JEE Main 24 February 2021 Shift 1: 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 · Mathematical Reasoning · Single correct
The statement among the following that is a tautology is:
Answer: (d)
Given $A \land (\sim A \lor B) \to B$. $$= [(A \land \sim A) \lor (A \land B)] \to B$$ $$= (A \land B) \to B$$ $$= \sim A \lor \sim B \lor B$$ $$= t$$
Maths · Straight Lines and Pair of Straight Lines · Single correct
A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is $\frac{1}{4}$. Three stones $A$, $B$ and $C$ are placed at the points $(1,1)$, $(2,2)$ and $(4,4)$ respectively. Then which of these stones is/are on the path of the man?
Answer: (a)
$\dfrac{x}{a}+\dfrac{y}{b}=1$ $\dfrac{h}{a}+\dfrac{k}{b}=1 \quad \ldots\ (1)$ $\dfrac{\frac{1}{a}+\frac{1}{b}}{2}=\dfrac{1}{4} \quad \ldots\ (2)$ $\therefore\ \dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{2}$ $\therefore$ Line passes through fixed point $B(2,2)$ (from (1) and (2))
Maths · Three Dimensional Geometry · Single correct
The equation of the plane passing through the point $(1,2,-3)$ and perpendicular to the planes $3x + y - 2z = 5$ and $2x - 5y - z = 7$, is:
Answer: (c)
Normal vector of required plane is $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 2 & -5 & -1 \end{vmatrix} = -11\hat{i} - \hat{j} - 17\hat{k}$. Therefore, $11(x - 1) + (y - 2) + 17(z + 3) = 0$. $11x + y + 17z + 38 = 0$
Maths · Differential Equations · Single correct
The population $P = P(t)$ at time 't' of a certain species follows the differential equation $\frac{dP}{dt} = 0.5P - 450$. If $P(0) = 850$, then the time at which population becomes zero is:
Answer: (b)
Given $\($ $\frac{dp}{dt}$ = $\frac{p - 900}{2}$ $\)$. Integrating both sides, we have: $$ \int_{850}^{0} \frac{dp}{p - 900} = \int_{0}^{t} \frac{dt}{2} $$ This simplifies to: $$ \ln | P - 900 | \bigg|_{850}^{0} = \frac{t}{2} $$ Evaluating the limits, we get: $$ \ln |900| - \ln |50| = \frac{t}{2} $$ Thus, $$ \frac{t}{2} = \ln |18| $$ Therefore, $$ t = 2 \ln 18 $$
Maths · Determinants · Single correct
The system of linear equations $$3x - 2y - kz = 10$$ $$2x - 4y - 2z = 6$$ $$x + 2y - z = 5m$$ is inconsistent if :
Answer: (d)
Given $$\Delta = \begin{vmatrix} 3 & -2 & -k \\ 1 & -4 & -2 \\ 1 & 2 & -1 \end{vmatrix} = 0$$ Calculate $$3(4 + 4) + 2(-2 + 2) - k(4 + 4) = 0$$ This implies $$k = 3$$ Next, $$\Delta_x = \begin{vmatrix} 10 & -2 & -3 \\ 6 & -4 & -2 \\ 5m & 2 & -1 \end{vmatrix} \neq 0$$ Calculate $$10(4 + 4) + 2(-6 + 10m) - 3(12 + 20m) \neq 0$$ Simplify $$80 - 12 + 20m - 36 - 60m \neq 0$$ This implies $$40m \neq 32 \Rightarrow m \neq \frac{4}{5}$$ Next, $$\Delta_y = \begin{vmatrix} 3 & 10 & -3 \\ 2 & 6 & -2 \\ 1 & 5m & -1 \end{vmatrix} \neq 0$$ Calculate $$3(-6 + 10m) - 10(-2 + 2) - 3(10m - 6) \neq 0$$ Simplify $$-18 + 30m - 30m + 18 \neq 0 \Rightarrow 0 \neq 0$$ Next, $$\Delta_z = \begin{vmatrix} 3 & -2 & 10 \\ 2 & -4 & 6 \\ 1 & 2 & 5m \end{vmatrix} \neq 0$$ Calculate $$3(-20m - 12) + 2(10m - 6) + 10(4 + 4) - 40m + 32 \neq 0$$ This implies $$m \neq \frac{4}{5}$$
Maths · Continuity and Differentiability · Single correct
If $f : \mathbb{R} \to \mathbb{R}$ is a function defined by $f(x) = \lfloor x - 1 \rfloor \cos\left(\frac{2x-1}{2}\right)\pi$, where $\lfloor . \rfloor$ denotes the greatest integer function, then $f$ is :
Answer: (d)
Doubtful points are $x = n$, $n \in I$. L.H.L = $$\lim_{x \to n^-} [x - 1] \cos\left(\frac{2x - 1}{2}\right) \pi = (n - 2) \cos\left(\frac{2n - 1}{2}\right) \pi = 0$$ R.H.L = $$\lim_{x \to n^+} [x - 1] \cos\left(\frac{2x - 1}{2}\right) \pi = (n - 1) \cos\left(\frac{2n - 1}{2}\right) \pi = 0$$ $f(n) = 0$ Hence continuous.
Maths · Three Dimensional Geometry · Single correct
The distance of the point $(1,1,9)$ from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x+y+z=17$ is:
Answer: (a)
Given $\($ $\frac{x-3}{1}$ = $\frac{y-4}{2}$ = $\frac{z-5}{2}$ = $\lambda$ $\)$ $\($ $\Rightarrow$ x = $\lambda$ + 3, $\;$ y = 2$\lambda$ + 4, $\;$ z = 2$\lambda$ + 5 $\)$ Which lines on given plane hence $\($ $\Rightarrow$ $\lambda$ + 3 + 2$\lambda$ + 4 + 2$\lambda$ + 5 = 17 $\)$ $\($ $\Rightarrow$ $\lambda$ = $\frac{5}{5}$ = 1 $\)$ Hence, point of intersection is $\($ Q(4, 6, 7) $\)$ Therefore, required distance $\($ = PQ $\)$ $\($ = $\sqrt{9 + 25 + 4}$ $\)$ $\($ = $\sqrt{38}$ $\)$
Maths · Applications of Derivatives · Single correct
If the tangent to the curve $y = x^3$ at the point $P(t, t^3)$ meets the curve again at $Q$, then the ordinate of the point which divides $PQ$ internally in the ratio $1:2$ is:
Answer: (a)
Equation of tangent at $P \left( t, t^3 \right)$ $$(y - t^3) = 3t^2(x - t) \ldots (i)$$ Now solve the above equation with $$y = x^3 \ldots (ii)$$ By (i) and (ii), $$x^3 - t^3 = 3t^2(x - t)$$ $$x^2 + xt + t^2 = 3t^2$$ $$x^2 + xt - 2t^2 = 0$$ $$(x - t)(x + 2t) = 0$$ $$\Rightarrow x = -2t \Rightarrow Q \left( -2t, -8t^3 \right)$$ Ordinate of required point = $$\frac{2t^3 + (-8t^3)}{3} = -2t^3$$
Maths · Integrals · Single correct
If $\int\frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}}$ dx = a $\sin^{-1}(\frac{\sin x + \cos x}{b})$ + c, where c is a constant of integration, then the ordered pair (a, b) is equal to :
Answer: (b)
Put $\sin x + \cos x = t$. $$\Rightarrow 1 + \sin 2x = t^2$$ $$\Rightarrow (\cos x - \sin x) dx = dt$$ $$\therefore I = \int \frac{dt}{\sqrt{8 - (t^2 - 1)}} = \int \frac{dt}{\sqrt{9 - t^2}}$$ $$= \sin^{-1} \left( \frac{t}{3} \right) + C = \sin^{-1} \left( \frac{\sin x + \cos x}{3} \right) + c$$ $$\Rightarrow a = 1 , b = 3 $$
Maths · Binomial Theorem · Single correct
The value of \[ -{}^{15}\mathrm{C}_1 + 2\,{}^{15}\mathrm{C}_2 - 3\,{}^{15}\mathrm{C}_3 + \cdots - 15\,{}^{15}\mathrm{C}_{15} + {}^{14}\mathrm{C}_1 + {}^{14}\mathrm{C}_3 + {}^{14}\mathrm{C}_5 + \cdots + {}^{14}\mathrm{C}_{11} \] is:
Answer: (d)
Given $$S_1 = -^{15}C_1 + 2 \cdot ^{15}C_2 - \ldots - 15^{15}C_{15}$$ We have $$S_1 = \sum_{r=1}^{15} (-1)^r \cdot r \cdot ^{15}C_r = 15 \sum_{r=1}^{15} (-1)^r \cdot ^{14}C_{r-1}$$ This simplifies to $$= 15 \left( -^{14}C_0 + ^{14}C_1 - \ldots - ^{14}C_{14} \right) = 15(0) = 0$$ Next, $$S_2 = ^{14}C_1 + ^{14}C_3 + \ldots + ^{14}C_{11}$$ This is equal to $$= \left( ^{14}C_1 + ^{14}C_3 + \ldots + ^{14}C_{11} \right) - ^{14}C_{13}$$ Which simplifies to $$= 2^{13} - 14$$ Therefore, $$S_1 + S_2 = 2^{13} - 14$$
Maths · Applications of Derivatives · Single correct
The function $f(x) = \frac{4x^3 - 3x^2}{6} - 2 \sin x + (2x - 1) \cos x$:
Answer: (a)
$f'(x)=(2x-1)(x-\sin x)$ $\Rightarrow f'(x)\geq 0$ for $x\in\left[\frac{1}{2},\infty\right)$
Maths · Relations and Functions · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = 2x - 1$ and $g : \mathbb{R} - \{1\} \to \mathbb{R}$ be defined as $g(x) = \frac{x - \frac{1}{2}}{x - 1}$. Then the composition function $f(g(x))$ is :
Answer: (d)
Given $$f(g(x)) = 2g(x) - 1$$ Simplifying, we have: $$= 2 \left( \frac{x - \frac{1}{2}}{x - 1} \right) = \frac{x}{x - 1}$$ Thus, $$f(g(x)) = 1 + \frac{1}{x - 1}$$ The function is one-one and into.
Maths · Probability · Single correct
An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is :
Answer: (b)
P (odd no. twice) = P (even no. thrice) $$\Rightarrow nC_2 \left( \frac{1}{2} \right)^n = nC_3 \left( \frac{1}{2} \right)^n \Rightarrow n = 5$$ Success is getting an odd number then P (odd successes) = P(1) + P(3) + P(5) $$= \binom{5}{1} \left( \frac{1}{2} \right)^5 + \binom{5}{3} \left( \frac{1}{2} \right)^5 + \binom{5}{5} \left( \frac{1}{2} \right)^5$$ $$= \frac{16}{2^5} = \frac{1}{2}$$
Maths · Permutations and Combinations · Single correct
A scientific committee is to formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed is:
Answer: (c)
(2I, 4F) + (3I, 6F) + (4I, 8F) $$= \binom{6}{2} \binom{8}{4} + \binom{6}{3} \binom{8}{6} + \binom{6}{4} \binom{8}{8}$$ $$= 15 \times 70 + 20 \times 28 + 15 \times 1$$ $$= 1050 + 560 + 15 = 1625$$
Maths · Applications of Integrals · Single correct
The area (in sq. units) of the part of the circle $x^2 + y^2 = 36$, which is outside the parabola $y^2 = 9x$, is :
Answer: (d)
The curves intersect at point $(3, \pm 3\sqrt{3})$. Required area $$= \pi r^2 - 2 \left[ \int_0^3 \sqrt{9x} \, dx + \int_3^6 \sqrt{36 - x^2} \, dx \right]$$ $$= 36\pi - 12\sqrt{3} - 2 \left( \frac{x}{2} \sqrt{36 - x^2} + 18 \sin^{-1} \left( \frac{x}{6} \right) \right)_3^6$$ $$= 36\pi - 12\sqrt{3} - 2 \left( 9 - \left( \frac{9\sqrt{3}}{2} + 3\pi \right) \right) = 24\pi - 3\sqrt{3}$$
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $p$ and $q$ be two positive number such that $p + q = 2$ and $p^4 + q^4 = 272$. Then $p$ and $q$ are roots of the equation:
Answer: (d)
$(p^2 + q^2)^2 - 2p^2q^2 = 272$ $((p + q)^2 - 2pq)^2 - 2p^2q^2 = 272$ $16 + 16pq + 2p^2q^2 = 272$ $(pq)^2 - 8pq - 128 = 0$ $pq = \dfrac{8 \pm 24}{2} = 16,\,-8$ $pq = 16$ Now $x^2 - (p + q)x + pq = 0$ $x^2 - 2x + 16 = 0$
Maths · Heights and Distances · Single correct
Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:
Answer: (d)
Given $\tan \theta = \frac{h}{75} = \frac{75}{3h}$. Therefore, $$h^2 = \frac{(75)^2}{3}$$ which gives $$h = 25 \sqrt{3} \, \mathrm{m}.$$
Maths · Integrals · Single correct
$\displaystyle \lim_{x\to 0}\frac{\int_0^{x^2}(\sin\sqrt{t})\,dt}{x^3}$ is equal to :
Answer: (a)
Given the limit $$\lim_{x \to 0} \frac{\int_0^{x^2} \sin \sqrt{t} \, dt}{x^3}$$ we simplify it as follows: $$= \lim_{x \to 0} \frac{(\sin |x|) 2x}{3x^2}$$ This can be further simplified to: $$= \lim_{x \to 0} \left( \frac{\sin x}{x} \right) \times \frac{2}{3}$$ Finally, the limit evaluates to: $$= \frac{2}{3}$$
Maths · Sequences and Series · Single correct
If $e^{(\cos^2 x + \cos^4 x + \cos^6 x + \ldots \infty)} \log_e 2$ satisfies the equation $t^2 - 9t + 8 = 0$, then the value of $$\frac{2 \sin x}{\sin x + \sqrt{3} \cos x} \left(0 < x < \frac{\pi}{2}\right)$$ is
Answer: (c)
Given $e^{(\cos^2 x + \cos^4 x + \ldots \ldots \infty)} \ln 2 = 2^{\cos^2 x + \cos^4 x + \ldots \ldots \infty}$. This simplifies to $2^{\cot^2 x}$. The equation $t^2 - 9t + 8 = 0$ gives $t = 1, 8$. Thus, $2^{\cot^2 x} = 1, 8$ implies $\cot^2 x = 0, 3$. Therefore, $S(0) \Rightarrow \frac{2 \sin x}{\sin x + \sqrt{3} \cos x} = \frac{2}{1 + \sqrt{3} \cot x} = \frac{2}{4} = \frac{1}{2}$.
Maths · Conic Sections · Single correct
The locus of the mid-point of the line segment joining the focus of the parabola $y^2 = 4ax$ to a moving point of the parabola, is another parabola whose directrix is:
Answer: (b)
Given $h = \frac{at^2 + a}{2}$, $k = \frac{2at + 0}{2}$. Therefore, $t^2 = \frac{2h - a}{a}$ and $t = \frac{k}{a}$. Thus, $\frac{k^2}{a^2} = \frac{2h - a}{a}$. The locus of $(h, k)$ is $y^2 = a(2x - a)$. This simplifies to $y^2 = 2a \left( x - \frac{a}{2} \right)$. Its directrix is $x - \frac{a}{2} = -\frac{a}{2} \Rightarrow x = 0$.
Maths · Complex Numbers and Quadratic Equations · Numerical
If the least and the largest real values of $\alpha$, for which the equation $z + \alpha |z - 1| + 2i = 0$ ($z \in \mathbb{C}$ and $i = \sqrt{-1}$) has a solution, are $p$ and $q$ respectively; then $4 \left( p^2 + q^2 \right)$ is equal to
Answer: 10
Given the equation $x + iy + \alpha \sqrt{(x-1)^2 + y^2} + 2i = 0$. Therefore, $y + 2 = 0$ and $x + \alpha \sqrt{(x-1)^2 + y^2} = 0$. Substituting $y = -2$ into the equation gives $x^2 = \alpha^2 (x^2 - 2x + 1 + 4)$. This implies $\alpha^2 = \frac{x^2}{x^2 - 2x + 5} \Rightarrow x^2 (\alpha^2 - 1) - 2x\alpha^2 + 5\alpha^2 = 0$. Since $x \in \mathbb{R} \Rightarrow D \geq 0$, we have $\alpha \alpha^4 - 4(\alpha^2 - 1)5\alpha^2 \geq 0$. Thus, $\alpha^2 [4\alpha^2 - 2\alpha^2 + 20] \geq 0$. Simplifying gives $\alpha^2 [-16\alpha^2 + 20] \geq 0$. Therefore, $\alpha^2 \left[ \alpha^2 - \frac{5}{4} \right] \leq 0$. This implies $0 \leq \alpha^2 \leq \frac{5}{4}$. Thus, $\alpha^2 \in \left[ 0, \frac{5}{4} \right]$. Therefore, $\alpha \in \left[ -\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{2} \right]$. Then $4 \left[ (q)^2 + (p)^2 \right] = 4 \left[ \frac{5}{4} + \frac{5}{4} \right] = 10$.
Maths · Integrals · Numerical
If $\displaystyle \int_{-a}^{a}(|x|+|x-2|)\,dx=22$, $(a>2)$ and $[x]$ denotes the greatest integer $\leq x$, then $\displaystyle \int_{a}^{-a}(x+[x])\,dx$ is equal to _____.
Answer: 3
Given $$\int_{-a}^{0}(-2x + 2) \, dx + \int_{0}^{2}(x + 2 - x) \, dx + \int_{2}^{a}(2x - 2) \, dx = 22$$ Calculate each integral: $$x^2 - 2x \bigg|_{-a}^{0} + 2x \bigg|_{0}^{2} + x^2 - 2x \bigg|_{2}^{a} = 22$$ Simplify: $$a^2 + 2a + 4 + a^2 - 2a - (4 - 4) = 22$$ Combine terms: $$2a^2 = 18 \Rightarrow a = 3$$ Evaluate the integral: $$\int_{3}^{-3}(x + [x]) \, dx = - \left( \int_{-3}^{3}(x + [x]) \, dx \right) = - \left( \int_{-3}^{3}[x] \, dx \right)$$ Calculate the sum: $$= -(-3 - 2 - 1 + 0 + 1 + 2) = 3$$
Maths · Sets · Numerical
Let $$A = \{ n \in N : n is a 3-digit number \}$$ $$B = \{ 9k + 2 : k \in N \}$$ and $$C : \{ 9k + \ell : k \in N \} for some \ell (0 < \ell < 9)$$ If the sum of all the elements of the set $A \cap (B \cup C)$ is $274 \times 400$, then $\ell$ is equal to
Answer: 5
3 digit numbers of the form $9K + 2$ are $\{101, 109, \ldots, 992\}$. Sum equal to $\frac{100}{2} (1093) = s_1 = 54650$. $$274 \times 400 = s_1 + s_2$$ $$274 \times 400 = \frac{100}{2} [101 + 992] + s_2$$ $$274 \times 400 = 50 \times 1093 + s_2$$ $$s_2 = 109600 - 54650$$ $$s_2 = 54950$$ $$s_2 = 54950 = \frac{100}{2} [(99 + \ell) + (990 + \ell)]$$ $$1099 = 2\ell + 1089$$ $$\ell = 5$$
Maths · Matrices · Numerical
Let $M$ be any $3 \times 3$ matrix with entries from the set $\{0,1,2\}$. The maximum number of such matrices, for which the sum of diagonal elements of $M^\top M$ is seven, is
Answer: 540
Given the matrices: $$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \end{bmatrix}$$ The equation is: $$a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 + i^2 = 7$$ Case I: Seven (1's) and two (0's) $$^9C_2 = 36$$ Case II: One (2) and three (1's) and five (0's) $$\frac{9!}{5!3!} = 504$$ Therefore, Total = 540
Maths · Conic Sections · Numerical
If one of the diameters o of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$ is a chord of another circle $C'$ whose center is at $(2, 1)$, then its radius is
Answer: 3
The distance between $(1,3)$ and $(2,1)$ is $\sqrt{5}$. Therefore, $$(\sqrt{5})^2 + (2)^2 = r^2$$ which implies $$r = 3$$
Maths · Applications of Derivatives · Numerical
The minimum value of $\alpha$ for which the equation $\frac{4}{\sin x} + \frac{1}{1 - \sin x} = \alpha$ has at least one solution in $\left(0, \frac{\pi}{2}\right)$ is
Answer: 9
Given $f(x) = \frac{4}{\sin x} + \frac{1}{1 - \sin x}$. Let $\sin x = t$. Therefore, $x \in \left(0, \frac{\pi}{2}\right) \Rightarrow 0 < t < 1$. Then $f(t) = \frac{4}{t} + \frac{1}{1-t}$. The derivative is $f'(t) = \frac{-4}{t^2} + \frac{1}{(1-t)^2}$. Simplifying, $$f'(t) = \frac{t^2 - 4(1-t)^2}{t^2(1-t)^2}$$ $$= \frac{(t - 2(1-t))(t + 2(1-t))}{t^2(1-t)^2}$$ $$= \frac{(3t - 2)(2-t)}{t^2(1-t)^2}$$ The minimum $f_{\min}$ occurs at $t = \frac{2}{3}$. Thus, $\alpha_{\min} = f\left(\frac{2}{3}\right) = \frac{4}{\frac{2}{3}} + \frac{1}{1 - \frac{2}{3}}$. This simplifies to $= 6 + 3$. Therefore, $= 9$.
Maths · Inverse Trigonometric Functions · Subjective
$\displaystyle \lim_{n \to \infty} \tan \left\{ \sum_{r=1}^{n} \tan^{-1}\left(\frac{1}{1+r+r^2}\right) \right\}$ is equal to
Answer: 1
Given $$\tan\left(\lim_{n \to \infty} \sum_{r=1}^{n} \left[ \tan^{-1}(r+1) - \tan^{-1}(r) \right]\right)$$ This simplifies to $$= \tan\left(\lim_{n \to \infty} \left( \tan^{-1}(n+1) - \frac{\pi}{4} \right)\right)$$ Finally, $$= \tan\left(\frac{\pi}{4}\right) = 1$$
Maths · Vector Algebra · Numerical
Let three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ be such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}, \vec{a} \cdot \vec{c} = 7$ and $\vec{b}$ is perpendicular to $\vec{c}$, where $\vec{a} = -\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{k}$, then the value of $2 \left| \vec{a} + \vec{b} + \vec{c} \right|^2$ is
Answer: 75
Given $\vec{C} = \lambda (\vec{b} \times (\vec{a} \times \vec{b}))$. $$= \lambda ((\vec{b} \cdot \vec{b}) \vec{b} - (\vec{b} \cdot \vec{a}) \vec{b})$$ $$= \lambda (5(-\hat{i} + \hat{j} + \hat{k}) + 2 \hat{i} + \hat{k})$$ $$= \lambda (-3 \hat{i} + 5 \hat{j} + 6 \hat{k})$$ Given $\vec{C} \cdot \vec{a} = 7 \Rightarrow 3 \lambda + 5 \lambda + 6 \lambda = 7$. $$\lambda = \frac{1}{2}$$ Therefore, $$: 2 \left| \left( \frac{-3}{2} - 1 + 2 \right) \hat{i} + \left( \frac{5}{2} + 1 \right) \hat{j} + (3 + 1 + 1) \hat{k} \right|$$ $$= 2 \left( \frac{1}{4} + \frac{49}{4} + 25 \right) = 25 + 50 = 75$$
Maths · Probability · Numerical
Let $B_i (i = 1, 2, 3)$ be three independent events in a sample space. The probability that only $B_1$ occur is $\alpha$ , only $B_2$ occurs is $\beta$ and only $B_3$ occurs is $\gamma$. Let $p$ be the probability that none of the events $B_i$ occurs and these 4 probabilities satisfy the equations ($\alpha$ - 2$\beta$)p = $\alpha$ $\beta$ and ($\beta$ - 3$\gamma$)p = 2$\beta$ $\gamma$ (All the probabilities are assumed to lie in the interval (0,1)$\)$. Then $\frac{\mathrm{P}(B_1)}{\mathrm{P}(B_3)}$ is equal to
Answer: 6
Let $x,\;y,\;z$ be probability of $B_1,\;B_2,\;B_3$ respectively. $\Rightarrow x(1-y)(1-z)=\alpha$ $\Rightarrow y(1-z)(1-x)=\beta$ $\Rightarrow z(1-x)(1-y)=\gamma$ $\Rightarrow (1-x)(1-y)(1-z)=p$ $(\alpha-2\theta p)=\alpha\theta$ $\Big(x(1-y)(1-z)-2y(1-x)(1-z)\Big)(1-x)(1-y)(1-z)=xy(1-x)(1-y)(1-z)$ $\Rightarrow x-xy-2y+2xy=xy$ $\Rightarrow x=2y \qquad \ldots (1)$ Similarly, $(\beta-3\eta p)=2\beta\eta$ $\Rightarrow y=3z \qquad \ldots (2)$ From (1) and (2), $x=6z$ Now, $\dfrac{x}{z}=6$
Maths · Matrices · Numerical
Let $P = \begin{pmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{pmatrix}$, where $\alpha \in \mathbb{R}$. Suppose $Q = [q_{ij}]$ is a matrix satisfying $PQ = kI_3$ for some non-zero $k \in \mathbb{R}$. If $q_{23} = -\frac{k}{8}$ and $|Q| = \frac{k^2}{2}$, then $\alpha^2 + k^2$ is equal to
Answer: 17
As $PQ = KI \Rightarrow Q = kP^{-1}I$. Now $Q = \frac{k}{|P|}(adjP)I$. Therefore, $$Q = \frac{k}{(20 + 12\alpha)} \begin{bmatrix} - & - & - \\ - & - & (-3\alpha - 4) \\ - & - & - \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ Thus, $q_{23} = \frac{-k}{8}$. Therefore, $$\frac{k}{(20 + 12\alpha)}(-3\alpha - 4) = \frac{-k}{8}$$ $$\Rightarrow 2(3\alpha + 4) = 5 + 3\alpha$$ $$3\alpha = -3 \Rightarrow \alpha = -1$$ Also, $|Q| = \frac{k^3 |I|}{|P|}$. Therefore, $$\frac{k^2}{2} = \frac{k^3}{(20 + 12\alpha)}$$ $$(20 + 12\alpha) = 2k \Rightarrow 8 = 2k$$ $$\Rightarrow k = 4$$
Physics · Thermodynamics · Single correct
n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes - A $\to$ B : Isothermal expansion at temperature T so that the volume is doubled from $V_1$ to $V_2 = 2 V_1$ and pressure changes from $P_1$ to $P_2$. B $\to$ C : Isobaric compression at pressure $P_2$ to initial volume $V_1$. C $\to$ A : Isochoric change leading to change of pressure from $P_2$ to $P_1$. Total workdone in the complete cycle ABCA is -
Answer: (a)
$A\rightarrow B=$ isothermal process $B\rightarrow C=$ isobaric process $C\rightarrow A=$ isochoric process Also, $V_{2}=2V_{1}$ Work done by gas in the complete cycle $ABCA$ is $W=W_{AB}+W_{BC}+W_{CA} \qquad \cdots (1)$ $\Rightarrow W_{CA}=0$ as it is an isochoric process $\Rightarrow W_{AB} =P_{1}V_{1}\ln\left(\frac{V_{2}}{V_{1}}\right) =nRT\ln(2)$ $\Rightarrow W_{BC} =P_{2}(V_{1}-V_{2}) =P_{2}(V_{1}-2V_{1}) =-P_{2}V_{1} =-nRT$ Now put the values of $W_{AB},\ W_{BC}$ and $W_{CA}$ in equation (1), we get $w=nRT\ln(2)-nRT+0$ $\Rightarrow w=nRT\,[\ln(2)-1]$ $\Rightarrow w=nRT\left[\ln(2)-\frac{1}{2}\right]$
Physics · Ray Optics and Optical Instruments · Single correct
The focal length $f$ is related to the radius of curvature $r$ of the spherical convex mirror by -
Answer: (c)
So, \[ \frac{R}{2} = f. \] Hence, \[ f = +\frac{R}{2}. \]
Physics · Wave Optics · Single correct
In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
Answer: (a)
Given: $\omega_2 = 3\omega_1$. Also, $A \propto \omega$. $$\frac{\omega_1}{\omega_2} = \frac{1}{3} \cdots (1)$$ Assume $\omega_1 = x$, $\omega_2 = 3x$. We know that $$I_{\max} = (A_1 + A_2)^2, I_{\min} = (A_1 - A_2)^2$$ $$\frac{A_1}{A_2} = \frac{\omega_1}{\omega_2} \cdots (2)$$ From equation (2) we can say that $A_1 = A$ and $A_2 = 3A$. Now, $$\frac{I_{\max}}{I_{\min}} = \frac{(A + 3A)^2}{(A - 3A)^2} = \frac{16A^2}{4A^2} = \frac{4}{1}$$ $$\Rightarrow \frac{I_{\max}}{I_{\min}} = \frac{4}{1}$$
Physics · Gravitation · Single correct
Two stars of masses $m$ and $2m$ at a distance $d$ rotate about their common centre of mass in free space. The period of revolution is -
Answer: (a)
Given $G(m)(2m) \over d^2 = m \omega^2 \times \frac{2d}{3}$. Therefore, $$\frac{2Gm}{d^2} = \omega^2 \times \frac{2d}{3}$$ which implies $$\omega^2 = \frac{3Gm}{d^3}$$ and thus $$\omega = \sqrt{\frac{3Gm}{d^3}}$$. We know that $\omega = \frac{2\pi}{T}$ so $T = \frac{2\pi}{\omega}$. Therefore, $$T = \frac{2\pi}{\sqrt{\frac{3Gm}{d^3}}} \Rightarrow T = 2\pi \sqrt{\frac{d^3}{3Gm}}.$$
Physics · Current Electricity · Single correct
A current through a wire depends on time as $i = \alpha_0 t + \beta t^2$ where $\alpha_0 = 20 \, \mathrm{A/s}$ and $\beta = 8 \, \mathrm{As^{-2}}$. Find the charge crossed through a section of the wire in 15 s.
Answer: (d)
Given: $i = \alpha t + \beta t^2$ $\alpha = 20 \, \mathrm{A/s}$ and $\beta = 8 \, \mathrm{As^{-2}}$ $t = 15 \, \mathrm{sec}$ We know that, $i = \frac{dq}{dt}$ implies $\int_0^t i \, dt = \int_0^Q dq$ $$\int_0^{15} (\alpha t + \beta t^2) \, dt = \int_0^Q dq$$ $$\Rightarrow Q = \left[ \frac{\alpha t^2}{2} + \frac{\beta t^3}{3} \right]_0^{15}$$ $$\Rightarrow Q = \frac{20 \times 15 \times 15}{2} + \frac{8 \times 15 \times 15 \times 15}{3} - 0$$ $$\Rightarrow Q = 11250 \, \mathrm{c}$$
Physics · System of Particles and Rotational Motion · Single correct
Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as - $I_1$ = M.I. of thin circular ring about its diameter, $I_2$ = M.I. of circular disc about an axis perpendicular to disc and going through the centre, $I_3$ = M.I. of solid cylinder about its axis and $I_4$ = M.I. of solid sphere about its diameter. Then :-
Answer: (d)
Given $I_1 = M.I.$ of thin circular ring about its diameter, $I_2 = M.I.$ circular disc about an axis perpendicular to disc and going through the centre. $I_3 = M.I.$ of solid cylinder about its axis. $I_4 = M.I.$ of solid sphere about its diameter. We know that, $$I_1 = \frac{MR^2}{2}, \ I_2 = \frac{MR^2}{2}, \ I_3 = \frac{MR^2}{2}$$ $$I_4 = \frac{2}{5}MR^2$$ So, $I_1 = I_2 = I_3 > I_4$
Physics · Dual Nature of Radiation and Matter · Single correct
Given below are two statements: Statement-I: Two photons having equal linear momenta have equal wavelengths. Statement-II: If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease. In the light of the above statements, choose the correct answer from the options given below.
Answer: (d)
By theory
Physics · Oscillations · Single correct
In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be -
Answer: (a)
We know that $\omega = \sqrt{\frac{k}{m}}$ and $\omega_i = \sqrt{\frac{k}{M}}$, $A_i = A$. Also, momentum is conserved just before and just after the block of mass $(m)$ is placed because there is no impulsive force. So - $$MA_i \omega_i = (M + m)v'$$ $$v' = \frac{MA_i \omega_i}{(M+m)} \Rightarrow v' = A_f \omega_f$$ $$\frac{MA_i \omega_i}{(M+m)} = A_f \sqrt{\frac{K}{M}}$$ $$\Rightarrow \frac{MA_i}{M+m} \times \sqrt{\frac{M+m}{K}} = A_f$$ $$\Rightarrow A_f = A \sqrt{\frac{M}{(M+m)}}$$
Physics · Mechanical Properties of Solids · Single correct
If $Y$, $K$ and $\eta$ are the values of Young's modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.
Answer: (a)
Given $y = 3k(1 - 2\sigma)$. $$\Rightarrow \sigma = \frac{1}{2} \left(1 - \frac{y}{3k}\right) \cdots (1)$$ Given $y = 2\eta(1 + \sigma)$. $$\Rightarrow \sigma = \frac{y}{2\eta} - 1 \cdots (2)$$ By comparing equation (1) and (2), we get $$\Rightarrow \frac{y}{2\eta} - 1 = \frac{1}{2} \left(1 - \frac{y}{3k}\right)$$ $$\Rightarrow \frac{y}{\eta} - 2 = 1 - \frac{y}{3k}$$ $$\Rightarrow \frac{y}{\eta} = 1 + 2 - \frac{y}{3k} \Rightarrow \frac{y}{\eta} = 3 - \frac{y}{3k}$$ $$\Rightarrow \frac{y}{3k} = 3 - \frac{y}{\eta} \Rightarrow \frac{y}{3k} = \frac{3\eta - y}{\eta}$$ $$\Rightarrow k = \frac{\eta y}{9\eta - 3y}$$
Physics · Atoms · Single correct
In the given figure, the energy levels of hydrogen atom have been shown along with some transitions marked A, B, C, D and E. The transitions A, B and C respectively represents -
Answer: (a)
A $\rightarrow$ series limit of Lyman. B $\rightarrow$ 3^{rd} member of Balmer series. C $\rightarrow$ 2^{nd} member of Paschen series.
Physics · Gravitation · Single correct
Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be
Answer: (a)
By resolving force $F_2$, we get $F_1 + F_2 \cos 45^\circ + F_2 \cos 45^\circ$. $F_1 + 2F_2 \cos 45^\circ = F_c$. $F_c =$ centripetal force $= \frac{MV^2}{R}$. $$\frac{GM^2}{(2R)^2} + \left[ \frac{2GM^2}{(\sqrt{2}R)^2} \cos 45^\circ \right] = \frac{MV^2}{R}$$ $$\frac{GM^2}{4R^2} + \frac{2GM^2}{2\sqrt{2}R^2} = \frac{MV^2}{R}$$ $$\frac{GM}{4R} + \frac{GM}{\sqrt{2}R} = V^2$$ $$V = \sqrt{\frac{GM}{4R} + \frac{GM}{\sqrt{2}R}}$$ $$V = \sqrt{\frac{GM}{R} \left[ \frac{1+2\sqrt{2}}{4} \right]}$$ $$V = \frac{1}{2} \sqrt{\frac{GM}{R} (1 + 2\sqrt{2})}$$ (given: mass $= 1 \, \mathrm{kg}$, radius $= 1 \, \mathrm{m}$) $$\Rightarrow V = \frac{1}{2} \sqrt{G(1 + 2\sqrt{2})}$$
Physics · Motion in a Straight Line · Single correct
If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?
Answer: (a)
Given $a = \frac{dv}{dt} = slope of (v-t) curve$. If $m = +ve$, then equation of straight line is $y = mx + c \Rightarrow v = mt + c$ (for MB). If $m = -ve$, then equation of straight line is $y = -mx + c \Rightarrow v = -mt + c$ (for AM). If we differentiate equation (1) and (2), we get $a_{MB} = +ve = m$ and $a_{AM} = -ve = -m$, so graph of $(a-t)$ will be
Physics · Electrostatic Potential and Capacitance · Single correct
Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be -
Answer: (b)
Given that first connection $$\frac{1}{C_{12}} = \frac{1}{C} + \frac{1}{C} \Rightarrow C_{12} = \frac{C}{2}$$ Second connection $$C_{34} = C + C = 2C$$ Now, the ratio of equivalent capacities in the two cases will be $$\Rightarrow \frac{C_{12}}{C_{34}} = \frac{C/2}{2C} \Rightarrow \frac{C_{12}}{C_{34}} = \frac{1}{4}$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
If an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of $\beta$ will be -
Answer: (a)
Given: $\Delta I_E = 4 \, \mathrm{mA}$ $\Delta I_C = 3.5 \, \mathrm{mA}$ We know that, $\alpha = \frac{\Delta I_C}{\Delta I_E}$ $$\Rightarrow \alpha = \frac{3.5}{4} = \frac{7}{8}$$ Also, $\beta = \frac{\alpha}{1 - \alpha}$, so $$\beta = \frac{\frac{7}{8}}{1 - \frac{7}{8}} = \frac{7}{1}$$ $$\beta = 7$$
Physics · Thermodynamics · Single correct
Match List I with List II List-I (a) Isothermal (b) Isochoric (c) Adiabatic (d) Isobaric List-II (i) Pressure constant (ii) Temperature constant (iii) Volume constant (iv) Heat content is constant Choose the correct answer from the options given below -
Answer: (b)
By theory in isothermal process, temperature is constant. In isochoric process, volume is constant. In adiabatic process, heat content is constant. In isobaric process, pressure is constant.
Physics · Thermal Properties of Matter · Single correct
Each side of a box made of metal sheet in cubic shape is 'a' at room temperature 'T', the coefficient of linear expansion of the metal sheet is ' $\alpha$ '. The metal sheet is heated uniformly, by a small temperature $\Delta$ T, so that its new temperature is T + $\Delta$ T. Calculate the increase in the volume of the metal box-
Answer: (c)
Volume expansion $\gamma = 3\alpha$. $$\frac{\Delta V}{V} = \gamma \Delta T$$ $$\Delta V = V \cdot \gamma \Delta T$$ $$\Delta V = a^3 \cdot 3\alpha \Delta T$$
Physics · Current Electricity · Single correct
A cell $E_1$ of emf $6 \, \mathrm{V}$ and internal resistance $2\Omega$ is connected with another cell $E_2$ of emf $4 \, \mathrm{V}$ and internal resistance $8\Omega$ (as shown in the figure). The potential difference across points $X$ and $Y$ is -
Answer: (c)
The emf of $E_1 = 6\, \mathrm{V}$. $r_1 = 2\, \Omega$ The emf of $E_2 = 4\, \mathrm{V}$. $r_2 = 8\, \Omega$ $|v_x - v_y|$ is the potential difference across points $x$ and $y$. $E_{eff} = 6 - 4 = 2\, \mathrm{V}$ $R_{eq} = 2 + 8 = 10\, \Omega$ So, the current in the circuit will be $$I = \frac{E_{eff}}{R_{eq}} \Rightarrow I = \frac{2}{10} = 0.2\, \mathrm{A}$$ Now, the potential difference across points $X$ and $Y$ is $|v_x - v_y| = E + iR$. $$\Rightarrow |v_x - v_y| = 4 + 0.2 \times 8 = 5.6\, \mathrm{V}$$ $$\Rightarrow |v_x - v_y| = 5.6\, \mathrm{V}$$
Physics · Electric Charges and Fields · Single correct
A cube of side 'a' has point charges +Q located at each of its vertices except at the origin where the charge is -Q. The electric field at the centre of cube is:
Answer: (c)
If only $+Q$ charges are placed at the corners of a cube of side $a$, then the electric field at the centre of the cube will be zero. But in the given condition, one $(-Q)$ is placed at one corner of the cube, so here $E_1 = E_6$, $E_2 = E_5$ and $E_3 = E_8$ (So it will cancel out each other, so electric field at centre is due to $Q_4$ and $Q_7$). Here electric field at centre $= 2(E. f)_4$. As, $|E_4| = |E_7|$ $$(E. F)_C = \frac{2kQ}{\left(\frac{\sqrt{3}a}{2}\right)^2} = \frac{8\, kQ}{3a^2} \left\{\cdot K = \frac{1}{4\pi \varepsilon_0}\right\}$$ $$(E. F)_C = \frac{2Q}{3a^2 \varepsilon_0}$$ In vector form $\Rightarrow \vec{E} = \frac{-2Q}{3a^2 \varepsilon_0} \times \left(\frac{\hat{x} + \hat{y} + \hat{z}}{\sqrt{3}}\right)$$
Physics · Gravitation · Single correct
Consider two satellites $S_1$ and $S_2$ with periods of revolution 1 hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite $S_1$ to the angular velocity of satellite $S_2$ is -
Answer: (a)
We know that $\omega = \frac{2\pi}{T}$. Given: Ratio of time period $$\frac{T_1}{T_2} = \frac{1}{8}$$ $$\Rightarrow \omega \propto \frac{1}{T}$$ $$\Rightarrow \frac{\omega_1}{\omega_2} = \frac{T_2}{T_1}$$ $$\Rightarrow \frac{\omega_1}{\omega_2} = \frac{8}{1}$$ $$\Rightarrow \omega_1 : \omega_2 = 8 : 1$$
Physics · Physical World, Units and Measurements · Single correct
The workdone by a gas molecule in an isolated system is given by, $W = \alpha \beta^2 e^{-\frac{x^2}{\alpha k T}}$, where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature. $\alpha$ and $\beta$ are constants. Then the dimensions of $\beta$ will be -
Answer: (c)
Given: work = $\alpha$ $\cdot$ $\beta$^2 $\cdot$ e^{-$\frac{x^2}{\alpha \cdot k \cdot T}$}. $k$ = Boltzmann constant. $T$ = temperature, $x$ = displacement. We know that, $\frac{x^2}{\alpha \cdot k \cdot T}$ = dimensionless. $$\left[ \frac{x^2}{\alpha \cdot k \cdot T} \right] = \left[ M^0 L^0 T^0 \right]$$ $$[\alpha] = \left[ \frac{L^2}{K \cdot T} \right]$$ $$\Rightarrow [K] = \left[ M^1 L^2 T^{-2} K^{-1} \right]$$ $$[T] = [K]$$ $$\Rightarrow [\alpha] = \left[ \frac{L^2}{M^1 L^2 T^{-2} K^{-1} \times K} \right] \Rightarrow [\alpha] = \left[ M^{-1} T^2 \right]$$ $$\Rightarrow \omega = \alpha \cdot \beta^2$$ $$\Rightarrow \left[ \frac{M^1 L^1 T^{-2}}{M^{-1} T^2} \right] = [\beta^2] = \left[ M^2 L^2 T^{-4} \right]$$ $$[\beta] = \left[ M L T^{-2} \right]$$
Physics · Laws of Motion · Numerical
The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be N [g = 10 $\mathrm{ms}^{-2}$]
Answer: 25
Given: $\mu_s = 0.2$ $m = 0.5 \, \mathrm{kg}$ $g = 10 \, \mathrm{m/s^2}$ we know that $$f_s = \mu N \ldots (1)$$ To keep the block adhere to the wall here $N = F \ldots (2)$ $$f_s = mg \ldots (3)$$ from equation (1), (2), and (3), we get $$mg = \mu F$$ $$\Rightarrow F = \frac{mg}{\mu} \Rightarrow F = \frac{0.5 \times 10}{0.2}$$ $$F = 25 \, \mathrm{N}$$
Physics · Alternating Current · Numerical
A resonance circuit having inductance and resistance $2 \times 10^{-4} \, \mathrm{H}$ and $6.28 \, \Omega$ respectively oscillates at $10 \, \mathrm{MHz}$ frequency. The value of quality factor of this resonator is $[\pi = 3.14]$
Answer: 2000
Given: $R = 6.28 \, \Omega$ $f = 10 \, \mathrm{MHz}$ $L = 2 \times 10^{-4} \, \mathrm{Henry}$ We know that quality factor $Q$ is given by $$\Rightarrow Q = \frac{X_L}{R} = \frac{\omega L}{R}$$ Also, $\omega = 2 \pi f$, so $$\Rightarrow Q = \frac{2 \pi f L}{R}$$ $$\Rightarrow Q = \frac{2 \pi \times 10 \times 10^6 \times 2 \times 10^{-4}}{6.28} = 2000$$ $Q = 2000$
Physics · Mechanical Properties of Fluids · Numerical
A hydraulic press can lift 100 kg when a mass 'm' is placed on the smaller piston. It can lift ____ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass 'm' on the smaller piston.
Answer: 25600
Atmospheric pressure $P_0$ will be acting on both the limbs of hydraulic lift. Applying Pascal's law for same liquid level $$\Rightarrow P_0 + \frac{mg}{A_1} = P_0 + \frac{(100)g}{A_2}$$ $$\Rightarrow \frac{Mg}{A_1} = \frac{(100)g}{A_2} \Rightarrow \frac{m}{100} = \frac{A_1}{A_2} \ldots (1)$$ Diameter of piston on side of 100 kg is increased by 4 times so new area $= 16 \, A_2$ Diameter of piston on side of $(m)\, \mathrm{kg}$ is decreasing $$A_1 = \frac{A_1}{16}$$ (In order to increasing weight lifting capacity, diameter of smaller piston must be reduced) Again, $$\frac{mg}{\left(\frac{A_1}{16}\right)} = \frac{M'g}{16A_2} \Rightarrow \frac{256m}{M'} = \frac{A_1}{A_2}$$ From equation (1) $= \frac{256m}{M'} = \frac{m}{100} \Rightarrow \therefore M' = 25600 \, \mathrm{kg}$
Physics · Laws of Motion · Numerical
An inclined plane is bent in such a way that the vertical cross-section is given by $y = \frac{x^2}{4}$ where $y$ is in vertical and $x$ in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction $\mu = 0.5$, the maximum height in cm at which a stationary block will not slip downward is ____ cm.
Answer: 25
Given $y = \frac{x^2}{4}$ and $\mu = 0.5$. The condition for the block not to slip downward is $mg \sin \theta = \mu mg \cos \theta$. This implies $\tan \theta = \mu$. We know that $\tan \theta = \frac{dy}{dx}$. Therefore, $$\frac{dy}{dx} = \mu \Rightarrow \frac{x}{2} = 0.5$$ which gives $x = 1$. Substituting $x = 1$ in the equation $y = \frac{x^2}{4}$, we get $$y = \frac{(1)^2}{4} \Rightarrow y = \frac{1}{4} \Rightarrow y = 0.25$$ Thus, $y = 25 \, \mathrm{cm}$.
Physics · Electromagnetic Waves · Numerical
An electromagnetic wave of frequency $5 \, \mathrm{GHz}$, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is ____ $\times 10^7 \, \mathrm{m/s}$
Answer: 15
Given: $f = 5 \, \mathrm{GHz}$ $\varepsilon_r = 2$ $\mu_r = 2$ Velocity of wave $\Rightarrow v = \frac{c}{n}$ ...(1) Where, $n = \sqrt{\mu_r \varepsilon_r}$ and $c = speed of light = 3 \times 10^8 \, \mathrm{m/s}$ $n = \sqrt{2 \times 2} = 2$ Put the value of $n$ in we get $$\Rightarrow v = \frac{3 \times 10^8}{2} = 15 \times 10^7 \, \mathrm{m/s}$$ $$\Rightarrow X \times 10^7 = 15 \times 10^7$$ $$X = 15$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Numerical
In connection with the circuit drawn below, the value of current flowing through $2\,\mathrm{k}\Omega$ resistor is ____ $\times 10^{-4}\,\mathrm{A}$
Answer: 25
In zener diode there will be no change in current after 5 V zener diode breakdown $\Rightarrow i = \dfrac{5}{2 \times 10^3}$ $\Rightarrow i = 2.5 \times 10^{-3}\,\text{A}$ $\Rightarrow i = 25 \times 10^{-4}\,\text{A}$
Physics · Communication Systems · Numerical
An audio signal $v_m = 20 \sin 2\pi (1500t)$ amplitude modulates a carrier $v_c = 80 \sin 2\pi (100,000t)$. The value of percent modulation is
Answer: 25
Given: $v_m = 20 \sin \left[ 100 \pi t + \frac{\pi}{4} \right]$ $v_C = 80 \sin \left[ 10^4 \pi t + \frac{\pi}{6} \right]$ We know that, modulation index $= \frac{A_m}{A_c}$ From given equations, $A_m = 20$ and $A_c = 80$ Percentage modulation index $= \frac{A_m}{A_c} \times 100$ $$\Rightarrow \frac{20}{80} \times 100 = 25\%$$ The value of percentage modulation index is $$= 25$$
Physics · Work, Energy and Power · Numerical
A ball with a speed of $9 \, \mathrm{m/s}$ collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of $30^\circ$ with the original direction. The ratio of velocities of the balls after collision is $x : y$, where $x$ is
Answer: 1
Momentum is conserved just before and just after the collision in both $x$-$y$ direction. In $y$-direction $p_i = 0$ $$P_f = m \times \frac{1}{2} v_1 - m \times \frac{1}{2} v_2$$ $p_i = P_f$, so $$\frac{mv_1}{2} - \frac{mv_2}{2} = 0$$ $$\Rightarrow \frac{mv_1}{2} = \frac{mv_2}{2} \Rightarrow v_1 = v_2$$ $$\frac{v_1}{v_2} = 1$$
Physics · Alternating Current · Numerical
A common transistor radio set requires 12 V (D.C.) for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220 V (A.C.) on standard domestic A.C. supply. The number of turns of secondary coil are 24, then the number of turns of primary are ____
Answer: 440
Given Primary voltage, $V_p = 220 \, \mathrm{V}$ Secondary voltage, $v_s = 12 \, \mathrm{V}$ No. of turns in secondary coil is $N_s = 24$ No. of turns in primary coil, $N_p = ?$ We know that for a transformer $$\frac{N_p}{N_s} = \frac{V_p}{V_s}$$ $$\Rightarrow N_p = \frac{V_p \times N_s}{V_s} = \frac{220 \times 24}{12}$$ $$\Rightarrow N_p = 440$$
Physics · Wave Optics · Numerical
An unpolarized light beam is incident on the polarizer of a polarization experiment and the intensity of light beam emerging from the analyzer is measured as 100 Lumens. Now, if the analyzer is rotated around the horizontal axis (direction of light) by $30^\circ$ in clockwise direction, the intensity of emerging light will be ____ Lumens.
Answer: 75
Given: $I_0 = 100$ lumens, $\theta = 30$. $I_{net} = I_0 \cos^2 \theta$ $$I_{net} = 100 \times \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{100 \times 3}{4}$$ $I_{net} = 75$ lumens
Chemistry · Haloalkanes and Haloarenes · Single correct
The product formed in the first step of the reaction of with excess Mg/Et$_2$O (Et = C$_2$H$_5$) is :
Answer: (c)
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Consider the elements Mg, Al, S, P and Si, the correct increasing order of their first ionization enthalpy is:
Answer: (a)
Order of IE, in 3rd period is Na $\mathrm{Al}$ $\mathrm{S}$ < $\mathrm{Cl}$ < $\mathrm{Ar}$ Na < $\mathrm{Al}$ < $\mathrm{Mg}$ < $\mathrm{Si}$ < $\mathrm{S}$ < $\mathrm{P}$ < $\mathrm{Cl}$ < $\mathrm{Ar}$
Chemistry · Amines · Single correct
'A' and 'B' in the following reactions are :
Answer: (c)
The given reaction sequence involves the conversion of aniline to a diazonium salt using $\mathrm{NaNO_2/HCl}$. The diazonium salt is then reacted with $\mathrm{KCN}$ to form a benzonitrile. This benzonitrile is then reduced using $\mathrm{SnCl_2/HCl}$ in the presence of $\mathrm{H_3O^+}$ to form a benzaldehyde. This sequence of reactions is known as Stephen's reaction.
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Which of the following ore is concentrated using group 1 cyanide salt?
Answer: (a)
Concentration of sphalerite, first by cyanide salt as a depressant to remove the impurity of galena. $$\mathrm{ZnS + PbS + NaCN \rightarrow Na_2[Zn(CN)_4] + PbS \uparrow}$$
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
$Al_2O_3$ was leached with alkali to get X. The solution of X on passing of gas Y, forms Z. X, Y and Z respectively are:
Answer: (a)
Given the reactions: (1) $\mathrm{Al_2O_3} + \mathrm{NaOH} \rightarrow \mathrm{Na[Al(OH)_4]}$ (2) $\mathrm{Na[Al(OH)_4]} \xrightarrow{CO_2} \mathrm{Al(OH)_3}$ or $\mathrm{Al_2O_3, xH_2O}$
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Which of the following are isostructural pairs? (A) $\mathrm{SO_4^{2-}}$ and $\mathrm{CrO_4^{2-}}$ (B) $\mathrm{SiCl_4}$ and $\mathrm{TiCl_4}$ (C) $\mathrm{NH_3}$ and $\mathrm{NO_3^-}$ (D) $\mathrm{BCl_3}$ and $\mathrm{BrCl_3}$
Answer: (b)
(a) $\mathrm{SO_4^{2-}}$ and $\mathrm{CrO_4^{2-}}$ both have tetrahedral structure. (b) $\mathrm{SiCl_4}$ and $\mathrm{TiCl_4}$ both have tetrahedral structure also.
Chemistry · Alcohols, Phenols and Ethers · Single correct
Which is the final product (major) 'A' in the given reaction?
Answer: (c)
The reaction begins with the protonation of the alcohol group, forming a good leaving group. The molecule then loses water, resulting in the formation of a carbocation. A 1,2-hydride shift occurs to form a more stable carbocation. Finally, chloride ion attacks the carbocation, resulting in the final product, which is the most stable carbocation.
Chemistry · Hydrocarbons · Single correct
In the following reaction the reason why meta-nitro product also formed is:
Answer: (a)
In acidic medium the $\mathrm{-NH_2}$ group in aniline converts into anilinium ion which is meta directing.
Chemistry · Surface Chemistry · Single correct
In Freundlich adsorption isotherm, slope of AB line is
Answer: (a)
Freundlich adsorption isotherm is: $$\frac{x}{m} = kp^{1/n}$$ where $x$ is the mass of adsorbate, $m$ is the mass of adsorbent, and $P$ is the equilibrium pressure. The equation can be expressed as: $$k_{1n} = \frac{1}{n} \log p + \log k$$ Comparing with $y = mx + c$, we have: $$m = \frac{1}{n} = slope \left[ \frac{1}{n} = 0 to 1 \right]$$ where $n > 1$.
Chemistry · Redox Reactions · Single correct
(A) $\mathrm{HOCl} + \mathrm{H_2O_2} \rightarrow \mathrm{H_3O^+} + \mathrm{Cl^-} + \mathrm{O_2}$ (B) $\mathrm{I_2} + \mathrm{H_2O_2} + 2\mathrm{OH^-} \rightarrow 2\mathrm{I^-} + 2\mathrm{H_2O} + \mathrm{O_2}$ Choose the correct option.
Answer: (c)
When $\mathrm{H_2O_2}$ acts as a reducing agent it liberates the $\mathrm{O_2}$. $$\mathrm{H_2O_2} \rightleftharpoons 2\mathrm{H^+} + \mathrm{O_2} + 2e^-$$
Chemistry · Hydrocarbons · Single correct
What is the major product formed by HI on reaction with
Answer: (c)
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Which of the following reagent is used for the following reaction? $$\mathrm{CH_3CH_2CH_3} \xrightarrow{?} \mathrm{CH_3CH_2CHO}$$ The reaction given is: $$\mathrm{CH_3 - CH_2 - CH_3 \xrightarrow{MO_2O_3}} {CH_3 - CH_2 - CH = O}$$
Answer: (b)
The reaction given is: $$\mathrm{CH_3 - CH_2 - CH_3 \xrightarrow{MO_2O_3} CH_3 - CH_2 - CH = O}$$
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Given below are two statements: Statement I: Colourless cupric metaborate is reduced to cuprous metaborate in a luminous flame. Statement II: Cuprous metaborate is obtained by heating boric anhydride and copper sulphate in a non-luminous flame. In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (d)
Both are False (1) Copper sulphate form copper meta boric with beric an hydride $\mathrm{CuSO_4} \rightarrow \mathrm{CuO} + \mathrm{SO_3}$ $$\mathrm{CuO} + \mathrm{B_2O_3} \rightarrow \mathrm{Cu(BO_2)_2}$$ blue in cold oxidising flame (non luminous flame) (2) Blue coloured metal borate is reduced to copper in a luminous flame.
Chemistry · Biomolecules · Single correct
Out of the following, which type of interaction is responsible for the stabilisation $\alpha$-helix structure of proteins?
Answer: (b)
The $\alpha$-helix is stabilized by hydrogen bond between the NH and CO group of the main chain.
Chemistry · Polymers · Single correct
Match List I with List II. \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List I} & \multicolumn{2}{c|}{List II} \\ \multicolumn{2}{|c|}{(Monomer Unit)} & \multicolumn{2}{c|}{(Polymer)} \\ \hline (a) & Caprolactum & (i) & Natural rubber \\ \hline (b) & 2-Chloro-1,3-butadiene & (ii) & Buna-N \\ \hline (c) & Isoprene & (iii) & Nylon 6 \\ \hline (d) & Acrylonitrile & (iv) & Neoprene \\ \hline \end{tabular} Choose the correct answer from the options given below :
Answer: (a)
1. Polymer of caprolactum is nylon-6. 2. Polymer of 2-chloro-1,3-butadiene is neoprene. 3. Polymer of isoprene is natural rubber. 4. Polymer of acrylonitrile and 1,3-butadiene is buna-N.
Chemistry · Environmental Chemistry · Single correct
The gas released during anaerobic degradation of vegetation may lead to:
Answer: (a)
Biogas is the mixture of gases produced by the breakdown of organic matter in the absence of oxygen (anaerobically), primarily consisting of methane and carbon dioxide. Biogas can be produced from raw material such as agricultural waste, manure, municipal waste, plant material, sewage, green waste or food waste. Due to release of $\mathrm{CH_4}$ gas during anaerobic vegetative degradation which causes global warming and cancer.
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The major components in "Gun Metal" are:
Answer: (b)
Gun metal is an alloy of copper with tin and zinc.
Chemistry · Electrochemistry · Single correct
The electrode potential of $\mathrm{M}^{2+}/\mathrm{M}$ of 3 $d$ - series elements shows positive value for:
Answer: (d)
(A) Zn $-0.76$ (B) CO $-0.28$ (C) Fe $-0.44$ (D) Cu $+0.34$
Chemistry · Hydrocarbons · Single correct
Identify products A and B.
Answer: (b)
The given reaction sequence involves two steps. In the first step, the methyl group on the cyclopentene is oxidized using dilute $\mathrm{KMnO_4}$ at $273 \, \mathrm{K}$ to form a diol. In the second step, the diol is further oxidized using $\mathrm{CrO_3}$ to form a ketone.
Chemistry · Alcohols, Phenols and Ethers · Single correct
Which of the following compound gives pink colour on reaction with phthalic anhydride in conc. $\mathrm{H_2SO_4}$ followed by treatment with $\mathrm{NaOH}$ ?
Answer: (b)
The reaction involves phenolphthalein reacting with a compound containing an OH group and a CH3 group. The product formed is a compound with a pink color.
Chemistry · Solutions · Numerical
When 9.45 g of $ClCH_2COOH$ is added to 500 $\mathrm{mL}$ of water, its freezing point drops by $0.5^\circ$ $\mathrm{C}$. The dissociation constant of $ClCH_2COOH$ is $x \times 10^{-3}$. The value of $x$ is ____ (Rounded off to the nearest integer) [ $K_{f(H_2O)}$ = 1.86 $\mathrm{K}$ \, $\mathrm{kg}$ \, $\mathrm{mol^{-1}}$ ]
Answer: 35
$\mathrm{ClCH_2COOH \rightleftharpoons ClCH_2COO^- + H^+}$ At $t=0$ $\mathrm{ClCH_2COOH}=c,\qquad \mathrm{ClCH_2COO^-}=0,\qquad H^+=0.$ At $t=t$ $\mathrm{ClCH_2COOH}=c-c\alpha$ $\mathrm{ClCH_2COO^-}=c\alpha$ $H^+=c\alpha$ Total no. of moles $c+c\alpha=c(1+\alpha)$ $i=\dfrac{\text{observed}}{\text{calculated}}$ $=\dfrac{c(1+\alpha)}{c}$ $=(1+\alpha)$ M.W. $=94.5$ $\Delta T_f=i\times k_f\times m$ $\Delta T_f=0.5^\circ\mathrm{C}$ $i=\dfrac{1+\alpha}{1}$ $m=\dfrac{\frac{45}{94.5}}{\frac{500}{1000}}$ $=\dfrac{\text{mole}}{\text{kg (Solvent)}}$ $m=1.86\ \mathrm{kg/mol}$ $0.5=(1+\alpha)\times1.86$ $(1+\alpha)=\dfrac{2.5}{1.86}$ $\alpha=\dfrac{2.5}{1.86}-1=\dfrac{32}{93}$ $K_a=\dfrac{c\alpha^2}{1-\alpha}$ $=\dfrac{0.2\times1024}{93\times93\times\frac{61}{93}}$ $K_a=0.0351=35.1\times10^{-3}$
Chemistry · Some Basic Concepts of Chemistry · Numerical
$4.5\,\mathrm{g}$ of compound A ($\mathrm{MW}=90$) was used to make $250\,\mathrm{mL}$ of its aqueous solution. The molarity of the solution in $\mathrm{M}$ is $x \times 10^{-1}$. The value of $x$ is \underline{\hspace{1cm}}. (Rounded off to the nearest integer)
Answer: 2
Moles of $A = \frac{Weight}{M.w}$ $$= \frac{4.5}{90} = \frac{1}{20} = 0.05$$ Volume (Lit) $= \frac{250}{1000} = 0.250$ lit Moles of $A = \frac{Weight}{M.w}$ $$= \frac{4.5}{90} = \frac{1}{20} = 0.05$$ Volume (Lit) $= \frac{250}{1000} = 0.250$ lit Molarity $(M) = \frac{Mole}{(Lit) volume} = \frac{0.05}{0.250} = 0.2$ $$= 2 \times 10^{-1} \frac{mol}{Lit}$$ $x = 2$ Molarity $(M) = \frac{Mole}{(Lit) volume} = \frac{0.05}{0.250} = 0.2$ $$= 2 \times 10^{-1} \frac{mol}{Lit} x = 2$$
Chemistry · Equilibrium · Numerical
At 1990 K and 1 atm pressure, there are equal number of Cl_2 molecules and Cl atoms in the reaction mixture. The value of K_p for the reaction $\mathrm{Cl_2}_{(g)} = 2\mathrm{Cl}_{(g)}$ under the above conditions is $x \times 10^{-1}$. The value of $x$ is ____. (Rounded off to the nearest integer)
Answer: 5
Given the reaction $\mathrm{Cl_2} \rightleftharpoons 2\mathrm{Cl}$. Let's consider the moles at equilibrium. The partial pressures at equilibrium are given by: $$\frac{x}{2x} \times 1$$ $$\frac{x}{2x} \times 1$$ Both simplify to $\frac{1}{2}$. Therefore, the equilibrium constant $K_p$ is calculated as: $$K_p = \frac{[P_{\mathrm{Cl}}]^2}{[P_{\mathrm{Cl_2}}]} = \left[\frac{1}{2}\right]^2 \div \frac{1}{2} = \frac{1}{2} = 0.5 = 5 \times 10^{-1}$$ Thus, $X = 5$.
Chemistry · The s-Block Elements · Single correct
Number of amphoteric compounds among the following is
Answer: (b)
BeO and Be(OH)_2 are amphoteric in nature.
Chemistry · Redox Reactions · Numerical
The reaction of sulphur in alkaline medium is given below: $$\mathrm{S_{8(s)}} + a\mathrm{OH^-_{(aq)}} \longrightarrow b\mathrm{S^{2-}_{(aq)}} + c\mathrm{S_2O_3^{2-}(aq)} + d\mathrm{H_2O_{(\ell)}}$$ The values of $a$ is
Answer: 12
Given the reactions: $$\mathrm{S_8 + aOH^- \rightarrow bs^{-2} + CdS_2O_3^{-2} + dH_2O}$$ $$\mathrm{S_8 + bOH^- \rightarrow 4S^{-2} + 2S_2O_3^{-2} + dH_2O}$$ $$\mathrm{S_8 + 12OH^- \rightarrow 4S^{-2} + 2S_2O_3^{-2} + 6H_2O}$$ From the third equation, we find that $a = 12$.
Chemistry · Equilibrium · Numerical
For the reaction $\mathrm{A}_{(g)} \rightarrow \mathrm{B}_{(g)}$, the value of the equilibrium constant at $300 \, \mathrm{K}$ and $1 \, \mathrm{atm}$ is equal to $100.0$. The value of $\Delta_r G$ for the reaction at $300 \, \mathrm{K}$ and $1 \, \mathrm{atm}$ in $\mathrm{Jmol}^{-1}$ is $-xR$, where $x$ is ____ (Rounded off to the nearest integer) $\left[ R = 8.31 \, \mathrm{J \, mol}^{-1} \, \mathrm{K}^{-1} and \ln 10 = 2.3 \right]$
Answer: 1380
Given $\Delta G^\circ = -RT \ln Keq$. $$= -R \times 300 \times \ln(10^2)$$ $$= 300 \times 2 \times 2.3 \times (-R)$$ $$= -1380R$$ Therefore, $x = 1380$ ans.
Chemistry · Structure of Atom · Numerical
A proton and a $\mathrm{Li}^{3+}$ nucleus are accelerated by the same potential. If $\lambda_{\mathrm{Li}}$ and $\lambda_{\mathrm{p}}$ denote the de Broglie wavelengths of $\mathrm{Li}^{3+}$ and proton respectively, then the value of $\frac{\lambda_{\mathrm{Li}}}{\lambda_{\mathrm{p}}}$ is $x \times 10^{-1}$. The value of $x$ is ____. [Rounded off to the nearest integer] [Mass of $\mathrm{Li}^{3+} = 8.3$ mass of proton]
Answer: 2
De Broglie Wavelength $\lambda = \frac{h}{\sqrt{2 \, mk.E}}$. $$\frac{\lambda_{\mathrm{Li}+3}}{\lambda_{p}} = \sqrt{\frac{m_{p} \times (e - v)_{p}}{m_{\mathrm{Li}+3} \times 3 e_{p} V}}$$ Given $m_{\mathrm{Li}+3} = 8.3 m_{p}$. $$\frac{\lambda_{\mathrm{Li}+3}}{\lambda_{p}} = \sqrt{\frac{m_{p}}{3 \times 8.3 m_{p}}} = \sqrt{\frac{1}{25}}$$ $$= \frac{1}{5} = 0.2 = 2 \times 10^{-1}$$ Therefore, $x = 2$.
Chemistry · Equilibrium · Numerical
The stepwise formation of $[\mathrm{Cu(NH_3)_4}]^{2+}$ is given below: $$\mathrm{Cu^{2+} + NH_3 \xrightleftharpoons{K_1} [Cu(NH_3)_2]^{2+}}$$ $$\mathrm{[Cu(NH_3)_2]^{2+} + NH_3 \xrightleftharpoons{K_2} [Cu(NH_3)_3]^{2+}}$$ $$\mathrm{[Cu(NH_3)_3]^{2+} + NH_3 \xrightleftharpoons{K_3} [Cu(NH_3)_4]^{2+}}$$ The value of stability constants $K_1$, $K_2$, $K_3$ and $K_4$ are $10^4$, $1.58 \times 10^2$, $5 \times 10^2$ and $10^2$ respectively. The overall equilibrium constants for dissociation of $[\mathrm{Cu(NH_3)_4}]^{2+}$ is $x \times 10^{-12}$. The value of $x$ is ____. (Rounded off to the nearest integer)
Answer: 1
For this: $$\mathrm{Cu^{+2} + NH_3 \overset{k_1}{\rightleftharpoons} k_1\mathrm{Cu(NH_3)}^{+2}} \ldots (1)$$ $$\mathrm{[Cu(NH_3)]^{+2} + NH_3 \overset{k_2}{\rightleftharpoons} [Cu(NH_3)_2]^{+2}} \ldots (2)$$ $$\mathrm{[Cu(NH_3)_2]^{+2} + NH_3 \overset{k_3}{\rightleftharpoons} Cu(NH_3)_3^{+2}} \ldots (3)$$ $$\mathrm{[Cu(NH_3)_3]^{+2} + NH_3 \overset{k_4}{\rightleftharpoons} [Cu(NH_3)_4]^{+2}} \ldots (4)$$ (1) + (2) + (3) + (4) $$\mathrm{Cu^{+2} + 4NH_3 \overset{k_1,k_2,k_3,k_4}{\rightleftharpoons} [C_4(NH_3)_4]^{+4}} \ldots (B)$$ So for (A) $$K = \frac{1}{k_1 \cdot K_2 \cdot K_3 \cdot K_4}$$ Putting the value of $k_1$, $k_2$, $k_3$ and $k_4$. $$K = \frac{1}{(10)^4 \cdot (1.58 \times 10^3) \cdot (5 \times 10^2) \cdot (10)^2} = 1.26 \times 10^{-12}$$ $$x = 1$$
Chemistry · The Solid State · Numerical
The coordination number of an atom in a body-centered cubic structure is ____ [Assume that the lattice is made up of atoms]
Answer: 8
Fact
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
Gaseous cyclobutene isomerizes to butadiene in a first order process which has a 'k' value of $3.3 \times 10^{-4} \, \mathrm{s}^{-1}$ at $153^\circ \mathrm{C}$. The time in minutes it takes for the isomerization to proceed 40$\%$ to completion at this temperature is ____ (Rounded off to the nearest integer)
Answer: 1
For first order reaction: $$t = \frac{2.303}{k} \log \left[ \frac{100}{100-x} \right]$$ Given $X = 40$, $k = 3.3 \times 10^{-4}$ Substitute the values: $$t = \frac{2.303}{3.3 \times 10^{-4}} \log \left[ \frac{100}{60} \right]$$ Calculate the logarithm: $$t = \frac{2.303}{3.3 \times 10^{-4}} \times 0.22$$ Simplify: $$t = 0.1535 \times 3 \times 10^{4}$$ Calculate $t$: $$t = 1535 sec$$ Convert to minutes: $$t = 1535 sec = 25.6 Min$$