JEE Main 24 January 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 · Conic Sections · Single correct
The equation of the chord, of the ellipse $\($ $\frac{x^2}{25}$ + $\frac{y^2}{16}$ = 1 $\)$, whose mid-point is $\($(3, 1)$\)$ is :
Answer: (a)
Equation of chord with given middle point $$T = S_1$$ $$\frac{3x}{25} + \frac{y}{16} - 1 = \frac{9}{25} + \frac{1}{16} - 1$$ $$48x + 25y = 144 + 25$$ $$48x + 25y = 169 .$$
Maths · Relations and Functions · Single correct
The function $f : (-\infty, \infty) \to (-\infty, 1)$, defined by $f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$ is:
Answer: (d)
Given $f(x) = \frac{2^{2x} - 1}{2^{2x} + 1}$. $$= 1 - \frac{2^{2x} + 1}{2}$$ The derivative is $$f'(x) = \frac{2 \cdot 2^{2x} \cdot \ln 2}{(2^{2x} + 1)^2},$$ which is always positive. So $f(x)$ is an increasing function. Therefore, $f(-\infty) = -1$ and $f(\infty) = 1$. Thus, $f(x) \in (-1, 1) \neq$ co-domain, so the function is one-one but not onto.
Maths · Inverse Trigonometric Functions · Single correct
If $\alpha > \beta > \gamma > 0$, then the expression $\cot^{-1}\left\{ \beta + \frac{(1+\beta^2)}{(\alpha-\beta)} \right\} + \cot^{-1}\left\{ \gamma + \frac{(1+\gamma^2)}{(\beta-\gamma)} \right\} + \cot^{-1}\left\{ \alpha + \frac{(1+\alpha^2)}{(\gamma-\alpha)} \right\}$ is equal to:
Answer: (a)
Given $$\cot^{-1}\left(\frac{\alpha \beta + 1}{\alpha - \beta}\right) + \cot^{-1}\left(\frac{\beta \gamma + 1}{\beta - \gamma}\right) + \cot^{-1}\left(\frac{\alpha \gamma + 1}{\gamma - \alpha}\right)$$ This implies $$\tan^{-1}\left(\frac{\alpha - \beta}{1 + \alpha \beta}\right) + \tan^{-1}\left(\frac{\beta - \gamma}{1 + \beta \gamma}\right) + \pi + \tan^{-1}\left(\frac{\gamma - \alpha}{1 + \gamma \alpha}\right)$$ This simplifies to $$(\tan^{-1} \alpha - \tan^{-1} \beta) + (\tan^{-1} \beta - \tan^{-1} \gamma) + (\pi + \tan^{-1} \gamma - \tan^{-1} \alpha)$$ Therefore, the result is $$\pi$$
Maths · Differential Equations · Single correct
Let $f : (0, \infty) \to \mathbb{R}$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2 f'(x) = 2x f(x) + 3$, with $f(1) = 4$. Then $2f(2)$ is equal to:
Answer: (a)
Given $x^2 f'(x) - 2x f(x) = 3$. (1) $$\left( x^2 f'(x) - 2x f(x) \right) = \frac{3}{(x^2)^2}$$ $$\Rightarrow \frac{d}{dx} \left( \frac{f(x)}{x^2} \right) = \frac{3}{x^4}$$ Integrating both sides $$\frac{f(x)}{x^2} = -\frac{1}{x^3} + C$$ $$f(x) = -\frac{1}{x} + Cx^2$$ Put $x = 1$ $$4 = -1 + C \Rightarrow C = 5$$ $$f(x) = -\frac{1}{x} + 5x^2$$ Now $2 \times f(2) = 2 \times \left[ -\frac{1}{2} + 5 \times 2^2 \right]= 39$
Maths · Sets · Single correct
Let $A=\left\{x\in(0,\pi)-\left\{\frac{\pi}{2}\right\}:\log_{\frac{2}{\pi}}|\sin x|+\log_{\frac{2}{\pi}}|\cos x|=2\right\}$ and\ $B=\left\{x\geq0:\sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\right\}$.\ Then $n(A\cup B)$ is equal to:
Answer: (b)
Given $\log_{2\pi} |\sin x| + \log_{2\pi} |\cos x| = 2$. This implies $\log_{2\pi} (|\sin x \cdot \cos x|) = 2$. Therefore, $|\sin 2x| = \frac{8}{\pi^2}$. The number of solutions is 4. For part B, let $\sqrt{x} = t 2$. Then $t^2 - 4t - 3t + 6 + 6 = 0$ simplifies to $t^2 - 7t + 12 = 0$. Solving gives $t = 3, 4$, so $x = 9, 16$. The total number of solutions is $n(A \cup B) = 4 + 4 = 8$.
Maths · Vector Algebra · Single correct
Let the position vectors of three vertices of a triangle be $4\vec{p} + \vec{q} - 3\vec{r}$, $-5\vec{p} + \vec{q} + 2\vec{r}$ and $2\vec{p} - \vec{q} + 2\vec{r}$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p} + \vec{q} + \vec{r}}{4}$ and $\alpha \vec{p} + \beta \vec{q} + \gamma \vec{r}$ respectively, then $\alpha + 2\beta + 5\gamma$ is equal to:
Answer: (a)
We know that $$O \left( \frac{\vec{p} + \vec{q} + \vec{r}}{4} \right)$$ C (circum centre) $\alpha \vec{p} + \beta \vec{q} + \gamma \vec{r}$ $$C (centroid) = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$$ By relation $$\Rightarrow 2(\alpha \vec{p} + \beta \vec{q} + \gamma \vec{r}) + \frac{\vec{p} + \vec{q} + \vec{r}}{4} = 3 \left( \frac{\vec{p} + \vec{q} + \vec{r}}{3} \right)$$ $$\Rightarrow 8(\alpha \vec{p} + \beta \vec{q} + \gamma \vec{r}) = 3(\vec{p} + \vec{q} + \vec{r})$$ $$\Rightarrow 8\alpha = 3, 8\beta = 3, 8\gamma = 3$$ $$\alpha = \frac{3}{8}, \beta = \frac{3}{8}, \gamma = \frac{3}{8}$$ $$\therefore \alpha + \beta + \gamma$$ $$\frac{3}{8} + \frac{6}{8} + \frac{15}{8} = \frac{24}{8} = 3$$
Maths · Continuity and Differentiability · Single correct
Let $[x]$ denote the greatest integer function, and let $m$ and $n$ respectively be the numbers of the points, where the function $f(x) = [x] + |x - 2|, -2 < x < 3$, is not continuous and not differentiable. Then $m + n$ is equal to:
Answer: (b)
Given $f(x) = [x] + |x - 2|$, $-2 < x < 3$. Therefore, $$f(x) = \begin{cases} -x, & -2 < x < -1 \\ 1 - x, & -1 \leq x < 0 \\ 2 - x, & 0 \leq x < 1 \\ 3 - x, & 1 \leq x < 2 \\ x, & 2 \leq x < 3 \end{cases}$$ It is clearly discontinuous at 4 points and nondifferentiable at 4 points. Therefore, $m + n = 8$.
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let the points $\left( \frac{11}{2}, \alpha \right)$ lie on or inside the triangle with sides $x + y = 11, x + 2y = 16$ and $2x + 3y = 29$. Then the product of the smallest and the largest values of $\alpha$ is equal to:
Answer: (c)
Clearly, $x = \frac{11}{2}$ intersect $x + y - 11 = 0$ at $\left( \frac{11}{2}, \frac{11}{2} \right)$ and $2x + 3y - 29 = 0$ at $\left( \frac{11}{2}, 6 \right)$ which implies $\alpha = \left[ \frac{11}{2}, 6 \right]$. Therefore, $\alpha_{\min} \cdot \alpha_{\max} = \frac{11}{2} \cdot 6 = 33$.
Maths · Sequences and Series · Single correct
In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57$, then $S_{30} - S_{10}$ is equal to:
Answer: (c)
Given $S_{40} = 1030$, we have $$\frac{40}{2} [2a + 39d] = 1030$$ which implies $$2a + 39d = \frac{103}{2} \ldots (1)$$ For $S_{12} = 57$, we have $$\frac{12}{2} [2a + 11d] = 57$$ which implies $$2a + 11d = \frac{57}{6} \ldots (2)$$ Subtracting equation (2) from equation (1), we get $$28d = \frac{103}{2} - \frac{57}{6}$$ $$28d = \frac{309 - 57}{6}$$ $$d = \frac{3}{2}$$ Therefore, $$a = -\frac{7}{2}$$ Now, $$S_{30} - S_{10} = \frac{30}{2} [2a + 29d] - \frac{10}{2} [2a + 9d]$$ $$= 15[2a + 29d] - 5[2a + 9d]$$ $$= 5[6a + 87d - 2a - 9d]$$ $$= 5[4a + 78d]$$ $$= 5[-14 + 117]$$ $$= 515$$
Maths · Sequences and Series · Single correct
If $7 = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \frac{1}{7^3}(5 + 3\alpha) + \ldots \infty$, then the value of $\alpha$ is:
Answer: (b)
Given $S = a + (a + d)r + (a + 2d)r^2 + \ldots$ Then $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$, $|r| < 1$ Since, $r = \frac{1}{7}$ and $a = 5$, $d = \alpha$ $$7 = \frac{5}{1 - \frac{1}{7}} + \alpha \cdot \frac{1}{7} \left(1 - \frac{1}{7}\right)^2$$ $$\Rightarrow \alpha = 6$$
Maths · Determinants · Single correct
If the system of equations $$x + 2y - 3z = 2$$ $$2x + \lambda y + 5z = 5$$ $$14x + 3y + \mu z = 33$$ has infinitely many solutions, then $\lambda + \mu$ is equal to:
Answer: (c)
Given $$\Delta = \begin{vmatrix} 1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu \end{vmatrix} = 0 \Rightarrow \lambda \mu + 42 \lambda - 4 \mu + 107 = 0$$ $$\Delta_1 = \begin{vmatrix} 2 & 2 & -3 \\ 5 & \lambda & 5 \\ 33 & 3 & \mu \end{vmatrix} = 0 \Rightarrow 2 \lambda \mu + 99 \lambda - 10 \mu + 255 = 0$$ $$\Delta_2 = \begin{vmatrix} 1 & 2 & -3 \\ 2 & 5 & 5 \\ 14 & 33 & \mu \end{vmatrix} = 0 \Rightarrow \mu = 13$$ Also, $\lambda = -1$ Hence, $\lambda + \mu = 13 - 1 = 12$
Maths · Applications of Derivatives · Single correct
Let (2, 3) be the largest open interval in which the function $f(x) = 2 \log_e (x - 2) - x^2 + ax + 1$ is strictly increasing and $(b, c)$ be the largest open interval, in which the function $g(x) = (x - 1)^3(x + 2 - a)^2$ is strictly decreasing. Then $100(a + b - c)$ is equal to:
Answer: (b)
Given $f'(x) = \frac{2}{x-2} - 2x + a \geq 0$. The second derivative is $f''(x) = \frac{-2}{(x-2)^2} - 2 < 0$. Since $f'(x)$ is decreasing, we have $f'(3) \geq 0$. This gives $2 - 6 + a \geq 0$, so $a \geq 4$. Thus, $a_{\min} = 4$. Consider $g(x) = (x-1)^3(x+2-a)^2$. Substituting $a = 4$, we have $g(x) = (x-1)^3(x-2)^2$. The derivative is $g'(x) = (x-1)^3 2(x-2) + (x-2)^2 3(x-1)^2$. Simplifying, $g'(x) = (x-1)^2(x-2)(2x-2 + 3x-6)$. This simplifies to $g'(x) = (x-1)^2(x-2)(5x-8) < 0$. Thus, $x \in \left(\frac{8}{5}, 2\right)$. Finally, $100(a+b-c) = 100 \left(4 + \frac{8}{5} - 2\right) = 360$.
Maths · Binomial Theorem · Single correct
Suppose A and B are the coefficients of $30^{\mathrm{th}}$ and $12^{\mathrm{th}}$ terms respectively in the binomial expansion of $(1+x)^{2n-1}$. If $2A=5B$, then $n$ is equal to:
Answer: (c)
Given $\($ A = $\binom{2n-1}{29}$ $\)$ and $\($ B = $\binom{2n-1}{11}$ $\)$. $\[$ $\binom{2n-1}{29}$ = 5 $\binom{2n-1}{11}$ $\]$ $\[$ $\frac{2(2n-1)!}{29!(2n-30)!}$ = $\frac{5(2n-1)!}{(2n-12)!11!}$ $\]$ $\[$ $\frac{2}{29 \cdots 12 \cdot 5}$ = $\frac{5}{(2n-12)(2n-13) \cdots (2n-29)2}$ $\]$ $\[$ $\frac{1}{30 \cdot 29 \cdots 12}$ = $\frac{1}{(2n-12)(2n-13) \cdots (2n-29)12}$ $\]$ $\($ 2n - 12 = 30 $\)$ $\($ n = 21 $\)$
Maths · Vector Algebra · Single correct
Let $\vec{a} = 3\hat{i} - \hat{j} + 2\hat{k}$, $\vec{b} = \vec{a} \times (\hat{i} - 2\hat{k})$ and $\vec{c} = \vec{b} \times \hat{k}$. Then the projection of $\vec{c} - 2\hat{j}$ on $\vec{a}$ is:
Answer: (a)
Given $\vec{b} = \vec{a} \times (\hat{i} - 3\hat{k})$. $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{vmatrix} = 2\hat{i} + 8\hat{j} + \hat{k}$$ $\vec{c} = \vec{b} \times \hat{k} = 8\hat{i} - 2\hat{j}$ $\vec{c} - 2\hat{j} = 8\hat{i} - 4\hat{j}$ Projection of $(\hat{i} - 2\hat{j})$ on $\vec{a}$ $$(\vec{c} - 2\hat{j}) \cdot \hat{a} = \langle 8, -4, 0 \rangle \cdot \langle 3, -1, 2 \rangle$$ $$= \frac{28}{\sqrt{14}}$$ $$= 2\sqrt{14}$$
Maths · Determinants · Single correct
For some $a, b$, let $$f(x) = \begin{vmatrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{vmatrix}, x \neq 0,$$ $$\lim_{x \to 0} f(x) = \lambda + \mu a + \nu b.$$ Then $(\lambda + \mu + \nu)^2$ is equal to:
Answer: (a)
Given $$\lim_{x \to 0} \begin{vmatrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{vmatrix} = \lambda + \mu a + vb$$ At $\lim_{x \to 0}$, $$f(x) = \begin{vmatrix} a + 1 & 1 & b \\ a & 1 + 1 & b \\ a & 1 & b + 1 \end{vmatrix} = \lambda + \mu a + vb$$ Performing row operations: $$R_1 \to R_1 - R_2$$ $$R_2 \to R_2 - R_3$$ $$\begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ a & 1 & b + 1 \end{vmatrix} = \lambda + \mu a + vb$$ Performing column operation: $$C_2 \to C_1 - C_2$$ $$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ a & a + 1 & b + 1 \end{vmatrix} = \lambda + \mu a + vb$$ Thus, $$a + b + 2 = \lambda + \mu a + vb$$ Given $\lambda = 2$, $\mu = 1$, $v = 1$ $$(\lambda + \mu + v) = (2 + 1 + 1)^2 = 16$$
Maths · Permutations and Combinations · Single correct
Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:
Answer: (c)
The table shows the distribution of boys (B) and girls (G) in Groups A and B, along with the number of ways to choose them. For Group A with 4 boys and 1 girl, and Group B with 0 boys and 3 girls, the number of ways is: $$^7C_4 \cdot ^3C_1 \cdot ^6C_0 \cdot ^5C_3$$ For Group A with 3 boys and 2 girls, and Group B with 1 boy and 2 girls, the number of ways is: $$^7C_3 \cdot ^3C_2 \cdot ^6C_1 \cdot ^5C_2$$ For Group A with 2 boys and 3 girls, and Group B with 2 boys and 1 girl, the number of ways is: $$^7C_2 \cdot ^3C_3 \cdot ^6C_2 \cdot ^5C_1$$ The total number of ways is: $$30 \cdot ^7C_4 + 180 \cdot ^7C_3 + 75 \cdot ^7C_2 = 8925$$
Maths · Applications of Integrals · Single correct
The area of the region enclosed by the curves $y = e^x$, $y = |e^x - 1|$ and $y$-axis is:
Answer: (a)
Given $e^x = 1 - e^x$, we have $2e^x = 1$. Therefore, $e^x = \frac{1}{2}$. This implies $x = \ln \frac{1}{2}$. The integral is given by $$\int_{\ln(1/2)}^0 \left[ e^x - (1 - e^x) \right] \, dx$$ This simplifies to $$= \int_{\ln 2}^0 (2e^x - 1) \, dx = 2e^x - x \bigg|_{-\ln 2}^0$$ Evaluating the integral, we get $$= 2 - (1 + \ln 2)$$ Thus, $$= 1 - \log_e 2$$
Maths · Complex Numbers and Quadratic Equations · Single correct
The number of real solution(s) of the equation $x^2 + 3x + 2 = \min\{|x - 3|, |x + 2|\}$ is:
Answer: (c)
Given the equation $x^2 + 3x + 2 = \min\{|x - 3|, |x + 2|\}$. We start with the equation: $$y = x^2 + 3x + 2$$ Complete the square: $$y = x^2 + 2 \left(\frac{3}{2}\right)x + \frac{9}{4} - \frac{9}{4} + 2$$ This simplifies to: $$y = \left(x + \frac{3}{2}\right)^2 - \frac{1}{4}$$ Rewriting gives: $$y + \frac{1}{4} = \left(x + \frac{3}{2}\right)^2$$ Thus, the parabola vertex is: $$\left(-\frac{3}{2}, -\frac{1}{4}\right)$$ By graph 2, solution possible.
Maths · Matrices · Single correct
Let $A = \begin{bmatrix} a_{ij} \end{bmatrix}$ be a square matrix of order 2 with entries either 0 or 1. Let $E$ be the event that $A$ is an invertible matrix. Then the probability $P(E)$ is:
Answer: (c)
Given $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}_{2 \times 2}$ and entries are 0 or 1. Since $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = 0$, it implies $ad - bc = 0$. Case I: $ad = bc = 1$ Therefore, $a = b = c = d = 1$. Case II: $ad = bc = 0$ $a = 0, d = 0 b = 0, c = 0$ $a = 0, d = 1 b = 0, c = 1$ $a = 1, d = 0 b = 1, c = 0$ Therefore, there are a total of 10 cases when the matrix is non-invertible. The total possible matrices are $2^4 = 16$. The required probability of invertible is $$= \frac{16 - 10}{16} = \frac{6}{16} = \frac{3}{8}$$
Maths · Conic Sections · Single correct
If the equation of the parabola with vertex V ( $\frac{3}{2}$, 3) and the directrix $x+2y=0$ is $$\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$$, then $\alpha + \beta + \gamma$ is equal to:
Answer: (b)
The equation of the parabola is $PS = (1)PM$. Therefore, $PS^2 = PM^2$. $$(x - 3)^2 + (y - 6)^2 = \left( \frac{x + 2y}{\sqrt{5}} \right)^2$$ Expanding and simplifying, we have: $$5x^2 - 30x + 5y^2 - 60y + 225 = x^2 + 4y^2 + 4xy$$ Rearranging terms gives: $$4x^2 + y^2 - 4xy - 30x - 60y + 225 = 0$$ We get: $\alpha = 4$, $\beta = 1$, $\gamma = 4$. Therefore, $\alpha + \beta + \gamma = 9$.
Maths · Relations and Functions · Numerical
Number of functions $f : \{1, 2, \ldots, 100\} \to \{0, 1\}$, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to _______.
Answer: 392
$$98 \times 2 \times 2 = 392.$$
Maths · Three Dimensional Geometry · Numerical
Let P be the image of the point Q(7, -2, 5) in the line L : $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $\triangle PQR$ is _______.
Answer: 957
Let $R(2\lambda + 1, 3\lambda - 1, 4\lambda)$. $2\lambda + 1 = 5$ $\lambda = 2$ $R(5, 5, 8)$ Let $T(2\lambda + 1, 3\lambda - 1, 4\lambda)$. $$\overrightarrow{QT} = (2\lambda - 6)\hat{i} + (3\lambda + 1)\hat{j} + (4\lambda - 5)\hat{k}$$ $$\overrightarrow{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$ $$\overrightarrow{QT} \cdot \overrightarrow{b} = 0$$ $4\lambda - 12 + 9\lambda + 3 + 16\lambda - 20 = 0$ $\lambda = 1$ $T(3, 2, 4)$ $QT = \sqrt{33}$ $RT = \sqrt{29}$ $$\left(area of \triangle PQR\right)^2 = \left(\frac{1}{2} \sqrt{29} \cdot 2 \sqrt{33}\right)^2$$ $= 957$
Maths · Differential Equations · Numerical
Let $y = y(x)$ be the solution of the differential equation $2 \cos x \frac{dy}{dx} = \sin 2x - 4y \sin x, x \in \left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{3}\right) = 0$, then $y'\left(\frac{\pi}{4}\right) + y\left(\frac{\pi}{4}\right)$ is equal to ________.
Answer: 1
$\dfrac{dy}{dx}+2y\tan x=\sin x$ I.F. $=e^{\int 2\tan x\,dx}=\sec^2x$ $\therefore\ y\sec^2x=\int\dfrac{\sin x}{\cos^2x}\,dx$ $=\int\tan x\,\sec x\,dx$ $=\sec x+C$ $C=-2$ $\therefore\ y=\cos x-2\cos^2x$ $y\left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt2}-1$ $y'=-\sin x+4\cos x\sin x$ $y'\left(\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt2}+2$ $y'\left(\dfrac{\pi}{4}\right)+y\left(\dfrac{\pi}{4}\right)=1$
Maths · Conic Sections · Numerical
Let $\mathrm{H}_1 : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\mathrm{H}_2 : -\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ be two hyperbolas having length of latus rectums $15\sqrt{2}$ and $12\sqrt{5}$ respectively. Let their eccentricities be $e_1 = \sqrt{\frac{5}{2}}$ and $e_2$ respectively. If the product of the lengths of their transverse axes is $100\sqrt{10}$, then $25e_2^2$ is equal to _______.
Answer: 55
Given the equations $\($ $\frac{x^2}{a^2}$ - $\frac{y^2}{2b^2}$ = 1 $\)$ and $\($ $\frac{a}{b^2}$ = 15$\sqrt{2}$ $\ldots$ (i) $\)$. Also, $\($ $\sqrt{1 + \frac{b^2}{a^2}}$ = $\sqrt{\frac{5}{2}}$ $\ldots$ (ii) $\)$. From (i) and (ii), $\($ a = 5$\sqrt{2}$ $\)$ and $\($ b^2 = 75 $\)$. The equation $\($ $\frac{x^2}{A^2}$ - $\frac{y^2}{B^2}$ = -1 $\)$ gives $\($ $\frac{2A^2}{B}$ = 12$\sqrt{5}$ $\ldots$ (iii) $\)$. Since the product of the transverse axis is $\($ 100$\sqrt{10}$ $\)$, $\($ (2A) $\cdot$ (2B) = 100$\sqrt{10}$ $\)$. From (iii) and (iv), $\($ A^2 = 150 $\)$ and $\($ B = 5$\sqrt{5}$ $\)$. The eccentricity $\($ e_2 = $\sqrt{1 + \frac{A^2}{B^2}}$ = $\sqrt{\frac{11}{5}}$ $\)$. Therefore, $\($ 25e_2^2 = 25 $\left$( $\frac{11}{5}$ $\right$) = 55 $\)$.
Maths · Integrals · Fill in the blank
If $\int \dfrac{2x^2+5x+9}{\sqrt{x^2+x+1}}\,dx = x\sqrt{x^2+x+1} + \alpha\sqrt{x^2+x+1}$ $+ \beta\log_e\left|x + \dfrac{1}{2} + \sqrt{x^2+x+1}\right| + C$, where $C$ is the constant of integration, then $\alpha + 2\beta$ is equal to $\underline{\hspace{1cm}}$.
Answer: 16
Given $2x^2 + 5x + 9 = A(x^2 + x + 1) + B(2x + 1) + C$. $A = 2$, $B = \frac{3}{2}$, $C = \frac{11}{2}$. $$2 \int \sqrt{x^2 + x + 1} \, dx + \frac{3}{2} \int \frac{2x + 1}{\sqrt{x^2 + x + 1}} \, dx + \frac{11}{2} \int \frac{dx}{\sqrt{x^2 + x + 1}}$$ $$2 \int \sqrt{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, dx$$ $$2 \left( x + \frac{1}{2} \right) \sqrt{x^2 + x + 1} + \frac{3}{8} \ln \left( x + \frac{1}{2} + \sqrt{x^2 + x + 1} \right)$$ $$+ \frac{11}{2} \ln \left( x + \frac{1}{2} + \sqrt{x^2 + x + 1} \right) + C$$ $\alpha = \frac{7}{2}$, $\beta = \frac{25}{4}$ $\alpha + 2\beta = 16$
Physics · Wave Optics · Single correct
Young's double slit inteference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm. The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm. The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m, will be:
Answer: (a)
Given $\Delta W = \frac{\lambda D}{d \cdot \mu} = \frac{690 \times 10^{-9} \times 0.72}{1.5 \times 10^{-3} \times 1.44}$. Therefore, $\Delta W = 2.3 \times 10^{-4} \, \mathrm{m} = 0.23 \, \mathrm{mm}$.
Physics · Electromagnetic Waves · Single correct
Arrange the following in the ascending order of wavelength ($\lambda$): (A) Microwaves ($\lambda_1$) (B) Ultraviolet rays ($\lambda_2$) (C) Infrared rays ($\lambda_3$) (D) X-rays ($\lambda_4$) Choose the most appropriate answer from the options given below:
Answer: (d)
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) : A electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path.\ Reason (R): The magnetic field in that region is along the direction of velocity of the electron.\ In the light of the above statements, choose the correct answer from the options given below :
Answer: (c)
If the electron's velocity is along the direction of the magnetic field, then the magnetic force on the electron is zero and it will not accelerate.
Physics · System of Particles and Rotational Motion · Single correct
A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is:
Answer: (c)
Given $\mathrm{KE}_{(T)} = \frac{1}{2} mv^2$. $\mathrm{KE}_{(R)} = \frac{1}{2} \cdot \frac{2}{5} mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{2} mv^2 \left( \frac{2}{5} \right)$. So, $\frac{\mathrm{KE}_{(T)}}{\mathrm{KE}_{(R)}} = \frac{5}{2}$.
Physics · Moving Charges and Magnetism · Single correct
A long straight wire of a circular cross-section with radius ' $a$ ' carries a steady current $I$. The current $I$ is uniformly distributed across this cross-section. The plot of magnitude of magnetic field $B$ with distance $r$ from the centre of the wire is given by
Answer: (d)
We know inside the wire $$B = \frac{\mu_0 I}{2 \pi R^2} \cdot r (0 < r < R)$$ And $$B = \frac{\mu_0 I}{2 \pi r} for \ (R < r)$$
Physics · Kinetic Theory · Single correct
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases. Reason (R): Free expansion of an ideal gas is an irreversible and an adiabatic process. In the light of the above statements, choose the correct answer from the options given below :
Answer: (b)
If the container is insulated then temperature is expected to increase in adiabatic compression. So (A) is wrong. And free expansion is irreversible and adiabatic process. So (R) is correct.
Physics · Electric Charges and Fields · Single correct
In the first configuration (1) as shown in the figure, four identical charges ($q_0$) are kept at the corners $A$, $B$, $C$ and $D$ of square of side length $'a'$. In the second configuration (2), the same charges are shifted to mid points $G$, $E$, $H$ and $F$, of the square, If $K = \frac{1}{4\pi\varepsilon_0}$, the difference between the potential energies of configuration (2) and (1) is given by:
Answer: (d)
Given $$u_\oplus = \left( \frac{2Kq_0}{a} + \frac{Kq_0}{\sqrt{2a}} \right) q_0 \times 2$$ $$u_0 = \left( \frac{2Kq_0\sqrt{2}}{a} + \frac{Kq_0}{a} \right) q_0 \times 2$$ So, $$\Delta u = u_2 - u_1 = 2q_0 \frac{kq_0}{a} \left[ 2\sqrt{2} + 1 - 2 - \frac{1}{\sqrt{2}} \right]$$ $$\Rightarrow \Delta u = \frac{2q_0^2}{4\pi \varepsilon_0 a} \left[ 4 - \sqrt{2} - 1 \right] = \frac{2q_0^2 (3 - \sqrt{2})}{4\pi \varepsilon_0 a \sqrt{2}}$$ $$\Rightarrow \Delta u = \frac{2kq_0^2}{a} \left[ \frac{3 - \sqrt{2}}{\sqrt{2}} \right] = \frac{kq_0^2}{a} (3\sqrt{2} - 2)$$
Physics · Motion in a Plane · Single correct
The position vector of a moving body at any instant of time is given as $\vec{r} = \left( 5t^2 \hat{i} - 5t \hat{j} \right) \, \mathrm{m}$. The magnitude and direction of velocity at $t = 2 \, \mathrm{s}$ is,
Answer: (d)
Given $\vec{r} = 5t^2 \hat{i} - 5t \hat{j}$. At $t = 2 sec$, $\vec{v} = 10 \hat{i} - 5 \hat{j}$ and $\vec{v} = 20 \hat{i} - 5 \hat{j}$. The components are $v_x = 20$ and $v_y = -5$. $$\tan \theta = \frac{20}{5} = 4$$ $$\theta = \tan^{-1} 4$$ From the negative Y-axis.
Physics · System of Particles and Rotational Motion · Single correct
A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be $t_1$ and $t_2$, respectively, then
Answer: (c)
The time $t$ is given by the equation $$t = \sqrt{\frac{2\ell}{a_{cm}}}.$$ The acceleration $a_{cm}$ is given by $$a_{cm} = \frac{g \sin \theta}{1 + \frac{I_{cm}}{MR^2}}.$$ For a solid object, $$a_1 = a_{cm1} = \frac{5g \sin \theta}{7} Solid.$$ For a hollow object, $$a_2 = a_{cm2} = \frac{3g \sin \theta}{5} Hollow.$$ It follows that $a_1 > a_2$ and $t_1 < t_2$.
Physics · Thermal Properties of Matter · Single correct
Which of the following figure represents the relation between Celsius and Fahrenheit temperatures?
Answer: (d)
Given $\($ $\frac{C}{5}$ = $\frac{F - 32}{9}$ $\)$, we have $\($ C = $\frac{5}{9}$F - $\frac{160}{9}$ $\)$.
Physics · Moving Charges and Magnetism · Single correct
N equally spaced charges each of value $q$, are placed on a circle of radius $R$. The circle rotates about its axis with an angular velocity $\omega$ as shown in the figure. A bigger Amperian loop $B$ encloses the whole circle where as a smaller Amperian loop $A$ encloses a small segment. The difference between enclosed currents, $I_A - I_B$, for the given Amperian loops is
Answer: (d)
The current at point A is given by: $$I_A = \frac{Nq}{2\pi}$$ The current at point A due to angular velocity $\omega$ is: $$I_A = \frac{Nq\omega}{2\pi}$$ The current at point B is: $$I_B = 0$$ Thus, the currents at points A and B are equal: $$I_A = I_B = \frac{Nq\omega}{2\pi}$$
Physics · Dual Nature of Radiation and Matter · Single correct
In photoelectric effect, the stopping potential $(V_0)$ vs frequency $(\nu)$ curve is plotted.\ $(h$ is the Planck's constant and $\phi_0$ is work function of metal$)$\ \ (A) $V_0$ vs $\nu$ is linear.\ (B) The slope of $V_0$ vs $\nu$ curve $= \dfrac{\phi_0}{h}$.\ (C) $h$ constant is related to the slope of $V_0$ vs $\nu$ line.\ (D) The value of electric charge of electron is not required to determine $h$ using the $V_0$ vs $\nu$ curve.\ (E) The work function can be estimated without knowing the value of $h$.\ Choose the correct answer from the options given below:
Answer: (b)
Given $h\nu = \phi + \mathrm{KE}_{\max}$. $\newline$ $\mathrm{KE}$_{$\max$} = eV_0 $\newline$ V_0 = $\frac{h\nu - \phi}{e}$ $\newline$ (A) $\;$ V_0 V/s V $\;$ is linear correct $\newline$ (B) $\;$ Slope $\newline$ v_0 = $\left$( $\frac{h}{e}$ $\right$) v - $\frac{\phi}{e}$ $\;$ Wrong $\newline$ Slope $\;$ $\frac{h}{e}$ $\newline$ ($C$) $\;$ Correct $\newline$ (D) $\;$ Incorrect $\newline$ (E) $\;$ Correct
Physics · Thermal Properties of Matter · Single correct
The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure) is (in SI unit):
Answer: (a)
The work done $W$ is given by the formula $W = \frac{1}{2} \pi R^2$. Substituting the values, we have: $$W = \frac{1}{2} \times \pi \times \left( \frac{200}{2} \times 10^3 \right) \times \frac{200}{2} \times 10^{-6}$$ Simplifying, we get: $$W = \frac{10\pi}{2} = 5\pi \, \mathrm{J}$$
Physics · Ray Optics and Optical Instruments · Single correct
A photograph of a landscape is captured by a drone camera at a height of $18\,\mathrm{km}$. The size of the camera film is $2\,\mathrm{cm}\times2\,\mathrm{cm}$, and the area of the landscape photographed is $400\,\mathrm{km^2}$. The focal length of the lens in the drone camera is:
Answer: (a)
Given: $H = 18 \, \mathrm{km}$ Size of camera film $= 2 \, \mathrm{cm} \times 2 \, \mathrm{cm}$ $A_{image} = 400 \, \mathrm{km^2}$ $x = 20 \times 10^3 \, \mathrm{m} = 2 \times 10^4 \, \mathrm{m}$ $y = 2 \times 10^{-2} \, \mathrm{m}$ $x = \frac{18 \, \mathrm{Km}}{y} = \frac{10^6}{f}$ $f = 18 \times 10^{-3} \, \mathrm{m} = 18 \, \mathrm{mm}$ $f = 1.8 \, \mathrm{cm}$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The output of the circuit is low (zero) for: \ $(A)$ $X=0,\ Y=0$\ $(B)$ $X=0,\ Y=1$\ $(C)$ $X=1,\ Y=0$\ $(D)$ $X=1,\ Y=1$\ Choose the correct answer from the options given below:
Answer: (a)
\begin{tabular}{|c|c|c|} \hline x & y & \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 0 \\ \hline 1 & 1 & 0 \\ \hline \end{tabular}
Physics · Thermal Properties of Matter · Single correct
The temperature of a body in air falls from $40^\circ \mathrm{C}$ to $24^\circ \mathrm{C}$ in 4 minutes. The temperature of the air is $16^\circ \mathrm{C}$. The temperature of the body in the next 4 minutes will be:
Answer: (d)
Given $\Delta T = T_2 - T_1 = 16^\circ \mathrm{C}$ and $T_0 = 16^\circ \mathrm{C}$. $$\frac{\Delta T}{t_1} = -k(32 - 16^\circ) \ldots (i)$$ $$\frac{(24 - T_3)}{4} = -k \left( \frac{24 + T_3}{2} - 16 \right) \ldots (ii)$$ $$\frac{16}{4} = -k(16)$$ $$\Rightarrow \frac{(24 - T_3)}{4} = -k \left( 12 + \frac{T_3}{2} - 16 \right)$$ $$\Rightarrow \frac{16}{24 - T_3} = \frac{16}{T_2}$$ $$\Rightarrow \frac{T_3 - 4}{2} = 24 - T_3$$ $$\Rightarrow \frac{3T_3}{2} = 28$$ $$\Rightarrow T_3 = \frac{56}{3}^\circ \mathrm{C}$$
Physics · Dual Nature of Radiation and Matter · Single correct
The energy $E$ and momentum $p$ of a moving body of mass $m$ are related by some equation. Given that $c$ represents the speed of light, identify the correct equation
Answer: (d)
We need to check the dimensions only. With momentum the dimension $$E^2 = p^2 C^2$$ And with mass $$E^2 = m^2 c^4$$ So $$E^2 = p^2 C^2 + m^2 c^4$$ (dimensionally)
Physics · Electric Charges and Fields · Single correct
A small uncharged conducting sphere is placed in contact with an identical sphere but having $4 \times 10^{-8} \, \mathrm{C}$ charge and then removed to a distance such that the force of repulsion between them is $9 \times 10^{-3} \, \mathrm{N}$. The distance between them is (Take $\frac{1}{4\pi\varepsilon_0}$ as $9 \times 10^9$ in SI units)
Answer: (b)
Given $Q = 4 \times 10^{-8}$. The force $F$ is given by $$F = k \left( \frac{\theta}{2} \right) \left( \frac{\theta}{2} \right) \frac{1}{r^2}$$ Substituting the values, $$9 \times 10^{-3} = 9 \times 10^9 \times (4 \times 10^{-8}) \times 4 \times 10^{-8} \frac{1}{4 \times r^2}$$ Solving for $r^2$, $$r^2 = \frac{9 \times 10^9 \times 16 \times 10^{-16}}{4 \times 9 \times 10^{-3}} = 4 \times 10^{-4}$$ Therefore, $$r = 2 \times 10^{-2} \, \mathrm{m} \Rightarrow 2 \, \mathrm{cm}$$
Physics · Oscillations · Single correct
A particle oscillates along the $x$-axis according to the law, $x(t) = x_0 \sin^2 \left( \frac{t}{2} \right)$ where $x_0 = 1 \, \mathrm{m}$. The kinetic energy (K) of the particle as a function of $x$ is correctly represented by the graph
Answer: (d)
Given $x(t) = x_0 \sin^2\left(\frac{t}{2}\right) = \frac{x_0}{2} (1 - \cos t)$. Clearly $\frac{x_0}{2}$ is mean position.
Physics · Wave Optics · Single correct
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of $P_1$ and $P_2$ are orthogonal to each other. The polarizer $P_3$ covers both the slits with its transmission axis at $45^\circ$ to those of $P_1$ and $P_2$. An unpolarized light of wavelength $\lambda$ and intensity $I_0$ is incident on $P_1$ and $P_2$. The intensity at a point after $P_3$ where the path difference between the light waves from $s_1$ and $s_2$ is $\frac{\lambda}{3}$, is
Answer: (b)
After passing through the third polariser, the intensity of both the waves must be $\frac{I_0}{4}$. Now, at a point where the path difference is $\frac{\lambda}{3}$, the phase difference is $$\Delta \phi = 2K \left( \frac{\Delta x}{\lambda} \right) = \frac{2\pi}{3}.$$ Therefore, $$I_{res} = \sqrt{\left( \frac{I_0}{4} \right)^2 + \left( \frac{I_0}{4} \right)^2 + 2 \left( \frac{I_0}{4} \right)^2 \cos \frac{2\pi}{3}} = \frac{I_0}{4}.$$
Physics · Moving Charges and Magnetism · Numerical
A tightly wound long solenoid carries a current of $1.5\,\mathrm{A}$. An electron is executing uniform circular motion inside the solenoid with a time period of $75\,\mathrm{ns}$. The number of turns per metre in the solenoid is $\underline{\hspace{1cm}}$. [Take mass of electron $m_e=9\times10^{-31}\,\mathrm{kg}$, charge of electron $|q_e|=1.6\times10^{-19}\,\mathrm{C}$, $\mu_0=4\pi\times10^{-7}\,\mathrm{N\,A^{-2}}$, and $1\,\mathrm{ns}=10^{-9}\,\mathrm{s}$.]
Answer: 250
Since the time period of a revolving charge is $\frac{2\pi m}{qB}$. Where $B =$ magnetic field due to a solenoid $= \mu_0 n I$. Therefore, $T = \frac{2\pi \, \mathrm{m}}{q (\mu_0 n I)}$. $$75 \times 10^{-9} = \frac{(2\pi) (9 \times 10^{-31})}{1.6 \times 10^{-19} \times 4\pi \times 10^{-7} \times n \times 1.5}$$ $N = 250$
Physics · Laws of Motion · Fill in the blank
A string of length $L$ is fixed at one end and carries a mass of $M$ at the other end. The mass makes $\left( \frac{3}{\pi} \right)$ rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is $\ldots$$\ldots$ ML.
Answer: 36
Given $\omega = \frac{3}{\pi} \times 2\pi = 6 \, \mathrm{rad/s}$. $R = L \sin \theta$ and $T = M \sqrt{g^2 + \omega^4 R^2}$. Also, $T \sin \theta = M \omega^2 \cdot L \sin \theta$. Therefore, $$T = M(36)L$$ Thus, $$T = 36ML$$
Physics · Dual Nature of Radiation and Matter · Fill in the blank
The ratio of the power of a light source $S_1$ to that the light source $S_2$ is 2. $S_1$ is emitting $2 \times 10^{15}$ photons per second at 600 nm. If the wavelength of the source $S_2$ is 300 nm, then the number of photons per second emitted by $S_2$ is $\ldots \times 10^{14}$.
Answer: 5
Since power emitting by a source is given as total energy emitted over time, we have: $$= \frac{(E_1 photon) \times Number of photons (N)}{t}$$ $$P_1 = (E_1) n$$ $$P_1 = (E_1) n_1 = \left( \frac{hC}{\lambda_1} \right) n_1$$ $$P_2 = (E_2) n_2 = \left( \frac{hC}{\lambda_2} \right) n_2$$ $$\frac{P_1}{P_2} = \left( \frac{\lambda_2}{\lambda_1} \right) \frac{n_1}{n_2}$$ Substituting the given values: $$2 = \left( \frac{300}{600} \right) \times \frac{2 \times 10^{15}}{n_2}$$ $$n_2 = \frac{1}{2} \times 10^{15} = 5 \times 10^{14} Photon/sec$$
Physics · Mechanical Properties of Solids · Numerical
The increase in pressure required to decrease the volume of a water sample by 0.2$\%$ is $P \times 10^5 \, \mathrm{Nm}^{-2}$. Bulk modulus of water is $2.15 \times 10^9 \, \mathrm{Nm}^{-2}$. The value of $P$ is $\ldots$ $\ldots$
Answer: 43
Since bulk modulus is given as $$B = \frac{-\Delta P}{\left( \frac{\Delta V}{V} \right)}$$ $$2.15 \times 10^9 = \frac{-\Delta P}{\left( \frac{0.2}{100} \right)}$$ $$\Delta P = 2.15 \times 10^9 \times 2 \times 10^{-3}$$ $$= 4.3 \times 10^6 = 43 \times 10^5 \, \mathrm{N/m^2}$$
Physics · Gravitation · Numerical
Acceleration due to gravity on the surface of earth is $'g'$. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is
Answer: 9
Acceleration due to gravity on the surface is given by $$g = \frac{GM}{R_e^2}$$ Now since diameter is reduced to $1/3^{rd}$, radius also reduces to $1/3^{rd}$, keeping mass constant. New value of acceleration due to gravity on Earth's surface is $$g' = \frac{GM}{\left(\frac{R_e}{3}\right)^2} = 9 \frac{GMe}{R_e^2} = 9g$$
Chemistry · Electrochemistry · Single correct
Based on the data given below : $E^\circ_{\mathrm{Cr_2O_7^{2-}/Cr^{3+}}}=1.33\,\mathrm{V}\qquad\qquad E^\circ_{\mathrm{Cl_2/Cl^-}}=1.36\,\mathrm{V}$ $E^\circ_{\mathrm{MnO_4^-/Mn^{2+}}}=1.51\,\mathrm{V}\qquad\qquad E^\circ_{\mathrm{Cr^{3+}/Cr}}=-0.74\,\mathrm{V}$ the strongest reducing agent is :
Answer: (a)
Given $E^0_{\mathrm{Cr_2O_7^{2-}/Cr^{3+}}} = 1.33 \, \mathrm{V}$ and $E^0_{\mathrm{C_2/Cr}} = 1.36 \, \mathrm{V}$. $E^0_{\mathrm{MnO_4^-/Mn^{2+}}} = 1.51 \, \mathrm{V}$ and $E^0_{\mathrm{Cr^{3+}/Cr}} = -0.74 \, \mathrm{V}$. The species which has the most negative value of standard reduction potential will be the strongest reducing agent. Since $\mathrm{Cr^{3+}/Cr}$ has SRP value of $-0.74 \, \mathrm{V}$, $\mathrm{Cr}$ is the strongest reducing agent.
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
Given below are two statements: In the light of the above statements, choose the correct answer from the options given below:
Answer: (a)
For a first order reaction, $t_{1/2} = \frac{\ln 2}{k}$. The graph of $t_{1/2}$ versus $[R]_0$ is a horizontal line. For a first order reaction, the plot of $\log \frac{[R]_0}{[R]}$ versus time is a straight line with a slope of $\frac{k}{2.303}$. The equation is given by $$\log \frac{[R]_0}{[R]} = \frac{1}{2.303} kt$$ or $$\log \frac{[R]_0}{[R]} = \left( \frac{k}{2.303} \right) \times t.$$
Chemistry · Amines · Single correct
For reaction The correct order of set of reagents for the above conversion is :
Answer: (b)
The reaction begins with aniline ($\mathrm{NH_2}$) reacting with concentrated sulfuric acid ($\mathrm{H_2SO_4}$) to form anilinium hydrogen sulfate. This intermediate is heated to $453 - 473 \, \mathrm{K}$ to produce sulfanilic acid ($\mathrm{NH_2}$ and $\mathrm{SO_3H}$ groups on the benzene ring). Acetic anhydride ($\mathrm{Ac_2O}$) is then used to acetylate the amino group, forming acetanilide with a $\mathrm{SO_3H}$ group. Bromination with $\mathrm{Br_2}$ in water and heat ($\Delta$) introduces a bromine atom on the benzene ring. Finally, treatment with sodium hydroxide ($\mathrm{NaOH}$) removes the acetyl group, regenerating the amino group and yielding the final product with $\mathrm{NH_2}$ and $\mathrm{Br}$ groups on the benzene ring.
Chemistry · Structure of Atom · Single correct
For hydrogen atom, the orbital/s with lowest energy is/are: $(A)\ 4s$ $(B)\ 3p_x$ $(C)\ 3d_{x^2-y^2}$ $(D)\ 3d_{z^2}$ $(E)\ 4p_z$ Choose the correct answer from the options given below :
Answer: (b)
For hydrogen atom and one electron species, the energy of orbitals is decided by the value of principal quantum number. Higher the value of principal quantum number, higher will be the energy of orbital. (A) $4s$ $n = 4$ (B) $3p_x$ $n = 3$ (C) $3d_{z^2-y^2}$ $n = 3$ (D) $3d_{z^2}$ $n = 3$ (E) $4p_z$ $n = 4$ Therefore, (B), (C) and (D) have orbitals with the lowest energy.
Chemistry · Chemical Bonding and Molecular Structure · Single correct
In the given structure, number of $sp$ and $sp^2$ hybridized carbon atoms present respectively are :
Answer: (b)
Number of $sp$ and $sp^2$ hybridised carbon atom are 3 and 5.
Chemistry · Thermodynamics · Single correct
Which of the following mixing of 1 M base and 1 M acid leads to the largest increase in temperature?
Answer: (c)
The rise in temperature of neutralization reaction will be maximum for maximum number of moles of strong acid and strong base neutralized and lower volume of final solution. $$\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$$ $$mmol 30 30$$ Final volume of solution = 60 mL Option (1) and (2) have weak acids and in option (4) only 20 mmol of HCl will be neutralized with 70 mL final volume.
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Given below are two statements : Statement (I): Experimentally determined oxygen-oxygen bond lengths in the $\mathrm{O}_3$ are found to be same and the bond length is greater than that of a $\mathrm{O} = \mathrm{O}$ (double bond) but less than that of a single $(\mathrm{O} - \mathrm{O})$ bond. Statement (II) : The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact that the bond length in ozone is smaller than that of a double bond $(\mathrm{O} = \mathrm{O})$ but more than that of a single bond $(\mathrm{O} - \mathrm{O})$. In the light of the above statements, choose the correct answer from the options given below :
Answer: (c)
Bond length is proportional to $\frac{1}{Bond order}$. Order of $\mathrm{O} - \mathrm{O}$ bond length is $\mathrm{O} = \mathrm{O} < \mathrm{O}_3 < \mathrm{O} - \mathrm{O}$. Therefore, Statement I is true. Lone pair-lone pair repulsion between O-atoms is not solely responsible for the correct order of O-O bond length. Bond order also should be considered. Therefore, Statement II is false.
Chemistry · The d-and f-Block Elements · Single correct
Find the compound '$A$' from the following reaction sequence. $A \xrightarrow{\text{aqua-regia}} B \xrightarrow[(2)\ \mathrm{AcOH}]{(1)\ \mathrm{KNO_2/NH_4OH}} \text{yellow ppt}$
Answer: (a)
Compound (A) in the given reaction sequence is likely to be CoS. (1) $\mathrm{CoS} + \mathrm{HNO_3} + 3\mathrm{HCl} \rightarrow \mathrm{Co^{2+}} + \mathrm{S} \downarrow + \mathrm{NOCl} \uparrow + 2\mathrm{Cl^-} + 2\mathrm{H_2O}$ The above solution is neutralised with $\mathrm{NH_4OH}$. To a neutral solution of $\mathrm{Co^{2+}}$, acetic acid and saturated solution of $\mathrm{KNO_2}$ are added which results in the formation of yellow precipitate of $K_3[Co(NO_2)_6]$ $Co^{2+}$ + 7$NO_2^-$ + 2$H^+$ + 3 $K^+$ $\rightarrow$ $K_3[Co(NO_2)_6]$ + $NO$ $\uparrow$ + $H_2O$
Chemistry · Equilibrium · Single correct
For the reaction, $$\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$$ Attainment of equilibrium is predicted correctly by:
Answer: (b)
The reaction is given by $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$. Concentrations of $\mathrm{H_2(g)}$ and $\mathrm{I_2(g)}$ decrease with time while the concentration of $\mathrm{HI(g)}$ increases with time. At equilibrium, $\mathrm{H_2(g)}$, $\mathrm{I_2(g)}$, and $\mathrm{HI(g)}$ attain constant values. The correct plot of molar concentration with time is
Chemistry · The d-and f-Block Elements · Single correct
Match List - I with List - II. Choose the correct answer from the options given below :
Answer: (a)
Sc^{+3} = 3$\,$d^0 $\therefore$ $\mu$_{spin} = 0 $\\$ V^{+2} = 3$\,$d^3 $\therefore$ $\mu$_{spin} = 3.87 $\,$ B.M. $\\$ Ni^{+2} = 3$\,$d^8 $\therefore$ $\mu$_{spin} = 2.84 $\,$ B.M. $\\$ Ti^{+3} = 3$\,$d^1 $\therefore$ $\mu$_{spin} = 1.73 $\,$ B.M.
Chemistry · Some Basic Concepts of Chemistry · Single correct
The elemental composition of a compound is 54.2$\%$C, 9.2$\%$H and 36.6$\%$O. If the molar mass of the compound is 132 $\mathrm{g \, mol^{-1}}$, the molecular formula of the compound is : [Given : The relative atomic mass of C : H : O = 12 : 1 : 16 ]
Answer: (d)
Empirical formula of compound is $\mathrm{C_2H_4O}$. Molecular mass of compound $= 132 \, \mathrm{g \, mol^{-1}}$. Molecular formula of compound is $(\mathrm{C_2H_4O})_n$. $$n = \frac{Molecular mass}{EF mass} = \frac{132}{44} = 3$$ Therefore, molecular formula of compound is $\mathrm{C_6H_{12}O_3}$.
Chemistry · Co-ordination Compounds · Single correct
When Ethane-1,2-diamine is added progressively to an aqueous solution of Nickel (II) chloride, the sequence of colour change observed will be:
Answer: (c)
Q19. $[\mathrm{Ni(H_2O)_6}]^{+2}_{(aq)} + \mathrm{en}_{(aq)} \rightarrow [\mathrm{Ni(H_2O)_4(en)}]^{+2}_{(aq)} + 2\mathrm{H_2O}$ Green $[\mathrm{Ni(H_2O)_4(en)}]^{+2}_{(aq)} + \mathrm{en}_{(aq)} \rightarrow [\mathrm{Ni(H_2O)_2(en)_2}]^{+2}_{(aq)} + 2\mathrm{H_2O}$ Pale Blue $[\mathrm{Ni(H_2O)_2(en)_2}]^{+2}_{(aq)} + \mathrm{en}_{(aq)} \rightarrow [\mathrm{Ni(en)_3}]^{+2}_{(aq)} + 2\mathrm{H_2O}$ Blue / purple
Chemistry · Co-ordination Compounds · Single correct
The conditions and consequence that favours the $t_{2g}^{3}e_g^{1}$ configuration in a metal complex are:
Answer: (b)
The conditions and consequence that favour $t_{2g}^3 e_g^1$ configuration in a metal complex are (i) weak field ligand, and (ii) high spin complex. For weak field ligands, splitting energy ($\Delta_0$) is lower than pairing energy ($P$). As a result, distribution of electrons for $3\,d^4$ will be $t_{2g}^3 e_g^1$. It results in high spin complex due to maximum number of unpaired electrons.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Identify correct statement/s : (A) $-\mathrm{OCH}_3$ and $-\mathrm{NHCOCH}_3$ are activating group. (B) $-\mathrm{CN}$ and $-\mathrm{OH}$ are meta directing group. (C) $-\mathrm{CN}$ and $-\mathrm{SO}_3\mathrm{H}$ are meta directing group. (D) Activating groups act as ortho - and para directing groups. (E) Halides are activating groups. Choose the correct answer from the options given below :
Answer: (d)
-OCH$_3$, NHCOCH$_3$ and -OH are activating groups because the atom directly bonded to benzene ring activates the ring by +R effect using its lone pair of electrons. Activating groups are ortho- and para directing groups. -CN and -SO$_3$H are deactivating groups because the atom directly bonded to benzene is bonded to more electronegative atom through multiple bonds and they are meta directing groups due to -R effect. Halides are deactivating groups due to -I effect.
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Given below are two statements: Statement (I): The first ionization energy of Pb is greater than that of Sn. Statement (II): The first ionization energy of Ge is greater than that of Si. In the light of the above statements, choose the correct answer from the options given below:
Answer: (b)
First ionization energy of Lead $= 715 \, \mathrm{kJ \, mol^{-1}}$. First ionization energy of Tin $= 708 \, \mathrm{kJ \, mol^{-1}}$. $(\mathrm{IE_1})$ of Lead is greater than that of Tin due to ineffective shielding of d- and f-electrons. Therefore Statement-I is true. First ionization energy of Germanium $= 761 \, \mathrm{kJ \, mol^{-1}}$. First ionization energy of Silicon $= 786 \, \mathrm{kJ \, mol^{-1}}$. $(\mathrm{IE_1})$ of Germanium is lower than that of Silicon as the effect of higher atomic radius of Ge outweighs the increase in nuclear charge from Si to Ge and effective shielding of inner electrons. Therefore Statement-II is false.
Chemistry · Thermodynamics · Single correct
$S(g) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) + 2x \mathrm{kcal}$ \ $SO_2(g) + \frac{1}{2} O_2(g) \rightarrow SO_3(g) + y \mathrm{kcal}$ \ The heat of formation of $SO_2(g)$ is given by:
Answer: (d)
(i) $\mathrm{S(g)} + \frac{3}{2} \mathrm{O_2(g)} \rightarrow \mathrm{SO_3(g)} + 2x kcal \Delta H_1$ (ii) $\mathrm{SO_2(g)} + \frac{1}{2} \mathrm{O_2(g)} \rightarrow \mathrm{SO_3(g)} + y kcal \Delta H_2$ (i) - (ii) $\mathrm{S(g)} + \mathrm{O_2(g)} \rightarrow \mathrm{SO_2(g)} \Delta H$ $$\Delta H = \Delta H_1 - \Delta H_2$$ $$= -2x - (-y) = (y - 2x) kcal$$
Chemistry · Co-ordination Compounds · Single correct
Match List - I with List - II.
Answer: (d)
The Stephen reaction involves the conversion of a nitrile (RCN) to an aldehyde (RCHO) using $\mathrm{SnCl_2}$ and $\mathrm{HCl}$ followed by hydrolysis. The Rosenmund reduction involves the reduction of an acyl chloride to an aldehyde using $\mathrm{H_2}$ and $\mathrm{Pd-BaSO_4}$. The Etard reaction involves the oxidation of a methyl group attached to an aromatic ring to an aldehyde using $\mathrm{CrO_2Cl_2}$ and $\mathrm{CS_2}$ followed by hydrolysis with $\mathrm{H_3O^+}$. The Gattermann-Koch reaction involves the formylation of an aromatic ring using $\mathrm{CO}$ and $\mathrm{HCl}$ in the presence of anhydrous $\mathrm{AlCl_3/CuCl}$.
Chemistry · Haloalkanes and Haloarenes · Single correct
The structure of the major product formed in the following reaction is:
Answer: (c)
Haloalkanes react with $\mathrm{AgCN}$ to give isocyanide as the major product and haloarenes do not react with $\mathrm{AgCN}$.
Chemistry · Biomolecules · Single correct
Match List - I with List - II. Choose the correct answer from the options given below:
Answer: (b)
The correct matching of compounds with their structures is as follows: Adenine corresponds to structure III, Cytosine corresponds to structure IV, Thymine corresponds to structure II, and Uracil corresponds to structure I.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The successive 5 ionisation energies of an element are 800, 2427, 3658, 25024 and 32824 $\mathrm{kJ/mol}$, respectively. By using the above values predict the group in which the above element is present:
Answer: (a)
The successive 5 ionisation energies of an element are 800, 2427, 3658, 25024 and 32824 kJ/mol, respectively. $$\frac{\mathrm{IE}_2}{\mathrm{IE}_1} = \frac{2427}{800} = 3.03$$ $$\frac{\mathrm{IE}_3}{\mathrm{IE}_2} = \frac{3658}{2427} = 1.51$$ $$\frac{\mathrm{IE}_4}{\mathrm{IE}_3} = \frac{25024}{3658} = 6.84$$ $$\frac{\mathrm{IE}_5}{\mathrm{IE}_4} = \frac{32824}{25024} = 1.31$$ Since $\left( \frac{\mathrm{IE}_4}{\mathrm{IE}_3} \right)$ value is maximum, the element belongs to group 13.
Chemistry · Solutions · Numerical
The observed and normal molar masses of compound $MX_2$ are 65.6 and 164 respectively. The percent degree of ionisation of $MX_2$ is $\ldots$ $\ldots$ $\%$ (Nearest integer)
Answer: 75
Normal molar mass of MX_2 = 164.0 $\,$ $\mathrm{g \, mol^{-1}}$. Observed molar mass of MX_2 = 65.6 $\,$ $\mathrm{g \, mol^{-1}}$. Van’t Hoff factor (i) = $\frac{Normal molar mass}{Observed molar mass}$ = $\frac{164}{65.6}$ = 2.5. If $\alpha$ is the degree of ionisation, then $$\mathrm{MX_2 \rightleftharpoons \frac{\mathrm{M^{2+}}}{1-\alpha} + \frac{2\mathrm{X^-}}{2\alpha}}$$ $$i = 1 - \alpha + \alpha + 2\alpha = 1 + 2\alpha$$ $$1 + 2\alpha = 2.5$$ $$\alpha = 0.75$$ Therefore, percent degree of ionisation = 75$\%$.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical
The possible number of stereoisomers for 5-phenylpent-4-en-2-ol is $\ldots$ $\ldots$
Answer: 4
The given compound is 5-phenylpent-4-en-2-ol. Possible stereoisomers are trans (±) and cis (±). Therefore, there are 4 possible stereoisomers.
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
Consider a complex reaction taking place in three steps with rate constants $k_1$, $k_2$ and $k_3$ respectively. The overall rate constant $k$ is given by the expression $k = \sqrt{\frac{k_1 k_3}{k_2}}$. If the activation energies of the three steps are $60$, $30$ and $10 \, \mathrm{kJ \, mol^{-1}}$ respectively, then the overall energy of activation in $\mathrm{kJ \, mol^{-1}}$ is $\ldots$$\ldots$ (Nearest integer)
Answer: 20
Given $k_1$, $k_2$ and $k_3$ are given as rate constants of three steps of a complex reaction. Rate constant $(k)$ of the overall reaction is given as $$k = \sqrt{\frac{k_1 k_3}{k_2}}$$ Activation energies of the three steps are given as $E_{a_1} = 60 \, \mathrm{kJ \, mol^{-1}}$, $E_{a_2} = 30 \, \mathrm{kJ \, mol^{-1}}$, $E_{a_3} = 10 \, \mathrm{kJ \, mol^{-1}}$. From Arrhenius equation, we know that $$k = A e^{-E_a/RT}$$ If $E_a$ is the activation energy of the overall reaction, then $$E_a = \frac{1}{2} \left[ E_{a_1} + E_{a_3} - E_{a_2} \right]$$ $$= \frac{1}{2} \left[ 60 + 10 - 30 \right] = 20 \, \mathrm{kJ \, mol^{-1}}$$
Chemistry · Hydrocarbons · Numerical
The hydrocarbon (X) with molar mass $80 \, \mathrm{g \, mol^{-1}}$ and $90\%$ carbon has $\ldots$ degree of unsaturation.
Answer: 3
Therefore, the empirical formula is $\mathrm{C_3H_4}$. Molecular mass $= 80 \, \mathrm{g \, mol^{-1}}$. Molecular formula $= (\mathrm{C_3H_4})_n$. $$n = \frac{Molecular mass}{EF mass} = \frac{80}{40} = 2$$ Therefore, the molecular formula is $\mathrm{C_6H_8}$. Degree of unsaturation $= \frac{2 \times 6 + 2 - 8}{2} = 3$
Chemistry · Analytical Chemistry · Numerical
In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is $\ldots \times 10^{-1}\%$ (Nearest integer). (Given : Molar mass of Ag is 108 and Br is 80 $\mathrm{g}\;\mathrm{mol}^{-1}$)
Answer: 255
Mass of organic compound = 0.25 g. Mass of AgBr = 0.15 g. Number of moles of Br = Number of moles of AgBr = $\($ $\frac{0.15}{188}$ $\)$. Mass of Br = $\($ $\frac{0.15 \times 80}{188}$ $\)$ g. Percentage of Br = $\($ $\frac{0.15 \times 80 \times 100}{188 \times 0.25}$ $\)$. $\($ = 25.5$\%$ $\)$. $\($ = 255 $\times$ 10^{-1}$\%$ $\)$.