JEE Main 3 April 2025 Shift 2: all 75 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 · Applications of Derivatives · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined by $f(x) = ||x + 2| - 2|x||$. If $m$ is the number of points of local minima and $n$ is the number of points of local maxima of $f$, then $m + n$ is
Answer: (b)
Given $f(x) = ||x + 2| - 2|x||$. Critical points are $0$, $-2$, $2$, $-\frac{2}{3}$. The number of maxima is $1$. The number of minima is $2$. Option $(2)$.
Maths · Three Dimensional Geometry · Single correct
Each of the angles $\beta$ and $\gamma$ that a given line makes with the positive $y$- and $z$-axes, respectively, is half of the angle that this line makes with the positive $x$-axes. Then the sum of all possible values of the angle $\beta$ is
Answer: (a)
Given $\beta = \frac{\alpha}{2}$, $\gamma = \frac{\alpha}{2}$. $$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$$ Substituting, we have: $$\cos^2 \alpha + 2 \cos^2 \frac{\alpha}{2} = 1$$ Expanding, we get: $$\cos^2 \alpha + \cos \alpha = 0$$ Factoring gives: $$\cos \alpha (\cos \alpha + 1) = 0$$ Thus, $\cos \alpha = 0, -1$. Therefore, $\alpha = \frac{\pi}{2}, \pi$. Now $\beta = \frac{\alpha}{2} \implies \frac{\pi}{4}, \frac{\pi}{2}$. So the sum is $\frac{3\pi}{4}$.
Maths · Conic Sections · Single correct
If the four distinct points $(4, 6)$, $(-1, 5)$, $(0, 0)$ and $(k, 3k)$ lie on a circle of radius $r$, then $10k + r^2$ is equal to
Answer: (d)
Given $m_1 m_2 = -1$, so the right angle equation circle is $$(x - 4)(x - 0) + (y - 6)(y - 0) = 0$$ Expanding, we get $$x^2 + y^2 - 4x - 6y = 0$$ The point $(k, 3k)$ lies on it, so $$k^2 + 9k^2 - 4k - 18k = 0$$ Simplifying, $$10k^2 - 22k = 0$$ Solving for $k$, we get $$k = 0, \frac{11}{5}$$ Since $k = 0$ is not possible, we have $k = \frac{11}{5}$. Also, $r = \sqrt{4 + 9} = \sqrt{13}$. Thus, $$10k + r^2 = 10 \cdot \frac{11}{5} + (\sqrt{13})^2 = 35$$
Maths · Statistics · Single correct
Let the Mean and Variance of five observations $x_1 = 1, x_2 = 3, x_3 = a, x_4 = 7$ and $x_5 = b, a > b,$ be 5 and 10 respectively. Then the Variance of the observations $n + x_n, n = 1, 2, \ldots 5$ is
Answer: (d)
Given $\($ $\bar{x}$ = $\frac{\sum x_i}{n}$ = $\frac{1 + 3 + a + 7 + b}{5}$ = 5 $\)$. Therefore, $\($ a + b = 14 $\)$. The variance $\($ $\sigma$^2 = $\frac{\sum x_i^2}{n}$ - ($\bar{x}$)^2 $\)$. $\[$ $\Rightarrow$ $\frac{1^2 + 3^2 + a^2 + 7^2 + b^2}{5}$ - 25 = 10 $\]$ $\($ a^2 + b^2 = 116 $\)$. Since $\($ a > b $\)$, let $\($ a = 10 $\)$ and $\($ b = 4 $\)$. The numbers are $\($ n + x_n : 2, 5, 13, 11, 9 $\)$. $\[$ $\sigma$^2 = $\frac{2^2 + 5^2 + 13^2 + 11^2 + 9^2}{5}$ - $\left$( $\frac{2 + 5 + 13 + 11 + 9}{5}$ $\right$)^2 $\]$ $\[$ = 80 - 64 = 16 $\]$ option 4
Maths · Straight Lines and Pair of Straight Lines · Single correct
Consider the lines $x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5$, $\lambda$ being a parameter, all passing through a point $P$. One of these lines (say $L$) is farthest from the origin. If the distance of $L$ from the point $(3, 6)$ is $d$, then the value of $d^2$ is
Answer: (a)
Given $x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5$ and $(x + 2y - 5) + \lambda(3x + 7y - 17) = 0$. The intersection of the family of lines is $P(1, 2)$. Let $Q(3, 6)$. The distance $d = PQ = \sqrt{2^2 + 4^2} = \sqrt{20}$. Therefore, $d^2 = 20$. The correct option is option (1).
Maths · Sets · Single correct
Let $A = \{-2, -1, 0, 1, 2, 3\}$. let $R$ be a relation on $A$ defined by $xRy$ if and only if $y = \max\{x, 1\}$. Let $l$ be the number of elements in $R$. Let $m$ and $n$ be the minimum number of elements required to be added in $R$ to make it reflexive and symmetric relations, respectively. Then $l + m + n$ is equal to
Answer: (a)
Given $A = \{-2, -1, 0, 1, 2, 3\}$. The relation $R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$. We have $\ell = 6$, $m = 3$, $n = 3$. Therefore, $\ell + m + n = 12$.
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let the equation $x(x + 2)(12 - k) = 2$ have equal roots. Then the distance of the point $\left(k, \frac{k}{2}\right)$ from the line $3x + 4y + 5 = 0$ is
Answer: (a)
Given $\left( x^2 + 2x \right) \left( 12 - k \right) = 2$. (1) $\lambda x^2 + 2 \lambda x - 2 = 0$ where $k \neq 12$. Let $12 - k = \lambda$. $D = 0$ $$4 \lambda^2 + 8 \lambda = 0$$ $$\lambda = 0 or \lambda = -2$$ $$\Rightarrow 12 - k = -2$$ $$k = 14$$ So $P \left( k, \frac{k}{2} \right) = (14, 7)$ $$d = \left| \frac{3 \times 14 + 4 \times 7 + 5}{5} \right| = 15$$ option (1)
Maths · Permutations and Combinations · Single correct
Line $L_1$ of slope 2 and line $L_2$ of slope $\frac{1}{2}$ intersect at the origin $O$. In the first quadrant, $P_1, P_2, \ldots, P_{12}$ are 12 points on line $L_1$ and $Q_1, Q_2, \ldots, Q_9$ are 9 points on line $L_2$. Then the total number of triangles, that can be formed having vertices at three of the 22 points $O, P_1, P_2, \ldots, P_{12}, Q_1, Q_2, \ldots, Q_9$, is:
Answer: (b)
Total number of $\Delta$ are $$= \binom{9}{1} \binom{12}{2} + \binom{9}{2} \binom{12}{1} + \binom{1}{1} \binom{9}{1} \binom{12}{1}$$ $$= 594 + 432 + 108$$ $$= 1134$$
Maths · Integrals · Single correct
The integral $$\int_0^\pi \frac{8x \, dx}{4 \cos^2 x + \sin^2 x}$$ is equal to
Answer: (a)
\[ I = \int_0^{\pi} \frac{8x\,dx}{4\cos^2 x + \sin^2 x} \] \[ I = \int_0^{\pi} \frac{8(\pi-x)\,dx}{4\cos^2 x + \sin^2 x} \] \[ 2I = 8\pi \int_0^{\pi} \frac{dx}{4\cos^2 x + \sin^2 x} \] \[ 2I = 8\pi \times 2\int_0^{\pi/2} \frac{\sec^2 x}{4+\tan^2 x}\,dx \] \[ I = 8\pi \int_0^{\infty} \frac{dt}{4+t^2} = 8\pi \times \frac{1}{2}\left[\tan^{-1}\frac{t}{2}\right]_0^{\infty} \] \[ = 4\pi \times \frac{\pi}{2} = 2\pi^2 \] \[ \text{option (1)} \]
Maths · Relations and Functions · Single correct
Let $f$ be a function such that $f(x) + 3f\left(\frac{24}{x}\right) = 4x, x \neq 0$. Then $f(3) + f(8)$ is equal to
Answer: (a)
Given $f(x) + 3f\left(\frac{24}{x}\right) = 4x$. Put $x = 3$, $f(3) + 3f(8) = 12$. Put $x = 8$, $f(8) + 3f(3) = 32$. Add both $4(f(3) + f(8)) = 44$. $f(3) + f(8) = 11$.
Maths · Applications of Integrals · Single correct
The area of the region $\{(x, y) : |x - y| \leq y \leq 4\sqrt{x}\}$ is
Answer: (b)
Given $|x - y| \leq y \leq 4\sqrt{x}$. Now $y = |x - y|$. So, $y^2 = (x - y)^2$. This implies $y = \frac{x}{2}$ and $x = 0$. Now the area is $$\int_{0}^{64} \left(4\sqrt{x} - \frac{x}{2}\right) \, dx$$ which equals $$\left[ \frac{4x^{3/2}}{3/2} - \frac{x^2}{4} \right]_{0}^{64} = \frac{8}{3} \cdot 8^3 - \frac{64^2}{4} = 64^2 \left( \frac{1}{12} \right)$$ which simplifies to $$\frac{1024}{3}$$.
Maths · Relations and Functions · Single correct
If the domain of the function $f(x) = \log_7 \left( 1 - \log_4 (x^2 - 9x + 18) \right)$ is $(\alpha, \beta) \cup (\gamma, \delta)$, then $\alpha + \beta + \gamma + \delta$ is equal to
Answer: (a)
Domain $1 - \log_4(x^2 - 9x + 18) > 0$. Also $x^2 - 9x + 18 > 0$. $$(x - 3)(x - 6) > 0$$ $$x \in (-\infty, 3) \cup (6, \infty) \ldots (1)$$ Also $x^2 - 9x + 18 < 4$. $$x^2 - 9x + 14 < 0$$ $$x \in (2, 7) \ldots (2)$$ $$(1) \cap (2) (2, 3) \cup (6, 7) = (\alpha, \beta) \cup (\gamma, \delta)$$ $$\Rightarrow \alpha + \beta + \gamma + \delta = 18$$
Maths · Probability (Advanced) · Single correct
If the probability that the random variable X takes the value x is given by $P(X = x) = k(x + 1)3^{-x}$, $x = 0, 1, 2, 3 \ldots$, where k is a constant, then $P(X \geq 3)$ is equal to
Answer: (d)
Given $$\sum_{x=0}^{\infty} k(x+1)3^{-x} = 1$$ $$\Rightarrow \frac{1}{k} = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots (i)$$ $$\frac{1}{3k} = 1 + \frac{2}{3} + \frac{3}{3^2} + \ldots (ii)$$ Subtracting (ii) from (i) gives: $$\frac{1}{k} - \frac{1}{3k} = 1 + \frac{1}{3} + \frac{1}{3^2} + \ldots$$ $$\Rightarrow k = \frac{4}{9}$$ The probability $$P(x \geq 3) = 1 - P(x = 0) - P(x = 1) - P(x = 2)$$ $$= 1 - k \left( 1 + \frac{2}{3} + \frac{3}{9} \right) = \frac{1}{9}$$
Maths · Differential Equations · Single correct
Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 3 \left( \tan^2 x \right) y + 3y = \sec^2 x$ $y(0) = \frac{1}{3} + e^3$. Then $y \left( \frac{\pi}{4} \right)$ is equal to
Answer: (b)
Given $\dfrac{dy}{dx} + 3(\sec^2 x)\, y = \sec^2 x,\quad y(0) = \dfrac{1}{3} + e^3$. If $e^3 \int \sec^2 x\, dx = e^{3\tan x}$. Therefore, the solution is \[ e^{3\tan x}\, y = \int e^{3\tan x} \sec^2 x\, dx \] \[ e^{3\tan x}\, y = \frac{e^{3\tan x}}{3} + c \] \[ \therefore\, y(0) = \frac{1}{3} + e^3 \Rightarrow c = e^3 \] \[ \therefore\, y\!\left(\frac{\pi}{4}\right) = \frac{e^3 + e^3}{e^3} = \frac{4}{3} \]
Maths · Complex Numbers and Quadratic Equations · Single correct
If $z_1, z_2, z_3 \in \mathbb{C}$ are the vertices of an equilateral triangle, whose centroid is $z_0$, then $\sum_{k=1}^{3} (z_k - z_0)^2$ is equal to
Answer: (a)
Given $z_1 + z_2 + z_3 = 3z_0$. $(z_1 + z_2 + z_3)^2 = 9z_0^2$ $\($ $\Rightarrow$ z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1) = 9z_0^2 $\)$ $\($ $\Rightarrow$ z_1^2 + z_2^2 + z_3^2 = 3z_0^2 $\)$ $\($ $\sum$_{k=1}^{3} (z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2 $\)$ $\($ = z_1^2 + z_2^2 + z_3^2 + 3z_0^2 - 2(z_1 + z_2 + z_3)z_0 $\)$ $\($ = 6z_0^2 - 6z_0^2 $\)$ $\($ = 0 $\)$
Maths · Trigonometric Functions · Single correct
The number of solutions of equation $(4 - \sqrt{3}) \sin x - 2\sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}, \; x \in \left[ -2\pi, \frac{5\pi}{2} \right]$ is
Answer: (d)
Given $ (4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^2 x = \frac{-4}{1 + \sqrt{3}}, x \in \left[ -2\pi, \frac{5\pi}{2} \right] $. Therefore, $$ (4 - \sqrt{3}) \sin x - 2 \sqrt{3} \left( 1 - \sin^2 x \right) = 2(1 - \sqrt{3}) $$ This simplifies to $$ 2 \sqrt{3} \sin^2 x + 4 \sin x - \sqrt{3} \sin x - 2 = 0 $$ Further simplifying, $$ (2 \sin x - 1)(\sqrt{3} \sin x + 2) = 0 $$ Thus, $$ \sin x = \frac{1}{2} $$ Therefore, the number of solutions is $5$.
Maths · Conic Sections · Single correct
Let $C$ be the circle of minimum area enclosing the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentricity $\frac{1}{2}$ and foci $(\pm 2, 0)$. Let $PQR$ be a variable triangle, whose vertex $P$ is on the circle $C$ and the side $QR$ of length $29$ is parallel to the major axis of $E$ and contains the point of intersection of $E$ with the negative $y$-axis. Then the maximum area of the triangle $PQR$ is :
Answer: (d)
Area of $\triangle PQR$ is given by $$\frac{1}{2} (2a)(a \sin \theta + b).$$ Therefore, the maximum area is $a(a + b)$. $$= 4(4 + 2\sqrt{3}) = 8(2 + \sqrt{3}).$$
Maths · Applications of Derivatives · Single correct
The shortest distance between the curves $y^2 = 8x$ and $x^2 + y^2 + 12y + 35 = 0$ is :
Answer: (d)
Equation of normal to parabola $y^2 = 8x$ is $y = mx - 4m - 2m^3$. Passes through $(0, -6)$ we get $$-6 = -4m - 2m^3$$ $$\Rightarrow m^3 + 2m - 3 = 0$$ $$\Rightarrow (m - 1)(m^2 + m + 3) = 0 \Rightarrow m = -1$$ $P = (am^2, -2am) = (2, -4)$ Therefore, shortest distance $= PC - r$ $$= (2\sqrt{2} - 1)$$
Maths · Three Dimensional Geometry · Single correct
The distance of the point (7, 10, 11) from the line $\frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3}$ along the line $\frac{x-9}{2} = \frac{y-13}{3} = \frac{z-17}{6}$ is
Answer: (b)
Line $PQ$ is parallel to the line $\($ $\frac{x-9}{2}$ = $\frac{y-3}{3}$ = $\frac{z-17}{6}$ $\)$. Therefore, $\($ $\frac{\lambda - 3}{2}$ = $\frac{-6}{3}$ = $\frac{3\lambda - 9}{6}$ $\)$. This implies $\($ $\lambda$ = -1 $\)$. Thus, $\($ Q = (3, 4, -1) $\)$. Therefore, $\($ PQ = $\sqrt{16 + 36 + 144}$ = 14 $\)$.
Maths · Sequences and Series · Single correct
The sum $1 + \frac{1+3}{2!} + \frac{1+3+5}{3!} + \frac{1+3+5+7}{4!} + \ldots$ upto $\infty$ terms, is equal to
Answer: (d)
Given $$S = 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \ldots$$ This can be written as $$= \sum_{r=1}^{\infty} \frac{r^2}{r!}$$ Rewriting the terms, $$= \sum_{r=1}^{\infty} \frac{(r - 1 + 1)}{(r - 1)!} = \sum_{r=2}^{\infty} \frac{1}{(r - 2)!} + \sum_{r=1}^{\infty} \frac{1}{(r - 1)!}$$ This simplifies to $$= 2e$$
Maths · Matrices · Numerical
Let I be the identity matrix of order 3 $\times$ 3 and for the matrix A = \begin{bmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{bmatrix}, |A| = -1. Let B be the inverse of the matrix $\operatorname{adj}\!\left(A \operatorname{adj}(A^2)\right)$. Then |($\lambda$ B + 1)| is equal to
Answer: 38
Given $|A| = \begin{vmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{vmatrix} = -1$. $\lambda(16) - 2(-34) + 3(-39) = -1$. $16\lambda = 48 \Rightarrow \lambda = 3$. $B^{-1} = adj \left( A \cdot adj \left( A^2 \right) \right)$. Let $C = A \cdot adj \left( A^2 \right)$. $AC = A^2 adj \left( A^2 \right) = |A|^2 \cdot I = I \Rightarrow C = A^{-1}$. Now $B^{-1} = adj \left( A^{-1} \right) = B = adj(A)$. Now $\lambda B + I \Rightarrow 3B + I$. Let $P = 3B + I$. $P = 3 adj(A) + I$. $AP = 3A adj(A) + A$. $AP = 3|A| \cdot I + A$. $AP = A - 3I$. $|AP| = |A - 3I| = \begin{vmatrix} 0 & 2 & 3 \\ 4 & 2 & 6 \\ 7 & -1 & -1 \end{vmatrix} = 38$. $|A| \cdot |P| = 38$. $|P| = -38$.
Maths · Binomial Theorem · Numerical
Let $(1+x+x^2)^{10}=a_0+a_1x+a_2x^2+\cdots+a_{20}x^{20}$. If $(a_1+a_3+a_5+\cdots+a_{19})-11a_2=121k$, then $k$ is equal to ______.
Answer: 239
Given $ (1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + \ldots + a_{20} x^{20} $. Therefore, $3^{10} = a_0 + a_1 + a_2 + \ldots + a_{20} \ldots (i)$. $1 = a_0 - a_1 + a_2 \ldots + a_{20} \ldots (ii)$. Subtracting (ii) from (i) gives $a_1 + a_3 + \ldots + a_{19} = \frac{3^{10} - 1}{2} = 29524$. Also, $\{1 + x(1 + x)\}^{10} = 1 + \binom{10}{1} x(1 + x) + \binom{10}{2} x^2 (1 + x)^2 + \ldots$. Therefore, $a_2 = \binom{10}{1} + \binom{10}{2} = 55$. Thus, $\frac{(a_1 + a_3 + \ldots + a_{19}) - 11 a_2}{121} = 239$.
Maths · Limits and Derivatives · Numerical
If $\lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p$, then $96 \log_e p$ is equal to ______
Answer: 32
Given $P = \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{x^2}$. Therefore, $P = e^{\lim_{x \to 0} \left( \frac{\tan x - x}{x^3} \right)}$. This simplifies to $e^{\lim_{x \to 0} \left( \frac{x^3 + \frac{2x^5}{15} + \ldots - x}{x^3} \right)}$. Thus, $P = e^{1/3}$. Therefore, $96 \log_e P = 96 \times \frac{1}{3} = 32$.
Maths · Vector Algebra · Fill in the blank
Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$, $\vec{b}=3\hat{i}-3\hat{j}+3\hat{k}$, $\vec{c}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{d}$ be a vector such that $\vec{b}\times\vec{d}=\vec{c}\times\vec{d}$ and $\vec{a}\cdot\vec{d}=4$. Then $|(\vec{a}\times\vec{d})|^2$ is equal to ______.
Answer: 128
Given $\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$ and $\vec{a} \cdot \vec{d} = 4$. Therefore, $\vec{d} = \lambda (\vec{b} - \vec{c}) = \lambda (\hat{i} - 2\hat{j} + \hat{k})$. Since $\vec{a} \cdot \vec{d} = 4$, it follows that $\lambda = -2$. Also, $|\vec{a} \times \vec{d}|^2 + |\vec{a}|^2 |\vec{d}|^2$. Thus, $|\vec{a} \times \vec{d}|^2 = 6 \times 4 \times 6 - 16 = 128$.
Maths · Conic Sections · Fill in the blank
If the equation of the hyperbola with foci (4, 2) and (8, 2) is $3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$, then $\alpha + \beta + \gamma$ is equal to _____.
Answer: 141
Equation of hyperbola is $$\frac{(x-6)^2}{a^2} - \frac{(y-2)^2}{4-a^2} = 1$$ which implies $$\left(4-a^2\right)(x-6)^2 - a^2(y-2)^2 = a^2(4-a^2)$$ comparing with $$3x^2 - y^2 - \alpha x + \beta y + \gamma = 0,$$ we get $a^2 = 1$ and $\alpha = 36$, $\beta = 4$ and $\gamma = 101$. Therefore, $$\alpha + \beta + \gamma = 141$$
Physics · Moving Charges and Magnetism · Single correct
A magnetic dipole experiences a torque of $80\sqrt{3} \, \mathrm{N \, m}$ when placed in uniform magnetic field in such a way that dipole moment makes angle of $60^\circ$ with magnetic field. The potential energy of the dipole is :
Answer: (d)
Given $\tau = \mathbf{M} \times \mathbf{B} = MB \sin 60 = \frac{\sqrt{3}}{2} MB = 80 \sqrt{3}$. $MB = 160$. $U = -\mathbf{M} \cdot \mathbf{B} = -MB \cos 60$. $U = -160 \times 1/2 = -80 \, \mathrm{J}$.
Physics · Waves · Single correct
In the resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is:
Answer: (d)
Fundamental frequency in close/organ pipe $$f = \frac{v}{4\ell}$$ $$f_1 = \frac{v}{4\ell_1} \& f_2 = \frac{v}{4\ell_2}$$ Beat = $(f_1 - f_2) = \frac{v}{4} \left( \frac{1}{\ell_1} - \frac{1}{\ell_2} \right)$ $$15 = \frac{v}{4} \left( \frac{1}{1} - \frac{1}{1.2} \right)$$ $$v = \left( \frac{15 \times 4 \times 1.2}{0.2} \right) = 60 \times 6 = 360 \, \mathrm{m/s}$$
Physics · Mechanical Properties of Fluids · Single correct
Two cylindrical vessels of equal cross sectional area of $2 \, \mathrm{m^2}$ contain water up to height $10 \, \mathrm{m}$ and $6 \, \mathrm{m}$, respectively. If the vessels are connected at their bottom then the work done by the force of gravity is: (Density of water is $10^3 \, \mathrm{kg/m^3}$ and $g = 10 \, \mathrm{m/s^2}$)
Answer: (d)
Given the initial potential energy $U_1 = (\rho A \times 10)g \times 5 + (\rho A 6)g \times 3$. The intermediate potential energy $U_i = \rho A g (50 + 18)$. Therefore, $U_i = 68 \rho A g$. The final potential energy $U_f = (\rho A \times 16)g \times 4$. Simplifying, $U_f = (\rho A g) \times 64$. The work done $\omega = \Delta U = 4 \times \rho A g$. Calculating, $= 4 \times 1000 \times 2 \times 10 = 8 \times 10^4 \, \mathrm{J}$.
Physics · Wave Optics · Single correct
Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :
Answer: (b)
Given $I \propto width$, we have $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$. Since $I_1 = I_0, I_2 = 2I_0$, we find $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$. Calculating: $$\frac{I_{max}}{I_{min}} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2} \Rightarrow \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}.$$
Physics · Kinetic Theory · Single correct
An ideal gas exists in a state with pressure $P_0$ volume $V_0$. It is isothermally expanded to 4 times of its initial volume ($V_0$), then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is:
Answer: (a)
Given the processes in the diagram, we have the following calculations: $$\omega_1 = P_0 v_0 \ln 4$$ $$\omega_2 = \frac{P_0}{4} (-3v_0) = -\frac{3P_0 v_0}{4}$$ $$\omega_3 = 0$$ The total work done is given by: $$Q_T = \Delta U_{cyclic} + \omega$$ Since the internal energy change over a cycle is zero, $\Delta U_{cyclic} = 0$, we have: $$Q_T = \omega$$ Thus, $$Q_T = P_0 v_0 \left( \ln 4 - \frac{3}{4} \right)$$ Simplifying further: $$Q_T = P_0 v_0 (2 \ln 2 - 0.75)$$
Physics · Wave Optics · Single correct
Two monochromatic light beams have intensities in the ratio 1:9. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is
Answer: (d)
Given $\($ $\frac{I_{max}}{I_{min}}$ = $\left$( $\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}$ $\right$)^2 $\)$. This implies $\($ $\left$( $\frac{4}{2}$ $\right$)^2 $\)$ which simplifies to $\($ $\frac{16}{4}$ = 4 $\)$.
Physics · Atoms · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The Bohr model is applicable to hydrogen and hydrogen-like atoms only. Reason R: The formulation of Bohr model does not include repulsive force between electrons. In the light of the above statements, choose the correct answer from the options given below:
Q1. Conceptual
Physics · Electrostatic Potential and Capacitance · Single correct
Using a battery, a 100 $\mathrm{pF}$ capacitor is charged to 60 $\mathrm{V}$ and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 $\mathrm{V}$, its capacitance is : (in pF)
Answer: (b)
New potential $= \frac{C_0V_0}{C_0+C} = \frac{V_0}{3}$ $3C_0V_0 = C_0V_0 + CV_0$ $2C_0V_0 = CV_0$ $C \Rightarrow 2C_0$
Physics · Wave Optics · Single correct
A monochromatic light of frequency $5 \times 10^{14} \, \mathrm{Hz}$ travelling through air, is incident on a medium of refractive index ' 2 '. Wavelength of the refracted light will be :
Answer: (a)
Given $f \lambda = v$. The wavelength in the medium $\lambda_{medium}$ is given by $\lambda_{medium} = \frac{\lambda_{vacuum}}{\mu}$. For $\lambda_{medium}$: $$\lambda_{medium} \Rightarrow \frac{3 \times 10^8}{2.5 \times 10^{14}} \Rightarrow 0.3 \times 10^{-6} \Rightarrow 300 \, nm$$
Physics · Work, Energy and Power · Single correct
Consider two blocks A and B of masses $m_1 = 10 \, \mathrm{kg}$ and $m_2 = 5 \, \mathrm{kg}$ that are placed on a frictionless table. The block A moves with a constant speed $v = 3 \, \mathrm{m/s}$ towards the block B kept at rest. A spring with spring constant $k = 3000 \, \mathrm{N/m}$ is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring)
Answer: (c)
Given $m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{cm}$. $$v_{cm} \Rightarrow \frac{10 \times 3}{10 + 5} \Rightarrow \frac{30}{15} = 2 \, \mathrm{m/s}$$ $$\frac{1}{2} k x^2 = \frac{1}{2} (10)(3)^2 - \left[ \frac{1}{2} (15)(2)^2 \right]$$ $$\Rightarrow 90 - 60 = 30 = 3000 x^2$$ $$x^2 \Rightarrow \frac{30}{3000} = \frac{1}{100}$$ $$x \Rightarrow \frac{1}{10} \, \mathrm{m}.$$
Physics · Motion in a Plane · Single correct
A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{nu^2}{2g}$, where value of $n$ is: (Given $'g'$ is the acceleration due to gravity).
Answer: (d)
Range $= 3H_{\max}$ $$\frac{u^2 \sin 2\theta}{g} = \frac{3u^2 \sin^2 \theta}{2g}$$ $$2 \sin \theta \cos \theta = \frac{3}{2} \sin^2 \theta$$ $$\tan \theta = \frac{4}{3} \Rightarrow \theta = 53^\circ$$ $$R = \frac{u^2 \left(2 \times \frac{3}{5} \times \frac{4}{5}\right)}{g} \Rightarrow \frac{24u^2}{25g}$$
Physics · Mechanical Properties of Fluids · Single correct
A solid steel ball of diameter $3.6\,\mathrm{mm}$ acquired terminal velocity $2.45\times10^{-2}\,\mathrm{m/s}$ while falling under gravity through an oil of density $925\,\mathrm{kg\,m^{-3}}$. Take density of steel as $7825\,\mathrm{kg\,m^{-3}}$ and $g$ as $9.8\,\mathrm{m/s^2}$. The viscosity of the oil in SI unit is
Answer: (d)
Given $v_T = \frac{2}{9} (\rho_0 - \rho_\ell) r^2 g$. $$\eta = \frac{2}{9} \left( 7825 - 925 \right) \left( 2.45 \times 10^{-2} \right) \times (1.8)^2 \times 10^{-6} \times 9.8$$ $$\eta \approx 1.99$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The truth table corresponding to the circuit given below is
Answer: (b)
The circuit consists of an OR gate followed by an AND gate. The expression for the output C is given by $C = A \cdot (A + B)$. The truth table is as follows: For $A = 0$ and $B = 0$, $A + B = 0$, so $C = 0 \cdot 0 = 0$. For $A = 1$ and $B = 0$, $A + B = 1$, so $C = 1 \cdot 1 = 1$. For $A = 0$ and $B = 1$, $A + B = 1$, so $C = 0 \cdot 1 = 0$. For $A = 1$ and $B = 1$, $A + B = 1$, so $C = 1 \cdot 1 = 1$.
Physics · Motion in a Straight Line · Single correct
A particle moves along the $x$-axis and has its displacement $x$ varying with time $t$ according to the equation $x = c_0 \left(t^2 - 2\right) + c(t - 2)^2$ where $c_0$ and $c$ are constants of appropriate dimensions. Then, which of the following statements is correct?
Answer: (d)
Given $v = \frac{dx}{dt} = 2tC_0 + 2C(t - 2)$. Acceleration $a = \frac{dv}{dt} = 2C_0 + 2C$.
Physics · Alternating Current · Single correct
An electric bulb rated as 100 $\mathrm{W}$ - 220 $\mathrm{V}$ is connected to an ac source of rms voltage 220 $\mathrm{V}$. The peak value of current through the bulb is:
Answer: (a)
Given $P = v_{rms} i_{rms}$. $$(1) i_{rms} = \frac{100}{220}$$ $$i_0 = \sqrt{2} i_{rms} = 0.64 \, A$$
Physics · Physical World, Units and Measurements · Single correct
Match the LIST-I with LIST-II \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{\textbf{LIST-I}} & \multicolumn{2}{c|}{\textbf{LIST-II}} \\ \hline \multicolumn{2}{|c|}{} & \multicolumn{2}{c|}{} \\ \hline \textbf{A.} & Boltzmann constant & \textbf{I.} & $\mathrm{ML^2T^{-1}}$ \\ \hline \textbf{B.} & Coefficient of viscosity & \textbf{II.} & $\mathrm{MLT^{-3}K^{-1}}$ \\ \hline \textbf{C.} & Planck's constant & \textbf{III.} & $\mathrm{ML^2T^{-2}K^{-1}}$ \\ \hline \textbf{D.} & Thermal conductivity & \textbf{IV.} & $\mathrm{ML^{-1}T^{-1}}$ \\ \hline \end{tabular} Choose the correct answer from the options given below:
Answer: (a)
(A) $[k] = \frac{PV}{NT} = \frac{ML^2 \, T^{-2}}{K} = ML^2 \, T^{-2} \, K^{-1}$ (B) $[\eta] = \frac{F}{6 \pi r v} = \frac{MLT^{-2}}{L^2 \, T^{-1}} = M L^{-1} \, T^{-1}$ (C) $[h] = \frac{E}{f} = \frac{ML^2 \, T^{-2}}{T^{-1}} = ML^2 \, T^{-1}$ (D) $\frac{dQ}{dt} = k \frac{AdT}{dx}$ $k = \left(ML^2 \, T^{-3}\right) \frac{L}{L^2 \, K} = ML \, T^{-3} \, K^{-1}$
Physics · Kinetic Theory · Single correct
Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4$\%$ when heated by $1^\circ \mathrm{C}$. Its initial temperature must be:
Answer: (c)
Isochoric process $P \propto T$ $$\frac{\Delta P}{P} = \frac{\Delta T}{T}$$ $$\frac{0.4}{100} = \frac{1}{T}$$ Solving for $T$ gives: $$T = 250 \, \mathrm{K}$$
Physics · Current Electricity · Single correct
A motor operating on 100 V draws a current of 1 A. If the efficiency of the motor is 91.6$\%$, then the loss of power in units of cal/s is
Answer: (c)
Given $P_{input} = Vi = 100 \, W$. The efficiency $\eta$ is given by $$\eta = \frac{P_{out}}{P_{input}} = 0.916$$ Therefore, $$P_{out} = 91.6 \, W$$ The loss is calculated as $$Loss = 100 - 91.6 = 8.4 \, J/s = 2 \, cal/s$$
Physics · Work, Energy and Power · Single correct
A block of mass 1 kg , moving along x with speed $v_i = 10 \, \mathrm{m/s}$ enters a rough region ranging from $x = 0.1 \, \mathrm{m}$ to $x = 1.9 \, \mathrm{m}$. The retarding force acting on the block in this range is $F_r = -kxN$, with $k = 10 \, \mathrm{N/m}$. Then the final speed of the block as it crosses rough region is
Answer: (d)
Given $a = \frac{F}{m} = -10x$. $v \frac{dv}{dx} = -10x$. $$\int_{10}^{v} v \, dv = -10 \int_{0.1}^{1.9} x \, dx$$ $$\frac{v^2 - 100}{2} = -10 \left( \frac{1.9^2 - 0.1}{2} \right)^2$$ $v = 8 \, \mathrm{m/s}$
Physics · Moving Charges and Magnetism · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: If oxygen ion $\left( \mathrm{O}^{-2} \right)$ and Hydrogen ion $\left( \mathrm{H}^{+} \right)$ enter normal to the magnetic field with equal momentum, then the path of $\mathrm{O}^{-2}$ ion has a smaller curvature than that of $\mathrm{H}^{+}$. Reason R: A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly. In the light of the above statements, choose the correct answer from the options given below:
Answer: (a)
Given $r = \frac{mv}{qB} = \frac{p}{qB}$. Therefore, $r \propto \frac{1}{q}$. Assertion is true reason is false.
Physics · Ray Optics and Optical Instruments · Numerical
Light from a point source in air falls on a spherical glass surface (refractive index, $\mu = 1.5$ and radius of curvature $= 50 \, \mathrm{cm}$). The image is formed at a distance of $200 \, \mathrm{cm}$ from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is m.
Answer: 4
Given $\mu = 1.5$ and $R = 50 \, \mathrm{cm}$. Using the lens formula: $$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$ Substituting the values: $$\frac{1.5}{200} - \frac{1}{-x} = \frac{1.5 - 1}{50}$$ Simplifying: $$\frac{1}{200} + \frac{1}{x} = \frac{1}{100}$$ $$\frac{1}{x} = \frac{1}{100} - \frac{1}{200}$$ $$\frac{1}{x} = \frac{1}{200}$$ $$x = 200 \, \mathrm{cm}$$ Therefore, $x = 4 \, \mathrm{m}$.
Physics · Mechanical Properties of Fluids · Fill in the blank
The excess pressure inside a soap bubble A in air is half the excess pressure inside another soap bubble B in air. If the volume of the bubble A is $n$ times the volume of the bubble B, then, the value of $n$ is ___.
Answer: 8
Given $\Delta P = \frac{4T}{R}$. $\[$ $\frac{R_A}{R_B}$ = $\frac{\Delta P_B}{\Delta P_A}$ = 2 $\]$ $\[$ $\frac{V_A}{V_B}$ = $\left$( $\frac{R_A}{R_B}$ $\right$)^3 = 8 $\]$
Physics · Current Electricity · Numerical
Two cells of emf 1 V and 2 V and internal resistance 2$\,$$\Omega$ and 1$\,$$\Omega$, respectively, are connected in series with an external resistance of 6$\,$$\Omega$. The total current in the circuit is $I_1$. Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is $I_2$. The value of $\left( \frac{I_1}{I_2} \right)$ is $\frac{x}{3}$. The value of $x$ is _____.
Answer: 4
Given $\varepsilon_{eq} = 3$ and $R_{eq} = 9$. Calculate $i_1$ using the formula: $$i_1 = \frac{3}{9} = \frac{1}{3}$$ For $\varepsilon_{eq}$: $$\varepsilon_{eq} = \frac{\varepsilon_1 r_1 + \varepsilon_2 r_2}{r_1 + r_2}$$ Substitute the values: $$\varepsilon_{eq} = \frac{1 \times 1 + 2 \times 2}{1 + 2} = \frac{5}{3}$$ Calculate $r_{equ}$: $$r_{equ} = 2 \times 1 + 6 = 8$$ Calculate $i_2$: $$i_2 = \frac{1}{4} \Rightarrow i_1 = \frac{4}{3} \Rightarrow i_2 = 3$$
Physics · Atoms · Numerical
An electron in the hydrogen atom initially in the fourth excited state makes a transition to $n^{th}$ energy state by emitting a photon of energy 2.86 eV. The integer value of $n$ will be ____.
Answer: 2
Given the equation $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. We have $2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{5^2} \right)$. This simplifies to $\frac{1}{n^2} = 0.21 + \frac{1}{2.5}$. Solving gives $n^2 = 4$. Thus, $n = 2$.
Physics · Physical World, Units and Measurements · Numerical
A physical quantity C is related to four other quantities p, q, r and s as follows $C = \frac{pq^2}{r^3 \sqrt{s}}$ The percentage errors in the measurement of p, q, r and s are 1$\%$, 2$\%$, 3$\%$ and 2$\%$ respectively. The percentage error in the measurement of C will be _____$\%$.
Answer: 15
Given $C = P^1 q^2 r^{-3} s^{1/2}$. The maximum change in $\left( \frac{dC}{C} \right)_{max}$ is given by: $$\left( \frac{dC}{C} \right)_{max} = \frac{dP}{P} + \frac{2dq}{q} + \frac{3dr}{r} + \frac{1}{2} \frac{ds}{s}$$ Substituting the values, we have: $$= \left( 1 + 2 \times 2 + 3 \times 3 + \frac{1}{2} \times 2 \right) \%$$ This simplifies to: $$= 15\%$$ Therefore, the answer is 15.
Chemistry · Electrochemistry · Single correct
40 $\,$ $\mathrm{mL}$ of a mixture of $\mathrm{CH_3COOH}$ and $\mathrm{HCl}$ (aqueous solution) is titrated against 0.1 $\,$ $\mathrm{M}$ NaOH solution conductometrically. Which of the following statement is correct?
Answer: (b)
From the given graph 2 ml NaOH solution is used for neutralisation of HCl and 3 ml NaOH solution is used for neutralisation of CH_3COOH. Therefore, Moles of HCl = Moles of NaOH used $$M \times 40 = 0.1 \times 2$$ $$M = 0.005$$ Therefore, Moles of CH_3COOH = Moles of NaOH used $$M \times 40 = 0.1 \times 3$$ $$M = 0.0075$$ HCl is a strong acid and will be neutralised first.
Chemistry · Equilibrium · Single correct
10 $\,$ $\mathrm{mL}$ of 2 $\,$ $\mathrm{M}$ NaOH solution is added to 20 $\,$ $\mathrm{mL}$ of 1 $\,$ $\mathrm{M}$ HCl solution kept in a beaker. Now, 10 $\,$ $\mathrm{mL}$ of this mixture is poured into a volumetric flask of 100 $\,$ $\mathrm{mL}$ containing 2 moles of HCl and made the volume upto the mark with distilled water. The solution in this flask is:
Answer: (b)
When 10 ml, 2 M NaOH solution is added to 20 ml of 1 M HCl solution: $$\mathrm{NaOH} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}$$ Initial: $MV = 2 \times 0.1$ $MV = 1 \times 0.2$ $$= 0.2 mole = 0.2 mole$$ Final $0 0$ Therefore, the resulting solution becomes neutral. Now when 10 ml of the above solution is poured into a flask containing 2 mole HCl and made solution 100 ml with distilled water. Molarity of HCl = $$\frac{2}{100} \times 1000 = 20$$
Chemistry · Biomolecules · Single correct
Fat soluble vitamins are: A. Vitamin $\mathrm{B_1}$ B. Vitamin C C. Vitamin E D. Vitamin $\mathrm{B_{12}}$ E. Vitamin K Choose the correct answer from the options given below:
Answer: (d)
Vit D, E, K, A are fat soluble vitamins.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
\begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{LIST-I} & \multicolumn{2}{c|}{LIST-II} \\ \multicolumn{2}{|c|}{(Family)} & \multicolumn{2}{c|}{(Symbol of Element)} \\ \hline A. & Pnicogen (group 15) & I. & Ts \\ \hline B. & Chalcogen & II. & Og \\ \hline C. & Halogen & III. & Lv \\ \hline D. & Noble gas & IV. & Mc \\ \hline \end{tabular}
Answer: (b)
(A) Pnictogen $\Rightarrow \mathrm{Mc}$ (Moscovium), Atomic No. $= 115$ (B) Chalcogen $\Rightarrow \mathrm{Lv}$ (Livermorium), Atomic No. $= 116$ $(C)$ Halogen $\Rightarrow \mathrm{Ts}$ (Tennessine), Atomic No. $= 117$ (D) Noble gas $\Rightarrow \mathrm{Og}$ (Oganesson), Atomic No. $= 118$
Chemistry · Structure of Atom · Single correct
For electron in '2$\,$s' and '2$\,$p' orbitals, the orbital angular momentum values, respectively are:
Answer: (d)
Orbital angular momentum $= \sqrt{\ell(\ell + 1)} \frac{h}{2\pi}$. For 2 s orbital: $\ell = 0$. Orbital angular momentum $= 0$. For 2p orbital: $\ell = 1$. Orbital angular momentum $= \sqrt{1(1 + 2)} \frac{h}{2\pi}$. $= \sqrt{2} \frac{h}{2\pi}$.
Chemistry · Analytical Chemistry · Single correct
Compounds that should not be used as primary standards in titrimetric analysis are: A. $\mathrm{Na_2Cr_2O_7}$ B. Oxalic acid C. $\mathrm{NaOH}$ D. $\mathrm{FeSO_4 \cdot 6H_2O}$ E. Sodium tetraborate Choose the most appropriate answer from the options given below:
Answer: (d)
The primary standard is a highly pure stable compound with a known exact composition that can be accurately weighed and dissolved to create a solution of known concentration. NaOH is hygroscopic and can't be used. FeSO_4 $\cdot$ 6H_2O is unstable and can be easily oxidised. Na_2Cr_2O_7 is hygroscopic and can't be used.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The major product $(P)$ in the following reaction is :
Answer: (b)
The reaction begins with the deprotonation of the terminal alkyne by $\mathrm{KOH}$, forming an acetylide ion. This ion attacks the carbonyl carbon of the ketone, resulting in the formation of an alkoxide intermediate. The rate-determining step (rds) involves the rearrangement of this intermediate to form the final product.
Chemistry · Haloalkanes and Haloarenes · Single correct
In the following series of reactions identify the major products A $\&$ B respectively.
Answer: (b)
Chemistry · Electrochemistry · Single correct
The standard cell potential ($E^\circ_{cell}$) of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as 1.21 V. The standard half cell reduction potential for $O_2$ ($E^\circ_{O_2/H_2O}$) is 1.229 V. Choose the correct statement:
Answer: (a)
Given $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$. $$1.21 = 1.229 - E^\circ_{anode}$$ Fuel cell involves oxidation of methanol which will occur at anode and reduction of $\mathrm{O_2}$ will occur at cathode.
Chemistry · Co-ordination Compounds · Single correct
Identify the diamagnetic octahedral complex ions from below ; A. $\left[ \mathrm{Mn(CN)}_6 \right]^{3-}$ B. $\left[ \mathrm{Co(NH}_3)_6 \right]^{3+}$ C. $\left[ \mathrm{Fe(CN)}_6 \right]^{4-}$ D. $\left[ \mathrm{Co(H}_2\mathrm{O)}_3 \mathrm{F}_3 \right]$ Choose the correct answer from the options given below :
Answer: (d)
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is (Given : Aqueous tension at 300 K = 15 $\mathrm{mmHg}$)
Answer: (a)
Pressure of $\mathrm{N_2}$ gas evolved $= 715 - 15 = 700 \, \mathrm{mmHg} = \frac{700}{760} \, \mathrm{atm}$. Therefore, mole of $\mathrm{N_2}$ evolved $= \frac{PV}{RT} = \frac{700 \times 60 \times 10^{-3}}{760 \times 0.0821 \times 300} = 0.0022 \, mole$. Therefore, wt. $\%$ of nitrogen in compound $= \frac{wt. of nitrogen}{wt. of compound} \times 100 = \frac{0.063}{0.4} \times 100 = 15.71\%$.
Chemistry · Some Basic Concepts of Chemistry · Single correct
Mass of magnesium required to produce 220 $\,$ $\mathrm{mL}$ of hydrogen gas at STP on reaction with excess of dil. HCl is Given : Molar mass of Mg is 24 $\,$ $\mathrm{g \, mol^{-1}.}$
Answer: (a)
The reaction is given by $\($ $\mathrm{Mg}$ + 2$\mathrm{HCl}$ $\rightarrow$ $\mathrm{MgCl_2}$ + $\mathrm{H_2}$ $\)$. The volume of $\($ $\mathrm{H_2}$ $\)$ evolved is 220 ml. Mole of $\($ $\mathrm{H_2}$ = $\frac{220 \times 10^{-3}}{22.4}$ = $\)$ mole of $\($ $\mathrm{Mg}$ $\)$ used. Therefore, the mass of $\($ $\mathrm{Mg}$ $\)$ used is $$ \frac{220 \times 10^{-3}}{22.4} \times 24 $$ $$ = 235.7 \times 10^{-3} \mathrm{gm} $$ $$ = 235.7 \mathrm{mg} $$
Chemistry · Biomolecules · Single correct
Given below are two statements: Statement I: Wet cotton clothes made of cellulose based carbohydrate takes comparatively longer time to get dried than wet nylon polymer based clothes. Statement II: Intermolecular hydrogen bonding with water molecule is more in nylon-based clothes than in the case of cotton clothes. In the light of above statements, choose the Correct answer from the options given below
Answer: (b)
Cellulose derivative has more number of hydroxy groups, so more H-bonding is present with water in cellulose derivatives cotton cloths.
Chemistry · The d-and f-Block Elements · Single correct
Given below are two statements: Statement I: $\mathrm{CrO}_3$ is a stronger oxidizing agent than $\mathrm{MoO}_3$ Statement II: Cr(VI) is more stable than Mo(VI) In the light of the above statements, choose the correct answer from the options given below
Answer: (b)
Statement-I is true but statement II is false. $\mathrm{Cr(VI)}$ is less stable than $\mathrm{Mo(VI)}$. Hence, $\mathrm{CrO_3}$ easily reduce into $\mathrm{Cr^{+3}}$ as compared to $\mathrm{MoO_3}$ and show stronger oxidizing nature.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Given below are two statements: Statement I : Hyperconjugation is not a permanent effect. Statement II : In general, greater the number of alkyl groups attached to a positively charged C -atom, greater is the hyperconjugation interaction and stabilization of the cation. In the light of the above statements, choose the correct answer from the options given below
Answer: (c)
Hyper conjugation is a permanent effect because an external reagent is not required, so Statement-I is false and Statement-II is true. Because more alkyl group, more $\alpha - H$, so more hyperconjugation which results in more stability of carbocation.
Chemistry · Thermodynamics · Single correct
Given below are two statements: Statement I: When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the system until whole ice gets melted. Statement II: At melting point of ice, there is absorption of heat in order to overcome intermolecular forces of attraction within the molecules of water in ice and kinetic energy of molecules is not increased at melting point. In the light of the above statements, choose the correct answer from the options given below:
Answer: (c)
At melting point when ice melts, supplied heat is utilised to overcome intermolecular attraction within the molecules so temperature remains constant.
Chemistry · Amines · Single correct
The sequence from the following that would result in giving predominantly 3, 4, 5-Tribromoaniline is :
Answer: (c)
The reaction sequence starts with the nitration of aniline to form a nitroaniline. The first step involves bromination with excess $\mathrm{Br_2}$ in acetic acid, leading to the formation of 2,4,6-tribromoaniline. Next, diazotization occurs using $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ to form a diazonium salt. The Sandmeyer reaction with $\mathrm{CuBr}$ follows, resulting in the replacement of the diazonium group with a bromine atom. Finally, reduction with $\mathrm{SN}$ and $\mathrm{HCl}$ yields 3,4,5-tribromoaniline.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The correct orders among the following are Atomic radius : B < Al < Ga < In < Tl Electronegativity : Al < Ga < In < Tl < B Density : Tl < In < Ga < Al < B 1st Ionisation Energy : In < Al < Ga < Tl < B Choose the correct answer from the options given below :
Answer: (a)
\begin{tabular}{|l|c|c|c|c|c|} \hline & \textbf{B} & \textbf{Al} & \textbf{Ga} & \textbf{In} & \textbf{Tl} \\ \hline Atomic radius (pm) & 88 & 143 & 135 & 167 & 170 \\ \hline Electronegativity & 2.0 & 1.5 & 1.6 & 1.7 & 1.8 \\ \hline Density (g/cm$^3$) & 2.35 & 2.70 & 5.90 & 7.31 & 11.85 \\ \hline Ionisation Energy (kJ/mol) & 801 & 577 & 579 & 558 & 589 \\ \hline \end{tabular}
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
What is the correct IUPAC name of
Answer: (a)
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
Consider the following statements related to temperature dependence of rate constants. Identify the correct statements, A. The Arrhenius equation holds true only for an elementary homogenous reaction. B. The unit of A is same as that of k in Arrhenius equation. C. At a given temperature, a low activation energy means a fast reaction. D. A and Ea as used in Arrhenius equation depend on temperature. E. When Ea $\gg$ RT. A and Ea become interdependent. Choose the correct answer from the options given below :
Answer: (c)
Arrhenius equation holds true for elementary as well as complex reactions. Unit of $A$ is the same as the unit of $k$. Rate of reaction is high if activation energy is low, $A$ and $E_a$ are temperature independent.
Chemistry · Amines · Numerical
$X$ g of nitrobenzene on nitration gave $4.2$ g of m-dinitrobenzene. $X=$ ______ g. (nearest integer) [Given: molar mass (in $\mathrm{g\,mol^{-1}}$) C: 12, H: 1, O: 16, N: 14]
Answer: 3
Given $\mathrm{C_6H_5NO_2}$ with molecular weight $123$ and $\mathrm{C_6H_4N_2O_4}$ with molecular weight $168$. Therefore, $\frac{4.2}{168} = 0.025$ mol. Thus, required grams of nitrobenzene $$= 123 \times 0.025$$ $$= 3.075$$ Therefore, the nearest integer is $3$.
Chemistry · Thermodynamics · Numerical
A perfect gas (0.1 mol) having $\overline{C}_v = 1.50R$ (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is (−) ________ J (nearest integer) [ Given : $R = 0.082 \, $\mathrm{L \, atm \, K}^{-1}$ \, \mathrm{mol}^{-1}$ ]
Answer: 304
Given $$W_{1 \to 2} = 0$$ $$W_{2 \to 3} = -P \Delta V$$ $$= -3[2 - 1]$$ $$= -3 \, atm - \ell$$ $$W_{3 \to 4} = 0$$ Total work done $$= -3 \, atm - \ell$$ $$= -3 \times 101.3 \, Joule$$ $$= -304 \, Joule$$
Chemistry · Thermodynamics · Numerical
A sample of n-octane $(1.14\,\mathrm{g})$ was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is $5\,\mathrm{kJ\,K^{-1}}$. As a result of combustion reaction, the temperature of the calorimeter is increased by $5\,\mathrm{K}$. The magnitude of the heat of combustion of octane at constant volume is ______ $\mathrm{kJ\,mol^{-1}}$
Answer: 2500
Mole of octane $= \frac{1.14}{114} = 0.01 mole$ Heat evolved $= C \times \Delta T$ $= 5 \times 5 \, kJ$ $= 25 \, kJ$ Therefore, Magnitude of Heat of combustion $= \frac{25}{0.01} = 2500 \, kJ/mole$
Chemistry · The d-and f-Block Elements · Fill in the blank
Among, Sc, Mn, Co and Cu, identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is _____ BM (in nearest integer).
Answer: 4
\begin{tabular}{|l|c|c|c|c|} \hline & Sc & Mn & Co & Cu \\ \hline Enthalpy of Atomisation (kJ/mole) & 326 & 281 & 425 & 339 \\ \hline \end{tabular} Highest : Co $Co^{2+}=(Ar)\,3d^7$ $n=3$ $\mu=\sqrt{15}=3.87$ Nearest integer $=4$
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical
The total number of structural isomers possible for the substituted benzene derivatives with the molecular formula $\mathrm{C_9H_{12}}$ is .
Answer: 8