98$.
Question 23
Maths · Matrices · Numerical
Let $A$ be a matrix of order $2\times2$, whose entries are from the set $\{0,1,2,3,4,5\}$. If the sum of all the entries of $A$ is a prime number $p$, $2<p<8$, then the number of such matrices $A$ is ______.
Answer: 180
Solution
Let $A= \begin{bmatrix} a&b\\ c&d \end{bmatrix}$ $a,b,c,d\in\{0,1,2,3,4,5\}$ $a+b+c+d=p$ $p\in\{3,5,7\}$ Case-(i) $a+b+c+d=3$ $a,b,c,d\in\{0,1,2,3\}$ No. of ways $={}^{3+4-1}C_{4-1}$ $={}^{6}C_3$ $=20$ ...(1) Case-(ii) $a+b+c+d=5$ $a,b,c,d\in\{0,1,2,3,4,5\}$ No. of ways $={}^{5+4-1}C_{4-1}$ $={}^{8}C_3$ $=56$ ...(2) Case-(iii) $a+b+c+d=7$ No. of ways = total ways when $a,b,c,d\in\{0,1,2,3,4,5,6,7\}$ − total ways when $a,b,c,d\notin\{6,7\}$ No. of ways $={}^{10}C_3-16$ $=120-16$ $=104$ ...(3) Hence total no. of ways $=20+56+104$ $=180$
Question 24
Maths · Binomial Theorem · Numerical
If the sum of the coefficients of all the positive powers of x, in the binomial expansion of $$\left( x^n + \frac{2}{x^5} \right)^7$$ is 939, then the sum of all the possible integral values of n is :
Maths · Limits and Derivatives · Fill in the blank
Let [t] denote the greatest integer $\leq t$ and $\{$t$\}$ denote the fractional part of t. Then integral value of $\alpha$ for which the left hand limit of the function $$f(x) = [1 + x] + \frac{\alpha^{2[x] + \{x\}} + [x] - 1}{2[x] + \{x\}}$$ at $x = 0$ is equal to $$\alpha - \frac{4}{3}$$ is ____
Answer: 3
Solution
Given $$f(x) = [1 + x] + \frac{\alpha^{2[x] + \{x\}} + [x] - 1}{2[x] + \{x\}}$$ Taking the limit as $x$ approaches $0$ from the negative side: $$\lim_{x \to 0^-} f(x) = \alpha - \frac{4}{3} \implies 0 + \frac{\alpha^{-1} - 2}{-1}$$ $$=\alpha - \frac{4}{3}$$ This simplifies to: $$2 - \frac{1}{\alpha} = \alpha - \frac{4}{3}$$ Rearranging gives: $$\alpha + \frac{1}{\alpha} = \frac{10}{3}$$ Solving for $\alpha$ gives: $$\alpha = 3; \alpha \in \mathbb{I}$$
Question 26
Maths · Continuity and Differentiability · Numerical
If $y(x)=(x^{x^x})$, $x>0$ then $\frac{d^2x}{dy^2}+20$ at $x=1$ is equal to:
If the area of the region $$\left\{(x, y) : x^{\frac{2}{3}} + y^{\frac{2}{3}} \leq 1, x + y \geq 0, y \geq 0 \right\}$$ is $A$, then $$\frac{256A}{\pi}$$ is
Answer: 36
Solution
The area $A$ is given by the integral $$A = \frac{3}{2} \int_0^1 (1-x^{2/3})^{3/2} \, dx.$$ Let $x = \sin^3 \theta$. Then, $$A = \frac{3}{2} \int_0^{\pi/2} \left(1-\sin^2 \theta \right)^{3/2} \cdot 3 \sin^2 \theta \cos \theta \, d \theta.$$ This simplifies to $$= \frac{3}{2} \int_0^{\pi/2} 3 \sin^2 \theta \cos^4 \theta \, d \theta$$ $$= \frac{9}{2} \int_0^{\pi/2} \sin^2 \theta \cos^4 \theta \, d \theta.$$ The area $A$ is then $$A = \frac{9}{2} \times \frac{1 \cdot 3.1}{(2+4)(4)(2)} \cdot \frac{\pi}{2}.$$ This gives $$\Rightarrow A = \frac{9 \pi}{64} \Rightarrow \frac{64 A}{\pi} = 9.$$ Therefore, $$\frac{256 A}{\pi} = 36 Ans.$$
Question 28
Maths · Differential Equations · Numerical
Let $\nu$ be the solution of the differential equation $$(1-x^2)dy = \left(xy+(x^3+2)\sqrt{1-x^2}\right)dx, -1<x<1$$ and $y(0) = 0$ if $$\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-x^2} \, y(x) \, dx = k$$ then $k^{-1}$ is equal to :
Answer: 320
Solution
Given $$(1-x^2) \frac{dy}{dx} = xy + (x^3 + 2) \sqrt{1-x^2}$$ This implies $$\frac{dy}{dx} + \left(\frac{-x}{1-x^2}\right) y = \frac{x^3 + 2}{\sqrt{1-x^2}}$$ The integrating factor is $$\mathrm{IF} = e^{\int \frac{-x}{1-x^2} \, dx} = \sqrt{1-x^2}$$ Thus, $$y(x) \cdot \sqrt{1-x^2} = \frac{x^4}{4} + 2x + c$$ Given $y(0) = 0$, it follows that $c = 0$. Therefore, $$\sqrt{1-x^2} \, y(x) = \frac{x^4}{4} + 2x$$ The required value is $$\int_{-1/2}^{1/2} \left( \frac{x^4}{4} + 2x \right) \, dx - \frac{1}{4} \cdot 2 \int_0^{1/2} x^4 \, dx$$ This evaluates to $$= \frac{1}{10} \left( x^5 \right)_{0}^{1/2} = \frac{1}{320}$$ Thus, $$k^{-1} = 320$$
Question 29
Maths · Conic Sections · Numerical
Let a circle C of radius 5 lie below the x-axis. The line $L_1 = 4x + 3y - 2$ passes through the centre P of the circle C and intersects the line $L_2: 3x - 4y - 11 = 0$ at Q. The line $L_2$ touches C at the point Q. Then the distance of P from the line $5x - 12y + 51 = 0$ is
Answer: 11
Solution
The equations of the lines are $4x + 3y + 2 = 0$ and $3x - 4y - 11 = 0$. Using the triangle, we have $\cos \theta = \frac{3}{5}$ and $\sin \theta = \frac{4}{5}$. The parametric equations are $\frac{x}{-25} = \frac{y}{50} = \frac{1}{-25}$. Solving for $x$ and $y$, we have: $$x - 1 = \pm \frac{3}{5}$$ $$y + 2 = \pm \frac{4}{5}$$ For $y = -2 + 5 \left( -\frac{4}{5} \right) = -6$ and $x = 1 + 5 \left( \frac{3}{5} \right) = 4$. The required distance is: $$\frac{|5(4) - 12(-6) + 51|}{13}$$ Calculating: $$= \frac{|20 + 72 + 51|}{13}$$ $$= \frac{143}{13} = 11$$
Question 30
Maths · Probability · Numerical
Let $S=\{E_1,E_2,\ldots,E_8\}$ be a sample space of random experiment such that $P(E_n)=\frac{n}{36}$ for every $n = 1, 2, \ldots, 8$. Then the number of elements in the set $\left\{ A \subseteq S: P(A) \geq \frac{4}{5} \right\}$ is
Answer: 19
Solution
Given $P(A^\prime) < \frac{1}{5} = \frac{36}{180}$. 5 times the sum of missing numbers should be less than 36. If 1 digit is missing = 7 If 2 digits are missing = 9 If 3 digits are missing = 2 If 0 digits are missing = 1 Alternate A is a subset of S hence A can have elements: type 1: $\{ \}$ type 2: $\{E_1\}, \{E_2\}, \ldots \{E_8\}$ type 3: $\{E_1, E_2\}, \{E_1, E_3\}, \ldots \{E_7, E_8\}$ ... type 6: $\{E_1, E_2, \ldots E_5\}, \ldots \{E_4, E_5, E_6, E_7, E_8\}$ type 7: $\{E_1, E_2, \ldots E_6\}, \ldots \{E_3, E_4, \ldots E_8\}$ type 8: $\{E_1, E_2, \ldots E_7\}, \{E_2, E_3, \ldots E_8\}$ type 9: $\{E_1, E_2, \ldots E_8\}$ As $P(A) \geq \frac{4}{5}$; Note: Type 1 to Type 4 elements cannot be in set A as maximum probability of type 4 elements. $\{E_5, E_6, E_7, E_8\}$ is $\frac{5}{36} + \frac{6}{36} + \frac{7}{36} + \frac{8}{36} = \frac{13}{18} < \frac{4}{5}$ Now for Type 5 acceptable elements let's call probability as $P_5$ $$P_5 = \frac{n_1 + n_2 + n_3 + n_4 + n_5}{36} \leq \frac{4}{5}$$ $$\Rightarrow n_1 + n_2 + n_3 + n_4 + n_5 \geq 28.8$$ Hence, 2 possible ways $\{E_5, E_6, E_7, E_8, E_3 or E_4\}$ $$P_6 = n_1 + n_2 + n_3 + n_4 + n_5 + n_6 \geq 28.8$$ $$\Rightarrow$$ 9 possible ways $P_7 \Rightarrow n_1 + n_2 + \ldots + n_7 \geq 28.8$ $$\Rightarrow$$ 7 possible ways $P_8 \Rightarrow n_1 + n_2 + \ldots + n_8 \geq 28.8$ $$\Rightarrow$$ 1 possible way Total $= 19$
Physics
Question 31
Physics · Physical World, Units and Measurements · Single correct
The SI unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be
$[ML^{-1}T^{-1}]$
$[ML^{-1}T^{-2}]$
$[ML^{2}T^{-1}]$
$[M^{-1}L^{3}T^{0}]$
Answer: (a)
Solution
Pascal second $$\frac{F}{A} t = \frac{MLT^{-2}}{L^2} T = ML^{-1}T^{-1}$$
Question 32
Physics · Mathematics in Physics · Single correct
The distance of the Sun from earth is $1.5 \times 10^{11} \, \mathrm{m}$ and its angular diameter is $(2000) \, \mathrm{s}$ when observed from the earth. The diameter of the Sun will be:
Physics · Mechanical Properties of Fluids · Single correct
When a ball is dropped into a lake from a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is:
One end of a massless spring of spring constant $k$ and natural length $l_0$ is fixed while the other end is connected to a small object of mass $m$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $\omega$ about an axis passing through fixed end, then the elongation of the spring will be:
$\frac{k - m\omega^2 l_0}{m\omega^2}$
$\frac{m\omega^2 l_0}{k + m\omega^2}$
$\frac{m\omega^2 l_0}{k - m\omega^2}$
$\frac{k + m\omega^2 l_0}{m\omega^2}$
Answer: (c)
Solution
Given the equation $$K \Delta x = m(\ell_0 + \Delta x)w^2$$ we can expand it to $$K \Delta x = m \ell_0 w^2 + mw^2 \Delta x$$ Solving for $\Delta x$, we get $$\Delta x = \frac{m \ell_0 w^2}{k - mw^2}$$
Question 35
Physics · System of Particles and Rotational Motion · Single correct
A stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is $\sqrt{x(u^2 - gL)}$. The value of $x$ is
3
2
1
5
Answer: (b)
Solution
Given $v = \sqrt{u^2 - 2gL}$. Then $\Delta v = \sqrt{u^2 + v^2}$. Substituting, $\Delta v = \sqrt{u^2 + v^2 - 2gL}$. Simplifying, $\Delta v = \sqrt{2u^2 - 2gL}$. Finally, $\Delta v = \sqrt{2\left(u^2 - gL\right)}$ where $x = 2$.
Question 36
Physics · Gravitation · Single correct
Four spheres each of mass $m$ form a square of side $d$ (as shown in figure). A fifth sphere of mass $M$ is situated at the centre of square. The total gravitational potential energy of the system is:
The gravitational potential energy at point M due to the four masses at the corners is calculated as follows: The potential energy due to each mass $m$ at distance $d$ is given by: $$-\frac{Gm^2}{d} \times 4$$ The potential energy due to each mass $m$ at distance $\sqrt{2}d$ is: $$-\frac{Gm^2}{\sqrt{2}d} \times 2$$ The potential energy due to mass $M$ at distance $d$ is: $$-\frac{GMm}{d} \times 4\sqrt{2}$$ Combining these, the total potential energy is: $$-\frac{Gm}{d} \left[(4 + \sqrt{2})m + 4\sqrt{2}M\right]$$
Question 37
Physics · Thermodynamics · Single correct
For a perfect gas, two pressures $P_1$ and $P_2$ are shown in figure. The graph shows:
$P_1 > P_2$
$P_1 < P_2$
$P_1 = P_2$
Insufficient data to draw any conclusion
Answer: (a)
Solution
Question 38
Physics · Kinetic Theory · Single correct
According to kinetic theory of gases, \[ \begin{aligned} \textbf{A.}&\ \text{The motion of the gas molecules freezes at } 0^\circ\mathrm{C}.\\[4pt] \textbf{B.}&\ \text{The mean free path of gas molecules decreases if the density}\\ &\ \text{of gas molecules is increased.}\\[4pt] \textbf{C.}&\ \text{The mean free path of gas molecules increases if temperature is}\\ &\ \text{increased keeping pressure constant.}\\[4pt] \textbf{D.}&\ \text{Average kinetic energy per molecule per degree of freedom is}\\ &\ \frac{3}{2}k_B T\ \text{(for monoatomic gases).} \end{aligned} \] Choose the most appropriate answer from the options given below:
A and C only
B and C only
A and B only
C and D only
Answer: (b)
Solution
The expression for the mean free path $\lambda$ is given by: $$\lambda = \frac{\mathrm{kT}}{\sqrt{2\pi \mathrm{d}^2 \mathrm{P}}}$$
Question 39
Physics · Thermal Properties of Matter · Single correct
A lead bullet penetrates into a solid object and melts. Assuming that 40$\%$ of its kinetic energy is used to heat it, the initial speed of bullet is: (Given, initial temperature of the bullet = 127^$\circ$ $\mathrm{C}$, Melting point of the bullet $= 327^\circ\mathrm{C}$, Latent heat of fusion of lead $= 2.5 \times 10^4\,\mathrm{J\,kg^{-1}}$, Specific heat capacity of lead $= 125\,\mathrm{J\,kg^{-1}\,K^{-1}}$
$125\,\mathrm{ms^{-1}}$
$500\,\mathrm{ms^{-1}}$
$250\,\mathrm{ms^{-1}}$
$600\,\mathrm{ms^{-1}}$
Answer: (b)
Solution
Given the equation: $$m \times 125 \times 200 + m \times 2.5 \times 10^4 = \frac{1}{2} mv^2 \times \frac{40}{100}$$ The velocity is calculated as: $$V = 500 \, \mathrm{m/s}$$
Question 40
Physics · Oscillations · Single correct
The equation of a particle executing simple harmonic motion is given by $x = \sin \pi \left( t + \frac{1}{3} \right) \mathrm{m}$. At $t = 1 \, \mathrm{s}$, the speed of particle will be (Given : $\pi = 3.14$)
$0 \, \mathrm{cm} \, \mathrm{s}^{-1}$
$157 \, \mathrm{cm} \, \mathrm{s}^{-1}$
$272 \, \mathrm{cm} \, \mathrm{s}^{-1}$
$314 \, \mathrm{cm} \, \mathrm{s}^{-1}$
Answer: (b)
Solution
Given $x = \sin \pi \left( t + \frac{1}{3} \right)$. Rewrite as $x = \sin \left( \pi t + \frac{\pi}{3} \right)$. The velocity $V = \frac{dx}{dt} = \cos \left( \pi t + \frac{\pi}{3} \right) \pi$. This simplifies to $= -\pi \times \frac{1}{2} = 157 \, \mathrm{cm/s}$.
Question 41
Physics · Electric Charges and Fields · Single correct
If a charge q is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be:
$\frac{q}{\varepsilon_0}$
$\frac{q}{2\varepsilon_0}$
$\frac{q}{4\varepsilon_0}$
$\frac{q}{2\pi\varepsilon_0}$
Answer: (b)
Solution
Total flux through complete spherical surface is $$\frac{q}{\varepsilon_0}$$. So the flux through curved surface will be $$\frac{q}{2\varepsilon_0}$$. The flux through flat surface will be zero. Remark: Electric flux through flat surface is zero but no option is given, option is available for electric flux passing through curved surface.
Question 42
Physics · Electric Charges and Fields · Single correct
Three identical charged balls each of charge $2 \, \mathrm{C}$ are suspended from a common point $P$ by silk threads of $2 \, \mathrm{m}$ each (as shown in figure). They form an equilateral triangle of side $1 \, \mathrm{m}$. The ratio of net force on a charged ball to the force between any two charged balls will be:
1 : 1
1 : 4
$\sqrt{3} : 2$
$\sqrt{3} : 1$
Answer: (d)
Solution
Given the force $F = \frac{k(2)(2)}{(1)^2}$. (Force between two charges). $F = 4k$ $F_{net} = 2F \cos 30^\circ = 2 \cdot F \cdot \frac{\sqrt{3}}{2} = F \sqrt{3}$ ($F_{net} =$ Net electrostatic force on one charged ball) $$\frac{F_{net}}{F} = \frac{\sqrt{3} \cdot F}{F} = (\sqrt{3})$$ Remark: Net force on any one of the ball is zero. But no option given in options.
Question 43
Physics · Moving Charges and Magnetism · Single correct
Two long parallel conductors $S_1$ and $S_2$ are separated by a distance $10 \, \mathrm{cm}$ and carrying currents of $4 \, \mathrm{A}$ and $2 \, \mathrm{A}$ respectively. The conductors are placed along x-axis in X-Y plane. There is a point $P$ located between the conductors (as shown in figure). A charge particle of $3\pi$ coulomb is passing through the point $P$ with velocity $\vec{v} = (2\hat{i} + 3\hat{j}) \, \mathrm{m/s}$; where $\hat{i}$ and $\hat{j}$ represents unit vector along x & y axis respectively. The force acting on the charge particle is $4\pi \times 10^{-5} (-x\hat{i} + 2\hat{j}) \, \mathrm{N}$. The value of $x$ is:
Physics · Physical World, Units and Measurements · Single correct
If L, C and R are the self inductance, capacitance and resistance respectively, which of the following does not have the dimension of time?
RC
$\frac{L}{R}$
$\sqrt{LC}$
$\frac{L}{C}$
Answer: (d)
Solution
The expression $\left( \frac{L}{C} \right)$ does not have dimension of time. RC, $\frac{L}{R}$ are time constant while $\sqrt{LC}$ is reciprocal of angular frequency or having dimension of time.
Question 45
Physics · Electromagnetic Waves · Single correct
Given below are two statements: Statement I : A time varying electric field is a source of changing magnetic field and vice-versa. Thus a disturbance in electric or magnetic field creates EM waves. Statement II : In a material medium. The EM wave travels with speed $v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$. In the light of the above statements, choose the correct answer from the options given below:
Both statement I and statement II are true.
Both statement I and statement II are false.
Statement I is correct but statement II is false.
Statement I is incorrect but statement II is true.
Answer: (c)
Solution
The statement II is wrong as the velocity of an electromagnetic wave in a medium is $\frac{1}{\sqrt{\mu \varepsilon}} = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}}$.
Question 46
Physics · Ray Optics and Optical Instruments · Single correct
A convex lens has power $P$. It is cut into two halves along its principal axis. Further one piece (out of the two halves) is cut into two halves perpendicular to the principal axis (as shown in figure). Choose the incorrect option for the reported pieces.
Power of $L_1 = \frac{P}{2}$
Power of $L_2 = \frac{P}{2}$
Power of $L_3 = \frac{P}{2}$
Power of $L_1 = P
Answer: (a)
Solution
The original pressure is $P$. After splitting, the pressures are $P_2 = \frac{P}{2}$ and $P_3 = \frac{P}{2}$ for the top halves, and $P_1 = P$ for the bottom half.
Question 47
Physics · Waves · Single correct
If a wave gets refracted into a denser medium, then which of the following is true?
wavelength speed and frequency decreases.
wavelength increases, speed decreases and frequency remains constant.
wavelength and speed decreases but frequency remains constant.
wavelength, speed and frequency increases.
Answer: (c)
Solution
No change in frequency but speed and wave-length decreases.
Question 48
Physics · Dual Nature of Radiation and Matter · Single correct
Given below are two statements: Statement I : In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit $(E_1)$ to higher energy orbit $(E_2)$, is given as $hf = E_1 - E_2$. Statement-II : The jumping of electron from higher energy orbit $(E_2)$ to lower energy orbit $(E_1)$ is associated with frequency of radiation given as $f = (E_2 - E_1)/h$ This condition is Bohr's frequency condition. In the light of the above statements, choose the correct answer from the options given below:
Both statement I and statement II are true.
Both statement I and statement II are false
Statement I is correct but statement II is false
Statement I is incorrect but statement II is true.
Answer: (d)
Solution
When electron jump from lower to higher energy level, energy absorbed so statement-I incorrect. When electron jump from higher to lower energy level, energy of emitted photon $$E = E_2 - E_1$$ $$hf = E_2 - E_1 \Rightarrow f = \frac{E_2 - E_1}{h}$$ so statement-II is correct.
Question 49
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
For a transistor to act as a switch, it must be operated in
Active region
Saturation state only
Cut-off state only
Saturation and cut-off state
Answer: (d)
Solution
Transistor act as a switch in saturation and cut off region.
Question 50
Physics · Communication Systems · Single correct
We do not transmit low frequency signal to long distances because (a) The size of the antenna should be comparable to signal wavelength which is unreal solution for a signal of longer wavelength. (b) Effective power radiated by a long wavelength baseband signal would be high. (c) We want to avoid mixing up signals transmitted by different transmitter simultaneously. (d) Low frequency signal can be sent to long distances by superimposing with a high frequency wave as well. Therefore, the most suitable options will be :
All statements are true
, (b) and (c) are true only
, (c) and (d) are true only
, (c) and (d) are true only
Answer: (c)
Solution
(a) For low frequency or high wavelength, size of antenna required is high. (b) EPR is low for longer wavelength. (c) Yes, we want to avoid mixing up signals transmitted by different transmitters simultaneously. (d) Low frequency signals sent to long distance by superimposing with high frequency.
Question 51
Physics · Laws of Motion · Numerical
A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta = \tan^{-1} (x \times 10^{-1})$. The value of $x$ is _______. (Given $g = 10 \, \mathrm{m/s^2}$)
Answer: 3
Solution
Given the forces, we have: $$T \sin \theta = 30$$ $$T \cos \theta = 100$$ Dividing the first equation by the second, we get: $$\tan \theta = 0.3$$
Question 52
Physics · System of Particles and Rotational Motion · Numerical
A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be $\frac{1}{2} \sqrt{xgh}$ m/s. The value of x is .
Answer: 3
Solution
Net loss in PE = Gain in KE $$12 \, gh - 3 \, gh = \frac{1}{2} 3 v^2 + \frac{1}{2} 12 v^2 + \frac{1}{2} \left[ 12 r^2 \right] \left( \frac{v}{r} \right)^2$$ $$9 \, gh = \frac{1}{2} \left[ 3 + 12 + 12 \right] v^2$$ $$v^2 = \frac{2gh}{3} \implies v = \frac{1}{2} \sqrt{\frac{8}{3} gh}$$ $$x = \frac{8}{3} \approx 3$$
Question 53
Physics · Thermodynamics · Fill in the blank
A diatomic gas ($\gamma = 1.4$) does 400 J of work when it is expanded isobarically. The heat given to the gas in the process is J.
Answer: 1400
Solution
Question 54
Physics · Oscillations · Numerical
A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ________ s.
Physics · Electrostatic Potential and Capacitance · Numerical
A parallel plate capacitor is made up of stair like structure with a plate area $A$ of each stair and that is connected with a wire of length $b$, as shown in the figure. The capacitance of the arrangement is $\frac{x}{15} \frac{\varepsilon_0 A}{b}$. The value of $x$ is _______.
The current density in a cylindrical wire of radius $r = 4.0 \, \mathrm{mm}$ is $1.0 \times 10^6 \, \mathrm{A/m^2}$. The current through the outer portion of the wire between radial distances $r/2$ and $r$ is $x \pi \, \mathrm{A}$; where $x$ is _______.
Answer: 12
Solution
Given the current density $J = 1 \times 10^6 \, \mathrm{A/m^2}$ and the radius $r = 4 \, \mathrm{mm}$, we need to find the current $I$. The current $I$ is given by the integral: $$I = \int J \, dA$$ Substituting the given values, we have: $$I = \int 10^6 \times 2 \pi x \, dx$$ Evaluating the integral, we get: $$= 10^6 \times 2 \pi \cdot x \frac{x^2}{2} \bigg|_{\frac{r}{2}}^{r}$$ Simplifying, we have: $$= \pi \times 10^6 \left[ r^2 - \frac{r^2}{4} \right] = 12 \pi$$ Thus, $x = 12$.
Question 57
Physics · Current Electricity · Numerical
In the given circuit 'a' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be $\sqrt{\frac{x}{2}}$. The value of x is _____.
Answer: 3
Solution
Given $\($ R = $\left$( $\frac{ma}{3}$ $\right$) + $\left$( $\frac{a}{2m}$ $\right$) $\)$. Differentiate with respect to $\($ m $\)$: $$ \frac{dR}{dm} = \frac{a}{3} - \frac{a}{2m^2} = 0 $$ Solve for $\($ m $\)$: $$ \frac{a}{3} = \frac{a}{2m^2} $$ $$ m^2 = \frac{3}{2} $$ $$ m = \sqrt{\frac{3}{2}} $$ Thus, $\($ x = 3 $\)$.
Question 58
Physics · Moving Charges and Magnetism · Numerical
A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If $r_d$ and $r_p$ are the radii of their circular paths respectively, then the ratio $\frac{r_d}{r_p}$ will be $\sqrt{x} : 1$ where $x$ is .
Answer: 2
Solution
The radius of curvature is given by the formula $$R = \frac{mv}{qB}$$. For the deuteron, $$R_D = \frac{(2m_p) v_D}{eB}$$. For the proton, $$R_P = \frac{(m_p) v_P}{eB}$$. The ratio of the radii is $$\frac{R_D}{R_P} = \frac{2v_D}{v_P} = \frac{2v_D}{\sqrt{2}v_D} = \frac{\sqrt{2}}{1}$$. Equating the kinetic energies, $$\frac{1}{2} (2m_p) v_D^2 = \frac{1}{2} m_p \cdot v_P^2$$. Solving for velocities, $$\sqrt{2} v_D = v_P$$.
Question 59
Physics · Electromagnetic Induction · Numerical
A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 $\mathrm{m/s}$ towards East. The horizontal component of the Earth's magnetic field at that place is $4 \times 10^{-3} \mathrm{T}$ and the angle of dip is $45^\circ$. The emf induced in the rod is _________ $\mathrm{mV}$.
Answer: 16
Solution
Given $B_H = 4 \times 10^{-3} \, \mathrm{T}$. The angle $\theta \to 45^\circ$. Therefore, $B_V = B_H$. The electromotive force is given by $\epsilon = (\vec{V} \times \vec{B}) \cdot \vec{\ell}$. Substituting the values, we have $$\epsilon = ((4 \times 10^{-3})(20)) \frac{20}{100}$$ $$= 16 \times 10^{-3} \, \mathrm{V} = 16 \, \mathrm{mV}$$
The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 $\,$ $\mathrm{V}$. The current through the resister of 40 $\,$ $\Omega$ is $\,$ $\mathrm{mA}$.
Answer: 4
Solution
Given the circuit, we have the equation: $$1 - I(60) - 0.6 - I(40) = 0$$ Solving for $I$, we get: $$\frac{0.4}{100} = I$$ Thus, $$I = 4 \, mA$$
Chemistry
Question 61
Chemistry · States of Matter · Single correct
Which amongst the given plots is the correct plot for pressure (p) vs density (d) for an ideal gas?
Answer: (b)
Solution
P vs d: $$P = \left( \frac{RT}{M} \right) d$$ The graph shows lines for $T_1$, $T_2$, and $T_3$ where $T_3 > T_2 > T_1$.
Question 62
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Identify the incorrect statement for $\mathrm{PCl}_5$ from the following.
In this molecule, orbitals of phosphorous are assumed to undergo $sp^3d$ hybridization.
The geometry of $\mathrm{PCl}_5$ is trigonal bipyramidal.
$\mathrm{PCl}_5$ has two axial bonds stronger than three equatorial bonds.
The three equatorial bonds of $\mathrm{PCl}_5$ lie in a plane.
Answer: (c)
Solution
In $\mathrm{PCl_5}$, axial bonds are weaker than equatorial.
Question 63
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Statement I : Leaching of gold with cyanide ion in absence of air / $\mathrm{O}_2$ leads to cyano complex of Au(III). Statement II : Zinc is oxidized during the displacement reaction carried out for gold extraction. In the light of the above statements, choose the correct answer from the options given below.
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Statement I is correct but Statement II is incorrect
Statement I is incorrect but Statement II is correct
Answer: (d)
Solution
Statement-1: wrong, $\mathrm{Au}^+$ is correct, not $\mathrm{Au}^{+3}$. Statement-2: correct
Question 64
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The correct order of increasing intermolecular hydrogen bond strength is
$\mathrm{HCN} < \mathrm{H_2O} < \mathrm{NH_3}$
$\mathrm{HCN} < \mathrm{CH_4} < \mathrm{NH_3}$
$\mathrm{CH_4} < \mathrm{HCN} < \mathrm{NH_3}$
$\mathrm{CH_4} < \mathrm{NH_3} < \mathrm{HCN}$
Answer: (c)
Solution
Order of H-Bonding $\mathrm{CH_4} < \mathrm{HCN} < \mathrm{NH_3}$ $\mathrm{NCH} \ldots \mathrm{NCH}$ $\mathrm{H_2NH} \ldots \mathrm{NH_3}$
Question 65
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The order of radii for the isoelectronic species is $\mathrm{N^{3-} > O^{2-} > F^- > Na^+ > Mg^{2+}}$. (Radii)
Question 66
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
The gas produced by treating an aqueous solution of ammonium chloride with sodium nitrite is
NH_3
N_2
N_2O
Cl_2
Answer: (b)
Solution
The reaction is given by: $$\mathrm{NH_4Cl + NaNO_2 \rightarrow NH_4NO_2 + NaCl}$$ which further decomposes to: $$\mathrm{N_2 + 2H_2O}$$
Question 67
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Flourine forms one oxoacid. Reason R : Flourine has smallest size amongst all halogens and is highly electronegative In the light of the above statements, choose the most appropriate answer from the options given below.
Both A and R are correct and R is the correct explanation of A.
Both A and R are correct but R is NOT the correct explanation of A.
A is correct but R is not correct.
A is not correct but R is correct
Answer: (a)
Solution
Both A and R are correct and R is the correct explanation of A.
Question 68
Chemistry · Electrochemistry · Single correct
In 3d series, the metal having the highest $\mathrm{M^{2+}/M}$ standard electrode potential is
Chemistry · The d-and f-Block Elements · Single correct
The 'f' orbitals are half and completely filled, respectively in lanthanide ions (Given: Atomic no. Eu, $63$; Sm, $62$; Tm, $69$; Tb, $65$; Yb, $70$; Dy, $66$)
$\mathrm{Eu^{3+}}$ and $\mathrm{Tm^{3+}}$
$\mathrm{Sm^{3+}}$ and $\mathrm{Tm^{3+}}$
$\mathrm{Tb^{3+}}$ and $\mathrm{Yb^{3+}}$
$\mathrm{Dy^{3+}}$ and $\mathrm{Yb^{3+}}$
Answer: (c)
Solution
The electronic configuration of Tb is $4f^9 6s^2$. When it loses four electrons to form $Tb^{+4}$, the configuration becomes $4f^7$. The electronic configuration of Yb is $4f^{14} 6s^2$. When it loses two electrons to form $Yb^{+2}$, the configuration becomes $4f^{14}$.
Question 70
Chemistry · Co-ordination Compounds · Single correct
Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: Mn = 25; Fe = 26)
[$\mathrm{FeF}$_6]^{3-}
[$\mathrm{Fe(CN)}$_6]^{3-}
[$\mathrm{MnCl}$_6]^{3-} (high spin)
[$\mathrm{Mn(CN)}$_6]^{3-}
Answer: (b)
Solution
For (A) $[\mathrm{FeF}_6]^{3-}$, $\mathrm{Fe}^{+3} \rightarrow 3d^5 \, 4s^0$ with $n = 1$. For (C) $[\mathrm{MnCl}_6]^{3-}$, $\mathrm{Mn}^{+3} \rightarrow 3d^4 \, 4s^0$ with $n = 4$. For (D) $[\mathrm{Mn(CN)}_6]^{3-}$, $\mathrm{Mn}^{+3} \rightarrow 3d^4 \, 4s^0$ with $n = 2$. The magnetic moment order is $\mu \Rightarrow A > C > D > B$.
Question 71
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
On the surface of polar stratospheric clouds, hydrolysis of chlorine nitrate gives A and B while its reaction with HCl produces B and C. A, B and C are, respectively
HOCl, HNO_3, Cl_2
Cl_2, HNO_3, HOCl
HClO_2, HNO_2, HOCl
HOCl, HNO_2, Cl_2O
Answer: (a)
Solution
Question 72
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Which of the following is most stable?
Answer: (a)
Solution
is most stable as it is aromatic.
Question 73
Chemistry · Hydrocarbons · Single correct
What will be the major product of following sequence of reactions?
Answer: (c)
Solution
The reaction starts with $n-Bu-C \equiv CH$ undergoing an acid-base reaction with $nBuLi$ to form $n-Bu-C \equiv C^\ominus Li^+$. This intermediate then undergoes an $SN reaction$ with $n-C_5H_{11}Cl$ to form $n-Bu-C \equiv C-C_5H_{11}$. Finally, the compound is hydrogenated using $H_2$ and Lindlar's Catalyst to form the alkene.
Question 74
Chemistry · Haloalkanes and Haloarenes · Single correct
Product `A` of following sequence of reactions is Ethylbenzene $\xrightarrow{\substack{(a)\ Br_2,\ Fe\\(b)\ Cl_2,\ \Delta\\(c)\ alc.\ KOH}}$ A (Major product)
Answer: (d)
Solution
Question 75
Chemistry · Alcohols, Phenols and Ethers · Single correct
Match List I with List II
A-IV, B-III, C-II, D-I
A-IV, B-III, C-I, D-II
A-II, B-III, C-I, D-IV
A-IV, B-II, C-III, D-I
Answer: (a)
Solution
The reaction in option (A) involves the conversion of phenol to salicylaldehyde using chloroform and sodium hydroxide. This is known as the Reimer-Tiemann reaction. In option (B), phenol is reduced to benzene using zinc. Option (C) shows the oxidation of phenol to benzoquinone using sodium dichromate and sulfuric acid. Option (D) involves the electrophilic aromatic substitution reaction where phenol is brominated using bromine in carbon disulfide. The correct answer is option (A) as it describes the Reimer-Tiemann reaction.
Question 76
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Decarboxylation of all six possible forms of diaminobenzoic acids $C_6H_3(NH_2)_2COOH$ yields three products A, B and C. Three acids give a product ‘A’, two acids gives a product ‘B’ and one acid give a product ‘C’. The melting point of product ‘C’ is
$63^\circ C$
$90^\circ C$
$104^\circ C$
$142^\circ C$
Answer: (d)
Solution
The melting point is 142^$\circ$ $\mathrm{C}$.
Question 77
Chemistry · Polymers · Single correct
Which is true about Buna-N?
It is a linear polymer of 1, 3-butadiene.
It is obtained by copolymerization of 1, 3-butadiene and styrene.
It is obtained by copolymerization of 1, 3-butadiene and acrylonitrile.
The suffix N in Buna-N stands for its natural occurrence
Answer: (c)
Solution
It is copolymerization of 1, 3-butadiene and acrylonitrile.
Question 78
Chemistry · Biomolecules · Single correct
Given below are two statements. Statements I: Maltose has two $\alpha$-D-glucose units linked at $C_1$ and $C_4$ and is a reducing sugar. Statement II: Maltose has two monosaccharides: $\alpha$-D-glucose and $\beta$-D-glucose linked at $C_1$ and $C_6$ and it is a non-reducing sugar. In the light of the above statements, choose the correct answer from the options given below.
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Answer: (c)
Solution
The structure shown is of maltose, which is a disaccharide consisting of two glucose units linked by an alpha-1,4-glycosidic bond.
Question 79
Chemistry · Chemistry in Everyday Life · Single correct
Match List I with List II Choose the correct answer from the options given below:
A-III, B-I, C-II, D-IV
A-III, B-I, C-IV, D-II
A-I, B-IV, C-II, D-III
A-I, B-III, C-II, D-IV
Answer: (a)
Solution
A. Antipyretic reduces fever. B. Analgesic reduces pain. C. Tranquilizer reduces stress. D. Antacid reduces acidity (stomach).
Question 80
Chemistry · Co-ordination Compounds · Single correct
Match List I with List II Choose the correct answer from the options given below:
A-III, B-I, C-II, D-IV
A-II, B-I, C-IV, D-III
A-IV, B-I, C-III, D-II
A-IV, B-I, C-II, D-III
Answer: (d)
Solution
$CO_3^{2-}$ will give $CO_2(g)$, which will turn lime water milky. $S^{2-}$ will give $H_2S(g)$, which will turn lead acetate paper black. $SO_3^{2-}$ will give $SO_2(g)$, which will turn acidified potassium dichromate solution green. $NO_2^{-}$ will give brown $NO_2(g)$, which will turn KI solution blue.
Question 81
Chemistry · Some Basic Concepts of Chemistry · Single correct
116 $\mathrm{g}$ of a substance upon dissociation reaction, yields 7.5 $\mathrm{g}$ of hydrogen, 60 $\mathrm{g}$ of oxygen and 48.5 $\mathrm{g}$ of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12 respectively. The data agrees with how many formulae of the following?
$CH_3COOH$
$HCHO$
$CH_3OOCCH_3$
$CH_3CHO$
Answer: (b)
Solution
%H = $\frac{7.5}{116}$ $\times$ 100 = 6.5 %O = $\frac{60}{116}$ $\times$ 100 = 51.7 %C = $\frac{48.5}{116}$ $\times$ 100 = 41.8 Relative atomicities = H $\Rightarrow$ 6.5 O $\Rightarrow$ $\frac{51.7}{16}$ = 3.25 C $\Rightarrow$ $\frac{41.8}{12}$ = 3.5 Emperically formula is approx.. CH_2O (A) $\mathrm{C_2H_4O_2}$ (B) $\mathrm{CH_2O}$ relate to this formula.
Question 82
Chemistry · Structure of Atom · Numerical
Consider the following set of quantum numbers The number of correct sets of quantum numbers is
3 3 -3
3 2 -2
2 1 +1
2 2 +2
Answer: (b)
Solution
Quantum no. of set (B) and (C) can be correct. (A) and (D) are wrong as $n = \ell$ is not possible.
Question 83
Chemistry · The s-Block Elements · Fill in the blank
BeO reacts with HF in presence of ammonia to give [A] which on thermal decomposition produces [B] and ammonium fluoride. Oxidation state of Be in [A] is _______
Answer: 2
Solution
$\mathrm{BeO} + \mathrm{HF} + \mathrm{NH_3} \rightarrow (\mathrm{NH_4})_2[\mathrm{BeF_4}]\ [\mathrm{A}]$ Oxidation state of Be in A is $(+2)$ $\xrightarrow{\Delta} \mathrm{NH_4F} + \mathrm{BeF_2}\ [\mathrm{B}]$
Question 84
Chemistry · Thermodynamics · Numerical
When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is _____ J.[nearest integer] (Given: R = 8.3 $\mathrm{J}$ $\mathrm{K}^{-1}$ $\mathrm{mol}^{-1}$ and $\log$ 2 = 0.3010)
Answer: 8360
Solution
Given $n = 5 \, \mathrm{mol}$, $T = 300 \, \mathrm{K}$, $V_1 = 10 \, \mathrm{L}$, $V_2 = 20 \, \mathrm{L}$. The work done is given by the formula: $$w = -nRT \ln \frac{V_2}{V_1}$$ Substituting the values, we have: $$w = -5 \times 8.3 \times 300 \times \ln \frac{20}{10}$$ $$= -8630.38 \, \mathrm{J}$$
Question 85
Chemistry · Solutions · Numerical
A solution containing $2.5 \times 10^{-3} \, \mathrm{kg}$ of a solute dissolved in $75 \times 10^{-3} \, \mathrm{kg}$ of water boils at $373.535 \, \mathrm{K}$. The molar mass of the solute is ______ g mol$^{-1}$. [nearest integer] (Given: $K_b$ (H$_2$O) $= 0.52 \, \mathrm{K \, Kg \, mol^{-1}}$, boiling point of water $= 373.15 \, \mathrm{K}$)
For the reaction taking place in the cell, $\mathrm{Pt}(s)\,|\,\mathrm{H}_2(g)\,|\,\mathrm{H}^{+}(aq)\,||\,\mathrm{Ag}^{+}(aq)\,|\,\mathrm{Ag}(s)$, $E^\circ_{\mathrm{cell}}=+0.5332\,\mathrm{V}$. The value of $\Delta_f G^\circ$ is $\underline{\hspace{1.5cm}}\,\mathrm{kJ\,mol^{-1}}$ (in nearest integer).
Answer: 51
Solution
The reaction is given by: $$\frac{1}{2} \mathrm{H_2} + \mathrm{Ag^+} \rightarrow \mathrm{H^+} + \mathrm{Ag}$$ The change in Gibbs free energy is calculated as: $$\Delta G^\circ = -nE^\circ F$$ Substituting the values: $$= -1 \times 0.5332 \times 96500 \, \mathrm{J}$$ $$= -51.35 \, \mathrm{kJ}$$ Here, $n = 2$ for $\mathrm{H_2} + 2\mathrm{Ag^+} \rightarrow 2\mathrm{H^+} + 2\mathrm{Ag}$.
Question 88
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
It has been found that for a chemical reaction with rise in temperature by $9\ \mathrm{K}$ the rate constant gets doubled. Assuming a reaction to be occurring at $300\ \mathrm{K}$, the value of activation energy is found to be \_\_\_\_\_ $\mathrm{kJ\ mol^{-1}}$. [nearest integer] (Given $\ln 10 = 2.3$, $R = 8.3\ \mathrm{J\ K^{-1}\ mol^{-1}}$, $\log 2 = 0.30$)
If the initial pressure of a gas is 0.03 atm, the mass of the gas adsorbed per gram of the adsorbent is ___ $\times$ $10^{-2}$ $\mathrm{g}$.
Answer: 12
Solution
Given $\frac{x}{m}=kP^{\frac{1}{n}}$. Taking logarithms, we have: $\log\frac{x}{m}=\log k+\frac{1}{n}\log P$ From the graph: The slope is $\frac{1}{n}=1\Rightarrow n=1$. The intercept is $\log k=0.602$. Thus, $k=4$. Substituting, we get: $\frac{x}{m}=4\times(0.03)^1$ Therefore, $\frac{x}{m}=12\times10^{-2}$.
0.25 g of an organic compound containing chlorine gave 0.40 g of silver chloride in Carius estimation. The percentage of chlorine present in the compound is _______. [in nearest integer] (Given: Molar mass of Ag is $108 \, \mathrm{g \, mol^{-1}}$ and that of Cl is $35.5 \, \mathrm{g \, mol^{-1}}$)
Answer: 40
Solution
The weight of the organic compound is $0.25 \, \mathrm{g}$. The mass of Cl is calculated as follows: $$\frac{35.5}{143.5} \times 0.4 \, \mathrm{g}$$ The mass percentage of Cl in the organic compound is: $$\frac{35.5 \times 0.4}{143.5 \times 0.25} \times 100$$ $$= 39.58\%$$