JEE Main 1 February 2024 Shift 1 question paper with solutions
JEE Main 1 February 2024 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Probability · Single correct
A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:
$\frac{2}{5}$
$\frac{2}{7}$
$\frac{1}{7}$
$\frac{1}{5}$
Answer: (b)
Solution
The probability $P(4W4B/2W2B)$ is given by \[ \frac{ P(4W4B)\times P(2W2B/4W4B) }{ \begin{matrix} P(2W6B)\times P(2W2B/2W6B)+P(3W5B)\times P(2W2B/3W5B)\\ +\cdots+P(6W2B)\times P(2W2B/6W2B) \end{matrix} } \] This simplifies to $$= \frac{\frac{1}{5} \times \frac{4 \binom{C_2}{4} \binom{C_2}{4}}{8 \binom{C_4}{4}}}{\frac{1}{5} \times \frac{2 \binom{C_2}{6} \binom{C_2}{2}}{8 \binom{C_4}{4}} + \frac{1}{5} \times \frac{3 \binom{C_2}{5} \binom{C_2}{2}}{8 \binom{C_4}{4}} + \ldots + \frac{1}{5} \times \frac{6 \binom{C_2}{2} \binom{C_2}{2}}{8 \binom{C_4}{4}}}$$ Finally, we have $$= \frac{2}{7}$$
Question 2
Maths · Integrals · Single correct
The value of the integral $$\int_{0}^{\frac{\pi}{4}} \frac{xdx}{\sin^4(2x) + \cos^4(2x)}$$ equals :
$\($ $\frac{\sqrt{2}\pi^2}{8}$ $\)$
$\($ $\frac{\sqrt{2}\pi^2}{16}$ $\)$
$\($ $\frac{\sqrt{2}\pi^2}{32}$ $\)$
$\($ $\frac{\sqrt{2}\pi^2}{64}$ $\)$
Answer: (c)
Solution
Let $2x = t$ then $dx = \frac{1}{2} dt$. $$I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{tdt}{\sin^4 t + \cos^4 t}$$ $$I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \left( \frac{\pi}{2} - t \right) \frac{dt}{\sin^4 \left( \frac{\pi}{2} - t \right) + \cos^4 \left( \frac{\pi}{2} - t \right)}$$ $$I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{\sin^4 t + \cos^4 t}{\sin^4 t + \cos^4 t} - I$$ $$2I = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t}$$ $$2I = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{\sec^4 t dt}{\tan^4 t + 1}$$ Let $\tan t = y$ then $\sec^2 t dt = dy$. $$2I = \frac{\pi}{8} \int_0^{\infty} \frac{(1 + y^2) \, dy}{1 + y^4}$$ $$= \frac{\pi}{16} \int_0^{\infty} \frac{1 + \frac{1}{y^2}}{y^2 + \frac{1}{y^2}} \, dy$$ Put $y - \frac{1}{y} = p$. $$I = \frac{\pi}{16} \int_{-\infty}^{\infty} \frac{dp}{p^2 + (\sqrt{2})^2}$$ $$= \frac{\pi}{16\sqrt{2}} \left[ \tan^{-1} \left( \frac{p}{\sqrt{2}} \right) \right]_{-\infty}^{\infty}$$ $$I = \frac{\pi^2}{16\sqrt{2}}$$
Question 3
Maths · Matrices · Single correct
If $A=\begin{bmatrix} \sqrt{2} & 1\\ -1 & \sqrt{2} \end{bmatrix}$, $B=\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}$, $C=ABA^{T}$ and $X=A^{T}C^{2}A$, then $\det X$ is equal to:
If $\tan A = \frac{1}{\sqrt{x(x^2+x+1)}}$, $\tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan C = \left(x^{-3} + x^{-2} + x^{-1}\right)^{\frac{1}{2}}$, $0 < A, B, C < \frac{\pi}{2}$, then $A + B$ is equal to :
C
$\pi$ - C
2$\pi$ - C
$\frac{\pi}{2}$ - C
Answer: (a)
Solution
Finding $\tan(A + B)$ we get $$\Rightarrow \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{1}{\sqrt{x(x^2 + x + 1)}} + \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}}{1 - \frac{1}{x^2 + x + 1}}$$ $$\Rightarrow \tan(A + B) = \frac{(1 + x) \left( \sqrt{x^2 + x + 1} \right)}{(x^2 + x)(\sqrt{x})}$$ $$\frac{(1 + x) \left( \sqrt{x^2 + x + 1} \right)}{(x^2 + x)(\sqrt{x})}$$ $$\tan(A + B) = \frac{\sqrt{x^2 + x + 1}}{x \sqrt{x}} = \tan C$$ Therefore, $A + B = C$.
Question 5
Maths · Permutations and Combinations · Single correct
If n is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then n is equal to:
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $S = \{ z \in \mathbb{C} : |z - 1| = 1 and (\sqrt{2} - 1)(z + \bar{z}) - i(z - \bar{z}) = 2\sqrt{2} \}$. Let $z_1, \ z_2 \in S$ be such that $|z_1| = \max_{z \in S} |z|$ and $|z_2| = \min_{z \in S} |z|$. Then $\left| \sqrt{2} z_1 - z_2 \right|^2$ equals:
1
4
3
2
Answer: (d)
Solution
Let $Z = x + iy$. Then $(x - 1)^2 + y^2 = 1 \rightarrow [OPTION1]$ and $\, (\sqrt{2} - 1)(2x) - i(2iy) = 2\sqrt{2}$ implies $(\sqrt{2} - 1)x + y = \sqrt{2} \rightarrow (2)$. Solving (1) and (2) we get either $x = 1$ or $x = \frac{1}{2 - \sqrt{2}} \rightarrow (3)$. On solving (3) with (2) we get for $x = 1 \Rightarrow y = 1 \Rightarrow Z_2 = 1 + i$ and for $x = \frac{1}{2 - \sqrt{2}} \Rightarrow y = \sqrt{2} - \frac{1}{\sqrt{2}} \Rightarrow Z_1 = \left(1 + \frac{1}{\sqrt{2}}\right) + \frac{i}{\sqrt{2}}$. Now $$\left|\sqrt{2}z_1 - z_2\right|^2$$ $$= \left|\left(\frac{1}{\sqrt{2}} + 1\right)\sqrt{2} + i - (1 + i)\right|^2$$ $$= (\sqrt{2})^2$$ $$= 2$$
Question 7
Maths · Statistics · Single correct
Let the median and the mean deviation about the median of 7 observation 170, 125, 230, 190, 210, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is :
31
28
30
32
Answer: (c)
Solution
Median $= 170 \Rightarrow 125, a, b, 170, 190, 210, 230$ Mean deviation about Median $= \frac{0 + 45 + 60 + 20 + 40 + 170 - a + 170 - b}{7} = \frac{205}{7}$ $\Rightarrow a + b = 300$ Mean $= \frac{170 + 125 + 230 + 190 + 210 + a + b}{7} = 175$ Mean deviation About mean $= \frac{50 + 175 - a + 175 - b + 5 + 15 + 35 + 55}{7} = 30$
Question 8
Maths · Vector Algebra · Single correct
Let $\vec{a}$ = -5$\hat{i}$ + $\hat{j}$ - 3$\hat{k}$, $\vec{b}$ = $\hat{i}$ + 2$\hat{j}$ - 4$\hat{k}$ and $\vec{c}$ = ((($\vec{a}$ $\times$ $\vec{b}$) $\times$ $\hat{i}$) $\times$ $\hat{i}$) $\times$ $\hat{i}$ . Then $\vec{c}$ $\cdot$ (-$\hat{i}$ + $\hat{j}$ + $\hat{k}$) is equal to
-12
-10
-13
-15
Answer: (a)
Solution
Given $\vec{a} = -5\hat{i} + \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 4\hat{k}$. The expression $(\vec{a} \times \vec{b}) \times \hat{i}$ is calculated as follows: $$(\vec{a} \cdot \hat{i}) \vec{b} - (\vec{b} \cdot \hat{i}) \vec{a}$$ This simplifies to: $$-5\vec{b} - \vec{a}$$ Further simplification gives: $$((( -5\vec{b} - \vec{a}) \times \hat{i}) \times \hat{i})$$ This becomes: $$((-11\hat{j} + 23\hat{k}) \times \hat{i}) \times \hat{i}$$ Which simplifies to: $$(11\hat{k} + 23\hat{j}) \times \hat{i}$$ Finally, this results in: $$(11\hat{j} - 23\hat{k})$$ The dot product $\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$ is calculated as: $$11 - 23 = -12$$
Question 9
Maths · Sequences and Series · Single correct
Let $S = \left\{x \in \mathbb{R} : \left(\sqrt{3} + \sqrt{2}\right)^x + \left(\sqrt{3} - \sqrt{2}\right)^x = 10\right\}$. Then the number of elements in $S$ is:
4
0
2
1
Answer: (c)
Solution
Given $\left( \sqrt{3} + \sqrt{2} \right)^x + \left( \sqrt{3} - \sqrt{2} \right)^x = 10$. Let $\left( \sqrt{3} + \sqrt{2} \right)^x = t$. Then $t + \frac{1}{t} = 10$. Solving the quadratic equation $t^2 - 10t + 1 = 0$, we find $$t = \frac{10 \pm \sqrt{100 - 4}}{2} = 5 \pm 2\sqrt{6}.$$ Therefore, $\left( \sqrt{3} + \sqrt{2} \right)^x = \left( \sqrt{3} \pm \sqrt{2} \right)^2$. Thus, $x = 2$ or $x = -2$. Number of solutions = 2.
Question 10
Maths · Applications of Integrals · Single correct
The area enclosed by the curves $xy + 4y = 16$ and $x + y = 6$ is equal to:
28 - 30 $\log$_2 2
30 - 28 $\log$_2 2
30 - 32 $\log$_2 2
32 - 30 $\log$_2 2
Answer: (c)
Solution
Given the equations $xy + 4y = 16$ and $x + y = 6$. Rewriting the first equation as $y(x + 4) = 16$ and the second as $x + y = 6$. On solving equations (1) and (2), we get $x = 4$ and $x = -2$. The area is given by the integral: $$Area = \int_{-2}^{4} \left( (6 - x) - \left( \frac{16}{x + 4} \right) \right) \, dx$$ which evaluates to $30 - 32 \ln 2$.
Question 11
Maths · Relations and Functions · Single correct
Let f : $\mathbb{R}$ $\to$ $\mathbb{R}$ and g : $\mathbb{R}$ $\to$ $\mathbb{R}$ be defined as $$f(x) = \begin{cases} \log_e x , & x > 0 \\ e^{-x} , & x \leq 0 \end{cases}$$ and $$g(x) = \begin{cases} x , & x \geq 0 \\ e^x , & x < 0 \end{cases}.$$ Then, gof : $\mathbb{R}$ $\to$ $\mathbb{R}$ is :
one-one but not onto
neither one-one nor onto
onto but not one-one
both one-one and onto
Answer: (b)
Solution
Given the function $g(f(x))$ defined as follows: $$g(f(x)) = \begin{cases} f(x), & f(x) \geq 0 \\ e^{f(x)}, & f(x) < 0 \end{cases}$$ And another definition: $$g(f(x)) = \begin{cases} e^{-x}, & (-\infty, 0] \\ e^{\ln x}, & (0, 1) \\ \ln x, & [1, \infty) \end{cases}$$ The graph of $g(f(x))$ is shown. $g(f(x))$ is many one into.
Question 12
Maths · Determinants · Single correct
If the system of equations $$2x + 3y - z = 5$$ $$x + \alpha y + 3z = -4$$ $$3x - y + \beta z = 7$$ has infinitely many solutions, then $13\alpha\beta$ is equal to
1110
1120
1210
1220
Answer: (b)
Solution
Using family of planes $$2x + 3y - z - 5 = k_1(x + \alpha y + 3z + 4) + k_2(3x - y + \beta z - 7)$$ $$2 = k_1 + 3k_2, 3 = k_1 \alpha - k_2, -1 = 3k_1 + \beta k_2, -5 = 4k_1 - 7k_2$$ On solving we get $$k_2 = \frac{13}{19}, k_1 = \frac{-1}{19}, \alpha = -70, \beta = \frac{-16}{13}$$ $$13 \alpha \beta = 13(-70) \left(\frac{-16}{13}\right)$$ $$= 1120$$
Question 13
Maths · Conic Sections · Single correct
For $0 < \theta < \pi/2$, if the eccentricity of the hyperbola $x^2 - y^2 \operatorname{cosec}^2\theta = 5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^2 \operatorname{cosec}^2\theta + y^2 = 5$, then the value of $\theta$ is :
Let y = y(x) be the solution of the differential equation $\frac{dy}{dx}$ = 2x(x+y)^3 - x(x+y) - 1, y(0) = 1. Then, ( $\frac{1}{\sqrt{2}}$ + y( $\frac{1}{\sqrt{2}}$ ) )^2 equals:
$\frac{4}{4+\sqrt{e}}$
$\frac{3}{3-\sqrt{e}}$
$\frac{2}{1+\sqrt{e}}$
$\frac{1}{2-\sqrt{e}}$
Answer: (d)
Solution
Given $\($ $\frac{dy}{dx}$ = 2x(x+y)^3 - x(x+y) - 1 $\)$. Let $\($ x+y = t $\)$. Then $\($ $\frac{dt}{dx}$ - 1 = 2xt^3 - xt - 1 $\)$. Rearranging gives $\($ $\frac{dt}{2t^3 - t}$ = xdx $\)$. This implies $\($ $\frac{tdt}{2t^4 - t^2}$ = xdx $\)$. Let $\($ t^2 = z $\)$. Then $\($ $\int$ $\frac{dz}{2(2z^2 - z)}$ = $\int$ xdx $\)$. This simplifies to $\($ $\int$ $\frac{dz}{4z\left(z - \frac{1}{2}\right)}$ = $\int$ xdx $\)$. Integrating gives $\($ $\ln$ $\left$| $\frac{z - \frac{1}{2}}{z}$ $\right$| = x^2 + k $\)$. Solving for $\($ z $\)$ gives $\($ z = $\frac{1}{2 - \sqrt{e}}$ $\)$.
Question 15
Maths · Continuity and Differentiability · Single correct
Let f : $\mathbb{R} \to \mathbb{R}$ be defined as $$f(x) = \begin{cases} \frac{a - b \cos 2x}{x^2} & ; x 1 \end{cases}$$ If $f$ is continuous everywhere in $\mathbb{R}$ and $m$ is the number of points where $f$ is NOT differential then $m + a + b + c$ equals :
1
4
3
2
Answer: (d)
Solution
At $x = 1$, $f(x)$ is continuous therefore, $f(1^-) = f(1) = f(1^+)$. $f(1) = 3 + c$. $$f(1^+) = \lim_{h \to 0} 2(1 + h) + 1$$ $$f(1^+) = \lim_{h \to 0} 3 + 2h = 3$$ from (1) and (2), $c = 0$. At $x = 0$, $f(x)$ is continuous therefore, $f(0^-) = f(0) = f(0^+)$. $f(0) = f(0^+) = 2$. $f(0^-)$ has to be equal to 2. $$\lim_{h \to 0} \frac{a - b \cos(2h)}{h^2}$$ $$\lim_{h \to 0} \frac{a - b \left\{ 1 - \frac{4h^2}{2!} + \frac{16h^4}{4!} + \ldots \right\}}{h^2}$$ $$\lim_{h \to 0} \frac{a - b + b \left\{ 2h^2 - \frac{3}{2} h^4 \ldots \right\}}{h^2}$$ for limit to exist $a - b = 0$ and limit is $2b$. From (3), (4) and (5), $a = b = 1$. Checking differentiability at $x = 0$. LHD: $$\lim_{h \to 0} \frac{\frac{1 - \cos 2h}{h^2} - 2}{-h}$$ $$1 - \left( 1 - \frac{4h^2}{2!} + \frac{16h^4}{4!} \cdots \right) - 2h^2$$ $$\lim_{h \to 0} \frac{-h^3}{-h} = 0$$ RHD: $$\lim_{h \to 0} \frac{(0 + h)^2 + 2 - 2}{h} = 0$$ Function is differentiable at every point in its domain. Therefore, $m = 0$. $m + a + b + c = 0 + 1 + 1 + 0 = 2$
Question 16
Maths · Conic Sections · Single correct
Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a > b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is:
Let 3, a, b, c be in A.P. and 3, a - 1, b + 1, c + 9 be in G.P. Then, the arithmetic mean of a, b and c is :
-4
-1
13
11
Answer: (d)
Solution
Given the sequence $3, a, b, c$ is an arithmetic progression (A.P), we have $3, 3 + d, 3 - 2d, 3 + 3d$. The sequence $3, a - 1, b + 1, c + 9$ is a geometric progression (G.P), giving $3, 2 + d, 4 + 2d, 12 + 3d$. Solving for $a$, $b$, and $c$, we find $a = 3 + d$, $b = 3 + 2d$, and $c = 3 + 3d$. The possible values for $d$ are $4$ and $-2$. If $d = 4$, the G.P becomes $3, 6, 12, 24$. Then $a = 7$, $b = 11$, and $c = 15$. The average is given by $$\frac{a + b + c}{3} = 11.$$
Question 18
Maths · Conic Sections · Single correct
Let $C : x^2 + y^2 = 4$ and $C' : x^2 + y^2 - 4\lambda x + 9 = 0$ be two circles. If the set of all values of $\lambda$ so that the circles $C$ and $C'$ intersect at two distinct points, is $\mathbb{R} - [a, b]$, then the point $(8a + 12, 16b - 20)$ lies on the curve :
Maths · Three Dimensional Geometry · Single correct
If the shortest distance between the lines $\frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}$ is 1, then the sum of all possible values of $\lambda$ is:
If $x = x(t)$ is the solution of the differential equation $(t + 1) dx = \left(2x + (t + 1)^4\right) dt, x(0) = 2$, then, $x(1)$ equals
Answer: 14
Solution
(t+1)dx = $\left$(2x + (t+1)^4$\right$) dt $\frac{dx}{dt}$ = $\frac{2x + (t+1)^4}{t+1}$ $\frac{dx}{dt}$ - $\frac{2x}{t+1}$ = (t+1)^3 I $\cdot$ F = e^{-$\int$ $\frac{2}{t+1}$ dt} = e^{-2 $\ln$(t+1)} = $\frac{1}{(t+1)^2}$ $\frac{x}{(t+1)^2}$ = $\int$ $\frac{1}{(t+1)^2}$ (t+1)^3 dt + c $\frac{x}{(t+1)^2}$ = $\frac{(t+1)^2}{2}$ + c $\Rightarrow$ c = $\frac{3}{2}$ x = $\frac{(t+1)^4}{2}$ + $\frac{3}{2}$ (t+1)^2 put, t = 1 x = 2^3 + 6 = 14
Question 22
Maths · Sets · Numerical
The number of elements in the set $$S = \{(x, y, z) : x, y, z \in \mathbb{Z}, x + 2y + 3z = 42, x, y, z \geq 0\}$$ equals $\ldots$.
Answer: 169
Solution
Question 23
Maths · Binomial Theorem · Numerical
If the Coefficient of $x^{30}$ in the expansion of $$\left(1 + \frac{1}{x}\right)^6 (1 + x^2)^7 (1 - x^3)^8$$ ; $x \neq 0$ is $\alpha$, then $|\alpha|$ equals
Let 3, 7, 11, 15, $\ldots$, 403 and 2, 5, 8, 11, $\ldots$, 404 be two arithmetic progressions. Then the sum, of the common terms in them, is equal to
Answer: 6699
Solution
The sequences are 3, 7, 11, 15, $\ldots$, 403 and 2, 5, 8, 11, $\ldots$, 404. The least common multiple of 4 and 3 is 12. The sequence 11, 23, 35, $\ldots$ is considered, ending at 403. The equation is given by: $$403 = 11 + (n - 1) \times 12$$ Solving for $n$: $$\frac{392}{12} = n - 1$$ $$33 = n$$ The sum is calculated as: $$Sum = \frac{33}{2} (22 + 32 \times 12)$$ $$= 6699$$
Question 25
Maths · Limits and Derivatives · Numerical
Let $\{$x$\}$ denote the fractional part of $x$ and $f(x) = \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}$, $x \neq 0$. If $L$ and $R$ respectively denotes the left hand limit and the right hand limit of $f(x)$ at $x = 0$, then $\frac{32}{\pi^2} \left(L^2 + R^2\right)$ is equal to
Let the line $L : \sqrt{2}x + y = \alpha$ pass through the point of the intersection $P$ (in the first quadrant) of the circle $x^2 + y^2 = 3$ and the parabola $x^2 = 2y$. Let the line $L$ touch two circles $C_1$ and $C_2$ of equal radius $2\sqrt{3}$. If the centres $Q_1$ and $Q_2$ of the circles $C_1$ and $C_2$ lie on the y-axis, then the square of the area of the triangle $PQ_1Q_2$ is equal to
Answer: 72
Solution
Given $x^2 + y^2 = 3$ and $x^2 = 2y$. $y^2 + 2y - 3 = 0 \Rightarrow (y + 3)(y - 1) = 0$ $y = -3$ or $y = 1$ $y = 1x = \sqrt{2} \Rightarrow P(\sqrt{2}, 1)$ $p$ lies on the line $\sqrt{2}x + y = \alpha$ $\sqrt{2}(\sqrt{2}) + 1 = \alpha$ $\alpha = 3$ For circle $C_1$ $Q_1$ lies on $y$ axis Let $Q_1(0, \alpha)$ coordinates $R_1 = 2\sqrt{3}$ (Given) Line $L$ act as tangent Apply $P = r$ (condition of tangency) $$\Rightarrow \left| \frac{\alpha - 3}{\sqrt{3}} \right| = 2\sqrt{3}$$ $$\Rightarrow |\alpha - 3| = 6$$ $\alpha - 3 = 6$ or $\alpha - 3 = -6$ $$\Rightarrow \alpha = 9 \alpha = -3$$ $$\triangle PQ_1Q_2 = \frac{1}{2} \begin{vmatrix} \sqrt{2} & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1 \end{vmatrix}$$ $$= \frac{1}{2} (\sqrt{2}(12)) = 6\sqrt{2}$$ $$(\Delta PQ_1Q_2)^2 = 72$$
Question 27
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $\mathbb{P} = \{ z \in \mathbb{C} : |z + 2 - 3i| \leq 1 \}$ and $\mathbb{Q} = \{ z \in \mathbb{C} : z(1+i) + \bar{z}(1-i) \leq -8 \}$. Let in $\mathbb{P} \cap \mathbb{Q}$, $|z - 3 + 2i|$ be maximum and minimum at $z_1$ and $z_2$ respectively. If $|z_1|^2 + 2|z_2|^2 = \alpha + \beta\sqrt{2}$, where $\alpha$, $\beta$ are integers, then $\alpha + \beta$ equals
Answer: 36
Solution
Clearly for the shaded region $z_1$ is the intersection of the circle and the line passing through $P$ ($L_1$) and $z_2$ is the intersection of line $L_1$ and $L_2$. Circle: $(x + 2)^2 + (y - 3)^2 = 1$. $L_1$: $x + y - 1 = 0$. $L_2$: $x - y + 4 = 0$. On solving circle and $L_1$ we get $$z_1 : \left(-2 - \frac{1}{\sqrt{2}}, 3 + \frac{1}{\sqrt{2}}\right)$$ On solving $L_1$ and $z_2$ is intersection of line $L_1$ and $L_2$ we get $$z_2 : \left(-\frac{3}{2}, \frac{5}{2}\right)$$ $$|z_1|^2 + 2|z_2|^2 = 14 + 5\sqrt{2} + 17$$ $$= 31 + 5\sqrt{2}$$ $$\alpha = 31$$ So $\beta = 5$. $\alpha + \beta = 36$
Question 28
Maths · Integrals · Numerical
If $$\int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2} \cos x \, dx}{(1+e^{\sin x})(1+\sin^4 x)} = \alpha \pi + \beta \log_e(3 + 2\sqrt{2})$$, where $\alpha$, $\beta$ are integers, then $\alpha^2 + \beta^2$ equals
Let the line of the shortest distance between the lines $$L_1 : \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k})$$ and $$L_2 : \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu (\hat{i} + \hat{j} - \hat{k})$$ intersect $L_1$ and $L_2$ at $P$ and $Q$ respectively. If $(\alpha, \beta, \gamma)$ is the midpoint of the line segment $PQ$, then $$2(\alpha + \beta + \gamma)$$ is equal to
Answer: 21
Solution
Given $\($ $\vec{b}$ = $\hat{i}$ - $\hat{j}$ + $\hat{k}$ $\)$ (DR's of $\($ L_1 $\)$) and $\($ $\vec{d}$ = $\hat{i}$ + $\hat{j}$ - $\hat{k}$ $\)$ (DR's of $\($ L_2 $\)$). $\($ $\vec{b}$ $\times$ $\vec{d}$ = $\begin{vmatrix}$ $\hat{i}$ & $\hat{j}$ & $\hat{k}$ $\\$ 1 & -1 & 1 $\\$ 1 & 1 & -1 $\end{vmatrix}$ $\)$ $\($ = 0 $\hat{i}$ + 2 $\hat{j}$ + 2 $\hat{k}$ $\)$ (DR's of Line perpendicular to $\($ L_1 $\)$ and $\($ L_2 $\)$) DR of AB line $\($ = (0, 2, 2) = $\left$( $\frac{3 + \mu - \lambda}{0}$, $\frac{3 + \mu + \lambda}{2}$, $\frac{3 - \mu - \lambda}{2}$ $\right$) $\)$ Solving above equation we get $\($ $\mu$ = -$\frac{3}{2}$ $\)$ and $\($ $\lambda$ = $\frac{3}{2}$ $\)$ Point A = $\($ $\left$( $\frac{5}{2}$, $\frac{1}{2}$, $\frac{9}{2}$ $\right$) $\)$ Point B = $\($ $\left$( $\frac{5}{2}$, $\frac{7}{2}$, $\frac{15}{2}$ $\right$) $\)$ Point of AB = $\($ $\left$( $\frac{5}{2}$, 2, 6 $\right$) = ($\alpha$, $\beta$, $\gamma$) $\)$ $\($ 2($\alpha$ + $\beta$ + $\gamma$) = 5 + 4 + 12 = 21 $\)$
Question 30
Maths · Relations and Functions · Numerical
Let A = {1, 2, 3, $\ldots$, 20}. Let $R_1$ and $R_2$ two relation on A such that $R_1$ = $\{$(a, b) : b is divisible by a$\}$ $R_2$ = $\{$(a, b) : a is an integral multiple of b$\}$. Then, number of elements in $R_1 - R_2$ is equal to $\ldots$.
Answer: 46
Solution
Physics
Question 31
Physics · Mechanical Properties of Solids · Single correct
With rise in temperature, the Young's modulus of elasticity
changes erratically
decreases
increases
remains unchanged
Answer: (b)
Solution
Q1 Conceptual questions
Question 32
Physics · Gravitation · Single correct
If $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g = \pi^2 \, \mathrm{m/s^2}$, then the length of the second's pendulum at a height $h = 2R$ from the surface of earth will be,:
$\frac{2}{9} \, \mathrm{m}$
$\frac{1}{9} \, \mathrm{m}$
$\frac{4}{9} \, \mathrm{m}$
$\frac{8}{9} \, \mathrm{m}$
Answer: (b)
Solution
Given $g' = \frac{GMe}{(3R)^2} = \frac{1}{9} g$. The time period $T = 2\pi \sqrt{\frac{\ell}{g'}}$. Since the time period of the second pendulum is 2 sec, $T = 2 sec$. $2 = 2\pi \sqrt{\frac{\ell}{9g}}$. Therefore, $\ell = \frac{1}{9} m$.
Question 33
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
In the given circuit if the power rating of Zener diode is 10 $\mathrm{mW}$, the value of series resistance $R_s$ to regulate the input unregulated supply is:
5k$\Omega$
10$\Omega$
1k$\Omega$
None of these
Answer: (d)
Solution
Pd across $R_s$ $V_1 = 8 - 5 = 3 \, \mathrm{V}$ Current through the load resistor $$I = \frac{5}{1 \times 10^3} = 5 \, \mathrm{mA}$$ Maximum current through Zener diode $$I_{z max.} = \frac{10}{5} = 2 \, \mathrm{mA}$$ And minimum current through Zener diode $I_{z min.} = 0$ Therefore, $I_s max. = 5 + 2 = 7 \, \mathrm{mA}$ And $R_s min. = \frac{V_1}{I_s max.} = \frac{3}{7} \, \mathrm{k}\Omega$ Similarly $I_s min. = 5 \, \mathrm{mA}$ And $R_s max. = \frac{V_1}{I_s min.} = \frac{3}{5} \, \mathrm{k}\Omega$ Therefore, $\frac{3}{7} \, \mathrm{k}\Omega < R_s < \frac{3}{5} \, \mathrm{k}\Omega$
Question 34
Physics · Current Electricity · Single correct
The reading in the ideal voltmeter (V) shown in the given circuit diagram is :
5 \, $\mathrm{V}$
10 \, $\mathrm{V}$
0 \, $\mathrm{V}$
3 \, $\mathrm{V}$
Answer: (c)
Solution
Given $\($ i = $\frac{E_{eq}}{r_{eq}}$ = $\frac{8 \times 5}{8 \times 0.2}$ $\)$. $\($ I = 25 \, A $\)$ $\($ V = E - ir $\)$ $\($ = 5 - 0.2 $\times$ 25 $\)$ $\($ = 0 $\)$
Question 35
Physics · Electrostatic Potential and Capacitance · Single correct
Two identical capacitors have same capacitance $C$. One of them is charged to the potential $V$ and other to the potential $2 \, V$. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :
$\frac{1}{4} CV^2$
$2CV^2$
$\frac{1}{2} CV^2$
$\frac{3}{4} CV^2$
Answer: (a)
Solution
Given $V_C = \frac{q_{net}}{C_{net}} = \frac{CV + 2CV}{2C}$. $V_C = \frac{3V}{2}$. Loss of energy $$= \frac{1}{2}CV^2 + \frac{1}{2}C(2V)^2 - \frac{1}{2}2C\left(\frac{3V}{2}\right)^2$$ $$= \left(\frac{CV^2}{4}\right)$$
Question 36
Physics · Kinetic Theory · Single correct
Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :
$\frac{9}{4} R$
$\frac{7}{4} R$
$\frac{3}{2} R$
$\frac{5}{2} R$
Answer: (a)
Solution
Given \[ C_V=\frac{n_1C_{v_1}+n_2C_{v_2}}{n_1+n_2} \] \[ =\frac{2\times\frac{3}{2}R+6\times\frac{5}{2}R}{2+6} \] \[ =\frac{9}{4}R \]
Question 37
Physics · System of Particles and Rotational Motion · Single correct
A ball of mass 0.5 kg is attached to a string of length 50 cm. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is 400 N. The maximum possible value of angular velocity of the ball in rad/s is,;
1600
40
1000
20
Answer: (b)
Solution
Given $T = m \omega^2 \ell$. $$400 = 0.5 \omega^2 \times 0.5$$ $$\omega = 40 \, \mathrm{rad/s}.$$
Question 38
Physics · Alternating Current · Single correct
A parallel plate capacitor has a capacitance $C = 200 \, \mathrm{pF}$. It is connected to $230 \, \mathrm{V}$ ac supply with an angular frequency $300 \, \mathrm{rad/s}$. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :
$1.38 \, \mu \mathrm{A}$ and $1.38 \, \mu \mathrm{A}$
$14.3 \, \mu \mathrm{A}$ and $143 \, \mu \mathrm{A}$
$13.8 \, \mu \mathrm{A}$ and $138 \, \mu \mathrm{A}$
$13.8 \, \mu \mathrm{A}$ and $13.8 \, \mu \mathrm{A}$
Answer: (d)
Solution
Given the formula for current, we have: $$I = \frac{V}{X_C} = 230 \times 300 \times 200 \times 10^{-12} = 13.8 \, \mu A$$
Question 39
Physics · Thermodynamics · Single correct
The pressure and volume of an ideal gas are related as $PV^{3/2} = K$ (Constant). The work done by gas when the gas is taken from state $A (P_1, V_1, T_1)$ to state $B (P_2, V_2, T_2)$ is :
$2 (P_1 V_1 - P_2 V_2)$
$2 (P_2 V_2 - P_1 V_1)$
$2 (\sqrt{P_1} V_1 - \sqrt{P_2} V_2)$
$2 (P_2 \sqrt{V_2} - P_1 \sqrt{V_1})$
Answer: (a)
Solution
For $PV^x = constant$ If work done by gas is asked then $$W = \frac{nR \Delta T}{1-x}$$ Here $x = \frac{3}{2}$ Therefore, $$W = \frac{P_2 V_2 - P_1 V_1}{-\frac{1}{2}}$$ $$= 2 (P_1 V_1 - P_2 V_2) \ldots \ldots Option (1) is correct$$ If work done by external is asked then $W = -2 (P_1 V_1 - P_2 V_2) \ldots \ldots Option (2) is correct$
Question 40
Physics · Current Electricity · Single correct
A galvanometer has a resistance of $50\,\Omega$ and it allows maximum current of $5\,\mathrm{mA}$. It can be converted into voltmeter to measure upto $100\,\mathrm{V}$ by connecting in series a resistor of resistance
$5975\,\Omega$
$20050\,\Omega$
$19950\,\Omega$
$19500\,\Omega$
Answer: (c)
Solution
The resistance is calculated as follows: $$R = \frac{V}{I_g} - R_g = \frac{100}{5 \times 10^{-3}} - 50$$ $$= 20000 - 50$$ $$= 19950 \, \Omega$$
Question 41
Physics · Dual Nature of Radiation and Matter · Single correct
The de Broglie wavelengths of a proton and an $\alpha$ particle are $\lambda$ and $2\lambda$ respectively. The ratio of the velocities of proton and $\alpha$ particle will be:
1 : 8
1 : 2
4 : 1
8 : 1
Answer: (d)
Solution
Given $\lambda = \frac{h}{p} = \frac{h}{mv}$, we have $v = \frac{h}{m\lambda}$. The ratio of velocities is given by $$\frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} \times \frac{\lambda_\alpha}{\lambda_p}$$ Calculating this gives $$= 4 \times 2 = 8$$
Question 42
Physics · Experimental Physics · Single correct
10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is:
In series LCR circuit, the capacitance is changed from C to 4C. To keep the resonance frequency unchanged, the new inductance should be:
reduced by $\frac{1}{4} L$
increased by $2 L$
reduced by $\frac{3}{4} L$
increased to $4 L$
Answer: (c)
Solution
Given $\omega' = \omega$. $$\frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{LC}}$$ Therefore, $L'C' = LC$. $$L'(4C) = LC$$ $$L' = \frac{L}{4}$$ Thus, inductance must be decreased by $\frac{3L}{4}$.
Question 44
Physics · Mathematics in Physics · Single correct
The radius $(r)$, length $(l)$ and resistance $(R)$ of a metal wire was measured in the laboratory as $r = (0.35 \pm 0.05)\, \mathrm{cm}$ $R = (100 \pm 10)\, \mathrm{ohm}$ $l = (15 \pm 0.2)\, \mathrm{cm}$ The percentage error in resistivity of the material of the wire is:
25.6%
39.9%
37.3%
35.6%
Answer: (b)
Solution
Given $\rho = R \frac{\rho}{\ell}$. The relative change in $\rho$ is given by $$\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta r}{r} + \frac{\Delta \ell}{\ell}$$ Substituting the given values, $$= \frac{10}{100} + 2 \times \frac{0.05}{0.35} + \frac{0.2}{15}$$ Simplifying, $$= \frac{1}{10} + \frac{2}{7} + \frac{1}{75}$$ Therefore, $$\frac{\Delta \rho}{\rho} = 39.9\%$$
Question 45
Physics · Physical World, Units and Measurements · Single correct
The dimensional formula of angular impulse is:
[M L^{-2} T^{-1}]
[M L^2 T^{-2}]
[M L T^{-1}]
[M L^2 T^{-1}]
Answer: (d)
Solution
Question 46
Physics · System of Particles and Rotational Motion · Single correct
A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass $10^{-2} \, \mathrm{kg}$ moving with a speed of $2 \times 10^2 \, \mathrm{ms^{-1}}$. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use $g = 10 \, \mathrm{m/s^2}$)
0.30 m
0.20 m
0.35 m
0.40 m
Answer: (b)
Solution
Given the equation for momentum conservation: $$mu = (M + m)V$$ Substituting the given values: $$10^{-2} \times 2 \times 10^2 \cong 1 \times V$$ Solving for $V$: $$V \cong 2 \, \mathrm{m/s}$$ The height $h$ is given by: $$h = \frac{V^2}{2g} = 0.2 \, \mathrm{m}$$
Question 47
Physics · Motion in a Plane · Single correct
A particle moving in a circle of radius $R$ with uniform speed takes time $T$ to complete one revolution. If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to $4R$. The angle of projection $\theta$ is then given by:
Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in $\mathrm{ms^{-2}}$ is: (Consider that the string is massless and unstretchable and the pulley is also massless and frictionless):
3
4
2
1.2
Answer: (c)
Solution
Given $f_k = \mu N = 0.04 \times 20 \, g = 8 \, Newton$. $$a = \frac{60 - 8}{26} = 2 \, \mathrm{m/s^2}$$
Question 49
Physics · Atoms · Single correct
The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly:
1.5 $\mathrm{eV}$
13.6 $\mathrm{eV}$
1.9 $\mathrm{eV}$
12.1 $\mathrm{eV}$
Answer: (d)
Solution
Transition from $n = 1$ to $n = 3$ with $\Delta E = 12.1 \, \mathrm{eV}$.
Question 50
Physics · Wave Optics · Single correct
A monochromatic light of wavelength $6000\,\mathrm{\AA}$ is incident on the single slit of width $0.01\,\mathrm{mm}$. If the diffraction pattern is formed at the focus of the convex lens of focal length $20\,\mathrm{cm}$, the linear width of the central maximum is:
Physics · Moving Charges and Magnetism · Numerical
A regular polygon of 6 sides is formed by bending a wire of length $4\pi$ meter. If an electric current of $4\pi\sqrt{3}$ A is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be $x \times 10^{-7}$ T. The value of $x$ is_____
Answer: 72
Solution
The magnetic field B is calculated as follows: $$B = 6 \left( \frac{\mu_0 I}{4 \pi r} \right) (\sin 30^\circ + \sin 30^\circ)$$ Substituting the values: $$= 6 \times \frac{10^{-7} \times 4 \pi \sqrt{3}}{\left( \frac{\sqrt{3} \times 4 \pi}{2 \times 6} \right)}$$ Simplifying gives: $$= 72 \times 10^{-7} \, \mathrm{T}$$
Question 52
Physics · Electromagnetic Induction · Numerical
A rectangular loop of sides $12\,\mathrm{cm}$ and $5\,\mathrm{cm}$, with its sides parallel to the x-axis and y-axis respectively moves with a velocity of $5\,\mathrm{cm/s}$ in the positive x axis direction, in a space containing a variable magnetic field in the positive z direction. The field has a gradient of $10^{-3}\,\mathrm{T/cm}$ along the negative x direction and it is decreasing with time at the rate of $10^{-3}\,\mathrm{T/s}$. If the resistance of the loop is $6\,\mathrm{m\Omega}$, the power dissipated by the loop as heat is $\underline{\hspace{1cm}}\times10^{-9}\,\mathrm{W}$.
Answer: 216
Solution
Given $B_0$ is the magnetic field at origin. $$\frac{dB}{dx} = \frac{-10^{-3}}{10^{-2}}$$ $$\int_{B_0}^{B} dB = -\int_{0}^{x} 10^{-1} dx$$ $$B - B_0 = -10^{-1} x$$ $$B = \left( B_0 - \frac{x}{10} \right)$$ Motional emf in $AB = 0$. Motional emf in $CD = 0$. Motional emf in $AD = \varepsilon_1 = B_0 \ell v$. Magnetic field on rod $BC$ is $$\left( B_0 - \frac{(-12 \times 10^{-2})}{10} \right)$$ Motional emf in $BC = \varepsilon_2 = \left( B_0 + \frac{12 \times 10^{-2}}{10} \right) \ell \times v$. $$\varepsilon_{eq} = \varepsilon_2 - \varepsilon_1 = 300 \times 10^{-7} \, \mathrm{V}$$ For time variation: $$(\varepsilon_{eq})' = A \frac{dB}{dt} = 60 \times 10^{-7} \, \mathrm{V}$$ $$(\varepsilon_{eq})_{net} = \varepsilon_{eq} + (\varepsilon_{eq})' = 360 \times 10^{-7} \, \mathrm{V}$$ Power $= \frac{(\varepsilon_{eq})_{net}^2}{R} = 216 \times 10^{-9} \, \mathrm{W}$$
Question 53
Physics · Ray Optics and Optical Instruments · Fill in the blank
The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20cm. The focal length of the lens used is ____ cm
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is $1.5 \, \mathrm{g/cc}$, the dielectric constant of water will be _____ (Take density of water $= 1 \, \mathrm{g/cc}$)
Answer: 3
Solution
In air, $\tan \frac{\theta}{2} = \frac{F}{mg} = \frac{q^2}{4\pi \varepsilon_0 r^2 mg}$. In water, $\tan \frac{\theta}{2} = \frac{F'}{mg'} = \frac{q^2}{4\pi \varepsilon_0 \varepsilon_r r^2 mg_{eff}}$. Equate both equations: $$\varepsilon_0 \, g = \varepsilon_0 \varepsilon_r g \left[ 1 - \frac{1}{1.5} \right]$$ $$\varepsilon_r = 3$$
Question 55
Physics · Nuclei · Numerical
The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is $\frac{1000}{x}$, where $x$ is ____.
Physics · System of Particles and Rotational Motion · Numerical
The identical spheres each of mass $2M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $4 \, \mathrm{m}$ each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is $\frac{4\sqrt{2}}{x}$, where the value of $x$ is _____
A tuning fork resonates with a sonometer wire of length $1\,\mathrm{m}$ stretched with a tension of $6\,\mathrm{N}$. When the tension in the wire is changed to $54\,\mathrm{N}$, the same tuning fork produces $12$ beats per second with it. The frequency of the tuning fork is ___ Hz.
Answer: 6
Solution
The frequency is given by the formula $$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}.$$ For the first frequency, $$f_1 = \frac{1}{2} \sqrt{\frac{6}{\mu}}.$$ For the second frequency, $$f_2 = \frac{1}{2} \sqrt{\frac{54}{\mu}}.$$ The ratio of the frequencies is $$\frac{f_1}{f_2} = \frac{1}{3}.$$ The difference between the frequencies is $$f_2 - f_1 = 12.$$ Therefore, $$f_1 = 6 \, \mathrm{Hz}.$$
Question 58
Physics · Mechanical Properties of Fluids · Numerical
A plane is in level flight at constant speed and each of its two wings has an area of $40 \, \mathrm{m}^2$. If the speed of the air is $180 \, \mathrm{km/h}$ over the lower wing surface and $252 \, \mathrm{km/h}$ over the upper wing surface, the mass of the plane is _____ kg. (Take air density to be $1 \, \mathrm{kg} \, \mathrm{m}^{-3}$ and $g = 10 \, \mathrm{ms}^{-2}$)
The current in a conductor is expressed as $I = 3t^2 + 4t^3$, where $I$ is in Ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor during $t = 1 \, \mathrm{s}$ to $t = 2 \, \mathrm{s}$ is_____ C.
Answer: 22
Solution
Given $$q = \int_1^2 i \, dt = \int_1^2 (3t^2 + 4t^3) \, dt$$ Evaluate the integral: $$q = \left( t^3 + t^4 \right) \bigg|_1^2$$ Substitute the limits: $$q = 22C$$
Question 60
Physics · Motion in a Straight Line · Numerical
A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position (x) with time (t) is given as $x = -3t^3 + 18t^2 + 16t$, where $x$ is in $m$ and $t$ is in $s$. The velocity of the particle when its acceleration becomes zero is_____ m/s.
Answer: 52
Solution
Given $$x = 3t^3 + 18t^2 + 16t$$ Velocity is given by $$v = -9t^2 + 36 + 16$$ Acceleration is given by $$a = -18t + 36$$ Setting acceleration to zero at $t = 2 \, \mathrm{s}$: $$a = 0 at t = 2 \, \mathrm{s}$$ Substitute $t = 2$ into the velocity equation: $$v = -9(2)^2 + 36 \times 2 + 16$$ Calculate the velocity: $$v = 52 \, \mathrm{m/s}$$
Chemistry
Question 61
Chemistry · Biomolecules · Single correct
If one strand of a DNA has the sequence ATGCTTCA, sequence of the bases in complementary strand is:
CATTAGCT
TACGAAGT
GTACTTAC
ATGCGACT
Answer: (b)
Solution
Adenine base pairs with thymine with 2 hydrogen bonds and cytosine base pairs with guanine with 3 hydrogen bonds.
Question 62
Chemistry · Haloalkanes and Haloarenes · Single correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason $(R)$. Assertion (A): Haloalkanes react with KCN to form alkyl cyanides as a main product while with AgCN form isocyanide as the main product. Reason $(R)$: KCN and AgCN both are highly ionic compounds. In the light of the above statement, choose the most appropriate answer from the options given below:
(A) is correct but $(R)$ is not correct
Both (A) and $(R)$ are correct but $(R)$ is not the correct explanation of (A)
(A) is not correct but $(R)$ is correct
Both (A) and $(R)$ are correct and $(R)$ is the correct explanation of (A)
Answer: (a)
Solution
When $\mathrm{KCN}$ reacts with $\mathrm{R-X}$, the major product is $\mathrm{R-CN}$ due to the ionic nature of $\mathrm{KCN}$. When $\mathrm{AgCN}$ reacts with $\mathrm{R-X}$, the major product is $\mathrm{R-NC}$ due to the covalent nature of $\mathrm{AgCN}$. AgCN is mainly covalent in nature and nitrogen is available for attack, so alkyl isocyanide is formed as the main product.
Question 63
Chemistry · Redox Reactions · Single correct
In acidic medium, $K_2Cr_2O_7$ shows oxidising action as represented in the half reaction $$Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$ $X, Y, Z$ and $A$ are respectively are:
$8, 6, 4$ $and$ $Cr_2O_3$
$14, 7, 6$ $and$ $Cr^{3+}$
$8, 4, 6$ $and$ $Cr_2O_3$
$14, 6, 7$ $and$ $Cr^{3+}$
Answer: (d)
Solution
The balanced reaction is, $\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6e^- \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O}$ X = 14 Y = 6 A = 7
Question 64
Chemistry · Redox Reactions · Single correct
Which of the following reactions are disproportionation reactions? (A) $\mathrm{Cu}^+ \rightarrow \mathrm{Cu}^{2+} + \mathrm{Cu}$ (B) $3\mathrm{MnO}_4^{2-} + 4\mathrm{H}^+ \rightarrow 2\mathrm{MnO}_4^- + \mathrm{MnO}_2 + 2\mathrm{H}_2\mathrm{O}$ (C) $2\mathrm{KMnO}_4 \rightarrow \mathrm{K}_2\mathrm{MnO}_4 + \mathrm{MnO}_2 + \mathrm{O}_2$ (D) $2\mathrm{MnO}_4^- + 3\mathrm{Mn}^{2+} + 2\mathrm{H}_2\mathrm{O} \rightarrow 5\mathrm{MnO}_2 + 4\mathrm{H}^+$ Choose the correct answer from the options given below:
(A), (B)
(B), (C), (D)
(A), (B), (C)
(A), (D)
Answer: (a)
Solution
When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation. $$\mathrm{Cu^+ \rightarrow Cu^{2+}}$$ $+$ $Cu$ $$3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O}$$
Question 65
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
In case of isoelectronic species the size of $\mathrm{F}^-$, Ne and $\mathrm{Na}^+$ is affected by:
Principal quantum number (n)
None of the factors because their size is the same
Electron-electron interaction in the outer orbitals
Nuclear charge (z)
Answer: (d)
Solution
In $\mathrm{F}^-$, Ne, $\mathrm{Na}^+$ all have $1s^2, 2s^2, 2p^6$ configuration. They have different size due to the difference in nuclear charge.
Question 66
Chemistry · Structure of Atom · Single correct
According to the wave-particle duality of matter by de-Broglie, which of the following graph plot presents most appropriate relationship between wavelength of electron ($\lambda$) and momentum of electron (p) ?
Answer: (a)
Solution
Given $\lambda = \frac{h}{p} \left[ \lambda \propto \frac{1}{p} \right]$. Therefore, $\lambda p = h$ (constant). So, the plot is a rectangular hyperbola.
Question 67
Chemistry · Co-ordination Compounds · Single correct
Given below are two statements: Statement (I): A solution of $[\mathrm{Ni(H_2O)_6}]^{2+}$ is green in colour. Statement (II): A solution of $[\mathrm{Ni(CN)_4}]^{2-}$ is colourless. In the light of the above statements, choose the most appropriate answer from the options given below:
Both Statement I and Statement II are incorrect
Both Statement I and Statement II are correct
Statement I is incorrect but Statement II is correct
Statement I is correct but Statement II is incorrect
Answer: (b)
Solution
The complex $[\mathrm{Ni(H_2O)_6}]^{+2}$ forms a green colour solution due to $d-d$ transition. The complex $[\mathrm{Ni(CN)_4}]^{-2}$ is diamagnetic and it is colourless.
Question 68
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason $(R)$. Assertion (A) : PH$_3$ has lower boiling point than NH$_3$. Reason $(R)$ : In liquid state NH$_3$ molecules are associated through vander waal's forces, but PH$_3$ molecules are associated through hydrogen bonding. In the light of the above statements, choose the most appropriate answer from the options given below:
Both (A) and $(R)$ are correct and $(R)$ is not the correct explanation of (A)
is not correct but $(R)$ is correct
Both (A) and $(R)$ are correct but $(R)$ is the correct explanation of (A)
is correct but $(R)$ is not correct
Answer: (d)
Solution
Unlike $\mathrm{NH_3}$, $\mathrm{PH_3}$ molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of $\mathrm{PH_3}$ is lower than $\mathrm{NH_3}$.
Question 69
Chemistry · Haloalkanes and Haloarenes · Single correct
Identify $A$ and $B$ in the following sequence of reaction
Answer: (b)
Solution
The reaction sequence starts with toluene, which is converted to benzal chloride by the action of $\mathrm{Cl_2}$ in the presence of light ($\mathrm{hv}$). Benzal chloride is then converted to benzaldehyde by hydrolysis with water at $373 \, \mathrm{K}$.
Question 70
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Given below are two statements: Statement (I) : Aminobenzene and aniline are same organic compounds. Statement (II) : Aminobenzene and aniline are different organic compounds. In the light of the above statements, choose the most appropriate answer from the options given below:
Both Statement I and Statement II are correct
Statement I is correct but Statement II is incorrect
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are incorrect
Answer: (b)
Solution
Aniline is also known as amino benzene.
Question 71
Chemistry · Co-ordination Compounds · Single correct
Which of the following complex is homoleptic?
$[\mathrm{Ni(CN)_4}]^{2-}$
$[\mathrm{Ni(NH_3)_2Cl_2}]$
$[\mathrm{Fe(NH_3)_4Cl_2}]^{+}$
$[\mathrm{Co(NH_3)_4Cl_2}]^{+}$
Answer: (a)
Solution
In Homoleptic complex all the ligand attached with the central atom should be the same. Hence $[\mathrm{Ni(CN)_4}]^{2-}$ is a homoleptic complex.
Question 72
Chemistry · Hydrocarbons · Single correct
Which of the following compound will most easily be attacked by an electrophile?
Answer: (d)
Solution
Higher the electron density in the benzene ring more easily it will be attacked by an electrophile. Phenol has the highest electron density amongst all the given compounds.
Question 73
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Ionic reactions with organic compounds proceed through: (A) Homolytic bond cleavage (B) Heterolytic bond cleavage (C) Free radical formation (D) Primary free radical (E) Secondary free radical Choose the correct answer from the options given below:
only
$(C)$ only
only
and (E) only
Answer: (c)
Solution
Heterolytic cleavage of a bond leads to the formation of ions.
Question 74
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Arrange the bonds in order of increasing ionic character in the molecules. $LiF, K_2O, N_2, SO_2$ and $ClF_3.$
ClF_3 < N_2 < SO_2 < K_2O < LiF
LiF < K_2O < ClF_3 < SO_2 < N_2
N_2 < SO_2 < ClF_3 < K_2O < LiF
N_2 < ClF_3 < SO_2 < K_2O < LiF
Answer: (c)
Solution
Increasing order of ionic character $\mathrm{N_2} < \mathrm{SO_2} < \mathrm{ClF_3} < \mathrm{K_2O} < \mathrm{LiF}$. Ionic character depends upon difference of electronegativity (bond polarity).
Question 75
Chemistry · Solutions · Single correct
We have three aqueous solutions of $\mathrm{NaCl}$ labelled as 'A', 'B' and 'C' with concentration $0.1\,\mathrm{M}$, $0.01\,\mathrm{M}$ $\&$ $0.001\,\mathrm{M}$, respectively. The value of van t' Hoff factor $(i)$ for these solutions will be in the order.
$i_A < i_B < i_C$
$i_A < i_C < i_B$
$i_A = i_B = i_C$
$i_A > i_B > i_C$
Answer: (a)
Solution
The table shows the values of $i$ for NaCl at different concentrations. As the solution becomes very dilute, the value of $i$ approaches 2.
Question 76
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
In Kjeldahl's method for estimation of nitrogen, $CuSO_4$ acts as:
Reducing agent
Catalytic agent
Hydrolysis agent
Oxidising agent
Answer: (b)
Solution
Kjeldahl's method is used for estimation of Nitrogen where $\mathrm{CuSO_4}$ acts as a catalyst.
Question 77
Chemistry · Redox Reactions · Single correct
Given below are two statements: Statement (I): Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution. Statement (II): In this titration phenolphthalein can be used as indicator. In the light of the above statements, choose the most appropriate answer from the options given below:
Both Statement I and Statement II are correct
Statement I is correct but Statement II is incorrect
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are incorrect
Answer: (a)
Solution
Statement (I): Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution as it is economical and its concentration does not change with time. Phenolphthalein can act as an indicator in acid-base titration as it shows colour in pH range 8.3 to 10.1.
Question 78
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Match List - I with List -II. Choose the correct answer from options given below:
(A)-(III), (B)-(IV), \text(C)-(I), (D)-(II)
(A)-(IV), (B)-(II), \text(C)-(I), (D)-(III)
(A)-(IV), (B)-(II), \text(C)-(III), (D)-(I)
(A)-(III), (B)-(IV), \text(C)-(II), (D)-(I)
Answer: (b)
Solution
The sequence of reactions is as follows: 1. $\mathrm{CH_3(CH_2)_5COOC_2H_5}$ is treated with DIBAL-H and $\mathrm{H_2O}$ to form $\mathrm{CH_3(CH_2)_5CHO}$. 2. $\mathrm{C_6H_5COC_6H_5}$ is treated with $\mathrm{Zn(Hg)}$ and conc. $\mathrm{HCl}$ to form $\mathrm{C_6H_5CH_2C_6H_5}$. 3. $\mathrm{C_6H_5CHO}$ is treated with $\mathrm{CH_3MgBr}$ followed by $\mathrm{H_2O}$ to form $\mathrm{C_6H_5CH(OH)CH_3}$. 4. $\mathrm{CH_3COCH_2COOC_2H_5}$ is treated with $\mathrm{NaBH_4}$ and $\mathrm{H^+}$ to form $\mathrm{CH_3CH(OH)CH_2COOC_2H_5}$.
Question 79
Chemistry · Thermodynamics · Single correct
Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following:
$q = 0, \Delta T \neq 0, w = 0$
$q = 0, \Delta T < 0, w \neq 0$
$q \neq 0, \Delta T = 0, w = 0$
$q = 0, \Delta T = 0, w = 0$
Answer: (d)
Solution
During free expansion of an ideal gas under adiabatic condition $q = 0$, $\Delta T = 0$, $w = 0$.
Question 80
Chemistry · Hydrocarbons · Single correct
Given below are two statements: Statement (I) : The $\mathrm{NH}_2$ group in Aniline is ortho and para directing and a powerful activating group. Statement (II) : Aniline does not undergo FriedelCraft's reaction (alkylation and acylation). In the light of the above statements, choose the most appropriate answer from the options given below :
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Statement I is incorrect but Statement II is correct
Statement I is correct but Statement II is incorrect
Answer: (a)
Solution
The $\mathrm{NH_2}$ group in Aniline is ortho and para directing and a powerful activating group as $\mathrm{NH_2}$ has strong $+M$ effect. Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation) as Aniline will form complex with $\mathrm{AlCl_3}$ which will deactivate the benzene ring.
Number of optical isomers possible for 2-chlorobutane
Answer: 2
Solution
There is one chiral centre present in the given compound. So, total optical isomers = 2.
Question 82
Chemistry · Electrochemistry · Fill in the blank
The potential for the given half cell at 298 K is $(-) \ldots \times 10^{-2} \mathrm{V}$. $2\mathrm{H}^+_{(aq)} + 2\mathrm{e}^- \rightarrow \mathrm{H}_2(g)$ $[\mathrm{H}^+] = 1\mathrm{M}, \;P_{\mathrm{H}_2} = 2 \mathrm{atm}$ (Given: $2.303RT/F = 0.06 \mathrm{V}, \log 2 = 0.3$)
Answer: 1
Solution
Given $$E = E^\circ_{\mathrm{H^+/H_2}} - \frac{0.06}{2} \log \frac{P_{\mathrm{H_2}}}{[\mathrm{H^+}]^2}$$ Substituting the values, we have $$E = 0.00 - \frac{0.06}{2} \log \frac{2}{[1]^2}$$ Simplifying, $$E = -0.03 \times 0.3 = -0.9 \times 10^{-2} \, \mathrm{V}$$
Question 83
Chemistry · Analytical Chemistry · Single correct
The number of white coloured salts among the following is \begin{enumerate} \item[(A)] $SrSO_4$ \item[(B)] $Mg(NH_4)PO_4$ \item[(C)] $BaCrO_4$ \item[(D)] $Mn(OH)_2$ \item[(E)] $PbSO_4$ \item[(F)] $PbCrO_4$ \item[(G)] $AgBr$ \item[(H)] $PbI_2$ \item[(I)] $CaC_2O_4$ \item[(J)] $[Fe(OH)_2(CH_3COO)]$ \end{enumerate}
SrSO_4
$\mathrm{Mg(NH_4)PO_4}$
BaCrO_4
Mn(OH)_2
Answer: e
Solution
Q1 $\mathrm{SrSO_4}$ - white $\mathrm{Mg(NH_4)PO_4}$ - white $\mathrm{BaCrO_4}$ - yellow $\mathrm{Mn(OH)_2}$ - white $\mathrm{PbSO_4}$ - white $\mathrm{PbCrO_4}$ - yellow $\mathrm{AgBr}$ - pale yellow $\mathrm{PbI_2}$ - yellow $\mathrm{CaC_2O_4}$ - white $[\mathrm{Fe(OH)_2(CH_3COO)}]$ - Brown Red
Question 84
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The ratio of $\frac{^{14}C}{^{12}C}$ in a piece of wood is $\frac{1}{8}$ part that of atmosphere. If half life of $^{14}C$ is 5730 years, the age of wood sample is _____ years.
Answer: 17190
Solution
Given $$ \lambda t = \ln \frac{(^{14}\mathrm{C}/^{12}\mathrm{C})_{atmosphere}}{(^{14}\mathrm{C}/^{12}\mathrm{C})_{wood sample}} $$ As per the question, $$ \frac{(^{14}\mathrm{C}/^{12}\mathrm{C})_{wood}}{(^{14}\mathrm{C}/^{12}\mathrm{C})_{atmosphere}} = \frac{1}{8} $$ So, $$ \lambda t = \ln 8 $$ $$ \frac{\ln 2}{t_{1/2}} t = \ln 8 $$ $$ t = 3 \times t_{1/2} = 17190 years $$
Question 85
Chemistry · Chemical Bonding and Molecular Structure · Numerical
The number of molecules/ion/s having trigonal bipyramidal shape is ........ . $\mathrm{PF}_5$, $\mathrm{BrF}_5$, $\mathrm{PCl}_5$, $[\mathrm{PtCl}_4]^{2-}$, $\mathrm{BF}_3$, $\mathrm{Fe(CO)}_5$
Total number of deactivating groups in aromatic electrophilic substitution reaction among the following is
Answer: 2
Solution
The groups are classified based on their mesomeric effect. The first group $-\mathrm{OCH_3}$ is a $-M$ group. The second group $-\mathrm{NH}$ is a $+M$ group. The third group $-\mathrm{CH_3}$ is a $+M$ group. The fourth group $-\mathrm{C \equiv N}$ is a $-M$ group. The fifth group $-\mathrm{OCH_3}$ is a $+M$ group.
Question 87
Chemistry · Classification of Elements and Periodicity in Properties · Numerical
Lowest Oxidation number of an atom in a compound $\mathrm{A_2 B}$ is $-2$. The number of an electron in its valence shell is
Answer: 6
Solution
Given $\mathrm{A_2B \rightarrow 2A^+ + B^{2-}}$, $\mathrm{B^{2-}}$ has complete octet in its dianionic form, thus in its atomic state it has 6 electrons in its valence shell. As it has negative charge, it has acquired two electrons to complete its octet.
Question 88
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
Among the following oxide of p-block elements, number of oxides having amphoteric nature is $\mathrm{Cl_2O_7}$, CO, $\mathrm{PbO_2}$, $\mathrm{N_2O}$, NO, $\mathrm{Al_2O_3}$, $\mathrm{SiO_2}$, $\mathrm{N_2O_5}$, $\mathrm{SnO_2}$
Chemistry · Some Basic Concepts of Chemistry · Numerical
Consider the following reaction: $$3\mathrm{PbCl}_2 + 2(\mathrm{NH}_4)_3\mathrm{PO}_4 \rightarrow \mathrm{Pb}_3(\mathrm{PO}_4)_2 + 6\mathrm{NH}_4\mathrm{Cl}$$ If $72\,\mathrm{mmol}$ of $\mathrm{PbCl}_2$ is mixed with $50\,\mathrm{mmol}$ of $(\mathrm{NH}_4)_3\mathrm{PO}_4$, then amount of $\mathrm{Pb}_3(\mathrm{PO}_4)_2$ formed is ___ mmol. (nearest integer)
Answer: 24
Solution
Limiting Reagent is $\mathrm{PbCl_2}$. mmol of $\mathrm{Pb_3(PO_4)_2}$ formed is given by $$\frac{mmol of \mathrm{PbCl_2} reacted}{3}$$ $$= 24 \, \mathrm{mmol}$$
Question 90
Chemistry · Equilibrium · Numerical
$K_a$ for $\mathrm{CH_3COOH}$ is $1.8 \times 10^{-5}$ and $K_b$ for $\mathrm{NH_4OH}$ is $1.8 \times 10^{-5}$. The pH of ammonium acetate solution will be