JEE Main 28 July 2022 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 · Basics Of Mathematics · Single correct
Let $S=\left\{x\in[-6,3]-\{-2,2\}:\frac{|x+3|-1}{|x|-2}\ge0\right\}$ and $T=\left\{x\in\mathbb{Z}:x^2-7|x|+9\le0\right\}$. Then the number of elements in $S\cap T$ is
Answer: (d)
The intersection of sets $S$ and $T$ is given by: $$S \cap T = \{-5, -4, 3\}$$
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $\alpha$, $\beta$ be the roots of the equation $x^2 - \sqrt{2}x + \sqrt{6} = 0$ and $\frac{1}{\alpha^2} + 1, \frac{1}{\beta^2} + 1$ be the roots of the equation $x^2 + ax + b = 0$. Then the roots of the equation $x^2 - (a + b - 2)x + (a + b + 2) = 0$ are:
Answer: (b)
Given $$a = \frac{-1}{\alpha^2} - \frac{1}{\beta^2} - 2$$ $$b = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + 1 + \frac{1}{\alpha^2 \beta^2}$$ $$a + b = \frac{1}{(\alpha \beta)^2} - 1 = \frac{1}{6} - 1 = -\frac{5}{6}$$ $$x^2 - \left( -\frac{5}{6} - 2 \right)x + \left( 2 - \frac{5}{6} \right) = 0$$ $$6x^2 + 17x + 7 = 0$$ $$x = -\frac{7}{3}, \; x = -\frac{1}{2}$$ are the roots. Both roots are real and negative.
Maths · Matrices · Single correct
Let A and B be any two 3 $\times$ 3 symmetric and skew symmetric matrices respectively. Then which of the following is $\underline{NOT}$ true?
Answer: (c)
(A) $C=A^4-B^4$ $C^T=(A^4-B^4)^T$ $=(A^T)^4-(B^T)^4$ $=A^4-B^4=C$ (B) $C=AB-BA$ $C^T=(AB-BA)^T$ $=(AB)^T-(BA)^T$ $=B^TA^T-A^TB^T$ $=-BA+AB=C$ \text{(C)} $C=B^5-A^5$ $C^T=(B^5-A^5)^T$ $=(B^T)^5-(A^T)^5$ $=-B^5-A^5$ $\ne C$ (D) $C=AB+BA$ $C^T=(AB+BA)^T$ $=(AB)^T+(BA)^T$ $=-BA-AB$ $=-C$ Option C is not true.
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $f(x)=ax^2+bx+c$ be such that $f(1)=3$, $f(-2)=\lambda$ and $f(3)=4$. If $f(0)+f(1)+f(-2)+f(3)=14$, then $\lambda$ is equal to
Answer: (d)
Given $f(0) + 3 + \lambda + 4 = 14$. Therefore, $f(0) = 7 - \lambda = c$. $f(1) = a + b + c = 3 \ldots (i)$ $f(3) = 9a + 3b + c = 4 \ldots (ii)$ $f(-2) = 4a - 2b + c = \lambda \ldots (iii)$ Subtracting equation (iii) from (ii): $$a + b = \frac{4 - \lambda}{5}$$ Substitute in equation (i): $$\frac{4 - \lambda}{5} + 7 - \lambda = 3$$ Solving gives $6 \lambda = 24$; therefore, $\lambda = 4$.
Maths · Continuity and Differentiability · Single correct
The function $f : \mathbb{R} \to \mathbb{R}$ defined by $$f(x) = \lim_{n \to \infty} \frac{\cos(2\pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} - x^{2n}}$$ is continuous for all $x$ in
Answer: (b)
n should be given as a natural number. $$f(x) = \begin{cases} \frac{-\sin(x-1)}{x-1} & x 1 \end{cases}$$ $f(x)$ is discontinuous at $x = -1$ and $x = 1$
Maths · Applications of Derivatives · Single correct
The function $f(x) = xe^{x(1-x)}$, $x \in \mathbb{R}$, is
Answer: (a)
Given $f(x) = x \, e^{x(1-x)}$. The derivative is $f'(x) = -e^{x(1-x)} (2x + 1)(x - 1)$. The function $f(x)$ is increasing in $\left( -\frac{1}{2}, 1 \right)$.
Maths · Applications of Derivatives · Single correct
The sum of the absolute maximum and absolute minimum values of the function $$f(x) = \tan^{-1}(\sin x - \cos x)$$ in the interval $[0, \pi]$ is
Answer: (c)
Given $f(x) = \tan^{-1}(\sin x - \cos x)$. The derivative is $f'(x) = \frac{\cos x + \sin x}{(\sin x - \cos x)^2 + 1} = 0$. Therefore, $x = \frac{3\pi}{4}$. The table of values is: $\[$ $\begin{array}{c|c|c|c}$ x & 0 & $\frac{3\pi}{4}$ & $\pi$ $\\$ $\hline$ f(x) & -$\frac{\pi}{4}$ & $\tan$^{-1}$\sqrt{2}$ & $\frac{\pi}{4}$ $\\$ $\end{array}$ $\]$ The maximum value of $f(x)$ is $\tan^{-1}\sqrt{2}$ and the minimum value is $-\frac{\pi}{4}$. The sum is $\tan^{-1}\sqrt{2} - \frac{\pi}{4}$. This equals $\cos^{-1}\frac{1}{\sqrt{3}} - \frac{\pi}{4}$.
Maths · Applications of Derivatives · Single correct
Let $x(t) = 2\sqrt{2} \cos t \sqrt{\sin 2t}$ and $y(t) = 2\sqrt{2} \sin t \sqrt{\sin 2t}$, $t \in \left(0, \frac{\pi}{2}\right)$. Then $$ 1 + \frac{\left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}} $$ at $t=\frac{\pi}{4}$ is equal to
Answer: (d)
Given $x = 2\sqrt{2} \cos t \sqrt{\sin 2t}$. $$\frac{dx}{dt} = \frac{2\sqrt{2} \cos 3t}{\sqrt{\sin 2t}}$$ $y(t) = 2\sqrt{2} \sin t \sqrt{\sin 2t}$. $$\frac{dy}{dt} = \frac{2\sqrt{2} \sin 3t}{\sqrt{\sin 2t}}$$ $$\frac{dy}{dx} = \tan 3t$$ $$\frac{dy}{dx} = -1 at t = \frac{\pi}{4}$$ $$\frac{d^2y}{dx^2} = \frac{3}{2\sqrt{2}} \sec^3 3t \cdot \sqrt{\sin 2t} = -3 at t = \frac{\pi}{4}$$ Therefore, $$\frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}} = \frac{1 + 1}{-3} = -\frac{2}{3}$$
Maths · Integrals · Single correct
Let $I_n(x) = \int_0^x \frac{1}{(t^2 + 5)^n} \, dt$, $n = 1, 2, 3, \ldots$ Then
Answer: (a)
Given $I_n(x) = \int_0^x \frac{dt}{(t^2 + 5)^n}$. Applying integral by parts: $$I_n(x) = \left[ \frac{t}{(t^2 + 5)^n} \right]_0^x - \int_0^x n(t^2 + 5)^{-n-1} \cdot 2t^2$$ $$I_n(x) = \frac{x}{(x^2 + 5)^n} + 2n \int_0^x \frac{t^2}{(t^2 + 5)^{n+1}} \, dt$$ $$I_n(x) = \frac{x}{(x^2 + 5)^n} + 2n \int_0^x \frac{(t^2 + 5) - 5}{(t^2 + 5)^{n+1}} \, dt$$ $$I_n(x) = \frac{x}{(x^2 + 5)^n} + 2n \, I_n(x) - 10n \, I_{n+1}(x)$$ $$10n \, I_{n+1}(x) + (1 - 2n)I_n(x) = \frac{x}{(x^2 + 5)^n}$$ Put $n = 5$
Maths · Applications of Integrals · Single correct
The area enclosed by the curves $y = \log_e (x + e^2)$, $x = \log_e \left( \frac{2}{y} \right)$ and $x = \log_e 2$, above the line $y = 1$ is
Answer: (b)
Required area is $$= \int_{e-e^2}^{0} \ln(x+e^2) - 1 \, dx + \int_{0}^{\ln 2} 2e^{-x} - 1 \, dx = 1 + e - \ln 2$$
Maths · Differential Equations · Single correct
Let y = y(x) be the solution curve of the differential equation $\frac{dy}{dx} + \frac{1}{x^2 - 1} y = \left( \frac{x-1}{x+1} \right)^{\frac{1}{2}}$, $x > 1$ passing through the point $\left( 2, \sqrt{\frac{1}{3}} \right)$. Then $\sqrt{7} y(8)$ is equal to
Answer: (d)
Given $\($ $\frac{dy}{dx}$ + $\frac{1}{x^2 - 1}$ y = $\left$( $\frac{x-1}{x+1}$ $\right$)^{$\frac{1}{2}$} $\)$, $\($ $\frac{dy}{dx}$ + Py = Q $\)$ I.F. = $\($ e^{$\int$ P $\,$ dx} = $\left$( $\frac{x-1}{x+1}$ $\right$)^{$\frac{1}{2}$} $\)$ $\($ y $\left$( $\frac{x-1}{x+1}$ $\right$)^{$\frac{1}{2}$} = $\int$ $\left$( $\frac{x-1}{x+1}$ $\right$)^{1} $\,$ dx $\)$ $\($ = x - 2 $\log$_e |x+1| + C $\)$ Curve passes through $\($ $\left$( 2, $\frac{1}{\sqrt{3}}$ $\right$) $\)$ $\($ $\Rightarrow$ C = 2 $\log$_e 3 - $\frac{5}{3}$ $\)$ at $\($ x = 8, $\)$ $\($ $\sqrt{7}$ y(8) = 19 - 6 $\log$_e 3 $\)$
Maths · Differential Equations · Single correct
The differential equation of the family of circles passing through the points (0, 2) and (0, -2) is
Answer: (a)
Equation of circle passing through $(0, -2)$ and $(0, 2)$ is $$x^2 + (y^2 - 4) + \lambda x = 0, \; (\lambda \in \mathbb{R})$$ Divided by $x$ we get $$\frac{x^2 + (y^2 - 4)}{x} + \lambda = 0$$ Differentiating with respect to $x$ $$x \left[ 2x + 2y \cdot \frac{dy}{dx} \right] - \left[ x^2 + y^2 - 4 \right] \cdot 1 \over x^2 = 0$$ $$\Rightarrow 2xy \cdot \frac{dy}{dx} + (x^2 - y^2 + 4) = 0$$
Maths · Conic Sections · Single correct
Let the tangents at two points A and B on the circle $x^2 + y^2 - 4x + 3 = 0$ meet at origin O (0, 0). Then the area of the triangle of OAB is
Answer: (b)
Given the circle $C: (x-2)^2 + y^2 = 1$. The equation of chord $AB$ is $2x = 3$. The lengths $OA = OB = \sqrt{3}$. The length $AM = \frac{\sqrt{3}}{2}$. The area of triangle $OAB$ is given by $$\frac{1}{2} (2AM)(OM) = \frac{3\sqrt{3}}{4} sq. units.$$
Maths · Conic Sections · Single correct
Let the hyperbola H : $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ pass through the point $\left(2\sqrt{2}, -2\sqrt{2}\right)$. A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?
Answer: (b)
Given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The foci are $S (ae, 0)$ and $S' (-ae, 0)$. The foot of the directrix of the parabola is $(-ae, 0)$. The focus of the parabola is $(ae, 0)$. Now, the semi-latus rectum of the parabola is $|SS'| = 2ae$. Given, $4ae = e \left( \frac{2b^2}{a} \right)$. $$\Rightarrow b^2 = 2a^2 \ldots (1)$$ Given, $\left( 2\sqrt{2}, -2\sqrt{2} \right)$ lies on $H$. $$\Rightarrow \frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{8} \ldots (2)$$ From (1) and (2), $$a^2 = 4, \; b^2 = 8$$ Therefore, $b^2 = a^2 (e^2 - 1)$. Thus, $e = \sqrt{3}$. Therefore, the equation of the parabola is $y^2 = 8\sqrt{3}x$.
Maths · Three Dimensional Geometry · Single correct
Let the lines $\($ $\frac{x-1}{\lambda}$ = $\frac{y-2}{1}$ = $\frac{z-3}{2}$ $\)$ and $\($ $\frac{x+26}{-2}$ = $\frac{y+18}{3}$ = $\frac{z+28}{\lambda}$ $\)$ be coplanar and P be the plane containing these two lines. Then which of the following points does NOT lies on P?
Answer: (d)
Given, $L_1 : \frac{x-1}{\lambda} = \frac{y-2}{1} = \frac{z-3}{2}$ and $L_2 : \frac{x+26}{-2} = \frac{y+18}{3} = \frac{z+28}{\lambda}$ are coplanar. $$\begin{vmatrix} 27 & 20 & 31 \\ \lambda & 1 & 2 \\ -2 & 3 & \lambda \end{vmatrix} = 0$$ $$\Rightarrow \lambda = 3$$ Now, normal of plane $P$, which contains $L_1$ and $L_2$, $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ -2 & 3 & 3 \end{vmatrix} = -3\hat{i} - 13\hat{j} + 11\hat{k}$$ $$\Rightarrow Equation of required plane P :$$ $$3x + 13y - 11z + 4 = 0$$ $(0, 4, 5)$ does not lie on plane $P$.
Maths · Three Dimensional Geometry · Single correct
A plane P is parallel to two lines whose direction ratios are $-2, 1, -3$, and $-1, 2, -2$ and it contains the point $(2, 2, -2)$. Let P intersect the co-ordinate axes at the points $A, B, C$ making the intercepts $\alpha, \beta, \gamma$. If $V$ is the volume of the tetrahedron $OABC$, where $O$ is the origin and $p = \alpha + \beta + \gamma$, then the ordered pair $(V, p)$ is equal to
Answer: (b)
Normal of plane P: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & -3 \\ -1 & 2 & -2 \end{vmatrix} = 4\hat{i} - \hat{j} - 3\hat{k}$$ Equation of plane P which passes through (2, 2, -2) is $4x - y - 3z - 12 = 0$. Now, $A (3, 0, 0)$, $B (0, -12, 0)$, $C (0, 0, -4)$. $$\Rightarrow \alpha = 3, \beta = -12, \gamma = -4$$ $$\Rightarrow p = \alpha + \beta + \gamma = -13$$ Now, volume of tetrahedron OABC $$V = \left| \frac{1}{6} \overrightarrow{OA} \cdot (\overrightarrow{OB} \times \overrightarrow{OC}) \right| = 24$$ $$(V, p) = (24, -13)$$
Maths · Vector Algebra · Single correct
Let S be the set of all $a \in \mathbb{R}$ for which the angle between the vectors $\vec{u} = a (\log_e b) \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{v} = (\log_e b) \hat{i} + 2 \hat{j} + 2a (\log_e b) \hat{k}, (b > 1)$ is acute. Then S is equal to
Answer: (b)
For angle to be acute $$\mathbf{u} \cdot \mathbf{v} > 0$$ $$\Rightarrow a (\log_e b)^2 - 12 + 6a (\log_e b) > 0$$ $$\forall \, b > 1$$ Let $\log_e b = t \Rightarrow t > 0$ as $b > 1$ $$y = at^2 + 6at - 12 and y > 0, \forall \, t > 0$$ $$\Rightarrow a \in \phi$$
Maths · Heights and Distances · Single correct
A horizontal park is in the shape of a triangle OAB with AB = 16. A vertical lamp post OP is erected at the point O such that $\angle PAO = \angle PBO = 15^\circ$ and $\angle PCO = 45^\circ$, where C is the midpoint of AB. Then $(OP)^2$ is equal to
Answer: (b)
Given $\($ $\frac{OP}{OA}$ = $\tan$ 15^$\circ$ $\)$ $\($ $\Rightarrow$ OA = OP $\cot$ 15^$\circ$ $\)$ $\($ $\frac{OP}{OC}$ = $\tan$ 45^$\circ$ $\Rightarrow$ OP = OC $\)$ Now, $\($ OP = $\sqrt{OA^2 - 8^2}$ $\)$ $\($ $\Rightarrow$ OP^2 = (OP)^2 $\cot$^2 15^$\circ$ - 64 $\)$ $\($ $\Rightarrow$ OP^2 = $\frac{32}{\sqrt{3}}$ (2 - $\sqrt{3}$) $\)$
Maths · Probability (Advanced) · Single correct
Let A and B be two events such that $\mathrm{P}(B|A)=\frac{2}{5}$, $\mathrm{P}(A|B)=\frac{1}{7}$ and $\mathrm{P}(A\cap B)=\frac{1}{9}$. Consider (S1) $\mathrm{P}(A'\cup B)=\frac{5}{6}$, (S2) $\mathrm{P}(A'\cap B')=\frac{1}{18}$. Then
Answer: (a)
Given $\mathrm{P(A \mid B)} = \frac{1}{7} \Rightarrow \frac{\mathrm{P(A \cap B)}}{\mathrm{P(B)}} = \frac{1}{7}$. Therefore, $\mathrm{P(B)} = \frac{7}{9}$. $\mathrm{P(B \mid A)} = \frac{2}{5} \Rightarrow \frac{\mathrm{P(A \cap B)}}{\mathrm{P(A)}} = \frac{2}{5}$. Therefore, $\mathrm{P(A)} = \frac{5}{18}$. Now, $\mathrm{P(A' \cup B)} = 1 - \mathrm{P(A \cup B)} + \mathrm{P(B)} = 1 - \mathrm{P(A)} + \mathrm{P(A \cap B)} = \frac{5}{6}$. $\mathrm{P(A' \cap B')} = 1 - \mathrm{P(A \cup B)} = 1 - \mathrm{P(A)} - \mathrm{P(B)} + \mathrm{P(A \cap B)} = \frac{1}{18}$. Therefore, both (S1) and (S2) are true.
Maths · Mathematical Reasoning · Single correct
Let $\textbf{p}$ : Ramesh listens to music. $\textbf{q}$ : Ramesh is out of his village. $\textbf{r}$ : It is Sunday. $\textbf{s}$ : It is Saturday. Then the statement “Ramesh listens to music only if he is in his village and it is Sunday or Saturday” can be expressed as
Answer: (d)
Let $p$ be "Ramesh listens to music". Let $\sim q$ be "He is in village". Let $r \lor s$ be "Saturday or Sunday". The expression is $p \Rightarrow ((\sim q) \land (r \lor s))$.
Maths · Binomial Theorem · Numerical
Let the coefficients of the middle terms in the expansion of \[ \left(\frac{1}{\sqrt{6}}+\beta x\right)^4,\quad (1-3\beta x)^2 \quad\text{and}\quad \left(1-\frac{\beta}{2}x\right)^6, \] where $\beta>0$, respectively form the first three terms of an A.P. If $d$ is the common difference of this A.P., then \[ 50-\frac{2d}{\beta^2} \] is equal to
Answer: 57
The terms $^4C_2 \times \frac{\beta^2}{6}$, $-6\beta$, $^{-6}C_3 \times \frac{\beta^3}{8}$ are in A.P. $$\beta^2 - \frac{5}{2} \beta^3 = -12\beta$$ $$\beta = \frac{12}{5} or \beta = -2 \therefore \beta = \frac{12}{5}$$ $$d = -\frac{72}{5} - \frac{144}{25} = -\frac{504}{25}$$ Therefore, $50 - \frac{2d}{\beta^2} = 57$
Maths · Permutations and Combinations · Numerical
A class contains $b$ boys and $g$ girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then $b + 3g$ is equal to
Answer: 17
Given $$^bC_3 \times ^gC_2 = 168$$. We have $$b(b-1)(b-2)(g)(g-1) = 8 \times 7 \times 6 \times 3 \times 2$$. Also, $$b + 3g = 17$$.
Maths · Conic Sections · Numerical
Let the tangents at the points P and Q on the ellipse $\frac{x^2}{2} + \frac{y^2}{4} = 1$ meet at the point $R(\sqrt{2}, 2\sqrt{2} - 2)$. If S is the focus of the ellipse on its negative major axis, then $SP^2 + SQ^2$ is equal to
Answer: 13
Ellipse is $$\frac{x^2}{2} + \frac{y^2}{4} = 1; e = \frac{1}{\sqrt{2}}; S \equiv (0, -\sqrt{2})$$ Chord of contact is $$\frac{x}{\sqrt{2}} + \frac{(2\sqrt{2} - 2)y}{4} = 1$$ $$\Rightarrow \frac{x}{\sqrt{2}} = 1 - \frac{(\sqrt{2} - 1)y}{2} solving with ellipse$$ $$\Rightarrow y = 0, \sqrt{2} \therefore x = \sqrt{2}, 1$$ $$\mathbf{P} \equiv (1, \sqrt{2}) \mathbf{Q} \equiv (\sqrt{2}, 0)$$ $$\therefore (SP)^2 + (SQ)^2 = 13$$
Maths · Binomial Theorem · Numerical
If $1+\left(2+{}^{49}C_1+{}^{49}C_2+\ldots+{}^{49}C_{49}\right)\left({}^{50}C_2+{}^{50}C_4+\ldots+{}^{50}C_{50}\right)$ is equal to $2^n\cdot m$, where $m$ is odd, then $n+m$ is equal to
Answer: 99
Given $$1 + (1 + 2^{49})(2^{49} - 1) = 2^{98}$$ Let $m = 1$, $n = 98$. Then $m + n = 99$.
Maths · Conic Sections · Numerical
Two tangent lines $l_1$ and $l_2$ are drawn from the point $(2, 0)$ to the parabola $2y^2 = -x$. If the lines $l_1$ and $l_2$ are also tangent to the circle $(x - 5)^2 + y^2 = r$, then $17r$ is equal to
Answer: 9
Given $y^2 = -\frac{x}{2}$. The equation of the tangent is $y = mx - \frac{1}{8m}$. This tangent passes through $(2, 0)$. Solving for $m$, we get $m = \pm \frac{1}{4}$, i.e., one tangent is $x - 4y - 2 = 0$. Finally, $17r = 9$.
Maths · Binomial Theorem · Fill in the blank
If $\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^9}+\ldots+\frac{10240}{3}=2^n\cdot m$, where $m$ is odd, then $m\cdot n$ is equal to ______.
Answer: 12
Given the expression: $$\frac{6}{3^{12}} + 10 \left( \frac{1}{3^{11}} + \frac{2}{3^{10}} + \frac{2^2}{3^9} + \frac{2^3}{3^8} + \ldots + \frac{2^{10}}{3} \right)$$ This simplifies to: $$\frac{6}{3^{12}} + \frac{10}{3^{11}} \left( \frac{6^{11} - 1}{6 - 1} \right)$$ Finally, we have: $$= 2^{12} \cdot 1; \; m \cdot n = 12$$
Maths · Trigonometric Functions · Fill in the blank
Let $S = \left[ -\pi, \frac{\pi}{2} \right) - \left\{ -\frac{\pi}{2}, -\frac{\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4} \right\}$. Then the number of elements in the set $A = \left\{ \theta \in S : \tan \theta \left( 1 + \sqrt{5} \tan (2\theta) \right) = \sqrt{5} - \tan (2\theta) \right\}$ is ______.
Answer: 5
Given $\tan \theta + \sqrt{5} \tan 2\theta \tan \theta = \sqrt{5} - \tan 2\theta$. $\tan 3\theta = \sqrt{5}$ $\theta = \frac{n\pi}{3} + \frac{\alpha}{3}$; $\tan \alpha = \sqrt{5}$ Five solution
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $z = a + ib$, $b \neq 0$ be complex numbers satisfying $z^2 = \overline{z} \cdot 2^{1 - |z|}$. Then the least value of $n \in \mathbb{N}$, such that $z^n = (z+1)^n$, is equal to
Answer: 6
Given $|z^2| = |\bar{z}| \cdot 2^{1 - |z|} \Rightarrow |z| = 1$. $z^2 = \bar{z} \Rightarrow z^3 = 1 \therefore z = \omega or \omega^2$. $\omega^n = (1 + \omega)^n = (-\omega^2)^n$. Least natural value of $n$ is 6.
Maths · Probability · Numerical
A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If $\sigma^2$ is the variance of X, then $100 \sigma^2$ is equal to
Answer: 56
\begin{tabular}{|l|l|l|l|l|} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & $\frac16$ & $\frac12$ & $\frac3{10}$ & $\frac1{30}$ \\ \hline \end{tabular} Given the probability distribution: The variance is calculated as: $$\sigma^2 = \sum X^2 P(X) - \left(\sum X P(X)\right)^2 = \frac{56}{100}$$ Thus, $$100 \sigma^2 = 56$$
Maths · Integrals · Numerical
The value of the integral $$\int_{0}^{\frac{\pi}{2}} \frac{60 \sin(6x)}{\sin x} \, dx$$ is equal to
Answer: 104
Given $$I = 60 \int_{0}^{\pi/2} \left( \frac{\sin 6x - \sin 4x}{\sin x} + \frac{\sin 4x - \sin 2x}{\sin x} + \frac{\sin 2x}{\sin x} \right) \, dx$$ Simplifying, we have $$I = 60 \int_{0}^{\pi/2} \left( 2 \cos 5x + 2 \cos 3x + 2 \cos x \right) \, dx$$ Further simplifying, $$I = 60 \left( \frac{2}{5} \sin 5x + \frac{2}{3} \sin 3x + 2 \sin x \right) \bigg|_{0}^{\pi/2} = 104$$
Physics · Thermodynamics · Single correct
Consider the efficiency of Carnot's engine is given by $\eta = \frac{\alpha \beta}{\sin \theta} \log_e \frac{\beta x}{kT}$, where $\alpha$ and $\beta$ are constants. If $T$ is temperature, $k$ is Boltzman constant, $\theta$ is angular displacement and $x$ has the dimensions of length. Then, choose the $\textit{incorrect}$ option.
Answer: (d)
The dimensions of $\alpha \beta$, $\eta$, and $\sin \theta$ are dimensionless. Therefore, $$[\alpha \beta] = [\eta] = [\sin \theta] = Dimensionless$$ and $$[\eta^{-1} \sin \theta] = [\alpha \beta] = D.L.$$
Physics · Motion in a Plane · Single correct
At time $t = 0$ a particle starts travelling from a height $7\hat{z} \, \mathrm{cm}$ in a plane keeping $z$ coordinate constant. At any instant of time it's position along the $x$ and $y$ directions are defined as $3t$ and $5t^3$ respectively. At $t = 1 \, \mathrm{s}$ acceleration of the particle will be
Answer: (b)
Given $\vec{r} = 3t\hat{i} + 5t^3\hat{j} + 7\hat{k}$. The second derivative is given by $$\frac{d^2 \vec{r}}{dt^2} = 30t\hat{j}.$$ At $t = 1$, $$\frac{d^2 \vec{r}}{dt^2} = 30\hat{j}.$$
Physics · Mechanical Properties of Fluids · Single correct
A pressure-pump has a horizontal tube of cross-sectional area $10 \, \mathrm{cm}^2$ for the outflow of water at a speed of $20 \, \mathrm{m/s}$. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is: [given : density of water = $1000 \, \mathrm{kg/m}^3$]
Answer: (d)
Given $F = \rho a v^2 = 10^3 \times 10^{-4} \times 20 \times 20$. $F = 400$
Physics · Laws of Motion · Single correct
A uniform metal chain of mass $m$ and length ‘$L$’ passes over a massless and frictionless pulley. It is released from rest with a part of its length ‘$l$’ is hanging on one side and rest of its length ‘$L - l$’ is hanging on the other side of the pulley. At a certain point of time, when $l = \frac{L}{x}$, the acceleration of the chain is $\frac{g}{2}$. The value of $x$ is ………
Answer: (d)
The acceleration is given by $$a = \frac{(m_2 - m_1)}{(m_2 + m_1)} g$$. We have $$\frac{g}{2} = \frac{(\lambda (L - \ell) - \lambda \ell) g}{\lambda L} \Rightarrow L = \frac{L}{4} = \frac{L}{x}$$. Therefore, $$x = 4$$.
Physics · Work, Energy and Power · Single correct
A bullet of mass $200 \, \mathrm{g}$ having initial kinetic energy $90 \, \mathrm{J}$ is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to $40 \, \mathrm{J}$ within $1 \, \mathrm{s}$, the minimum length of the pool, the bullet has a to travel so that it completely comes to rest is
Answer: (a)
Using $mv = \sqrt{2mk}$. $$u = \frac{1}{0.2} \sqrt{2 \times 0.2 \times 90} = 30 \, \mathrm{m/s}$$ $$v = \frac{1}{0.2} \sqrt{2 \times 0.2 \times 40} = 20 \, \mathrm{m/s}$$ $$a = \frac{20 - 30}{1} = -10 \, \mathrm{m/s^2}$$ $$s = \frac{-u^2}{2a} = 45 \, \mathrm{m}$$
Physics · Gravitation · Single correct
Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height h above the earth surface. Clock-1 and Clock-2 operate at time periods 4s and 6s respectively. Then the value of h is - (consider radius of earth $R_E = 6400 \, \mathrm{km}$ and $g$ on earth $10 \, \mathrm{m/s^2}$)
Answer: (c)
Given $t \propto \frac{1}{\sqrt{g}}$ and $g \propto \frac{1}{(R+h)^2}$. $$\frac{t_1}{t_2} = \sqrt{\frac{g'}{g}} = \sqrt{\frac{R^2}{(R+h)^2}}$$ $$\frac{t_1}{t_2} = \frac{4}{6} = \frac{R}{(R+h)} \Rightarrow h = 3200 \, \mathrm{km}$$
Physics · Mechanical Properties of Fluids · Single correct
Consider a cylindrical tank of radius 1 m is filled with water. The top surface of water is at 15 m from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of 5 m from the bottom. A force of $5 \times 10^5$ N is applied an the top surface of water using a piston. The speed of efflux from the hole will be : (given atmospheric pressure $P_A = 1.01 \times 10^5$ Pa, density of water $\rho_w = 1000$ kg/m$^3$ and gravitational acceleration $g = 10$ m/s$^2$)
Answer: (c)
Apply Bernoulli’s theorem between Piston and hole $P_A + \rho gh = P_0 + \frac{1}{2} \rho v_e^2$. Assuming there is no atmospheric pressure on piston $$\frac{5 \times 10^5}{\pi} + 10^3 \times 10 \times 10 = 1.01 \times 10^5 + \frac{1}{2} \times 10^3 \times v_e^2$$ $v_e = 17.8 \, \mathrm{m/s}$
Physics · Kinetic Theory · Single correct
A vessel contains $14 \, \mathrm{g}$ of nitrogen gas at a temperature of $27^\circ \mathrm{C}$. The amount of heat to be transferred to the gas to double the r.m.s. speed of its molecules will be: (Take $R = 8.32 \, \mathrm{J \, mol^{-1} \, }K^{-1}$)
Answer: (c)
Given $v_{rms} \propto \sqrt{T}$. $v_{rms} \propto \sqrt{300 \, \mathrm{K}}$, $v_{rms_f} = 2v_{rms_i}$. $v_{rms_f} \propto \sqrt{1200 \, \mathrm{K}}$. $T_f = 1200 \, \mathrm{K}$, $T_i = 300 \, \mathrm{K}$, $n = \frac{14}{28} = \frac{1}{2}$. $Q = nC_v \Delta T = \frac{1}{2} \times \frac{5R}{2} \times 900$. $Q = 9360 \, \mathrm{J}$
Physics · Electrostatic Potential and Capacitance · Single correct
A slab of dielectric constant $K$ has the same sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} d$, where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be: (Given $C_0$ = capacitance of capacitor with air as medium between plates.)
Answer: (a)
Given the diagram, we have the following equations: $$x + y + \frac{3d}{4} = d$$ Simplifying, we get: $$x + y = \frac{d}{4}$$ The initial capacitance is given by: $$\frac{A \varepsilon_0}{d} = C_0$$ The potential difference is: $$\Delta V = Ex + \frac{E}{k} \times \frac{3d}{4} + Ey$$ Simplifying, we have: $$= \frac{3Ed}{4k} + E(x + y)$$ Thus, the potential difference becomes: $$\Delta V = E \left[ \frac{3d}{4k} + \frac{d}{4} \right]$$ Further simplifying: $$\Delta V = \frac{\sigma}{\varepsilon_0} \left[ \frac{3d + dk}{4k} \right] = \frac{Qd}{A \varepsilon_0} \left[ \frac{3 + k}{4k} \right]$$ The capacitance is: $$\frac{Q}{\Delta V} = C = \frac{A \varepsilon_0}{d} \left[ \frac{4k}{3 + k} \right] = \frac{4kC_0}{k + 3}$$
Physics · Moving Charges and Magnetism · Single correct
A uniform electric field $E = \left(\frac{8m}{e}\right) \, \mathrm{V/m}$ is created between two parallel plates of length $1 \, \mathrm{m}$ as shown in figure, (where $m =$ mass of electron and $e =$ charge of electron). An electron enters the field symmetrically between the plates with a speed of $2 \, \mathrm{m/s}$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be
Answer: (b)
Given $a_y = \frac{F_y}{m} = \frac{e(E)}{m} = \frac{e\left(\frac{8m}{e}\right)}{m} = 8 \, \mathrm{m/s^2}$. $s_x = u_x t$ $1 = 2 \times t$ $t = \frac{1}{2} \, \mathrm{sec}$ $v_y = u_y + a_y t$ $v_y = 0 + 8 \times \frac{1}{2}$ $v_y = 4 \, \mathrm{m/s}$ $\tan \theta = \frac{v_y}{v_x} = \frac{4}{2} = 2 \implies \theta = \tan^{-1}(2)$
Physics · Current Electricity · Single correct
Given below are two statements : Statement I : A uniform wire of resistance 80 $\Omega$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5 $\Omega$. Statement II : Two resistance 2R and 3R are connected in parallel in a electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio 3 : 2. In the light of the above statements, choose the most appropriate answer from the options given below
Answer: (c)
Statement 1 - $R = 80 \, \Omega$. $R_1 = R_2 = R_3 = R_4 = 20 \, \Omega$. In parallel $R_{eq} = \frac{20}{4} = 5 \, \Omega$. Statement 2 - $P_{th} = \frac{V^2}{R}$. $$\frac{P_1}{P_2} = \left( \frac{R_2}{R_1} \right) = \frac{2}{3} (where P is power)$$
Physics · Moving Charges and Magnetism · Single correct
A triangular shaped wire carrying 10 A current is placed in a uniform magnetic field of 0.5 T, as shown in the figure. The magnetic force on segment CD is (Given $BC = CD = BD = 5\,\mathrm{cm}$).
Answer: (c)
F_M(CD) = BI\ell_{eff} = 0.5 \times 10 \times \left(5\sin 60^\circ \times 10^{-2}\right) = 0.216\,\mathrm{N}
Physics · Magnetism and Matter · Single correct
The magnetic field at the center of current carrying circular loop is $B_1$. The magnetic field at a distance of $\sqrt{3}$ times radius of the given circular loop from the center on its axis is $B_2$. The value of $B_1/B_2$ will be
Answer: (c)
The magnetic field at point $B_1$ is given by $$B_1 = \frac{\mu_0 I}{2R}.$$ The magnetic field at point $B_2$ is given by $$B_2 = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{1}{8} \left( \frac{\mu_0 I}{2R} \right) = \frac{B_1}{8}.$$ Therefore, the ratio of the magnetic fields is $$\frac{B_1}{B_2} = \frac{8}{1}.$$
Physics · Electromagnetic Induction · Single correct
A transformer operating at primary voltage 8 kV and secondary voltage 160 V serves a load of 80 kW. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be
Answer: (c)
Given $\($ $\frac{(8 \times 10^3)^2}{R_P}$ = 80 $\times$ 10^3 $\)$. $\($ R_P = 800 $\Omega$ $\)$. Given $\($ $\frac{(160)^2}{R_S}$ = 80 $\times$ 10^3 $\)$. $\($ R_S = 0.32 $\Omega$ $\)$.
Physics · Electromagnetic Waves · Single correct
Sun light falls normally on a surface of area 36 $\mathrm{cm}^2$ and exerts an average force of $7.2 \times 10^{-9} \, \mathrm{N}$ within a time period of $20 \, \mathrm{minutes}$. Considering a case of complete absorption, the energy flux of incident light is
Answer: (d)
$\dfrac{I}{C} \times \text{area} = \text{force}$ $\dfrac{I}{C} \times 36 \times 10^{-4} = 7.2 \times 10^{-9}$ $I = \dfrac{7.2 \times 10^{-9} \times 3 \times 10^{8}}{36 \times 10^{-9} \times 10}$ $= \dfrac{6 \times 10^{-1}}{10^{-3}}$ $I = 6 \times 10^{2}\,\dfrac{\text{W}}{\text{m}^2}$ $= 0.06\,\dfrac{\text{W}}{\text{cm}^2}$
Physics · Ray Optics and Optical Instruments · Single correct
The power of a lens (biconvex) is 1.25 $\mathrm{m}^{-1}$ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 $\mathrm{cm}$ and 40 $\mathrm{cm}$ respectively. The refractive index of surrounding medium:
Answer: (d)
P = $\frac{\mu_2}{f}$ = ($\mu$_1 - $\mu$_2) $\left$( $\frac{1}{R_1}$ - $\frac{1}{R_2}$ $\right$) (For this formula refer to NCERT Part-2, Chapter-9, Page no. 328, solved example 8) $\newline$ ($\mu$_1 is refractive index of lens and $\mu$_2 is of surrounding medium) $\newline$ 1.25 = (1.5 - $\mu$_2) $\left$( $\frac{1}{0.2}$ + $\frac{1}{0.4}$ $\right$) $\newline$ $\frac{1.25 \times 0.08}{0.6}$ = (1.5 - $\mu$_2) $\newline$ $\Rightarrow$ $\mu$_2 = $\frac{4}{3}$
Physics · Dual Nature of Radiation and Matter · Single correct
Two streams of photons, possessing energies to five and ten times the work function of metal are incident on the metal surface successively. The ratio of the maximum velocities of the photoelectron emitted, in the two cases respectively, will be
Answer: (c)
Given the equations: $$\frac{1}{2}mv_1^2 = 4\phi$$ and $$\frac{1}{2}mv_2^2 = 9\phi$$ We find the ratio of velocities: $$\frac{v_1}{v_2} = \frac{2}{3}$$
Physics · Nuclei · Single correct
A radioactive sample decays $\frac{7}{4}$ times its original quantity in 15 minutes. The half-life of the sample is
Answer: (a)
Given $\($ $\frac{I}{C}$ $\times$ area = force $\)$. $\($ $\frac{I}{C}$ $\times$ 36 $\times$ 10^{-4} = 7.2 $\times$ $10^{-9}$ $\)$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
An n.p.n transistor with current gain $\beta = 100$ in common emitter configuration is shown in figure. The output voltage of the amplifier will be
Answer: (b)
$$\frac{v_{out}}{v_{in}} = \beta \frac{R_{out}}{R_{in}}$$ $$v_{out} = \frac{100 \times 10 \times 10^{3}}{10^{3}} \times 10^{-3}$$ $$= 1 \ \text{V}$$
Physics · Communication Systems · Single correct
A FM Broadcast transmitter, using modulating signal of frequency $20 \, \mathrm{kHz}$ has a deviation ratio of $10$. The Bandwidth required for transmission is:
Answer: (d)
Given FM broadcast Modulating frequency $= 20 \, \mathrm{kHz} = f$ Deviation ratio $= \frac{frequency deviation}{modulating frequency} = \frac{\Delta f}{f}$ Therefore, frequency deviation $\Delta f = f \times 10$ $= 20 \, \mathrm{kHz} \times 10 = 200 \, \mathrm{kHz}$ Therefore, Bandwidth $= 2(f + \Delta f)$ $= 2 (20 + 200) \, \mathrm{kHz}$ $= 440 \, \mathrm{kHz}$
Physics · Motion in a Straight Line · Numerical
A ball is thrown vertically upwards with a velocity of $19.6 \, \mathrm{ms^{-1}}$ from the top of a tower. The ball strikes the ground after $6 \, \mathrm{s}$. The height from the ground up to which the ball can rise will be $\left( \frac{k}{5} \right) \, \mathrm{m}$. The value of $k$ is ..... (use $g = 9.8 \, \mathrm{m/s^2}$)
Answer: 392
Given $t_a = \frac{u}{g} = \frac{19.6}{9.8} = 2 \, \mathrm{s}$. $t_d = 6 - 2 \, \mathrm{s} = \sqrt{\frac{2h_{\max}}{g}}$. Therefore, $h_{\max} = \frac{16 \times 9.8}{2} = \frac{392}{5}$.
Physics · System of Particles and Rotational Motion · Numerical
The distance of centre of mass from end A of a one dimensional rod (AB) having mass density $$\rho = \rho_0 \left( 1 - \frac{x^2}{L^2} \right)$$ kg/m and length $L$ (in meter) is $$\frac{3L}{\alpha}$$ m. The value of $\alpha$ is ........ (where $x$ is the distance from end A)
Answer: 8
Given $\mathrm{dm} = \lambda \cdot \mathrm{dx} = \lambda_0 \left( 1 - \frac{x^2}{\ell^2} \right)$. The center of mass $X_{\mathrm{cm}}$ is given by $$X_{\mathrm{cm}} = \frac{\int x \, \mathrm{dm}}{\int \mathrm{dm}_\ell}$$ Substituting the expression for $\mathrm{dm}$, we have $$= \frac{\lambda_0 \int_0^\ell x \left( 1 - \frac{x^2}{\ell^2} \right) \mathrm{dx}}{\int_0^\ell \lambda_0 \left( 1 - \frac{x^2}{\ell^2} \right) \mathrm{dx}}$$ Evaluating the integrals, we get $$= \frac{\frac{\ell^2}{2} - \frac{\ell^4}{4\ell^2}}{\ell - \frac{\ell^3}{3\ell^2}} = \frac{3\ell}{8}$$
Physics · Mechanical Properties of Solids · Numerical
A string of area of cross-section $4 \, \mathrm{mm}^2$ and length $0.5$ is connected with a rigid body of mass $2 \, \mathrm{kg}$. The body is rotated in a vertical circular path of radius $0.5 \, \mathrm{m}$. The body acquires a speed of $5 \, \mathrm{m/s}$ at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is $\ldots\ldots \times 10^{-5}$. (Use Young’s modulus $10^{11} \, \mathrm{N/m}^2$ and $g = 10 \, \mathrm{m/s}^2$)
Answer: 30
Strain is given by the formula $\frac{F}{AY}$. Substituting the values, we have: $$\frac{mg + \frac{mv^2}{R}}{AY}$$ Substituting the given values: $$\frac{20 + \frac{2(5)^2}{0.5}}{3 \times 10^{-6} \times 10^{11}} = 30 \times 10^{-5}$$
Physics · Kinetic Theory · Numerical
At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J of work when it expands under constant pressure. The amount of heat absorbed by the gas will be ........ J.
Answer: 750
Given $W = nR \Delta T = 150 \, \mathrm{J}$. $$Q = \left( \frac{f}{2} + 1 \right) nR \Delta T = \left( \frac{8}{2} + 1 \right) 150 = 750 \, \mathrm{J}$$
Physics · Oscillations · Numerical
The potential energy of a particle of mass 4 kg in motion along the x-axis is given by $U = 4(1 - \cos 4x) \, \mathrm{J}$. The time period of the particle for small oscillation $(\sin \theta \approx 0)$ is $\left( \frac{\pi}{K} \right) \, \mathrm{s}$. The value of $K$ is ……..
Answer: 2
$U=4(1-\cos4x)$ $F=-\dfrac{dU}{dx}$ $=-4(\sin4x)\cdot4$ $=-16\sin(4x)$ For small $\theta$ $\sin\theta\approx\theta$ $F=-64x$ $a=\dfrac{F}{m}=-16x$ $\omega^2=16$ $T=\dfrac{2\pi}{\omega}=\dfrac{\pi}{2}$
Physics · Current Electricity · Numerical
An electrical bulb rated $220 \,$ $\mathrm{V}$, $100 \,$ $\mathrm{W}$, is connected in series with another bulb rated $220 \,$ $\mathrm{V}$, $60 \,$ $\mathrm{W}$. If the voltage across combination is $220 \,$ $\mathrm{V}$, the power consumed by the $100 \,$ $\mathrm{W}$ bulb will be about $\ldots$ $\ldots$ $\ldots$ $\ldots$ $\mathrm{W}$.
Answer: 14
Given $R_1 = \frac{V^2}{P} = \frac{220^2}{100} = 484$. $R_2 = \frac{V^2}{P} = \frac{220^2}{60} = 484 \left( \frac{10}{6} \right)$. $I = \frac{220}{484 + 484 \times \frac{10}{6}}$. $P_1 = I^2 R_1 = 14.06 \, \mathrm{W}$.
Physics · Current Electricity · Numerical
For the given circuit the current through battery of 6 $\mathrm{V}$ just after closing the switch 'S' will be .......... $\mathrm{A}$.
Answer: 1
Just after closing the switch S, inductor behaves like an open circuit. $$I = \frac{6}{2+4} = 1 \, \mathrm{A}$$
Physics · Ray Optics and Optical Instruments · Numerical
An object ‘o’ is placed at a distance of 100 cm in front of a concave mirror of radius of curvature 200 cm as shown in the figure. The object starts moving towards the mirror at a speed 2 cm/s. The position of the image from the mirror after 10s will be at …… cm.
Answer: 400
After 10 sec. $u = -80 \, \mathrm{cm}$ $f = -100 \, \mathrm{cm}$ $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ $v = 400 \, \mathrm{cm}$
Physics · Experimental Physics · Numerical
In an experiment with a convex lens. The plot of the image distance $(v')$ against the object distance $(\mu')$ measured from the focus gives a curve $v' \mu' = 225$. If all the distances are measured in cm. The magnitude of the focal length of the lens is ……….. cm.
Answer: 15
Given $vu = f^2$ (by Newton's formula). $$f^2 = 225$$ Therefore, $f = 15 \, \mathrm{cm}$.
Physics · Mathematics in Physics · Numerical
In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1s resolution. If measured value of length is 10 cm known to 1mm accuracy. The accuracy in the determination of g is found to be x %. The value of x is
Answer: 5
Given the formula for the period, $T = 2\pi \sqrt{\frac{\ell}{g}}$. Solving for $g$, we have $$g = \frac{1}{4\pi^2} \frac{T^2}{\ell}.$$ The relative error in $g$ is given by $$\frac{\Delta g}{g} = \frac{2 \Delta T}{T} + \frac{\Delta \ell}{\ell}.$$ Substituting the given values, $$\frac{\Delta g}{g} = 2 \cdot \frac{1}{100 \times 0.5} + \frac{1 \, \mathrm{mm}}{10 \, \mathrm{cm}}.$$ Simplifying, $$\frac{\Delta g}{g} = \frac{5}{100}.$$
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R Assertion A: Zero orbital overlap is an out of phase overlap. Reason: It results due to different orientation/direction of approach of orbitals. In the light of the above statements. Choose the correct answer from the options given below
Answer: (a)
Zero overlapping or zero overlapping due to improper orientation of orbitals.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The correct decreasing order for metallic character is
Answer: (a)
Across a period metallic character decreases
Chemistry · Thermodynamics · Single correct
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state. Reason R : The value of $\Delta G^\Theta$ becomes more on negative side as entropy is higher in liquid state than solid state. In the light of the above statements. Choose the most appropriate answer from the options given below
Answer: (a)
Given $\Delta G = \Delta H - T \Delta S$. Therefore, the entropy of liquid is more than solid. Thus, on melting the entropy increases and $\Delta G$ becomes more negative and hence it becomes easier to reduce metal.
Chemistry · The s-Block Elements · Single correct
The products obtained during treatment of hard water using Clark's method are:
Answer: (c)
In Clark's method lime water is used $$\mathrm{Ca(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 + 2H_2O}$$ $$\mathrm{Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 + Mg(OH)_2 + 2H_2O}$$
Chemistry · The s-Block Elements · Single correct
Statement I: An alloy of lithium and magnesium is used to make aircraft plates. Statement II: The magnesium ions are important for cell-membrane integrity. In the light the above statements, choose the correct answer from the options given below
Answer: (b)
Alloy of Li and Mg is used to make armour plates and not aircraft plates. Calcium plays important roles in neuromuscular function, interneuronal transmission and cell membrane integrity.
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
White phosphorus reacts with thionyl chloride to give
Answer: (b)
The reaction is given by: $$\mathrm{P_4} + 8\mathrm{SOCl_2} \rightarrow 4\mathrm{PCl_3} + 4\mathrm{SO_2} + 2\mathrm{S_2Cl_2}$$
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Concentrated $\mathrm{HNO_3}$ reacts with Iodine to give
Answer: (c)
The balanced chemical equation is: $$\mathrm{I_2} + 10\mathrm{HNO_3(conc)} \Rightarrow 2\mathrm{HIO_3} + 10\mathrm{NO_2} + 4\mathrm{H_2O}$$
Chemistry · The d-and f-Block Elements · Single correct
Which of the following pair is not isoelectronic species? (At. no. Sm, $62$; Er, $68$; Yb, $70$; Lu, $71$; Eu, $63$; Tb, $65$; Tm, $69$)
Answer: (d)
$\mathrm{Sm^{2+}} \rightarrow$ electron $= 60$ $\mathrm{Er^{3+}} \rightarrow$ electron $= 65$ $\mathrm{Tb^{2+}} \rightarrow$ electron $= 63$ $\mathrm{Tm^{4+}} \rightarrow$ electron $= 65$ (not isoelectronic)
Chemistry · Redox Reactions · Single correct
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : Permanganate titrations are not performed in presence of hydrochloric acid. Reason R : Chlorine is formed as a consequence of oxidation of hydrochloric acid. In the light of the above statements, choose the correct answer from the options given below
Answer: (a)
The reaction is given by: $$2\mathrm{KMnO_4} + 16\mathrm{HCl} \rightarrow 2\mathrm{MnCl_2} + 2\mathrm{KCl} + 8\mathrm{H_2O} + \mathrm{Cl_2}$$ HCl gets oxidised by KMnO$_4$ into Cl$_2$.
Chemistry · Co-ordination Compounds · Single correct
Match $\textbf{List I}$ with $\textbf{List II}$ \begin{tabular}{|c|c|c|c|} \hline & List I (Complex) & & List II (Hybridization) \\ \hline A & $\mathrm{Ni(CO)_4}$ & I & $sp^3$ \\ \hline B & $\mathrm{[Ni(CN)_4]^{2-}}$ & II & $sp^3d^2$ \\ \hline C & $\mathrm{[Co(CN)_6]^{3-}}$ & III & $d^2sp^3$ \\ \hline D & $\mathrm{[CoF_6]^{3-}}$ & IV & $dsp^2$ \\ \hline \end{tabular} Choose the correct answer from the options given below:
Answer: (b)
$\mathrm{Ni(CO)_4}$ Hybridisation $sp^3$ $\mathrm{[Ni(CN)_4]^{2-}}$ Hybridisation $dsp^2$ $\mathrm{[Co(CN)_6]^{3-}}$ Hybridisation $d^2sp^3$ $\mathrm{[Co(F)_6]^{3-}}$ Hybridisation $sp^3d^2$
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Dinitrogen and dioxygen, the main constituents of air do not react with each other in atmosphere to form oxides of nitrogen because
Answer: (d)
$N_2+O_2 \overset{(1483\text{--}2000\,K)}{\rightleftharpoons} 2NO$ The process is endothermic and feasible at high temperature.
Chemistry · Alcohols, Phenols and Ethers · Single correct
The Most major product in the given reaction is:
Answer: (b)
Chemistry · Hydrocarbons · Single correct
Arrange the following in increasing order of reactivity towards nitration A. p-xylene B. bromobenzene C. mesitylene D, nitrobenzene E. benzene Choose the correct answer from the options given below
Answer: (b)
$-\mathrm{NO_2}$ is strongly deactivating $-\mathrm{Br}$ $-$ deactivating $-\mathrm{CH_3}$ $-$ activating group $D < B < E < A < C$
Chemistry · Alcohols, Phenols and Ethers · Single correct
Compound I is heated with Conc. HI to give a hydroxy compound A which is further heated with Zn dust to give compound B. Identify A and B.
Answer: (d)
The reaction involves the cleavage of the ether bond using concentrated HI and heat ($\Delta$). The phenol group is then reduced to benzene using zinc and heat ($\Delta$).
Chemistry · Amines · Single correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Aniline on nitration yields ortho, meta $\&$ para nitro derivatives of aniline. Reason R : Nitrating mixture is a strong acidic mixture. In the light of the above statements, choose the correct answer from the options given below
Answer: (a)
Due to formation of anilinium ion in acidic medium, meta product is also obtained in significant amount.
Chemistry · Polymers · Single correct
Match List I with List II Choose the correct answer from the options given below:
Answer: (b)
Neoprene is elastomer. Nylon-6, 6 is fiber. PVC is thermoplastic. Novolac is thermosetting.
Chemistry · Chemistry in Everyday Life · Single correct
Two statements in respect of drug-enzyme interaction are given below Statement I : Action of an enzyme can be blocked only when an inhibitor blocks the active site of the enzyme. Statement II : An inhibitor can form a strong covalent bond with the enzyme. In the light of the above statements. Choose the correct answer from the options given below
Answer: (d)
Some drugs do not bind to active sites. These bind to different site of enzyme called allosteric sites.
Chemistry · Surface Chemistry · Single correct
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : Thin layer chromatography is an adsorption chromatography. Reason : A thin layer of silica gel is spread over a glass plate of suitable size in thin layer chromatography which acts as an adsorbent. In the light of the above statements, choose the correct answer from the options given below
Answer: (a)
Theory based Thin layer chromatography (TLC) is another type of adsorption chromatography, which involves separation of substances of a mixture over a thin layer of an adsorbent coated on a glass plate. A thin layer (about 0.2 mm thick) of an adsorbent (silica gel) or (Alumina) is spread over a glass plate of suitable size. Hence Assertion (A) is correct and Reason ($R$) is correct explanation of (A).
Chemistry · Biomolecules · Single correct
The formulas of A and B for the following reaction sequence are Fructose
Answer: (a)
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Find out the major product for the above reaction.
Answer: (c)
Chemistry · Redox Reactions · Numerical
$2\,\mathrm{L}$ of $0.2\,\mathrm{M}$ $\mathrm{H_2SO_4}$ is reacted with $2\,\mathrm{L}$ of $0.1\,\mathrm{M}$ $\mathrm{NaOH}$ solution, the molarity of the resulting product $\mathrm{Na_2SO_4}$ in the solution is \_\_\_\_ millimolar. (Nearest integer).
Answer: 25
The reaction is $\mathrm{H_2SO_4} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2SO_4} + 2\mathrm{H_2O}$. Initially, there are $0.4 \, \mathrm{mol}$ of $\mathrm{H_2SO_4}$ and $0.2 \, \mathrm{mol}$ of $\mathrm{NaOH}$. After the reaction, $0.3 \, \mathrm{mol}$ of $\mathrm{H_2SO_4}$ and $0.1 \, \mathrm{mol}$ of $\mathrm{Na_2SO_4}$ remain. The molarity of $\mathrm{Na_2SO_4}$ is calculated as $\frac{0.1}{4} = 0.025 \, \mathrm{M}$. Therefore, the molarity is $25 \, \mathrm{mM}$.
Chemistry · The Solid State · Numerical
Metal M crystallizes into a FCC lattice with the edge length of $4.0 \times 10^{-8} \, \mathrm{cm}$. The atomic mass of the metal is ______ g/mol. (Nearest integer). (Use : $N_A = 6.02 \times 10^{23} \, \mathrm{mol^{-1}}$, density of metal, $M = 9.03 \, \mathrm{g \, cm^{-3}}$)
Answer: 87
Given $a = 4 \times 10^{-8} \, \mathrm{cm}$ and $d = 9.03 \, \mathrm{g/ml}$. The formula for density is given by $$d = \frac{ZM}{N_A a^3}$$ Substituting the values, we have $$M = \frac{9.03 \times 6.02 \times 10^{23} \times 64 \times 10^{-24}}{4} = 86.97$$
Chemistry · Structure of Atom · Numerical
If the wavelength for an electron emitted from H-atom is $3.3 \times 10^{-10} \, \mathrm{m}$, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is _____ times. (Nearest integer). [Given : $h = 6.626 \times 10^{-34} \, \mathrm{Js}$, Mass of electron $= 9.1 \times 10^{-31}$]
Answer: 2
Given $\lambda=\frac{h}{\sqrt{2mK}}$. $K=\frac{h^2}{2m\lambda^2}$ $K=\frac{h^2}{2m\lambda^2}=\frac{43.9\times10^{-68}}{2\times9.1\times10^{-31}\times10.89\times10^{-20}}$ $K=2.215\times10^{-18}$ $E_{abs}=E_{req}+K$ $\frac{E_{abs}}{E_{req}}=1+\frac{K}{E_{req}}=1+\frac{2.215\times10^{-18}}{13.6\times1.602\times10^{-19}}=2.0166$
Chemistry · Solutions · Numerical
A gaseous mixture of two substances A and B, under a total pressure of 0.8 atm is in equilibrium with an ideal liquid solution. The mole fraction of substance A is 0.5 in the vapour phase and 0.2 in the liquid phase. The vapour pressure of pure liquid A is _______ atm. (Nearest integer)
Answer: 2
Given $Y_A = 0.5 \Rightarrow Y_B = 0.5$. $P_A = P_B = 0.4 \, \mathrm{atm}$. $P_A = P_A^0 X_A$. $P_A^0 = 2$.
Chemistry · Equilibrium · Numerical
At 600K, 2 mol of NO are mixed with 1 mol of $O_2$. $$2\mathrm{NO}_{(g)} + \mathrm{O}_2{(g)} \rightleftharpoons 2\mathrm{NO}_2{(g)}$$ The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that 0.6 mol of oxygen are present at equilibrium. The equilibrium constant for the reaction is _______. (Nearest integer).
Answer: 2
The reaction is given as $2\mathrm{NO} + \mathrm{O_2} \rightarrow 2\mathrm{NO_2}$. The initial moles are 2 for $\mathrm{NO}$ and 1 for $\mathrm{O_2}$. At equilibrium, the moles are $2-2x$ for $\mathrm{NO}$, $1-x$ for $\mathrm{O_2}$, and $2x$ for $\mathrm{NO_2}$. The equilibrium moles are 1.2 for $\mathrm{NO}$, 0.6 for $\mathrm{O_2}$, and 0.8 for $\mathrm{NO_2}$. The equilibrium constant $K_p$ is calculated as: $$K_p = \frac{\left(\frac{0.8}{2.6}\right)^2}{\left(\frac{1.2}{2.6}\right)^2 \left(\frac{0.6}{2.6}\right)} = 1.925$$
Chemistry · Some Basic Concepts of Chemistry · Numerical
A sample of 0.125 g of an organic compound when analysed by Duma’s method yields 22.78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mm Hg. The percentage of nitrogen in the given organic compound is ____. (Nearest integer). (a) The vapour pressure of water at $280\ \mathrm{K}$ is $14.2\ \mathrm{mm\ Hg}$ (b) $R=0.082\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$
Answer: 22
Given $V = 22.78 \, \mathrm{ml}$, $T = 280 \, \mathrm{K}$. The total pressure $P_{total} = 759 \, \mathrm{mmHg}$. The pressure of nitrogen $P_{\mathrm{N_2}} = 759 - 14.2 = 744.8 \, \mathrm{mmHg}$. The number of moles of nitrogen $n_{\mathrm{N_2}}$ is calculated as $$n_{\mathrm{N_2}} = \frac{744.8 \times 22.78}{760 \times 1000 \times 0.082 \times 280} = 0.00097.$$ The weight of nitrogen $W_{Nitrogen} = 0.02716$. The percentage of nitrogen is $$\%N = \frac{0.02716}{0.125} \times 1000 = 21.728.$$
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
On reaction with stronger oxidizing agent like $\mathrm{KIO}_4$, hydrogen peroxide oxidizes with the evolution of $\mathrm{O}_2$. The oxidation number of I in $\mathrm{KIO}_4$ changes to .
Answer: 5
The chemical reaction is given by the equation: $$\mathrm{IO_4^- + H_2O_2 \rightarrow IO_3^- + O_2}$$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For a reaction, given below is the graph of $\ln k$ vs $\frac{1}{T}$. The activation energy for the reaction is equal to ________ cal mol$^{-1}$. (Nearest integer). (Given : $R = 2$ cal K$^{-1}$ mol$^{-1}$)
Answer: 8
Given the equation $K = A e^{-E_a/RT}$. Taking the natural logarithm, we have $$\ln k = \frac{-E_a}{RT} + \ln A.$$ The slope is given by $$Slope = \frac{E_a}{R} = \frac{20}{5}.$$ Therefore, $$E_a = 4R = 8 \, Cal/mol.$$
Chemistry · Surface Chemistry · Single correct
Among the following the number of curves not in accordance with Freundlich adsorption isotherm is
Answer: (c)
Given $\($ $\frac{X}{m}$ = KP^{$\frac{1}{n}$} $\)$. Taking the logarithm of both sides, we have: $$ \log \frac{x}{m} = \frac{1}{n} \log p + \log k $$ The graph of $\($ $\log$ $\frac{x}{m}$ $\)$ versus $\($ $\log$ p $\)$ shows a slope of $\($ $\frac{1}{n}$ $\)$ and an intercept of $\($ $\log$ k $\)$.
Chemistry · Thermodynamics · Fill in the blank
Among the following the number of state variable is _____. Internal energy (U), Volume (V), Heat (q), Enthalpy (H)
Answer: (c)
Internal energy, volume, enthalpy are state variables.