JEE Main 22 January 2025 Shift 1: 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 · Sets · Single correct
The number of non-empty equivalence relations on the set $\{1,2,3\}$ is:
Answer: (b)
Let $R$ be the required relation. $A = \{(1,1), (2,2), (3,3)\}$ (i) $|R| = 3$, when $R = A$ (ii) $|R| = 5$, e.g. $R = A \cup \{(1,2), (2,1)\}$ Number of $R$ can be $[3]$ (iii) $R = \{1,2,3\} \times \{1,2,3\}$ Ans. (5)
Maths · Applications of Integrals · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be a twice differentiable function such that $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$. If $f'(0) = 4a$ and $f$ satisfies $f''(x) - 3a f'(x) - f(x) = 0$, $a > 0$, then the area of the region $R = \{(x, y) \mid 0 \leq y \leq f(ax), 0 \leq x \leq 2\}$ is:
Answer: (a)
Given $f(x+y) = f(x) \cdot f(y)$. (1) $\Rightarrow f(x) = e^{\lambda x}$, $f'(0) = 4a$ $\Rightarrow f'(x) = \lambda e^{\lambda x} \Rightarrow \lambda = 4a$ So, $f(x) = e^{4ax}$ $f''(x) - 3af'(x) - f(x) = 0$ $\Rightarrow \lambda^2 - 3a\lambda - 1 = 0$ $\Rightarrow 16a^2 - 12a^2 - 1 = 0 \Rightarrow 4a^2 = 1 \Rightarrow a = \frac{1}{2}$ $F(x) = e^{2x}$ Area $= \int_0^2 e^x \, dx = e^2 - 1$
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let the triangle PQR be the image of the triangle with vertices (1,3), (3,1) and (2,4) in the line $x + 2y = 2$. If the centroid of $\triangle PQR$ is the point $(\alpha, \beta)$, then $15(\alpha - \beta)$ is equal to:
Answer: (d)
The centroid $G''(\alpha, \beta)$ of $\triangle PQR$ is the image of the centroid of the given triangle $P'Q'R'$. Centroid of $\triangle P'Q'R' = \left( \frac{1+3+2}{3}, \frac{3+1+4}{3} \right) = G' \left( 2, \frac{8}{3} \right)$. Image of $G \left( 2, \frac{8}{3} \right)$ with respect to the line $x + 2y = 2$ is $(\alpha, \beta)$. Then $$\frac{\alpha - 2}{1} = \frac{\beta - \frac{8}{3}}{2} = \frac{-2 \left( 2 + \frac{16}{3} - 2 \right)}{1 + 4}$$ Therefore, $$\frac{\alpha - 2}{1} = \frac{\beta - \frac{8}{3}}{2} = \frac{-32}{15}$$ Thus, $$\alpha = -\frac{2}{15}$$ and $$\beta = -\frac{8}{5}$$ Then $$15(\alpha - \beta) = 15 \left( -\frac{2}{15} + \frac{24}{15} \right) = 22$$
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $z_1$, $z_2$ and $z_3$ be three complex numbers on the circle $|z| = 1$ with $\arg(z_1) = \frac{-\pi}{4}$, $\arg(z_2) = 0$ and $\arg(z_3) = \frac{\pi}{4}$. If $\left| z_1 \overline{z_2} - z_2 \overline{z_3} + z_3 \overline{z_1} \right|^2 = \alpha + \beta \sqrt{2}$, $\alpha, \beta \in \mathbb{Z}$, then the value of $\alpha^2 + \beta^2$ is:
Answer: (b)
Given $|z| = 1$. For $\arg(z_1) = -\frac{\pi}{4}$, $\arg(z_2) = 0$, $\arg(z_3) = \frac{\pi}{4}$, we have: $$z_1 = |1| e^{-\frac{\pi}{4}i} = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$$ $$z_2 = 1 + 0i$$ $$z_3 = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$$ $$z_1 \overline{z_2} = 1 - i$$ $$z_2 \overline{z_3} = \frac{1 - i}{\sqrt{2}}$$ $$(1 + i)^2$$ $$z_3 \overline{z_1} = 2$$ $$z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1} = \sqrt{2} + i(1 - \sqrt{2})$$ $$|z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}|^2 = 5 - 2\sqrt{2}$$ $$\alpha = 5, \beta = -2$$ $$\alpha^2 + \beta^2 = 29$$
Maths · Inverse Trigonometric Functions · Single correct
Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16\left((\sec^{-1}x)^2 + (\csc^{-1}x)^2\right)$ is:
Answer: (b)
Given $16(\sec^{-1} x)^2 + (\csc^{-1} x)^2$. Let $\sec^{-1} x = a \in [0, \pi] - \left\{ \frac{\pi}{2} \right\}$. Then $\csc^{-1} x = \frac{\pi}{2} - a$. This becomes: $$= 16 \left[ a^2 + \left( \frac{\pi}{2} - a \right)^2 \right] = 16 \left[ 2a^2 - \pi a + \frac{\pi^2}{4} \right]$$ For $\max|_{a=\pi}$: $$= 16 \left[ 2\pi^2 - \pi^2 + \frac{\pi^2}{4} \right] = 20\pi^2$$ For $\min|_{a=\frac{\pi}{4}}$: $$= 16 \left[ 2 \times \frac{\pi^2}{16} - \frac{\pi^2}{4} \right] = 2\pi^2$$ Sum $= 22\pi^2$
Maths · Probability (Advanced) · Single correct
A coin is tossed three times. Let $X$ denote the number of times a tail follows a head. If $\mu$ and $\sigma^2$ denote the mean and variance of $X$, then the value of $64 \left( \mu + \sigma^2 \right)$ is:
Answer: (d)
The mean $\mu$ is calculated as $\mu = \sum x_i P_i = \frac{1}{2}$. The variance $\sigma^2$ is calculated as $\sigma^2 = \sum x_i^2 P_i - \mu^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$. Therefore, $64 (\mu + \sigma^2) = 64 \left[ \frac{1}{2} + \frac{1}{4} \right] = 64 \times \frac{3}{4} = 48$.
Maths · Sequences and Series · Single correct
Let $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive terms. If $a_1 a_5 = 28$ and $a_2 + a_4 = 29$, then $a_6$ is equal to
Answer: (d)
Given $a_1 \cdot a_5 = 28 \Rightarrow a \cdot ar^4 = 28 \Rightarrow a^2 r^4 = 28 \ldots (1)$ $a_2 + a_4 = 29 \Rightarrow ar + ar^3 = 29$ $\Rightarrow ar(1 + r^2) = 29$ $\Rightarrow a^2 r^2 (1 + r^2)^2 = (29)^2 \ldots (2)$ By Eq. (1) and (2) $$\frac{r^2}{(1 + r^2)^2} = \frac{28}{29 \times 29}$$ $$\Rightarrow \frac{r}{1 + r^2} = \frac{\sqrt{28}}{29} \Rightarrow r = \sqrt{28}$$ $$\therefore a^2 r^4 = 28 \Rightarrow a^2 \times (28)^2 = 28$$ $$\Rightarrow a = \frac{1}{\sqrt{28}}$$ $$\therefore a_6 = ar^5 = \frac{1}{\sqrt{28}} \times (28)^{\frac{5}{2}} = 784$$
Maths · Three Dimensional Geometry · Single correct
Let $L_1 : \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $L_2 : \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?
Answer: (a)
Given the lines: $$L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ $$L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$ Points on the lines are: $$P(2\lambda + 1, 3\lambda + 2, 4\lambda + 3)$$ $$Q(3\mu + 2, 4\mu + 4, 5\mu + 5)$$ Direction ratios of $PQ$ are $ $. The vector $PQ$ is given by: $$PQ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = -\hat{i} + 2\hat{j} - \hat{k}$$ Equating the direction ratios: $$2\lambda - 3\mu - 1 = -1$$ $$3\lambda - 4\mu - 2 = 2$$ $$4\lambda - 5\mu - 2 = -1$$ Solving these equations, we find: $$\lambda = 3, \mu = 6$$ Thus, the points are: $$P \left( 5, 13, 3 \right)$$ $$Q \left( 3, 10, 25 \right)$$ Direction ratios of $PQ$ are $ $. Therefore, the line equation is: $$\frac{y - 5}{1} = \frac{y - 3}{-2} = \frac{y - 13}{1}$$
Maths · Continuity and Differentiability · Single correct
The product of all solutions of the equation $e^{5(\log_e x)^2 + 3} = x^8, x > 0$, is:
Answer: (a)
Given $e^{5(\ln x)^2 + 3} = x^8$. Taking natural logarithm on both sides, we have: $$\ln e^{5(\ln x)^2 + 3} = \ln x^8$$ This simplifies to: $$5(\ln x)^2 + 3 = 8 \ln x$$ Let $\ln x = t$. Then: $$5t^2 - 8t + 3 = 0$$ The sum of the roots $t_1 + t_2 = \frac{8}{5}$. The product of the roots $\ln x_1 x_2 = \frac{8}{5}$. Thus, $x_1 x_2 = e^{8/5}$.
Maths · Sequences and Series · Single correct
If $\sum_{r=1}^{n} T_r = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$, then $\lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{1}{T_r} \right)$ is equal to :
Answer: (b)
Given $T_n = S_n - S_{n-1}$. Therefore, $T_n = \frac{1}{8} (2n - 1)(2n + 1)(2n + 3)$. Thus, $\frac{1}{T_n} = \frac{8}{(2n - 1)(2n + 1)(2n + 3)}$. The limit is $\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{T_r} = \lim_{n \to \infty} 8 \sum_{r=1}^{n} \frac{1}{(2n - 1)(2n + 1)(2n + 3)}$. This equals $\lim_{n \to \infty} \frac{8}{4} \sum \left( \frac{1}{(2n - 1)(2n + 1)} - \frac{1}{(2n + 1)(2n + 3)} \right)$. Therefore, $= \lim_{n \to \infty} 2 \left[ \left( \frac{1}{1.3} - \frac{1}{3.5} \right) + \left( \frac{1}{3.5} - \frac{1}{5.7} \right) + \cdots \right]$. This simplifies to $= \frac{2}{3}$.
Maths · Permutations and Combinations · Single correct
From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ' M ', is :
Answer: (a)
Selection of two letters before M and selection of two letters after M. $$= \binom{12}{2} \times \binom{13}{2} = 5148$$
Maths · Differential Equations · Single correct
Let $x = x(y)$ be the solution of the differential equation $y^2 \, dx + \left( x - \frac{1}{y} \right) \, dy = 0$. If $x(1) = 1$, then $x\left( \frac{1}{2} \right)$ is:
Answer: (c)
Given $y^2 dx + \left( x - \frac{1}{y} \right) dy = 0$. Rearrange to $y^2 dx = \left( \frac{1}{y} - x \right) dy$. This implies $\frac{y^2 dx}{dy} = \frac{1}{y} - x$. Rearrange to $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$. The integrating factor is $I.F. = e^{\int \frac{1}{y^2} dy} = e^{-\frac{1}{y}}$. Therefore, the solution is $xe^{-\frac{1}{y}} = \int e^{-\frac{1}{y}} \times \frac{1}{y^3} dy + C$. Let $-\frac{1}{y} = t$, then $\frac{1}{y^2} dy = dt$. This implies $xe^{-\frac{1}{y}} = -\int e^t dt + C$. Therefore, $xe^{-\frac{1}{y}} = -e^t (t - 1) + C$. This implies $xe^{-\frac{1}{y}} = -e^{-\frac{1}{y}} \left( -\frac{1}{y} - 1 \right) + C$. Given $x(1) = 1$, $e^{-1} = -e^{-1}(-2) + C$. This implies $C = -e^{-1}$. Therefore, $x = \frac{1}{y} - 1 - e^{-1 + \frac{1}{y}}$. Finally, $x\left( \frac{1}{2} \right) = 3 - e$.
Maths · Conic Sections · Single correct
Let the parabola $y = x^2 + px - 3$, meet the coordinate axes at the points $P$, $Q$ and $R$. If the circle $C$ with centre at $(-1, -1)$ passes through the points $P$, $Q$ and $R$, then the area of $\triangle PQR$ is:
Answer: (c)
Given $y = x^2 + px - 3$. Let $P(\alpha, 0)$, $Q(\beta, 0)$, $R(0, -3)$. Circle with centre $(-1, -1)$ is $(x + 1)^2 + (y + 1)^2 = r^2$. Passes through $(0, -3)$. $$1^2 + (-2)^2 = r^2$$ $$r^2 = 5$$ $$(x + 1)^2 + (y + 1)^2 = 5$$ Put $y = 0$. $$(x + 1)^2 = 5 - 1$$ $$(x + 1)^2 = 4$$ $$x + 1 = \pm 2$$ $$x = 1 or x = -3$$ Therefore, $P(1,0)$ and $Q(-3,0)$. Area of $\triangle PQR = \frac{1}{2} \begin{vmatrix} 1 & 0 & 1 \\ -3 & 0 & 1 \\ 0 & -3 & 1 \end{vmatrix} = 6$
Maths · Conic Sections · Single correct
A circle $C$ of radius 2 lies in the second quadrant and touches both the coordinate axes. Let $r$ be the radius of a circle that has centre at the point (2, 5) and intersects the circle $C$ at exactly two points. If the set of all possible values of $r$ is the interval $(\alpha, \beta)$, then $3\beta - 2\alpha$ is equal to:
Answer: (b)
The distance between $C_1$ and $C_2$ is calculated as follows: $$C_1C_2 = \sqrt{(2 + 2)^2 + (5 - 2)^2}$$ $$= \sqrt{16 + 9}$$ $$= 5$$ Given $r + 2 > 5$, we have: $$r > 3$$ And given $r < 5 + 2$, we have: $$r < 7$$ Therefore, $\alpha = 3$ and $\beta = 7$. Calculating $3\beta - 2\alpha$ gives: $$3\beta - 2\alpha = 3(7) - 2(3) = 21 - 6 = 15$$
Maths · Integrals · Single correct
Let for $f(x) = 7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$, $I_1 = \int_0^{\pi/4} f(x) \, dx$ and $I_2 = \int_0^{\pi/4} x f(x) \, dx$. Then $7I_1 + 12I_2$ is equal to:
Answer: (b)
Given $f(x) = 7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$. This can be rewritten as $7 \tan^6 x (1 + \tan^2 x) - 3 \tan^2 x (1 + \tan^2 x)$. Simplifying, we have $(7 \tan^6 x - 3 \tan^2 x) (1 + \tan^2 x)$. This further simplifies to $(7 \tan^6 x - 3 \tan^2 x) \sec^2 x$. Let $I_1 = \int_0^{\pi/4} f(x) dx = \int_0^{\pi/4} (7 \tan^6 x - 3 \tan^2 x) \sec^2 x \, dx$. Evaluating, we get: $$= \left( \frac{7 \tan^7 x}{7} - \frac{3 \tan^3 x}{3} \right) \bigg|_0^{\pi/4} = 1 - 1 = 0.$$ Now, let $I_2 = \int_0^{\pi/4} x f(x) dx = \int_0^{\pi/4} x (7 \tan^6 x - 3 \tan^2 x) \sec^2 x \, dx$. This becomes: $$= x (\tan^7 x - \tan^3 x) \bigg|_0^{\pi/4} - \int_0^{\pi/4} 1 \cdot (\tan^7 x - \tan^3 x) \, dx.$$ Simplifying further: $$= 0 - \int_0^{\pi/4} \tan^3 x (\tan^2 x - 1) (\tan^2 x + 1) \, dx.$$ This evaluates to: $$= \int_0^{\pi/4} (\tan^3 x - \tan^5 x) \sec^2 x \, dx = \frac{\tan^4 x}{4} - \frac{\tan^6 x}{6} \bigg|_0^{\pi/4}.$$ Finally, we have: $$= \frac{1}{12}.$$ Hence $7I_1 + 12I_2 = 1$.
Maths · Continuity and Differentiability · Single correct
Let $f(x)$ be a real differentiable function such that $f(0) = 1$ and $f(x+y) = f(x)f'(y) + f'(x)f(y)$ for all $x, y \in \mathbb{R}$. Then $\sum_{n=1}^{100} \log_e f(n)$ is equal to:
Answer: (a)
Given $f(x+y) = f(x) \cdot f(y) + f(x) \cdot f(y)$ for all $x, y \in \mathbb{R}$. And $f(0) = 1$. Now replace $x$ by zero and $y$ by zero, we get $$f(0) = f(0)f(0) + f(0)f(0)$$ $$1 = f(0) + f(0)$$ Therefore, $f'(0) = \frac{1}{2}$. Now replace $y$ by zero in equation (i), we get $$f(x) = \frac{1}{2} f(x) + f'(x)$$ or, $\frac{1}{2} f(x) = f'(x)$ then $f'(x) = \frac{1}{2}$ $f(x) = \frac{1}{2}$ Hence $\ln |f(x)| = \frac{x}{2} + c$ Put $x = 0$, we get $c = 0$ Therefore, $\ln |f(x)| = \frac{x}{2}$ Then $\sum_{n=1}^{100} \ln(f(\eta)) = \left( \frac{1}{2} + \frac{2}{2} + \frac{3}{2} + \ldots + \frac{100}{2} \right)$ $$= \frac{5050}{2} = 2525$$
Maths · Sets · Single correct
Let $A = \{1, 2, 3, \ldots, 10\}$ and $B = \left\{ \frac{m}{n} : m, n \in A, m < n and \gcd(m, n) = 1 \right\}$. Then $n(B)$ is equal to:
Answer: (b)
Given $A = \{1, 2, \ldots, 10\}$. Define $B = \left\{ \frac{m}{n} \mid m, n \in A, m < n, \gcd(m, n) = 1 \right\}$. Calculate $n(B)$. For $n = 2$, we have $\left\{ \frac{1}{2} \right\}$. For $n = 3$, we have $\left\{ \frac{1}{3}, \frac{2}{3} \right\}$. For $n = 4$, we have $\left\{ \frac{1}{4}, \frac{3}{4} \right\}$. For $n = 5$, we have $\left\{ \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \right\}$. For $n = 6$, we have $\left\{ \frac{1}{6}, \frac{5}{6} \right\}$. For $n = 7$, we have $\left\{ \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7} \right\}$. For $n = 8$, we have $\left\{ \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8} \right\}$. For $n = 9$, we have $\left\{ \frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9} \right\}$. For $n = 10$, we have $\left\{ \frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10} \right\}$. Thus, $n(B) = 31$.
Maths · Applications of Integrals · Single correct
The area of the region, inside the circle $(x - 2\sqrt{3})^2 + y^2 = 12$ and outside the parabola $y^2 = 2\sqrt{3}x$ is:
Answer: (b)
Required area is given by $$2 \int_{0}^{2\sqrt{3}} \left( \sqrt{4\sqrt{3}x - x^2} - \sqrt{2\sqrt{3}x} \right) \, dx$$ This simplifies to $$= 2 \int_{0}^{2\sqrt{3}} \left( \sqrt{12 - (x - 2\sqrt{3})^2} - \sqrt{2\sqrt{3}x} \right) \, dx$$ Evaluating the integral, we have $$= 2 \left[ \frac{x - 2\sqrt{3}}{2} \sqrt{12 - (x - 2\sqrt{3})^2} + \frac{12}{2} \sin^{-1} \left( \frac{x - 2\sqrt{3}}{2\sqrt{3}} \right) - \frac{\sqrt{2\sqrt{3}x^3}}{3/2} \right]_{0}^{2\sqrt{3}}$$ This results in $$= 2 \{ 3\pi - 8 \}$$ Finally, the area is $$= 6\pi - 16 sq. units.$$
Maths · Probability · Single correct
Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac{m}{n}$, where $\gcd(m, n) = 1$, then $m + n$ is equal to:
Answer: (b)
Bag contains 4 white and 6 black balls. $A$: first ball selected is black. $B$: Second ball is also black. $$P\left( \frac{A}{B} \right) = \frac{\frac{6}{10} \times \frac{5}{9}}{\frac{4}{10} \times \frac{6}{9} + \frac{6}{10} \times \frac{5}{9}} = \frac{30}{24 + 30}$$ $$= \frac{30}{54} = \frac{5}{9}$$ $m + n = 5 + 9 = 14$
Maths · Conic Sections · Single correct
Let the foci of a hyperbola be (1, 14) and (1, -12). If it passes through the point (1, 6), then the length of its latus-rectum is:
Answer: (d)
Given $be = 13$, $b = 5$. $a^2 = b^2 (e^2 - 1)$ $= b^2 e^2 - b^2$ $= 169 - 25 = 144$ The length of the latus rectum $\ell (LR) = \frac{2a^2}{b} = \frac{2 \times 144}{5} = \frac{288}{5}$.
Maths · Continuity and Differentiability · Numerical
Let the function, $$f(x) = \begin{cases} -3ax^2 - 2, & x 1$, $b \in \mathbb{R}$. If the area of the region enclosed by $y = f(x)$ and the line $y = -20$ is $\alpha + \beta \sqrt{3}$, $\alpha, \beta \in \mathbb{Z}$, then the value of $\alpha + \beta$ is
Answer: 34
Given that $f(x)$ is continuous and differentiable. At $x = 1$, $LHL = RHL$, $LHD = RHD$ $$-3a - 2 = a^2 + b, -6a = b$$ $$a = 2; b = -12$$ $$f(x) = \begin{cases} -6x^2 - 2, & x < 1 \\ 4 - 12x, & x \geq 1 \end{cases}$$ Area $= \int_{-\sqrt{3}}^1 (-6x^2 - 2 + 20) \, dx + \int_1^2 (4 - 12x + 20) \, dx$ $$= 16 + 12\sqrt{3} + 6 = 22 + 12\sqrt{3}$$ Therefore, $\alpha + \beta = 34$
Maths · Binomial Theorem · Fill in the blank
If $\sum_{r=0}^{5} \frac{{^{11}C_{2r+1}}}{2r+2} = \frac{m}{n}, \gcd(m, n) = 1$, then $m - n$ is equal to
Answer: 2035
(1 + x)^{11} = $\binom{11}{0}$ + $\binom{11}{1}$x + $\binom{11}{2}$x^2 + $\cdots$ + $\binom{11}{11}$x^{11} $\int$_0^1 (1 + x)^{11} $\,$ dx = $\int$_0^1 ($\binom{11}{0}$ + $\binom{11}{1}$x + $\binom{11}{2}$x^2 + $\cdots$ + $\binom{11}{11}$x^{11}) $\,$ dx $\frac{(1-x)^{12}}{12}$ $\bigg$|_0^1 = $\left$[ $\binom{11}{0}$x + $\frac{\binom{11}{1}x^2}{2}$ + $\frac{\binom{11}{2}x^3}{3}$ + $\cdots$ + $\frac{\binom{11}{11}x^{12}}{12}$ $\right$]_0^1 2^{12} - 1 = $\binom{11}{0}$ + $\frac{\binom{11}{1}}{2}$ + $\frac{\binom{11}{2}}{3}$ + $\cdots$ + $\frac{\binom{11}{11}}{12}$ $\cdots$ (1) Now, $\int$_{-1}^0 (1 + x)^{11} $\,$ dx = $\int$_{-1}^0 ($\binom{11}{0}$ + $\binom{11}{1}$x + $\binom{11}{2}$x^2 + $\cdots$ + $\binom{11}{11}$x^{11}) $\,$ dx $\frac{(1+x)^{12}}{12}$ $\bigg$|_{-1}^0 = $\left$[ $\binom{11}{0}$x + $\frac{\binom{11}{1}x^2}{2}$ + $\frac{\binom{11}{2}x^3}{3}$ + $\cdots$ + $\frac{\binom{11}{11}x^{12}}{12}$ $\right$]_{-1}^0 $\frac{1}{12}$ = $\binom{11}{0}$ - $\frac{\binom{11}{1}}{2}$ + $\frac{\binom{11}{2}}{3}$ $\cdots$ (1) - (2) = $\frac{2^{12} - 2}{12}$ = 2 $\left$[ $\frac{\binom{11}{1}}{2}$ + $\frac{\binom{11}{3}}{4}$ + $\cdots$ $\right$] $\Rightarrow$ $\sum$_{r=0}^5 $\frac{\binom{11}{2r+1}}{2r+2}$ = $\frac{2^{11} - 1}{12}$ = $\frac{2047}{12}$ = $\frac{m}{n}$ = 2047 - 12 = 2035
Maths · Determinants · Fill in the blank
Let $A$ be a square matrix of order 3 such that $\det(A) = -2$ and $\det(3 \mathrm{adj}(-6 \mathrm{adj}(3A))) = 2^{m+n} \cdot 3^{mn}$, $m > n$. Then $4m + 2n$ is equal to ______.
Answer: 34
As $A adj A = |A| I$, $\det(\lambda A) = \lambda^n \det A$. $$\det(3 adj(-6 adj(3A))) = 3^3 \det(adj(-6 adj(3A)))$$ $$= 3^3 (-6 adj(3A))^2$$ $$= 3^3 (-6)^6 |3A|^4$$ $$= 3^9 2^6 \cdot 3^{12} \cdot (-2)^4$$ $$= 3^{21} \cdot 2^{10}$$ Now comparing with given condition $$2^{m+n} 3^{mn} = 2^{10} \cdot 3^{21}$$ $$m+n = 10, \; mn = 21$$ $$\Rightarrow \; m = 7, \; n = 3 \; (m > n)$$ $$\therefore \; 4m + 2n = 28 + 6 = 34$$
Maths · Three Dimensional Geometry · Numerical
Let $L_1 : \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$ and $L_2 : \frac{x-2}{2} = \frac{y}{0} = \frac{z+4}{\alpha}$, $\alpha \in \mathbb{R}$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1, 1, -1)$ on $L_2$, then the value of $26\alpha(PB)^2$ is ________
Answer: 216
Point $B$ $$(3\lambda + 1, -\lambda + 1, -1) \equiv (2\mu + 2, 0, \alpha \mu - 4)$$ $$3\lambda + 1 = 2\mu + 2$$ $$-\lambda + 1 = 0$$ $$-1 = \alpha \mu - 4$$ $$\lambda = 1, \mu = 1, \alpha = 3$$ $$B(4, 0, -1)$$ Let Point $P'$ is $(2\delta + 2, 0, 3\delta - 4)$ Dr's of $AP$ $ $ $$AP \perp L_2 \Rightarrow \delta = \frac{7}{13}$$ $$P \left( \frac{40}{13}, 0, \frac{-31}{13} \right)$$ $$\therefore 26\alpha (PB)^2 = 26 \times 3 \times \left( \frac{144}{169} + \frac{324}{169} \right)$$ $$= 216$$
Maths · Vector Algebra · Numerical
Let $\vec{c}$ be the projection vector of $\vec{b} = \lambda \hat{i} + 4 \hat{k}, \lambda > 0$, on the vector $\vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k}$. If $|\vec{a} + \vec{c}| = 7$, then the area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is _______
Answer: 16
Given $\vec{c} = \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{b}|} \right) \frac{\vec{a}}{|\vec{a}|} = \left( \frac{\lambda + 8}{9} \right) (\hat{i} + 2\hat{j} + 2\hat{k})$. The magnitude $|\vec{a} + \vec{c}| = 7$. This implies: $$\left( \frac{\lambda + 8}{9} + 1 \right) \hat{i} + \left( \frac{2(\lambda + 8)}{9} + 2 \right) \hat{j} + \left( \frac{2(\lambda + 8)}{9} + 2 \right) \hat{k} \bigg| = 7$$ Squaring both sides: $$\left( \frac{\lambda + 8}{9} + 1 \right)^2 + \left( \frac{2(\lambda + 8)}{9} + 2 \right)^2 + \left( \frac{2(\lambda + 8)}{9} + 2 \right)^2 = 49$$ Solving gives $\lambda = 4$. Therefore, $\vec{c} = \frac{4}{3} \hat{i} + \frac{8}{3} \hat{j} + \frac{8}{3} \hat{k}$. The area of the parallelogram is given by: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{4}{3} & \frac{8}{3} & \frac{8}{3} \\ 4 & 8 & 4 \end{vmatrix} = 16$$
Physics · Experimental Physics · Single correct
Given below are two statements: Statement I: In a vernier callipers, one vernier scale division is always smaller than one main scale division. Statement II: The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions. In the light of the above statements, choose the correct answer from the options given below.
Answer: (a)
In general one vernier scale division is smaller than one main scale division but in some modified cases it may be not correct. Also least count is given by one main scale division / number of vernier scale division for normal vernier calliper.
Physics · Electric Charges and Fields · Single correct
A line charge of length $'\frac{a}{2}'$ is kept at the center of an edge $BC$ of a cube $ABCDEFGH$ having edge length $'a'$ as shown in the figure. If the density of line charge is $\lambda$ C per unit length, then the total electric flux through all the faces of the cube will be _______. (Take, $\varepsilon_0$ as the free space permittivity)
Answer: (d)
Charge of the line charge $= \frac{a \lambda}{2}$ Portion of wire inside cube $= \frac{1}{4}$ Therefore, $q_{en} = \frac{1}{4} \left( \frac{a \lambda}{2} \right) = \frac{a \lambda}{8}$ $$\phi = \frac{q_{en}}{\varepsilon_0} = \frac{a \lambda}{8 \varepsilon_0}$$
Physics · Current Electricity · Single correct
Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $R_p = 1\Omega$ as shown in the figure. An external resistance of $R_e = 2\Omega$ is connected via the sliding contact.
Answer: (d)
The equivalent resistance $R_{eq}$ is calculated as follows: $$R_{eq} = 0.5 + \frac{0.5 \times 2}{2 + 0.5} = \left( \frac{5}{10} + \frac{10}{25} \right) \, \Omega$$ Simplifying, we have: $$= \frac{45}{50} = \frac{9}{10} = 0.9$$ Therefore, the current $i$ is: $$i = \frac{0.9}{0.9} = 1 \, \mathrm{A}$$
Physics · Wave Optics · Single correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion-(A): If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer. Reason-(R): The speed of light reduces in an optically denser medium than air while its frequency does not change. In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (b)
Given, $\beta$ (fringe width) $=\dfrac{\lambda D}{d}$ In a denser medium, $\lambda \downarrow \Rightarrow \beta \downarrow$ $\Rightarrow$ fringes come closer. Also, $\mu=\dfrac{c}{v}$ $\Rightarrow v=\dfrac{c}{\mu}$ Frequency remains the same. $\Rightarrow \mu=\dfrac{\lambda_{\mathrm{vac}}\,f}{\lambda_{\mathrm{med}}\,f}$ $\Rightarrow \lambda_{\mathrm{med}}=\dfrac{\lambda_{\mathrm{vac}}}{\mu}$.
Physics · Thermal Properties of Matter · Single correct
Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K. If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding temperature to be negligible),
Answer: (b)
Given $\($ $\frac{d \theta}{dt}$ = $\sigma$ $\varepsilon$ A T^4 $\Rightarrow$ P $\propto$ A T^4 $\)$. $\($ P_{smaller} = (0.2)^2 $\times$ 800^4 $\)$ $\($ P_{larger} = (0.8)^2 $\times$ 400^4 $\)$ $\($ $\frac{1}{16}$ $\times$ 16 = 1 $\)$ Therefore, $\($ P_{larger} = P_{smaller} $\)$.
Physics · Thermal Properties of Matter · Single correct
An amount of ice of mass $10^{-3} \, \mathrm{kg}$ and temperature $-10^\circ \mathrm{C}$ is transformed to vapour of temperature $110^\circ \mathrm{C}$ by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice $= 2100 \, \mathrm{Jkg^{-1} \, K^{-1}}$, specific heat of water $= 4180 \, \mathrm{Jkg^{-1} \, K^{-1}}$, specific heat of steam $= 1920 \, \mathrm{Jkg^{-1} \, K^{-1}}$, Latent heat of ice $= 3.35 \times 10^5 \, \mathrm{Jkg^{-1}}$ and Latent heat of steam $= 2.25 \times 10^6 \, \mathrm{Jkg^{-1}}$)
Answer: (a)
Given: $$\Delta Q_1 = m \times S_1 \times \Delta T = 10^{-3} \times 2100 \times 10 = 21 \, \mathrm{J}$$ $$\Delta Q_2 = m \times L_f = 10^{-3} \times 3.35 \times 10^5 = 335 \, \mathrm{J}$$ $$\Delta Q_3 = m \times S_w \times \Delta T = 10^{-3} \times 4180 \times 100 = 418 \, \mathrm{J}$$ $$\Delta Q_4 = m \times L_v = 10^{-3} \times 2.25 \times 10^6 = 2250 \, \mathrm{J}$$ $$\Delta Q_5 = m \times S_v \times \Delta T = 10^{-3} \times 1920 \times 10 = 19.2 \, \mathrm{J}$$ $$\Delta Q_{net} = 3043.2 \, \mathrm{J}$$
Physics · Dual Nature of Radiation and Matter · Single correct
An electron in the ground state of the hydrogen atom has the orbital radius of $5.3 \times 10^{-11} \, \mathrm{m}$ while that for the electron in third excited state is $8.48 \times 10^{-10} \, \mathrm{m}$. The ratio of the de Broglie wavelengths of electron in the excited state to that in the ground state is
Answer: (b)
Given $\lambda = \frac{h}{mv}$. $mvr = \frac{nh}{2\pi}$ $mv = \frac{2\pi nh}{2\pi r}$ $\lambda = \frac{2\pi rh}{nh} = \frac{r}{n}$ $\lambda \propto \frac{r}{n}$ $\frac{\lambda_1}{\lambda_4} = \frac{r_1 n_4}{n_1 r_4} = \frac{5.3 \times 10^{-11} \times 4}{1 \times 84.8 \times 10^{-11}}$ $\frac{\lambda_1}{\lambda_4} = \frac{1}{4}$
Physics · Ray Optics and Optical Instruments · Single correct
In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to $|R_1|$ and $|R_2|$, i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is
Answer: (d)
The equivalent power is given by $p_{eq} = p_1 + p_2 + p_3$. For $p_1$, we have: $$p_1 = \left( \frac{4}{3} - 1 \right) \left( 1 \over \infty - \frac{1}{|R_1|} \right)$$ Simplifying, we get: $$p_1 = \left( \frac{1}{3|R_1|} \right)$$ For $p_2$, we have: $$p_2 = \left( \frac{1}{2} \right) \left( 1 \over -|R_1| - \frac{1}{|R_2|} \right)$$ Simplifying, we get: $$p_2 = \frac{1}{2} \left( \frac{1}{|R_2|} - \frac{1}{|R_1|} \right)$$ For $p_3$, we have: $$p_3 = \left( \frac{1}{3} \right) \left( 1 \over -|R_2| - \frac{1}{\infty} \right) = -\frac{1}{3|R_2|}$$ Thus, the equivalent power is: $$p_{eq} = \frac{1}{3} \left( \frac{1}{|R_1|} \right) - \frac{1}{2} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)$$ Simplifying further: $$p_{eq} = -\frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)$$
Physics · Electric Charges and Fields · Single correct
An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity $10^6 \, \mathrm{m/s}$. If the magnitude of the electric field between the plates is $9.1 \, \mathrm{V/cm}$, then the vertical component of velocity of electron is (mass of electron $= 9.1 \times 10^{-31} \, \mathrm{kg}$ and charge of electron $= 1.6 \times 10^{-19} \, \mathrm{C}$)
Answer: (c)
Given $t = \frac{\ell}{V_x} = \frac{10 \times 10^{-2}}{10^6} = 10^{-7}$. Therefore, $V_y = u_y + a_y t$. Since $V_y = 0 + \frac{eE}{m} \times 10^{-7}$, we have $V_y = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \times 9.1 \times 10^{-2} \times 10^{-7}$. Thus, $V_y = 16 \times 10^6$.
Physics · Current Electricity · Single correct
Which of the following resistivity ( $\rho$ ) v/s temperature (T) curves is most suitable to be used in wire bound standard resistors?
Answer: (d)
Resistivity is independent of temperature for wire bound resistors.
Physics · Waves · Single correct
A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities $\rho_1$ and $\rho_2'$, respectively. The frequency of $9^{th}$ harmonic of closed tube is identical with $4^{th}$ harmonic of open tube. If the length of the closed tube is $10 \, \mathrm{cm}$ and the density ratio of the gases is $\rho_1 : \rho_2 = 1 : 16$, then the length of the open tube is :
Answer: (d)
The 9th harmonic of a closed pipe is given by $9 V_1 = 4 \ell_1$. The 4th harmonic of an open pipe is given by $2 V_2 = \ell_2$. Therefore, $$\frac{9}{4 \ell_1} \sqrt{\frac{B}{\rho_1}} = \frac{2}{\ell_2} \sqrt{\frac{B}{\rho_2}} \implies \frac{\ell_2}{\ell_1} = \frac{8}{9} \sqrt{\frac{\rho_1}{\rho_2}}.$$ Thus, $$\ell_2 = \ell_1 \times \frac{8}{9} \times \frac{1}{4} = \frac{20}{9} \, cm.$$
Physics · System of Particles and Rotational Motion · Single correct
A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius $R/2$ is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.
Answer: (d)
Step 1: Moment of Inertia of the Original Disc The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about its central axis (perpendicular to its plane) is given by: $$I_{original} = \frac{1}{2} M R^2$$ Step 2: Moment of Inertia of the Removed Part The removed part is a smaller disc of radius $R/2$. Since, the original disc has uniform mass distribution, the mass of the smaller disc (proportional to its area) is: $$M_{removed} = M \times \frac{\pi (R/2)^2}{\pi R^2} = M \times \frac{1}{4} = \frac{M}{4}$$ The moment of inertia of a smaller disc about its own center is: $$I_{removed, center} = \frac{1}{2} M_{removed} \left( \frac{R}{2} \right)^2$$ $$I_{removed, center} = \frac{1}{2} \times \frac{M}{4} \times \frac{R^2}{4} = \frac{1}{32} M R^2$$ $$I_{removed} = I_{removed, center} + M_{removed} d^2$$ $$I_{removed} = \frac{1}{32} M R^2 + \left( \frac{M}{4} \times \frac{R^2}{4} \right)$$ $$I_{removed} = \frac{1}{32} M R^2 + \frac{1}{16} M R^2$$ $$I_{removed} = \frac{1}{32} M R^2 + \frac{2}{32} M R^2 = \frac{3}{32} M R^2$$ Step 3: Moment of Inertia of the Remaining Part The moment of inertia of the remaining part is: $$I_{remaining} = I_{original} - I_{removed}$$ $$I_{remaining} = \frac{1}{2} M R^2 - \frac{3}{32} M R^2$$ $$I_{remaining} = \frac{16}{32} M R^2 - \frac{3}{32} M R^2$$ $$I_{remaining} = \frac{13}{32} M R^2$$
Physics · Gravitation · Single correct
A small point of mass $m$ is placed at a distance $2R$ from the centre $'O'$ of a big uniform solid sphere of mass $M$ and radius $R$. The gravitational force on '$m$' due to $M$ is $F_1$. A spherical part of radius $R/3$ is removed from the big sphere as shown in the figure and the gravitational force on $m$ due to remaining part of $M$ is found to be $F_2$. The value of ratio $F_1 : F_2$ is
Answer: (a)
Given $$F_1 = \frac{GMm}{(2R)^2} \cdots (1)$$ $$F_2 = \frac{GMm}{(2R)^2} - \left( G \left( \frac{M}{27} \right) m \left( \frac{4R}{3} \right)^2 \right)$$ $$F_2 = \frac{11}{48} \frac{GMm}{R^2} \cdots (2)$$ The ratio $F_1 : F_2 = 12 : 11$.
Physics · Dual Nature of Radiation and Matter · Single correct
The work functions of cesium (Cs) and lithium (Li) metals are 1.9 $\mathrm{eV}$ and 2.5 $\mathrm{eV}$, respectively. If we incident a light of wavelength 550 $\mathrm{nm}$ on these two metal surfaces, then photo-electric effect is possible for the case of
Answer: (c)
Step 1: Calculate the Energy of the Incident Photon. The energy of a photon is given by the equation: $E = \frac{hc}{\lambda}$ where: $h = 6.626 \times 10^{-34} \, \mathrm{J \cdot s}$ (Planck's constant), $c = 3.0 \times 10^8 \, \mathrm{m/s}$ (speed of light), $\lambda = 550 \, \mathrm{nm} = 550 \times 10^{-9} \, \mathrm{m}$. First, calculate the photon energy in joules: $$E = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{550 \times 10^{-9}}$$ $$E = \frac{1.9878 \times 10^{-25}}{550 \times 10^{-9}}$$ $$E = 3.615 \times 10^{-19} \, \mathrm{J}$$ Convert this to electron volts (eV) using $1 \, \mathrm{eV} = 1.6 \times 10^{-19} \, \mathrm{J}$: $$E = \frac{3.615 \times 10^{-19}}{1.6 \times 10^{-19}}$$ $$E \approx 2.26 \, \mathrm{eV}$$ Step 2: Compare Photon Energy with Work Functions. Cesium ($\phi_{\mathrm{Cs}} = 1.9 \, \mathrm{eV}$). Since $E_{\mathrm{photon}} = 2.26 \, \mathrm{eV}$ is greater than $\phi_{\mathrm{Cs}} = 1.9 \, \mathrm{eV}$, photoelectric emission occurs. Lithium ($\phi_{\mathrm{Li}} = 2.5 \, \mathrm{eV}$). Since $E_{\mathrm{photon}} = 2.26 \, \mathrm{eV}$ is less than $\phi_{\mathrm{Li}} = 2.5 \, \mathrm{eV}$, photoelectric emission does not occur. Conclusion: Photoelectric effect is possible only for Cesium (Cs), but not for Lithium (Li).
Physics · Physical World, Units and Measurements · Single correct
If $B$ is magnetic field and $\mu_0$ is permeability of free space, then the dimensions of $(B/\mu_0)$ is
Answer: (c)
For a current carrying loop at centre $$B = \frac{\mu_0 i}{2R}$$ Therefore, $$\frac{B}{\mu_0} \equiv \frac{i}{R} \equiv \left[ AL^{-1} \right]$$
Physics · Work, Energy and Power · Single correct
A bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity $v_0$ at the point 'A', the string becomes slack when, the bob reaches at the point 'D'. The ratio of the kinetic energy of the bob at the points $B$ and $C$ is ______.
Answer: (b)
Given $\($ $\frac{1}{2}$ mv_A^2 = $\frac{1}{2}$ mv_B^2 + mgh $\)$ $\($ $\Rightarrow$ $\frac{1}{2}$ m(5g $\ell$) = $\frac{1}{2}$ mv_B^2 + mg $\frac{\ell}{2}$ $\)$ $\($ $\Rightarrow$ $\frac{5mg\ell}{2}$ - $\frac{mg\ell}{2}$ = $\mathrm{KE}$_B $\)$ $\($ $\Rightarrow$ $\mathrm{KE}$_B = 2mg$\ell$ $\)$ $\($ $\frac{1}{2}$ mv_C^2 = $\frac{1}{2}$ mv_D^2 + mg $\frac{\ell}{2}$ $\)$ $\($ $\Rightarrow$ $\mathrm{KE}$_C = $\frac{1}{2}$ mg$\ell$ + mg $\frac{\ell}{2}$ = mg$\ell$ $\)$ $\($ $\Rightarrow$ $\frac{\mathrm{KE}_B}{\mathrm{KE}_C}$ = 2 $\)$
Physics · Current Electricity · Single correct
Given below are two statements: Statement-I: The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs. Statement-II: The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries. In the light of the above statements, choose the correct answer from the options given below.
Answer: (b)
In parallel connections $$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}$$ $$E_{eq} = \frac{E_1}{r_1} + \frac{E_2}{r_2}$$ If $E_1 = E_2$ and $r_1 = r_2$, then $E_{eq} = E_1 = E_2$. Therefore, Statement I is false. $r_{eq}$ is less than both $r_1$ and $r_2$. Therefore, Statement II is true.
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Which of the following circuits represents a forward biased diode? Choose the correct answer from the options given below:
Answer: (d)
For forward bias, the potential of the $p$ side should be higher than the $n$ side.
Physics · Electrostatic Potential and Capacitance · Single correct
A parallel-plate capacitor of capacitance $40 \, \mu \mathrm{F}$ is connected to a $100 \, \mathrm{V}$ power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant $K = 2$. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are
Answer: (a)
Given $\Delta q = (KC - C)V$ $$= 40 \times 10^{-6} \times 100$$ $$= 4000 \times 10^{-3} = 4 \, \mathrm{mC}$$ $\Delta U = \frac{1}{2} C'V^2 - \frac{1}{2} CV^2 = \frac{1}{2} (K - 1)CV^2$ $$= \frac{1}{2} CV^2 (2 - 1)$$ $$= \frac{1}{2} CV^2 = \frac{1}{2} \times 40 \times 10^{-6} \times 10000$$ $$= 0.2 \, \mathrm{J}$$
Physics · Ray Optics and Optical Instruments · Single correct
Given is a thin convex lens of glass (refractive index $\mu$) and each side having radius of curvature $R$. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself?
Answer: (d)
Given the equation: $$ -\frac{1}{f_{eq}} = \frac{2}{f_l} - \frac{1}{f_m} $$ Substituting the values, we have: $$ = 2(\mu - 1) \frac{2}{R} + \frac{2}{R} $$ Simplifying further: $$ -\frac{1}{f_{eq}} = \frac{2(2\mu - 1)}{R} $$ Thus, $$ f_{eq} = -\frac{R}{2(2\mu - 1)} $$ For a concave mirror, the object should be at $2f$ for the image to be at the same point. Therefore, the distance is: $$ Distance = \frac{R}{(2\mu - 1)} $$
Physics · Mechanical Properties of Fluids · Numerical
Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is _______.
Answer: 4
Given $$R = \frac{R_2 R_1}{R_2 - R_1} = \frac{4 \times 2}{2} = 4 \, \mathrm{cm}$$
Physics · Ray Optics and Optical Instruments · Numerical
The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature $R = 2 \, \mathrm{m}$. Another car approaches him from behind with a uniform speed of $90 \, \mathrm{km/hr}$. When the car is at a distance of $24 \, \mathrm{m}$ from him, the magnitude of the acceleration of the image of the car in the side view mirror is ' $a$ '. The value of $100 \, a$ is ____ $\mathrm{m/s^2}$.
Answer: 8
Image distance, $v = \frac{uf}{u-f} = \frac{-24 \cdot 1}{-24 - 1} = \frac{24}{25}$. Magnification, $m = -\frac{v}{u} = -\frac{24}{25(-24)} = \frac{1}{25}$. Mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. Differentiating with respect to time gives, $$-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$$ here, $\frac{dv}{dt} = v_1; \frac{du}{dt} = v_0$. $$\Rightarrow v_1 = -\frac{v^2}{u^2} v_0 = -\frac{24^2}{25} \times 25 = -\frac{1}{25}$$ Again, differentiating with respect to time gives, $$\Rightarrow a_I = -2 \frac{vv'u - vu'v}{u^2} v_0 = -2 \frac{vv_1u - v v_0}{u^2}$$ $$\Rightarrow a_I = -2 \frac{24}{25} - \frac{1}{25} - 24 - \frac{24}{25} 25$$ $$\Rightarrow a_I = -2 \frac{-24^2}{25}$$ $$\Rightarrow a_I = -\frac{2}{25} \, \mathrm{ms^{-2}}$$ Thus, $100 a_I = 100 \times \frac{2}{25} = 8$
Physics · Thermal Properties of Matter · Numerical
Three conductors of same length having thermal conductivity $k_1$, $k_2$ and $k_3$ are connected as shown in figure. Area of cross sections of $1^{st}$ and $2^{nd}$ conductor are same and for $3^{rd}$ conductor it is double of the $1^{st}$ conductor. The temperatures are given in the figure. In steady state condition, the value of $\theta$ is ______ $^{\circ}\mathrm{C}$. (Given : $k_1 = 60\,\mathrm{J}\,\mathrm{s}^{-1}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$, $k_2 = 120\,\mathrm{J}\,\mathrm{s}^{-1}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$, $k_3 = 135\,\mathrm{J}\,\mathrm{s}^{-1}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$)
Answer: 40
Given the resistances: $$R_1 = \frac{2L}{K_1 A}$$ $$R_2 = \frac{2L}{K_2 A}$$ $$R_3 = \frac{L}{K_3 A}$$ Using the equation: $$\frac{\theta - 100}{\frac{R_1 R_2}{R_1 + R_2}} + \frac{\theta - 0}{R_3} = 0$$ Solving for $\theta$, we find: $$\theta = 40$$
Physics · System of Particles and Rotational Motion · Numerical
The position vectors of two 1 kg particles, (A) and (B), are given by $\vec{r}_A = \left( \alpha_1 t^2 \hat{i} + \alpha_2 t \hat{j} + \alpha_3 t \hat{k} \right) \, \mathrm{m}$ and $\vec{r}_B = \left( \beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t \hat{k} \right) \, \mathrm{m}$, respectively; $(\alpha_1 = 1 \, \mathrm{m/s^2}, \alpha_2 = 3 \, \mathrm{m/s}, \alpha_3 = 2 \, \mathrm{m/s}, \beta_1 = 2 \, \mathrm{m/s}, \beta_2 = -1 \, \mathrm{m/s^2}, \beta_3 = 4p \, \mathrm{m/s})$, where $t$ is time, $n$ and $p$ are constants. At $t = 1 \, \mathrm{s}$, $|\vec{V}_A| = |\vec{V}_B|$ and velocities $\vec{V}_A$ and $\vec{V}_B$ of the particles are orthogonal to each other. At $t = 1 \, \mathrm{s}$, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is $\sqrt{L} \, \mathrm{kg \, m^2 \, s^{-1}}$. The value of $L$ is ________.
Answer: 90
At $t = 1$. $$r_{AB} = -1\hat{i} + (3n + 1)\hat{j} + (2 - 4p)\hat{k}$$ At $t = 1$: $$v_A = 2\hat{i} + 3n\hat{j} + 2\hat{k}$$ $$v_B = 2\hat{i} - 2\hat{j} + 4p\hat{k}$$ $$\vec{v}_A - \vec{v}_B = 0, 4 - 6n + 8p = 0$$ $$|v_A| = |v_B| (3n)^2 + 4 = 4 + 16p^2$$ $$3n = -4p$$ $$4 + 16p = 0$$ $$p = -\frac{1}{4}, n = \frac{1}{3}$$ $$r_{AB} = -\hat{i} + 2\hat{j} + 3\hat{k}$$ $$v_A = 2\hat{i} + \hat{j} + 2\hat{k}$$ Therefore, $L = m |\vec{r}_{AB} \times \vec{v}_A| = 90$
Physics · Motion in a Plane · Numerical
A particle is projected at an angle of $30^\circ$ from horizontal at a speed of $60 \, \mathrm{m/s}$. The height traversed by the particle in the first second is $h_0$ and height traversed in the last second, before it reaches the maximum height, is $h_1$. The ratio $h_0 : h_1$ is ____ [Take, $g = 10 \, \mathrm{m/s^2}$]
Answer: 5
Given $60 \sin 30^\circ = 30$. Calculate $S_1$ as follows: $$S_1 = 30 \times 1 - \frac{1}{2} \times 10 \times 1 = 25$$ Calculate $S_3$ as follows: $$S_3 = 30 + \left( \frac{-10}{2} \right) \times (2 \times 3 - 1) = 5$$ Finally, we have: $$S_1 = \frac{25}{5} = 5$$ $$S_3 = 5$$
Chemistry · Electrochemistry · Single correct
A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A. The amount of the aluminium deposited at the cathode is [Given : molar mass of aluminium and chlorine are $27 \, \mathrm{g \, mol^{-1}}$ and $35.5 \, \mathrm{g \, mol^{-1}}$ respectively. Faraday constant $= 96500 \, \mathrm{C \, mol^{-1}}$]
Answer: (b)
The reaction is $\mathrm{Al^{1+++} + 3e^- \rightarrow Al}$. Moles of electron $= \frac{2 \times 30 \times 60}{96500}$ $$= \frac{36}{965}$$ Moles of $\mathrm{Al} = \frac{36}{3 \times 965}$ $$= \frac{12}{965}$$ Mass of $\mathrm{Al} = \frac{12}{965} \times 27$ $$= 0.336 \, \mathrm{gm}$$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
Which of the following statement is not true for radioactive decay?
Answer: (a)
For radioactive decay, decay constant does not depend upon temperature because for radioactive decay activation energy is zero.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
How many different stereoisomers are possible for the given molecule? $$\mathrm{CH_3-CH-CH=CH-CH_3}$$ $$\mathrm{|}$$ $$\mathrm{OH}$$
Answer: (c)
The compound has 4 stereoisomers: $R$ cis, $R$ trans, $S$ cis, $S$ trans.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Which of the following electronegativity order is incorrect?
Answer: (d)
The electronegativity (E.N.) values on the Pauling scale are given for the elements: Li: $1$, Be: $1.5$, B: $2$, C: $2.5$, N: $3$, O: $3.5$, F: $4.0$. For another set of elements: Na: $0.9$, Mg: $1.2$, Al: $1.5$, Si: $1.8$, P: $2.1$, S: $2.5$, Cl: $3.0$. The correct order of electronegativity is $Mg < Al < B < N$.
Chemistry · The d-and f-Block Elements · Single correct
Lanthanoid ions with $4f^7$ configuration are : (A) Eu$^{2+}$ (B) Gd$^{3+}$ ($C$) Eu$^{3+}$ (D) Tb$^{3+}$ (E) Sm$^{2+}$ Choose the correct answer from the options given below :
Answer: (c)
The electronic configurations are given as follows: For $^{63}\mathrm{Eu}^{2+}$: $$[\mathrm{Xe}]4f^76s^0$$ For $_{64}\mathrm{Gd}^{3+}$: $$[\mathrm{Xe}]4f^56d^06s^0$$ For $_{63}\mathrm{Eu}^{3+}$: $$[\mathrm{Xe}]4f^66s^0$$ For $_{65}\mathrm{Tb}^{3+}$: $$[\mathrm{Xe}]4f^86s^0$$ For $_{62}\mathrm{Sm}^{2+}$: $$[\mathrm{Xe}]4f^66s^0$$ The correct answer is $\mathrm{Eu}^{2+}$ and $\mathrm{Gd}^{3+}$.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Match List-I with List-II. Choose the correct answer from the options given below :
Answer: (a)
Ionic radii: $\mathrm{Al^{3+} < Mg^{2+} < Na^+ < F^-}$. Ionisation energy: $\mathrm{B < C < O < N}$. Metallic character: $\mathrm{B < Al < Mg < K}$. Electron negativity: $\mathrm{Si < P < S < Cl}$.
Chemistry · Biomolecules · Single correct
Which of the following acids is a vitamin?
Answer: (b)
Vitamin-C is Ascorbic acid.
Chemistry · Thermodynamics · Single correct
A liquid when kept inside a thermally insulated closed vessel at $25^\circ \mathrm{C}$ was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters?
Answer: (d)
Thermally insulated implies $q = 0$ from the first law. $$\Delta U = q + w$$ $$\Delta U = w$$ Since $w > 0$, $\Delta U > 0$.
Chemistry · Structure of Atom · Single correct
Radius of the first excited state of Helium ion is given as : $a_0 \rightarrow$ radius of first stationary state of hydrogen atom.
Answer: (c)
Given $\($ r = a_0 $\left$( $\frac{n^2}{z}$ $\right$) $\)$. Substituting $\($ n = 2 $\)$ and $\($ z = 2 $\)$, we have: $$ r = a_0 (2)^2 \left( \frac{1}{2} \right) $$ Simplifying, we get: $$ r = 2a_0 $$
Chemistry · Haloalkanes and Haloarenes · Single correct
Given below are two statements: Statement I: $CH_3 - O - CH_2 - Cl$ will undergo $S_N1$ reaction though it is a primary halide. Statement II: will not undergo $S_N2$ reaction very easily though it is a primary halide. In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (b)
Both statement-I and statement-II are correct. Statement - I: $CH_3-O-\overset{\oplus}{CH_2}$ is highly stable carbocation. Statement - II: Due to hindrance, $\mathrm{S_N2}$ will not take place easily.
Chemistry · Hydrocarbons · Single correct
Given below are two statements: Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of $\mathrm{H}_2$ gas. Statement II: Four g of propyne reacts with $\mathrm{NaNH}_2$ to liberate $\mathrm{NH}_3$ gas which occupies $224 \, \mathrm{mL}$ at STP. In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (c)
Statement-I is correct. Moles of $\mathrm{C_3H_4} = \frac{4}{40} = 0.1$ mole. $$\mathrm{CH_3 - C \equiv CH + NaNH_2 \rightarrow CH_3 - C \equiv C^-Na^+ + NH_3}$$ $0.1$ mole $0.1$ mole Volume of $\mathrm{NH_3} = (0.1)(22.4) = 2.24 \, \mathrm{L}$ Statement-II is incorrect.
Chemistry · Equilibrium · Single correct
A vessel at 1000 K contains $\mathrm{CO}_2$ with a pressure of 0.5 atm. Some of $\mathrm{CO}_2$ is converted into $\mathrm{CO}$ on addition of graphite. If total pressure at equilibrium is 0.8 atm, then $K_p$ is:
Answer: (b)
The reaction is given as $\mathrm{CO_2(g) + C(s) \rightleftharpoons 2CO(g)}$. Initially, the concentration of $\mathrm{CO_2}$ is $0.5$. At equilibrium, the concentration of $\mathrm{CO_2}$ is $0.5 - x$ and the concentration of $\mathrm{CO}$ is $2x$. The total pressure is given by $P_{total} = 0.5 + x = 0.8$. Solving for $x$, we find $x = 0.3$. The equilibrium constant $K_P$ is calculated as $$K_P = \frac{(0.6)^2}{0.2} = 1.8.$$
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The IUPAC name of the following compound is :
Answer: (c)
The compound shown is 6-Methoxycarbonyl-2,5-dimethylhexanoic acid.
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Which of the following electrolyte can be used to obtain $\mathrm{H_2 S_2 O_8}$ by the process of electrolysis?
Answer: (d)
Theory based. At anode: $$2\mathrm{HSO}_4^- \rightarrow \mathrm{H}_2\mathrm{S}_2\mathrm{O}_8 + 2e^-$$
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The compounds which give positive Fehling's test are : Choose the correct answer from the options given below :
Answer: (b)
Aliphatic aldehyde, $\alpha$-hydroxy ketone gives Fehlings solution test.
Chemistry · Co-ordination Compounds · Single correct
In which of the following complexes the CFSE, $\Delta_o$ will be equal to zero?
Answer: (d)
For the complex $\mathrm{K_3[Fe(SCN)_6]}$, the electronic configuration of $\mathrm{Fe^{3+}}$ is $[\mathrm{Ar}]3d^5$. The weak field ligand (W.F.L.) causes the splitting of the $d$ orbitals into $e_g^2$ and $t_{2g}^3$. Calculation of CFSE: $$= (-0.4 \times 3 + 0.6 \times 2) \Delta_0$$ $$= 0 \Delta_0$$
Chemistry · Solutions · Single correct
Arrange the following solutions in order of their increasing boiling points. (i) $10^{-4} \mathrm{M}$ NaCl (ii) $10^{-4} \mathrm{M}$ Urea (iii) $10^{-3} \mathrm{M}$ NaCl (iv) $10^{-2} \mathrm{M}$ NaCl
Answer: (d)
For $10^{-4} \, M \mathrm{NaCl}$, $i = 2$. For $10^{-4} \, M$ Urea, $i = 1$. For $10^{-3} \, M \mathrm{MgCl_2}$, $i = 3$. For $10^{-2} \, M \mathrm{NaCl}$, $i = 2$. More the value of $i$, more will be the elevation in boiling point hence increasing order of boiling point is $10^{-4} \, M$ Urea $< 10^{-4} \, M \mathrm{NaCl} < 10^{-3} \, M \mathrm{MgCl_2} < 10^{-2} \, M \mathrm{NaCl}$.
Chemistry · Amines · Single correct
The products formed in the following reaction sequence are:
Answer: (c)
The reaction sequence starts with the nitration of benzene to form nitrobenzene. The first step involves bromination using $\mathrm{Br_2/AcOH}$ to give bromonitrobenzene. Next, reduction with $\mathrm{Sn/HCl}$ converts the nitro group to an amino group, forming bromoaniline. Diazotization with $\mathrm{NaNO_2 + HCl}$ produces the diazonium salt. Finally, the reaction with ethanol ($\mathrm{EtOH}$) yields the final product, bromoethoxybenzene.
Chemistry · Co-ordination Compounds · Single correct
From the magnetic behaviour of $[NiCl_4]^{2-}$ (paramagnetic) and $[Ni(CO)_4]$ (diamagnetic), choose the correct geometry and oxidation state.
Answer: (c)
For $[NiCl_4]^{2-}$, $Ni^{+2}$ has the configuration $[Ar] 3d^8 4s^0 \rightarrow sp^3$, Tetrahedral. The number of unpaired electrons is 2, making it paramagnetic. For $[Ni(CO)_4]$, $Ni(0)$ rearranges to $[Ar] 3d^{10} 4s^0$. There are no unpaired electrons, making it $sp^3$, Tetrahedral, and Diamagnetic.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The incorrect statements regarding geometrical isomerism are : (A) Propene shows geometrical isomerism. (B) Trans isomer has identical atoms/groups on the opposite sides of the double bond. (C) Cis-but-2-ene has higher dipole moment than trans-but-2-ene. (D) 2-methylbut-2-ene shows two geometrical isomers. (E) Trans-isomer has lower melting point than cis isomer. Choose the correct answer from the options given below :
Answer: (b)
Q13. (A) $\mathrm{CH_3 - CH = CH_2}$. GI is not possible. (B) Trans isomer has identical atoms/groups on the opposite side of double bond. (C) $>$ (dipole moment only) (D) $\mathrm{H_3C - C = CH - CH_3}$ $\newline$ $\mathrm{CH_3}$$ \newline 2-methylbut-2-ene (does not show GI) (E) $>$ (Melting point)
Chemistry · Some Basic Concepts of Chemistry · Numerical
Some $CO_2$ gas was kept in a sealed container at a pressure of 1 atm and at 273 K. This entire amount of $CO_2$ gas was later passed through an aqueous solution of $Ca(OH)_2$. The excess unreacted $Ca(OH)_2$ was later neutralized with 0.1 M of 40 mL HCl. If the volume of the sealed container of $CO_2$ was $x$, then $x$ is ______ cm$^3$ (nearest integer). [Given : The entire amount of $CO_2 (g)$ reacted with exactly half the initial amount of $Ca(OH)_2$ present in the aqueous solution.]
Answer: 22400
Let moles of $\mathrm{CO_2} = n$ moles of $\mathrm{Ca(OH)_2}$ total initially $= 2n$ excess $\mathrm{Ca(OH)_2} = n$ gm equivalent of $\mathrm{(45) \ Ca(OH)_2} = gm equivalent of HCl$. $$n \times 2 = 0.1 \times \frac{40}{1000} \times 1$$ $$n = 2 \times 10^{-3}$$ Volume of $\mathrm{CO_2} = 2 \times 10^{-3} \times 22400 = 44.8 \, \mathrm{cm^3}$$
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical
In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl. The percentage composition of chlorine in the compound is _______ %. (Given : molar mass in g mol$^{-1}$ of Ag : 108, Cl : 35.5)
Answer: 20
Given $\($ n_{Cl} = n_{AgCl} = $\frac{143.5 \times 10^{-3}}{143.5}$ = 10^{-3} $\)$. $\($ $\%$ Cl = $\frac{10^{-3} \times 35.5}{180 \times 10^{-3}}$ $\times$ 100 = 19.72 $\)$.
Chemistry · Chemical Bonding and Molecular Structure · Numerical
\[ \mathrm{A} \rightarrow \mathrm{B} \] The number of molecules/ions that show linear geometry among the following is: $\mathrm{SO_2},\ \mathrm{BeCl_2},\ \mathrm{CO_2},\ \mathrm{N_3^-},\ \mathrm{NO_2},\ \mathrm{F_2O},\ \mathrm{XeF_2},\ \mathrm{NO_2^+},\ \mathrm{I_3^-},\ \mathrm{O_3}$.
Answer: 6
For $\mathrm{Cl} - \mathrm{Be} - \mathrm{Cl}$ and $\mathrm{N} \equiv \mathrm{N} : \rightarrow \mathrm{O}$, the hybridization is $sp$, linear. For $\mathrm{N}^- = \mathrm{N}^+ = \mathrm{N}^-$ and $\mathrm{O} = \mathrm{N}^+ = \mathrm{O}$, the hybridization is $sp$, linear. Like $\mathrm{I}_3^-$, $\mathrm{XeF}_2 \rightarrow sp^3d_1$, Linear. Answer is 4.
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The molecule A changes into its isomeric form B by following first-order kinetics at a temperature of $1000\,\mathrm{K}$. If the energy barrier with respect to the reactant energy for this isomeric transformation is $191.48\,\mathrm{kJ\,mol^{-1}}$ and the frequency factor is $10^{20}\,\mathrm{s^{-1}}$, then the time required for $50\%$ of the molecules of A to become B is $\underline{\hspace{1cm}}$ picoseconds (nearest integer). [Given: $R=8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$]
Answer: 69
Given $t_{1/2} = \frac{0.693}{K}$. $K = A e^{-E_a/RT}$ $$= 10^{20} \times e^{-\frac{191.48 \times 10^3}{8.314 \times 1000}}$$ $$= 10^{20} \times e^{-23.031} = 10^{20} \times e^{-\ln 10 \times 10}$$ $$= \frac{10^{20}}{10^{10}} = 10^{10} sec.$$ $t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11}$ $$= 69.3 \times 10^{-12} sec.$$
Chemistry · Amines · Numerical
Consider the following sequence of reactions: Molar mass of the product formed (A) is _____ gmol$^{-1}$.
Answer: 154
The reaction sequence involves the conversion of a nitro group to an amine group using $\mathrm{Sn + HCl}$. This is followed by diazotization with $\mathrm{NaNO_2 + HCl}$ at $0^\circ \mathrm{C}$ to form a diazonium salt. The diazonium salt is then reacted with $\mathrm{Cu_2Cl_2}$ to form a chlorobenzene derivative. Finally, the reaction with sodium in dry ether leads to the formation of biphenyl ($\mathrm{C_{12}H_{10}}$). The molar mass is calculated as follows: $$Molar mass = 120 \times 12 + 10 \times 1 = 154$$