JEE Main 2 April 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 · Permutations and Combinations · Single correct
The largest $n \in \mathbb{N}$ such that $3^n$ divides $50!$ is:
Answer: (b)
Given $2^\alpha \cdot 3^\beta \cdot 5^\gamma$. $B = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{3^2} \right\rfloor + \left\lfloor \frac{50}{3^3} \right\rfloor + \left\lfloor \frac{50}{3^4} \right\rfloor$ $= 16 + 5 + 1$ $= 2$ Maximum value of $n$ is 22.
Maths · Conic Sections · Single correct
Let one focus of the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be at $(\sqrt{10}, 0)$ and the corresponding directrix be $x = \frac{9}{\sqrt{10}}$. If $e$ and $l$ respectively are the eccentricity and the length of the latus rectum of $H$, then $9 \left(e^2 + l\right)$ is equal to:
Answer: (c)
Given $ae = \sqrt{10}$ and $\frac{a}{e} = \frac{9}{10}$. Therefore, $a^2 = 9$ and $e = \frac{\sqrt{10}}{3}$. $$(ae)^2 = a^2 + b^2$$ $$10 = 9 + b^2 \Rightarrow b^2 = 1$$ Now, $$\ell = \frac{2\, b^2}{a} = \frac{2(1)}{3}$$ Therefore, $$9 \left(e^2 + \ell \right)$$ $$= 9 \left(\frac{10}{9} + \frac{2}{3} \right)$$ $$= 10 + 6$$ $$= 16$$
Maths · Permutations and Combinations · Single correct
The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to
Answer: (c)
No. of sequences = $$\frac{10!}{5!3!2!} = 2520$$ Note: Sequence can start with 0.
Maths · Continuity and Differentiability · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be a twice differentiable function such that $(\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y))$, for all $x, y \in \mathbb{R}$. If $f'(0) = \frac{1}{2}$, then the value of $24f''\left(\frac{5\pi}{3}\right)$ is:
Answer: (b)
Given $\sin x \cos y (f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)$. Let $f(2x + 2y) + f(2x - 2y)$. Then $f(2x + 2y)(\sin(x - y)) = f(2x - 2y) \sin(x + y)$. Therefore, $f(2x + 2y) = f(2x - 2y)$. We have $\sin(x + y) = \sin(x - y)$. Put $2x + 2y = m, 2x - 2y = n$. Then $f(m) = f(n) = K$. Thus, $f(m) = K \sin\left(\frac{m}{2}\right)$. Therefore, $f(x) = K \sin\left(\frac{x}{2}\right)$. The derivative is $f'(x) = \frac{K}{2} \cos\left(\frac{x}{2}\right)$. Put $x = 0$; $1 = \frac{K}{2} \Rightarrow K = 1$. Therefore, $f'(x) = \frac{1}{2} \cos\left(\frac{x}{2}\right)$. The second derivative is $f''(x) = -\frac{1}{4} \sin\left(\frac{x}{2}\right)$. Then $4f''\left(\frac{5\pi}{3}\right) = \left(-\frac{1}{4} \sin\left(\frac{5\pi}{6}\right)\right) 24 = -\frac{24}{8} = -3$.
Maths · Matrices · Single correct
Let $A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix}$, $\alpha > 0$, such that $\det(A) = 0$ and $\alpha + \beta = 1$. If $I$ denotes $2 \times 2$ identity matrix, then the matrix $(1 + A)^8$ is:
Answer: (d)
Given $|A| = 0$. $\alpha \beta + 6 = 0$ $\alpha \beta = -6$ $\alpha + \beta = 1$ Therefore, $\alpha = 3$, $\beta = -2$. $A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}$ $A^2 = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}$ Thus, $A^2 = A$. $A = A^2 = A^3 = A^4 = A^5$ $(I + A)^8$ $= I + \binom{8}{1} A^7 + \binom{8}{2} A^6 + \ldots + \binom{8}{8} A^8$ $= I + A \left( \binom{8}{1} + \binom{8}{2} + \ldots + \binom{8}{8} \right)$ $= I + A (2^8 - 1)$ $= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 765 & -255 \\ 1530 & -510 \end{bmatrix}$ $= \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix}$
Maths · Binomial Theorem · Single correct
The term independent of $x$ in the expansion of $$\left( \frac{(x+1)}{(x^{2/3}+1-x^{1/3})} - \frac{(x+1)}{(x-x^{1/2})} \right)^{10}, \ x > 1$$ is:
Answer: (a)
Given $$\left( \frac{(x+1)}{x^3+1-x^\frac{1}{3}} - \frac{(x-1)}{x-x^\frac{1}{2}} \right)^{10}$$. This simplifies to $$\left( \frac{x^\frac{1}{3} + 1}{\sqrt{x} + 1} \right)^{10}$$. Further simplification gives $$\left( \frac{x^\frac{1}{3} - 1}{\sqrt{x}} \right)^{10}$$. The term $$T_{r+1} = \binom{10}{r} \left( x^\frac{1}{3} \right)^{10-r} (-1)^r (x)^\frac{-r}{2}$$. Setting the exponents equal: $$10 - r - \frac{r}{2} = 0$$ Solving gives: $$3 - 2r = 0$$ Thus, $$(20 - 2r) - 3r = 0$$ Solving for $$r$$ gives $$r = 4$$. Therefore, $$\binom{10}{4}(-1)^4 = 210$$.
Maths · Trigonometric Functions · Single correct
If $\theta \in [-2\pi, 2\pi]$, then the number of solutions of $2\sqrt{2} \cos^2 \theta + (2 - \sqrt{6}) \cos \theta - \sqrt{3} = 0$, is equal to:
Answer: (c)
Given the equation $2\sqrt{2} \cos^2 \theta + 2 \cos \theta - \sqrt{6} \cos \theta - \sqrt{3} = 0$. Rewriting, we have: $$(2 \cos \theta - \sqrt{3})(\sqrt{2} \cos \theta + 1) = 0$$ Solving for $\cos \theta$, we get: $$\cos \theta = \frac{\sqrt{3}}{2}, \frac{-1}{\sqrt{2}}$$ Number of solutions = 8.
Maths · Sequences and Series · Single correct
Let $a_1, a_2, a_3 \ldots$ be in an A.P. such that $\sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1, a_1 \neq 0$. If $\sum_{k=1}^{n} a_k = 0$, then $n$ is:
Answer: (a)
Let $a_1 = a$, common difference $= d$. $a_1 + a_3 + a_5 + \ldots + a_{23} = -\frac{72}{5} a$ $$\frac{12}{2} [2a + 11 \times 2d] = -\frac{72}{5} a$$ $$12a + 132d = -\frac{72}{5} a$$ $$132a + 132 \times 5d = 0$$ $$a = -5d$$ $$\frac{n}{2} (2a + (n-1)d) = 0 \Rightarrow -10d + nd - d = 0$$ $$n = 11$$
Maths · Applications of Derivatives · Single correct
If the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$, where $a > 0$, attains its local maximum and local minimum values at $p$ and $q$, respectively, such that $p^2 = q$, then $f(3)$ is equal to:
Answer: (d)
Given $f'(x) = 6x^2 - 18ax + 12a^2$. We have $f'(x) = 6 \left( x^2 - 3ax + 2a^2 \right)$. The roots are $a, 2a$. Given $p^2 = q \Rightarrow a^2 = 2a$. Solving gives $a = 2$. Therefore, $f(x) = 2x^3 - 18x^2 + 48x + 1$. Evaluating at $x = 3$, we find $f(3) = 37$.
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $z$ be a complex number such that $|z| = 1$. If $\frac{2+k^2z}{k+\overline{z}} = kz$, $k \in \mathbb{R}$, then the maximum distance of $k + ik^2$ from the circle $|z - (1 + 2i)| = 1$ is:
Answer: (a)
Given the equation $$2 + \frac{k^2}{\overline{z}} = kz$$ and $$k + \overline{z}$$. We have $$|z|^2 k = 2$$ and $$k = 2$$. The point is $$p(2, 4)$$ and the center is $$(1, 2)$$. The distance from the circle $$(x - 1)^2 + (y - 2)^2 = 1$$ is maximum. If $$(OP + r) = \sqrt{1 + 4 + 1} = \sqrt{5 + 1}$$.
Maths · Vector Algebra · Single correct
If $\vec{a}$ is nonzero vector such that its projections on the vectors $2\hat{i} - \hat{j} + 2\hat{k}$, $\hat{i} + 2\hat{j} - 2\hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\vec{a}$ is:
Answer: (c)
Let $\mathbf{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$. $a_1^2 + a_2^2 + a_3^2 = 1$ Let $\mathbf{b} = 2\hat{i} - \hat{j} + 2\hat{k}$, $\mathbf{c} = \hat{i} - 2\hat{j} - 2\hat{k}$, $\mathbf{d} = \hat{k}$. $$\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} = \frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{c}|} = \frac{\mathbf{a} \cdot \mathbf{d}}{|\mathbf{d}|}$$ $$\frac{2a_1 - a_2 + 2a_3}{3} = \frac{a_1 + 2a_2 - 2a_3}{3} = a_3$$ By solving $a_1 = \frac{7}{\sqrt{155}}$, $a_2 = \frac{9}{\sqrt{155}}$, $a_3 = \frac{5}{\sqrt{155}}$
Maths · Sets · Single correct
Let A be the set of all functions $f : \mathbb{Z} \to \mathbb{Z}$ and R be a relation on A such that $R = \{(f, g) : f(0) = g(1) and f(1) = g(0)\}$. Then R is:
Answer: (b)
Given $R = \{(f, g) : f(0) = g(1) and f(1) = g(0)\}$. Reflexive: $(f, f) \in R$ $\Rightarrow f(0) = f(1) and f(1) = f(0) \rightarrow$ must hold $\Rightarrow$ but this is not true for all functions so not reflexive. Symmetric: If $(f, g) \in R \Rightarrow (g, f) \in R$ Now, $g(0) = f(1)$ and $g(1) = f(0) \rightarrow$ true $\therefore$ symmetric. Transitive: If $(f, g) \in R$ and $(g, h) \in R$ $\Rightarrow (f, h) \in R$ Now $(f, g) \in R \Rightarrow f(0) = g(1) and f(1) = g(0)$ $(g, h) \in R \Rightarrow g(0) = h(1) and g(1) = h(0)$ For $(f, h) \in R$ we need $f(0) = h(1) and f(1) = h(0)$ Now $f(0) = g(1) = h(0)$ and $f(1) = g(0) = h(1)$ Hence not transitive.
Maths · Limits and Derivatives · Single correct
For $\alpha, \beta, \gamma, \in \mathbb{R}$, if $\lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1) e^{x^2}}{\sin 2x - \beta x} = 3$, then $\beta + \gamma - \alpha$ is equal to:
Answer: (a)
Given $x^2 (\alpha x) + (\gamma - 1) \left( 1 + \frac{x^2}{1} \right) = 3$. $$\lim_{x \to 10} \frac{2x - \frac{8x^3}{6} - \beta x}{(\gamma - 1) + (\gamma - 1)x^2 + \alpha x^3} = 3$$ $$\lim_{x \to 0} \frac{(2 - \beta)x - \frac{4}{3}x^3}{\gamma - 1} = 3$$ Given $\gamma - 1, \beta = 2, \frac{-3\alpha}{4} = +3 \Rightarrow \alpha = -4$. Also, $\beta + \gamma - \alpha = 7$.
Maths · Determinants · Single correct
If the system of linear equations $$3x + y + \beta z = 3$$ $$2x + \alpha y - z = -3$$ $$x + 2y + z = 4$$ has infinitely many solutions, then the value of $22\beta - 9\alpha$ is :
Answer: (b)
Given the determinant equation: $$\Delta = \begin{vmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0$$ Expanding, we get: $$3\alpha + 4\beta - \alpha \beta + 3 = 0$$ Now consider the determinant: $$\Delta_3 = \begin{vmatrix} 3 & 1 & 3 \\ 2 & \alpha & -3 \\ 1 & 2 & 4 \end{vmatrix} = 0$$ This simplifies to: $$9\alpha + 19 = 0$$ Solving for $\alpha$, we find: $$\alpha = \frac{-19}{9}$$ For $\beta$, we have: $$\beta = \frac{6}{11}$$ Thus, the equation becomes: $$\Rightarrow 22\beta - 9\alpha = 31$$
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $P_n = \alpha^n + \beta^n, n \in \mathbb{N}$. If $P_{10} = 123, P_9 = 76, P_8 = 47$ and $P_1 = 1$, then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:
Answer: (b)
Given $\alpha^{10} + \beta^{10} = 123$. We have $\alpha + \beta = 1$. Also, $\alpha^9 + \beta^9 = 76$ and $\alpha^8 + \beta^8 = 47$. Using $P_{10} = P_9 + P_8$, we find $x^2 = x + 1 \Rightarrow x^2 - x - 1 = 0$. Therefore, $\alpha + \beta = 1$, $\alpha \beta = -1$. Thus, $\($ $\frac{1}{\alpha}$ + $\frac{1}{\beta}$ = $\frac{\alpha + \beta}{\alpha \beta}$ = $\frac{1}{-1}$ = -1 $\)$.
Maths · Conic Sections · Single correct
If $S$ and $S'$ are the foci of the ellipse $\frac{x^2}{18} + \frac{y^2}{9} = 1$ and $P$ be a point on the ellipse, then $\min (SP \cdot S'P) + \max (SP \cdot S'P)$ is equal to:
Answer: (d)
Given $PS + PS' = 2 \times 3\sqrt{2}$. $b^2 = a^2 (1 - e^2) \Rightarrow 9 = 18 (1 - e^2)$ $\Rightarrow e = \frac{1}{\sqrt{2}}$. Directrix $x = \frac{a}{e} = \frac{3\sqrt{2}}{\frac{1}{\sqrt{2}}} = 6$. $PS \cdot PS' = \left| \frac{1}{\sqrt{2}} (3\sqrt{2} \cos \theta - 6) \frac{1}{\sqrt{2}} (3\sqrt{2} \cos \theta + 6) \right|$ $= \frac{1}{2} |18 \cos^2 \theta - 36|$. $(PS \cdot PS')_{\max} = 18; \ (PS \cdot PS')_{\min} = 9$. sum $= 27$
Maths · Three Dimensional Geometry · Single correct
Let the vertices $Q$ and $R$ of the triangle $PQR$ lie on the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$, $QR = 5$ and the coordinates of the point $P$ be $(0, 2, 3)$. If the area of the triangle $PQR$ is $\frac{m}{n}$ then:
Answer: (b)
M is given by the coordinates $M(5\lambda - 3, 2\lambda + 1, 3\lambda - 4)$. The direction ratios of $PM$ are $5\lambda - 3, 2\lambda - 1, 3\lambda - 7$. The direction ratios of line $L$ are $5, 2, 3$. Since $PM \perp L$, we have: $$ (5\lambda - 3)5 + (2\lambda - 1)2 + (3\lambda - 7)3 = 0 $$ Solving gives $\lambda = 1$. Therefore, $M(2, 3, -1)$. The length $PM$ is $\sqrt{4 + 1 + 16} = \sqrt{21}$. The area is $\frac{1}{2} \times 5 \times \sqrt{21} = \frac{m}{n}$. Solving $2m - 5\sqrt{21}n = 0$.
Maths · Vector Algebra · Single correct
Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the $\triangle$ BCD is equal to:
Answer: (c)
Ar($\triangle$ BCD) = $\sqrt{(Ar(\triangle ABC))^2 + (Ar(\triangle ACD))^2 + (Ar(\triangle ADB))^2}$ = $\sqrt{5^2 + 6^2 + 7^2}$ = $\sqrt{110}$
Maths · Determinants · Single correct
Let $a \in \mathbb{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det(A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det((a + 1) adj((a - 1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \ldots 20\}$, then $m + n$ is equal to:
Answer: (d)
Given $\($ A = $\begin{bmatrix}$ 1 & a & 1 $\\$ 2 & 1 & 0 $\\$ a & 1 & 2 $\end{bmatrix}$ $\)$ and $\($ I = $\begin{bmatrix}$ 0 & a & 1 $\\$ 2 & 0 & 0 $\\$ a & 1 & 1 $\end{bmatrix}$ $\)$. $\($|A| = -4 $\Rightarrow$ 2 - 2a = -4 $\Rightarrow$ a = 3$\)$ $\($|(a+1) adj(a-1)A| = |4 adj 3A|$\)$ $\($= 4^3 | adj 3A|$\)$ $\($= 4^3 $\times$ |3A|^{3-1} = 64|3A|^2$\)$ $\($= 64 $\times$ (3^3)^2 |A|^2$\)$ $\($= 2^6 $\times$ 3^6 $\times$ 16$\)$ $\($2^m $\times$ 3^n = 2^{10} $\times$ 3^6$\)$ Therefore, $\($ m = 10, n = 6 $\)$ $\($$\Rightarrow$ m + n = 16$\)$
Maths · Conic Sections · Single correct
Let the focal chord $PQ$ of the parabola $y^2 = 4x$ make an angle of $60^\circ$ with the positive $x$-axis, where $P$ lies in the first quadrant. If the circle, whose one diameter is $PS$, $S$ being the focus of the parabola, touches the $y$-axis at the point $(0, \alpha)$, then $5\alpha^2$ is equal to :
Answer: (a)
Given $\tan 60^\circ = \frac{2t - 0}{t^2 - 1} = \sqrt{3}$, we have $t = \sqrt{3}$. Therefore, $P(3, 2\sqrt{3})$. For the circle: $$(x - 1)(x - 3) + (y - 0)(y - 2\sqrt{3}) = 0$$ At $x = 0$: $$3 + y^2 - 2\sqrt{3}y = 0$$ This implies $y = \sqrt{3} = \alpha$. Therefore, $5\alpha^2 = 15$.
Maths · Integrals · Numerical
Let [.] denote the greatest integer function. If $$\int_{0}^{e^3} \left[ \frac{1}{e^x - 1} \right] \, dx = \alpha - \log_e 2$$, then $\alpha^3$ is equal to ________.
Answer: 8
Given $f(x) = \frac{1}{e^{x-1}} = e^{1-x}$. We have $f(x) = 2$ and $f(x) = 1$. Solving $\frac{1}{e^{x-1}} = 2$, we get $x = 1 - \ln 2$. Solving $x = 1$, we have $f(0) = e^1 = 2.71$. Also, $f(e^3) = e^{1-e^3} \in (0, 1)$. The integral $I = \int_0^{1-\ln 2} 2 \, dx + \int_{1-\ln 2}^1 1 \, dx + \int_1^{e^3} 0 \, dx$. This simplifies to $2(1 - \ln 2 - 0) + 1(1 - 1 + \ln 2) + 0$. Therefore, $\alpha - \ln 2 = 2 - \ln 2$, giving $\alpha = 2$. Thus, $\alpha^3 = 8$.
Maths · Differential Equations · Numerical
Let $f : \mathbb{R} \to \mathbb{R}$ be a thrice differentiable odd function satisfying $f'(x) \geq 0$, $f'(x) = f(x)$, $f(0) = 0$, $f'(0) = 3$. Then $9f\left(\log_e 3\right)$ is equal to ______.
Answer: 36
Given $f''(x) = f(x)$. This implies $f'(x) \cdot f''(x) = f'(x) \cdot f(x)$. Therefore, $$\frac{(f'(x))^2}{2} = \frac{(f(x))^2}{2} + C$$ which implies $$(f'(x))^2 = (f(x))^2 + C'.$$ Given $f(0) = 0$, $f'(0) = 3$, we find $C' = 9$. Thus, $$(f'(x))^2 = (f(x))^2 + 9.$$ So, $f'(x) = \sqrt{(f(x))^2 + 9}$, and $f'(x) \geq 0$. Integrating, $$\int \frac{dy}{\sqrt{y^2 + 9}} = \int dx \Rightarrow \ln\left|y + \sqrt{y^2 + 9}\right| = x + C.$$ Given $f(0) = 0$, we find $C = \ln 3$. Thus, $$y + \sqrt{y^2 + 9} = 3e^x.$$ At $x = \ln 3$, $y = 4$. Therefore, $9f(\ln 3) = 36$.
Maths · Applications of Integrals · Numerical
If the area of the region $$\{(x, y) : |4 - x^2| \leq y \leq x^2, y \leq 4, x \geq 0\}$$ is $$\left(\frac{80\sqrt{2}}{\alpha} - \beta\right)$$, $\alpha, \beta \in \mathbb{N}$, then $\alpha + \beta$ is equal to _______.
Answer: 22
The area $A$ is given by the integral $$A = \int_0^4 \sqrt{4 + y} \, dy - \int_0^2 \sqrt{4 - y} \, dy - \int_2^4 \sqrt{y} \, dy.$$ This evaluates to $$= \left( \frac{(4+y)^\frac{3}{2}}{\frac{3}{2}} \right)_0^4 + \left( \frac{(4-y)^\frac{3}{2}}{\frac{3}{2}} \right)_0^2 - \left( \frac{y^\frac{3}{2}}{\frac{3}{2}} \right)_2^4.$$ Simplifying, we get $$= \frac{80\sqrt{2}}{3} - 16 = \frac{40\sqrt{2}}{3} - 16.$$ Therefore, $\alpha = 6$, $\beta = 16$, and $\alpha + \beta = 22.$
Maths · Probability · Numerical
Three distinct numbers are selected randomly from the set \{1, 2, 3, $\ldots$, 40$\}$. If the probability, that the selected numbers are in an increasing G.P. is $\frac{m}{n}$ , $\gcd$(m, n) = 1, then $\($ m + n $\)$ is equal to _____.
Answer: 4949
Given $1 \leq a < ar < ar^2 \leq 40$. If $r \in \mathbb{N}$: If $r = 2$ $$1 \leq a < 2a < 4a \leq 40$$ $a \in \{1, \ldots, 10\}$ gives 10 GP. If $r = 3$ $$1 \leq a < 3a < 9a \leq 40$$ $a \in \{1, 2, 3, 4\}$ gives 4 GP. If $r = 4$ $$1 \leq a < 4a < 16a \leq 40$$ $a \in \{1, 2\}$ gives 2 GP. If $r = 5$ $$1 \leq a < 5a < 25a \leq 40$$ $a \in \{1\}$ gives 1 GP. If $r = 6$ $$1 \leq a < 6a < 36a \leq 40$$ $a \in \{1\}$ gives 1 GP. $$P = \frac{18}{9880} = \frac{9}{4940}$$ as per NTA for $r \in \mathbb{N}$. $m + n = 4949$ If $r \notin \mathbb{N}$ (also possible): $r = \frac{3}{2}$ $$ar^2 = \frac{9a}{4}; \ a = 4k$$ $(4, 6, 9)$ $(8, 12, 18)$ $(12, 18, 27)$ $(16, 24, 36)$ gives 4 GP. $r = \frac{5}{2}$ $$ar^2 = \frac{25a}{4}; \ a = 4k$$ $(4, 10, 25)$ gives 1 GP. $r = \frac{4}{3}$ $$ar^2 = \frac{16a}{9} \rightarrow a = 9k$$ $(9, 12, 16), (18, 24, 32)$ gives 2 GP. $r = \frac{5}{3}$ $$ar^2 = \frac{25a}{9}; \ a = 9k$$ $(9, 15, 25)$ gives 1 GP. $r = \frac{5}{4}$ $$ar^2 = \frac{25a}{16}; \ a = 16k$$ $(16, 20, 25)$ gives 1 GP. $r = \frac{6}{5}$ $$ar^2 = \frac{36a}{25}; \ a = 25k$$ $(25, 30, 36)$ gives 1 GP. Total = $18 + 10 = 28$ $$P = \binom{28}{3} = \frac{28}{9880} = \frac{7}{2470}$$ $m + n = 2477$
Maths · Conic Sections · Fill in the blank
The absolute difference between the squares of the radii of the two circles passing through the point $(-9, 4)$ and touching the lines $x + y = 3$ and $x - y = 3$, is equal to ______.
Answer: 768
Centre $(a, 0)$ $$r = \left| \frac{a - 0 - 3}{\sqrt{2}} \right|$$ circle $(x - a)^2 + y^2 = \left( \frac{a - 3}{\sqrt{2}} \right)^2$ passes through $(-9, 4)$ $$2 \left( a^2 + 18a + 81 + 16 \right) = (a^2 - 6a + 9)$$ $$a^2 + 42a + 185 = 0$$ $$(a + 37)(a + 5) = 0$$ $$\Rightarrow a = -37, -5$$ $$r_1 = \left| \frac{-37 - 3}{\sqrt{2}} \right| = 20 \sqrt{2}$$ $$r_2 = \left| \frac{-5 - 3}{\sqrt{2}} \right| = 4 \sqrt{2}$$ $$|r_1^2 - r_2^2| = |800 - 32| = 768$$
Physics · Wave Optics · Single correct
A light wave is propagating with plane wave fronts of the type $x + y + z = constant$. The angle made by the direction of wave propagation with the $x$-axis is:
Answer: (a)
The direction of propagation of light is perpendicular to the wave front and is symmetric about x, y and z axis. Therefore, the angle made by the light with x, y, and z axis is the same. Thus, $\cos \alpha = \cos \beta = \cos \gamma$ (where $\alpha, \beta, \gamma$ are angles made by light with x, y, and z axis respectively). Also, $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$ [Sum of direction cosines]. Therefore, $\alpha = \cos^{-1} \frac{1}{\sqrt{3}}$.
Physics · Physical World, Units and Measurements · Single correct
The equation for real gas is given by $\left( P + \frac{a}{V^2} \right) (V - b) = RT$, where $P$, $V$, $T$ and $R$ are the pressure, volume, temperature and gas constant, respectively. The dimension of $ab^{-2}$ is equivalent to that of:
Answer: (d)
Given $\[$ $\left$[ P + $\frac{a}{V^2}$ $\right$] (V - b) = RT $\]$ Therefore, $\($[a] = [P][V^2] = ML^{-1}T^{-2}L^6 = ML^5T^{-2}$\)$ $\($[b] = [V] = L^3$\)$ $\($[ab^{-2}] = ML^5T^{-2}L^{-6} = ML^{-1}T^{-2}$\)$ Dimension of energy density.
Physics · System of Particles and Rotational Motion · Single correct
A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m, would be:
Answer: (a)
Given $W_F = 20 \times 1 = 20 \, \mathrm{J}$. Therefore, $\Delta \mathrm{KE} = 20 \, \mathrm{J} = \frac{1}{2} I \omega^2$. $I = MR^2 = 10 \times 0.1^2 = 0.1 \, \mathrm{kg} \, \mathrm{m}^2$. Therefore, $20 = \frac{1}{2} \times 0.1 \times \omega^2$. Thus, $\omega = 20 \, \mathrm{rad/sec}$.
Physics · Ray Optics and Optical Instruments · Single correct
A slanted object $AB$ is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is:
Answer: (b)
Location of image of A: $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{60} \Rightarrow v = 60 \, \mathrm{cm}$$ Therefore, $m = 2$. Since size of object is small with respect to the location, hence $$dv = m^2 du \Rightarrow dv = 4 \times 1 = 4 \, \mathrm{cm}$$ $$h_i = mh_0 \Rightarrow h_i (dy) = 2 \times 2 = 4 \, \mathrm{cm}$$ Therefore, angle made with principle axis is $-45^\circ$.
Physics · Electric Charges and Fields · Single correct
Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density $+\sigma$ and $-2\sigma$. The force experienced by a point charge $+q$ placed at the mid point between two plates will be:
Answer: (b)
Final charge distribution will be Therefore, $F_{net} = \frac{3 \sigma}{2 \varepsilon_0} q$
Physics · Motion in a Plane · Single correct
A river is flowing from west to east direction with speed of $9 \, \mathrm{km \, h^{-1}}$. If a boat capable of moving at a maximum speed of $27 \, \mathrm{km \, h^{-1}}$ in still water, crosses the river in half a minute, while moving with maximum speed at an angle of $150^\circ$ to direction of river flow, then the width of the river is:
Answer: (b)
Therefore, $V_\perp = river flow = 27 \times \cos 60^\circ = \frac{27}{2} \, km/hr$. Time taken $= 30 \, sec$. Therefore, $S = Vt = \frac{27}{2} \times \frac{5}{18} \times 30 \, m = 112.5 \, m$
Physics · Electric Charges and Fields · Single correct
A point charge +q is placed at the origin. A second point charge +9 q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is :
Answer: (b)
Let $E_P = 0$. Therefore, $\frac{kq}{x^2} = \frac{k9q}{(d-x)^2}$ $\Rightarrow \frac{d-x}{x} = 3 \Rightarrow x = \frac{d}{4}$ Therefore, the coordinate of $P$ is $\left(\frac{d}{4},\,0,\,0\right)$.
Physics · Current Electricity · Single correct
The battery of a mobile phone is rated as $4.2 \, \mathrm{V}$, $5800 \, \mathrm{mAh}$. How much energy is stored in it when fully charged?
Answer: (c)
Given $V = 4.2$ volt. Energy supplied by battery $$vq = 4.2 \times 5800 \times 3600 \times 10^{-3} \, \mathrm{J} = 87.696 \, \mathrm{kJ}$$ Energy stored in the battery when fully charged $$= 87.696 \, \mathrm{kJ} \approx 87.7 \, \mathrm{kJ}$$
Physics · Oscillations · Single correct
A particle is subjected to two simple harmonic motions as: $$ x_1=\sqrt{7}\sin(5t)\,\mathrm{cm} $$ and $$ x_2=2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right)\,\mathrm{cm} $$ where $x$ is displacement and $t$ is time in seconds. The maximum acceleration of the particle is $\left(x\times10^{-2}\,\mathrm{m\,s^{-2}}\right)$. The value of $x$ is:
Answer: (a)
Given $x_1 = \sqrt{7} \sin 5t$ and $x_2 = 2\sqrt{7} \sin \left( 5t + \frac{\pi}{3} \right)$. From phasor, the amplitude of resultant SHM is $7$. $$\phi = \tan^{-1} \frac{2\sqrt{7} \times \sqrt{3}/2}{\sqrt{7} + 2\sqrt{7} \times \frac{1}{2}} = \tan^{-1} \frac{\sqrt{21}}{2\sqrt{7}} = \tan^{-1} \frac{\sqrt{3}}{2}$$ Thus, $X_R = 7 \sin(5t + \phi)$. The acceleration $a_R = -7 \times 25 \sin(5t + \phi)$. Therefore, $a_{\max} = 175 \, \mathrm{cm/sec}^2 = 175 \times 10^{-2} \, \mathrm{m/sec}^2$.
Physics · Magnetism and Matter · Single correct
The relationship between the magnetic susceptibility ($\chi$) and the magnetic permeability ($\mu$) is given by : ($\mu_0$ is the permeability of free space and $\mu_r$ is relative permeability)
Answer: (a)
We have $$\mu_r = (1 + \chi) \Rightarrow \chi = (\mu_r - 1)$$ $$\mu = \mu_0 \mu_r \Rightarrow \mu_r = \frac{\mu}{\mu_0}$$ Therefore, $$\chi = \left( \frac{\mu}{\mu_0} - 1 \right)$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 $\Omega$ resistor connected in series, the zener current is found to be 4 times load current. The load current $I_L$ and load resistance $R_L$ are:
Answer: (d)
From the circuit diagram, $$5i = \frac{20}{400} = \frac{1}{20} A$$ Therefore, $i = \frac{1}{100} A = 10 mA = Load current$ Also, $V_L = 5 V$ Therefore, $R_L = \frac{5}{10 \times 10^{-3}} \, \Omega = 500 \, \Omega$
Physics · Kinetic Theory · Single correct
In an adiabatic process, which of the following statements is true?
Answer: (c)
For adiabatic process, $dQ = 0$. Molar heat capacity $= 0$. $dQ = 0 \Rightarrow dU = -dW$ Also $dU = \frac{f}{2} nR dT$ Only option (3) is correct.
Physics · System of Particles and Rotational Motion · Single correct
A square Lamina OABC of length 10 cm is pivoted at 'O'. Forces act at Lamina as shown in figure. If Lamina remains stationary, then the magnitude of F is :
Answer: (c)
Since the lamina is in equilibrium. Therefore, $\mathbf{F}_{net} = 0$ and $\tau_{net} = 0$. $$T_o = 10\ell - F\ell \Rightarrow F = 10 \, \mathrm{N}$$
Physics · Moving Charges and Magnetism · Single correct
Let $B_1$ be the magnitude of magnetic field at center of a circular coil of radius $R$ carrying current $I$. Let $B_2$ be the magnitude of magnetic field at an axial distance $'x'$ from the center. For $x : R = 3 : 4$, $\frac{B_2}{B_1}$ is:
Answer: (c)
Given $B_1 = \frac{\mu_0 i}{2R}$ and $B_2 = B_1 \sin^3 \theta$. Therefore, $$\frac{B_2}{B_1} = \sin^3 \theta = \left(\frac{4}{5}\right)^3 = \frac{64}{125}.$$
Physics · Atoms · Single correct
Considering Bohr's atomic model for hydrogen atom: (A) the energy of H atom in ground state is same as energy of $\mathrm{He}^{+}$ ion in its first excited state. (B) the energy of H atom in ground state is same as that for $\mathrm{Li}^{2+}$ ion in its second excited state. (C) the energy of H atom in its ground state is same as that of $\mathrm{He}^{+}$ ion in its ground state. (D) the energy of $\mathrm{He}^{+}$ ion in its first excited state is same as that for $\mathrm{Li}^{2+}$ ion in its ground state. Choose the correct answer from the options given below:
Answer: (b)
Given $E \propto \frac{Z}{n^2}$. $Z_\mathrm{H} = 1$, $Z_{\mathrm{He}^+} = 2$, $Z_{\mathrm{Li}^{+2}} = 3$. 1st excited state $\Rightarrow n = 2$. 2nd excited state $\Rightarrow n = 3$. From the given statements only A $\&$ B are correct.
Physics · System of Particles and Rotational Motion · Single correct
Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is $\alpha$. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is:
Answer: (b)
Given $\alpha = \frac{M \ell^2}{12}$. $$\alpha' = 2 \left[ \frac{M}{2} \left( \frac{\ell}{2} \right)^2 \right]$$ $$\alpha' = \frac{M \ell^2}{48} = \frac{\alpha}{4}$$ Correct option is (2)
Physics · Ray Optics and Optical Instruments · Single correct
A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object ' O ', is : ( C is the center of curvature of the spherical surface and R is the radius of curvature)
Answer: (b)
Given $\($ $\frac{\mu_2}{v}$ - $\frac{\mu_1}{u}$ = $\frac{\mu_2 - \mu_1}{R}$ $\)$. $\($ $\frac{1.5}{v}$ - $\frac{1}{(-0.2)}$ = $\frac{1.5 - 1}{0.4}$ $\)$ $\($ $\frac{1.5}{v}$ = $\frac{0.5}{0.4}$ - 0.2 $\)$ $\($ $\frac{1.5}{v}$ = $\frac{1.5}{0.4}$ $\)$ $\($ v = -0.4 $\,$ $\mathrm{m}$ $\)$
Physics · Physical World, Units and Measurements · Single correct
Match List-I with List-II. \[ \begin{array}{ll} \text{List-I} & \text{List-II} \\[4pt] (A)\ \text{Coefficient of viscosity} & (I)\ [M L^{0} T^{-3}] \\[4pt] (B)\ \text{Intensity of wave} & (II)\ [M L^{-2} T^{-2}] \\[4pt] (C)\ \text{Pressure gradient} & (III)\ [M^{-1} L T^{2}] \\[4pt] (D)\ \text{Compressibility} & (IV)\ [M L^{-1} T^{-1}] \end{array} \] Choose the correct answer from the options given below:
Answer: (b)
The coefficient of viscosity $[\eta] = \left[ M^1 L^{-1} T^{-1} \right]$. Intensity $[I] = \left[ M^1 L^0 T^{-3} \right]$. Pressure gradient $= \left[ M L^{-2} T^{-2} \right]$. Compressibility $[K] = \left[ M^{-1} L^1 T^2 \right]$.
Physics · Electric Charges and Fields · Single correct
A small bob of mass 100 mg and charge +10 $\mu$ C is connected to an insulating string of length 1 m. It is brought near to an infinitely long nonconducting sheet of charge density ' $\sigma$ ' as shown in figure. If string subtends an angle of $45^\circ$ with the sheet at equilibrium the charge density of sheet will be : (Given, $\varepsilon$_0 = 8.85 $\times$ 10^{-12} $\mathrm{F/m}$ and acceleration due to gravity, g = 10 $\mathrm{m/s^2}$)
Answer: (d)
Given $qE = mg$. $$q \left[ \frac{\sigma}{2\varepsilon_0} \right] = mg$$ Solving for $\sigma$: $$\sigma = \frac{2\varepsilon_0 mg}{q}$$ Substitute the values: $$\sigma = \frac{2 \times 8.85 \times 10^{-12} \times 100 \times 10^{-6} \times 10}{10 \times 10^{-6}}$$ Calculate: $$\sigma = 17.7 \times 10^{-10} \, \mathrm{C/m^2}$$ Convert to nanocoulombs per square meter: $$\sigma = 1.77 \, \mathrm{nC/m^2}$$
Physics · Dual Nature of Radiation and Matter · Single correct
A monochromatic light is incident on a metallic plate having work function $\phi$. An electron, emitted normally to the plate from a point $A$ with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point $B$. The distance between $A$ and $B$ is: (Given: The magnitude of charge of an electron is $e$ and mass is $m$, $h$ is Planck's constant and $c$ is velocity of light. Take the magnetic field exists throughout the path of electron)
Answer: (c)
Given $\mathrm{KE_{max}} = \frac{hc}{\lambda} - \phi$. $p = \sqrt{2m \mathrm{K_{max}}}$. Substituting for $\mathrm{K_{max}}$, we have: $$p = \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)}.$$ The distance $d_{\mathrm{A-B}} = 2R$ is given by: $$d_{\mathrm{A-B}} = 2 \left[ \frac{p}{qB} \right].$$ Substituting for $p$, we get: $$d_{\mathrm{AB}} = \frac{2 \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)}}{eB} = \frac{\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)}}{eB}.$$
Physics · Mechanical Properties of Fluids · Numerical
A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of 100 $\mathrm{cm}^2$ with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density $1.5 \times 10^3 \, \mathrm{kg/m}^3$. What force one needs to apply on the hinged door so that it does not get opened ? (Acceleration due to gravity = $10 \, \mathrm{m/s}^2$)
Answer: 150
In equilibrium, $F_{ext} + F_w = F_\ell$. Therefore, $F_{ext} = F_\ell - F_w$. This gives $F_{ext} = (P_0 + \rho_\ell gh)A - (P_0 + \rho_w gh)A$. Simplifying, $F_{ext} = (\rho_\ell - \rho_w)ghA$. Substituting the values, $F_{ext} = (1500 - 1000) \times 10 \times 3 \times (100 \times 10^{-4})$. Therefore, $F_{ext} = 150 \, \mathrm{m}$.
Physics · Mechanical Properties of Solids · Numerical
A steel wire of length 2 m and Young's modulus $2.0 \times 10^{11} \, \mathrm{Nm}^{-2}$ is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and $10^{-3}$ respectively, then the elastic potential energy density of the wire is _______ $\times 10^5$ (in SI units)
Answer: 25
Given $\ell = 2 \, \mathrm{m}$; $Y = 2 \times 10^{11} \, \mathrm{N/m^2}$. $$\mu = - \left( \frac{\Delta r}{r} \right) \left( \frac{\Delta \ell}{\ell} \right) \implies \frac{\Delta \ell}{\ell} = \frac{1}{\mu} \times \left( \frac{\Delta r}{r} \right)$$ $$= \frac{1}{0.2} \times (10^{-3})$$ $$\implies \frac{\Delta \ell}{\ell} = 5 \times 10^{-3}$$ $$u = \frac{1}{2} Y \varepsilon_{\ell}^2 = \frac{1}{2} \times 2 \times 10^{11} \times [5 \times 10^{-3}]^2$$ $$= 25$$
Physics · Wave Optics · Numerical
If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is $30^\circ$ in a single slit diffraction pattern recorded using $628 \, \mathrm{nm}$ light, then the width of the slit is _____ $\mu \mathrm{m}$.
Answer: 6
Given $\theta_1 = \sin^{-1}\left(\frac{2\lambda}{a}\right)$ and $\theta_2 = \sin^{-1}\left(\frac{3\lambda}{a}\right)$. Therefore, $\theta_1 + \theta_2 = 30^\circ$. This implies $$\sin^{-1}\left(\frac{2\lambda}{a}\right) + \sin^{-1}\left(\frac{3\lambda}{a}\right) = \frac{\pi}{6}$$ $$\Rightarrow \frac{2\lambda}{a} \sqrt{1 - \left(\frac{3\lambda}{a}\right)^2} + \frac{3\lambda}{a} \sqrt{1 + \left(\frac{2\lambda}{a}\right)^2} = \sin \frac{\pi}{6}$$ Here $\lambda = 628 \, \mathrm{nm}$. After solving, $A = 6.07\mu \mathrm{m}$. Approximate Method: $\theta = \theta_1 + \theta_2$. $$\Rightarrow \frac{\pi}{6} = \frac{2\lambda}{a} + \frac{3\lambda}{a}$$ $$\Rightarrow \frac{\pi}{6} = \frac{5}{a} (628 \, \mathrm{nm})$$ $$\Rightarrow a = 6\mu \mathrm{m}$$
Physics · Kinetic Theory · Numerical
$\gamma_A$ is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. $\gamma_B$ is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If $\frac{\gamma_A}{\gamma_B} = \left( 1 + \frac{1}{n} \right)$, then the value of $n$ is
Answer: 3
Given $\($ $\gamma$_A = $\frac{f_A + 2}{f_B}$ $\)$ and $\($ $\gamma$_B = $\frac{f_A}{f_B + 2}$ $\)$. $\($ = $\frac{3 + 2}{3}$ $\times$ $\frac{6}{(6 + 2) + 2}$ $\)$ $\($ = $\frac{5}{3}$ $\times$ $\frac{8}{10}$ = $\frac{40}{30}$ $\)$ $\($ $\therefore$ $\frac{40}{30}$ = 1 + $\frac{1}{n}$ $\)$ $\($ $\Rightarrow$ 30 - 1 = n $\)$ $\($ $\Rightarrow$ n = 3 $\)$
Physics · Motion in a Straight Line · Numerical
A person travelling on a straight line moves with a uniform velocity $v_1$ for a distance $x$ and with a uniform velocity $v_2$ for the next $\frac{3}{2}x$ distance. The average velocity in this motion is $\frac{50}{7}\,\mathrm{m/s}$. If $v_1$ is $5\,\mathrm{m/s}$ then $v_2=$ $\underline{\hspace{2cm}}$ $\mathrm{m/s}$.
Answer: 10
Given $\($ v_{avg} = $\frac{x_1 + x_2}{t_1 + t_2}$ $\)$. $\[$ $\Rightarrow$ $\frac{50}{7}$ = $\frac{x + \frac{3x}{2}}{\frac{x}{5} + \frac{3}{2v_2}}$ $\]$ $\[$ $\Rightarrow$ $\frac{50}{7}$ = $\frac{5/2}{\frac{1}{5} + \frac{3}{2v_2}}$ $\]$ $\[$ $\Rightarrow$ $\frac{1}{5}$ + $\frac{3}{2v_2}$ = $\frac{7}{20}$ $\]$ $\[$ $\Rightarrow$ $\frac{3}{2v_2}$ = $\frac{7}{20}$ - $\frac{1}{5}$ = $\frac{7 - 4}{20}$ $\]$ $\[$ $\Rightarrow$ $\frac{3}{2v_2}$ = $\frac{3}{20}$ $\]$ $\[$ $\Rightarrow$ 2v_2 = 20 $\]$ $\[$ $\Rightarrow$ v_2 = 10 $\,$ m/s $\]$
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Designate whether each of the following compounds is aromatic or not aromatic.
Answer: (d)
a, c, d, e, h follow Huckel's rule. b, f, g, are not aromatic, these compounds do not follow Huckel's rule.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
An optically active alkyl halide $\mathrm{C_4 H_9 Br}$ [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic $\mathrm{NaNH_2}$. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [E] is:
Answer: (c)
The reaction starts with compound (A), which is treated with $\mathrm{KOH}$ in ethanol and heat to form compound (B). Compound (B) is then reacted with $\mathrm{Br_2}$ to form compound (C). Compound (C) is treated with $\mathrm{NaNH_2}$ in alcohol to form compound (D), which is $\mathrm{CH_3CH_2C \equiv CH}$. Finally, compound (D) is reacted with $\mathrm{H_2O}$ and $\mathrm{HgSO_4}$ in the presence of $\mathrm{H^+}$ to form compound (E), which is Butan-2-one.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The property/properties that show irregularity in first four elements of group-17 is/are: (A) Covalent radius (B) Electron affinity $(C)$ Ionic radius (D) First ionization energy Choose the correct answer from the options given below:
Answer: (c)
The order of first four elements of group-17 are as follows. $\mathrm{F} \mathrm{F} > \mathrm{Br} > \mathrm{I}$ (Electron affinity) $\mathrm{F}^- \mathrm{Cl} > \mathrm{Br} > \mathrm{I}$ (1$\textsuperscript{st}$ ionization energy) Electron affinity order is irregular.
Chemistry · Solutions · Single correct
Which of the following graph correctly represents the plots of $K_H$ at 1 bar gases in water versus temperature?
Answer: (d)
Chemistry · Structure of Atom · Single correct
According to Bohr's model of hydrogen atom, which of the following statement is incorrect?
Answer: (c)
Given $r \propto n^2$. (1) $\frac{r_3}{r_1} = \frac{9}{1}$ (2) $\frac{r_8}{r_4} = \frac{64}{16} = 4$ (3) $\frac{r_6}{r_4} = \left(\frac{6}{4}\right)^2 = \frac{9}{4}$ (4) $\frac{r_4}{r_2} = \left(\frac{4}{2}\right)^2 = 4$
Chemistry · Thermodynamics · Single correct
Two vessels A and B are connected via stopcock. The vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and is allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?
Answer: (d)
It is free expansion of gas, therefore $P_{ext} = 0$. Where $w = 0$, $q = 0$ and $\Delta U = 0$.
Chemistry · Solutions · Single correct
A solution is made by mixing one mole of volatile liquid $A$ with 3 moles of volatile liquid $B$. The vapour pressure of pure $A$ is $200 \, \mathrm{mmHg}$ and that of the solution is $500 \, \mathrm{mmHg}$. The vapour pressure of pure $B$ and the least volatile component of the solution, respectively, are:
Answer: (d)
Given $P_S = P_A^\circ \cdot X_A + P_B^\circ \cdot X_B$. $$500 = 200 \times \frac{1}{4} + P_B^\circ \cdot \frac{3}{4}$$ $P_B^\circ = 600 \, \mathrm{mmHg}$ As $P_A^\circ < P_B^\circ \Rightarrow A$ is least volatile.
Chemistry · Some Basic Concepts of Chemistry · Single correct
$\mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)}$ Consider the above reaction, what mass of $\mathrm{CaCl_2}$ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of $\mathrm{CaCO_3}$? (Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol$^{-1}$, respectively)
Answer: (c)
The reaction is given by: $$\mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O}$$ Moles of $\mathrm{CaCO_3}$ are calculated as: $$\frac{1000}{100} = 10$$ Moles of HCl are calculated as: $$0.76 \times \frac{250}{1000} = 0.19 (L.R.)$$ Moles of $\mathrm{CaCl_2}$ formed are: $$\frac{0.19}{2}$$ Mass of $\mathrm{CaCl_2}$ is: $$\frac{0.19}{2} \times 111 = 10.545 gm$$
Chemistry · Equilibrium · Single correct
If equal volumes of $\mathrm{AB}_2$ and $\mathrm{XY}$ (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of $\mathrm{AY}_2$ at $300 \, \mathrm{K}$? (Given $K_{\mathrm{sp}}$ (at $300 \, \mathrm{K}$) for $\mathrm{AY}_2 = 5.2 \times 10^{-7}$)
Answer: (c)
When equal volumes are mixed, molarity reduces to half. For precipitation, $Q_{SP} = \left[ A^{+2} \right] \left[ Y^{-} \right]^2 > K_{SP}$. (1) $Q_{SP} = \left( 1.8 \times 10^{-3} \right) \left( \frac{5}{2} \times 10^{-4} \right)^2 K_{SP}$ (4) $Q_{SP} = \left( \frac{1.5}{2} \times 10^{-4} \right) \left( \frac{1.5}{2} \times 10^{-3} \right)^2 < K_{SP}$
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Among $\mathrm{SO_2}$, $\mathrm{NF_3}$, $\mathrm{NH_3}$, $\mathrm{XeF_2}$, $\mathrm{ClF_3}$ and $\mathrm{SF_4}$, the hybridization of the molecule with non-zero dipole moment and highest number of lone-pairs of electrons on the central atom is
Answer: (d)
\begin{tabular}{|l|c|c|c|} \hline \textbf{Molecule} & \textbf{Hybridisation} & \textbf{Dipole Moment} & \textbf{Lone pair on the central atom} \\ \hline SO$_2$ & $sp^2$ & Non-zero & 1 \\ \hline NF$_3$ & $sp^3$ & Non-zero & 1 \\ \hline NH$_3$ & $sp^3$ & Non-zero & 1 \\ \hline XeF$_2$ & $sp^3d$ & Zero & 3 \\ \hline ClF$_3$ & $sp^3d$ & Non-zero & 2 \\ \hline SF$_4$ & $sp^3d$ & Non-zero & 1 \\ \hline \end{tabular}
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Given below are two statements:
Answer: (b)
Phenolic group soluble in NaOH. Benzaldehyde derivative react with Tollen's reagent. Vanillin does not give self-aldol reaction due to lack of acidic H for condensation.
Chemistry · Biomolecules · Single correct
Identify the correct statement among the following:
Answer: (d)
Isoleucine has 2 chiral centres. Glycine is optically inactive. Aspartic acid also contains a COOH group at the side chain. Cysteine easily dimerizes due to the free SH group.
Chemistry · Amines · Single correct
The correct order of basic nature on aqueous solution for the bases $\mathrm{NH}_3$, $\mathrm{H}_2\mathrm{N} - \mathrm{NH}_2$, $\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2$, $(\mathrm{CH}_3\mathrm{CH}_2)_2\mathrm{NH}$ and $(\mathrm{CH}_3\mathrm{CH}_2)_3\mathrm{N}$ is:
Answer: (d)
Basic strength of amine depends on hydrogen bonding and electronic inductive effect. $$\mathrm{NH(Et)_2 > N(Et)_3 > NH_2Et > \underset{..}{N}H_3 > NH_2 - NH_2}$$
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Given below are two statements: Statement (I): The metallic radius of Al is less than that of Ga. Statement (II): The ionic radius of $\mathrm{Al}^{3+}$ is less than that of $\mathrm{Ga}^{3+}$. In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (b)
The metallic radius order of Al and Ga is $$\mathrm{B} < \mathrm{Ga} < \mathrm{Al} < \mathrm{In} < \mathrm{Tl}$$ (due to poor shielding of d-subshell electrons) The ionic radius order of $\mathrm{Al^{+3}}$ and $\mathrm{Ga^{+3}}$ is $\mathrm{B^{+3}} < \mathrm{Al^{+3}} < \mathrm{Ga^{+3}} < \mathrm{In^{+3}} < \mathrm{Tl^{+3}}$
Chemistry · Co-ordination Compounds · Single correct
Given below are two statements: Statement (I): In octahedral complexes, when $\Delta_o P$ low spin complexes are formed. Statement (II): In tetrahedral complexes because of $\Delta_t < P$, low spin complexes are rarely formed. In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (d)
In octahedral complex (CN = 6) If $\Delta_0 P.E.$, then low spin complexes are formed. But in tetrahedral complex (CN = 4) $\Delta_t < P.E.$, then mainly high spin complexes are formed and rarely low spin complexes are formed.
Chemistry · Analytical Chemistry · Single correct
Choose the correct tests with respective observations. (A) $\mathrm{CuSO_4}$ (acidified with acetic acid) $+ \mathrm{K_4[Fe(CN)_6]} \rightarrow$ Chocolate brown precipitate. (B) $\mathrm{FeCl_3} + \mathrm{K_4[Fe(CN)_6]} \rightarrow$ Prussian blue precipitate. $(C)$ $\mathrm{ZnCl_2} + \mathrm{K_4[Fe(CN)_6]}$, neutralised with $\mathrm{NH_4OH} \rightarrow$ White or bluish white precipitate. (D) $\mathrm{MgCl_2} + \mathrm{K_4[Fe(CN)_6]} \rightarrow$ Blue precipitate. (E) $\mathrm{BaCl_2} + \mathrm{K_4[Fe(CN)_6]}$, neutralised with NaOH $\rightarrow$ White precipitate. Choose the correct answer from the options given below:
Answer: (c)
$2CuSO_4 + K_4[Fe(CN)_6]$ $\xrightarrow{CH_3COOH}$ $Cu_2[Fe(CN)_6] + 2K_2SO_4$ (Chocolate brown ppt.) $4FeCl_3 + 3K_4[Fe(CN)_6]$ $\rightarrow$ $Fe_4[Fe(CN)_6]_3 + 12KCl$ (Prussian Blue ppt.) $3ZnCl_4 + 2K_4[Fe(CN)_6]$ $\xrightarrow{NH_4OH}$ $K_2Zn_3[Fe(CN)_6]_2 + 6KCl$ (White or bluish white ppt.)
Chemistry · Some Basic Concepts of Chemistry · Single correct
On complete combustion $1.0\,\mathrm{g}$ of an organic compound (X) gave $1.46\,\mathrm{g}$ of $\mathrm{CO}_2$ and $0.567\,\mathrm{g}$ of $\mathrm{H}_2 \mathrm{O}$. The empirical formula mass of compound (X) is $\ldots \mathrm{g}$. (Given molar mass in g mol^{-1} C : 12, H : 1, O : 16)
Answer: (a)
Moles of 'C' = n_{$\mathrm{CO_2}$} = $\frac{1.46}{44}$ = 0.033. Moles of 'C' = W_c = 0.033 $\times$ 12. Moles of 'H' = 2 $\times$ n_{$\mathrm{H_2O}$} = 2 $\times \frac{0.567}{18}$ = 0.063. Mass of 'H' = 0.0063. Mass of Oxygen (O) = 1 - (W_c + W_H) = 1 - (0.033 $\times$ 12 + 0.063 $\times$ 1) = $0.541\,\mathrm{gm}$. Moles of 'O' = $\frac{0.541}{16}$ = 0.033. Empirical formula = $\mathrm{CH_2O}$. Empirical formula mass = 30.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Consider the following compound (X) The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding $C - H$ bond are:
Answer: (d)
II most stable carbon radical due to resonance stabilize. I least stable carbon radical due to no stabilising factor.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Consider the following molecules: The correct order of rate of hydrolysis is:
Answer: (d)
Rate of hydrolysis is proportional to leaving group ability. The order of rate of hydrolysis is:
Chemistry · Chemical Bonding and Molecular Structure · Single correct
A molecule with the formula $\mathrm{AX}_4\mathrm{Y}$ has all it's elements from p-block. Element A is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among A, X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:
Answer: (a)
Given A is rarest, monoatomic, non-radioactive p-block element and form $AX_4Y$ type of molecule. (1) Therefore, it is concluded that it is Xe. It is given the electronegativity of A is less than X and Y. It is given the electronegativity of X and Y is highest and second highest respectively among all elements. Therefore, X and Y are F and O. Therefore, compound is considered as $\mathrm{XeOF_4}$ with square pyramidal shape.
Chemistry · Co-ordination Compounds · Numerical
A transition metal (M) among Mn, Cr, Co and Fe has the highest standard electrode potential $\left( \mathrm{M}^{3+}/\mathrm{M}^{2+} \right)$. It forms a metal complex of the type $[\mathrm{M(CN)}_6]^{4-}$. The number of electrons present in the $e_g$ orbital of the complex is _______.
Answer: 1
Co has the highest standard electrode potential $\left(\mathrm{M^{+3}/M^{+2}}\right)$ among Mn, Cr, Co, Fe. Therefore, the complex is $[\mathrm{Co(CN)_6}]^{4-}$ and its splitting is as follows. The electron in $e_g$ orbital is one.
Chemistry · Electrochemistry · Numerical
Consider the following electrochemical cell at standard condition. Au(s) | $\mathrm{QH_2, Q| NH_4X(0.01M)|}$ | $\mathrm{Ag^+(1M) | Ag(s)}$ E_{cell} = +0.4 $\,$ $\mathrm{V}$ The couple $\mathrm{QH_2 / Q}$ represents quinhydrone electrode, the half cell reaction is given below The $pK_b$ value of the ammonium halide salt $(\mathrm{NH_4 X})$ used here is . (nearest integer)
Answer: 6
The reaction is given by $\mathrm{QH_2 + 2Ag^+ \rightarrow 2Ag + Q + 2H^+}$. The equation for the cell potential is $$E = E^\circ - \frac{0.06}{2} \log [\mathrm{H^+}]^2$$ which simplifies to $$E = E^\circ - 0.06 \times \log [\mathrm{H^+}]$$ The pH is calculated as $$\mathrm{pH} = -\log (\mathrm{H^+}) = \frac{E - E^\circ}{0.06} = \frac{0.4 - 0.1}{0.06}$$ which gives $$= \frac{0.3}{0.06} = 5$$ For the buffer solution, $$\mathrm{pH} + \mathrm{NH_4X} = 7 - \frac{1}{2} \mathrm{pK_b} - \frac{1}{2} \log C$$ Substituting the values, $$5 = 7 - \frac{1}{2} \times \mathrm{pK_b} - \frac{1}{2} \log (10^{-2})$$ Solving for $\mathrm{pK_b}$ gives $$\mathrm{pK_b} = 6$$
Chemistry · Some Basic Concepts of Chemistry · Numerical
0.1 mol of the following given antiviral compound (P) will weigh ________ $\times 10^{-1}$ g
Answer: 372
Molar mass = 372 $\,$ $\mathrm{gm}$. Therefore, 0.1 mole has = 372 $\times$ 10^{-1} $\,$ $\mathrm{gm}$.
Chemistry · Equilibrium · Numerical
Consider the following equilibrium, $$\mathrm{CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)}$$ $0.1\,\mathrm{mol}$ of CO along with a catalyst is present in a $2\,\mathrm{dm^3}$ flask maintained at $500\,\mathrm{K}$. Hydrogen is introduced into the flask until the pressure is $5\,\mathrm{bar}$ and $0.04\,\mathrm{mol}$ of $\mathrm{CH_3OH}$ is formed. The $K_p^0$ is $\underline{\hspace{1cm}}\times10^{-3}$ (nearest integer). Given: $R=0.08\,\mathrm{dm^3\,bar\,K^{-1}\,mol^{-1}}$ Assume only methanol is formed as the product and the system follows ideal gas behaviour.
Answer: 74
At time $t = 0$, the concentrations are as follows: $$\mathrm{CO(g)} + 2\mathrm{H_2(g)} \rightleftharpoons \mathrm{CH_3OH(g)}$$ Initial concentrations: - $\mathrm{CO(g)}$: $0.1 \, \mathrm{mol}$ - $2\mathrm{H_2(g)}$: $a \, \mathrm{mol}$ - $\mathrm{CH_3OH(g)}$: $-$ At equilibrium: - $\mathrm{CO(g)}$: $0.1 - x$ - $2\mathrm{H_2(g)}$: $a - 2x$ - $\mathrm{CH_3OH(g)}$: $x = 0.04$ Calculations: $$0.1 - x = 0.06$$ $$a - 2x = a - 0.08$$ $$= 0.23 - 0.08$$ $$= 0.15 \, \mathrm{mole}$$ Given: - $V = 2 \, \mathrm{L}$ - $T = 500 \, \mathrm{K}$ - $P_{total} = 5 \, \mathrm{bar}$ Total moles: $$n_{Total} = 0.25 = \frac{1}{4} \, \mathrm{mol}$$ Pressure relation: $$P_{total} = n_{total} \times \frac{RT}{V}$$ Solving: $$5 = (0.06 + a - 0.08 + 0.04) \times \frac{0.08 \times 500}{2}$$ $$10 = (0.02 + a) \times 0.08 \times 500$$ $$a = 0.25 - 0.02 = 0.23 \, \mathrm{mol}$$ Equilibrium constant $K_P$: $$K_P = \frac{X_{\mathrm{CH_3OH}}}{X_{\mathrm{CO}} \times X_{\mathrm{H_2}}^2} \times \frac{1}{(P_T)^2} = \frac{0.04}{0.06 \times (0.15)^2} \times \left[ \frac{1/4}{5} \right]^2$$ Calculations: $$= \frac{4}{6 \times (0.15)^2 \times 16} \times \frac{1}{25}$$ $$= \frac{100 \times 100}{24 \times 225 \times 25} = \frac{100 \times 100}{135000}$$ $$= 0.074 = 74 \times 10^{-3}$$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For the reaction A $\rightarrow$ products. The concentration of A at 10 minutes is ___ $\times 10^{-3}$ $\mathrm{mol\ L}^{-1}$ (nearest integer). The reaction was started with 2.5 $\mathrm{mol\ L}^{-1}$ of A.
Answer: 2435
Given $t_{1/2} \propto [A]_0 \Rightarrow Order = zero$. $$t_{1/2} = \frac{A_0}{2 \, K} \Rightarrow Slope = \frac{1}{2 \, K} = 76.92$$ $$K = \frac{1}{2 \times 76.92}$$ $$[A]_{10} = -Kt + A_0 = -\frac{1}{2 \times 76.92} \times 10 + 2.5 = 2.435$$ $$= 2435 \times 10^{-3} \, mol/L$$