JEE Main 27 July 2022 Shift 1 question paper with solutions
JEE Main 27 July 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 · Relations and Functions · Single correct
Let $\mathcal{R}_1$ and $\mathcal{R}_2$ be two relations defined on $\mathbb{R}$ by $a \mathcal{R}_1 b \iff ab \geq 0$ and $a \mathcal{R}_2 b \iff a \geq b$, then
$\mathcal{R}_1$ is an equivalence relation but not $\mathcal{R}_2$
$\mathcal{R}_2$ is an equivalence relation but not $\mathcal{R}_1$
both $\mathcal{R}_1$ and $\mathcal{R}_2$ are equivalence relations
neither $\mathcal{R}_1$ nor $\mathcal{R}_2$ is an equivalence relation
Answer: (d)
Solution
Given $R_1 = \{ xy \geq 0, x, y \in \mathbb{R} \}$. For reflexive $x \times x \geq 0$ which is true. For symmetric, if $xy \geq 0 \Rightarrow yx \geq 0$. If $x = 2$, $y = 0$ and $z = -2$, then $x \cdot y \geq 0$ and $y \cdot z \geq 0$ but $x \cdot z \geq 0$ is not true. Therefore, it is not a transitive relation. Thus, $R_1$ is not an equivalence relation. For $R_2$, if $a \geq b$ it does not imply $b \geq a$. Therefore, $R_2$ is not an equivalence relation. Hence, the answer is $D$.
Question 2
Maths · Relations and Functions · Single correct
Let $f, g: \mathbb{N} - \{1\} \to \mathbb{N}$ be functions defined by $f(a) = \alpha$, where $\alpha$ is the maximum of the powers of those primes $p$ such that $p^\alpha$ divides $a$, and $g(a) = a + 1$, for all $a \in \mathbb{N} - \{1\}$. Then, the function $f + g$ is
one-one but not onto
onto but not one-one
both one-one and onto
neither one-one nor onto
Answer: (d)
Solution
Given $f : \mathbb{N} - \{1\} \to \mathbb{N}$ and $f(a) = \alpha$. Where $\alpha$ is the maximum of powers of prime $P$ such that $P^\alpha$ divides $a$. Also $g(a) = a + 1$. Therefore, $f(2) = 1$, $g(2) = 3$, $f(3) = 1$, $g(3) = 4$, $f(4) = 2$, $g(4) = 5$, $f(5) = 1$, $g(5) = 6$. Thus, $f(2) + g(2) = 4$, $(f(3) + g(3)) = 5$, $f(4) + g(4) = 7$, $f(5) + g(5) = 7$. Therefore, many $f(x) + g(x)$ does not contain $1$. Hence, the function is neither one-one nor onto.
Question 3
Maths · Complex Numbers and Quadratic Equations · Single correct
Let the minimum value $v_0$ of $v = |z|^2 + |z - 3|^2 + |z - 6i|^2$, $z \in \mathbb{C}$ is attained at $z = z_0$. Then $\left| 2z_0^2 - \overline{z_0}^3 + 3 \right|^2 + v_0^2$ is equal to
Let $A = \begin{pmatrix} 1 & 2 \\ -2 & -5 \end{pmatrix}$. Let $\alpha, \beta \in \mathbb{R}$ be such that $\alpha A^2 + \beta A = 2I$. Then $\alpha + \beta$ is equal to -
-10
-6
6
10
Answer: (d)
Solution
Characteristic equation of matrix A $$|A - \lambda I| = 0$$ $$\begin{vmatrix} 1 - \lambda & 2 \\ -2 & -5 - \lambda \end{vmatrix} = 0$$ $$\Rightarrow \lambda^2 + 4\lambda = 1$$ $$\Rightarrow A^2 + 4A = I$$ $$\Rightarrow 2A^2 + 8A = 2I \ldots (1)$$ Given that $\alpha A^2 + \beta A = 2I \ldots (2)$ Comparing equation (1) and (2) we get $$\alpha = 2, \beta = 8$$ Therefore, $\alpha + \beta = 10$ Ans. (D) (10)
Question 5
Maths · Binomial Theorem · Single correct
The remainder when $(2021)^{2022} + (2022)^{2021}$ is divided by 7 is
Suppose $a_1, a_2, \ldots, a_n, \ldots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms of the sum of first nine terms of the progression is $5 : 17$ and $110 < a_{15} < 120$, then the sum of the first ten terms of the progression is equal to -
Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined as $$f(x) = a \sin \left( \frac{\pi [x]}{2} \right) + [2 - x], \ a \in \mathbb{R},$$ where $[t]$ is the greatest integer less than or equal to $t$. If $$\lim_{x \to -1} f(x)$$ exists, then the value of $$\int_{0}^{4} f(x) \, dx$$ is equal to :
The equation of the circle is $x^2 + y^2 + 4\sqrt{3}x - 4 = 0$. The equation of the parabola is $y^2 = 8x + 4$. The points of intersection are $(0, 2)$ and $(0, -2)$. Both are symmetric about the x-axis. The area is given by: $$Area = 2 \int_{0}^{2} \left( \sqrt{16 - y^2 - 2\sqrt{3}} - \frac{y^2 - 4}{8} \right) \, dy$$ On solving, the area is: $$Area = \frac{1}{3} \left[ 8\pi + 4 - 12\sqrt{3} \right]$$
Question 10
Maths · Differential Equations · Single correct
Let $y = y_1(x)$ and $y = y_2(x)$ be two distinct solutions of the differential equation $\frac{dy}{dx} = x + y$, with $y_1(0) = 0$ and $y_2(0) = 1$ respectively. Then, the number of points of intersection of $y = y_1(x)$ and $y = y_2(x)$ is
0
1
2
3
Answer: (a)
Solution
$$ \frac{dy}{dx}=x+y \Rightarrow \frac{dy}{dx}-y=x $$ If $$ I=e^{-x} $$ Therefore, solution is $$ ye^{-x}=\int xe^{-x}\,dx $$ $$ ye^{-x}=-xe^{-x}-e^{-x}+c $$ $$ y=-x-1+ce^x $$ For $$ y(0)=0 $$ $$ 0=-0-1+c \Rightarrow c=1 $$ $$ y_1=-x-1+e^x \qquad ...(1) $$ For $$ y_2(0)=1 \Rightarrow c=2 $$ $$ y_2=-x-1+2e^x \qquad ...(2) $$ Now $$ y_2-y_1=e^x>0 $$ $$ \therefore y_2\ne y_1 $$ Therefore, number of points of intersection of \(y_1\) and \(y_2\) is zero.
Question 11
Maths · Conic Sections · Single correct
Let P $(a, b)$ be a point on the parabola $y^2 = 8x$ such that the tangent at P passes through the centre of the circle $x^2 + y^2 - 10x - 14y + 65 = 0$. Let A be the product of all possible values of $a$ and B be the product of all possible values of $b$. Then the value of $A + B$ is equal to:
0
25
40
65
Answer: (d)
Solution
P(a, b) is a point on $y^2 = 8x$, such that the tangent at P passes through the center of $x^2 + y^2 - 10x - 14y + 65 = 0$, i.e. (5, 7). The tangent at $P(at^2, 2at)$ is $ty = x + at^2$. Let $A = 2$ and it passes through (5, 7). $$7t = 5 + 2t^2$$ Therefore, $t = 1$, $t = \frac{5}{2}$. Thus, $P(at^2, 2at) \Rightarrow (2, 4)$ when $t = 1$ and $\left( \frac{25}{2}, 10 \right)$ when $t = \frac{5}{2}$. Therefore, $A = 2 \times \frac{25}{2} = 25$. $B = 4 \times 10 = 40$. Thus, $A + B = 65$.
Question 12
Maths · Vector Algebra · Single correct
Let $\vec{a} = \alpha \hat{i} + \hat{j} + \beta \hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + 4\hat{k}$ be two vectors, such that $\vec{a} \times \vec{b} = -\hat{i} + 9\hat{i} + 12\hat{k}$. Then the projection of $\vec{b} - 2\vec{a}$ on $\vec{b} + \vec{a}$ is equal to
Let $\vec{a}$ = 2$\hat{i}$ - $\hat{j}$ + 5$\hat{k}$ and $\vec{b}$ = $\alpha$ $\hat{i}$ + $\beta$ $\hat{j}$ + 2$\hat{k}$ . If $((\vec{a}\times\vec{b})\times\hat{i})\cdot\hat{k}=\frac{23}{2}$, then $\left|\vec{b}\times2\hat{j}\right|$ is equal to ______.
Maths · Permutations and Combinations · Single correct
Let S be the sample space of all five digit numbers. If $p$ is the probability that a randomly selected number from $S$, is a multiple of 7 but not divisible by 5, then $9p$ is equal to
1.0146
1.2085
1.0285
1.1521
Answer: (c)
Solution
Given $n(S)$ = all 5 digit numbers $= 9 \times 10^4$. Let $A$ be the set of numbers that are multiples of 7 but not divisible by 5. The smallest 5 digit number divisible by 7 is 10003. The largest 5 digit number divisible by 7 is 99995. Therefore, $99995 = 10003 + (n - 1) \times 7$, giving $n = 12857$. Numbers divisible by 35: $99995 = 10010 + (P-1) \times 35 \Rightarrow P = 2572$. Therefore, numbers divisible by 7 but not by 35 are $12857 - 2572 = 10285$. Thus, $P = \frac{10285}{90000}$, so $9P = 1.0285$.
Question 15
Maths · Heights and Distances · Single correct
Let a vertical tower AB of height $2h$ stands on a horizontal ground. Let from a point P on the ground a man can see upto height $h$ of the tower with an angle of elevation $2\alpha$. When from $P$, he moves a distance $d$ in the direction of $\overrightarrow{AP}$, he can see the top $B$ of the tower with an angle of elevation $\alpha$. If $d = \sqrt{7}h$, then $\tan \alpha$ is equal to
$\sqrt{5} - 2$
$\sqrt{3} - 1$
$\sqrt{7} - 2$
$\sqrt{7} - \sqrt{3}$
Answer: (c)
Solution
Given $\tan 2\alpha = \frac{h}{x}$ and $\tan \alpha = \frac{2h}{x + \sqrt{7}h}$. We have $$\tan \alpha = \frac{2h}{h \cot 2\alpha + \sqrt{7}h}$$ $$\tan \alpha = \frac{2}{\frac{\left(1 - \tan^2 \alpha\right)}{2 \tan \alpha} + \sqrt{7}}$$ Put $\tan \alpha = t$ and simplify $$\Rightarrow \tan \alpha = \sqrt{7} - 2$$
Question 16
Maths · Mathematical Reasoning · Single correct
$(p\land r)\leftrightarrow(p\land(\sim q))$ is equivalent to $(\sim p)$ when $r=q$
Maths · Three Dimensional Geometry · Single correct
If the plane P passes through the intersection of two mutually perpendicular planes $2x + ky - 5z = 1$ and $3kx - ky + z = 5$, $k < 3$ and intercepts a unit length on positive x-axis, then the intercept made by the plane P on the y-axis is
$\frac{1}{11}$
$\frac{5}{11}$
6
7
Answer: (d)
Solution
Two given planes mutually perpendicular $$2(3k) + k(-k) + (-5) \cdot 1 = 0$$ $$k = 1, 5$$ but $k < 3$ so $k = 1$ Plane passing through these planes is $$2x + y - 5z - 1 + \lambda (3x - y + z - 5) = 0$$ $$\frac{x}{5\lambda + 1} + \frac{y}{5\lambda + 1} + \frac{z}{5\lambda + 1} = 1$$ $$\frac{2 + 3\lambda}{1 - \lambda} = \frac{\lambda - 5}{1}$$ Given $$\frac{5\lambda + 1}{2 + 3\lambda} = 1 \implies \lambda = \frac{1}{2}$$ So intercept on y-axis is $$\frac{5\lambda + 1}{1 - \lambda} = 7$$
Question 18
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let $A(1,\ 1)$, $B(-4,\ 3)$, $C(-2,\ -5)$ be vertices of a triangle $ABC$, $P$ be a point on side $BC$, and $\Delta_1$ and $\Delta_2$ be the areas of triangle $APB$ and $ABC$. Respectively. If $\Delta_1 : \Delta_2 = 4 : 7$, then the area enclosed by the lines $AP$, $AC$ and the $x$-axis is
$\frac{1}{4}$
$\frac{3}{4}$
$\frac{1}{2}$
1
Answer: (c)
Solution
Given $\Delta_1 = \dfrac{1}{2} \begin{vmatrix} x & y & 1 \\ 1 & 1 & 1 \\ -4 & 3 & 1 \end{vmatrix}$ and $\Delta_2 = \dfrac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ -4 & 3 & 1 \\ -2 & -5 & 1 \end{vmatrix}$. Given $\dfrac{\Delta_1}{\Delta_2} = \dfrac{4}{7} \Rightarrow \dfrac{-2x - 5y + 7}{36} = \dfrac{4}{7}$ $$\Rightarrow 14x + 35y = -95 \hfill \ldots(1)$$ Equation of $BC$ is $4x + y = -13 \hfill \ldots(2)$ Solve equations (1) and (2). Point $P\left(-\dfrac{20}{7},\ -\dfrac{11}{7}\right)$ Here point $Q\left(-\dfrac{1}{2},\ 0\right)$ and $R\left(\dfrac{1}{2},\ 0\right)$ So the area of triangle $AQR = \dfrac{1}{2} \times 1 \times 1 = \dfrac{1}{2}$
Question 19
Maths · Conic Sections · Single correct
If the circle $x^2 + y^2 - 2gx + 6y - 19c = 0$, $g, c \in \mathbb{R}$ passes through the point $(6, 1)$ and its centre lies on the line $x - 2cy = 8$, then the length of intercept made by the circle on x-axis is
$\sqrt{11}$
4
3
$2\sqrt{23}$
Answer: (d)
Solution
Given circle $x^2 + y^2 - 2gx + 6y - 19c = 0$. Passes through $(6, 1)$. $$12g + 19c = 43 \ldots(1)$$ Centre $(g, -3)$ lies on given line. So, $g + 6c = 8 \ldots(2)$ Solve equation (1) and (2). $c = 1$ and $g = 2$. Equation of circle $x^2 + y^2 - 4x + 6y - 19 = 0$. Length of intercept on x-axis $$= 2\sqrt{g^2 - c} = 2\sqrt{23}$$
Question 20
Maths · Applications of Derivatives · Single correct
Let a function $f : \mathbb{R} \to \mathbb{R}$ be defined as : $$f(x) = \begin{cases} \int_{0}^{x} (5 - |t - 3|) \, dt, & x > 4 \\ x^2 + bx, & x \leq 4 \end{cases}$$ where $b \in \mathbb{R}$. If $f$ is continuous at $x = 4$, then which of the following statements is NOT true ?
$f$ is not differentiable at $x = 4$
$f'(3) + f'(5) = \frac{35}{4}$
$f$ is increasing in $\left(-\infty, \frac{1}{8}\right) \cup (8, \infty)$
$f$ has a local minima at $x = \frac{1}{8}$
Answer: (c)
Solution
Given $$f(x) = \begin{cases} \int_0^x (5 - |t - 3|) \, dt, & x > 4 \\ x^2 + bx, & x \leq 4 \end{cases}$$ $f(x)$ is continuous at $x = 4$ So $\lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4)$ So $16 + 4b = \int_0^3 (2 - t) \, dt + \int_3^4 (8 - t) \, dt$ At $x = 4$ LHD $= 2x + b = \frac{31}{4}$ RHD $= 5 - |x - 3| = 4$ LHD $\neq$ RHD Option (A) is true and $f'(3) + f'(5) = \frac{23}{4} + 3 = \frac{35}{4}$ Option (B) is true $\therefore f(x) = x^2 - \frac{x}{4}$ at $x \leq 4$ $f'(x) = 2x - \frac{1}{4}$ This function is not increasing. In the interval in $x \in \left(-\infty, \frac{1}{8}\right)$ Option (C) is NOT TRUE. This function $f(x)$ is also local minima at $x = \frac{1}{8}$
For $k\in\mathbb{R}$, let the solutions of the equation $\cos\!\left( \sin^{-1} \!\left( x\cot \!\left( \tan^{-1} (\cos(\sin^{-1}x)) \right) \right) \right)=k,$ $0<|x|<\dfrac1{\sqrt2}$ be $\alpha$ and $\beta$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $x^2-bx-5=0$ are $\dfrac1{\alpha^2} +\dfrac1{\beta^2}$ and $\dfrac{\alpha}{\beta}$, then $\dfrac{b}{k^2}$ is equal to
The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is __________.
Answer: 2
Solution
Given $$6^2 = \frac{\sum x_i^2}{10} - \left(\bar{x}\right)^2 = 15$$. This implies $$\sum_{i=1}^{10} x_i = 150$$. Therefore, $$\sum_{i=1}^{9} x_i + 25 = 150$$, which gives $$\sum_{i=1}^{9} x_i = 125$$. Adding 15 to the sum, $$\sum_{i=1}^{9} x_i + 15 = 140$$. The actual mean is $$\frac{140}{10} = 14 = \bar{x}_{new}$$. Now, $$\frac{\sum_{i=1}^{9} x_i^2 + 25^2 - 15^2}{10} = 15$$. This implies $$\sum_{i=1}^{9} x_i^2 + 625 = 2400$$, so $$\sum_{i=1}^{9} x_i^2 = 1775$$. The actual variance is $$6^2_{actual} = \frac{\left(\sum x_i^2\right)_{actual}}{10} - \left(\bar{x}_{new}\right)^2$$. Substituting the values, $$= \frac{2000}{10} - 14^2$$, which simplifies to $$= 200 - 196 = 4$$. Therefore, $$(S.D.)_{actual} = 6 = 2$$.
Question 23
Maths · Three Dimensional Geometry · Fill in the blank
Let the line $\frac{x-3}{7} = \frac{y-2}{-1} = \frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1} = \frac{y+1}{-2} = \frac{z}{1}$ and $4ax-y+5z-7a=0=2x-5y-z-3, a \in \mathbb{R}$ at the point $P(\alpha, \beta, \gamma)$. Then the value of $\alpha + \beta + \gamma$ equals ______.
Answer: 12
Solution
Equation of plane $4ax - y + 5z - 7a + \lambda \left(2x - 5y - z - 3\right) = 0$ this satisfy $(4, -1, 0)$ $$16a + 1 - 7a + \lambda (8 + 5 - 3) = 0$$ $$9a + 1 + 10\lambda = 0 \ldots(1)$$ Normal vector of the plane $A$ is $(4a + 2\lambda, -1 - 5\lambda, 5 - \lambda)$ vector along the line which contained the plane $A$ is $i - 2j + k$ Therefore, $4a + 2\lambda + 2 + 10\lambda + 5 - \lambda = 0$ $$11\lambda + 4a + 7 = 0 \ldots(2)$$ Solve (1) and (2) to get $a = 1$, $\lambda = -1$ Now equation of plane $$x + 2y + 3z - 2 = 0$$ Let the point in the line $\frac{x - 3}{7} = \frac{y - 2}{-1} = \frac{z - 3}{-4} = t$ is $(7t + 3, -t + 2, -4t + 3)$ satisfy the equation of plane $A$ $$7t + 3 - 2t + 4 + 9 - 12t - 2 = 0$$ $t = 2$ So $\alpha + \beta + \gamma = 2t + 8 = 12$
Question 24
Maths · Conic Sections · Numerical
An ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through the vertices of the hyperbola $H : \frac{x^2}{49} - \frac{y^2}{64} = -1$. Let the major and minor axes of the ellipse $E$ coincide with the transverse and conjugate axes of the hyperbola $H$. Let the product of the eccentricities of $E$ and $H$ be $\frac{1}{2}$. If $l$ is the length of the latus rectum of the ellipse $E$, then the value of $113l$ is equal to _______.
Answer: 1552
Solution
Hyperbola: $\dfrac{x^2}{64}-\dfrac{y^2}{49}=1$ An ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ passes through the vertices of the hyperbola $H:\dfrac{x^2}{64}-\dfrac{y^2}{49}=1$. So, $b^2=64$ $e=\sqrt{1+\dfrac{a^2}{b^2}} =\sqrt{1+\dfrac{49}{64}}$ For the ellipse, $e=\sqrt{1-\dfrac{b^2}{a^2}} =\sqrt{1-\dfrac{a^2}{64}}$ Hence, $8\sqrt{1-\dfrac{49}{64}} =\sqrt{64-a^2}$ $\sqrt{64-a^2}\times\sqrt{113}=32$ $64-a^2=\dfrac{32^2}{113}$ $a^2=64-\dfrac{32^2}{113}$ $l=\dfrac{2a^2}{b}$ $=\dfrac{2}{8}\left(64-\dfrac{32^2}{113}\right)$ $=\dfrac{1552}{113}$ $\boxed{\dfrac{1552}{113}}$
Question 25
Maths · Differential Equations · Numerical
Let y = y(x) be the solution curve of the differential equation $$\sin(2x^2) \log_e(\tan x^2) dy + \left(4xy - 4\sqrt{2}x \sin\left(x^2 - \frac{\pi}{4}\right)\right) dx = 0,$$ $$0 < x < \sqrt{\frac{\pi}{2}},$$ which passes through the point $$\left(\sqrt{\frac{\pi}{6}}, 1\right).$$ Then $$\left|y\left(\sqrt{\frac{\pi}{3}}\right)\right|$$ is equal to _______.
Answer: 1
Solution
Given the equation $$\sin(2x^2) \ln(\tan x^2) \, dy + \left(4xy - 4\sqrt{2}x \sin\left(\frac{x^2 - \pi}{4}\right)\right) dx = 0$$ we have $$\ln(\tan x^2) \, dy + \frac{4\sqrt{2} \, x \sin\left(x^2 - \frac{\pi}{4}\right)}{\sin(2x^2)} \, dx = 0$$ which simplifies to $$d\left(y \ln(\tan x^2)\right) - 4\sqrt{2}x \left(\frac{\sin x^2 - \cos x^2}{\sqrt{2} - \sin x^2 \cos x^2}\right) dx = 0.$$ Integrating, we get $$\Rightarrow \int d\left(y \ln(\tan x^2)\right) + 2 \int \frac{dt}{t^2 - 1} = \int 0$$ which leads to $$\Rightarrow y \ln(\tan x^2) + 2 \cdot \frac{1}{2} \ln\left|\frac{t-1}{t+1}\right| = c.$$ Therefore, $$y \ln(\tan x^2) + \ln\left(\frac{\sin x^2 + \cos x^2 - 1}{\sin x^2 + \cos x^2 + 1}\right) = c.$$ Put $y = 1$ and $x = \frac{\pi}{\sqrt{6}}$, we have $$1 \ln\left(\frac{1}{\sqrt{3}}\right) + \ln\left(\frac{\frac{1}{2} + \frac{\sqrt{3}}{2} - 1}{\frac{1}{2} + \frac{\sqrt{3}}{2} + 1}\right) = c.$$ Now, $$x = \frac{\pi}{\sqrt{3}} \Rightarrow y(\ln \sqrt{3}) + \ln\left(\frac{\frac{1}{2} + \frac{\sqrt{3}}{2} - 1}{\frac{1}{2} + \frac{\sqrt{3}}{2} + 1}\right) = \ln\left(\frac{1}{\sqrt{3}}\right) + \ln\left(\frac{\sqrt{3} - 1}{\sqrt{3} + 3}\right).$$ Thus, $$y(\ln \sqrt{3}) = \ln\left(\frac{1}{\sqrt{3}}\right)$$ which implies $$\Rightarrow y = -1$$ and $$|y| = 1.$$
Question 26
Maths · Applications of Derivatives · Fill in the blank
Let $M$ and $N$ be the number of points on the curve $y^5 - 9xy + 2x = 0$, where the tangents to the curve are parallel to $x$-axis and $y$-axis, respectively. Then the value of $M + N$ equals _______.
Answer: 2
Solution
Given $y^5 - 9xy + 2x = 0$. $$5y^4 \frac{dy}{dx} - 9x \frac{dy}{dx} - 9y + 2 = 0$$ $$\frac{dy}{dx} (5y^4 - 9x) = 9y - 2$$ $$\frac{dy}{dx} = \frac{9y - 2}{5y^4 - 9x} = 0 (for horizontal tangent)$$ $$y = \frac{2}{9} \Rightarrow Which does not satisfy the original equation \Rightarrow M = 0.$$ Now $5y^4 - 9x = 0$ (for vertical tangent) $$5y^4 (9y - 2) - 9y^5 = 0$$ $$y^4 [45y - 10 - 9y] = 0$$ $$y = 0 (Or) 36y = 10$$ $$y = \frac{5}{18}$$ $$y = 0 \Rightarrow x = 0 \& y = \frac{5}{18} \Rightarrow x =$$ $$(0, 0) \left(x, \frac{5}{18}\right)$$ $N = 2$ $M + N = 0 + 2 = 2$
Question 27
Maths · Sequences and Series · Fill in the blank
Let $f(x) = 2x^2 - x - 1$ and $S = \{ n \in \mathbb{Z} : |f(n)| \leq 800 \}$. Then, the value of $$ \sum_{n \in S} f(n) $$ is equal to ______.
Let S be the set containing all $3 \times 3$ matrices with entries from $\{$-1, 0, 1$\}$. The total number of matrices $A \in S$ such that the sum of all the diagonal elements of $A^T A$ is 6 is _________.
Answer: 5376
Solution
Given $\mathrm{Tr}\left(AA^T\right) = 6$. $$AA^T = \begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \end{bmatrix} \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$$ Now given $a^2 + d^2 + g^2 + b^2 + e^2 + h^2 + c^2 + f^2 + i^2 = 6$. $$= \binom{9}{3} \times 2^6$$ $$= 5376$$
Question 29
Maths · Conic Sections · Fill in the blank
If the length of the latus rectum of the ellipse $x^2 + 4y^2 + 2x + 8y - \lambda = 0$ is 4, and $l$ is the length of its major axis, then $\lambda + l$ is equal to .
Maths · Complex Numbers and Quadratic Equations · Fill in the blank
Let $S = \{z \in \mathbb{C} : z^2 + \bar{z} = 0\}$. Then $\displaystyle\sum_{z \in S}\left(\mathrm{Re}(z) + \mathrm{Im}(z)\right)$ is equal to $\underline{\phantom{xxxx}}$.
Answer: 0
Solution
Given the set $S = \{ z \in \mathbb{C} : z^2 + \overline{z} = 0 \}$. Let $z = x + iy$. Then $z^2 = x^2 - y^2 + 2ixy$ and $\overline{z} = x - iy$. Therefore, $$z^2 + \overline{z} = x^2 - y^2 + x + i(2xy - y) = 0.$$ This implies $$x^2 + x - y^2 = 0 and 2xy - y = 0.$$ Thus, $y = 0$ or $x = \frac{1}{2}$. If $y = 0$, then $x = 0, -1$. If $x = \frac{1}{2}$, then $y = \frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}$. The sum is $$\sum_{z \in S} \left( \mathrm{Re}(z) + \mathrm{Im}(z) \right) = \left( 0 - 1 + \frac{1}{2} + \frac{1}{2} \right) + 0 + 0 + \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \right).$$
Physics
Question 31
Physics · Physical World, Units and Measurements · Single correct
A torque meter is calibrated to reference standards of mass, length and time each with 5$\%$ accuracy. After calibration, the measured torque with this torque meter will have net accuracy of :
15$\%$
25$\%$
75$\%$
5$\%$
Answer: (b)
Solution
Dimensional formula for Torque $$[\tau] = [ML^2T^{-2}]$$ Now Percentage error in torque = % $\tau$ = % M + 2 $\%$ L + 2 $\%$ T % $\tau$ = 25$\%$
Question 32
Physics · System of Particles and Rotational Motion · Single correct
A bullet is shot vertically downwards with an initial velocity of 100 m/s from a certain height. Within 10 s, the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity-time curve for total time $t = 20 \, \mathrm{s}$ will be: (Take $g = 10 \, \mathrm{m/s^2}$)
Answer: (a)
Solution
Given the equation for velocity, $$V = -100 - 10t$$.
Question 33
Physics · Work, Energy and Power · Single correct
Sand is being dropped from a stationary dropper at a rate of $0.5 \, \mathrm{kg/s}^{-1}$ on a conveyor belt moving with a velocity of $5 \, \mathrm{m/s}^{-1}$. The power needed to keep belt moving with the same velocity will be:
1.25 W
2.5 W
6.25 W
12.5 W
Answer: (d)
Solution
$\text{Thrust} = \lambda V_{rel}$ $= 2.5\,\text{N}$ $\text{Now, Power} = F \times V = 12.5\,\text{W}$
Question 34
Physics · Laws of Motion · Single correct
A bag is gently dropped on a conveyor belt moving at a speed of $2 \, \mathrm{m/s}$. The coefficient of friction between the conveyor belt and bag is $0.4$. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is: [Take $g = 10 \, \mathrm{m/s}^{-2}$]
2 m
0.5 m
3.2 m
0.8 m
Answer: (b)
Solution
In frame of belt $a = \mu g = 4 \, \mathrm{m/s^2}$, $v = 2 \, \mathrm{m/s}$, $u = 0$ $$v^2 = u^2 + 2as$$ $$\Rightarrow s = 0.5 \, \mathrm{m}$$
Question 35
Physics · Mechanical Properties of Fluids · Single correct
Two cylindrical vessels of equal cross-sectional area $16 \, \mathrm{cm}^2$ contain water up to heights $100 \, \mathrm{cm}$ and $150 \, \mathrm{cm}$ respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take density of water $= 10^3 \, \mathrm{kg/m}^3$ and $g = 10 \, \mathrm{m/s}^2$]
0.25 J
1 J
8 J
12 J
Answer: (b)
Solution
Given $h = \frac{h_1 + h_2}{2}$. Now, $W = U_i - U_f$. $$W = (\rho A h_1) g \frac{h_1}{2} + (\rho A h_2) g \frac{h_2}{2} - \rho A (h_1 + h_2) g \left( \frac{h_1 + h_2}{4} \right)$$ $$W = \frac{\rho A g}{2} \left[ h_1^2 + h_2^2 - \left( \frac{h_1 + h_2}{2} \right)^2 \right]$$ $W = 1 \, \mathrm{J}$
Question 36
Physics · Gravitation · Single correct
Two satellites A and B having masses in the ratio 4:3 are revolving in circular orbits of radii 3r and 4r respectively around the earth. The ratio of total mechanical energy of A to B is :
9 : 16
16 : 9
1 : 1
4 : 3
Answer: (b)
Solution
Given that $\frac{m_1}{m_2} = \frac{4}{3}$, $\frac{r_1}{r_2} = \frac{3}{4}$. Now TE $= \frac{1}{2} mv^2 + \left( \frac{-GMm}{r} \right)$. But $\frac{mv^2}{r} = \frac{GMm}{r^2} \implies mv^2 = \frac{GMm}{r}$. Therefore, $TE = \frac{-GMm}{2r} \propto \frac{m}{r}$. $$\frac{TE_1}{TE_2} = \frac{m_1}{m_2} \cdot \frac{r_2}{r_1} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$$
Question 37
Physics · Thermal Properties of Matter · Single correct
If $K_1$ and $K_2$ are the thermal conductivities $L_1$ and $L_2$ are the lengths and $A_1$ and $A_2$ are the cross sectional areas of steel and copper rods respectively such that $\frac{K_2}{K_1} = 9$, $\frac{A_1}{A_2} = 2$, $\frac{L_1}{L_2} = 2$. Then, for the arrangement as shown in the figure. The value of temperature $T$ of the steel – copper junction in the steady state will be :
Read the following statements : A. When small temperature difference between a liquid and its surrounding is doubled the rate of loss of heat of the liquid becomes twice. B. Two bodies P and Q having equal surface areas are maintained at temperature 10^$\circ$ $\mathrm{C}$ and 20^$\circ$ $\mathrm{C}$. The thermal radiation emitted in a given time by P and Q are in the ratio 1 : 1.15 C. A carnot Engine working between 100 \, $\mathrm{K}$ and 400 \, $\mathrm{K}$ has an efficiency of 75$\%$ D. When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice. Choose the correct answer from the options given below :
A, B, C only
A, B only
A, C only
B, C, D only
Answer: (a)
Solution
Heat Transfer A. by Newton's law of cooling $\frac{d\theta}{dt} = \infty \, \Delta T$ B. $\mathrm{H} = \frac{d\theta}{dt} = \sigma e A T^4 \Rightarrow \frac{H_P}{H_Q} = \left( \frac{T_P}{T_Q} \right)^4 = \left( \frac{283}{293} \right)^4$ $H_P : H_Q = 1 \,(1.03)^4 = 1 : (1.03)^4 = 1 : 1.15$ $\Rightarrow$ B is correct C. $\eta = 1 - \frac{100}{400} = \frac{3}{4} = 75\%$ D. is wrong as $\frac{d\theta}{dt} \propto \Delta T$
Question 39
Physics · Kinetic Theory · Multiple correct
Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is $1:4$, then A. The r.m.s. velocity of gas molecules in two vessels will be the same. B. The ratio of pressure in these vessels will be $1:4$. C. The ratio of pressure will be $1:1$. D. The r.m.s. velocity of gas molecules in two vessels will be in the ratio of $1:4$.
A and C only
B and D only
A and B only
C and D only
Answer: (c)
Solution
KTG A. $V_{Rms} = \sqrt{\frac{3RT}{M_w}} \Rightarrow V_{Rms}$ is same B. $\frac{P_1}{P_2} = \frac{N_1}{N_2} \Rightarrow$ B is correct Ans [A $\&$ B only are correct]
Question 40
Physics · Oscillations · Single correct
Two identical positive charges $Q$ each are fixed at a distance of $'2a'$ apart from each other. Another point charge $q_0$ with mass $'m'$ is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge $q_0$ executes SHM. The time period of oscillation of charge $q_0$ will be:
$\sqrt{\frac{4\pi^3 \varepsilon_0 ma^3}{q_0 Q}}$
$\sqrt{\frac{q_0 Q}{4\pi^3 \varepsilon_0 ma^3}}$
$\sqrt{\frac{2\pi^2 \varepsilon_0 ma^3}{q_0 Q}}$
$\sqrt{\frac{8\pi^3 \varepsilon_0 ma^3}{q_0 Q}}$
Answer: (a)
Solution
Given the electrostatic setup, the force is given by: $$ F = m \cdot acc^n = \frac{KQq_0}{(a-x)^2} - \frac{KQq_0}{(a+x)^2} $$ Simplifying, we have: $$ m \cdot acc^n = \frac{KQq_0 [2a][2x]}{(a^2 - x^2)^2} $$ This implies: $$ \Rightarrow acc^n \approx \left( \frac{4kQq_0}{ma^3} \right) x $$ The period $T$ is given by: $$ T = 2\pi \sqrt{\frac{\pi \varepsilon_0 ma^3}{Qq_0}} $$ Simplifying further: $$ T = \sqrt{\frac{4\pi^3 \varepsilon_0 ma^3}{Qq_0}} $$
Question 41
Physics · Current Electricity · Single correct
Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are $r_1$ and $r_2$ ($r_1 > r_2$). If the potential difference across the source of internal resistance $r_1$ is zero then the value of $R$ will be
$r_1 - r_2$
$\frac{r_1 r_2}{r_1 + r_2}$
$\frac{r_1 + r_2}{2}$
$r_2 - r_1$
Answer: (a)
Solution
Given the circuit, we have the following equations. The current $I$ is given by: $$I = \frac{2E}{r_1 + r_2 + R}$$ The voltage across $R$ is: $$IR = E - Ir_2$$ Rearranging gives: $$I(R + r_2) = E$$ Solving for $I$ gives: $$I = \frac{E}{R + r_2}$$ Equating the two expressions for $I$: $$\frac{2E}{r_1 + r_2 + R} = \frac{E}{R + r_2}$$ Simplifying, we get: $$2R + 2r_2 = r_1 + r_2 + R$$ Solving for $R$ gives: $$R = r_1 - r_2$$
Question 42
Physics · Oscillations · Single correct
Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of 3 s and 4 s respectively. If their moments of inertia are in the ratio of 3 : 2 then the ratio of their magnetic moments will e :
2 : 1
8 : 3
1 : 3
27 : 16
Answer: (b)
Solution
The formula for the period is given by $$T = 2\pi \sqrt{\frac{I}{M B_H}}$$ For the first case, we have $$T_1 = 2\pi \sqrt{\frac{I_1}{M_1 B_H}}$$ For the second case, we have $$T_2 = 2\pi \sqrt{\frac{I_1}{M_2 B_H}}$$ Given that $$\sqrt{\frac{I_1}{I_2}} \times \frac{M_2}{M_1} = \frac{3}{4}$$ Simplifying, we have $$\sqrt{\frac{I_1}{I_2}} \times \sqrt{\frac{M_2}{M_1}} = \frac{3}{4}$$ This implies $$\sqrt{\frac{3}{2}} \times \sqrt{\frac{M_2}{M_1}} = \frac{3}{4}$$ Squaring both sides, we get $$\frac{3}{2} \times \frac{M_2}{M_1} = \frac{9}{16}$$ Solving for $\frac{M_1}{M_2}$, we find $$\frac{M_1}{M_2} = \frac{8}{3}$$
Question 43
Physics · Magnetism and Matter · Single correct
A magnet hung at $45^\circ$ with magnetic meridian makes an angle of $60^\circ$ with the horizontal. The actual value of the angle of dip is
A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of 3 $\Omega$ and 2 $\Omega$ respectively. The ratio of heat produced in the two resistances in same interval of time will be:
3 : 2
3 : 1
3 : 4
4 : 3
Answer: (b)
Solution
For DC, $H_1 = i^2 R_1 t$. For AC, $H_2 = i_{rms}^2 R_2 t$ where $i_{rms} = \frac{i_0}{\sqrt{2}}$. Calculating $H_1$: $$H_1 = 16(3)t$$ Calculating $H_2$: $$H_2 = \frac{i_0^2}{2} R_2 t$$ $$H_2 = 16t$$ The ratio $H_1 : H_2 = 3 : 1$.
Question 45
Physics · Electromagnetic Induction · Single correct
A beam of light travelling along X-axis is described by the electric field $E_y = 900 \sin \omega (t-x/c)$. The ratio of electric force to magnetic force on a charge $q$ moving along Y-axis with a speed of $3 \times 10^7 \, \mathrm{ms}^{-1}$ will be: [Given speed of light = $3 \times 10^8 \, \mathrm{ms}^{-1}$]
1 : 1
1 : 10
10 : 1
1 : 2
Answer: (c)
Solution
Given $E_y = 900 \sin \left( \omega t - \frac{\omega x}{c} \right)$. $E_0 = 900$. The forces are given by: $$F_E = qE_0$$ $$F_B = qvB_0$$ The ratio of the forces is: $$\frac{F_E}{F_B} = \frac{E_0}{vB_0} = \frac{c}{v} = \frac{3 \times 10^8}{3 \times 10^7} = 10:1$$
Question 46
Physics · Wave Optics · Single correct
A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is $\lambda$, calculate the change of microscope’s resolving power due to oil and choose the correct option
Resolving power will be $\frac{1}{4}$ in the oil than it was in the air
Resolving power will be twice in the oil than it was in the air.
Resolving power will be four times in the oil than it was in the air.
Resolving power will be $\frac{1}{2}$ in the oil than it was in the air.
Answer: (b)
Solution
The resolving power in air is given by $$(\mathrm{R.P})_{air} = \frac{2 \sin \theta}{1.22 \lambda}$$ The resolving power in oil is $$(\mathrm{R.P})_{oil} = \frac{2 \sin \theta}{1.22 \lambda_{oil}} = \frac{2 \sin \theta \times \mu}{1.22 \lambda}$$ Therefore, $$(\mathrm{R.P})_{oil} = (\mathrm{R.P})_{air} \times 2$$
Question 47
Physics · Atoms · Single correct
An electron (mass m) with an initial velocity $\vec{v} = v_0 \hat{i} (v_0 > 0)$ is moving in an electric field $\vec{E} = -E_0 \hat{i} (E_0 > 0)$ where $E_0$ is constant. If at $t = 0$ de Broglie wavelength is $\lambda_0 = \frac{h}{mv_0}$, then its de Broglie wavelength after time $t$ is given by
Given $\mathbf{e}$ with initial velocity $V_0$. The electric field is $\mathbf{E} = -E_0 \hat{i}$. The initial wavelength is $\lambda_0 = \frac{h}{mv_0}$. The velocity is $\mathbf{v} = \mathbf{v}_0 + \frac{eE_0 t}{m}$. The wavelength is $$\lambda = \frac{h}{mv} = \frac{h}{m \left( v_0 + \frac{eE_0 t}{m} \right)}.$$ This simplifies to $$\lambda' = \frac{h}{mv_0 \left( 1 + \frac{eE_0}{mv_0} t \right)}.$$ Thus, $$\lambda' = \frac{\lambda_0}{1 + \frac{eE_0}{mv_0} t}.$$
Question 48
Physics · Nuclei · Single correct
What is the half-life period of a radioactive material if its activity drops to $1/16^{th}$ of its initial value of 30 years?
9.5 years
8.5 years
7.5 years
10.5 years
Answer: (c)
Solution
Given $A = A_0 e^{-\lambda t}$. Therefore, $$-\lambda t = \ln \left( \frac{A}{A_0} \right)$$ which implies $$-\frac{\ln 2}{t_{1/2}} \times 30 = \ln \left( \frac{1}{16} \right)$$ leading to $$-\frac{\ln 2}{t_{1/2}} \times 30 = -4 \ln 2$$ Therefore, $$t_{1/2} = \frac{30}{4} = 7.5 yrs$$
Question 49
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
A logic gate circuit has two inputs A and B and output Y. The voltage waveforms of A, B and Y are shown below
AND gate
OR gate
NOR gate
NAND gate
Answer: (a)
Solution
By making Truth table Comparing with output of AND gate logic gate present is AND gate
Question 50
Physics · Communication Systems · Single correct
At a particular station, the TV transmission tower has a height of 100 m. To triple its coverage range, height of the tower should be increased to
200 m
300 m
600 m
900 m
Answer: (d)
Solution
Let $d$ be range $$d^2 = (h+R)^2 - R^2$$ $$= h^2 + R^2 + 2Rh - R^2$$ $$d^2 = h^2 + 2Rh$$ As $R >>>> h$ then $$d \approx \sqrt{2Rh} .... (1)$$ Now, if coverage is to be increased 3 times $$3d = \sqrt{2Rh'} .... (2)$$ Divide 2 and 1 $$\frac{3d}{d} = \sqrt{\frac{2Rh'}{2Rh}}$$ $$9 = \frac{h'}{h}$$ $$9h = h'$$ If $h = 100 \, \mathrm{m}$ then tower of height $900 \, \mathrm{m}$ is required
Question 51
Physics · Experimental Physics · Numerical
In meter bridge experiment for measuring unknown resistance ‘S’, the null point is obtained at a distance 30 cm from the left side as shown at point D. If R is 5.6 $\,$ $\mathrm{k\Omega}$ , then the value of unknown resistance ‘S’ will be
Answer: 2400
Solution
Given $\($ $\frac{S}{30}$ = $\frac{5.6 \times 10^3}{70}$ $\)$. Solving for $\($ S $\)$, we have: $$ S = \frac{3}{7} \times 5.6 \times 10^3 = 2400 $$
Question 52
Physics · Physical World, Units and Measurements · Numerical
The one division of the main scale of a Vernier callipers reads 1 mm and 10 divisions of the Vernier scale are equal to 9 divisions on the main scale. When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of the zero of the main scale and its 4$^{\text{th}}$ division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies between 4.1 $\mathrm{cm}$ and 4.2 $\mathrm{cm}$ and the 6$^{\text{th}}$ Vernier division coincides with a main scale division. The diameter of the bob will be ____ $\times$ 10$^{-2}$ $\mathrm{cm}$.
Answer: 412
Solution
10 VSD = 9 MSD 1 VST = 0.9 MSD L.C. = 0.1 mm = 0.01 cm +ve zero error = 0.4 mm = 0.04 cm Negative zero error = 4.1 cm + 6 $\times$ 0.01 = 4.12 cm = 412 $\times$ 10^{-2} cm
Question 53
Physics · Wave Optics · Numerical
Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are $\pi/2$ and $\pi/3$ at points A and B respectively. The difference between the resultant intensities at the two points is xI. The value of x will be _____.
To light a $50\,\mathrm{W}$, $100\,\mathrm{V}$ lamp is connected in series with a capacitor of capacitance $\dfrac{50}{\pi\sqrt{x}}\,\mu\mathrm{F}$, with $200\,\mathrm{V}$, $50\,\mathrm{Hz}$ AC source. The value of $x$ will be ____.
Answer: 3
Solution
Given $P = \frac{V^2}{R}$, we have $R = \frac{V^2}{P}$. For $V_R = 200 \, \Omega$ and $V_C$, with $V = 200 \, \mathrm{V}$ and $50 \, \mathrm{Hz}$, $$R = \frac{100 \times 10^2}{50} = R = 200 \, \Omega$$ The equation $V_R^2 + V_C^2 = V^2$ gives $$(100)^2 + V_C^2 = (200)^2$$ Solving for $i$, we have $$i = \frac{100}{200} = \frac{1}{2}$$ Thus, $V^2 = 40000$ and $V_C^2 = 30000$. For $V = I \times X_C$, we have $$V_C = 100 \sqrt{3}$$ $$X_C = 200 \sqrt{3}$$ Solving $200 \sqrt{3} = \frac{1}{\omega C}$, we find $$C = \frac{1}{20 \times 50 \times 20 \sqrt{3}} = \frac{50 \times 10^{-6}}{\sqrt{x}}$$
Question 55
Physics · Current Electricity · Numerical
A 1 m long copper wire carries a current of 1 A. If the cross section of the wire is 2.0 $\mathrm{mm^2}$ and the resistivity of copper is $1.7 \times 10^{-8} \, \Omega \, \mathrm{m}$, the force experienced by moving electron in the wire is $\times 10^{-23} \, \mathrm{N}$. (charge on electron $= 1.6 \times 10^{-19} \, \mathrm{C}$)
Physics · Electric Charges and Fields · Fill in the blank
A long cylindrical volume contains a uniformly distributed charge of density $\rho \, \mathrm{Cm}^{-3}$. The electric field inside the cylindrical volume at a distance $x = \frac{2 \varepsilon_0}{\rho}$ m from its axis is ______ $\mathrm{Vm}^{-1}$
Answer: 1
Solution
The integral of the electric field over the surface is given by $$\int E \, dS \cos 0 = \frac{q}{\varepsilon_0}$$ which implies $$E \cdot 2 \pi x h = \frac{\rho \times \pi x^2 h}{\varepsilon_0}$$ Therefore, $$E = \frac{\rho x}{2 \varepsilon_0}$$ Finally, $$E = \frac{\rho}{2 \varepsilon_0} \times \frac{2 \varepsilon_0}{\rho} = 1$$
Question 57
Physics · Oscillations · Numerical
A mass 0.9 $\mathrm{kg}$, attached to a horizontal spring, executes SHM with an amplitude $A_1$. When this mass passes through its mean position, then a smaller mass of 124 $\mathrm{g}$ is placed over it and both masses move together with amplitude $A_2$. If the ratio $\frac{A_1}{A_2}$ is $\frac{\alpha}{\alpha - 1}$, then the value of $\alpha$ will be .
Physics · Mechanical Properties of Solids · Numerical
A square aluminium (shear modulus is $25 \times 10^9 \, \mathrm{Nm}^{-2}$) slab of side $60 \, \mathrm{cm}$ and thickness $15 \, \mathrm{cm}$ is subjected to a shearing force (on its narrow face) of $18.0 \times 10^4 \, \mathrm{N}$. The lower edge is riveted to the floor. The displacement of the upper edge is _____ $\mu \, \mathrm{m}$.
Physics · System of Particles and Rotational Motion · Numerical
A pulley of radius 1.5 m is rotated about its axis by a force $F = (12t - 3t^2) \, \mathrm{N}$ applied tangentially (while $t$ is measured in seconds). If moment of inertia of the pulley about its axis of rotation is $4.5 \, \mathrm{kg \, m^2}$, the number of rotations made by the pulley before its direction of motion is reversed, will be $\frac{K}{\pi}$. The value of $K$ is ____.
A ball of mass m is thrown vertically upward. Another ball of mass 2 m is thrown an angle $\theta$ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $\frac{1}{x}$. The value of $x$ is _____.
Answer: 1
Solution
Time of flight is same. Therefore, vertical component of velocity is same. Hence, $H_{max}$ is same.
Chemistry
Question 61
Chemistry · Some Basic Concepts of Chemistry · Single correct
250 $\mathrm{g}$ solution of D-glucose in water contains 10.8$\%$ of carbon by weight. The molality of the solution is nearest to (Given: Atomic Weights are H, 1u ; C, 12u ; O, 16u)
1.03
2.06
3.09
5.40
Answer: (b)
Solution
$\mathrm{C_6H_{12}O_6}\rightarrow$ Glucose We know: $\frac{\text{mass of C}}{\text{mass of glucose}}=\frac{72}{180}$ Given: $\%C=10.8=\frac{\text{mass of C}}{\text{mass of solution}}\times100$ $\frac{10.8\times250}{100}=\text{mass of C}$ $\Rightarrow$ Mass of C $=27\,\mathrm{gm}$ $\therefore$ mass of glucose $=67.5\,\mathrm{gm}$ $\therefore$ moles of glucose $=0.375\,\mathrm{moles}$ Mass of solvent $=250-67.5\,\mathrm{gm}=182.5\,\mathrm{gm}$ $\therefore$ Molality $=\frac{0.375}{0.1825}=2.055\approx2.06$
Question 62
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Given below are two statements. \textbf{Statement I:} $O_2$, $Cu^{2+}$ and $Fe^{3+}$ are weakly attracted by magnetic field and are magnetized in the same direction as magnetic field. \textbf{Statement II:} $NaCl$ and $H_2O$ are weakly magnetized in opposite direction to magnetic field. In the light of the above statements, choose the \textbf{\textit{most appropriate}} 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: (a)
Solution
$O_2$, $\mathrm{Cu^{2+}}$ and $\mathrm{Fe^{3+}}$ are paramagnetic, $\therefore$ Weakly attracted by magnetic field. $\mathrm{NaCl}$ and $\mathrm{H_2O}$ are diamagnetic, $\therefore$ Weakly repelled by magnetic field.
Question 63
Chemistry · Structure of Atom · Single correct
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium. Reason R : Energies of the orbitals in the same subshell decrease with increase in the atomic number. In the light of the above statements, choose the correct answer from the options given below.
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is NOT the correct explanation of A.
A is true but R is false.
A is false but R is true.
Answer: (a)
Solution
Energy of orbitals decreases on increasing the atomic number.
Question 64
Chemistry · States of Matter · Single correct
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Activated charcoal adsorbs SO$_2$ more efficiently than CH$_4$. Reason R : Gases with lower critical temperatures are readily adsorbed by activated charcoal. In the light of the above statements, choose the correct 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: (c)
Solution
SO$_2$ is absorbed to a greater extent than CH$_4$ on activated charcoal under same conditions. Gases with higher critical temperature are readily absorbed by activated charcoal.
Question 65
Chemistry · Solutions · Single correct
Boiling point of a 2$\%$ aqueous solution of a non-volatile solute A is equal to the boiling point of 8$\%$ aqueous solution of a non-volatile solute B. The relation between molecular weights of A and B is.
$M_A = 4M_B$
$M_B = 4M_A$
$M_A = 8M_B$
$M_B = 8M_A$
Answer: (b)
Solution
For A: 100 gm solution $\rightarrow$ 2 gm solute A. Therefore, molality $= \frac{2}{M_A} / 0.098$. For B: 100 gm solution $\rightarrow$ 8 gm solute B. Therefore, molality $= \frac{8}{M_B} / 0.092$. Since $(\Delta T_B)_A = (\Delta T_B)_B$, the molality of A equals the molality of B. Therefore, $$\frac{2}{0.098 M_A} = \frac{8}{0.092 M_B}$$ $$\frac{2}{98} \times \frac{92}{8} = \frac{M_A}{M_B}$$ $$\frac{1}{4.261} = \frac{M_A}{M_B}$$ Thus, $M_B = 4.261 \times M_A$.
Question 66
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
The incorrect statement is
The first ionization enthalpy of K is less than that of Na and Li
Xe does not have the lowest first ionization enthalpy in its group
The first ionization enthalpy of element with atomic number 37 is lower than that of the element with atomic number 38.
The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Which of the following methods are not used to refine any metal? (A) Liquation (B) Calcination ($C$) Electrolysis (D) Leaching (E) Distillation Choose the correct answer from the options given below:
B and D only
A, B, D and E only
B, D and E only
A, C and E only
Answer: (a)
Solution
Calcination and leaching are the methods of concentration of ore and not that of refining.
Question 68
Chemistry · Hydrogen · Single correct
Given below are two statements: Statement I : Hydrogen peroxide can act as an oxidizing agent in both acidic and basic conditions. Statement II: Density of hydrogen peroxide at 298 $\mathrm{K}$ is lower than that of $\mathrm{D_2O}$. In the light of the above statements. Choose the correct answer from the options.
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
Depending on the nature of reducing agent $\mathrm{H_2O_2}$ can act as an oxidising agent in both acidic as well as basic medium. Density of $\mathrm{D_2O} = 1.1 \, \mathrm{g/cc}$ Density of $\mathrm{H_2O_2} = 1.45 \, \mathrm{g/cc}$
Question 69
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Given below are two statements: Statement I : The chlorides of Be and Al have Cl-bridged structure. Both are soluble in organic solvents and act as Lewis bases. Statement II: Hydroxides of Be and Al dissolve in excess alkali to give beryllate and aluminate ions. 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: (d)
Solution
Be$_2$Cl$_4$ is lewis acid and Al$_2$Cl$_6$ has complete octet. Be and Al are amphoteric metals therefore dissolve in acid as well as alkaline solution and form beryllate and aluminate ions in excess alkali.
Question 70
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Which oxoacid of phosphorous has the highest number of oxygen atoms present in its chemical formula?
Chemistry · The d-and f-Block Elements · Single correct
Given below are two statements: Statement I : Iron (III) catalyst, acidified $K_2Cr_2O_7$ and neutral $KMnO_4$ have the ability to oxidise $I^-$ to $I_2$ independently. Statement II: Manganate ion is paramagnetic in nature and involves $p\pi -p\pi$ bonding. In the light of the above statements, choose the correct answer from the options.
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: (b)
Solution
Neutral $\mathrm{KMnO_4}$ oxidises $\mathrm{I^-}$ to $\mathrm{IO_3^-}$. Manganate ion has $d\pi$-$p\pi$ bonding.
Question 72
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
The total number of $\mathrm{Mn = O}$ bonds in $\mathrm{Mn_2O_7}$ is
4
5
6
3
Answer: (c)
Solution
The structure shown is a dimer of manganese with bridging oxygen atoms. Each manganese is bonded to four oxygen atoms, two of which are double-bonded and one is a bridging oxygen.
Question 73
Chemistry · Environmental Chemistry · Single correct
Match List I with List II
A-IV, B-I, C-II, D-III
A-III, B-I, C-IV, D-II
A-II, B-IV, C-I, D-III
A-II, B-IV, C-III, D-I
Answer: (b)
Solution
A. Sulphate (>500 ppm) - Causes laxative effect that leads to dehydration. B. Nitrate (>50 ppm) - Causes methemoglobinemia, skin appears blue. C. Lead (> 50 ppb) – It damages kidney and RBC. D. Fluoride (>2 ppm) – It causes brown mottling of teeth.
Question 74
Chemistry · Hydrocarbons · Single correct
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : [6] Annulene. [8] Annulene and cis -[10] Annulene, are respectively aromatic, not-aromatic and aromatic. Reason R : Planarity is one of the requirements of aromatic systems. 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: (d)
Solution
If this annulene with five cis double bonds were planar, each internal angle would be $144^\circ$. Since a normal double bond has a bond angle of $120^\circ$, this would be far from ideal. This compound can be made but it does not adopt a planar conformation and therefore is not aromatic even though it has ten $\pi$ electrons.
Question 75
Chemistry · Alcohols, Phenols and Ethers · Single correct
In the above reaction product B is:
Answer: (a)
Solution
The reaction starts with the given compound reacting with $\mathrm{HCl}$ under heat $\Delta$ to form compound (A) with a chlorine substituent. Then, compound (A) undergoes an $\mathrm{S_N2}$ reaction with $\mathrm{NaI}$ to form compound (B) with an iodine substituent.
Question 76
Chemistry · Polymers · Single correct
Match List I with List II \begin{tabular}{|c|p{5.5cm}|c|l|} \hline \multicolumn{2}{|c|}{List I} & \multicolumn{2}{c|}{List II} \\ \multicolumn{2}{|c|}{Polymers} & \multicolumn{2}{c|}{Commercial names} \\ \hline A. & Phenol-formaldehyde resin & I. & Glyptal \\ \hline B. & Copolymer of 1,3-butadiene and styrene & II. & Novolac \\ \hline C. & Polyester of glycol and phthalic acid & III. & Buna-S \\ \hline D. & Polyester of glycol and terephthalic acid & IV. & Dacron \\ \hline \end{tabular} Choose the correct answer from the options given below:
A-II, B –III, C-IV, D-I
A-II, B –III, C-I, D-IV
A-II, B –I, C-III, D-IV
A-III, B –II, C-IV, D-I
Answer: (b)
Solution
Question 77
Chemistry · Biomolecules · Single correct
A sugar 'X' dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol gives the coloured product after sometime. Sugar 'X' is
Aldopentose
Aldotetrose
Oxalic acid
Ketotetrose
Answer: (a)
Solution
An aldopentose is converted to furfural in the presence of $H^+$. The furfural then reacts with resorcinol to form a cherry red product, which is indicative of Seliwanoff's test.
Question 78
Chemistry · Chemistry in Everyday Life · Single correct
Match List I with List II Choose the correct answer from the options given below:
A-IV, B-III, C-II, D-I
A-III, B-I, C-II, D-IV
A-III, B-IV, C-I, D-II
A-III, B-I, C-IV, D-II
Answer: (c)
Solution
Option (A) is morphine used for relief of pain, known for narcotic analgesic. Option (B) is chloroxylenol used as an antiseptic. Option (C) is phenelzine (Nardil) used as an antidepressant. Option (D) is saccharin, 550 times sweeter than cane sugar.
Question 79
Chemistry · Analytical Chemistry · Single correct
In Carius method of estimation of halogen. 0.45 g of an organic compound gave 0.36 g of AgBr. Find out the percentage of bromine in the compound. (Molar masses : AgBr = 188 g $mol^{-1}$: Br = 80 g $mol^{-1}$)
34.04$\%$
40.04$\%$
36.03$\%$
38.04$\%$
Answer: (a)
Solution
Mass of organic compound = 0.45 $\mathrm{\, gm}$ Mass of $\mathrm{AgBr}$ obtained = 0.36 $\mathrm{\, gm}$ $\therefore$ Moles of $\mathrm{AgBr}$ = $\frac{0.36}{188}$ $\therefore$ Mass of Bromine = $\frac{0.36}{188}$ $\times$ 80 = 0.1532 $\mathrm{\, gm}$ $\therefore$ $\%$ Br in compound = $\frac{0.1532}{0.45}$ $\times$ 100 = 34.04$\%$
Question 80
Chemistry · Amines · Single correct
Match List I with List II \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List I} & \multicolumn{2}{c|}{List II} \\ \hline A. & Benzenesulphonyl chloride & I. & Test for primary amines \\ \hline B. & Hoffmann bromamide reaction & II. & Anti Saytzeff \\ \hline C. & Carbylamine reaction & III. & Hinsberg reagent \\ \hline D. & Hofmann orientation & IV. & Known reaction of Isocyanates \\ \hline \end{tabular} Choose the correct answer from the options given below:
A-IV, B-III, C-II, D-I
A-IV, B-II, C-I, D-III
A-III, B-IV, C-I, D-II
A-IV, B-III, C-I, D-II
Answer: (c)
Solution
(A) Hinsberg reagent is benzen sulphonyl chloride. (B) Hoffmann bromamide reaction is a known reaction of isocyanates. $$R - \mathrm{CO} - \mathrm{NH}_2 + X_2 + 4 \mathrm{NaOH} \rightarrow R - \mathrm{NH}_2 + 2\mathrm{NaX} + \mathrm{Na_2CO_3} + 2\mathrm{H_2O}$$ Intermediate: $R - \mathrm{N} = \mathrm{C} = \mathrm{O}$ (isocyanate) (C) Carbylamine reaction is a test for primary amine. $$R - \mathrm{NH}_2 or \mathrm{Ar} - \mathrm{NH}_2 + \mathrm{CHCl}_3 + 3\mathrm{KOH} \rightarrow \mathrm{RNC} or \mathrm{Ar} - \mathrm{NC} + 3\mathrm{KCl} + 3\mathrm{H_2O}$$ (D) Hoffmann orientation is anti Saytzeff (formation of less substituted alkene as major product).
Question 81
Chemistry · Redox Reactions · Numerical
$20\,\mathrm{mL}$ of $0.02\,\mathrm{M}$ $\mathrm{K_2Cr_2O_7}$ solution is used for the titration of $10\,\mathrm{mL}$ of $\mathrm{Fe^{2+}}$ solution in the acidic medium. The molarity of $\mathrm{Fe^{2+}}$ solution is \_\_\_\_ $\times 10^{-2}\,\mathrm{M}$. (Nearest Integer)
Answer: 24
Solution
Eq. of $\mathrm{K_2Cr_2O_7} = Eq. of \mathrm{Fe^{2+}}$ $\Rightarrow (Molarity \times volume \times n.f) of \mathrm{K_2Cr_2O_7} = (molarity \times volume \times n.f) of \mathrm{Fe^{2+}}$ $$\Rightarrow 0.02 \times 20 \times 6 = M \times 10 \times 1$$ $$\Rightarrow M = 0.24$$ $$\Rightarrow Molarity = 24 \times 10^{-2}$$
Question 82
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
2NO + 2H₂ → N₂ + 2H₂O The above reaction has been studied at 800°C. The related data are given in the table below The order of the reaction with respect to NO is
Answer: 2
Solution
On decreasing pressure of NO by a factor of $2$, the rate of reaction decreases by a factor of $4$. Therefore, order of reaction with respect to NO is $2$.
Question 83
Chemistry · Chemical Bonding and Molecular Structure · Numerical
Amongst the following the number of oxide(s) which are paramagnetic in nature is $Na_2O, KO_2, NO_2, N_2O, ClO_2, NO, SO_2, Cl_2O$
Answer: 4
Solution
$KO_2, NO_2, ClO_2, NO$ are paramagnetic.
Question 84
Chemistry · Thermodynamics · Numerical
The molar heat capacity for an ideal gas at constant pressure is 20.785 $\mathrm{J}$ $\mathrm{K}^{-1}$ $\mathrm{mol}^{-1}$. The change in internal energy is 5000 $\mathrm{J}$ upon heating it from 300 $\mathrm{K}$ to 500 $\mathrm{K}$. The number of moles of the gas at constant volume is ___ [Nearest integer] (Given: R = 8.314 $\mathrm{J}$ $\mathrm{K}^{-1}$ $\mathrm{mol}^{-1}$)
Answer: 2
Solution
Question 85
Chemistry · Chemical Bonding and Molecular Structure · Numerical
According to MO theory, number of species/ions from the following having identical bond order is_____: $CN^-$, $\mathrm{NO}^+$, $\mathrm{O}_2$, $\mathrm{O}_2^+$, $\mathrm{O}_2^{2+}$
Answer: 3
Solution
$CN^-, NO^+, O_2^{2+}$ have bond order = 3
Question 86
Chemistry · Equilibrium · Numerical
At 310 $\mathrm{K}$, the solubility of $\mathrm{CaF_2}$ in water is 2.34 $\times 10^{-3}$ $\mathrm{g/100 \, mL}$. The solubility product of $\mathrm{CaF_2}$ is $\times 10^{-8}$ ($\mathrm{mol/L})^3$. (Given molar mass: $\mathrm{CaF_2}$ = 78 $\mathrm{g \, mol^{-1}}$)
The conductivity of a solution of complex with formula $\mathrm{CoCl}_3(\mathrm{NH}_3)_4$ corresponds to 1 : 1 electrolyte, then the primary valency of central metal ion is
Answer: 3
Solution
Primary valency = oxidation no. = +3
Question 88
Chemistry · The d-and f-Block Elements · Fill in the blank
In the titration of $\mathrm{KMnO_4}$ and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is _____
Answer: 1
Solution
Oxidation state of carbon changes from $+3$ to $+4$. $$2\mathrm{KMnO_4} + 5\mathrm{H_2C_2O_4} + 3\mathrm{H_2SO_4} (dil.) \rightarrow$$ $$\mathrm{K_2SO_4} + 2\mathrm{MnSO_4} + 10\mathrm{CO_2} + 8\mathrm{H_2O}$$
Optical activity of an enantiomeric mixture is $+12.6^\circ$ and the specific rotation of $(+)$ isomer is $+30^\circ$. The optical purity is $\%$
Answer: 42
Solution
% optical purity = $\frac{\text{observed rotation of mixture \times 100}}{\text{rotation of pure enantiomer}}$ = $\frac{+12.6^\circ}{+30^\circ}$ $\times$ 100 = 42
Question 90
Chemistry · Hydrocarbons · Numerical
In the following reaction The % yield for reaction I is 60% and that of reaction II is 50%. The overall yield of the complete reaction is ____% [nearest integer]
Answer: 30
Solution
Let initial moles of reactant taken = n. Total moles obtained for benzene sulphonic acid (with % yield = 60%) = 0.6n. Moles of benzene sulphonic acid before reaction II = 0.6n. Moles obtained for phenol (with % yield = 50%) = 0.6 $\times$ 0.5n = 0.3n. So overall % yield of complete reaction = $\frac{0.3n}{n}$ $\times$ 100 = 30.