JEE Main 25 July 2021 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 · Binomial Theorem · Single correct
The sum of all those terms which are rational numbers in the expansion of $\left(2^{1/3} + 3^{1/4}\right)^{12}$ is:
Answer: (d)
Given $T_{r+1} = {}^{12}C_r \left(2^{1/3}\right)^r \cdot \left(3^{1/4}\right)^{12-r}$. $T_{r+1}$ will be a rational number when $r = 0, 3, 6, 9, 12$. Therefore, $r = 0, 12$. $T_1 + T_{13} = 1 \times 3^3 + 1 \times 2^4 \times 1$ $= 24 + 16 = 43$
Maths · Statistics · Single correct
The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is:
Answer: (c)
Given $n_1 = 100$, $\overline{X}_1 = 15$, $V_1(x) = 9$. Also, $m = 250$, $\overline{X} = 15.6$, $Var(x) = 13.44$. The formula for $\sigma^2$ is $$\sigma^2 = \frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2} + \frac{n_1 n_2}{(n_1 + n_2)^2} (\overline{X}_1 - \overline{X}_2)^2.$$ Given $n_2 = 150$, $\overline{X}_2 = 16$, $V_2(x) = \sigma_2$. Substituting the values, $$13.44 = \frac{100 \times 9 + 150 \times \sigma_2^2}{250} + \frac{100 \times 150}{(250)^2} \times 1.$$ Solving this gives $$\sigma_2^2 = 16 \Rightarrow \sigma_2 = 4.$$
Maths · Integrals · Single correct
If $f(x) = \begin{cases} \int_0^x (5 + |1 - t|) \, \mathrm{dt}, & x > 2 \\ 5x + 1, & x \leq 2 \end{cases}$, then
Answer: (c)
Given $$f(x) = \int_0^1 (5 + (1-t)) dt + \int_1^x (5 + (t-1)) d$$ Simplifying, we have: $$= 6 - \frac{1}{2} + \left(4t + \frac{t^2}{2}\right) \bigg|_1^x$$ Evaluating the integrals: $$= \frac{11}{2} + 4x + \frac{x^2}{2} - 4 - \frac{1}{2}$$ Simplifying further: $$= \frac{x^2}{2} + 4x + 1$$ Now, evaluating at $x = 2^+$: $$f(2^+) = 2 + 8 + 1 = 11$$ And at $x = 2$: $$f(2) = f(2) = 5 \times 2 + 1 = 11$$ This implies continuity at $x = 2$. Clearly differentiable at $x = 1$, left derivative $Lf'(2) = 5$. Right derivative $Rf'(2) = 6$. Therefore, not differentiable at $x = 2$.
Maths · Binomial Theorem · Single correct
If the greatest value of the term independent of $x$ in the expansion of $(x \sin \alpha + a \frac{\cos \alpha}{x})^{10}$ is $\frac{10!}{(5!)^2}$, then the value of ' $a$ ' is equal to:
Answer: (d)
Given $$T_{r+1} = 10C_r (x \sin \alpha)^{10-r} \left( \frac{a \cos \alpha}{x} \right)^r$$ for $$r = 0, 1, 2, \ldots, 10$$. $$T_{i+1}$$ will be independent of $$x$$ when $$10 - 2r = 0 \Rightarrow r = 5$$. Thus, $$T_6 = 10C_5 (x \sin \alpha)^5 \times \left( \frac{a \cos \alpha}{x} \right)^5$$ $$= 10C_5 \times a^5 \times \frac{1}{25} (\sin 2\alpha)^5$$ will be greatest when $$\sin 2\alpha = 1$$. Therefore, $$\Rightarrow 10C_5 \frac{a^5}{25} = 10C_5 \Rightarrow a = 2$$.
Maths · Mathematical Reasoning · Single correct
Consider the statement "The match will be played only if the weather is good and ground is not wet". Select the correct negation from the following:
Answer: (c)
Given: $p$: weather is good $q$: ground is not wet $$\sim (p \land q) \equiv \sim p \lor \sim q$$ This is equivalent to: weather is not good or ground is wet.
Maths · Trigonometric Functions · Single correct
The value of $\cot \frac{\pi}{24}$ is:
Answer: (b)
Given $\($ $\cot$ $\theta$ = $\frac{1 + \cos 2\theta}{\sin 2\theta}$ = $\frac{1 + \left( \frac{\sqrt{3} + 1}{2\sqrt{2}} \right)}{\left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right)}$ $\)$ $\($ $\theta$ = $\frac{\pi}{24}$ $\)$ $\($ $\Rightarrow$ $\cot$ $\left$( $\frac{\pi}{24}$ $\right$) = $\frac{1 + \left( \frac{\sqrt{3} + 1}{2\sqrt{2}} \right)}{\left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right)}$ $\)$ $\($ = $\frac{(2\sqrt{2} + \sqrt{3} + 1)}{(\sqrt{3} - 1)}$ $\times$ $\frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)}$ $\)$ $\($ = $\frac{2\sqrt{6} + 2\sqrt{2} + 3 + \sqrt{3} + \sqrt{3} + 1}{2}$ $\)$ $\($ = $\sqrt{6}$ + $\sqrt{2}$ + $\sqrt{3}$ + 2 $\)$
Maths · Binomial Theorem · Single correct
The lowest integer which is greater than $\left(1 + \frac{1}{10^{100}}\right)^{10^{100}}$ is ___
Answer: (a)
Let $P = \left(1 + \frac{1}{10^{100}}\right)^{10^x}$, Let $x = 10^{100}$. Therefore, $$P = \left(1 + \frac{1}{x}\right)^x$$ implies $$P = 1 + (x)\left(\frac{1}{x}\right) + \frac{(x)(x-1)}{2} \cdot \frac{1}{x^2} + \frac{(x)(x-1)(x-2)}{3} \cdot \frac{1}{x^3} + \ldots$$ (up to $10^{100} + 1$ terms) implies $$P = 1 + 1 + \left(\frac{1}{2} - \frac{1}{2x^2}\right) + \left(\frac{1}{3} - \ldots\right) + \ldots$$ so on. Therefore, $$P = 2 + \left(Positive value less than \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots\right)$$ Also $e = 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots$ implies $$\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots = e - 2$$ Therefore, $$P = 2 + (positive value less than e - 2)$$ implies $P \in (2, 3)$. Therefore, the least integer value of $P$ is 3.
Maths · Integrals · Single correct
The value of the integral $$\int_{-1}^{1} \log \left( x + \sqrt{x^2 + 1} \right) \, dx$$ is:
Answer: (b)
Let $I = \int_{-1}^{1} \log \left( x + \sqrt{x^2 + 1} \right) \, dx$. Therefore, $\log \left( x + \sqrt{x^2 + 1} \right)$ is an odd function. Thus, $I = 0$.
Maths · Vector Algebra · Single correct
Let a, b and c be distinct positive numbers. If the vectors $a\hat{i} + a\hat{j} + c\hat{k}$, $\hat{i} + \hat{k}$ and $c\hat{i} + c\hat{j} + b\hat{k}$ are co-planar, then c is equal to:
Answer: (d)
Because vectors are coplanar Hence $$\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$$ $$\Rightarrow c^2 = ab \Rightarrow c = \sqrt{ab}$$
Maths · Sequences and Series · Single correct
If [x] be the greatest integer less than or equal to x, then $\sum_{n=8}^{100} \left[ \frac{(-1)^n n}{2} \right]$ is equal to:
Answer: (b)
The given expression is $$\sum_{n=8}^{100} \left[ \frac{(-1)^n \cdot n}{2} \right]$$ which simplifies to $$= 4 - 5 + 5 - 6 + 6 + \ldots - 50 + 50 = 4$$
Maths · Determinants · Single correct
The number of distinct real roots of $$\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$$ in the interval $$-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$$ is:
Answer: (b)
Given the determinant equation: $$\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0, -\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$$ Apply the row operations: $$R_1 \to R_1 - R_2 \& R_2 \to R_2 - R_3$$ The matrix becomes: $$\begin{vmatrix} \sin x - \cos x & \cos x - \sin x & 0 \\ 0 & \sin x - \cos x & \cos x - \sin x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$$ Simplifying further: $$\begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$$ This results in: $$(\sin x - \cos x)^2 (\sin x + 2 \cos x) = 0$$ Therefore, $$x = \frac{\pi}{4}$$
Maths · Vector Algebra · Single correct
If $|\vec{a}| = 2$, $|\vec{b}| = 5$ and $|\vec{a} \times \vec{b}| = 8$, then $|\vec{a} \cdot \vec{b}|$ is equal to:
Answer: (a)
Given $|\vec{a}| = 2$, $|\vec{b}| = 5$. The magnitude of the cross product is $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta = \pm 8$. Thus, $\sin \theta = \pm \frac{4}{5}$. Therefore, $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$. This gives $= 10 \cdot \left( \pm \frac{3}{5} \right) = \pm 6$. Hence, $|\vec{a} \cdot \vec{b}| = 6$.
Maths · Relations and Functions · Single correct
The number of real solutions of the equation, $x^2 - |x| - 12 = 0$ is:
Answer: (a)
Given $|x|^2 - |x| - 12 = 0$. $$(|x| + 3)(|x| - 4) = 0$$ $|x| = 4 \Rightarrow x = \pm 2$
Maths · Relations and Functions · Single correct
Consider function $f : A \rightarrow B$ and $g : B \rightarrow C (A, B, C \subseteq \mathbb{R})$ such that $(gof)^{-1}$ exists then:
Answer: (c)
Since $(gof)^{-1}$ exists, $gof$ is bijective. Therefore, $f'$ must be one-one and $g'$ must be onto.
Maths · Matrices · Single correct
If $\mathbf{P} = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix}$, then $\mathbf{P}^{50}$ is:
Answer: (a)
Given the matrix $P = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix}$. We calculate powers of $P$ as follows: $$P^2 = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$ $$P^3 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{3}{2} & 1 \end{bmatrix}$$ $$P^4 = \begin{bmatrix} 1 & 0 \\ \frac{3}{2} & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$$ Continuing this pattern, we find: $$\therefore P^{50} = \begin{bmatrix} 1 & 0 \\ 25 & 1 \end{bmatrix}$$
Maths · Probability · Single correct
Let x be a random variable such that the probability function of a distribution is given by $$P(X = 0) = \frac{1}{2}, P(X = j) = \frac{1}{3^j} \ (j = 1, 2, 3, \ldots, \infty)$$ Then the mean of the distribution and $P(X$ is positive and even$)$ respectively are:
Answer: (b)
mean $= \sum x_1 p_i = \sum_{r=0}^{\infty} r \cdot \frac{1}{3^r} = \frac{3}{4}$ $p(x is even) = \frac{1}{3^2} + \frac{1}{3^4} + \ldots \infty$ $= \frac{1}{9} \frac{1}{1 - \frac{1}{9}} = \frac{1/9}{8/9} = \frac{1}{8}$
Maths · Conic Sections · Single correct
If a tangent to the ellipse $x^2 + 4y^2 = 4$ meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point:
Answer: (a)
The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{1} = 1$. The equation of the tangent is $\cos \theta x + 2 \sin \theta y = 2$. The points are $B \left(-2, \frac{1+\cos \theta}{\sin \theta} \right)$ and $C \left(2, \frac{1-\cos \theta}{\sin \theta} \right)$. Alternatively, $B \left(-2, \cot \frac{\theta}{2} \right)$ and $C \left(2, \tan \frac{\theta}{2} \right)$. The equation of the circle is $(x+2)(x-2) + \left(y - \cot \frac{\theta}{2} \right) \left(y - \tan \frac{\theta}{2} \right) = 0$. Simplifying, $x^2 - 4 + y^2 - \left(\tan \frac{\theta}{2} + \cot \frac{\theta}{2} \right)y + 1 = 0$. Therefore, $(\sqrt{3}, 0)$ satisfies option (1).
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let the equation of the pair of lines, $y = px$ and $y = qx$, can be written as $(y - px)(y - qx) = 0$ Then the equation of the pair of the angle bisectors of the lines $x^2 - 4xy - 5y^2 = 0$ is:
Answer: (c)
Given $\dfrac{x^2-y^2}{1-(-5)}=\dfrac{xy}{-2}$ $\Rightarrow \dfrac{x^2-y^2}{6}=\dfrac{xy}{-2}$ $\Rightarrow x^2-y^2=-3xy$ $\Rightarrow x^2+3xy-y^2=0$
Maths · Permutations and Combinations · Single correct
If $^nP_r = \, ^nP_{r+1}$ and $^dC_r = \, ^nC_{r-1}$, then the value of $r$ is equal to:
Answer: (c)
Given $^nP_t = ^nP_{t+1}$, we have $$\frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!}$$ which implies $$(n-r) = 1 \ldots(1)$$ For $^nC_t = ^nC_{t-1}$, we have $$\frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r+1)!}$$ which implies $$\frac{1}{r(n-r)!} = \frac{1}{(n-r+1)(n-r)!}$$ Therefore, $$n-r+1 = r$$ which gives $$n+1 = 2r \ldots(2)$$ From equation (1), $2r - 1 - r = 1$ implies $r = 2$.
Maths · Differential Equations · Single correct
Let y = y(x) be the solution of the differential equation $x dy = \left( y + x^3 \cos x \right) dx$ with $y(\pi) = 0$, then $y\left( \frac{\pi}{2} \right)$ is equal to:
Answer: (a)
Given $x dy = \left( y + x^3 \cos x \right) dx$. Rewrite as $x dy = y dx + x^3 \cos x dx$. $$\frac{x dy - y dx}{x^2} = \frac{x^3 \cos x dx}{x^2}$$ Differentiate: $$\frac{d}{dx} \left( \frac{y}{x} \right) = \int x \cos x dx$$ Thus, $$\frac{y}{x} = x \sin x - \int 1 \cdot \sin x dx$$ $$\frac{y}{x} = x \sin x + \cos x + C$$ Given $0 = -1 + C \Rightarrow C = 1$, with $x = \pi$, $y = 0$. So, $$\frac{y}{x} = x \sin x + \cos x + 1$$ Therefore, $y = x^2 \sin x + x \cos x + x$ with $x = \frac{\pi}{2}$. Finally, $$y \left( \frac{\pi}{2} \right) = \frac{\pi^2}{4} + \frac{\pi}{2}$$
Maths · Binomial Theorem · Numerical
Let $n \in \mathbb{N}$ and $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$. If the sum of $(n+1)$ terms $\binom{n}{0}, \ 3 \cdot \binom{n}{1}, \ 5 \cdot \binom{n}{2}, \ 7 \cdot \binom{n}{3}, \ \ldots$ is equal to $2^{100} \cdot 101$ then $2 \left\lfloor \frac{n-1}{2} \right\rfloor$ is equal to
Answer: 98
1. $^1C_0 + 3 \cdot ^nC_1 + 5 \cdot ^nC_2 + \ldots + (2n + 1) \cdot ^nC_a$ $T_t = (2r + 1)^nC_t$ $S = \Sigma T_t$ $S = \Sigma (2r + 1)^nC_t = \Sigma 2r^nC_t + \Sigma ^nC_t$ $S = 2 \left( n \cdot 2^{n-1} \right) + 2^n = 2^n(n + 1)$ $2^a(n + 1) = 2^{100} \cdot 101 \Rightarrow n = 100$ $2 \left[ \frac{n-1}{2} \right] = 2 \left[ \frac{99}{2} \right] = 98$
Maths · Continuity and Differentiability · Numerical
Consider the function $f(x) = \frac{P(x)}{\sin(x-2)}$, $x \neq 2$ $$= 7, x = 2$$ Where $P(x)$ is a polynomial such that $P''(x)$ is always a constant and $P(3) = 9$. If $f(x)$ is continuous at $x = 2$, then $P(5)$ is equal to ___
Answer: 39
Given $$f(x) = \begin{cases} \frac{\mathrm{P}(x)}{\sin(x-2)}, & x \neq 2 \\ 7, & x = 2 \end{cases}$$ The second derivative of $\mathrm{P}(x)$ is constant, implying $\mathrm{P}(x)$ is a polynomial of degree 2. $f(x)$ is continuous at $x = 2$. Therefore, $f(2^+) = f(2^-)$. $$\lim_{x \to 2^+} \frac{\mathrm{P}(x)}{\sin(x-2)} = 7$$ $$\lim_{x \to 2^+} \frac{(x-2)(ax+b)}{\sin(x-2)} = 7 \Rightarrow 2a + b = 7$$ Assume $\mathrm{P}(x) = (x-2)(ax+b)$. Then $\mathrm{P}(3) = (3-2)(3a+b) = 9 \Rightarrow 3a + b = 9$. Solving these equations gives $a = 2$, $b = 3$. Finally, $\mathrm{P}(5) = (5-2)(2.5 + 3) = 3.13 = 39$.
Maths · Complex Numbers and Quadratic Equations · Numerical
The equation of a circle is $$\mathrm{Re}(z^2) + 2(\mathrm{Im}(z))^2 + 2\, \mathrm{Re}(z) = 0,$$ where $z = x + iy$ A line which passes through the center of the given circle and the vertex of the parabola, $$x^2 - 6x - y + 13 = 0,$$ has $y$-intercept equal to ___
Answer: 1
Equation of circle is $\left(x^2 - y^2\right) + 2y^2 + 2x = 0$ or $x^2 + y^2 + 2x = 0$. Centre: $(-1, 0)$. Parabola: $x^2 - 6x - y + 13 = 0$. $$(x - 3)^2 = y - 4$$ Vertex: $(3, 4)$. Equation of line $\equiv y - 0 = \frac{4 - 0}{3 + 1}(x + 1)$. $$y = x + 1$$ $y$-intercept $= 1$
Maths · Applications of Derivatives · Fill in the blank
If a rectangle is inscribed in an equilateral triangle of side length $2\sqrt{2}$ as shown in the figure, then the square of the largest area of such a rectangle is ___
Answer: 3
In $\triangle DBF$ $\tan60^\circ=\frac{2b}{2\sqrt{2}-\ell}$ $\Rightarrow b=\frac{\sqrt{3}(2\sqrt{2}-\ell)}{2}$ $A=\text{Area of rectangle}$ $=\ell\times b$ $=\ell\times\frac{\sqrt{3}}{2}(2\sqrt{2}-\ell)$ $\frac{dA}{d\ell}$ $=\frac{\sqrt{3}}{2}(2\sqrt{2}-\ell)-\frac{\ell\sqrt{3}}{2}$ $=0$ $\Rightarrow \ell=\sqrt{2}$ $A=\ell\times b$ $=\sqrt{2}\times\frac{\sqrt{3}}{2}(\sqrt{2})$ $=\sqrt{3}$ $\Rightarrow A^{2}=3$
Maths · Vector Algebra · Fill in the blank
If $(\vec{a}+3\vec{b})$ is perpendicular to $(7\vec{a}-5\vec{b})$ and $(\vec{a}-4\vec{b})$ is perpendicular to $(7\vec{a}-2\vec{b})$, then the angle between $\vec{a}$ and $\vec{b}$ (in degrees) is $\underline{\hspace{2cm}}$.
Answer: 60
Given $$ (\vec{a}+3\vec{b}) \perp (7\vec{a}-5\vec{b}) $$ $$ (\vec{a}+3\vec{b})\cdot(7\vec{a}-5\vec{b})=0 $$ $$ 7|\vec{a}|^2-15|\vec{b}|^2+16\,\vec{a}\cdot\vec{b}=0 \qquad \ldots (1) $$ $$ (\vec{a}-4\vec{b})\cdot(7\vec{a}-2\vec{b})=0 $$ $$ 7|\vec{a}|^2+8|\vec{b}|^2-30\,\vec{a}\cdot\vec{b}=0 \qquad \ldots (2) $$ From (1) and (2), $$ |\vec{a}|=|\vec{b}| $$ $$ \cos\theta=\frac{|\vec{b}|}{2|\vec{a}|} $$ $$ \therefore\ \theta=60^\circ $$
Maths · Differential Equations · Numerical
Let a curve $y = f(x)$ pass through the point $(2, (\log e \, 2)^2)$ and have slope $\frac{2y}{x \log_e x}$ for all positive real value of $x$. Then the value of $f(e)$ is equal to
Answer: 1
\[ y'=\frac{2y}{x\ln x} \] \[ \Rightarrow \frac{dy}{y} =\frac{2\,dx}{x\ln x} \] \[ \Rightarrow \ln |y| =2\ln |\ln x|+C \] \[ \text{put }x=2,\; y=(\ln 2)^2 \] \[ \Rightarrow c=0 \] \[ \Rightarrow y=(\ln x)^2 \] \[ \Rightarrow f(e)=1 \]
Maths · Complex Numbers and Quadratic Equations · Numerical
If $a + b + c = 1$, $ab + bc + ca = 2$ and $abc = 3$, then the value of $a^4 + b^4 + c^4$ is equal to ___
Answer: 13
Given $$a^2 + b^2 + c^2 = (a + b + c)^2 - 2 \Sigma ab = -3$$ $$(ab + bc + ca)^2 = \Sigma (ab)^2 + 2abc \Sigma a$$ Therefore, $$\Sigma (ab)^2 = -2$$ Now, $$a^4 + b^4 + c^4 = (a^2 + b^2 + c^2)^2 - 2 \Sigma (ab)^2$$ $$= 9 - 2(-2) = 13$$
Maths · Probability · Numerical
A fair coin is tossed n-times such that the probability of getting at least one head is at least 0.9. Then the minimum value of n is ___
Answer: 4
Given $\mathrm{P(Head)} = \frac{1}{2}$. $1 - \mathrm{P(All\ tail)} \geq 0.9$ $1 - \left(\frac{1}{2}\right)^n \geq 0.9$ $$\Rightarrow \left(\frac{1}{2}\right)^n \leq \frac{1}{10}$$ $$\Rightarrow n_{\min} = 4$$
Maths · Binomial Theorem · Numerical
If the co-efficient of $x^7$ and $x^8$ in the expansion of $\left(2 + \frac{x}{3}\right)^n$ are equal, then the value of $n$ is equal to ___
Answer: 55
Given $$\binom{n}{7} 2^{n-7} \frac{1}{3^7} = \binom{n}{8} 2^{n-8} \frac{1}{3^8}$$ This implies $$n - 7 = 48 \Rightarrow n = 55$$
Maths · Three Dimensional Geometry · Numerical
If the lines $\($ $\frac{x-k}{1}$ = $\frac{y-2}{2}$ = $\frac{z-3}{3}$ $\)$ and $\($ $\frac{x+1}{3}$ = $\frac{y+2}{2}$ = $\frac{z+3}{1}$ $\)$ are co-planar, then the value of $\($ k $\)$ is ___
Answer: 1
The determinant of the matrix is given by: $$\begin{vmatrix} k+1 & 4 & 6 \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{vmatrix} = 0$$ Expanding along the first row, we have: $$(k+1)(2-6) - 4(1-9) + 6(2-6) = 0$$ Simplifying, we get: $$(k+1)(-4) - 4(-8) + 6(-4) = 0$$ $$-4(k+1) + 32 - 24 = 0$$ $$-4k - 4 + 32 - 24 = 0$$ $$-4k + 4 = 0$$ Solving for $k$, we find: $$k = 1$$
Physics · Motion in a Straight Line · Single correct
The relation between time $t$ and distance $x$ for a moving body is given as $t = mx^2 + nx$, where $m$ and $n$ are constants. The retardation of the motion (When v stands for velocity)
Answer: (a)
Given $t = mx^2 + nx$. $$\frac{1}{v} = \frac{dt}{dx} = 2mx + n$$ $$v = \frac{1}{2mx+n}$$ $$\frac{dv}{dt} = -\frac{2m}{(2mx+n)^2} \left(\frac{dx}{dt}\right)$$ $$a = -(2m)v^3$$
Physics · Oscillations · Single correct
In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.
Answer: (b)
Given $$K = \frac{1}{2} m \omega^2 \left( A^2 - x^2 \right)$$ Simplifying, we have: $$= \frac{1}{2} m \omega^2 \left( A^2 - \frac{A^2}{4} \right)$$ $$= \frac{1}{2} m \omega^2 \left( \frac{3A^2}{4} \right)$$ Finally, $$K = \frac{3}{4} \left( \frac{1}{2} m \omega^2 A^2 \right)$$
Physics · Laws of Motion · Single correct
A force $\vec{F} = (40\hat{i} + 10\hat{j})\, \mathrm{N}$ acts on a body of mass $5\, \mathrm{kg}$. If the body starts from rest, its position vector $\vec{r}$ at time $t = 10\, \mathrm{s}$, will be:
Answer: (c)
Given $\($ $\frac{d \vec{v}}{dt}$ = $\vec{a}$ = $\frac{\vec{F}}{m}$ = (8$\hat{i}$ + 2$\hat{j}$) $\,$ $\mathrm{m/s^2}$ $\)$ $\($ $\frac{d \vec{r}}{dt}$ = $\vec{v}$ = (8t$\hat{i}$ + 2t$\hat{j}$) $\,$ $\mathrm{m/s}$ $\)$ $\($ $\vec{r}$ = (8$\hat{i}$ + 2$\hat{j}$) $\frac{t^2}{2}$ $\,$ $\mathrm{m}$ $\)$ At $\($ t = 10 $\mathrm{sec}$ $\)$ $\($ $\vec{r}$ = [(8$\hat{i}$ + 2$\hat{j}$)50] $\,$ $\mathrm{m}$ $\)$ $\($ $\Rightarrow$ $\vec{r}$ = (400$\hat{i}$ + 100$\hat{j}$) $\,$ $\mathrm{m}$ $\)$
Physics · Ray Optics and Optical Instruments · Single correct
A prism of refractive index $\mu$ and angle of prism $A$ is placed in the position of minimum angle of deviation. If minimum angle of deviation is also $A$, then in terms of refractive index
Answer: (a)
Given $$\mu = \frac{\sin \left( \frac{A + \delta_{\min}}{2} \right)}{\sin \left( \frac{A}{2} \right)}$$ Simplifying, $$\mu = \frac{\sin \left( \frac{A + A}{2} \right)}{\sin \left( \frac{A}{2} \right)}$$ This becomes, $$\mu = \frac{\sin A}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$$ Therefore, $$A = 2 \cos^{-1} \left( \frac{\mu}{2} \right)$$
Physics · Thermodynamics · Single correct
A heat engine has an efficiency of $\frac{1}{6}$. When the temperature of sink is reduced by $62^\circ \mathrm{C}$, its efficiency get doubled. The temperature of the source is :
Answer: (d)
Given the equation $$\eta = 1 - \frac{T_L}{T_H} \cdots (i)$$ We have $$2\eta = 1 - \frac{(T_L - 62)}{T_H} = 1 - \frac{T_L}{T_H} + \frac{62}{T_H}$$ This implies $$\eta = \frac{62}{T_H} \implies \frac{1}{6} = \frac{62}{T_H} \implies T_H = 6 \times 62 = 372 \, \mathrm{K}$$ In $^\circ \mathrm{C}$, $$\Rightarrow 372 - 273 = 99^\circ \mathrm{C}$$
Physics · Current Electricity · Single correct
In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 $\mathrm{cm}$. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 $\mathrm{cm}$. The ratio of the emf of two cells, $\frac{\varepsilon_1}{\varepsilon_2}$, is
Answer: (a)
Given $E_1 = k \ell_1$ (i) and $E_1 + E_2 = k \ell_2$ (ii). We have $$\frac{E_1}{E_1 + E_2} = \frac{\ell_1}{\ell_2} = \frac{250}{400} = \frac{5}{8}$$ $$8E_1 = 5E_1 + 5E_2$$ $$3E_1 = 5E_2$$ $$\frac{E_1}{E_2} = \frac{5}{3}$$
Physics · Moving Charges and Magnetism · Single correct
Two ions having same mass have charges in the ratio 1 : 2. They are projected normally in uniform magnetic field with their speeds in the ratio 2 : 3. The ratio of the radii of their circular trajectories is
Answer: (b)
Given $\mathbf{R} = \frac{mv}{qB}$, we have $\frac{R_1}{R_2} = \frac{\frac{mv_1}{q_1 B}}{\frac{mv_2}{q_2 B}} = \frac{v_1}{q_1} \times \frac{q_2}{v_2} = \frac{q_2}{q_1} \times \frac{v_1}{v_2}$. This simplifies to $$= \frac{2}{1} \times \left(\frac{2}{3}\right) = \frac{4}{3}$$
Physics · Alternating Current · Single correct
A 10$\Omega$ resistance is connected across 220$\mathrm{V}$ - 50$\mathrm{Hz}$ AC supply. The time taken by the current to change from its maximum value to the rms value is:
Answer: (a)
Given $V = 220 \, \mathrm{V}/50 \, \mathrm{Hz}$. Therefore, $i = i_0 \sin \omega t$. When $i = i_0$, $i_0 = i_0 \sin \omega t_1 \Rightarrow \omega t_1 = \frac{\pi}{2}$. When $i = \frac{i_0}{\sqrt{2}}$, $\frac{i_0}{\sqrt{2}} = i_0 \sin \omega t_2 \Rightarrow \omega t_2 = \frac{\pi}{4}$. Time taken by current from maximum value to rms value: $$\Rightarrow (t_1 - t_2) = \frac{\pi}{2\omega} - \frac{\pi}{4\omega} = \frac{\pi}{4\omega} = \frac{\pi}{4 \times 2 \pi f}$$ $$= \frac{1}{8 \times 50}$$ $$= \frac{1}{400} \, \mathrm{sec}$$ $$= 2.5 \, \mathrm{ms}$$
Physics · Motion in a Straight Line · Single correct
A balloon was moving upwards with a uniform velocity of 10 $\mathrm{m/s}$. An object of finite mass is dropped from the balloon when it was at a height of 75 $\mathrm{m}$ from the ground level. The height of the balloon from the ground when object strikes the ground was around (takes the value of g as 10 $\mathrm{m/s^2}$)
Answer: (c)
Object is projected as shown so as per motion under gravity $S = ut + \frac{1}{2} at^2$. $$-75 = +10t + \frac{1}{2}(-10)t^2 \Rightarrow t = 5 sec$$ Object takes $t = 5 s$ to fall on ground. Height of balloon from ground $H = 75 + ut$ $$= 75 + 10 \times 5 = 125 \, m$$
Physics · Electrostatic Potential and Capacitance · Single correct
If $q_f$ is the free charge on the capacitor plates and $q_b$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates, then bound charge $q_b$ can be expressed as:
Answer: (b)
When a dielectric is inserted in a capacitor due to free charge, $\vec{E} = \vec{E_0}$ only. After dielectric, $E' = \frac{E_0}{k}$. $q_B = q_f \left(1 - \frac{1}{k}\right)$
Physics · Gravitation · Single correct
Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is $W$, the weight of the same object on that planet will be:
Answer: (c)
Density is same $M = \frac{4}{3} \pi R^3 \rho$, $2 \, m = \frac{4}{3} \pi R^{33} \rho$. $$R' = 2^{1/3} R$$ $$\omega = \frac{GMm}{R^2}$$ $$\omega_2 = \frac{G2Mm}{R'^2}$$ $$\omega_2 = 2^{1/3} \omega$$
Physics · Electric Charges and Fields · Single correct
Two ideal electric dipoles A and B, having their dipole moment $p_1$ and $p_2$ respectively are placed on a plane with their centres at O as shown in the figure. At point C on the axis of dipole A, the resultant electric field is making an angle of $37^\circ$ with the axis. The ratio of the dipole moment of A and B, $\frac{p_1}{p_2}$ is : ( take $\sin 37^\circ = \frac{3}{5}$ )
Answer: (c)
Given the diagram, we have: $$\tan 37^\circ = \frac{3}{4} = \frac{\frac{kP_2}{r^3}}{\frac{2kP_1}{r^3}} = \frac{P_2}{2P_1} = \frac{3}{4}$$ Solving for $\frac{P_2}{P_1}$: $$\frac{P_2}{P_1} = \frac{3}{2}$$ Thus, the ratio $\frac{P_1}{P_2}$ is: $$\frac{P_1}{P_2} = \frac{2}{3}$$
Physics · Kinetic Theory · Single correct
Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:
Answer: (c)
The number of moles is conserved. $$n_1 + n_2 = n_3$$ $$P_1 V_1 + P_2 V_2 = P_3 V$$ $$\frac{4S}{r_1} \left( \frac{4}{3} \pi r_1^3 \right) + \frac{4S}{r_2} \left( \frac{4}{3} \pi r_2^3 \right) = \frac{4S}{r_3} \left( \frac{4}{3} \pi r_3^3 \right)$$ $$r_1^2 + r_2^2 = r_3^2$$ $$r_3 = \sqrt{r_1^2 + r_2^2}$$
Physics · Physical World, Units and Measurements · Single correct
The force is given in terms of time $t$ and displacement $x$ by the equation $$\mathbf{F} = A \cos Bx + C \sin Dt$$ The dimensional formula of $\frac{AD}{B}$ is:
Answer: (b)
Given the dimensions: $$[A] = [MLT^{-2}]$$ $$[B] = [L^{-1}]$$ $$[D] = [T^{-1}]$$ The dimension of $$\left[ \frac{AD}{B} \right]$$ is $$[ML^2T^{-3}]$$
Physics · Current Electricity · Single correct
The given potentiometer has its wire of resistance $10\Omega$. When the sliding contact is in the middle of the potentiometer wire, the potential drop across $2\Omega$ resistor is :
Answer: (c)
Given the circuit, we have the equation: $$\frac{20 - V_0}{5} + \frac{0 - V_0}{5} + \frac{20 - V_0}{2} = 0$$ Simplifying, we get: $$4 + 10 = \frac{2V_0}{5} + \frac{V_0}{2}$$ $$14 = \frac{4V_0 + 5V_0}{10}$$ Solving for $V_0$, we find: $$V_0 = \frac{140}{9} Volt$$ The potential difference across the $2\Omega$ resistor is $20 - V_0$. That is: $$\left(20 - \frac{140}{9}\right) Volt$$ Hence the answer is $$\left(\frac{40}{9}\right) Volt$$
Physics · Dual Nature of Radiation and Matter · Single correct
An electron moving with speed $v$ and a photon moving with speed $c$, have same $D$-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is:
Answer: (c)
Given $\lambda_e = \lambda_{\mathrm{Fh}}$. $$\frac{h}{p_e} = \frac{h}{p_{\mathrm{ph}}}$$ $$\sqrt{2m k_e} = \frac{E_{\mathrm{ph}}}{c}$$ $$2m k_e = \frac{(E_{\mathrm{ph}})^2}{c^2}$$ $$\frac{k_e}{E_{\mathrm{ph}}} = \frac{E_{\mathrm{ph}}}{c^2} \left( \frac{1}{2m} \right)$$ $$= \frac{p_{\mathrm{ph}}}{c} \left( \frac{1}{2m} \right)$$ $$= \frac{p_e}{c} \left( \frac{1}{2m} \right)$$ $$= \frac{mv}{c} \frac{1}{2m}$$ $$= \frac{v}{2c}$$
Physics · Motion in a Straight Line · Single correct
The instantaneous velocity of a particle moving in a straight line is given as $v = \alpha t + \beta t^2$, where $\alpha$ and $\beta$ are constants. The distance travelled by the particle between 1 s and 2 s is :
Answer: (b)
Given $V = \alpha t + \beta t^2$. $\($ $\frac{ds}{dt}$ = $\alpha$ t + $\beta$ t^2 $\)$ $$ \int_{S_1}^{S_2} ds = \int_1^2 (\alpha t + \beta t^2) \, dt $$ $$ S_2 - S_1 = \left[ \frac{\alpha t^2}{2} + \frac{\beta t^3}{3} \right]_1^2 $$ As particle is not changing direction, so distance = displacement. Distance = $$ \left[ \frac{\alpha [4 - 1]}{2} + \frac{\beta [8 - 1]}{3} \right] $$ $$ = \frac{3\alpha}{2} + \frac{7\beta}{3} $$
Physics · Ray Optics and Optical Instruments · Single correct
A ray of light entering from air into a denser medium of refractive index $\frac{4}{3}$, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle $\theta$ should be equal to:
Answer: (a)
At maximum angle $\theta$, the ray at point $B$ goes in grazing emergence; at all smaller values of $\theta$, total internal reflection (TIR) occurs. At point \(B\), \[ \frac{4}{3}\times \sin \theta'' = 1 \times \sin 90^\circ \] \[ \theta'' = \sin^{-1}\left(\frac{3}{4}\right) \] \[ \theta' = \left(\frac{\pi}{2} - \theta''\right) \] At point \(A\), \[ \sin \theta = \frac{4}{3}\times \sin \theta' \] \[ \sin \theta = \frac{4}{3}\times \sin\left(\frac{\pi}{2}-\theta''\right) \] \[ \sin \theta = \frac{4}{3}\cos\left[\cos^{-1}\left(\frac{\sqrt{7}}{4}\right)\right] \] \[ \sin \theta = \frac{4}{3}\times \frac{\sqrt{7}}{4} \] \[ \theta = \sin^{-1}\left(\frac{\sqrt{7}}{3}\right). \]
Physics · Dual Nature of Radiation and Matter · Single correct
When radiation of wavelength $\lambda$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is $4.8 \, \mathrm{V}$. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes $1.6 \, \mathrm{V}$. The threshold wavelength of the metal is:
Answer: (b)
Given $\($ V_S = h $\nu$ - $\phi$ $\ldots$ (i) $\)$ $\($ 4.8 = $\frac{hc}{\lambda}$ - $\phi$ $\ldots$ (ii) $\)$ $\($ 1.6 = $\frac{hc}{2\lambda}$ - $\phi$ $\ldots$ (iii) $\)$ Using above equation (i) - (ii) $\($ 3.2 = $\frac{hc}{\lambda}$ - $\frac{hc}{2\lambda}$ $\)$ $\($ 3.2 = $\frac{hc}{2\lambda}$ $\ldots$ (iii) $\)$ [ $\lambda$ = $\frac{hc}{6.4}$ $\]$ Put in equation (ii) $\($ $\phi$ = 1.6 $\)$ $\($ $\frac{hc}{\lambda_{th}}$ = 1.6 $\)$ $\($ $\lambda_{th}$ = $\frac{hc}{1.6}$ $\)$ $\($ = $\left$( $\frac{hc}{6.4}$ $\right$) $\times$ 4 = 4 $\lambda$ $\)$
Physics · Mathematics in Physics · Single correct
Two vectors $\vec{X}$ and $\vec{Y}$ have equal magnitude. The magnitude of $\left( \vec{X} - \vec{Y} \right)$ is $n$ times the magnitude of $\left( \vec{X} + \vec{Y} \right)$. The angle between $\vec{X}$ and $\vec{Y}$ is:
Answer: (b)
Given $X = Y$ $$\sqrt{X^2 + Y^2 - 2 \times Y \cos \theta} = n \sqrt{X^2 + Y^2 + 2 \times Y \cos \theta}$$ Square both sides $2X^2(1 - \cos \theta) = n^2 \cdot 2X^2(1 + \cos \theta)$ $$1 - \cos \theta = n^2 + n^2 \cos \theta$$ $$\cos \theta = \frac{1 - n^2}{1 + n^2}$$ $$\theta = \cos^{-1} \left[ \frac{n^2 - 1}{-n^2 - 1} \right]$$
Physics · Kinetic Theory · Numerical
A system consists of two types of gas molecules A and B having same number density $2 \times 10^{25} / \mathrm{m}^3$. The diameter of A and B are $10 \, \mathrm{A}$ and $5 \, \mathrm{A}$ respectively. They suffer collision at room temperature. The ratio of average distance covered by the molecule A to that of B between two successive collision is ____ $\times 10^{-2}$
Answer: 25
Therefore, mean free path $$\lambda = \frac{1}{\sqrt{2} \pi d_n^2 n}$$ $$\frac{\lambda_1}{\lambda_2} = \frac{d_2^2 n_2}{d_1^2 n_1}$$ $$= \left( \frac{5}{10} \right)^2 = 0.25 = 25 \times 10^{-2}$$
Physics · Dual Nature of Radiation and Matter · Numerical
A light beam of wavelength 500 $\mathrm{nm}$ is incident on a metal having work function of 1.25 $\mathrm{eV}$, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the magnetic field B, with maximum kinetic energy are bent into circular arc of radius 30 $\mathrm{cm}$. The value of B is $\times 10^{-7}$ $\mathrm{T}$ Given hc = 20 $\times$ $10^{-26}$ $\mathrm{J}$ - $\mathrm{m}$, mass of electron = 9 $\times 10^{-31}$ $\mathrm{kg}$
Answer: 125
By photoelectric equation $$\frac{hc}{\lambda} - \phi = k_{max}$$ $$k_{max} = \frac{1240}{500} - 1.25 \approx 1.25$$ $$r = \frac{\sqrt{2mk}}{eB}$$ $$B = \frac{\sqrt{2mk}}{er}$$ $$= 125 \times 10^{-7} \, T$$
Physics · Communication Systems · Numerical
A message signal of frequency 20$\mathrm{kHz}$ and peak voltage of 20$\mathrm{volt}$ is used to modulate a carrier wave of frequency 1$\mathrm{MHz}$ and peak voltage of 20$\mathrm{volt}$. The modulation index will be
Answer: 1
Modulation index $$\mu = \frac{A_m}{A_c} = \frac{20}{20} = 1$$
Physics · Current Electricity · Numerical
A 16$\Omega$ wire is bend to form a square loop. A 9 V supply having internal resistance of 1$\Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is ________$\times 10^{-1}$ V
Answer: 45
Here assume current as shown in the circuit. By KVL in the outer loop, $9 - 12i - 4i = 0$. Solving for $i$, we have $$8i = \frac{9}{2} = 4.5$$ which simplifies to $$i = 45 \times 10^{-1}$$
Physics · Alternating Current · Numerical
Two circuits are shown in the figure (a) $\&$ (b). At a frequency of rad/s the average power dissipated in one cycle will be same in both the circuits.
Answer: 500
For figure (a), $$P_{mg} = \frac{v_{rms}^2}{R}$$ $$\frac{v_{rms}^2}{Z^2} \times R = \frac{v_{rms}^2}{R} \times 1$$ $$R^2 = Z^2$$ $$25 = \left( \sqrt{(x_c - x_L)^2 + 5^2} \right)^2$$ $$25 = (x_c - x_L)^2 + 25$$ $$x_c = x_L \implies \frac{1}{\omega C} = \omega L$$ $$\omega^2 = \frac{1}{LC} = \frac{10^6}{0.1 \times 40}$$ $$\omega = 500$$
Physics · Nuclei · Numerical
From the given data, the amount of energy required to break the nucleus of aluminium $^{27}_{13}Al$ is $x \times 10^{-3}$ J Mass of neutron $= 1.00866\,u$ Mass of proton $= 1.00726\,u$ Mass of Aluminium nucleus $= 27.18846\,u$ (Assume $1\,u$ corresponds to $x\,J$ of energy) (Round off to the nearest integer)
Answer: 27
The change in mass $\Delta m$ is given by the equation: $$\Delta m = (Z m_p + (A - Z) m_n) - M_{Me}$$ Substituting the values: $$= (13 \times 1.00726 + 14 \times 1.00866) - 27.18846$$ $$= 27.21562 - 27.18846$$ $$= 0.02716 \, \mathrm{u}$$ The energy $E$ is calculated as: $$E = 27.16 \times 10^{-3} \, \mathrm{J}$$
Physics · Work, Energy and Power · Numerical
A force of $\mathbf{F} = (5y + 20)\hat{\mathbf{j}} \, \mathrm{N}$ acts on a particle. The work done by this force when the particle is moved from $y = 0 \, \mathrm{m}$ to $y = 10 \, \mathrm{m}$ is J
Answer: 450
Given $\mathbf{F} = (5y + 20) \hat{\mathbf{j}}$. $\omega = \int \mathbf{F} \, dy = \int_0^{10} (5y + 20) \, dy$ $$= \left( \frac{5y^2}{2} + 20y \right)_0^{10}$$ $$= \frac{5}{2} \times 100 + 20 \times 10$$ $$= 250 + 200 = 450 \, \mathrm{J}$$
Physics · System of Particles and Rotational Motion · Numerical
A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 $\mathrm{rpm}$, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in 10 s is______ $\pi\times10^{-1}\,\mathrm{N\,m}$
Answer: 4
Given $\tau = \frac{\Delta L}{\Delta t} = \frac{I(\omega_f - \omega_i)}{\Delta t}$. $$\tau = \frac{\frac{mR^2}{2} \times [0 - \omega]}{\Delta t}$$ $$= \frac{10 \times (20 \times 10^{-2})^2}{2} \times \frac{600 \times \pi}{30 \times 10}$$ $$= 0.4 \pi = 4 \pi \times 10^{-2}$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Numerical
In a semiconductor, the number density of intrinsic charge carriers at 27°C is $1.5 \times 10^{16}/\mathrm{m}^3$. If the semiconductor is doped with impurity atom, the hole density increases to $4.5 \times 10^{22}/\mathrm{m}^3$. The electron density in the doped semiconductor is $- \times 10^9/\mathrm{m}^3$.
Answer: 5
$n_e n_h = n_i^2$ $n_e = \dfrac{n_i^2}{n_h} = \dfrac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}}$ $= \dfrac{1.5 \times 1.5 \times 10^{32}}{4.5 \times 10^{22}}$ $= 5 \times 10^9/\text{m}^3$
Physics · Nuclei · Numerical
The nuclear activity of a radioactive element becomes $\left( \frac{1}{8} \right)^{th}$ of its initial value in 30 years. The half-life of radioactive element is _______ years
Answer: 10
Given $A = A_0 e^{-\lambda t}$. $$\frac{A_0}{8} = A_0 e^{-\lambda t} \implies \lambda t = \ln 8$$ $$\lambda t = 3 \ln 2$$ $$\frac{\ln 2}{\lambda} = \frac{t}{3} = \frac{30}{3} = 10 years$$
Chemistry · Chemical Bonding and Molecular Structure · Single correct
In the following the correct bond order sequence is:
Answer: (c)
$\mathrm{O_2}$ (16 electrons) $\sigma_{1s}^{2},\ \sigma_{1s}^{*2},\ \sigma_{2s}^{2},\ \sigma_{2s}^{*2},\ \sigma_{2p_z}^{2},\ \pi_{2p_x}^{2}=\pi_{2p_y}^{2},\ \pi_{2p_x}^{*1}=\pi_{2p_y}^{*1}$ Bond order of $\mathrm{O_2}$ $\Rightarrow 2$ Bond order of $\mathrm{O_2^-}$ $\Rightarrow 1.5$ Bond order of $\mathrm{O_2^{2-}}$ $\Rightarrow 1$ Bond order of $\mathrm{O_2^+}$ $\Rightarrow 2.5$
Chemistry · Polymers · Single correct
A biodegradable polyamide can be made from:
Answer: (c)
A biodegradable polyamide nylon-2-nylon-6 is made from glycine and amino caproic acid.
Chemistry · Co-ordination Compounds · Single correct
Match List I with List II : Choose the correct answer from the options given below :
Answer: (a)
(a) $\mathrm{CsI}$ salt is poor water soluble due to its low hydration energy. (b) $\mathrm{NaHCO_3}$ is used in fire extinguisher. (c ) $\mathrm{K}$ is most abundant element in cell fluid. (d) $\mathrm{Li_2CO_3}$ decomposes easily due to high covalent character caused by small size $\mathrm{Li^+}$ cation.
Chemistry · Co-ordination Compounds · Single correct
Which one of the following metal complexes is most stable?
Answer: (b)
Complex $[\mathrm{Co(en)_3}] \mathrm{Cl_2}$ is the most stable complex among the given complex compounds because more number of chelate rings are present in this complex as compared to others. (1) $[\mathrm{Co(en)(NH_3)_4}] \mathrm{Cl_2}$ 1 chelate ring (2) $[\mathrm{Co(en)_3}] \mathrm{Cl_2}$ 3 chelate rings (3) $[\mathrm{Co(en)_2(NH_3)_2}] \mathrm{Cl_2}$ 2 chelate rings (4) $[\mathrm{Co(NH_3)_6}] \mathrm{Cl_2}$ 0 chelate rings
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Match List I with List II : (Both having metallurgical terms) Choose the correct answer from the options given below :
Answer: (a)
(a) Concentration of Ag is performed by leaching with dilute NaCN solution. (b) Pig iron is formed in blast furnace. (c) Blister Cu is produced in Bessemer converter. (d) Froth floatation method is used for sulphide ores. Note: During extraction of Cu reverberatory furnace is involved.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The ionic radii of $\mathrm{F}^-$ and $\mathrm{O}^{2-}$ respectively are $1.33 \, \mathrm{A}$ and $1.4 \, \mathrm{A}$, while the covalent radius of $\mathrm{N}$ is $0.74 \, \mathrm{A}$. The correct statement for the ionic radius of $\mathrm{N}^{3-}$ from the following is:
Answer: (b)
$F^{-}$, $\mathrm{O}^{2-}$ and $\mathrm{N}^{3-}$ all are isoelectronic species in which $\mathrm{N}^{3-}$ have least number of protons due to which its size increases as least nuclear attraction is experienced by the outer shell electrons. Size order $\mathrm{N}^{3-}$ > $\mathrm{O}^{2-}$ > $\mathrm{F}^{-}$.
Chemistry · The Solid State · Single correct
The correct decreasing order of densities of the following compounds is:
Answer: (a)
The density order is (D) > (C) > (B).
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Consider the above reaction, the Product "P" is :
Answer: (b)
The reaction sequence involves the reduction of a nitro group to an amino group using $\mathrm{Sn + HCl}$. The amino group is then converted to an azo compound by reaction with $\mathrm{Ph{-}N_2Cl^{-}}$. The final product is a yellow colored azo dye.
Chemistry · Amines · Single correct
A reaction of benzonitrile with one equivalent $CH_3MgBr$ followed by hydrolysis produces a yellow liquid "P". The compound "P" will give positive
Answer: (a)
Benzonitrile reacts with $\mathrm{CH_3MgBr}$ to form an intermediate. Upon hydrolysis with $\mathrm{H_3O^+}$, it gives $\mathrm{C-CH_3 + NH_3}$. This product gives a positive Iodoform test.
Chemistry · Structure of Atom · Single correct
The spin only magnetic moments (in BM) for free $\mathrm{Ti}^{3+}$, $\mathrm{V}^{2+}$ and $\mathrm{Sc}^{3+}$ ions respectively are (At. No. $\mathrm{Sc}: 21$, $\mathrm{Ti}: 22$, $\mathrm{V}: 23$)
Answer: (b)
For $\mathrm{Ti}^{3+} = [\mathrm{Ar}]\,3d^1$, $n = 1$, $\mu = 1.73\,\mathrm{BM}$. For $\mathrm{V}^{2+} = [\mathrm{Ar}]\,3d^3$, $n = 3$, $\mu = 3.87\,\mathrm{BM}$. For $\mathrm{Sc}^{3+} = [\mathrm{Ar}]\,3d^0\,4s^0$, $n = 0$, $\mu = 0$. The formula for magnetic moment is given by $\mu = \sqrt{n(n+2)}\,\mathrm{BM}$.
Chemistry · Biomolecules · Single correct
Which one of the following is correct structure for cytosine?
Answer: (c)
The correct structure of cytosine is shown in the image with the chemical formula $\mathrm{C_4H_5N_3O}$. It consists of a pyrimidine ring with an amino group at position 4 and a keto group at position 2.
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Identify the species having one $\pi$-bond and maximum number of canonical forms from the following:
Answer: (d)
Among $\mathrm{SO_3}$, $\mathrm{O_2}$, $\mathrm{SO_2}$ and $\mathrm{CO_3^{2-}}$, only $\mathrm{O_2}$ and $\mathrm{CO_3^{2-}}$ has only one $\pi$-bond.
Chemistry · Hydrogen · Single correct
Which one of the following metals forms interstitial hydride easily?
Answer: (a)
Elements of group 7, 8, 9 do not form hydrides thus Cr will only form hydride among the given elements (Fe, Mn, Co)
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Maleic anhydride Maleic anhydride can be prepared by :
Answer: (b)
Cis but 2-enenoic acid is heated to form maleic anhydride.
Chemistry · Environmental Chemistry · Single correct
Given below are two statements: Statement I : Chlorofluoro carbons breakdown by radiation in the visible energy region and release chlorine gas in the atmosphere which then reacts with stratospheric ozone. Statement II : Atmospheric ozone reacts with nitric oxide to give nitrogen and oxygen gases, which add to the atmosphere. For the above statements choose the correct answer from the options given below :
Answer: (b)
Statement (1) CFCs are broken down by powerful UV radiation and releases chlorine free radical which reacts with ozone and start chain reaction. $$\mathrm{CF_2Cl_2(g) \xrightarrow{UV} \dot{Cl}(g) + \dot{CF_2Cl}(g)}$$ $$\mathrm{\dot{Cl}(g) + O_3(g) \rightarrow \dot{ClO}(g) + O_2(g)}$$ $$\mathrm{\dot{ClO}(g) + O(g) \rightarrow \dot{Cl}(g) + O_2(g)}$$ Statement (2) Atmosphere ozone reacts with nitric oxide to produce nitrogen dioxide and oxygen. $$\mathrm{NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)}$$
Chemistry · Haloalkanes and Haloarenes · Single correct
[where $\mathrm{Et} \Rightarrow -\mathrm{C_2H_5}$, $^\mathrm{t}\mathrm{Bu} \Rightarrow (\mathrm{CH_3})_3\mathrm{C}-$] Consider the above reaction sequence, Product "A" and Product "B" formed respectively are:
Answer: (a)
The reaction starts with $\mathrm{Br-CH_2-CH=O}$ treated with excess $\mathrm{EtOH}$ and dry $\mathrm{HCl}$ gas to form compound (A), $\mathrm{Br-CH_2-CH(OEt)_2}$, which is an acetal. Then, using a tertiary butoxide, an $\mathrm{E_2}$ elimination mechanism occurs, resulting in compound (B), $\mathrm{CH_2=C(OEt)_2}$.
Chemistry · Surface Chemistry · Single correct
Match List I with List II : \begin{tabular}{|c|c|} \hline List-I & List-II \\ \hline Example of colloids & Classification \\ \hline (a) Cheese & (i) dispersion of liquid in liquid \\ \hline (b) Pumice stone & (ii) dispersion of liquid in gas \\ \hline (c) Hair cream & (iii) dispersion of gas in solid \\ \hline (d) Cloud & (iv) dispersion of liquid in solid \\ \hline \end{tabular} Choose the most appropriate answer from the options given below.
Answer: (d)
Chemistry · Alcohols, Phenols and Ethers · Single correct
What is the major product "P" of the following reaction ?
Answer: (d)
The given reaction involves the conversion of a cyclohexylamine to a diazonium salt using $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ at $278 \, \mathrm{K}$. The diazonium salt then undergoes hydrolysis to form the major product, which is cyclohexanol with a methyl group, $\mathrm{CH_3CH_2OH}$.
Chemistry · Redox Reactions · Single correct
Identify the process in which change in the oxidation state is five:
Answer: (b)
Q6 $$\mathrm{MnO_4^- + 5e^- \rightarrow Mn^{2+}}$$
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Which among the following is the strongest acid?
Answer: (d)
The strongest acid is the cyclopentadiene because its conjugate base is aromatic.
Chemistry · Thermodynamics · Numerical
A system does 200 J of work and at the same time absorbs 150 J of heat. The magnitude of the change in internal energy is $\_$$\_$$\_$ J. (Nearest integer)
Answer: 50
Given $w = -200 \, \mathrm{J}$, $q = +150 \, \mathrm{J}$: $\Delta U = q + w$. $$\Delta U = 150 - 200 = -50 \, \mathrm{J}$$ Magnitude $= 50 \, \mathrm{J} = |\Delta U|$
Chemistry · Structure of Atom · Numerical
An accelerated electron has a speed of $5 \times 10^6 \, \mathrm{ms}^{-1}$ with an uncertainty of $0.02\%$. The uncertainty in finding its location while in motion is $x \times 10^{-9} \, \mathrm{m}$. The value of $x$ is . (Nearest integer) [Use mass of electron $m = 9.1 \times 10^{-31} \, \mathrm{kg}$, $h = 6.63 \times 10^{-34} \, \mathrm{Js}$, $\pi = 3.14$]
Answer: 58
Given $\Delta v = \frac{0.02}{100} \times 5 \times 10^6 = 10^3 \, \mathrm{m/s}$. $\Delta x \cdot \Delta v = \frac{h}{4 \pi m}$. $x \times 10^{-9} \times 10^3 = \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31}}$. $x \times 10^{-9} \times 10^3 = 0.058 \times 10^{-3}$. $x = \frac{0.058 \times 10^{-6}}{10^{-9}} = 58$
Chemistry · The d-and f-Block Elements · Numerical
Number of electrons present in 4f orbital of $\mathrm{Ho}^{3+}$ ion is (Given Atomic No. of $\mathrm{Ho} = 67$)
Answer: 10
Ho = $[\mathrm{Xe}]4f^{11}6s^2$ Ho$^{3+}$ = $[\mathrm{Xe}]4f^{10}$ so number of $e^-$ present in 4f is 10.
Chemistry · Hydrocarbons · Numerical
Consider the above chemical reaction. The total number of stereoisomers possible for Product 'P' is
Answer: 2
The total number of products possible is $2$.
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For a chemical reaction $\mathrm{A} \rightarrow \mathrm{B}$, it was found that concentration of $\mathrm{B}$ is increased by $0.2 \, \mathrm{mol \, L^{-1}}$ in $30 \, \mathrm{min}$. The average rate of the reaction is $\times$ $10^{-1}$ mol $L^{-1}$ $h^{-1}$. (in nearest integer)
Answer: 4
Average rate of reaction is given by $$Av. rate of reaction = -\frac{\Delta [A]}{\Delta t} = \frac{\Delta [B]}{\Delta t} = \frac{(0.2 - 0)}{\frac{1}{2}}$$ which simplifies to $$= 0.4 = 4 \times 10^{-1} \, mol/L \times hr$$
Chemistry · Some Basic Concepts of Chemistry · Numerical
The number of significant figures in 0.00340 is
Answer: 3
Number of significant figures = 3
Chemistry · Equilibrium · Numerical
Assuming that $\mathrm{Ba(OH)_2}$ is completely ionised in aqueous solution under the given conditions the concentration of $\mathrm{H_3O^+}$ ions in $0.005\, \mathrm{M}$ aqueous solution of $\mathrm{Ba(OH)_2}$ at $298\, \mathrm{K}$ is $\times 10^{-12}\, \mathrm{mol\, L^{-1}}$. (Nearest integer)
Answer: 1
The reaction is given by: $$\mathrm{Ba(OH)_2} \rightarrow \mathrm{Ba^{+2}} + 2\mathrm{OH}^-$$ The concentration of $\mathrm{OH}^-$ is calculated as: $$2 \times 0.005 = 0.01 = 10^{-2}$$ At 298 K, in aqueous solution: $$[\mathrm{H_3O^+}] [\mathrm{OH^-}] = 10^{-14}$$ Therefore, the concentration of $\mathrm{H_3O^+}$ is: $$[\mathrm{H_3O^+}] = \frac{10^{-14}}{10^{-2}} = 10^{-12}$$
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical
0.8 $\mathrm{g}$ of an organic compound was analysed by Kjeldahl's method for the estimation of nitrogen. If the percentage of nitrogen in the compound was found to be 42$\%$, then $\,$ $\mathrm{mL}$ of 1M $\mathrm{H_2SO_4}$ would have been neutralized by the ammonia evolved during the analysis.
Answer: 12
Organic compound: 0.8 gm wt. of N mole of N = $\frac{42 \times 0.8}{100 \times 14}$ = $\frac{2.4}{100}$ mol moles of NH_3 = $\frac{2.4}{100}$ 2$\mathrm{NH_3}$ + $\mathrm{H_2SO_4}$ $\rightarrow$ ($\mathrm{NH_4}$)_2$\mathrm{SO_4}$ $\downarrow$ $\frac{2.4}{100}$ mole $\frac{1.2}{100}$ mole $\frac{1.2}{100}$ = 1 $\times$ V($\ell$) $\Rightarrow$ V_{$\mathrm{H_2SO_4}$} = $\frac{1.2}{100}$ $\ell$ = 12 $\ell$
Chemistry · Solutions · Numerical
When $3.00\ \mathrm{g}$ of a substance $X'$ is dissolved in $100\ \mathrm{g}$ of $\mathrm{CCl_4}$, it raises the boiling point by $0.60\ \mathrm{K}$. The molar mass of the substance $X'$ is ______ $\mathrm{g\,mol^{-1}}$. (Nearest integer) [Given $K_b$ for $\mathrm{CCl_4}$ is 5.0 $\mathrm{K}$ \, $\mathrm{kg}$ \, $\mathrm{mol^{-1}}$]
Answer: 250
$\Delta T_b=K_b\times\text{molality}$ $0.60=5\times\left(\frac{3/M}{100/1000}\right)$ $M=250$
Chemistry · States of Matter · Numerical
An LPG cylinder contains gas at a pressure of $300\,\mathrm{kPa}$ at $27^\circ\mathrm{C}$. The cylinder can withstand the pressure of $1.2 \times 10^6\,\mathrm{Pa}$. The room in which the cylinder is kept catches fire. The minimum temperature at which the bursting of cylinder will take place is _____ $^\circ\mathrm{C}$. (Nearest integer)
Answer: 927
$\frac{P_1}{T_1} = \frac{P_2}{T_2}$ $\Rightarrow \frac{300 \times 10^3}{300} = \frac{1.2 \times 10^6}{T_2}$ $\Rightarrow T_2 = 1200\,\mathrm{K}$ $T_2 = 927^\circ\mathrm{C}$