JEE Main 28 January 2025 Shift 1: all 75 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
Register free to attempt this shift as a timed 180-minute test with instant scoring and chapter-wise analysis.
Maths · Permutations and Combinations · Single correct
The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is
Answer: (d)
Case I: $5\ldots\ldots 0$ Case II: $5\ 1$ $5\ 3$ $6\ 0$ $6\ 1$ $6\ 2$ $7\ 0$ $7\ 1$ $9 \times (8 \times 8 \times 8) = 4608$ but $50000$ is not included, so total numbers $4608 - 1 = 4607$
Maths · Conic Sections · Single correct
Let ABCD be a trapezium whose vertices lie on the parabola $y^2 = 4x$. Let the sides $AD$ and $BC$ of the trapezium be parallel to y-axis. If the diagonal $AC$ is of length $\frac{25}{4}$ and it passes through the point $(1, 0)$, then the area of $ABCD$ is
Answer: (a)
A $(at_1^2, 2at)$ and $C \left( \frac{a}{t_1^2}, -\frac{2a}{t_1} \right)$. Length $AC = a \left( t_1 + \frac{1}{t_1} \right)^2 = \frac{25}{4}$, $t_1 + \frac{1}{t_1} = \pm \frac{5}{2}$. Therefore, $t_1 = 2$ or $\frac{1}{2}$, $A \left( \frac{1}{2}, 1 \right)$, $D \left( \frac{1}{4}, -1 \right)$, $B(4, 4)$, $C(4, -4)$. So, area of trapezium $= \frac{1}{2} (8 + 2) \left( 4 - \frac{1}{4} \right) = \frac{75}{4}$.
Maths · Complex Numbers and Quadratic Equations · Single correct
Two number $k_1$ and $k_2$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $i^{k_1} + i^{k_2}$, $(i = \sqrt{-1})$ is non-zero, equals
Answer: (b)
Let $k_1 = 4\lambda_1 + r_1$, $r_1 \in \{0, 1, 2, 3\}$. $k_2 = 4\lambda_2 + r_2$. $$(i)^{k_1} + (i)^{k_2} = (i)^{r_1} + (i)^{r_2}$$ $$(i)^{r_1} \in \{1, i, -1, -i\}$$ Zero implies $1, (-1)$ pair implies $$\left\{ \begin{array}{cc} 1, & -1 \\ i, & -i \\ -i, & +i \\ -1, & 1 \end{array} \right\}$$ $i, (-i)$ pair. Zero probability is $$\frac{4}{\binom{4}{1} \cdot \binom{4}{1}} = \frac{1}{4}$$ Probability (non-zero) is $$1 - \frac{1}{4} = \frac{3}{4}$$
Maths · Relations and Functions · Single correct
If $f(x) = \frac{2^x}{2^x + \sqrt{2}}, x \in \mathbb{R}$, then $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)$ is equal to
Answer: (d)
Given $f(x) = \frac{2^x}{2^x + \sqrt{2}}$. We have $f(x) + f(1-x) = \frac{2^x}{2^x + \sqrt{2}} + \frac{2^{1-x}}{2^{1-x} + \sqrt{2}}$. This simplifies to $$= \frac{2^x}{2^x + \sqrt{2}} + \frac{2}{2 + \sqrt{2}2^x} = \frac{2^x + \sqrt{2}}{2^x + \sqrt{2}} = 1.$$ Now, $$\sum_{k=1}^{81} f\left(\frac{k}{82}\right) = f\left(\frac{1}{82}\right) + f\left(\frac{2}{82}\right) + \ldots + f\left(\frac{81}{82}\right).$$ This equals $$= f\left(\frac{1}{82}\right) + f\left(1 - \frac{2}{82}\right) + f\left(1 - \frac{1}{82}\right).$$ So, $$\left[f\left(\frac{1}{82}\right) + f\left(1 - \frac{1}{82}\right)\right] + \left[f\left(\frac{2}{82}\right) + f\left(1 - \frac{2}{82}\right)\right] + \ldots 40 cases + f\left(\frac{41}{82}\right).$$ This results in $$(1 + 1 + \ldots + 1) 40 times + 2^{1/2}.$$ Thus, $$40 + \frac{1}{2} = \frac{81}{2}.$$
Maths · Relations and Functions · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined by $$f(x) = (2 + 3a)x^2 + \left(\frac{a^2 + 2}{a - 1}\right)x + b, a \neq 1.$$ If $$f(x + y) = f(x) + f(y) + 1 - \frac{2}{7}xy,$$ then the value of $28 \sum_{i=1}^{5} |f(i)|$ is
Answer: (d)
Put $y = 0$. $$f(x) = f(0) + f(x) + 1 - 0$$ $$f(0) = -1$$ $$f(0) = 0 + 0 + b$$ $$\Rightarrow b = -1$$ $$f(-1 + 1) = f(-1) + f(1) + 1 + \frac{2}{7}$$ $$f(0) = f(-1) + f(1) + \frac{9}{7}$$ $$-1 = (2 + 3a) + \left( \frac{a + 2}{a - 1} \right)(-1) + b + (2 + 3a)$$ $$ + \frac{a + 2}{a - 1} + b + \frac{9}{7}$$ $$-1 = 4 + 6a - 2 + \frac{9}{7}$$ $$-1 = 2 + \frac{9}{7} + 6a$$ $$6a = -1 - 2 - \frac{9}{7}$$ $$a = -\frac{5}{7}$$ $$f(x) = -\frac{x^2}{7} + \frac{9}{7}x - 1$$ $$f(x) = -\frac{x^2}{7} - \frac{3}{4}x - 1$$ $$\sum_{i=1}^{5} f(i) = -\frac{1}{7} \left( \frac{5 \times 6 \times 11}{6} \right) - \frac{3}{4} \left( \frac{5 \times 6}{2} \right) - 5$$ $$= -\frac{55}{7} - \frac{45}{4} - 5$$ $$= \frac{675}{28}$$ $$\Rightarrow 28 \left| \sum_{i=1}^{5} f(i) \right| = 675$$
Maths · Three Dimensional Geometry · Single correct
Let A(x, y, z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, -1) and C = (2, 0, -2). Then among the statements (S1) : $\triangle ABC$ is an isosceles right angled triangle, and (S2) : the area of $\triangle ABC$ is $\frac{9\sqrt{2}}{2}$,
Answer: (c)
Let $P(0, 3, 2)$, $Q(2, 0, 3)$, $R(0, 0, 1)$. Given $AP = AQ = AR$. $$x^2 + (y - 3)^2 + (z - 2)^2 = (x - 2)^2 + y^2 + (z - 3)^2 = x^2 + y^2 + (z - 1)^2$$ In $xy$ plane $z = 0$. So, $x^2 - 4x + 4 + y^2 + 9 = x^2 + y^2 + 1$. $x = 3$. $$9 + y^2 - 6y + 9 + 4 = x^2 + y^2 + 1$$ So, $A(3, 2, 0)$ also $B(1, 4, -1)$ and $C(2, 0, -2)$. Now $AB = \sqrt{4 + 4 + 1} = 3$. $AC = \sqrt{1 + 4 + 4} = 3$. $BC = \sqrt{1 + 16 + 1} = \sqrt{18}$. $AB = AC$. Isosceles $\triangle ABC$ and $AB^2 + AC^2 = BC^2$. Right angle $\triangle$. Area of $\triangle ABC = \frac{1}{2} \times base \times height$. $$\frac{1}{2} \times 3 \times 3 = \frac{9}{2}$$ So only $S_1$ is true.
Maths · Sets · Single correct
The relation \[ R=\{(x,y):x,y\in\mathbb{Z}\text{ and }x+y\text{ is even}\} \] is:
Answer: (b)
For reflexive $(x,x) \in \mathbb{R},\ x \in \mathbb{Z}$ $\Rightarrow x + x = 2x \rightarrow$ even For symmetric of $(x,y) \in \mathbb{R}$ then $(y,x) \in \mathbb{R}$ when $x,y \in \mathbb{Z}$ $x + y \rightarrow$ even $\Rightarrow y + x \rightarrow$ even for transitive if $(x,y) \in \mathbb{R} \Rightarrow x + y \rightarrow$ even $(y,z) \in \mathbb{R} \Rightarrow y + z \rightarrow$ even $x + 2y + z \rightarrow$ even $\Rightarrow x + z$ is even $\Rightarrow (x,z) \in \mathbb{R}$ $\Rightarrow \mathbb{R}$ is an equivalence relation.
Maths · Conic Sections · Single correct
Let the equation of the circle, which touches $x$-axis at the point $(a, 0), a > 0$ and cuts off an intercept of length $b$ on $y$-axis be $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below $x$-axis, then the ordered pair $(2a, b^2)$ is equal to
Answer: (d)
By Pythagoras $r^2 = a^2 + \frac{b^2}{4} = P^2$. $$r = \sqrt{\frac{4a^2 + b^2}{4}}$$ Equation of circle is $(x - \alpha)^2 + (y - \beta)^2 = r^2$. $$x^2 + y^2 - 2ax - 2py + \alpha^2 + \beta^2 - r^2 = 0$$ Comparison $x^2 + y^2 - \alpha x + \beta y + r = 0$. $$-\alpha = -2a, \beta = -2p, r = a^2$$ $$\Rightarrow 2a = \alpha, 4a^2 + b^2 = 4p^2$$ $$a^2 + b^2 = 4p^2$$ $$\alpha^2 + \beta^2 = \beta^2$$ So, $(2a, b^2) = (\alpha, \beta^2 - 4r)$
Maths · Sequences and Series · Single correct
Let $\langle a_n \rangle$ be a sequence such that $a_0 = 0$, $a_1 = \frac{1}{2}$ and $2a_{n+2} = 5a_{n+1} - 3a_n$, $n = 0, 1, 2, 3, \ldots$. Then $\sum_{k=1}^{100} a_k$ is equal to
Answer: (b)
Given $a_0 = 0$, $a_1 = \frac{1}{2}$. The recurrence relation is $2a_{n+2} = 5a_{n+1} - 3a_n$. Solving the characteristic equation $2x^2 - 5x + 3 = 0$ gives $x = 1, \frac{3}{2}$. Therefore, $a_n = A1^n + B \left( \frac{3}{2} \right)^n$. For $n = 0$, $0 = A + B$. For $n = 1$, $\frac{1}{2} = A + \frac{3}{2} B$. Solving these gives $A = -1$, $B = 1$. Thus, $a_n = -1 + \left( \frac{3}{2} \right)^n$. The sum $\sum_{k=1}^{100} a_k = \sum_{k=1}^{100} (-1) + \left( \frac{3}{2} \right)^k$. This simplifies to $$= -100 + \left( \frac{3}{2} \right) \left( \left( \frac{3}{2} \right)^{100} - 1 \right)$$ $$= -100 + 3 \left( \left( \frac{3}{2} \right)^{100} - 1 \right)$$ $$= 3 \cdot (a_{100}) - 100$$
Maths · Inverse Trigonometric Functions · Single correct
$\cos\left(\sin^{-1}\dfrac{3}{5} + \sin^{-1}\dfrac{5}{13} + \sin^{-1}\dfrac{33}{65}\right)$ is equal to:
Answer: (b)
Given the expression: $$\cos \left( \sin^{-1} \frac{3}{5} + \sin^{-1} \frac{5}{13} + \sin^{-1} \frac{33}{65} \right)$$ We can rewrite it using tangent inverse: $$\cos \left( \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{33}{56} \right)$$ Using the identity for tangent addition: $$\cos \left( \tan^{-1} \left( \frac{\frac{3}{4} + \frac{5}{12}}{1 + \frac{3}{4} \cdot \frac{5}{12}} \right) + \tan^{-1} \frac{33}{56} \right)$$ Simplifying further: $$\cos \left( \tan^{-1} \frac{56}{33} + \cot^{-1} \frac{56}{33} \right)$$ This simplifies to: $$\cos \left( \frac{\pi}{2} \right) = 0$$
Maths · Sequences and Series · Single correct
Let $T_r$ be the $r^{th}$ term of an A.P. If for some $m$, $T_m = \frac{1}{25}$, $T_{25} = \frac{1}{20}$, and $20 \sum_{r=1}^{25} T_r = 13$, then $5 m \sum_{r=m}^{2m} T_r$ is equal to
Answer: (b)
Given $T_m = \frac{1}{25}$, $T_{25} = \frac{1}{20}$, $20 \sum_{r=1}^{25} T_r = 13$. $T_m = a + (m-1)d = \frac{1}{25}$ $\ldots$ (1) $T_{25} = a + 24d = \frac{1}{20}$ $20 \cdot \frac{25}{2} \left[ a + \frac{1}{20} \right] = 13 \Rightarrow a = \frac{1}{500}$ Also, $20 S_{25} = 20 \cdot \frac{25}{2} [2a + 24d] = 13 \Rightarrow d = \frac{1}{500}$ From (1) $\frac{1}{500} + \frac{m-1}{500} = \frac{1}{25} \Rightarrow m = 20$ Now, $$5m \sum_{r=m}^{2m} T_r = 100 \sum_{r=20}^{40} T_r = 126$$
Maths · Three Dimensional Geometry · Single correct
If the image of the point (4, 4, 3) in the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to
Answer: (a)
Let coordinate of $R = (2r + 1, r + 2, 3r + 1)$. Therefore, $PR$ is perpendicular to given line. Thus, $(2r - 3) \cdot 2 + (r - 2) \cdot 1 + (3r - 2) \cdot 3 = 0$. Therefore, $r = 1$. Thus, Coordinate of $R = (3, 3, 4)$. Therefore, $(\alpha, \beta, \gamma) = (2, 2, 5)$. Thus, $\alpha + \beta + \gamma = 9$
Maths · Integrals · Single correct
If $\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{96x^2\cos^2x}{1+e^x}\,dx=\pi\left(\alpha\pi^2+\beta\right),$ where $\alpha,\beta\in\mathbb{Z}$, then $(\alpha+\beta)^2$ equals ______.
Answer: (d)
Given $$I = \int_0^{\frac{\pi}{2}} \frac{96x^2 \cos^2 x}{1 + e^x} \, dx$$ We have $$2I = 2 \int_0^{\frac{\pi}{2}} 96x^2 \cos^2 x \, dx$$ Thus, $$I = 96 \int_0^{\frac{\pi}{2}} x^2 \cos^2 x \, dx$$ This simplifies to $$= 48 \int_0^{\frac{\pi}{2}} x^2 (1 + \cos 2x) \, dx$$ Evaluating, $$= 2\pi^2 + 48(0 - 0) - 48 \int_0^{\frac{\pi}{2}} x \sin 2x \, dx$$ Further simplifying, $$= 2\pi^2 - 12\pi + [0 - 0] = \pi \left(2\pi^2 - 12\right)$$ Therefore, $$= \pi \left(\alpha \pi^2 + \beta\right)$$ This implies $$(\alpha + \beta)^2 = 100$$
Maths · Applications of Derivatives · Single correct
The sum of all local minimum values of the function $f(x) = \begin{cases} 1-2x, & x 2 \end{cases}$ \ is
Answer: (a)
Given $$f(x) = \begin{cases} 1 - 2x, & x 2 \end{cases}$$ Therefore, local minimum values at A and B are calculated as follows: For point A: $$\frac{7}{3} - \frac{11}{72} = \frac{168 - 11}{72} = \frac{157}{72}$$
Maths · Complex Numbers and Quadratic Equations · Single correct
The sum, of the squares of all the roots of the equation $x^2 + |2x - 3| - 4 = 0$, is
Answer: (c)
For $x \geq \frac{3}{2}$ $$x^2 + 2x - 3 - 4 = 0$$ $$x^2 + 2x - 7 = 0$$ $$x = \frac{-2 \pm \sqrt{4 + 28}}{2} = -1 \pm 2\sqrt{2}$$ Only $2\sqrt{2} - 1$ is acceptable root. For $x < \frac{3}{2}$ $$x^2 - 2x + 3 - 4 = 0$$ $$x^2 - 2x - 1 = 0$$ $$x = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$$ Only $1 - \sqrt{2}$ is acceptable. Sum of the square = $(1 - \sqrt{2})^2 + (2\sqrt{2} - 1)^2$ $$= 6(2 - \sqrt{2})$$
Maths · Integrals · Single correct
Let for some function $y = f(x)$, $\int_{0}^{x} t f(t) dt = x^2 f(x)$, $x > 0$ and $f(2) = 3$. Then $f(6)$ is equal to
Answer: (a)
Given $\($ $\int$_0^x t f(t) dt = x^2 f(x) $\)$. This implies $\($ x f(x) = 2x f(x) + x^2 f'(x) $\)$. Therefore, $\($ x^2 $\cdot$ f'(x) = -x f(x) $\)$. Thus, $\($ f'(x) = -$\frac{1}{x}$ f(x) $\)$. Integrating, $\($ $\int$ $\frac{f'(x)}{f(x)}$ dx = -$\int$ $\frac{1}{x}$ dx $\)$. This gives $\($ $\ln$ f(x) = -$\ln$ x + C $\)$. Given $\($ f(2) = 3 $\)$, we have $\($ $\ln$ 3 = -$\ln$ 2 + C $\)$. Thus, $\($ C = $\ln$ 6 $\)$. Therefore, $\($ f(x) = $\frac{6}{x}$ $\)$ and $\($ f(6) = 1 $\)$.
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let $^nC_{r-1} = 28$, $^nC_r = 56$ and $^nC_{r+1} = 70$. Let $A(4 \cos t, 4 \sin t)$, $B(2 \sin t, -2 \cos t)$ and $C (3r - n, r^2 - n - 1)$ be the vertices of a triangle $ABC$, where $t$ is a parameter. If $(3x - 1)^2 + (3y)^2 = \alpha$, is the locus of the centroid of triangle $ABC$, then $\alpha$ equals
Answer: (d)
Given $\($ nC_{r-1} = 28 $\)$, $\($ nC_r = 56 $\)$, $\($ nC_{r+1} = 70 $\)$. $\[$ $\begin{aligned}$ & nC_{r-1} = $\frac{28}{56}$ $\implies$ $\frac{r}{n-r+1}$ = $\frac{1}{2}$ $\\$ & nC_r = $\frac{56}{70}$ $\implies$ $\frac{r+1}{n-r}$ = $\frac{70}{56}$ $\\$ & $\Rightarrow$ n = 8 = 3 $\end{aligned}$ $\]$ For the triangle with vertices $\($ A(4$\cos$ t, 4$\sin$ t) $\)$, $\($ B(2$\sin$ t, -2$\cos$ t) $\)$, $\($ C(1, 0) $\)$, the centroid $\($(h, k)$\)$ is given by: $\[$ h = $\frac{4\cos t + 2\sin t + 1}{3}$, k = $\frac{4\sin t - 2\cos t}{3}$ $\]$ Simplifying, we have: $\[$ 3h - 1 = 4$\cos$ t + 2$\sin$ t $\]$ $\[$ 3k - 1 = 4$\sin$ t - 2$\cos$ t $\]$ Using the identity: $\[$ (1)^2 + (2)^2 $\]$ The locus of the centroid is: $\[$ (3h - 1)^2 + (3k)^2 = 20 $\]$ Thus, the locus of the centroid is $\($(3x - 1)^2 + (3y)^2 = 20$\)$. $\($ $\Rightarrow$ $\alpha$ = 20 $\)$
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $O$ be the origin, the point $A$ be $z_1 = \sqrt{3} + 2\sqrt{2}i$, the point $B(z_2)$ be such that $\sqrt{3} |z_2| = |z_1|$ and $\arg(z_2) = \arg(z_1) + \frac{\pi}{6}$. Then
Answer: (b)
Given $OA = |z_1| = \sqrt{3} + 8 = \sqrt{11}$ and $OB = \frac{1}{\sqrt{3}} |z_1| = \sqrt{\frac{11}{3}}$. $$AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cos \frac{\pi}{6}$$ $$= 11 + \frac{11}{3} - 2 \cdot \frac{11}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2}$$ Therefore, $AB = \sqrt{\frac{11}{3}}$. Thus, the area of $\triangle ABD = \frac{1}{2} \cdot OA \cdot OB \cdot \sin \frac{\pi}{6}$ $$= \frac{11}{4\sqrt{3}} sq. units$$ Here $OB = AB$ and $\angle A = \frac{2\pi}{3}$. Therefore, $\triangle ABD$ is an obtuse angled isosceles triangle.
Maths · Probability · Single correct
Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If $x$ denote the number of defective oranges, then the variance of $x$ is
Answer: (a)
There are 3 bad oranges and 7 good oranges. Therefore, $X =$ number of bad oranges drawn. Variance $$= 0^2 \cdot \frac{{^7C_2}}{{^{10}C_2}} + 1^2 \cdot \left( \frac{{3 \times 7}}{{^{10}C_2}} \right) + 2^2 \left( \frac{3}{{^{10}C_2}} \right)$$ $$- \left( 0 + 1 \cdot \frac{{3 \times 7}}{{^{10}C_2}} + 2 \cdot \frac{3}{{^{10}C_2}} \right)^2$$ $$= \frac{28}{75}$$
Maths · Applications of Integrals · Single correct
The area (in sq. units) of the region $$\{(x, y) : 0 \leq y \leq 2|x| + 1, 0 \leq y \leq x^2 + 1, |x| \leq 3\}$$ is
Answer: (b)
The area is calculated as follows: $$Area = 2 \left[ \int_0^2 (x^2 + 1) \, dx + \frac{1}{2} [5 + 7] \times 1 \right]$$ This simplifies to: $$= \frac{64}{3}$$
Maths · Matrices · Numerical
Let $\mathrm{M}$ denote the set of all real matrices of order 3 $\times$ 3 and let $S = \{-3, -2, -1, 1, 2\}$. Let \[ S_1 = \left\{ A = [a_{ij}] \in M : A = A^{T} \text{ and } a_{ij} \in S,\; \forall i,j \right\} \] \[ S_2 = \left\{ A = [a_{ij}] \in M : A = -A^{T} \text{ and } a_{ij} \in S,\; \forall i,j \right\} \] \[ S_3 = \left\{ A = [a_{ij}] \in M : a_{11} + a_{22} + a_{33} = 0 \text{ and } a_{ij} \in S,\; \forall i,j \right\} \] \[ \text{If } n(S_1 \cup S_2 \cup S_3) = 125\alpha, \text{ then } \alpha \text{ equals } \underline{\hspace{1cm}} \]
Answer: 1613
The matrix is given as: $$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$ Number of elements in $S_1$: $A = A^T \Rightarrow 5^3 \times 5^3$ Number of elements in $A = -A^T \Rightarrow 0$ since no zero in 5. Number of elements in $S_3 \Rightarrow$ $$a_{11} + a_{22} + a_{33} = 0 \Rightarrow (1, 2, -3) \Rightarrow 31$$ or $$(1, 1, -2) \Rightarrow 3$$ or $$(-1, -1, 2) \Rightarrow 3$$ $$\Rightarrow 12 \times 5^6$$ $n(S_1 \cap S_3) = 12 \times 5^3$ $n(S_1 \cup S_2 \cup S_3) = 5^6(1 + 12) - 12 \times 5^3$ $$\Rightarrow 5^3 \times [13 \times 5^3 - 12] = 125\alpha$$ $$\alpha = 1613$$
Maths · Binomial Theorem · Numerical
If $\alpha = 1 + \sum_{r=1}^{6} (-3)^{r-1}\;{}^{12}C_{2r-1}$, then the distance of the point $(12, \sqrt{3})$ from the line $\alpha x - \sqrt{3}y + 1 = 0$ is
Answer: 5
Given $$\alpha = 1 + \sum_{r=1}^{6} (-1)^{r-1} 12C_{2r-1} 3^{r-1}$$ Let $i$ be iota, and let $\sqrt{3}i = x$. Then $$\alpha = 1 + \sum_{r=1}^{6} 12C_{2r-1} \left(\sqrt{3}i\right)^{2t-1}$$ $$= 1 + \frac{1}{\sqrt{3}i} \left(12C_1 x + 12C_3 x^3 + \ldots + 12C_{11} x^{11}\right)$$ $$= 1 + \frac{1}{\sqrt{3}i} \left((1 + \sqrt{3}i)^{12} - (1 - \sqrt{3}i)^{12}\right)$$ $$= 1 + \frac{1}{\sqrt{3}i} \left(\frac{(-2w)^2^{12} - (2w)^{12}}{2}\right) = 1$$ So the distance of $(12, \sqrt{3})$ from $x - \sqrt{3}y + 1 = 0$ is $$\frac{12 - 3 + 1}{2} = 5$$
Maths · Vector Algebra · Numerical
Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{d} = \vec{a} \times \vec{b}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = |\vec{c}|$, $|\vec{c} - 2\vec{a}|^2 = 8$ and the angle between $\vec{d}$ and $\vec{c}$ is $\frac{\pi}{4}$, then $\left| 10 - 3 \vec{b} \cdot \vec{c} \right| + \left| \vec{d} \times \vec{c} \right|^2$ is equal to
Answer: 6
Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$. Let $\vec{d} = \vec{a} \times \vec{b}$. Then $\vec{d} = -\hat{i} + \hat{j}$. We have $|\vec{c} - 2\vec{a}|^2 = 8$. Expanding, $|\vec{c}|^2 + 4|\vec{a}|^2 - 4\vec{a} \cdot \vec{c} = 8$. This simplifies to $|\vec{c}|^2 + 12 - 4|\vec{c}| = 8$. Further simplification gives $|\vec{c}|^2 - 4|\vec{c}| + 4 = 0$. Solving, $|\vec{c}| = 2$. Now, $\vec{d} = \vec{a} \times \vec{b}$ and $\vec{d} \times \vec{c} = (\vec{a} \times \vec{b}) \times \vec{c}$. Using the identity, $\left( |\vec{d}| \times |\vec{c}| \sin \frac{\pi}{4} \right)^2 = \left( (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} \right)^2$. This gives $4 = 4|\vec{b}|^2 + (\vec{b} \cdot \vec{c})^2 (|\vec{a}|^2) - 2(\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{b})$. Let $\vec{b} \cdot \vec{c} = x$. Then $4 = 36 + 3x^2 - 20x$. Solving $3x^2 - 20x + 32 = 0$, we find $x = \frac{8}{3}, 4$. Thus, $\vec{b} \cdot \vec{c} = \frac{8}{3}, 4$. Therefore, $\vec{b} \cdot \vec{c} = 8$. Now, $|10 - 3\vec{b} \cdot \vec{c}| + |\vec{d} \times \vec{c}|^2 = |10 - 8| + (2)^2 = 6$.
Maths · Continuity and Differentiability · Numerical
Let $f(x) = \begin{cases} 3x, & x 2, \end{cases}$ where $[.]$ denotes greatest integer function. If $\alpha$ and $\beta$ are the number of points, where $f$ is not continuous and is not differentiable, respectively, then $\alpha + \beta$ equals ____
Answer: 5
Given $$f(x) = \begin{cases} 3x & ; \ x 2 \end{cases}$$ The function is not continuous at $x \in \{1, 2\}$ which implies $\alpha = 2$. The function is not differentiable at $x \in \{0, 1, 2\}$ which implies $\beta = 3$. Therefore, $\alpha + \beta = 5$.
Maths · Conic Sections · Numerical
Let $\mathcal{E}_1 : \frac{x^2}{9} + \frac{y^2}{4} = 1$ be an ellipse. Ellipses $\mathcal{E}_1$'s are constructed such that their centres and eccentricities are the same as that of $\mathcal{E}_1$, and the length of minor axis of $\mathcal{E}_i$ is the length of major axis of $\mathcal{E}_{i+1}$ $(i \geq 1)$. If $A_i$ is the area of the ellipse $\mathcal{E}_i$, then $\frac{5}{\pi} \left( \sum_{i=1}^{\infty} A_i \right)$, is equal to
For $E_1$, $\frac{x^2}{9} + \frac{y^2}{4} = 1$ implies $e = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$. For $E_2$, $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$. Given $e = \frac{\sqrt{5}}{3} = \sqrt{1 - \frac{a^2}{4}}$, we have $\frac{5}{9} = 1 - \frac{a^2}{4}$. Solving for $a^2$, $a^2 = \frac{16}{9}$. Thus, $E_2$: $\frac{x^2}{\frac{16}{9}} + \frac{y^2}{4} = 1$. For $E_3$, $\frac{x^2}{\frac{16}{9}} + \frac{y^2}{b^2} = 1$. Given $e = \frac{\sqrt{5}}{3} = \sqrt{1 - \frac{16}{9b^2}}$, we find $b^2 = \frac{64}{81}$. Thus, $E_3$: $\frac{x^2}{\frac{16}{9}} + \frac{y^2}{\frac{64}{81}} = 1$. For $A_1 = \pi \times 3 \times 2 \Rightarrow 6\pi$. For $A_2 = \pi \times \frac{4}{3} \times 2 \Rightarrow \frac{8\pi}{3}$. For $A_3 = \pi \times \frac{4}{3} \times \frac{8}{9} \Rightarrow \frac{32\pi}{27}$. The sum $\sum_{i=1}^{\infty} A_i = 6\pi + \frac{8\pi}{3} + \frac{32\pi}{27} + \ldots \infty \Rightarrow \frac{6\pi}{1 - \frac{4}{9}} \Rightarrow 54\pi$. Thus, $\frac{5}{\pi} \sum_{i=1}^{\infty} A_i \Rightarrow \frac{5}{\pi} \times 54\pi \Rightarrow 54$.
Physics · Electrostatic Potential and Capacitance · Single correct
Two capacitors $C_1$ and $C_2$ are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are $U_1$ and $U_2$, respectively. Which of the given statements is true?
Answer: (d)
For a capacitor at steady state $q=CV$ and $U=\dfrac{1}{2}CV^2$ Since $C_1$ and $C_2$ are connected in parallel, $V_1=V_2$. Also, from the graph $q_1 C_1$ $\dfrac{U_1}{U_2} = \dfrac{C_1V_1^2}{C_2V_2^2} = \dfrac{C_1}{C_2} U_1$.
Physics · Mechanical Properties of Fluids · Single correct
In the experiment for measurement of viscosity ' $\eta$ ' of given liquid with a ball having radius $R$, consider following statements. A. Graph between terminal velocity $V$ and $R$ will be a parabola. B. The terminal velocities of different diameter balls are constant for a given liquid. C. Measurement of terminal velocity is dependent on the temperature. D. This experiment can be utilized to assess the density of a given liquid. E. If balls are dropped with some initial speed, the value of $\eta$ will change. Choose the correct answer from the options given below:
Answer: (c)
The terminal velocity of ball of radius $R$ inside a liquid of viscosity $\eta$ can be written as $V_T = \frac{2R^2 g}{3\eta} (\sigma - \rho)$, where $\sigma$ is the density of ball and $\rho$ is the density of the liquid. Hence, $A$ is correct since $V_T \propto R^2$ gives a parabola on a graph. $C$ is correct since $V_T \propto \frac{1}{\eta}$ and $\eta$ varies with temperature. $D$ is correct since $V_T \propto (\sigma - \rho)$ i.e., varies with density of liquid.
Physics · Mechanical Properties of Fluids · Single correct
Consider following statements: A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface, of a liquid. B. As the temperature of liquid rises, the coefficient of viscosity increases. C. As the temperature of gas increases, the coefficient of viscosity increases. D. The onset of turbulence is determined by Reynold's number. E. In a steady flow two stream lines never intersect. Choose the correct answer from the options given below:
Answer: (a)
Surface tension arises due to extra energy of the molecules at the surface as compared at the interior of a liquid. The coefficient of viscosity for a liquid decreases with rise in temperature whereas it increases for gases with increase in temperature. The flow is turbulent for a Reynold's number greater than 2000. Stream lines never intersect in a steady flow.
Physics · Electric Charges and Fields · Single correct
Three infinitely long wires with linear charge density $\lambda$ are placed along the $x$-axis, $y$-axis and $z$-axis respectively. Which of the following denotes an equipotential surface?
Answer: (c)
Potential due to an infinite wire is $V = 2k\lambda \ln r$, where $r$ is the distance from the wire. Taking the point in space $P(x, y, z)$ Distance from wire along $x$-axis is $r_x = \sqrt{y^2 + z^2}$ Distance from wire along $y$-axis is $r_y = \sqrt{x^2 + z^2}$ Distance from wire along $z$-axis is $r_z = \sqrt{x^2 + y^2}$ Therefore, potential at $P$ due to wire along $x$-axis is $$V_x = 2k\lambda \ln r_x$$ Potential at $P$ due to wire along $y$-axis is $$V_y = 2k\lambda \ln r_y$$ Potential at $P$ due to wire along $z$-axis is $$V_z = 2k\lambda \ln r_z$$ Thus, net potential at $P = V = V_x + V_y + V_z$ or $$V = 2k\lambda \ln r_x + 2k\lambda \ln r_y + 2k\lambda \ln r_z$$ i.e. $V = 2k\lambda \ln (r_x r_y r_z)$ or $$V = 2k\lambda \ln \left( \sqrt{y^2 + z^2} \sqrt{z^2 + x^2} \sqrt{x^2 + y^2} \right)$$ $$= k\lambda \ln (y^2 + z^2) (z^2 + x^2) (x^2 + y^2)$$ Therefore, for equipotential surface $$(x^2 + y^2) (y^2 + z^2) (z^2 + x^2) = constant$$
Physics · Ray Optics and Optical Instruments · Single correct
A hemispherical vessel is completely filled with a liquid of refractive index $\mu$. A small coin is kept at the lowest point (O) of the vessel as shown in figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point E (at the level of the vessel) is
Answer: (d)
For the rays from coin to reach the point $E$, the refracted rays must grazing the surface, i.e. they must be incident at critical angle $\theta_c$ inside the liquid. $$\mu = \frac{1}{\sin \theta_c}$$ $\mu$ is minimum when $\theta_c$ is maximum. Maximum value of $\theta_c = 45^\circ$ $$\Rightarrow \mu has a minimum value of \sqrt{2}.$$
Physics · Moving Charges and Magnetism · Single correct
Consider a long thin conducting wire carrying a uniform current I. A particle having mass " M " and charge " q " is released at a distance " a " from the wire with a speed $v_0$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance $x$ from the wire. The value of $x$ is [ $\mu_0$ is vacuum permeability]
Answer: (a)
For the path from $A \to B$, the velocity is $\vec{V} = -v_y \hat{j}$. The magnetic field is $\vec{B} = \frac{\mu_0 I}{2 \pi r} (-\hat{k})$. The force is $\vec{F} = q (\vec{V} \times \vec{B}) = \frac{\mu_0 I q}{2 \pi r} [-v_x \hat{j} - v_y \hat{i}]$. The acceleration in the $x$ direction is $a_x = -\frac{\mu_0 I q}{2 \pi m} \frac{v_y}{r}$ and in the $y$ direction is $a_y = -\frac{\mu_0 I q}{2 \pi m} \frac{v_x}{r}$. The equation $v_x dv_x = -\frac{\mu_0 I q}{2 \pi m} r dr$ is derived. Integrating from $0$ to $v_0$, we have $$\int_0^{v_0} v_x dv_x = -\frac{\mu_0 I q}{2 \pi m} \int_a^{x_1} r \, dr$$ which simplifies to $$\frac{v_0^2 - v_x^2}{2} = -\frac{\mu_0 I q}{2 \pi m} \left[ \frac{r^2}{2} \right]_a^{x_1}.$$ Let $z^2 = v_0^2 - v_x^2$, then $2z \, dz = -2v_x dv_x$. Substituting, we get $$\int_0^{v_0} dz = -\frac{\mu_0 I q}{2 \pi m} \ln \frac{x_1}{a}.$$ Solving for $v_0$, $$v_0 = -\frac{\mu_0 I q}{2 \pi m} \ln \frac{x_1}{a}.$$ Thus, $$x_1 = ae^{\frac{2 \pi m v_0}{\mu_0 I q}}.$$ For the path from $B \to C$, the velocity is $\vec{V} = -v_x \hat{i} - v_y \hat{j}$. The force is $\vec{F} = q (\vec{V} \times \vec{B}) = \frac{\mu_0 I q}{2 \pi r} (-v_x \hat{j} + v_y \hat{i})$. The acceleration in the $x$ direction is $a_x = +\frac{\mu_0 I q}{2 \pi m} \frac{v_y}{r}$ and in the $y$ direction is $a_y = -\frac{\mu_0 I q}{2 \pi m} \frac{v_x}{r}$. The equation $v_x dv_x = \frac{\mu_0 I q}{2 \pi m} v_y dr$ is derived. Integrating from $v_0$ to $0$, we have $$\int_{v_0}^0 v_x dv_x = \frac{\mu_0 I q}{2 \pi m} \int_{x_1}^x r \, dr.$$ Solving, $$\mu_0 I q \frac{x_1}{2 \pi m} = -\int_{v_0}^0 dz = -v_0.$$ Thus, $$x = x_1 e^{\frac{\mu_0 I q}{2 \pi m}}.$$ From equation 1 and 2, $$X = ae^{\frac{4 \pi m v_0}{\mu_0 I q}}.$$
Physics · Thermal Properties of Matter · Single correct
A Carnot engine (E) is working between two temperatures 473 K and 273 K. In a new system two engines - engine $E_1$ works between 473 K to 373 K and engine $E_2$ works between 373 K to 273 K. If $\eta_{12}$, $\eta_1$ and $\eta_2$ are the efficiencies of the engines $E$, $E_1$ and $E_2$, respectively, then
Answer: (d)
Efficiencies of a Carnot engine $\eta = 1 - \frac{T_{sink}}{T_{source}}$. $$\Rightarrow \eta_1 = 1 - \frac{373 \, \mathrm{K}}{473 \, \mathrm{K}} = \frac{100}{473}$$ $$\eta_2 = 1 - \frac{273 \, \mathrm{K}}{373 \, \mathrm{K}} = \frac{100}{373}$$ $$\eta_{12} = 1 - \frac{273 \, \mathrm{K}}{473 \, \mathrm{K}} = \frac{200}{473}$$ $$\eta_{12} - \eta_1 = \frac{200}{473} - \frac{100}{473} = \frac{100}{473} < \frac{100}{373}$$ Therefore, $\eta_{12} - \eta_1 < \eta_2$ or $\eta_{12} < \eta_1 + \eta_2$.
Physics · Waves · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: A sound wave has higher speed in solids than gases. Reason R: Gases have higher value of Bulk modulus than solids. In the light of the above statements, choose the correct answer from the options given below
Answer: (b)
Speed of sound in a medium depends on inertial and elastic properties as $v = \sqrt{\frac{B}{P}}$ for gases and $v = \sqrt{\frac{Y}{P}}$ for solids. Since the elastic property of solid happens to be many folds greater than that of gases, the speed of sound in solids is higher than in gases. Also, bulk modulus of gases varies between 0 and $\infty$ ($B = -v \frac{dP}{dv}$) hence reason is false.
Physics · Kinetic Theory · Single correct
For a particular ideal gas which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?
Answer: (b)
Given $V_{rms} = \sqrt{\frac{3RT}{M}}$. $$V_{rms}^2 = \frac{3RT}{M}$$ Hence we can conclude that $V_{rms}^2$ is directly proportional to temperature. $y = mx$ Therefore, the graph will be a straight line.
Physics · Work, Energy and Power · Single correct
A bead of mass ' m ' slides without friction on the wall of a vertical circular hoop of radius ' R ' as shown in figure. The bead moves under the combined action of gravity and a massless spring ( k ) attached to the bottom of the hoop. The equilibrium length of the spring is ' R '. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes ' R ', would be (spring constant is ' k ', g is acceleration due to gravity)
Answer: (a)
Work done by gravity is given by $mg(2R - R \, \cos 60^\circ)$. This simplifies to $3mgR$. Work done by the spring is $-\frac{1}{2}k(0^2 - R^2)$, which simplifies to $\frac{1}{2}kR^2$. The net work is equal to the change in kinetic energy. Thus, $\frac{3mgR}{2} + \frac{kR^2}{2} = \frac{1}{2}mv^2$. Solving for $v^2$, we get $v^2 = 3gR + \frac{kR^2}{m}$. Therefore, $v = \sqrt{3gR + \frac{kR^2}{m}}$.
Physics · Gravitation · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: In a central force field, the work done is independent of the path chosen. Reason R: Every force encountered in mechanics does not have an associated potential energy. In the light of the above statements, choose the most appropriate answer from the options given below
Answer: (b)
Assertion is correct as central forces are conservative in nature, i.e. work done is independent of path. Reason is true as some forces in mechanics like, friction are non-conservative because work done depends on path and only conservative forces have an associated potential energy. Also, reason does not explain assertion.
Physics · Nuclei · Single correct
Choose the correct nuclear process from the below options [p: proton, n: neutron, $e^{-}$: electron, $e^{+}$: positron, $\nu$: neutrino, $\bar{\nu}$: antineutrino]
Answer: (d)
For all nuclear processes, charge must be conserved. Also, when a release of an electron ($e^-$) is always accompanied by a release of an antineutrino ($\bar{\nu}$). Hence, $n \to p + e^- + \bar{\nu}$ is the correct answer.
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Which of the following circuits has the same output as that of the given circuit?
Answer: (b)
P = A $\cdot$ $\overline{B}$ Q = $\overline{A}$ $\cdot$ B Y = $\overline{\overline{P + Q}}$ = $\overline{A \cdot \overline{B} + \overline{A} \cdot B}$ = $\overline{A}$ $\cdot$ (B + $\overline{B}$) = $\overline{A}$ $\cdot$ I Y = $\overline{A}$
Physics · Current Electricity · Single correct
Find the equivalent resistance between two ends of the following circuit
Answer: (a)
All are in parallel. The equivalent resistance is given by $$R_{eq} = \frac{r/3}{3} = \frac{r}{9}.$$
Physics · Current Electricity · Single correct
A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is
Answer: (c)
For the wire bent into an equilateral triangle, each side has a resistance $\frac{R}{3}$. $$R_{eq} = \left(\frac{2R}{3}\right) \left(\frac{R}{3}\right) \div \left(\frac{2R}{3} + \frac{R}{3}\right) = \frac{2R}{9} = R_1 (lets say)$$ For the wire bent into a square, each side has a resistance $\frac{R}{4}$. $$R_{eq} = \left(\frac{3R}{4}\right) \left(\frac{R}{4}\right) \div \left(\frac{3R}{4} + \frac{R}{4}\right) = \frac{3R}{16} = R_3 (lets say)$$ $$\Rightarrow \frac{R_1}{R_3} = \frac{\frac{2R}{9}}{\frac{3R}{16}} = \frac{32}{27}$$
Physics · Electromagnetic Waves · Single correct
Due to presence of an em-wave whose electric component is given by $E = 100 \sin(\omega t - kx) \mathrm{NC}^{-1}$, a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as
Answer: (b)
Energy density of an $EM$ wave $= \frac{1}{2} \varepsilon E_0^2$, where $E_0$ is the amplitude of the wave. Since total energy is same for both cylinders $$\left( \frac{1}{2} \varepsilon E_1^2 \right) \pi R_1^2 L_1 = \left( \frac{1}{2} \varepsilon E_2^2 \right) \pi R_2^2 L_2$$ $$\Rightarrow E_1^2 R_1^2 L_1 = E_2^2 R_2^2 L_2$$ or $$E_2 = \frac{E_1 R_1}{R_2} \sqrt{\frac{L_1}{L_2}} = \frac{100 \ d}{(d/2)} \sqrt{\frac{L_1}{L_1}} = 200 \, \mathrm{N/C}$$ $$\left[ \therefore L_1 = L_2 = 200 \, \mathrm{cm} \right]$$ The amplitude of corresponding $EM$ wave is $200 \, \mathrm{N/C}$ or the wave is $E = 200 \sin(\omega t - kx) \mathrm{NC}^{-1}$
Physics · Electric Charges and Fields · Single correct
A particle of mass ' m ' and charge ' q ' is fastened to one end ' A ' of a massless string having equilibrium length $l$ , whose other end is fixed at point ' O '. The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x-axis is
Answer: (b)
The work done by all forces is equal to the change in kinetic energy. Therefore, $W_{all} = \Delta k$. The work done by the electric field is $W_e = k_f - k_i$. Given $qE \frac{\ell}{2} = \frac{1}{2} mv^2 - 0$, we can solve for $v$. Thus, $v = \sqrt{\frac{qE\ell}{m}}$.
Physics · Dual Nature of Radiation and Matter · Single correct
A proton of mass ' $m_p$ ' has same energy as that of a photon of wavelength ' $\lambda$ '. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.
Answer: (d)
Energy of photon $= E = \frac{hc}{\lambda}$. Therefore, wavelength of photon $= \lambda = \frac{hc}{E}$. Energy of proton $= E = \frac{1}{2} m_p v^2 = \frac{P^2}{2m_p}$. Therefore, linear momentum of proton $= P = \sqrt{2m_p E}$. Or de-Broglie wavelength of proton $= \lambda_p = \frac{h}{P} = \frac{h}{\sqrt{2m_p E}}$. Ratio $\frac{\lambda_p}{\lambda} = \frac{h}{\sqrt{2m_p E}} \times \frac{E}{hc} = \frac{1}{c} \sqrt{\frac{E}{2m_p}}$.
Physics · System of Particles and Rotational Motion · Single correct
The center of mass of a thin rectangular plate (fig - x) with sides of length $a$ and $b$, whose mass per unit area ($\sigma$) varies as $\sigma = \frac{\sigma_0 x}{ab}$ (where $\sigma_0$ is a constant), would be
Answer: (a)
Given $dm = \sigma dA$. (1) $d = \sigma (dx)(dy) = \frac{\sigma_0 x}{ab} (dx)(dy)$ $x_{com} = \frac{\int x \, dm}{\int dm} = \frac{\int x \left( \frac{\sigma_0 x}{ab} \right) (dx)(dy)}{\int_0^{\sigma_0} \frac{ab}{\sigma_0} (dx)(dy)}$ $$= \frac{\int_0^a x^2 dx \int_0^b dy}{\int_0^b x dx \int_0^b dy} = \frac{2a}{3}$$ $y_{com} = \frac{\int y \, dm}{\int dm} = \frac{\int y \left( \frac{\sigma_0 x}{ab} \right) (dx)(dy)}{\int_0^{\sigma_0} \frac{ab}{\sigma_0} (dx)(dy)}$ $$= \frac{\int_0^a x dx \int_0^b y dy}{\int_0^a x dx \int_0^b dy} = \frac{b}{2}$$ i.e., $\vec{r}_{com} \equiv \left( \frac{2a}{3}, \frac{b}{2} \right)$
Physics · Ray Optics and Optical Instruments · Single correct
A thin prism $P_1$ with angle $4^\circ$ made of glass having refractive index $1.54$, is combined with another thin prism $P_2$ made of glass having refractive index $1.72$ to get dispersion without deviation. The angle of the prism $P_2$ in degrees is
Answer: (a)
For dispersion without deviation, $$(\mu_1 - 1) A_1 = (\mu_2 - 1) A_2$$ which implies $$(1.54 - 1) 4^\circ = (1.72 - 1) A_2$$ Or $$A_2 = \frac{0.54}{0.72} \times 4 = 3^\circ$$
Physics · Experimental Physics · Numerical
A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm, respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be $\frac{x}{100}$ where $x$ is ________.
Answer: 3
Since least count of the instrument can be calculated as $$Least count = \frac{\text{pitch length}}{\text {No. of division on circular scale}}$$ $$= \frac{0.75}{15} = 0.05 \, mm.$$ Here we are provided $L = 5 \, mm$ and $W = 2.5 \, mm$. Therefore, we know that $$A = L \cdot W$$ For calculating fractional error, we can write $$\frac{dA}{A} = \frac{dL}{L} + \frac{dW}{W}$$ Here $dL = dW = 0.05 \, mm$ $$\frac{dA}{A} = \frac{0.05}{5} + \frac{0.05}{2.5}$$ $$\Rightarrow \frac{dA}{A} = \frac{1}{100} + \frac{2}{100} = \frac{3}{100}$$ So, $x = 3$
Physics · System of Particles and Rotational Motion · Numerical
The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is $n$ times higher than the moment of inertia of the given ring. Here, $n = \underline{\hspace{1cm}}$ (Consider all the bodies have equal masses)
Answer: 4
Given $I_1 = \frac{MR_1^2}{4}$, $I_2 = \frac{MR_2^2}{2}$, $I_3 = \frac{2MR_1^2}{5}$. According to the problem, $$\frac{I_1}{I_2} = 2.5 \implies \frac{\frac{MR_1^2}{4}}{\frac{MR_2^2}{2}} = \frac{5}{2} \implies \frac{R_1^2}{R_2^2} = 5.$$ Now we are provided with the information that $\frac{I_3}{I_2} = n$. $$\frac{\frac{2MR_1^2}{5}}{\frac{MR_2^2}{2}} = \frac{4R_1^2}{5R_2^2} = n.$$ From equations (1) and (2), $$\Rightarrow n = 4.$$
Physics · Physical World, Units and Measurements · Fill in the blank
In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of $[M^a L^b T^c]$. If $b = 3$, the value of $c$ is ________
Answer: 0
The modulus of elasticity is given by $[Modulus of elasticity] = M L^{-1} T^{-2}$. The torque is given by $[Torque] = M L^2 T^{-2}$. The modulus of elasticity per unit torque is calculated as follows: $$\frac{M L^{-1} T^{-2}}{M L^2 T^{-2}} = L^{-3}.$$
Physics · System of Particles and Rotational Motion · Fill in the blank
Two iron solid discs of negligible thickness have radii $R_1$ and $R_2$ and moment of inertia $I_1$ and $I_2$, respectively. For $R_2 = 2R_1$, the ratio of $I_1$ and $I_2$ would be $1/x$, where x = .
Answer: 16
Given $R_2 = 2R_1$ $$M_1 = \sigma \times \pi R_1^2 = M_o$$ $$M_2 = \sigma \times \pi R_2^2 = M_o$$ $$M_2 = \sigma \times \pi R_2^2 = \sigma \times \pi [2R_1]^2 = \sigma \times 4\pi R_1^2 = 4M_o$$ $$\frac{I_1}{I_2} = \frac{\frac{M_1 R_1^2}{2}}{\frac{M_2 R_2^2}{2}} = \frac{M_1 R_1^2}{M_2 R_2^2} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$$
Physics · Wave Optics · Numerical
A double slit interference experiment performed with a light of wavelength $600 \, \mathrm{nm}$ forms an interference fringe pattern on a screen with 10th bright fringe having its centre at a distance of $10 \, \mathrm{mm}$ from the central maximum. Distance of the centre of the same 10th bright fringe from the central maximum when the source of light is replaced by another source of wavelength $660 \, \mathrm{nm}$ would be _______ mm.
Answer: 11
Position of the $n^{th}$ bright fringe with respect to central maxima in a YDSE is $y_n = n \frac{\lambda D}{d}$. Therefore, $$\frac{y'_{10}}{y_{10}} = \frac{\lambda'}{\lambda}$$ or $$y'_{10} = \frac{\lambda'}{\lambda} y_{10}$$ $$= \left( \frac{660 \, \mathrm{nm}}{600 \, \mathrm{nm}} \right) 10 \, \mathrm{mm}$$ $$= 11 \, \mathrm{mm}$$
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The incorrect decreasing order of atomic radii is
Answer: (b)
Correct order of atomic radii: $\mathrm{Be} \mathrm{Al} > \mathrm{Si}$
Chemistry · Redox Reactions · Single correct
Given below are two statements: Statement I: In the oxalic acid vs $\mathrm{KMnO_4}$(in the presence of dil $\mathrm{H_2SO_4}$) titration the solution needs to be heated initially to $60^\circ\mathrm{C}$, but no heating is required in Ferrous ammonium sulphate (FAS) vs $\mathrm{KMnO_4}$ titration (in the presence of dil $\mathrm{H_2SO_4}$) Statement II: In oxalic acid vs $\mathrm{KMnO_4}$ titration, the initial formation of $\mathrm{MnSO_4}$ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs $\mathrm{KMnO_4}$, heating oxidizes $\mathrm{Fe^{2+}}$ into $\mathrm{Fe^{3+}}$ by oxygen of air and error may be introduced in the experiment. In the light of the above statements, choose the correct answer from the options given below
Heating is required in oxalic acid filtration due to high activation energy. Heating is not required in FAS vs KMnO_4 titration because $\mathrm{Fe^{2+}}$ will get converted into $\mathrm{Fe^{3+}}$ by oxygen of air and error may be introduced in the experiment. Both Statement-I and Statement-II are correct.
Chemistry · Co-ordination Compounds · Single correct
Match the LIST-I with LIST-II
Answer: (d)
(A) $\mathrm{CH_4(g) + 2O_2(g) \xrightarrow{\Delta} 2CO_2(g) + H_2O(l)}$ combination reaction (B) $\mathrm{2NaH(s) \xrightarrow{\Delta} 2Na(s) + H_2(g)}$ Decomposition reaction (C) $\mathrm{V_2O_5(s) + 5Ca(s) \xrightarrow{\Delta} 2V(s) + 5CaO(s)}$ Displacement reaction (D) $\mathrm{2H_2O_2(aq.) \xrightarrow{\Delta} 2H_2O(l) + O_2(g)}$ Disproportionation reaction
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Given below are two statements: In the light of the above statements, choose the most appropriate answer from the options given below
Answer: (c)
Rate of (a) is faster than rate of (b) because it is an intramolecular substitution.
Chemistry · Equilibrium · Single correct
A weak acid HA has degree of dissociation x. Which option gives the correct expression of (pH-$pK_a$)?
Answer: (b)
The equilibrium reaction is given as: $$\mathrm{H} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}$$ The concentrations are: $$\mathrm{C(1-x)} \mathrm{Cx} \mathrm{Cx}$$ The expression for $K_a$ is: $$K_a = \frac{[\mathrm{H^+}](x)}{C(1-x)}$$ Rearranging gives: $$[\mathrm{H^+}] = K_a \frac{(1-x)}{x}$$ Taking the logarithm: $$\log [\mathrm{H^+}] = \log K_a + \log \left( \frac{1-x}{x} \right)$$ This simplifies to: $$-\log [\mathrm{H^+}] = -\log K_a - \log \frac{1-x}{x}$$ Thus, the pH is: $$\mathrm{pH} = \mathrm{pK_a} - \log \frac{1-x}{x}$$ Finally: $$\mathrm{pH} - \mathrm{pK_a} = \log \frac{1-x}{x}$$
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Consider ' n ' is the number of lone pair of electrons present in the equatorial position of the most stable structure of $\mathrm{ClF_3}$. The ions from the following with ' n ' number of unpaired electrons are A. $\mathrm{V}^{3+}$ B. $\mathrm{Ti}^{3+}$ C. $\mathrm{Cu}^{2+}$ D. $\mathrm{Ni}^{2+}$ E. $\mathrm{Ti}^{2+}$ Choose the correct answer from the options given below:
Answer: (b)
The molecule $\mathrm{ClF_3}$ has a T-shaped structure with two lone pairs of electrons in the equatorial plane. This corresponds to $n = 2$ lone pairs. The options are given as follows: (A) $\mathrm{V^{+3}}$: $[\mathrm{Ar}]3d^2$ with 2 unpaired electrons. (B) $\mathrm{Ti^{3+}}$: $[\mathrm{Ar}]3d^1$ with 1 unpaired electron. ($C$) $\mathrm{Cu^{+2}}$: $[\mathrm{Ar}]3d^9$ with 1 unpaired electron. (D) $\mathrm{Ni^{+2}}$: $[\mathrm{Ar}]3d^8$ with 2 unpaired electrons. (E) $\mathrm{Ti^{+2}}$: $[\mathrm{Ar}]3d^2$ with 2 unpaired electrons.
Chemistry · Chemical Kinetics and Nuclear Chemistry · Multiple correct
For a given reaction R $\rightarrow$ P, $t_{1/2}$ is related to $[A]_0$ as given in table. Given: $\log 2 = 0.30$ Which of the following is true? A. The order of the reaction is 1/2. B. If $[A]_0$ is 1 M , then $t_{1/2}$ is $200\sqrt{10}$ min C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M. D. $t_{1/2}$ is 800 min for $[A]_0 = 1.6$M Choose the correct answer from the options given below: Options
Answer: (b)
Given, $t_{1/2} \propto (C_0)^{1-n}$ $\frac{t_1}{t_2}=\left(\frac{C_1}{C_2}\right)^{1-n}$ $\frac{200}{100}=\left(\frac{0.100}{0.025}\right)^{1-n}$ $2=(4)^{1-n}$ $(1-n)=\frac{1}{2}$ $n=\frac{1}{2}$ For $n=\frac{1}{2}$, $-\frac{dA}{dt}=k(A)^{1/2}$ $\int_{C_0}^{C}\frac{dA}{(A)^{1/2}}=-\int_{0}^{t}k\,dt$ $2(\sqrt{C}-\sqrt{C_0})=-kt$ $\sqrt{C}-\sqrt{C_0}=-\frac{kt}{2}$ For $C_0=0.1$ and $t_{1/2}=200$ min, $\frac{\sqrt{C_0}}{\sqrt{2}}-\sqrt{C_0}=-\frac{kt}{2}$ $kt=2\sqrt{C_0}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$ $t_{1/2}=\frac{2\sqrt{C_0}}{k}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$ $200=\frac{2\sqrt{0.1}}{k}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$ $k=\frac{\sqrt{0.1}}{100}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$ For $C_0=1\,\mathrm{M}$, $t_{1/2}=\frac{2\times100\sqrt{2}}{\sqrt{0.1}(\sqrt{2}-1)}\times\frac{\sqrt{2}-1}{\sqrt{2}}$ $t_{1/2}=200\sqrt{10}\,\text{min}$ Hence, $B$ is correct and $C$ is incorrect. For $C_0=1.6\,\mathrm{M}$, $t_{1/2}=\frac{2\sqrt{1.6}(\sqrt{2})}{\sqrt{0.1}(\sqrt{2}-1)}\times\frac{(\sqrt{2}-1)\times100}{\sqrt{2}}$ $t_{1/2}=\frac{400\times2}{2}\,\text{min}$ $t_{1/2}=800\,\text{min}$
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
A molecule ("P") on treatment with acid undergoes rearrangement and gives ("Q"). ("Q") on ozonolysis followed by reflux under alkaline condition gives ("R"). The structure of ("R") is given below. The structure of ("P") is
Answer: (d)
Chemistry · States of Matter · Single correct
Ice and water are placed in a closed container at a pressure of $\mathrm{1\,atm}$ and temperature $\mathrm{273.15\,K}$. If pressure of the system is increased $2$ times, keeping temperature constant, then identify correct observation from following
Answer: (d)
If pressure is made two times, then the mixture of ice and water will completely convert into water (liquid) form.
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The molecules having square pyramidal geometry are
Answer: (d)
$\mathrm{BrF_5}$: Square pyramidal $\mathrm{XeOF_4}$: Square pyramidal $\mathrm{SbF_5}$: Trigonal bipyramidal $\mathrm{PCl_5}$: Trigonal bipyramidal
Chemistry · Analytical Chemistry · Single correct
The metal ion whose electronic configuration is not affected by the nature of the ligand and which gives a violet colour in non-luminous flame under hot condition in borax bead test is
Answer: (c)
$\mathrm{Ni^{2+}}$ gives violet colour with borax bead test in non-luminous flame under hot conditions. $\mathrm{Ni^{2+}}$ has $d^8$ configuration, which does not depend on the nature of ligand present in octahedral field. $Ni^{2+}$ $\Rightarrow$ $t_{2g}^{6}e_g^{2}$
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Both acetaldehyde and acetone (individually) undergo which of the following reactions? A. Iodoform Reaction B. Cannizaro Reaction C. Aldol Condensation D. Tollen's Test E. Clemmensen Reduction Choose the correct answer from the options given below:
Answer: (c)
The table shows the reactions of acetaldehyde and acetone with different reagents. A. Iodoform reaction: Both acetaldehyde and acetone give a positive result (✓). B. Cannizaro reaction: Acetaldehyde gives a negative result (✗), while acetone is not applicable. C. Aldol reaction: Both acetaldehyde and acetone give a positive result (✓). D. Tollen's test: Acetaldehyde gives a positive result (✓), while acetone gives a negative result (✗). E. Clemmensen reduction: Acetaldehyde gives a negative result (✗), while acetone gives a positive result (✓).
Chemistry · Structure of Atom · Single correct
In a multielectron atom, which of the following orbitals described by three quantum numbers will have same energy in absence of electric and magnetic fields? A. n = 1, l = 0, $m_1$ = 0 B. n = 2, l = 0, $m_1$ = 0 C. n = 2, l = 1, $m_1$ = 1 D. n = 3, l = 2, $m_1$ = 1 E. n = 3, l = 2, $m_1$ = 0 Choose the correct answer from the options given below:
Answer: (d)
A: n = 1, $\ell$ = 0, m_$\ell$ = 0 orbital B: n = 2, $\ell$ = 0, m_$\ell$ = 0 1$\,$ s C: n = 3, $\ell$ = 1, m_$\ell$ = 1 2$\,$ s D: n = 3, $\ell$ = 2, m_$\ell$ = 1 3p E: n = 3, $\ell$ = 2, m_$\ell$ = 0 3$\,$ d 3$\,$ d In absence of electric and magnetic fields, all orbitals of 3d are degenerate
Chemistry · Haloalkanes and Haloarenes · Single correct
The products $A$ and $B$ in the following reactions, respectively are $$A \overset{Ag - NO_2}{\rightleftharpoons} CH_3 - CH_2 - CH_2 - Br \overset{AgCN}{\longrightarrow} B$$
Answer: (d)
Chemistry · Solutions · Single correct
What is the freezing point depression constant of a solvent, if $50\,\mathrm{g}$ of the solvent contains $1\,\mathrm{g}$ of a non-volatile solute (molar mass $256\,\mathrm{g\,mol^{-1}}$) and the decrease in freezing point is $0.40\,\mathrm{K}$?
Answer: (d)
Given $\Delta T_f = K_b \cdot m$. $$0.4 = K_b \frac{1}{256} \frac{50 \times 10^{-3}}{1}$$ $$K_b = 5.12 \, \mathrm{K \, kg/mol}$$
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Consider the following elements In, Tl, Al, Pb, Sn and Ge. The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are
Answer: (c)
Among Al, In, Tl, Ge, Sn, Pb, the metal having highest $\mathrm{IE}_1$ is Ge and lowest $\mathrm{IE}_1$ is In. Most stable oxidation state of Ge is $+4$ and In is $+3$.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The correct order of stability of following carbocations is :
Answer: (d)
C is aromatic due to the positive charge, hence it is most stable. A has more resonance structure. B has less resonance structure. D has only hyperconjugation. Consider first aromaticity > resonance > hyperconjugation. Ans. D < B < A < C
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The compounds that produce $\mathrm{CO}_2$ with aqueous $\mathrm{NaHCO}_3$ solution are:
Answer: (a)
A, C, D produce $\mathrm{CO_2}$ with aqueous $\mathrm{NaHCO_3}$ solution. A, C, D acids are stronger acid than $\mathrm{H_2CO_3}$ (Carbonic acid).
Chemistry · Redox Reactions · Multiple correct
Which of the following oxidation reactions are carried out by both $\mathrm{K_2Cr_2O_7}$ and $\mathrm{KMnO_4}$ in acidic medium? $\mathrm{A.\ I^- \rightarrow I_2}$ $\mathrm{B.\ S^{2-} \rightarrow S}$ $\mathrm{C.\ Fe^{2+} \rightarrow Fe^{3+}}$ $\mathrm{D.\ I^- \rightarrow IO_3^-}$ $\mathrm{E.\ S_2O_3^{2-} \rightarrow SO_4^{2-}}$ Choose the correct answer from the options given below:
Answer: (d)
The reactions given are as follows: $$\mathrm{I^- \xrightarrow{OH^-} IO_3^-}$$ $$\mathrm{S^{2-} \xrightarrow{H^+} S}$$ $$\mathrm{S_2O_3^{2-} \xrightarrow{OH^-} SO_4^{2-}}$$ $$\mathrm{Fe^{+2} \rightarrow Fe^{+3}}$$ $$\mathrm{S_2O_3^{2-} \xrightarrow{H^+} S \downarrow + SO_4^{2-}}$$
Chemistry · Biomolecules · Single correct
Given below are two statements: Statement I : D-glucose pentaacetate reacts with 2, 4-dinitrophenylhydrazine Statement II : Starch, on heating with concentrated sulfuric acid at $100^\circ\mathrm{C}$ and 2-3 atmosphere pressure produces glucose. In the light of the above statements, choose the correct answer from the options given below
Answer: (a)
D-glucose pentaacetate does not react with 2,4-DNP. Starch is converted to glucose using concentrated $\mathrm{H_2SO_4}$ under heat $\Delta$, at a pressure of $2 - 3 \, \mathrm{atm}$ and a temperature of $100^\circ \mathrm{C}$.
Chemistry · Electrochemistry · Numerical
Given below is the plot of the molar conductivity vs $\sqrt{concentration}$ for KCl in aqueous solution. If, for the higher concentration of KCl solution, the resistance of the conductivity cell is $100\, \Omega$, then the resistance of the same cell with the dilute solution is $x \, \Omega$. The value of $x$ is __________ (Nearest integer)
Answer: 150
Given $R = \rho \frac{\ell}{A}$. $$\kappa = G \cdot G^* G = \frac{1}{R} ; \kappa = \frac{1}{\rho}$$ $$G^* = \frac{\ell}{A}$$ $R$ = Resistance $\rho$ = Resistivity $\ell$ = cell constant ($G^*$) $A$ = area $$\kappa_c = \frac{R_d}{R_c} ; \lambda_m = \kappa \times 1000$$ $$\kappa_d = \frac{\lambda_m \cdot C}{R_c}$$ $c$ = concentrated solution. $d$ = diluted solution. $$100 \cdot (0.15)^2 = R_d$$ $$150 \cdot (0.1)^2 = 100$$ $$R_d = 150 \Omega$$
Chemistry · Some Basic Concepts of Chemistry · Fill in the blank
\[ \begin{aligned} &\text{Quantitative analysis of an organic compound (X) shows the following % composition.}\\ &\mathrm{C}: 14.5\%\\ &\mathrm{H}: 1.8\%\\ &\mathrm{Cl}: 64.46\%\\[4pt] &\text{(Empirical formula mass of the compound (X) is } \_\_\_\_\_\_\_\_ \times 10^{-1}\text{)}\\[4pt] &\text{(Given molar mass in g mol}^{-1}\text{ of C : 12, H : 1, O : 16, Cl : 35.5)} \end{aligned} \]
Answer: 1655
Given the percentage mass of elements: C: $14.5\%$, Cl: $64.46\%$, H: $1.8\%$, O: $19.24\%$. Calculate the molar ratio: $$\frac{14.5}{12} = 1.2, \frac{64.46}{35.5} = 1.8, \frac{1.8}{1} = 1.8, \frac{19.24}{16} = 1.2.$$ The minimum integral ratio is $2 : 3 : 3 : 2$. The empirical formula is $\mathrm{C_2H_3Cl_3O_2}$. Mass is $165.5$. Mass is $1655 \times 10^{-1}$.
Chemistry · Solutions · Numerical
The molarity of a 70$\%$ (mass / mass) aqueous solution of a monobasic acid (X) is (nearest integer) [Given: Density of aqueous solution of (X) is $1.25 \, \mathrm{g \, mL^{-1}}$ Molar mass of the acid is $70 \, \mathrm{g \, mol^{-1}}$]
Answer: 125
Moles of solute $= \frac{70}{70} = 1$ Volume of solution $= \frac{100}{1.25} = 80 \, \mathrm{mL}$ $$M = \frac{1}{80} \times 1000 = 12.5$$ $$M = 125 \times 10^{-1}$$
Chemistry · Haloalkanes and Haloarenes · Numerical
Consider the following sequence of reactions: 11.25 mg of chlorobenzene will produce __________ $\times 10^{-1}$ mg of product B. (Consider the reactions result in complete conversion.) [Given molar mass of C, H, O, N and Cl as 12, 1, 16, 14 and 35.5 g $mol^{-1}$ respectively]
Answer: 93
$$\frac{11.25 \times 10^{-3}}{112.5} = \frac{x \times 10^{-1} \times 10^{-3}}{93}$$ Solving for $x$: $$x \times 10^{-1} = 93 \times 0.1$$ $$x = 93 \, \mathrm{mg}$$
Chemistry · Thermodynamics · Numerical
The formation enthalpies, $\Delta H_f^\Theta$ for $\mathrm{H_{(g)}}$ and $\mathrm{O_{(g)}}$ are $220.0$ and $250.0 \, \mathrm{kJ \, mol^{-1}}$, respectively, at $298.15 \, \mathrm{K}$, and $\Delta H_f^\Theta$ for $\mathrm{H_2O_{(g)}}$ is $-242.0 \, \mathrm{kJ \, mol^{-1}}$ at the same temperature. The average bond enthalpy of the $\mathrm{O-H}$ bond in water at $298.15 \, \mathrm{K}$ is _________ $\mathrm{kJ \, mol^{-1}}$ (nearest integer).
Answer: 463
Given $\($ $\frac{1}{2}$ $\mathrm{H_2(g)}$ $\rightarrow$ $\mathrm{H(g)}$ $\)$; $\($ $\Delta$_f $\mathrm{H(H(g))}$ = 220 $\mathrm{KJ/mol}$ $\)$ and $\($ $\frac{1}{2}$ $\mathrm{O_2(g)}$ $\rightarrow$ $\mathrm{O(g)}$ $\)$; $\($ $\Delta$_f $\mathrm{H(O(g))}$ = 250 $\mathrm{KJ/mol}$ $\)$. The reaction is $\($ $\mathrm{H_2(g)}$ + $\frac{1}{2}$ $\mathrm{O_2(g)}$ $\rightarrow$ $\mathrm{H_2O(g)}$ $\)$ with $\($ $\Delta$_f $\mathrm{H(H_2O(g))}$ = -242 $\mathrm{KJ/mol}$ $\)$. Calculating the bond energy: $\[$ $\Delta$ $\mathrm{H_f(H_2O(l))}$ = -242 = 440 + 250 - 2($\mathrm{B.E.(O-H)}$) $\]$ Solving for bond energy: $\[$ $\mathrm{BE(O-H)}$ = 466 $\mathrm{KJ/mol}$ $\]$