JEE Main 31 January 2024 Shift 2: all 90 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 number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is
Answer: (d)
After giving 2 apples to each child 15 apples left now 15 apples can be distributed in $^{15+3-1}C_2 = ^{17}C_2$ ways $$= \frac{17 \times 16}{2} = 136$$
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let $A(a, b)$, $B(3, 4)$ and $(-6, -8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $P(2a + 3, 7b + 5)$ from the line $2x + 3y - 4 = 0$ measured parallel to the line $x - 2y - 1 = 0$ is
Answer: (c)
A(a, b), $\;$ B(3, 4), $\;$ C(-6, -8) $\frac{2:1}{C \; A}$ (-6, -8) (a, b) (3, 4) $\Rightarrow$ a = 0, $\;$ b = 0 $\Rightarrow$ P(3, 5) Distance from $\;$ P $\;$ measured $\;$ along $\;$ x - 2y - 1 = 0 $\Rightarrow$ x = 3 + r $\cos$ $\theta$, y = 5 + r $\sin$ $\theta$ Where $\;$ $\tan$ $\theta$ = $\frac{1}{2}$ r(2 $\cos$ $\theta$ + 3 $\sin$ $\theta$) = -17 $\Rightarrow$ r = $\left$| $\frac{-17 \sqrt{5}}{7}$ $\right$| = $\frac{17 \sqrt{5}}{7}$
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $z_1$ and $z_2$ be two complex number such that $z_1 + z_2 = 5$ and $z_1^3 + z_2^3 = 20 + 15i$. Then $|z_1^4 + z_2^4|$ equals-
Answer: (b)
Given $z_1 + z_2 = 5$. $z_1^3 + z_2^3 = 20 + 15i$. Using the identity $z_1^3 + z_2^3 = (z_1 + z_2)^3 - 3z_1z_2(z_1 + z_2)$, we have: $$z_1^3 + z_2^3 = 125 - 3z_1 \cdot z_2(5)$$ This implies $20 + 15i = 125 - 15z_1z_2$. Therefore, $3z_1z_2 = 25 - 4 - 3i$. Simplifying gives $3z_1z_2 = 21 - 3i$. Thus, $z_1 \cdot z_2 = 7 - i$. Now, $(z_1 + z_2)^2 = 25$. So, $z_1^2 + z_2^2 = 25 - 2(7 - i)$. This simplifies to $11 + 2i$. Then, $(z_1^2 + z_2^2)^2 = 121 - 4 + 44i$. Therefore, $z_1^4 + z_2^4 + 2(7 - i)^2 = 117 + 44i$. Simplifying further, $z_1^4 + z_2^4 = 117 + 44i - 2(49 - 1 - 14i)$. Finally, $|z_1^4 + z_2^4| = 75$.
Maths · Conic Sections · Single correct
Let a variable line passing through the centre of the circle $x^2 + y^2 - 16x - 4y = 0$, meet the positive co-ordinate axes at the point A and B. Then the minimum value of OA + OB, where O is the origin, is equal to
Answer: (b)
Given the equation of the line $(y - 2) = m(x - 8)$. To find the $x$-intercept, we solve: $$\left(\frac{-2}{m} + 8\right)$$ For the $y$-intercept, we have: $$(-8m + 2)$$ Therefore, $$\mathrm{OA} + \mathrm{OB} = \frac{-2}{m} + 8 - 8m + 2$$ Differentiating with respect to $m$, we get: $$f'(m) = \frac{2}{m^2} - 8 = 0$$ Solving for $m$, we find: $$m^2 = \frac{1}{4}$$ $$m = \frac{-1}{2}$$ Evaluating $f$ at $m = \frac{-1}{2}$ gives: $$f\left(\frac{-1}{2}\right) = 18$$ Therefore, the minimum value is 18.
Maths · Integrals · Single correct
Let $f, g : (0, \infty) \to \mathbb{R}$ be two functions defined by $f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} \, dt$ and $g(x) = \int_{0}^{x^2} t^{1/2} e^{-t} \, dt$. Then the value of $\left( f \left( \sqrt{\log_e 9} \right) + g \left( \sqrt{\log_e 9} \right) \right)$ is equal to
Answer: (c)
Given $$f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} \, dt$$ This implies $$f'(x) = 2 \cdot (|x| - x^2) e^{-x^2} \ldots (1)$$ Let $$g(x) = \int_{0}^{x^2} t^{\frac{1}{2}} e^{-t} \, dt$$ Then $$g'(x) = xe^{-x^2} (2x) - 0$$ Thus, $$f'(x) + g'(x) = 2xe^{-x^2} - 2x^2 e^{-x^2} + 2x^2 e^{-x^2}$$ Integrating both sides with respect to $x$, we have $$f(x) + g(x) = \int_{0}^{\alpha} 2xe^{-x^2} \, dx$$ Let $x^2 = t$, then $$\int_{0}^{\sqrt{\alpha}} e^{-t} \, dt = \left[ -e^{-t} \right]_{0}^{\sqrt{\alpha}}$$ This simplifies to $$-e^{(\log_{0}(9)^{-1})+1}$$ Therefore, $$9(f(x) + g(x)) = \left(1 - \frac{1}{9}\right) 9 = 8$$
Maths · Three Dimensional Geometry · Single correct
Let $(\alpha, \beta, \gamma)$ be mirror image of the point $(2, 3, 5)$ in the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$. Then $2\alpha + 3\beta + 4\gamma$ is equal to
Answer: (b)
Since $\overrightarrow{\mathrm{PR}} \perp (2, 3, 4)$, we have $\overrightarrow{\mathrm{PR}} \cdot (2, 3, 4) = 0$. Therefore, $(\alpha - 2, \beta - 3, \gamma - 5) \cdot (2, 3, 4) = 0$. This implies $2\alpha + 3\beta + 4\gamma = 4 + 9 + 20 = 33$.
Maths · Conic Sections · Single correct
Let $P$ be a parabola with vertex $(2, 3)$ and directrix $2x + y = 6$. Let an ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ of eccentricity $\frac{1}{\sqrt{2}}$ pass through the focus of the parabola $P$. Then the square of the length of the latus rectum of $E$, is
Answer: (d)
Slope of axis is $\frac{1}{2}$. $$y - 3 = \frac{1}{2}(x - 2)$$ $$\Rightarrow 2y - 6 = x - 2$$ $$\Rightarrow 2y - x - 4 = 0$$ $$2x + y - 6 = 0$$ $$4x + 2y - 12 = 0$$ $$\alpha + 1.6 = 4 \Rightarrow \alpha = 2.4$$ $$\beta + 2.8 = 6 \Rightarrow \beta = 3.2$$ Ellipse passes through $(2.4, 3.2)$. $$\Rightarrow \left( \frac{24}{10} \right)^2 \frac{1}{a^2} + \left( \frac{32}{10} \right)^2 \frac{1}{b^2} = 1$$ Also $1 - \frac{b^2}{a^2} = \frac{1}{2} = \frac{b^2}{a^2} = \frac{1}{2}$. $$\Rightarrow a^2 = 2b^2$$ Put in (1) $\Rightarrow b^2 = \frac{328}{25}$. $$\Rightarrow \left( \frac{2b^2}{a} \right)^2 = \frac{4b^2}{a^2} \times b^2 = 4 \times \frac{1}{2} \times \frac{328}{25} = \frac{656}{25}$$
Maths · Differential Equations · Single correct
The temperature $T(t)$ of a body at time $t = 0$ is $160^\circ \, F$ and it decreases continuously as per the differential equation $\frac{dT}{dt} = -K(T - 80)$, where $K$ is positive constant. If $T(15) = 120^\circ \, F$, then $T(45)$ is equal to
Answer: (c)
Given $\($ $\frac{dT}{dt}$ = -k(T - 80) $\)$. Integrating from 160 to $\($ T $\)$: $$ \int_{160}^{T} \frac{dT}{(T - 80)} = \int_{0}^{t} -k \, dt $$ $\($ [$\ln$ |T - 80|]_{160}^{T} = -kt $\)$. $\($ $\ln$ |T - 80| - $\ln$ 80 = -kt $\)$. $\($ $\ln$ $\left$| $\frac{T - 80}{80}$ $\right$| = -kt $\)$. $\($ T = 80 + 80e^{-kt} $\)$. Substituting $\($ T = 120 $\)$: $\($ 120 = 80 + 80e^{-k $\cdot$ 15} $\)$. $\($ $\frac{40}{80}$ = e^{-k $\cdot$ 15} = $\frac{1}{2}$ $\)$. Therefore, $\($ T(45) = 80 + 80e^{-k $\cdot$ 45} $\)$. $\($ = 80 + 80 $\left$( e^{-k $\cdot$ 15} $\right$)^3 $\)$. $\($ = 80 + 80 $\times$ $\frac{1}{8}$ $\)$. $\($ = 90 $\)$.
Maths · Sequences and Series · Single correct
Let $2^{nd}$, $8^{th}$ and $44^{th}$, terms of a non-constant A.P. be respectively the $1^{st}$, $2^{nd}$ and $3^{rd}$ terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to-
Answer: (c)
Given that $1 + d$, $1 + 7d$, $1 + 43d$ are in GP. $$(1 + 7d)^2 = (1 + d)(1 + 43d)$$ Expanding both sides, we have: $$1 + 49d^2 + 14d = 1 + 44d + 43d^2$$ Simplifying gives: $$6d^2 - 30d = 0$$ Solving for $d$, we get: $$d = 5$$ Now, calculate $S_{20}$: $$S_{20} = \frac{20}{2} [2 \times 1 + (20 - 1) \times 5]$$ Simplifying further: $$= 10[2 + 95]$$ $$= 970$$
Maths · Limits and Derivatives · Single correct
Let $f:\mathbb{R}\rightarrow(0,\infty)$ be a strictly increasing function such that \[ \lim_{x\to\infty}\frac{f(7x)}{f(x)}=1. \] Then, the value of \[ \lim_{x\to\infty}\left[\frac{f(5x)}{f(x)}-1\right] \] is equal to
Answer: (b)
Given $f : \mathbb{R} \to (0, \infty)$ and $\($ $\lim$_{x $\to$ $\infty$} $\frac{f(7x)}{f(x)}$ = 1 $\)$. Therefore, $f$ is increasing. Thus, $f(x) < f(5x) < f(7x)$. Therefore, $\($ $\frac{f(x)}{f(x)}$ < $\frac{f(5x)}{f(x)}$ < $\frac{f(7x)}{f(x)}$ $\)$. So, $\($ 1 < $\lim$_{x $\to$ $\infty$} $\frac{f(5x)}{f(x)}$ < 1 $\)$. Therefore, $\($ $\left$[ $\frac{f(5x)}{f(x)}$ - 1 $\right$] $\Rightarrow$ 1 - 1 = 0 $\)$.
Maths · Applications of Integrals · Single correct
The area of the region enclosed by the parabola $y = 4x - x^2$ and $3y = (x - 4)^2$ is equal to
Answer: (d)
$$\text{Area} = \int_1^4 \left[(4x - x^2) - \frac{(x-4)^2}{3}\right]dx$$ $$\text{Area} = \left.\frac{4x^2}{2} - \frac{x^3}{3} - \frac{(x-4)^3}{9}\right|_1^4$$ $$= \left(\frac{64}{2} - \frac{64}{3} - \frac{4}{2} + \frac{1}{3} - \frac{27}{9}\right)$$ $$\Rightarrow (27 - 21) = 6$$
Maths · Statistics · Single correct
Let the mean and the variance of 6 observation $a$, $b$, $68$, $44$, $48$, $60$ be $55$ and $194$, respectively if $a > b$, then $a + 3b$ is
Answer: (c)
Given the numbers $a, b, 68, 44, 48, 60$ with a mean of $55$ and variance $194$, where $a > b$. The mean is given by: $$\frac{a + b + 68 + 44 + 48 + 60}{6} = 55$$ This implies: $$220 + a + b = 330$$ Therefore, $a + b = 110 \ldots (1)$ Also, the variance is: $$\frac{\sum (x_i - \bar{x})^2}{n} = 194$$ This gives: $$(a - 55)^2 + (b - 55)^2 + (68 - 55)^2 + (44 - 55)^2 + (48 - 55)^2 + (60 - 55)^2 = 194 \times 6$$ Simplifying: $$(a - 55)^2 + (b - 55)^2 + 169 + 121 + 49 + 25 = 1164$$ Thus: $$(a - 55)^2 + (b - 55)^2 = 1164 - 364 = 800$$ Expanding: $$a^2 + 3025 - 110a + b^2 + 3025 - 110b = 800$$ Simplifying further: $$a^2 + b^2 = 800 - 6050 + 12100$$ Therefore: $$a^2 + b^2 = 6850 \ldots (2)$$ Solving equations (1) and (2): $a = 75$, $b = 35$. Thus, $a + 3b = 75 + (3 \times 35) = 75 + 105 = 180$
Maths · Relations and Functions · Single correct
If the function $f : (-\infty, -1] \to (a, b]$ defined by $f(x) = e^{x^3 - 3x + 1}$ is one-one and onto, then the distance of the point $P(2b + 4, a + 2)$ from the line $x + e^{-3}y = 4$ is :
Answer: (a)
Given $f(x) = e^{x^3 - 3x + 1}$. Differentiating, we have $$f'(x) = e^{x^3 - 3x + 1} \cdot (3x^2 - 3)$$ $$= e^{x^3 - 3x + 1} \cdot 3(x - 1)(x + 1)$$ For $f'(x) \geq 0$. Therefore, $f(x)$ is an increasing function. Thus, $a = e^{-\infty} = 0 = f(-\infty)$ $b = e^{-1 + 3 + 1} = e^3 = f(-1)$ $P(2b + 4, a + 2)$ Therefore, $P(2e^3 + 4, 2)$ For the diagram, $x + e^{-3} = 4$ and $y = 4$. The distance $d$ is given by $$d = \frac{(2e^3 + 4) + 2e^{-3} - 4}{\sqrt{1 + e^6}} = 2\sqrt{1 + e^6}$$
Maths · Continuity and Differentiability · Single correct
Consider the function $f : (0, \infty) \to \mathbb{R}$ defined by $f(x) = e^{-|\log_6 x|}$. If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m + n$ is
Answer: (c)
Given $f : (0, \infty) \to \mathbb{R}$, $f(x) = e^{-|\log_0 x|}$. $$f(x) = \frac{1}{e^{|\ln x|}} = \begin{cases} \frac{1}{e^{-\ln x}}; & 0 < x < 1 \\ \frac{1}{e^{\ln x}}; & x \geq 1 \end{cases}$$ $$\begin{cases} \frac{1}{\frac{1}{x}} = x; & 0 < x < 1 \\ \frac{1}{x}, & x \geq 1 \end{cases}$$ $m = 0$ (No point at which function is not continuous) $n = 1$ (Not differentiable) Therefore, $m + n = 1$
Maths · Trigonometric Functions · Single correct
The number of solutions, of the equation $e^{\sin x} - 2e^{-\sin x} = 2$ is
Answer: (d)
Take $e^{\sin x} = t (t > 0)$. $$\Rightarrow t - \frac{2}{t} = 2$$ $$\Rightarrow \frac{t^2 - 2}{t} = 2$$ $$\Rightarrow t^2 - 2t - 2 = 0$$ $$\Rightarrow t^2 - 2t + 1 = 3$$ $$\Rightarrow (t - 1)^2 = 3$$ $$\Rightarrow t = 1 \pm \sqrt{3}$$ $$\Rightarrow t = 1 \pm 1.73$$ $$\Rightarrow t = 2.73 or -0.73 (rejected as t > 0)$$ $$\Rightarrow e^{\sin x} = 2.73$$ $$\Rightarrow \log_e e^{\sin x} = \log_e 2.73$$ $$\Rightarrow \sin x = \log_e 2.73 > 1$$ So no solution.
Maths · Inverse Trigonometric Functions · Single correct
If $a = \sin^{-1}(\sin(5))$ and $b = \cos^{-1}(\cos(5))$, then $a^2 + b^2$ is equal to
Answer: (b)
Given $a = \sin^{-1}(\sin 5) = 5 - 2\pi$ and $b = \cos^{-1}(\cos 5) = 2\pi - 5$. Therefore, $$a^2 + b^2 = (5 - 2\pi)^2 + (2\pi - 5)^2$$ $$= 8\pi^2 - 40\pi + 50$$
Maths · Binomial Theorem · Single correct
If for some m, n; $^6C_m + 2 \left(^6C_{m+1}\right) + ^6C_{m+2} > ^8C_3$ and $n^{-1}P_3 : nP_4 = 1$, then $nP_{m+1} + n+1C_m$ is equal to
Answer: (d)
Given $^6C_m + 2 \left(^6C_{m+1}\right) + ^6C_{m+2} > ^8C_3$. $^7C_{m+1} + ^7C_{m+2} > ^8C_3$. $^8C_{m+2} > ^8C_3$. Therefore, $m = 2$. And $^{n-1}P_3 : ^nP_4 = 1 : 8$. $$\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)} = \frac{1}{8}$$ Therefore, $n = 8$. Thus, $^nP_{m+1} + ^{n+1}C_m = ^8P_3 + ^9C_2$. $$= 8 \times 7 \times 6 + \frac{9 \times 8}{2}$$ $$= 372$$
Maths · Probability · Single correct
A coin is based so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-
Answer: (a)
Let probability of tail is $\frac{1}{3}$. Therefore, probability of getting head $= \frac{2}{3}$. Probability of getting 2 tails and 1 head is $$\left( \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} \right) \times 3$$ $$= \frac{2}{27} \times 3$$ $$= \frac{2}{9}$$
Maths · Determinants · Single correct
Let A be a 3 $\times$ 3 real matrix such that $$A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, A \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = 4 \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 2 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}.$$ Then, the system $$(A - 3I) \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$ has
Answer: (a)
Let $A = \begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$. Given $A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix} \ldots (1)$, $x_1 + y_1 + z_1 = 2 \ldots (2)$, $x_2 + y_2 + z_2 = 0 \ldots (3)$, $x_3 + y_3 + z_3 = 0 \ldots (4)$. Given $A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 4 \\ 0 \end{bmatrix}$, $\Rightarrow \begin{bmatrix} -x_1 + y_1 \\ -x_2 + y_2 \\ -x_3 + y_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 4 \\ 0 \end{bmatrix}$, $\Rightarrow -x_1 + y_1 = 0 \ldots (5)$, $-x_2 + y_2 = 0 \ldots (6)$, $-x_3 + y_3 = 0 \ldots (7)$. Given $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}$, $\Rightarrow \begin{bmatrix} y_1 + z_1 \\ y_2 + z_2 \\ y_3 + z_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}$. $\therefore y_1 = 0, y_2 = 2, y_3 = 0$. From (2), (3), (4), (5), (6) and (7), $x_1 = 3x, x_2 = 0, x_3 = -1$, $y_1 = 0, y_2 = 2, y_3 = 0$, $z_1 = 1, z_2 = 0, z_3 = 3$. $\therefore A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 \end{bmatrix}$. Now $(A - 3I) \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix}$, $\Rightarrow \begin{bmatrix} -2 & 0 & 1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$, $\Rightarrow \begin{bmatrix} -z \\ -y \\ -x \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$, $\Rightarrow \begin{bmatrix} z \\ y \\ x \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \\ -3 \end{bmatrix}$. $[z = 1], [y = -2], [x = -3]$.
Maths · Three Dimensional Geometry · Single correct
The shortest distance between lines $L_1$ and $L_2$, where $L_1 : \frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+4}{2}$ and $L_2$ is the line passing through the points $A(-4, 4, 3) \cdot B(-1, 6, 3)$ and perpendicular to the line $\frac{x-3}{-2} = \frac{y}{3} = \frac{z-1}{1}$, is
Answer: (c)
Given the line equation $L_2 = \frac{x+4}{3} = \frac{y-4}{2} = \frac{z-3}{0}$. The direction ratios are given by the determinant: $$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{vmatrix}$$ Thus, the shortest distance (S.D) is: $$S.D = \frac{\begin{vmatrix} 5 & -5 & -7 \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{vmatrix}}{\left| \vec{n_1} \times \vec{n_2} \right|}$$ Calculating the determinant: $$= \frac{141}{\left| -4\hat{i} + 6\hat{j} + 13\hat{k} \right|}$$ The magnitude of the cross product is: $$= \frac{141}{\sqrt{16 + 36 + 169}}$$ Finally, the shortest distance is: $$= \frac{141}{\sqrt{221}}$$
Maths · Integrals · Subjective
\[ \left| \frac{120}{\pi^3}\int_{0}^{\pi} \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x}\,dx \right| \text{ is equal to } \underline{\hspace{2cm}}. \]
Answer: 15
Given the integral $$\int_0^\pi \frac{x^2 \sin x \cdot \cos x}{\sin^4 x + \cos^4 x} \, dx$$. This is equal to $$\int_0^\pi \frac{\sin x \cdot \cos x}{\sin^4 x + \cos^4 x} \left( x^2 - (\pi - x)^2 \right) \, dx$$. This simplifies to $$\int_0^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x \left( 2\pi x - \pi^2 \right)}{\sin^4 x + \cos^4 x} \, dx$$. This is equal to $$2\pi \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos 4x} \, dx - \pi^2 \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx$$. This simplifies to $$2\pi \cdot \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos^4 x}{\sin^4 x + \cos^4 x} \, dx - \pi^2 \int_0^{\frac{\pi}{2}} \frac{\sin x \cos^4 x}{\sin^4 x + \cos^4 x} \, dx$$. This is equal to $$-\frac{\pi^2}{2} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx$$. This simplifies to $$-\frac{\pi^2}{2} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x \, dx}{1 - 2 \sin^2 x \times \cos^2 x}$$. This is equal to $$-\frac{\pi^2}{2} \int_0^{\frac{\pi}{2}} \frac{\sin 2x}{2 - \sin^2 2x} \, dx$$. This simplifies to $$-\frac{\pi^2}{2} \int_0^{\frac{\pi}{2}} \frac{\sin 2x}{1 + \cos^2 2x} \, dx$$. Let $\cos 2x = t$.
Maths · Complex Numbers and Quadratic Equations · Numerical
Let a, b, c be the length of three sides of a triangle satisfying the condition $(a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0$. If the set of all possible values of $x$ is the interval $(\alpha, \beta)$, then $12(\alpha^2 + \beta^2)$ is equal to _______.
Answer: 36
$(a^2 + b^2)x^2 - 2b(a+c)x + b^2 + c^2 = 0$ $\Rightarrow a^2x^2 - 2abx + b^2 + b^2x^2 - 2bcx + c^2 = 0$ $\Rightarrow (ax - b)^2 + (bx - c)^2 = 0$ $\Rightarrow ax - b = 0,\quad bx - c = 0$ $\Rightarrow a + b > c,\quad b + c > a,\quad c + a > b$ $a + ax > bx$ $ax + bx > a$ $ax^2 + a > ax$ $a + ax > ax^2$ $ax + ax^2 > a$ $x^2 - x + 1 > 0$ (always true) $x^2 - x - 1 0$ $$\frac{1-\sqrt{5}}{2} \frac{-1+\sqrt{5}}{2}$$ $$\Rightarrow \frac{\sqrt{5}-1}{2} < x < \frac{\sqrt{5}+1}{2}$$ $\Rightarrow \alpha = \dfrac{\sqrt{5}-1}{2},\quad \beta = \dfrac{\sqrt{5}+1}{2}$ $12\left(\alpha^2 + \beta^2\right) = 12\left(\dfrac{(\sqrt{5}-1)^2 + (\sqrt{5}+1)^2}{4}\right) = 36$
Maths · Straight Lines and Pair of Straight Lines · Numerical
Let A(-2, -1), B(1, 0), C($\alpha$, $\beta$) and D($\gamma$, $\delta$) be the vertices of a parallelogram ABCD. If the point C lies on 2x - y = 5 and the point D lies on 3x - 2y = 6, then the value of |$\alpha$ + $\beta$ + $\gamma$ + $\delta$| is equal to _______.
Answer: 32
P is given by $$\left( \frac{\alpha - 2}{2}, \frac{\beta - 1}{2} \right) \equiv \left( \frac{\gamma + 1}{2}, \frac{\delta}{2} \right)$$ From this, we have: $$\frac{\alpha - 2}{2} = \frac{\gamma + 1}{2} and \frac{\beta - 1}{2} = \frac{\delta}{2}$$ This implies: $$\alpha - \gamma = 3 \ldots (1), \beta - \delta = 1 \ldots (2)$$ Also, $$(\gamma, \delta)$$ lies on $$3x - 2y = 6$$ Thus: $$3\gamma - 2\delta = 6 \ldots (3)$$ And $$(\alpha, \beta)$$ lies on $$2x - y = 5$$ Thus: $$2\alpha - \beta = 5 \ldots (4)$$ Solving equations (1), (2), (3), and (4), we get: $$\alpha = -3, \beta = -11, \gamma = -6, \delta = -12$$ Therefore, $$|\alpha + \beta + \gamma + \delta| = 32$$
Maths · Binomial Theorem · Numerical
Let the coefficient of $x^r$ in the expansion of $$(x+3)^{n-1} + (x+3)^{n-2}(x+2) + (x+3)^{n-3}(x+2)^2 + \ldots + (x+2)^{n-1}$$ be $\alpha_r$. If $\sum_{r=0}^{n} \alpha_r = \beta^n - \gamma^n$, $\beta, \gamma \in \mathbb{N}$, then the value of $\beta^2 + \gamma^2$ equals .
Answer: 25
$(x+3)^{n-1} + (x+3)^{n-2}(x+2) + (x+3)^{n-3}(x+2)^2 + \ldots + (x+2)^{n-1}$ $$\sum \alpha_r = 4^{n-1} + 4^{n-2} \times 3 + 4^{n-3} \times 3^2 + \ldots + 3^{n-1}$$ $$= 4^{n-1}\left[1 + \frac{3}{4} + \left(\frac{3}{4}\right)^2 + \ldots + \left(\frac{3}{4}\right)^{n-1}\right]$$ $$= 4^{n-1} \times \frac{1 - \left(\dfrac{3}{4}\right)^n}{1 - \dfrac{3}{4}}$$ $$= 4^n - 3^n = \beta^n - \gamma^n$$ $\beta = 4,\quad \gamma = 3$ $\beta^2 + \gamma^2 = 16 + 9 = 25$
Maths · Matrices · Numerical
Let $A$ be a $3\times3$ matrix and $\det(A)=2$. If $n= \det\left( \underbrace{ adj\left(adj\left(\cdots\left(adj\,A\right)\cdots\right)\right) }_{2024\ \text{times}} \right)$ Then the remainder when $n$ is divided by $9$ is equal to ______.
Answer: 7
Given $|A| = 2$. The expression $adj(adj(adj \ldots (a))))$ repeated 2024 times is equal to $|A|^{(n-1)^{2024}}$. This simplifies to $|A|^{2024}$. Therefore, $= |A|^{2024} = 2^{2024}$. We have $2^{2024} = (2^2)^{2022} = 4(8)^{674} = 4(9 - 1)^{674}$. Thus, $2^{2024} \equiv 4 \pmod{9}$. This implies $2^{2024} \equiv 9m + 4$, where $m \leftarrow even$. Now, $2^{9m+4} = 16 \cdot (2^3)^{3m} \equiv 16 \pmod{9}$. Finally, $\equiv 7$.
Maths · Vector Algebra · Numerical
Let $\vec{a}$ = 3$\hat{i}$ + 2$\hat{j}$ + $\hat{k}$ , $\vec{b}$ = 2$\hat{i}$ - $\hat{j}$ + 3$\hat{k}$ and $\vec{c}$ be a vector such that ($\vec{a}$ + $\vec{b}$) $\times$ $\vec{c}$ = 2($\vec{a}$ $\times$ $\vec{b}$) + 24$\hat{j}$ - 6$\hat{k}$ and ($\vec{a}$ - $\vec{b}$ + $\hat{i}$) $\cdot$ $\vec{c}$ = -3 . Then $|\vec{c}|^2$ is equal to .
Answer: 38
Given $\left( \vec{a} + \vec{b} \right) \times \vec{c} = 2 \left( \vec{a} \times \vec{b} \right) + 24 \hat{\jmath} - 6 \hat{k}$. $\left( 5 \hat{\imath} + \hat{\jmath} + 4 \hat{k} \right) \times \vec{c} = 2 (7 \hat{\imath} - 7 \hat{\kappa}) + 24 \hat{\jmath} - 6 \hat{k}$. $$\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z \end{vmatrix} = 14 \hat{\imath} + 10 \hat{\jmath} - 20 \hat{k}$$ $\Rightarrow \hat{\imath}(z - 4y) - \hat{\jmath}(5z - 4x) + \hat{k}(5y - x) = 14 \hat{\imath} + 10 \hat{\jmath} - 20 \hat{k}$. $z - 4y = 14, 4x - 5z = 10, 5y - x = -20$. $(a - b + i) \cdot \vec{c} = -3$. $(2 \hat{\imath} + 3 \hat{\jmath} - 2 \hat{k}) \cdot \vec{c} = -3$. $2x + 3y - 2z = -3$. Therefore, $x = 5, y = -3, z = 2$. $|\vec{c}|^2 = 25 + 9 + 4 = 38$.
Maths · Limits and Derivatives · Numerical
\[ \lim_{x\to 0} \frac{ ax^2e^x -b\log_e(1+x) +cxe^{-x} }{ x^2\sin x } =1 \] then \[ 16\left(a^2+b^2+c^2\right) \] is equal to
Answer: 81
Given $$ax^2 \left( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \right) - b \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \right)$$ $$+ cx \left( 1 - x + \frac{x^2}{x!} - \frac{x^3}{3!} + \ldots \right)$$ The limit is $$\lim_{x \to 0} \frac{x^3 \cdot \frac{\sin x}{x}}{x^3}$$ Simplifying, we have $$= \lim_{x \to \infty} \frac{(c-b)x + \left( \frac{b}{2} - c + a \right)x^2 + \left( a - \frac{b}{3} + \frac{c}{2} \right)x^3 + \ldots}{x^3} = 1$$ From this, we get the equations: $$c - b = 0,$$ $$\frac{b}{2} - c + a = 0$$ Solving these, we find: $$a - \frac{b}{3} + \frac{c}{2} = 1$$ $$a = \frac{3}{4}$$ $$b = c = \frac{3}{2}$$ Now, calculate $$a^2 + b^2 + c^2 = \frac{9}{16} + \frac{9}{4} + \frac{9}{4}$$ Finally, $$16 \left( a^2 + b^2 + c^2 \right) = 81$$
Maths · Three Dimensional Geometry · Numerical
A line passes through $A(4, -6, -2)$ and $B(16, -2, 4)$. The point $P(a, b, c)$ where $a, b, c$ are non-negative integers, on the line $AB$ lies at a distance of $21$ units, from the point $A$. The distance between the points $P(a, b, c)$ and $Q(4, -12, 3)$ is equal to .
Answer: 22
Given $\($ $\frac{x-4}{12}$ = $\frac{x+6}{4}$ = $\frac{z+2}{6}$ $\)$ and $\($ $\frac{x-4}{6}$ = $\frac{y+6}{2}$ = $\frac{z+2}{3}$ = 21 $\)$. $\[$ $\left$( 21 $\times$ $\frac{6}{7}$ + 4, $\frac{2}{7}$ $\times$ 21 - 6, $\frac{3}{7}$ $\times$ 21 - 2 $\right$) $\]$ $\($ = (22, 0, 7) = (a, b, c) $\)$ Therefore, $\($ $\sqrt{324 + 144 + 16}$ = 22 $\)$
Maths · Differential Equations · Numerical
Let $y = y(x)$ be the solution of the differential equation $$\sec^2 x \, dx + \left(e^{2y} \tan^2 x + \tan x \right) \, dy = 0$$ $$0 < x < \frac{\pi}{2}, \, y\left(\frac{\pi}{4}\right) = 0.$$ If $y\left(\frac{\pi}{6}\right) = \alpha,$ Then $e^{8\alpha}$ is equal to
Answer: 9
Given $\sec^2 x \frac{dx}{dy} + e^{2y} \tan^2 x + \tan x = 0$. Put $\tan x = t \Rightarrow \sec^2 x \frac{dx}{dy} = \frac{dt}{dy}$. $$\frac{dt}{dy} + e^{2y} \times t^2 + t = 0$$ $$\frac{dt}{dy} + t = -t^2 \cdot e^{2y}$$ $$\frac{1}{t^2} \frac{dt}{dy} + \frac{1}{t} = -e^{2y}$$ Put $\frac{1}{t} = u \Rightarrow -\frac{1}{t^2} \frac{dt}{dy} = \frac{du}{dy}$. $$-\frac{du}{dy} + u = -e^{2y}$$ $$\frac{du}{dy} - u = e^{2y}$$ I.F. $= e^{-\int dy} = e^{-y}$ $$ue^{-y} = \int e^{-y} \times e^{2y} dy$$ $$\frac{1}{\tan x} \times e^{-y} = e^y + c$$ $x = \frac{\pi}{4}, y = 0, c = 0$ $x = \frac{\pi}{6}, y = \alpha$ $$\sqrt{3} e^{-\alpha} = e^{\alpha} + 0$$ $$e^{2\alpha} = \sqrt{3}$$ $$e^{8\alpha} = 9$$
Maths · Relations and Functions · Numerical
Let A = {1, 2, 3, $\ldots$, 100}. Let $R$ be a relation on A defined by (x, y) $\in$ R if and only if 2x = 3y. Let $R_1$ be a symmetric relation on A such that $R$ $\subset$ $R_1$ and the number of elements in $R_1$ is n. Then, the minimum value of n is $\ldots$.
Answer: 66
Physics · Laws of Motion · Single correct
A light string passing over a smooth light fixed pulley connects two blocks of masses $m_1$ and $m_2$. If the acceleration of the system is $g/8$, then the ratio of masses is
Answer: (a)
Given the equation for acceleration: $$a = \frac{(m_1 - m_2)g}{(m_1 + m_2)} = \frac{g}{8}$$ Multiply both sides by $(m_1 + m_2)$: $$8m_1 - 8m_2 = m_1 + m_2$$ Rearrange the terms: $$7m_1 = 9m_2$$ Divide both sides by $m_2$: $$\frac{m_1}{m_2} = \frac{9}{7}$$
Physics · Moving Charges and Magnetism · Single correct
A uniform magnetic field of $2 \times 10^{-3} \, \mathrm{T}$ acts along positive Y-direction. A rectangular loop of sides $20 \, \mathrm{cm}$ and $10 \, \mathrm{cm}$ with current of $5 \, \mathrm{A}$ is Y - Z plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is :
Answer: (b)
The magnetic moment $\vec{M}$ is given by $\vec{M} = i \vec{A}$. Calculating $\vec{M}$: $$\vec{M} = 5 \times (0.2) \times (0.1)(-\hat{i})$$ $$= 0.1(-\hat{i})$$ The torque $\vec{\tau}$ is given by $\vec{\tau} = \vec{M} \times \vec{B}$. Calculating $\vec{\tau}$: $$\vec{\tau} = 0.1(-\hat{i}) \times (2 \times 10^{-3})(\hat{j})$$ $$= 2 \times 10^{-4}(-\hat{k}) \, \mathrm{N} \cdot \mathrm{m}$$
Physics · Physical World, Units and Measurements · Single correct
The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $N\%$. The value of $N$ is:
Answer: (c)
$T = 2\pi\sqrt{\frac{\ell}{g}}$ $g = \frac{4\pi^2\ell}{T^2}$ $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + \frac{2\Delta T}{T}$ $= \frac{0.2}{20} + 2\left(\frac{1}{40}\right)$ $= \frac{0.3}{20}$ Percentage change $= \frac{0.3}{20}\times100 = 6\%$
Physics · Electric Charges and Fields · Single correct
Force between two point charges $q_1$ and $q_2$ placed in vacuum at '$r$' cm apart is $F$. Force between them when placed in a medium having dielectric $K = 5$ at '$r/5$' cm apart will be:
Answer: (b)
In air $F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r_2}$. In medium $$F' = \frac{1}{4\pi (K \varepsilon_0)} \frac{q_1 q_2}{(r')^2} = \frac{25}{4\pi (5 \varepsilon_0)} \frac{q_1 q_2}{(r)^2} = 5 \, F$$
Physics · Alternating Current · Single correct
An AC voltage $V = 20 \sin 200 \pi t$ is applied to a series LCR circuit which drives a current $I = 10 \sin \left( 200 \pi t + \frac{\pi}{3} \right)$. The average power dissipated is:
Answer: (d)
The average power $\langle P \rangle$ is given by $IV \cos \phi$. $$= \frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60^\circ$$ $$= 50 \, \mathrm{W}$$
Physics · Wave Optics · Single correct
When unpolarized light is incident at an angle of $60^\circ$ on a transparent medium from air. The reflected ray is completely polarized. The angle of refraction in the medium is
Answer: (a)
By Brewster's law, at complete reflection, the refracted ray and reflected ray are perpendicular.
Physics · Waves · Single correct
The speed of sound in oxygen at S.T.P. will be approximately: (Given, $R = 8.3 \, \mathrm{J} \mathrm{K}^{-1}$, $\gamma = 1.4$)
Answer: (a)
Given $$v = \sqrt{\frac{\gamma RT}{M}} = \sqrt{\frac{1.4 \times 8.3 \times 273}{32 \times 10^{-3}}}$$ $$= 314.8541 \approx 315 \, \mathrm{m/s}$$
Physics · Kinetic Theory · Single correct
A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is
Answer: (c)
Given $U = n C_V T$. Therefore, $U = n_1 C_{V_1} T + n_2 C_{V_2} T$. This implies $8 \times \frac{3R}{2} \times T + 6 \times \frac{5R}{2} \times T$. Thus, $U = 27RT$.
Physics · Current Electricity · Single correct
The resistance per centimeter of a meter bridge wire is $r$, with $X\Omega$ resistance in left gap. Balancing length from left end is at $40 \, \mathrm{cm}$ with $25\Omega$ resistance in right gap. Now the wire is replaced by another wire of $2r$ resistance per centimeter. The new balancing length for same settings will be at
Answer: (d)
Given the circuit, we have the following equations: $$\frac{25}{r \ell_1} = \frac{X}{r \ell_2} ...(i)$$ $$\frac{25}{2r \ell_1'} = \frac{X}{2r \ell_2'} ...(ii)$$ From (i) and (ii), $$\ell_2' = \ell_2 = 40 \, cm$$
Physics · Electromagnetic Waves · Single correct
Given below are two statements: Statement I: Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields. Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface. In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (b)
Given $$\frac{1}{2} \varepsilon_0 \mathbf{E}^2 = \frac{\mathbf{B}^2}{2 \mu_0}$$. Therefore, $$\mathbf{E} = \mathrm{CB}$$ and $$\mathrm{C} = \frac{1}{\mu_0 \varepsilon_0}$$.
Physics · Dual Nature of Radiation and Matter · Single correct
In a photoelectric effect experiment a light of frequency 1.5 times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be:
Answer: (c)
Since $\frac{f}{2} < f_0$ i.e. the incident frequency is less than threshold frequency. Hence there will be no emission of photoelectrons. $\Rightarrow$ current = 0
Physics · Laws of Motion · Single correct
A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If $\vec{F}_1$ is the force required to just move the block up the inclined plane and $\vec{F}_2$ is the force required to just prevent the block from sliding down, then the value of $\left| \vec{F}_1 \right| - \left| \vec{F}_2 \right|$ is : [Use $g = 10 \, \mathrm{m/s^2}$ ] \[ \text{[We changed options. In official NTA paper no option was correct.]} \]
Answer: (b)
Given $f_k = \mu mg \cos \theta$. $$f_k = 0.1 \times \frac{50 \times \sqrt{3}}{2}$$ $$= 2.5 \sqrt{3} \, \mathrm{N}$$ $F_1 = mg \sin \theta + f_k$ $$= 25 + 2.5 \sqrt{3}$$ $F_2 = mg \sin \theta - f_k$ $$= 25 - 2.5 \sqrt{3}$$ Therefore, $F_1 - F_2 = 5 \sqrt{3} \, \mathrm{N}$.
Physics · Current Electricity · Single correct
By what percentage will the illumination of the lamp decrease if the current drops by 20$\%$ ?
Answer: (c)
Given $P = i^2 R$. $P_{int} = I_{int}^2 R$. $P_{final} = (0.8 I_{int})^2 R$. The percentage change in power is given by $$\frac{P_{final} - P_{int}}{P_{int}} \times 100 = (0.64 - 1) \times 100 = -36\%$$
Physics · Mathematics in Physics · Single correct
If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at an angle $\theta$, then
Answer: (c)
The magnitude of resultant vector $$R' = \sqrt{a^2 + b^2 + 2ab \cos \theta}$$ Here $a = b = R$ Then $$R' = \sqrt{R^2 + R^2 + 2R^2 \cos \theta}$$ $$= R \sqrt{2} \sqrt{1 + \cos \theta}$$ $$= \sqrt{2} R \sqrt{2 \cos^2 \frac{\theta}{2}}$$ $$= 2R \cos \frac{\theta}{2}$$
Physics · Nuclei · Single correct
The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is:
Answer: (a)
Given $R_1 = \frac{R_2}{2}$. $R_0 (A_1)^{1/3} = \frac{R_0}{2} (A_2)^{1/}$. $A_1 = \frac{1}{8} A_2$. $A_1 = \frac{192}{8} = 24$.
Physics · Gravitation · Single correct
The mass of the moon is $1/144$ times the mass of a planet and its diameter $1/16$ times the diameter of a planet. If the escape velocity on the planet is $v$, the escape velocity on the moon will be:
Answer: (a)
The escape velocity $V_{escape}$ is given by $$V_{escape} = \sqrt{\frac{2GM}{R}}.$$ The velocity of the planet $V_{planet}$ is $$V_{planet} = \sqrt{\frac{2GM}{R}} = V.$$ The velocity of the Moon $V_{Moon}$ is $$V_{Moon} = \sqrt{\frac{2GM \times 16}{144R}} = \frac{1}{3} \sqrt{\frac{2GM}{R}}.$$ Therefore, $$V_{Moon} = \frac{V_{Planet}}{3} = \frac{V}{3}.$$
Physics · Mechanical Properties of Fluids · Single correct
A small spherical ball of radius $r$, falling through a viscous medium of negligible density has terminal velocity '$v$'. Another ball of the same mass but of radius $2r$, falling through the same viscous medium will have terminal velocity:
Answer: (a)
Since density is negligible hence Buoyancy force will be negligible. At terminal velocity. $$Mg = 6 \pi \eta r v$$ $$v \propto \frac{1}{r} \quad \text{(as mass is constant)}$$ Now, $$\frac{v}{v'} = \frac{r'}{r}$$ $$r' = 2r$$ So, $$v' = \frac{v}{2}$$
Physics · Work, Energy and Power · Single correct
A body of mass 2 kg begins to move under the action of a time dependent force given by $\vec{F} = \left(6t\hat{i} + 6t^2\hat{j}\right) \, \mathrm{N}$. The power developed by the force at the time $t$ is given by:
Answer: (d)
Given $\vec{F} = (6t \hat{i} + 6t^2 \hat{j}) \, \mathrm{N}$. $\vec{F} = m \vec{a} = (6t \hat{i} + 6t^2 \hat{j})$. $\vec{a} = \frac{\vec{F}}{m} = (3t \hat{i} + 3t^2 \hat{j})$. $\vec{v} = \int_0^t \vec{a} \, dt = \frac{3t^2}{2} \hat{i} + t^3 \hat{j}$. $P = \vec{F} \cdot \vec{v} = (9t^3 + 6t^5) \, \mathrm{W}$.
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The output of the given circuit diagram is
Answer: (c)
If $A = 0$; $\overline{A} = 1$. $A = 1$; $\overline{A} = 0$. $B = 0$; $\overline{B} = 1$. $B = 1$; $\overline{B} = 0$. $Y = (A + \overline{B}) + (\overline{A} + B) = (1 + 1) = 0$.
Physics · Physical World, Units and Measurements · Single correct
Consider two physical quantities A and B related to each other as $E = \frac{B - x^2}{At}$ where $E$, $x$ and $t$ have dimensions of energy, length and time respectively. The dimension of $AB$ is
Answer: (b)
Physics · Current Electricity · Numerical
In the following circuit, the battery has an emf of $2 \, \mathrm{V}$ and an internal resistance of $\frac{2}{3} \, \Omega$. The power consumption in the entire circuit is _____ W.
Given $R_{eq} = \frac{4}{3} \, \Omega$. Therefore, $$P = \frac{V^2}{R_{eq}} = \frac{4}{4/3} = 3 \, \mathrm{W}$$
Physics · Ray Optics and Optical Instruments · Numerical
Light from a point source in air falls on a convex curved surface of radius 20 cm and refractive index 1.5. If the source is located at 100 cm from the convex surface, the image will be formed at _____ cm from the object.
Answer: 200
Given $\mu_1 = 1$, $\mu_2 = 1.5$, $R = 20 \, \mathrm{cm}$, $u = -100 \, \mathrm{cm}$. Using the lens formula: $$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$ Substitute the values: $$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$$ Solving for $v$: $$v = 100 \, \mathrm{cm}$$ Distance from object: $$= 100 + 100$$ $$= 200 \, \mathrm{cm}$$
Physics · Electromagnetic Induction · Numerical
The magnetic flux $\phi$ (in weber) linked with a closed circuit of resistance $8\,\Omega$ varies with time (in seconds) as $$\phi = 5t^2 - 36t + 1.$$ The induced current in the circuit at $t = 2\,\mathrm{s}$ is_____ A.
Answer: 2
Given $\($ $\varepsilon$ = - $\left$( $\frac{d\phi}{dt}$ $\right$) = 10t - 36 $\)$. At $\($ t = 2 $\)$, $\($ $\varepsilon$ = 16 $\mathrm{V}$ $\)$. $\($ i = $\frac{\varepsilon}{R}$ = $\frac{16}{8}$ = 2 $\mathrm{A}$ $\)$.
Physics · Mechanical Properties of Solids · Numerical
Two blocks of mass 2 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is $4.0 \times 10^{-5} \, \mathrm{m}$ and Young's modulus of the metal is $2.0 \times 10^{11} \, \mathrm{N/m^2}$. The longitudinal strain developed in the wire is $\frac{1}{\alpha \pi}$. The value of $\alpha$ is _____. [Use $g = 10 \, \mathrm{m/s^2}$]
Answer: 12
Given $$T = \left( \frac{2 \, m_1 \, m_2}{m_1 + m_2} \right) g = \frac{80}{3} \, \mathrm{N}$$ $$A = \pi r^2 = 16 \pi \times 10^{-10} \, \mathrm{m^2}$$ Strain is given by $$Strain = \frac{\Delta \ell}{\ell} = \frac{F}{AY} = \frac{T}{AY}$$ Substituting the values, $$\frac{80/3}{16 \pi \times 10^{-10} \times 2 \times 10^{11}} = \frac{1}{12 \pi}$$ Therefore, $$\alpha = 12$$
Physics · System of Particles and Rotational Motion · Numerical
A body of mass ' $m$ ' is projected with a speed ' $u$ ' making an angle of $45^\circ$ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $\frac{\sqrt{2}mu^3}{Xg}$. The value of ' $X$ ' is_____
Answer: 8
Given the projectile motion, the range $L$ is given by: $$L = mu \cos \theta \frac{u^2 \sin^2 \theta}{2g}$$ Simplifying, we have: $$= mu^3 \frac{1}{4\sqrt{2}g} \Rightarrow x = 8$$
Physics · Moving Charges and Magnetism · Numerical
Two circular coils P and Q of 100 turns each have same radius of $\pi \, \mathrm{cm}$. The currents in P and R are $1 \, \mathrm{A}$ and $2 \, \mathrm{A}$ respectively. P and Q are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is $\sqrt{x} \, \mathrm{mT}$, where x = [Use $\mu_0 = 4\pi \times 10^{-7} \, \mathrm{TmA^{-1}}$]
Answer: 20
Given $B_P = \frac{\mu_0 N i_1}{2r} = \frac{\mu_0 \times 1 \times 100}{2\pi} = 2 \times 10^{-3} \, \mathrm{T}$. $B_Q = \frac{\mu_0 N i_2}{2r} = \frac{\mu_0 \times 2 \times 100}{2\pi} = 4 \times 10^{-3} \, \mathrm{T}$. $B_{net} = \sqrt{B_P^2 + B_Q^2}$. $= \sqrt{20} \, \mathrm{mT}$. $x = 20$.
Physics · Electric Charges and Fields · Numerical
The distance between charges $+q$ and $-q$ is $2l$ and between $+2q$ and $-2q$ is $4l$. The electrostatic potential at point $P$ at a distance $r$ from centre $O$ is $-\alpha \left[ \frac{ql}{r^2} \right] \times 10^9 \, \mathrm{V}$, where the value of $\alpha$ is_____. (Use $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \mathrm{Nm^2C^{-2}}$)
Answer: 27
Given $\vec{P}_1 = 2q\ell$ and $\vec{P}_2 = 8q\ell$, the net dipole moment is $\vec{P}_{net} = 6q\ell$. The potential $V$ is given by $$V = \frac{K \vec{P} \cdot \vec{r}}{r^3} = \frac{9 \times 10^9 (6q\ell)}{r^2} \cos(120^\circ)$$ Simplifying, we have $$= -(27) \left( \frac{q\ell}{r^2} \right) \times 10^9 \mathrm{Nm^2c^{-2}}$$ Thus, $\alpha = 27$.
Physics · System of Particles and Rotational Motion · Numerical
Two identical spheres each of mass $2 \, \mathrm{kg}$ and radius $50 \, \mathrm{cm}$ are fixed at the ends of a light rod so that the separation between the centers is $150 \, \mathrm{cm}$. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is $\frac{x}{20} \, \mathrm{kg} \, \mathrm{m}^2$, where the value of $x$ is
Answer: 53
The moment of inertia is given by the formula: $$I = \left( \frac{2}{5} m R^2 + m d^2 \right) \times 2$$ Substituting the values: $$I = 2 \left( \frac{2}{5} \times 2 \times \left( \frac{1}{2} \right)^2 + 2 \times \left( \frac{3}{4} \right)^2 \right) = \frac{53}{20} \, \mathrm{kg \cdot m^2}$$ Therefore, $X = 53$.
Physics · Oscillations · Numerical
The time period of simple harmonic motion of mass $M$ in the given figure is $\pi \sqrt{\frac{\alpha M}{5K}}$, where the value of $\alpha$ is _____.
Answer: 12
Given $k_{eq} = \frac{2k \cdot k}{3k} + k = \frac{5k}{3}$. Angular frequency of oscillation $(\omega)$ is $\omega = \sqrt{\frac{k_{eq}}{m}}$. Therefore, $\omega = \sqrt{\frac{5k}{3m}}$. Period of oscillation $(\tau)$ is $\tau = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{3m}{5k}}$. This simplifies to $\tau = \pi \sqrt{\frac{12m}{5k}}$.
Physics · Nuclei · Numerical
A nucleus has mass number $A_1$ and volume $V_1$. Another nucleus has mass number $A_2$ and volume $V_2$. If relation between mass number is $A_2 = 4A_1$, then $\frac{V_2}{V_1}$ = _____
Answer: 4
For a nucleus, the volume is given by $V = \frac{4}{3} \pi R^3$. The radius $R$ is given by $R = R_0 (A)^{1/3}$. Therefore, the volume becomes $V = \frac{4}{3} \pi R_0^3 A$. This implies $$\frac{V_2}{V_1} = \frac{A_2}{A_1} = 4.$$
Chemistry · Co-ordination Compounds · Single correct
Match List I with List II Choose the correct answer from the options given below :
Answer: (d)
For $[\mathrm{Cr(H_2O)_6}]^{3+}$, it contains $\mathrm{Cr^{3+}}$: $[\mathrm{Ar}] \, 3d^3$: $t_{2g}^3 \, e_g^0$. For $[\mathrm{Fe(H_2O)_6}]^{3+}$, it contains $\mathrm{Fe^{3+}}$: $[\mathrm{Ar}] \, 3d^5$: $t_{2g}^3 \, e_g^2$. For $[\mathrm{Ni(H_2O)_6}]^{2+}$, it contains $\mathrm{Ni^{2+}}$: $[\mathrm{Ar}] \, 3d^8$: $t_{2g}^6 \, e_g^2$. For $[\mathrm{V(H_2O)_6}]^{3+}$, it contains $\mathrm{V^{3+}}$: $[\mathrm{Ar}] \, 3d^2$: $t_{2g}^2 \, e_g^0$.
Chemistry · Some Basic Concepts of Chemistry · Single correct
A sample of $CaCO_3$ and $MgCO_3$ weighed 2.21 $\mathrm{g}$ is ignited to constant weight of 1.152 $\mathrm{g}$. The composition of mixture is: (Given molar mass in $\mathrm{g/mol}^{-1}$ $CaCO_3$ : 100, $MgCO_3$ : 84)
Answer: (a)
\[ \mathrm{CaCO_3(s)} \xrightarrow{\Delta} \mathrm{CaO(s)} + \mathrm{CO_2(g)} \] \[ \mathrm{MgCO_3(s)} \xrightarrow{\Delta} \mathrm{MgO(s)} + \mathrm{CO_2(g)} \] Let the weight of $\mathrm{CaCO_3}$ be $x$ gm. $\therefore$, weight of $\mathrm{MgCO_3} = (2.21 - x)$ gm. Moles of $\mathrm{CaCO_3}$ decomposed = moles of $\mathrm{CaO}$ formed. $$\frac{x}{100} = moles of \mathrm{CaO} formed$$ Therefore, weight of $\mathrm{CaO}$ formed = $$\frac{x}{100} \times 56$$ Moles of $\mathrm{MgCO_3}$ decomposed = moles of $\mathrm{MgO}$ formed. $$\frac{(2.21-x)}{84} = moles of \mathrm{MgO} formed$$ Therefore, weight of $\mathrm{MgO}$ formed = $$\frac{2.21-x}{84} \times 40$$ $$\Rightarrow \frac{2.21-x}{84} \times 40 + \frac{x}{100} \times 56 = 1.152$$ Therefore, $x = 1.1886$ g = weight of $\mathrm{CaCO_3}$ and weight of $\mathrm{MgCO_3} = 1.0214$ g.
Chemistry · Hydrocarbons · Single correct
Identify A and B in the following reaction sequence.
Answer: (a)
The reaction starts with bromobenzene. Upon treatment with concentrated $\mathrm{HNO_3}$, nitration occurs, resulting in the formation of 2,4,6-tribromonitrobenzene. Next, treatment with $\mathrm{NaOH}$ leads to the formation of 2,4,6-trinitrophenol (picric acid). An acid-base reaction with $\mathrm{NaOH}$ forms the sodium salt of picric acid. Finally, treatment with $\mathrm{HCl}$ regenerates picric acid.
Chemistry · Redox Reactions · Single correct
Given below are two statements : Statement I: $S_8$ solid undergoes disproportionation reaction under alkaline conditions to form $S^{2-}$ and $S_2O_3^{2-}$ Statement II: $ClO_4^-$ can undergo disproportionation reaction under acidic condition. In the light of the above statements, choose the most appropriate answer from the options given below :
Answer: (a)
Statement 1: $S_8 + 12\mathrm{OH}^- \rightarrow 4\, S^{2-} + 2\, S_2O_3^{2-} + 6H_2O$ Statement 2: $\mathrm{ClO}_4^-$ cannot undergo disproportionation reaction as chlorine is present in its highest oxidation state.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Identify major product ' P ' formed in the following reaction.
Answer: (d)
The reaction begins with the acyl chloride reacting with $\mathrm{AlCl_3}$ to form an acylium ion and $\mathrm{AlCl_4^-}$. The acylium ion acts as an electrophile. This electrophile then reacts with benzene to form a complex. The complex rearranges, and after the removal of $\mathrm{H^+}$, the final product is formed.
Chemistry · Hydrocarbons · Single correct
Major product of the following reaction is -
Answer: (c)
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Identify structure of 2,3-dibromo-1-phenylpentane.
Answer: (c)
The compound shown is 2,3-dibromo-1-phenylpentane.
Chemistry · Co-ordination Compounds · Single correct
Select the option with correct property -
Answer: (d)
[$\mathrm{Ni(CO)_4}$] $\rightarrow$ diamagnetic, sp^3 hybridisation, number of unpaired electrons = 0 [$\mathrm{NiCl_4}$]^{2-} $\rightarrow$ paramagnetic, sp^3 hybridisation, number of unpaired electrons = 2
Chemistry · Amines · Single correct
The azo-dye $(Y)$ formed in the following reactions is Sulphanilic acid $+\ \mathrm{NaNO_2}$ $+\ \mathrm{CH_3COOH}$ $\longrightarrow X$
Answer: (d)
Chemistry · Amines · Single correct
Given below are two statements : Statement I: Aniline reacts with con. $\mathrm{H_2SO_4}$ followed by heating at $453 - 473 \, \mathrm{K}$ gives $p$-aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'. Statement II: In Friedel-Craft's alkylation and acylation reactions, aniline forms salt with the $\mathrm{AlCl_3}$ catalyst. Due to this, nitrogen of aniline aquires a positive charge and acts as deactivating group. In the light of the above statements, choose the correct answer from the options given below :
Answer: (d)
The reaction sequence involves the conversion of aniline to sulfanilic acid. Aniline ($\mathrm{NH_2}$) is treated with concentrated sulfuric acid ($\mathrm{H_2SO_4}$) to form anilinium hydrogen sulfate ($\mathrm{NH_3^+HSO_4^-}$). This is then heated to $453-473 \, \mathrm{K}$ to produce sulfanilic acid ($\mathrm{NH_2SO_3H}$). Lassaigne's test is performed, which results in the formation of $[\mathrm{Fe(SCN)}]^{2+}$, indicating a blood red color.
Chemistry · Equilibrium · Single correct
$\mathrm{A_{(g)}} \rightleftharpoons \mathrm{B_{(g)}} + \dfrac{C}{2}\mathrm{(g)}$. The correct relationship between $K_P$, $\alpha$ and equilibrium pressure $P$ is
Answer: (b)
For the reaction $\mathrm{A_{(g)} \rightleftharpoons B_{(g)} + \frac{C}{2} \,(g)}$, at equilibrium time $t = t_{eq}$, the concentrations are given by: $$(1 - \alpha)$$ $$\alpha$$ $$\frac{\alpha}{2}$$ The partial pressures are: $$P_B = \frac{\alpha}{\left(1 + \frac{\alpha}{2}\right)} \cdot P, P_A = \frac{(1 - \alpha)}{\left(1 + \frac{\alpha}{2}\right)} \cdot P, P_C = \frac{\frac{\alpha}{2}}{\left(1 + \frac{\alpha}{2}\right)} \cdot P$$ The equilibrium constant $K_P$ is given by: $$K_P = \frac{P_B \cdot P_C^{\frac{1}{2}}}{P_A}$$ Substituting the expressions for partial pressures: $$= \frac{\left(\alpha\right)^{\frac{3}{2}} (P)^{\frac{1}{2}}}{(1 - \alpha)(2 + \alpha)^{\frac{1}{2}}}$$
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Choose the correct statements from the following A. All group 16 elements form oxides of general formula $\mathrm{EO}_2$ and $\mathrm{EO}_3$ where $\mathrm{E} = \mathrm{S}, \mathrm{Se}, \mathrm{Te}$ and $\mathrm{Po}$. Both the types of oxides are acidic in nature. B. $\mathrm{TeO}_2$ is an oxidising agent while $\mathrm{SO}_2$ is reducing in nature. C. The reducing property decreases from $\mathrm{H}_2 \mathrm{S}$ to $\mathrm{H}_2 \mathrm{Te}$ down the group. D. The ozone molecule contains five lone pairs of electrons. Choose the correct answer from the options given below:
Answer: (d)
(A) All group 16 elements form oxides of the $\mathrm{EO_2}$ and $\mathrm{EO_3}$ type where $\mathrm{E} = \mathrm{S, Se, Te}$ or $\mathrm{Po}$. (B) $\mathrm{SO_2}$ is reducing while $\mathrm{TeO_2}$ is an oxidising agent. (C) The reducing property increases from $\mathrm{H_2S}$ to $\mathrm{H_2Te}$ down the group. (D) have six lone pairs.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Identify the name reaction.
Answer: (c)
The reaction shown is the Gatterman-Koch reaction.
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Which of the following is least ionic?
Answer: (b)
AgCl < $\mathrm{CoCl_2}$ < $\mathrm{BaCl_2}$ < $\mathrm{KCl}$ (ionic character) Reason: $\mathrm{Ag^+}$ has pseudo inert gas configuration.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is _______
Answer: (d)
Steam distillation technique is applied to separate substances which are steam volatile and are immiscible with water.
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Given below are two statements : Statement I: Group 13 trivalent halides get easily hydrolyzed by water due to their covalent nature. Statement II: AlCl$_3$ upon hydrolysis in acidified aqueous solution forms octahedral $\left[ \mathrm{Al(H_2O)_6} \right]^{3+}$ ion. In the light of the above statements, choose the correct answer from the options given below :
Answer: (d)
In trivalent state most of the compounds being covalent are hydrolysed in water. Trichlorides on hydrolysis in water form tetrahedral $[\mathrm{M(OH)_4}]^{-}$ species, the hybridisation state of element M is $\mathrm{sp}^3$. In case of aluminium, acidified aqueous solution forms octahedral $[\mathrm{Al(H_2O)_6}]^{3+}$ ion.
Chemistry · Structure of Atom · Single correct
The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are
Answer: (b)
The electronic configuration of potassium is $^{19}\mathrm{K} 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$. The outermost orbital of potassium is the $4s$ orbital. The quantum numbers are $n = 4$, $l = 0$, $m_1 = 0$, $s = \pm \frac{1}{2}$.
Chemistry · The d-and f-Block Elements · Single correct
Choose the correct statements from the following A. $\mathrm{Mn_2O_7}$ is an oil at room temperature B. $\mathrm{V_2O_4}$ reacts with acid to give $\mathrm{VO_2^{2+}}$ C. $\mathrm{CrO}$ is a basic oxide D. $\mathrm{V_2O_5}$ does not react with acid Choose the correct answer from the options given below :
Answer: (b)
(A) $\mathrm{Mn_2O_7}$ is green oil at room temperature. (B) $\mathrm{V_2O_4}$ dissolve in acids to give $\mathrm{VO_2^{2+}}$ salts. (C) $\mathrm{CrO}$ is basic oxide. (D) $\mathrm{V_2O_5}$ is amphoteric it reacts with acid as well as base.
Chemistry · Haloalkanes and Haloarenes · Single correct
The correct order of reactivity in electrophilic substitution reaction of the following compounds is :
Answer: (d)
−CH$_3$ shows $+M$ and $+I$. −Cl shows $+M$ and $−I$ but inductive effect dominates. −NO$_2$ shows $−M$ and $−I$. Electrophilic substitution $\alpha \frac{1}{−M and -I}$ $\alpha + M$ and $+ I$ Hence, order is B $>$ A $>$ C $>$ D.
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Consider the following elements. Which of the following is/are true about $A'$, $B'$, $C'$ and $D'$? (A) Order of atomic radii: $B'<A'<D'<C'$ (B) Order of metallic character: $B'<A'<D'<C'$ (C) Size of the element: $D'<C'<B'<A'$ (D) Order of ionic radii: $B'^{+}<A'^{+}<D'^{+}<C'^{+}$ Choose the correct answer from the options given below:
Answer: (b)
In general along the period from left to right, size decreases and metallic character decrease. In general down the group, size increases and metallic character increases. $\mathrm{B'} \mathrm{A'}$ (size) $\mathrm{D'} \mathrm{B'}$ (size) $\mathrm{B'} < \mathrm{A'}$ (metallic character) $\mathrm{D'} < \mathrm{C'}$ (metallic character) $\mathrm{B'}^{+} < \mathrm{A'}^{+}$ (size) $\mathrm{D'}^{+} < \mathrm{C'}^{+}$ (size) Therefore, C statement is incorrect.
Chemistry · Chemical Bonding and Molecular Structure · Numerical
A diatomic molecule has a dipole moment of 1.2D. If the bond distance is $1\,\mathrm{\AA}$, then fractional charge on each atom is $\times 10^{-1}\,\mathrm{esu}$. (Given 1D = $10^{-18}\,\mathrm{esu\,cm}$)
Answer: 0
Given $\mu = 1.2 \, \mathrm{D} = q \times d$. Therefore, $1.2 \times 10^{-10} \, \mathrm{esu\, \AA} = q \times 1 \, \mathrm{\AA}$. Thus, $q = 1.2 \times 10^{-10} \, \mathrm{esu}$.
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
r = k[A] for a reaction, 50$\%$ of A is decomposed in 120 minutes. The time taken for 90$\%$ decomposition of A is _______ minutes.
Answer: 399
Given $r = k[A]$. So, order of reaction $= 1$. $t_{1/2} = 120 \, min$. For 90$\%$ completion of reaction, $$k = \frac{2.303}{t} \log \left( \frac{a}{a-x} \right)$$ $$\frac{0.693}{t_{1/2}} = \frac{2.303}{t} \log \frac{100}{10}$$ Therefore, $t = 399 \, min$.
Chemistry · Biomolecules · Numerical
A compound (x) with molar mass $108 \, \mathrm{g \, mol^{-1}}$ undergoes acetylation to give product with molar mass $192 \, \mathrm{g \, mol^{-1}}$. The number of amino groups in the compound (x) is
Answer: 2
Gain in molecular weight after acylation with one $-\mathrm{NH_2}$ group is 42. Total increase in molecular weight = 84. Therefore, number of amino group in $x = \frac{84}{42} = 2$.
Chemistry · Hydrocarbons · Numerical
Number of isomeric products formed by monochlorination of 2-methylbutane in presence of sunlight is ________.
Answer: 6
The reaction of the given compound with $\mathrm{Cl_2/h\nu}$ produces several isomeric products. The products include chiral centers, leading to enantiomers. Therefore, the number of isomeric products is 6.
Chemistry · The d-and f-Block Elements · Numerical
Number of moles of $\mathrm{H}^+$ ions required by 1 mole of $\mathrm{MnO}_4^-$ to oxidise oxalate ion to $\mathrm{CO}_2$ is .
Answer: 8
The balanced chemical equation is: $$2\mathrm{MnO_4^-} + 5\mathrm{C_2O_4^{2-}} + 16\mathrm{H^+} \rightarrow 2\mathrm{Mn^{2+}} + 10\mathrm{CO_2} + 8\mathrm{H_2O}$$ Therefore, the number of moles of $\mathrm{H^+}$ ions required by 1 mole of $\mathrm{MnO_4^-}$ to oxidise oxalate ion to $\mathrm{CO_2}$ is 8.
Chemistry · The d-and f-Block Elements · Numerical
In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is (+)
Answer: 6
Given the reaction: $$\mathrm{K_2Cr_2O_7 (s) + 4KCl(s) + 6H_2SO_4 (conc.) \rightarrow 2CrO_2Cl_2 (g) + 6KHSO_4 + 3H_2O}$$ This reaction is called chromyl chloride test. Here oxidation state of Cr is +6.
Chemistry · Some Basic Concepts of Chemistry · Numerical
The molarity of 1 L orthophosphoric acid ($H_3PO_4$) having 70$\%$ purity by weight (specific gravity 1.54 $\mathrm{g \, cm^{-3}}$) is _______ M. (Molar mass of $H_3PO_4$ = 98 $\mathrm{g \, mol^{-1}}$)
Answer: 11
Specific gravity (density) $= 1.54 \, \mathrm{g/cc}$. Volume $= 1 \, \mathrm{L} = 1000 \, \mathrm{ml}$. Mass of solution $= 1.54 \times 1000 = 1540 \, \mathrm{g}$. % purity of $\mathrm{H_2SO_4}$ is $70\%$. So weight of $\mathrm{H_3PO_4} = 0.7 \times 1540 = 1078 \, \mathrm{g}$. Mole of $\mathrm{H_3PO_4} = \frac{1078}{98} = 11$. Molarity $= \frac{11}{1 \, \mathrm{L}} = 11$.
Chemistry · Electrochemistry · Numerical
The values of conductivity of some materials at 298.15 K in $\mathrm{Sm}^{-1}$ are $2.1 \times 10^3$, $1.0 \times 10^{-16}$, $1.2 \times 10$, $3.91$, $1.5 \times 10^{-2}$, $1 \times 10^{-7}$, $1.0 \times 10^3$. The number of conductors among the materials is .
Answer: 4
Conductivity $\left( \mathrm{Sm^{-1}} \right)$ $$\begin{cases} 2.1 \times 10^3 \\ 1.2 \times 10 \\ 3.91 \\ 1 \times 10^3 \end{cases} conductors at 298.15 K$$ $1 \times 10^{-16}$ Insulator at 298.15 K $$\begin{cases} 1.5 \times 10^{-2} \\ 1 \times 10^{-7} \end{cases} Semiconductor at 298.15 K$$ Therefore number of conductors is 4.
Chemistry · Biomolecules · Fill in the blank
From the vitamins $A, B, B_1, B_6, B_{12}, C, D, E$ and $K,$ the number vitamins that can be stored in our body is
Answer: 5
Vitamins A, D, E, K and $\mathrm{B_{12}}$ are stored in liver and adipose tissue.
Chemistry · Thermodynamics · Numerical
If 5 moles of an ideal gas expands from 10 L to a volume of 100 L at 300 K under isothermal and reversible condition then work, $w$, is $-x$ J. The value of $x$ is _____ (Given $R = 8.314 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$)
Answer: 28721
It is isothermal reversible expansion, so work done negative $$W = -2.303 nRT \log \left( \frac{V_2}{V_1} \right)$$ $$= -2.303 \times 5 \times 8.314 \times 300 \log \left( \frac{100}{10} \right)$$ $$= -28720.713 \, \mathrm{J}$$ $$\equiv -28721 \, \mathrm{J}$$