JEE Main 25 June 2022 Shift 1 question paper with solutions
JEE Main 25 June 2022 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Conic Sections · Single correct
Let a circle C touch the lines $L_1 : 4x - 3y + K_1 = 0$ and $L_2 : 4x - 3y + K_2 = 0$, $K_1, K_2 \in \mathbb{R}$. If a line passing through the centre of the circle C intersects $L_1$ at $(-1, 2)$ and $L_2$ at $(3, -6)$, then the equation of the circle C is
$(x - 1)^2 + (y - 2)^2 = 4$
$(x + 1)^2 + (y - 2)^2 = 4$
$(x - 1)^2 + (y + 2)^2 = 16$
$(x - 1)^2 + (y - 2)^2 = 16$
Answer: (c)
Solution
Given the lines $L_1 : 4x - 3y + K_1 = 0$ and $L_2 : 4x - 3y + K_2 = 0$. Now, $$-4 - 6 + K_1 = 0 \implies K_1 = 10$$ $$12 + 18 + K_2 = 0 \implies K_2 = -30$$ Tangent to the circle are $$4x - 3y + 10 = 0$$ $$4x - 3y - 30 = 0$$ Length of diameter $2r = \left| \frac{10 + 30}{5} \right| = 8$ Therefore, $r = 4$. Now, the center is the midpoint of $A$ and $B$. $x = 1, \; y = -2$ Equation of the circle $$(x - 1)^2 + (y + 2)^2 = 16$$
Question 2
Maths · Integrals · Single correct
The value of $$\int_{0}^{\pi} \frac{e^{\cos x} \sin x}{(1 + \cos^2 x)(e^{\cos x} + e^{-\cos x})} \, dx$$ is equal to
$\frac{\pi^2}{4}$
$\frac{\pi^2}{2}$
$\frac{\pi}{4}$
$\frac{\pi}{2}$
Answer: (c)
Solution
Given $$I = \int_0^\pi \frac{e^{\cos x} \sin x}{(1 + \cos^2 x)(e^{\cos x} + e^{-\cos x})} \, dx \ldots (1)$$ Use King's property $$I = \int_0^\pi \frac{e^{-\cos x} \sin x}{(1 + \cos^2 x)(e^{-\cos x} + e^{\cos x})} \, dx \ldots (2)$$ On adding equation (1) and (2), we get $$2I = \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \, dx = 2 \int_0^{\pi/2} \frac{\sin x}{1 + \cos^2 x} \, dx$$ On putting $\cos x = t$, we get $$I = \int_0^1 \frac{dt}{1 + t^2} = \left( \tan^{-1} t \right)_0^1 = \frac{\pi}{4}$$
Question 3
Maths · Properties of Triangles · Single correct
Let a, b and c be the length of sides of a triangle ABC such that $\frac{a+b}{7}$ = $\frac{b+c}{8}$ = $\frac{c+a}{9}$ . If r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of $\($ $\frac{R}{r}$ $\)$ is equal to
Let f : $\mathbb{N}$ $\to$ $\mathbb{R}$ be a function such that f(x+y) = 2f(x)f(y) for natural numbers x and y. If f(1) = 2, then the value of $\alpha$ for which $$\sum_{k=1}^{10} f(\alpha + k) = \frac{512}{3} (2^{20} - 1)$$ holds, is
Let A be a 3 $\times$ 3 real matrix such that $$A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}; A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$$ and $$A \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}.$$ If $$X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}^T$$ and I is an identity matrix of order 3, then the system $$(A - 2I)X = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}$$ has
Maths · Continuity and Differentiability · Single correct
Let f : $\mathbb{R}$ $\to$ $\mathbb{R}$ be defined as f(x) = x^3 + x - 5. If g(x) is a function such that f(g(x)) = x, $\forall$ x $\in$ $\mathbb{R}$, then g'(63) is equal to .
$\frac{1}{49}$
$\frac{3}{49}$
$\frac{43}{49}$
$\frac{91}{49}$
Answer: (a)
Solution
Given $f(x) = x^3 + x - 5$. Differentiating, we have $f'(x) = 3x^2 + 1$, which is an increasing function. Therefore, $f(x)$ is invertible. Let $g(x)$ be the inverse of $f(x)$. Then $g(f(x)) = x$. Differentiating, we get $g'(f(x))f'(x) = 1$. Given $f(x) = 63$, we solve $x^3 + x - 5 = 63$, which gives $x = 4$. Substituting $x = 4$, we have $g'(f(4))f'(4) = 1$. Thus, $g'(63) \times 49 = 1$ since $f'(4) = 49$. Therefore, $g'(63) = \frac{1}{49}$.
Question 7
Maths · Mathematical Reasoning · Single correct
Consider the following two propositions: P1 : $\sim (p \rightarrow \sim q)$ P2 : $(p \land \sim q) \land ((\sim p) \lor q)$ If the proposition $p \rightarrow ((\sim p) \lor q)$ is evaluated as FALSE, then:
P1 is TRUE and P2 is FALSE
P1 is FALSE and P2 is TRUE
Both P1 and P2 are FALSE
Both P1 and P2 are TRUE
Answer: (c)
Solution
The expression $p \to (\sim p \lor q)$ is false when $p$ is true and $q$ is false. From the table, $P1$ and $P2$ both are false.
Question 8
Maths · Sequences and Series · Single correct
If $\frac{1}{2^3 \cdot 3^{10}} + \frac{1}{2^2 \cdot 3^9} + \ldots + \frac{1}{2^{10} \cdot 3} = \frac{K}{2^{10} \cdot 3^{10}}$, then the remainder when $K$ is divided by $6$ is
Let $E_1$ and $E_2$ be two events such that the conditional probabilities $\mathrm{P}(E_1 | E_2) = \frac{1}{2}$, $\mathrm{P}(E_2 | E_1) = \frac{3}{4}$ and $\mathrm{P}(E_1 \cap E_2) = \frac{1}{8}$. Then:
Let $A= \begin{bmatrix} 0 & -2\\ 2 & 0 \end{bmatrix}$. If $M$ and $N$ are two matrices given by $M=\sum_{k=1}^{10}A^{2k}$ and $N=\sum_{k=1}^{10}A^{2k-1}$, then $MN^2$ is ______.
a non-identity symmetric matrix
a skew-symmetric matrix
neither symmetric nor skew-symmetric matrix
an identity matrix
Answer: (a)
Solution
Given $$A = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$$ We calculate $$A^2 = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix} = -4I$$ Then $$A^3 = -4A$$ $$A^4 = (-4I)(-4I) = (-4)^2 I$$ $$A^5 = (-4)^2 A, A^6 = (-4)^3 I$$ Now, $$M = \sum_{k=1}^{10} A^{2k} = A^2 + A^4 + \ldots + A^{20}$$ $$= [-4 + (-4)^2 + (-4)^3 + \ldots + (-4)^{20}]I$$ $$= -4\lambda I$$ Thus, $M$ is a symmetric matrix. For $N$, $$N = \sum_{k=1}^{10} A^{2k-1} = A + A^3 + \ldots + A^{19}$$ $$= A[1 + (-4) + (-4)^2 + \ldots + (-4)^9]$$ $$= \lambda A \implies skew symmetric$$ Therefore, $N^2$ is a symmetric matrix and $MN^2$ is a non-identity symmetric matrix.
Question 12
Maths · Applications of Integrals · Single correct
Let g : (0, $\infty$) $\to$ $\mathbb{R}$ be a differentiable function such that $\int$( $\frac{x(\cos x - \sin x)}{e^x + 1}$ + $\frac{g(x)(e^x + 1 - xe^x)}{(e^x + 1)^2}$) $\mathrm{d}$x = $\frac{xg(x)}{e^x + 1}$ + c, for all x > 0, where c is an arbitrary constant. Then.
g is decreasing in $\left(0, \frac{\pi}{4}\right)$
g' is increasing in $\left(0, \frac{\pi}{4}\right)$
g + g' is increasing in $\left(0, \frac{\pi}{2}\right)$
g - g' is increasing in $\left(0, \frac{\pi}{2}\right)$
Answer: (d)
Solution
On differentiating both sides with respect to $x$, we get $$\left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - xe^x)}{(e^x + 1)^2} \right) = \frac{(e^x + 1)(g(x) + xg'(x)) - e^x \cdot x \cdot g(x)}{(e^x + 1)^2}$$ $$(e^x + 1)x(\cos x - \sin x) + g(x)(e^x + 1 - xe^x)$$ $$= (e^x + 1)(g(x) + xg'(x)) - e^x \cdot x \cdot g(x)$$ $$\Rightarrow g'(x) = \cos x - \sin x$$ $$\Rightarrow g(x) = \sin x + \cos x + C$$ $g(x)$ is increasing in $(0, \pi/4)$ $$g''(x) = -\sin x - \cos x 0$$ $$\Rightarrow \phi$ is increasing Hence option D is correct.
Question 13
Maths · Applications of Derivatives · Single correct
Let f : $\mathbb{R} \to \mathbb{R}$ and g : $\mathbb{R} \to \mathbb{R}$ be two functions defined by $f(x) = \log_e(x^2 + 1) - e^{-x} + 1$ and $g(x) = \frac{1 - 2e^{2x}}{e^x}$. Then, for which of the following range of $\alpha$, the inequality $$f\left(g\left(\frac{(\alpha - 1)^2}{3}\right)\right) > f\left(g\left(\frac{\alpha - 5}{3}\right)\right)$$ holds?
(2, 3)
(-2, -1)
(1, 2)
(-1, 1)
Answer: (a)
Solution
Given $f(x) = \log_e(x^2 + 1) - e^{-x} + 1$. Differentiating, we have $$f'(x) = \frac{2x}{x^2 + 1} + e^{-x} > 0 \forall \; x \in \mathbb{R}$$ This implies $f$ is strictly increasing. Let $g(x) = \frac{1 - 2e^{2x}}{e^x} = e^{-x} - 2e^x$. Differentiating, we find $$g'(x) = -(2e^x + e^{-x}) f\left(g\left(\alpha - \frac{5}{3}\right)\right)$$ This implies $$g\left(\frac{(\alpha - 1)^2}{3}\right) > g\left(\alpha - \frac{5}{3}\right)$$ Thus, $$\frac{(\alpha - 1)^2}{3} < \alpha - \frac{5}{3}$$ Simplifying, we get $$\alpha^2 - 5\alpha + 6 < 0$$ Factoring, $$(\alpha - 2)(\alpha - 3) < 0$$ Therefore, $\alpha \in (2, 3)$.
Question 14
Maths · Vector Algebra · Single correct
Let \[ \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}, \] where $a_i>0,\ i=1,2,3$, be a vector which makes equal angles with the coordinate axes $\mathrm{OX}$, $\mathrm{OY}$ and $\mathrm{OZ}$. Also, let the projection of $\vec{a}$ on the vector $3\hat{i}+4\hat{j}$ be $7$. Let $\vec{b}$ be a vector obtained by rotating $\vec{a}$ through $90^\circ$. If $\vec{a}$, $\vec{b}$ and the $x$-axis are coplanar, then the projection of vector $\vec{b}$ on $3\hat{i}+4\hat{j}$ is equal to
Let y = y(x) be the solution of the differential equation (x + 1)y' - y = e^{3x}(x + 1)^2, with y(0) = $\frac{1}{3}$. Then, the point x = -$\frac{4}{3}$ for the curve y = y(x) is:
not a critical point
a point of local minima
a point of local maxima
a point of inflection
Answer: (b)
Solution
(x + 1)dy - y $\,$ dx = e^{3x}(x + 1)^2 $\frac{(x + 1)dy - y \, dx}{(x + 1)^2}$ = e^{3x} $\mathrm{d}$$\left$($\frac{y}{x+1}$$\right$) = e^{3x} $\implies$ $\frac{y}{x+1}$ = $\frac{e^{3x}}{3}$ + C $\left$(0, $\frac{1}{3}$$\right$) $\implies$ C = 0 $\implies$ y = $\frac{(x+1)e^{3x}}{3}$ $\frac{dy}{dx}$ = $\frac{1}{3}$((x+1)3e^{3x} + e^{3x}) = $\frac{3^{3x}}{3}$(3x + 4) Clearly, $\;$ x = $\frac{-4}{3}$ $\;$ is point of local minima
Question 16
Maths · Conic Sections · Single correct
If $y = m_1x + c_1$ and $y = m_2x + c_2$, $m_1 \neq m_2$ are two common tangents of circle $x^2 + y^2 = 2$ and parabola $y^2 = x$, then the value of $8|m_1m_2|$ is equal to
3 + 4$\sqrt{2}$
-5 + 6$\sqrt{2}$
-4 + 3$\sqrt{2}$
7 + 6$\sqrt{2}$
Answer: (c)
Solution
Given the equations $C_1: x^2 + y^2 = 2$ and $C_2: y^2 = x$. Let the tangent to the parabola be $y = mx + \frac{1}{4m}$. It is also a tangent of the circle, so the distance from the center of the circle $(0, 0)$ will be $\sqrt{2}$. $$\left| \frac{1}{\frac{4m}{\sqrt{1+m^2}}} \right| = \sqrt{2} \implies 1 = 32m^2 + 32m^4$$ By solving $$m^2 = \frac{3\sqrt{2} - 4}{8}, m^2 = \frac{-3\sqrt{2} - 4}{8} (rejected)$$ $$m = \pm \frac{\sqrt{3\sqrt{2} - 4}}{8}$$ So, $8 \left| m_1 m_2 \right| = 3\sqrt{2} - 4$
Question 17
Maths · Three Dimensional Geometry · Single correct
Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, -1, -1), parallel to the line PQ meets the plane S at R, then QR^2 is equal to:
2
5
7
11
Answer: (b)
Solution
Let parallel vector of L = $\vec{b}$. Mirror image of Q on given plane $x+y+z=5$: $$\frac{a-1}{1} = \frac{b-0}{1} = \frac{c-1}{1} = \frac{-2(2-5)}{3}$$ $a = 3$, $b = 2$, $c = 3$ $Q = (3, 2, 3)$ Therefore, $\vec{b} \parallel \overline{PQ}$ So, $\vec{b} = (1, 1, 1)$ Equation of line L: $$\frac{x-1}{1} = \frac{y+1}{1} = \frac{z+1}{1}$$ Let point R, $(\lambda + 1, \lambda - 1, \lambda - 1)$ lying on plane $x + y + z = 5$, so, $3\lambda - 1 = 5$ Therefore, $\lambda = 2$ Point R is $(3, 1, 1)$ $QR^2 = 5$ Ans.
Question 18
Maths · Differential Equations · Single correct
If the solution curve $y = y(x)$ of the differential equation $y^2 dx + (x^2 - xy + y^2) dy = 0$, which passes through the point $(1, 1)$ and intersects the line $y = \sqrt{3} \, x$ at the point $(\alpha, \sqrt{3} \, \alpha)$, then value of $\log_e(\sqrt{3} \, \alpha)$ is equal to
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$\frac{\pi}{12}$
$\frac{\pi}{6}$
Answer: (c)
Solution
Given $$y^2 dx - xy \, dy = - (x^2 + y^2) dy$$ Rearranging, $$y(y \, dx - x \, dy) = - (x^2 + y^2) \, dy$$ $$- y(x \, dx - y \, dx) = - (x^2 + y^2) dy$$ This implies $$\frac{x \, dy - y \, dx}{x^2} = \left(1 + \frac{y^2}{x^2}\right) \frac{dy}{y}$$ Therefore, $$\Rightarrow \frac{d(y/x)}{1 + \frac{y^2}{x^2}} = \frac{dy}{y}$$ Integrating both sides, $$\Rightarrow \tan^{-1}\left(\frac{y}{x}\right) = \ln y + C$$ Given the points $$(\alpha, \sqrt{3} \alpha) \Rightarrow \frac{\pi}{3} = \ln(\sqrt{3} \alpha) + \frac{\pi}{4}$$ Thus, $$\therefore \ln(\sqrt{3} \alpha) = \frac{\pi}{12}$$
Question 19
Maths · Conic Sections · Single correct
Let x = 2t, y = $\frac{t}{3}$ be a conic. Let S be the focus and B be the point on the axis of the conic such that SA $\perp$ BA, where A is any point on the conic. If k is the ordinate of the centroid of $\Delta$ SAB, then $\lim$_{t $\to$ 1} k is equal to
$\frac{17}{18}$
$\frac{19}{18}$
$\frac{11}{18}$
$\frac{13}{18}$
Answer: (d)
Solution
Parabola $x^2=12y$ $SA\perp SB$ so, $m_{AS}\cdot m_{AB}=-1$ $\left( \frac{3-t^2}{-2t} \right) \left( \frac{\alpha-t^2}{0-2t} \right) =-1$ By solving, $3\alpha= \frac{27t^2+t^4}{t^2-9}$ Ordinate of centroid of $\triangle SAB$ $=K$ $=\frac{\alpha+\frac{t^2}{3}+3}{3}$ $K= \frac{9+3\alpha+t^2}{9}$ $\lim_{t\to1}K$ $=\lim_{t\to1} \frac1{9} \left( 9+t^2+ \frac{27t^2+t^4}{t^2-9} \right)$ $=\frac{13}{18}$
Question 20
Maths · Complex Numbers and Quadratic Equations · Single correct
Let a circle C in complex plane pass through the points $z_1 = 3 + 4i$, $z_2 = 4 + 3i$ and $z_3 = 5i$. If $z(\neq z_1)$ is a point on C such that the line through $z$ and $z_1$ is perpendicular to the line through $z_2$ and $z_3$, then $\arg(z)$ is equal to:
$\tan^{-1}\left(\frac{2}{\sqrt{5}}\right) - \pi$
$\tan^{-1}\left(\frac{24}{7}\right) - \pi$
$\tan^{-1}(3) - \pi$
$\tan^{-1}\left(\frac{3}{4}\right) - \pi$
Answer: (b)
Solution
Slope of BC = $\frac{3 - 5}{4 - 0}$ = -$\frac{1}{2}$. Slope of AP = 2. Equation of AP: y - 4 = 2(x - 3). Therefore, y = 2(x - 1). P lies on circle x^2 + y^2 = 25. Therefore, x^2 + (2(x - 1))^2 = 25. Therefore, x = -$\frac{7}{5}$ and y = -$\frac{24}{5}$. Therefore, arg (z) = $\tan$^{-1}$\left$($\frac{24}{7}$$\right$) - $\pi$.
Question 21
Maths · Binomial Theorem · Numerical
Let $C_r$ denote the binomial coefficient of $x^r$ in the expansion of $(1 + x)^{10}$. If $\alpha, \beta \in \mathbb{R}$. $C_1 + 3 \cdot 2C_2 + 5 \cdot 3C_3 + \ldots$ upto 10 terms $$= \frac{\alpha \times 2^{11}}{2^\beta - 1} \left( C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots upto 10 terms \right)$$ then the value of $\alpha + \beta$ is equal to
Answer: 286
Question 22
Maths · Permutations and Combinations · Numerical
The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is _____.
Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$, where \[ |\vec{a}|=4,\qquad |\vec{b}|=3,\qquad \theta\in\left(\frac{\pi}{4},\frac{\pi}{3}\right). \] Then \[ \left|(\vec{a}-\vec{b})\times(\vec{a}+\vec{b})\right|^2 +4(\vec{a}\cdot\vec{b})^2 \] is equal to $\underline{\hspace{2cm}}$.
Let the abscissae of the two points P and Q be the roots of $2x^2 - rx + p = 0$ and the ordinates of P and Q be the roots of $x^2 - sx - q = 0$. If the equation of the circle described on PQ as diameter is $2(x^2 + y^2) - 11x - 14y - 22 = 0$, then $2r + s - 2q + p$ is equal to
Answer: 7
Solution
Given the equations $2x^2 - rx + p = 0$ with roots $x_1, x_2$ and $y^2 - sy - q = 0$ with roots $y_1, y_2$. The equation of the circle with $PQ$ as diameter is $2(x^2 + y^2) - rx - 2sy + p - 2q = 0$. On comparing with the given equation, we have $r = 11$, $s = 7$, and $p - 2q = -22$. Therefore, $2r + s - 2q + p = 22 + 7 - 22 = 7$.
Question 25
Maths · Trigonometric Functions · Numerical
The number of values of $x$ in the interval $\left( \frac{\pi}{4}, \frac{7\pi}{4} \right)$ for which $14 \csc^2 x - 2 \sin^2 x = 21 - 4 \cos^2 x$ holds, is _______
Answer: 4
Solution
Given $x \in \left( \frac{\pi}{4}, \frac{7\pi}{4} \right)$. $14 \csc^2 x - 2 \sin^2 x = 21 - 4 \cos^2 x$ $= 21 - 4 (1 - \sin^2 x)$ $= 17 + 4 \sin^2 x$ $14 \csc^2 x - 6 \sin^2 x = 17$ Let $\sin^2 x = p$. $\frac{14}{p} - 6p = 17 \Rightarrow 14 - 6p^2 = 17p$ $6p^2 + 17p - 14 = 0$ $p = -3.5, \frac{2}{3} \Rightarrow \sin^2 x = \frac{2}{3}$ $\Rightarrow \sin x = \pm \sqrt{\frac{2}{3}}$
Question 26
Maths · Sequences and Series · Numerical
For a natural number n, let $a_n = 19^n - 12^n$. Then, the value of $$\frac{31 \alpha_9 - \alpha_{10}}{57 \alpha_8}$$ is
Let f : $\mathbb{R}$ $\to$ $\mathbb{R}$ be a function defined by $$f(x) = 2 \left( 1 - \frac{x^{25}}{2} \right) \left( 2 + x^{25} \right)^{\frac{1}{50}}.$$ If the function $$g(x) = f\left(f\left(f(x)\right)\right) + f\left(f(x)\right),$$ the greatest integer less than or equal to $g(1)$ is
Let the lines $$L_1 : \vec{r} = \lambda (\hat{i} + 2\hat{j} + 3\hat{k}), \lambda \in \mathbb{R}$$ $$L_2 : \vec{r} = (\hat{i} + 3\hat{j} + \hat{k}) + \mu (\hat{i} + \hat{j} + 5\hat{k}); \mu \in \mathbb{R}$$ intersect at the point $S$. If a plane $ax + by - z + d = 0$ passes through $S$ and is parallel to both the lines $L_1$ and $L_2$, then the value of $a + b + d$ is equal to .
Answer: 5
Solution
Both the lines lie in the same plane. Therefore, equation of the plane $$\begin{vmatrix} x & y & z \\ 1 & 2 & 3 \\ 1 & 1 & 5 \end{vmatrix} = 0$$ $$\Rightarrow 7x - 2y - z = 0$$ Therefore, $a + b + d = 5$
Question 29
Maths · Matrices · Numerical
Let A be a $3 \times 3$ matrix having entries from the set $\{-1, 0, 1\}$. The number of all such matrices A having sum of all the entries equal to 5, is
Answer: 414
Solution
Case-I: 1 occurs 7 times and -1 occurs 2 times. The number of possible matrices is $$\frac{9!}{7! \, 2!} = 36$$. Case-II: 1 occurs 6 times, -1 occurs 1 time, and 0 occurs 2 times. The number of possible matrices is $$\frac{9!}{6! \, 2!} = 252$$. Case-III: 1 occurs 5 times, and 0 occurs 4 times. The number of possible matrices is $$\frac{9!}{5! \, 4!} = 126$$. Hence the total number of all such matrices A is 414.
Question 30
Maths · Sequences and Series · Numerical
The greatest integer less than or equal to the sum of first 100 terms of the sequence $$\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, \ldots$$ is equal to
Answer: 98
Solution
The series is given by $$\frac{1}{3} + \frac{5}{9} + \frac{19}{27} + \frac{65}{81} + \ldots$$ This can be rewritten as $$\left(1 - \frac{2}{3}\right) + \left(1 - \frac{4}{9}\right) + \left(1 - \frac{8}{27}\right) + \left(1 - \frac{16}{81}\right) \ldots 100 terms$$ This simplifies to $$100 - \left[\frac{2}{3} + \left(\frac{2}{3}\right)^2 + \ldots \right]$$ Using the formula for the sum of a geometric series, we have $$100 - \frac{\frac{2}{3} \left(1 - \left(\frac{2}{3}\right)^{100}\right)}{1 - \frac{2}{3}}$$ This simplifies to $$100 - 2 \left(1 - \left(\frac{2}{3}\right)^{100}\right)$$ Therefore, $$S = 98 + 2 \left(\frac{2}{3}\right)^{100}$$ Thus, $$\Rightarrow [S] = 98$$
Physics
Question 31
Physics · Mathematics in Physics · Single correct
If $Z = \frac{A^2 B^3}{C^4}$, then the relative error in $Z$ will be:
Physics · System of Particles and Rotational Motion · Single correct
If force $\mathbf{F} = 3\hat{i} + 4\hat{j} - 2\hat{k}$ acts on a particle having position vector $2\hat{i} + \hat{j} + 2\hat{k}$ then, the torque about the origin will be :-
The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given $g =$ acceleration due to gravity at the surface of earth)
$g/2$
$g/4$
$g/3$
$g/9$
Answer: (d)
Solution
Given $g = \frac{Gm}{r^2}$. For the new position, $g' = \frac{Gm}{(3r)^2}$. Simplifying, $g' = \frac{Gm}{9r^2}$. Therefore, $g' = \frac{g}{9}$.
Question 36
Physics · Mechanical Properties of Fluids · Single correct
The terminal velocity ($v_t$) of the spherical rain drop depends on the radius ($r$) of the spherical rain drop as:-
$r^{1/2}$
$r$
$r^2$
$r^3$
Answer: (c)
Solution
The terminal velocity $V_t$ is given by the equation $$V_t = \frac{2}{9} \frac{gr^2 (\rho_p - \rho_l)}{\eta};$$ where $V_t \propto r^2$.
Question 37
Physics · Kinetic Theory · Single correct
The relation between root mean square speed ($v_{rms}$) and most probable speed ($v_{p}$) for the molar mass $M$ of oxygen gas molecule at the temperature of 300 K will be :-
$v_{rms} = \sqrt{\frac{2}{3}} v_{p}$
$v_{rms} = \sqrt{\frac{3}{2}} v_{p}$
$v_{rms} = v_{p}$
$v_{rms} = \sqrt{\frac{1}{3}} v_{p}$
Answer: (b)
Solution
Given $v_{rms} = \sqrt{\frac{3RT}{M}}$ and $v_{mp} = \sqrt{\frac{2RT}{M}}$. Thus $v_{rms} = \sqrt{\frac{3}{2}} v_{mp}$.
Question 38
Physics · Electric Charges and Fields · Single correct
In the figure, a very large plane sheet of positive charge is shown. $P_1$ and $P_2$ are two points at distance $l$ and $2l$ from the charge distribution. If $\sigma$ is the surface charge density, then the magnitude of electric fields $E_1$ and $E_2$ at $P_1$ and $P_2$ respectively are:
As the sheet is very large, $\vec{E}$ is independent of distance from it. Thus $E_1 = E_2 = \frac{\sigma}{2 \varepsilon_0}$.
Question 39
Physics · Alternating Current · Single correct
Match List-I with List-II List-I (A) AC generator (B) Galvanometer (C) Transformer (D) Metal detector List-II (I) Detects the presence of current in the circuit (II) Converts mechanical energy into electrical energy (III) Works on the principle of resonance in AC circuit (IV) Changes an alternating voltage for smaller or greater value Choose the correct answer from the options given below :-
(A)–(II), (B)–(I), (C)–(IV), (D)–(III)
(A)–(II), (B)–(I), (C)–(III), (D)–(IV)
(A)–(III), (B)–(IV), (C)–(II), (D)–(I)
(A)–(III), (B)–(I), (C)–(II), (D)–(IV)
Answer: (a)
Solution
AC generator converts mechanical energy into electrical energy. Galvanometer shows deflection when current passes through it so it is used to show presence of current in any wire. Transformer is used to step up or step down the voltage. Metals detectors contain inductor coils and use principle of induction and resonance in AC circuit.
Question 40
Physics · Moving Charges and Magnetism · Single correct
A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r < R) from its centre will be :-
$B \propto r^2$
$B \propto r$
$B \propto \frac{1}{r^2}$
$B \propto \frac{1}{r}$
Answer: (b)
Solution
Question 41
Physics · Alternating Current · Single correct
If wattless current flows in the AC circuit, then the circuit is
Purely Resistive circuit
Purely Inductive circuit
LCR series circuit
RC series circuit only
Answer: (b)
Solution
Purely Inductive circuit $$\theta = \frac{\pi}{2}$$ $$\cos \frac{\pi}{2} = 0$$ Average power = 0
Question 42
Physics · Electromagnetic Waves · Single correct
The electric field in an electromagnetic wave is given by $\mathbf{E} = 56.5 \sin \omega (t - x/c) \, \mathrm{NC}^{-1}$. Find the intensity of the wave if it is propagating along $x$-axis in the free space. (Given $\varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C}^2 \mathrm{N}^{-1} \mathrm{m}^{-2}$)
The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point P and $\pi$ at point Q. Then the difference between the resultant intensities at P and Q will be:
A light wave travelling linearly in a medium of dielectric constant 4, incident on the horizontal interface separating medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be (Given : relative permeability of medium $\mu_r = 1$)
$10^\circ$
$20^\circ$
$30^\circ$
$60^\circ$
Answer: (d)
Solution
For total internal reflection, $i > \theta_C$. $$\Rightarrow \sin i > \sin \theta_C$$ $$\Rightarrow \sin i > \frac{\mu_R}{\mu_D} \ldots (1)$$ Also, $\mu = \sqrt{\mu_r \varepsilon_r}$. $$\frac{\mu_R}{\mu_D} = \frac{\sqrt{1 \times 1}}{\sqrt{4 \times 1}} = \frac{1}{2}$$ From (1), $\sin i > \frac{1}{2} \Rightarrow i > 30^\circ$, $i = 60^\circ$.
Question 45
Physics · Dual Nature of Radiation and Matter · Single correct
Given below are two statements :- Statement I : Davisson-Germer experiment establishes the wave nature of electrons. Statement II : If electrons have wave nature, they can interfere and show diffraction. In the light of the above statements choose the correct answer from the options given below:-
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Answer: (a)
Solution
In Davisson-Germer experiment the electrons exhibit diffraction thereby proving that electrons have wave nature. Hence both statements are correct. Both the options are correct by concept.
Question 46
Physics · Atoms · Single correct
The ratio for the speed of the electron in the $3^{rd}$ orbit of $\mathrm{He}^{+}$ to the speed of the electron in the $3^{rd}$ orbit of hydrogen atom will be :-
1 : 1
1 : 2
4 : 1
2 : 1
Answer: (d)
Solution
Given $v \propto \frac{Z}{n} \propto Z$ (where $n$ is constant). Therefore, $$\frac{v_{\mathrm{He}^+}}{v_{\mathrm{H}}} = \frac{Z_{\mathrm{He}^+}}{Z_{\mathrm{H}}} = \frac{2}{1}.$$
Question 47
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The photodiode is used to detect the optical signals. These diodes are preferably operated in reverse biased mode because.
fractional change in majority carriers produce higher forward bias current
fractional change in majority carriers produce higher reverse bias current
fractional change in minority carriers produce higher forward bias current
fractional change in minority carriers produce higher reverse bias current
A signal of $100\,\mathrm{THz}$ frequency can be transmitted with maximum efficiency by:
Coaxial cable
Optical fibre
Twisted pair of copper wires
Water
Answer: (b)
Solution
Question 49
Physics · Ray Optics and Optical Instruments · Single correct
The difference of speed of light in the two media A and B ($v_A - v_B$) is $2.6 \times 10^7 \, \mathrm{m/s}$. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is: (Given: speed of light in vacuum $c = 3 \times 10^8 \, \mathrm{ms^{-1}}$)
A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for $\frac{1}{3}$ deflection in the galvanometer. Which of the below is true for measuring value of G?
$\frac{1}{3}$ deflection method cannot be used for determining the resistance of the galvanometer.
$\frac{1}{3}$ deflection method can be used and in this case the G equals to twice the value of shunt resistance(s).
$\frac{1}{3}$ deflection method can be used and in this case, the G equals to three times the value of shunt resistance(s)
$\frac{1}{3}$ deflection method can be used and in this case the G value equals to the shunt resistance(s).
Answer: (b)
Solution
In galvanometer, $( (I - I_g) S = I_g G )$. $$\frac{I_g}{I} = \frac{S}{S+G}$$ $$\Rightarrow \frac{1}{3} = \frac{S}{S+G} \Rightarrow S + G = 3S \Rightarrow G = 2S$$
Question 51
Physics · Laws of Motion · Numerical
A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is $\mathrm{m}$.
Answer: 2
Solution
Mass per unit length is $\lambda$. The normal force $N$ is given by $N = mg = \lambda (L-x) g$. The maximum static friction $f_{s_{max}}$ is $f_{s_{max}} = \mu_s N$. Therefore, $f_{s_{max}} = (0.5)(\lambda)(L-x)g$. Also, $f_{s_{max}} = m_x g$. Thus, $0.5 \lambda (L-x) g = \lambda x g$. Solving for $x$, we have $$\frac{L-x}{2} = x$$ which gives $$\frac{L}{2} = \frac{3x}{2} \implies x = \frac{L}{3} = \frac{6}{3} = 2 \, m.$$
Question 52
Physics · Work, Energy and Power · Numerical
A 0.5 kg block moving at a speed of 12 $ms^{-1}$ compresses a spring through a distance 30 $cm$ when its speed is halved. The spring constant of the spring will be ______ $Nm^{-1}$.
Answer: 600
Solution
Given $U_i + K_i = U_f + K_f$. $$\Rightarrow 0 + \frac{1}{2} m (12)^2 = \frac{1}{2} K (0.3)^2 + \frac{1}{2} m (6)^2$$ $$\Rightarrow 0.5 (12^2 - 6^2) = K (0.3)^2$$ $K = 600 \, \mathrm{N/m}$
Question 53
Physics · Mechanical Properties of Fluids · Numerical
The velocity of upper layer of water in a river is $36 \, \mathrm{kmh}^{-1}$. Shearing stress between horizontal layers of water is $10^{-3} \, \mathrm{Nm}^{-2}$. Depth of the river is _________ m. (Co-efficiency of viscosity of water is $10^{-2} \, \mathrm{Pa.s}$)
Answer: 100
Solution
Given the equation $F = \eta A \frac{\Delta v_x}{\Delta y}$. Dividing both sides by $A$, we have $$\frac{F}{A} = \eta \frac{\Delta v_x}{\Delta y}$$ which implies $$10^{-3} = 10^{-2} \times \frac{36 \times 1000}{h \times 3600}$$ Solving for $h$, we get $$h = 10^{-2} \times \frac{36 \times 1000}{10^{-3} \times 3600} = 100 \, \mathrm{m}$$
Question 54
Physics · Thermal Properties of Matter · Numerical
A steam engine intakes $50\,\mathrm{g}$ of steam at $100^\circ\mathrm{C}$ per minute and cools it down to $20^\circ\mathrm{C}$. If latent heat of vaporization of steam is $540\,\mathrm{cal\,g^{-1}}$, then the heat rejected by the steam engine per minute is $x \times 10^3\,\mathrm{cal}$.
The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20 cm. The length of the open organ pipe is _____ cm.
Physics · Electrostatic Potential and Capacitance · Numerical
The equivalent capacitance between points A and B in below shown figure will be _____ $\mu$ $\mathrm{F}$.
Answer: 6
Solution
Two capacitors are short circuited. Finally, equivalent capacitance is calculated as follows: $$\frac{24 \times 8}{24 + 8} = \frac{24 \times 8}{32} = 6 \, \mu\mathrm{F}$$
Question 57
Physics · Current Electricity · Numerical
A resistor develops 300 J of thermal energy in 15s, when a current of 2A is passed through it. If the current increases to 3A, the energy developed in 10s is J.
Answer: 450
Solution
Given $H = i^2 R t$. $300 = 2^2 \times R \times 15$. $$\Rightarrow R = \frac{300}{60} = 5 \, \Omega$$ Now, for $i = 3 \, \mathrm{A}$, $t = 10 \, \mathrm{s}$, $R = 5 \, \Omega$. $H = 3^2 \times 5 \times 10 = 450 \, \mathrm{J}$
Question 58
Physics · Current Electricity · Numerical
The total current supplied to the circuit as shown in figure by the 5V battery is _________A
Answer: 2
Solution
Current supplied by 5V battery is given by $$\frac{5V}{2.5\, \Omega} = 2\, \mathrm{A}$$
Question 59
Physics · Electromagnetic Induction · Numerical
The current in a coil of self inductance 2.0 H is increasing according to $I = 2\sin(t^2)\,\mathrm{A}$. The amount of energy spent during the period when current changes from 0 to 2A is ______ J.
Answer: 4
Solution
Given $I = 2 \sin(t^2)$, we have $dI = 4t \sin(t^2) \, dt$. If $I = 0$, then $t = 0$. And if $I = 2$, then $2 = 2 \sin t^2$, which implies $t = \sqrt{\frac{\pi}{2}}$. The energy $E$ is given by the integral $$E = \int LI \, dI$$ which becomes $$= \int 2 \times 2 \sin(t^2) \times 4t \cos(t^2) \, dt$$ $$= 8 \int_0^{\sqrt{\pi/2}} t \sin(2t^2) \, dt$$ $$= 2 \left[ -\cos(2t^2) \right]_0^{\sqrt{\pi/2}}$$ $$= 2 \left[ -\cos \pi + \cos 0 \right] = 4.$$
Question 60
Physics · Laws of Motion · Numerical
A force on an object of mass 100g is $(10\hat{i} + 5\hat{j}) \, \mathrm{N}$. The position of that object at $t = 2 \, \mathrm{s}$ is $(a\hat{i} + b\hat{j}) \, \mathrm{m}$ after starting from rest. The value of $\frac{a}{b}$ will be
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?
NO
N_2
O_2
C_2
B_2
Answer: (c)
Solution
Bond strength is proportional to bond order. Removal of electron from antibonding MO increases B.O. $\mathrm{NO}$ and $\mathrm{O_2}$ have valence electrons in $\pi^*$ orbital.
Question 62
Chemistry · Surface Chemistry · Single correct
Incorrect statement for Tyndall effect is :-
The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.
The diameter of the dispersed particles is much smaller than the wavelength of the light used.
During projection of movies in the cinemas hall, Tyndall effect is noticed.
It is used to distinguish a true solution from a colloidal solution.
Answer: (b)
Solution
The diameter of dispersed particle should be somewhat below or near the wavelength of light.
Question 63
Chemistry · Structure of Atom · Single correct
The pair, in which ions are isoelectronic with $\mathrm{Al}^{3+}$ is :-
$\mathrm{Br}^{-}$ and $\mathrm{Be}^{2+}$
$\mathrm{Cl}^{-}$ and $\mathrm{Li}^{+}$
$\mathrm{S}^{2-}$ and $\mathrm{K}^{+}$
$\mathrm{O}^{2-}$ and $\mathrm{Mg}^{2+}$
Answer: (d)
Solution
Isoelectronic species have same number of electrons. $\mathrm{Al^{+3}}$, $\mathrm{O^{-2}}$, $\mathrm{Mg^{+2}}$ all have 10 electrons.
Question 64
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Leaching of gold with dilute aqueous solution of NaCN in presence of oxygen gives complex [A], which on reaction with zinc forms the elemental gold and another complex [B]. [A] and [B], respectively are :-
$[\mathrm{Au(CN)_4}]^{-}$ and $[\mathrm{Zn(CN)_2(OH)_2}]^{2-}$
$[\mathrm{Au(CN)_2}]^{-}$ and $[\mathrm{Zn(OH)_4}]^{2-}$
$[\mathrm{Au(CN)_2}]^{-}$ and $[\mathrm{Zn(CN)_4}]^{2-}$
$[\mathrm{Au(CN)_4}]^{2-}$ and $[\mathrm{Zn(CN)_6}]^{4-}$
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Number of electron deficient molecules among the following $\mathrm{PH_3}$, $\mathrm{B_2H_6}$, $\mathrm{CCl_4}$, $\mathrm{NH_3}$, $\mathrm{LiH}$ and $\mathrm{BCl_3}$ is
0
1
2
3
Answer: (c)
Solution
Electron deficient species have less than 8 electrons (or two electrons for H) in their valence (incomplete octet). $\mathrm{B_2H_6}$, $\mathrm{BCl_3}$ have incomplete octet.
Question 66
Chemistry · The s-Block Elements · Single correct
Which one of the following alkaline earth metal ions has the highest ionic mobility in its aqueous solution?
$\mathrm{Be}^{2+}$
$\mathrm{Mg}^{2+}$
$\mathrm{Ca}^{2+}$
$\mathrm{Sr}^{2+}$
Answer: (d)
Solution
Highest ionic mobility corresponds to lowest extent of hydration and highest size of gaseous ion. Hence $\mathrm{Sr^{2+}}$ has the highest ionic mobility in its aqueous solution.
Question 67
Chemistry · Co-ordination Compounds · Single correct
White precipitate of AgCl dissolves in aqueous ammonia solution due to formation of:
[$\mathrm{Ag(NH_3)_4}$]$\mathrm{Cl_2}$
[$\mathrm{Ag(Cl)_2(NH_3)_2}$]
[$\mathrm{Ag(NH_3)_2}$]$\mathrm{Cl}$
[$\mathrm{Ag(NH_3)Cl}$]$\mathrm{Cl}$
Answer: (c)
Solution
The reaction is given by: $$\mathrm{AgCl} + 2\mathrm{NH_3} \rightarrow [\mathrm{Ag(NH_3)_2}]^+\mathrm{Cl}^-$$ This compound is soluble.
Question 68
Chemistry · The d-and f-Block Elements · Single correct
Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it?
It will not prefer to undergo redox reactions.
It will prefer to gain electron and act as an oxidizing agent
It will prefer to give away an electron and behave as reducing agent
It acts as both, oxidizing and reducing agent.
Answer: (b)
Solution
Cerium exists in two different oxidation states +3, +4. $$\mathrm{Ce^{+4} + e^- \rightarrow Ce^{3+}} E^0 = +1.61 \, \mathrm{V}$$ $$\mathrm{Ce^{+3} + 3e^- \rightarrow Ce} E^0 = -2.336 \, \mathrm{V}$$ It shows $\mathrm{Ce^{+4}}$ acts as a strong oxidising agent and accepts electrons.
Question 69
Chemistry · Redox Reactions · Single correct
Among the following, which is the strongest oxidizing agent?
$Mn^{3+}$
$Fe^{3+}$
$Ti^{3+}$
$Cr^{3+}$
Answer: (a)
Solution
Strongest oxidising agent have highest reduction potential value $$E^0_{\mathrm{Mn}^{+3}/\mathrm{Mn}^{+2}} = 1.51 \, \mathrm{V} (highest)$$
Question 70
Chemistry · Environmental Chemistry · Single correct
The eutrophication of water body results in :
loss of Biodiversity
breakdown of organic matter
increase in biodiversity
decrease in BOD.
Answer: (a)
Solution
Eutrophication of water body results in loss of Biodiversity.
Question 71
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Phenol on reaction with dilute nitric acid, gives two products. Which method will be most effective for large scale separation?
Chromatographic separation
Fractional Crystallisation
Steam distillation
Sublimation
Answer: (c)
Solution
Para product has higher boiling point than ortho as intermolecular H-bond is possible in former, whereas intramolecular H-bond is possible in ortho product. Steam distillation can separate them as ortho product is steam volatile.
Question 72
Chemistry · Hydrocarbons · Single correct
In the following structures, which one is having staggered conformation with maximum dihedral angle?
Answer: (c)
Solution
Dihedral angle: It's the angle between 2 specified groups (–CH₃ here). Staggered form is given in option (C) and the angle is $180^\circ$.
Question 73
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The products formed in the following reaction.
Answer: (b)
Solution
Question 74
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The IUPAC name of ethylidene chloride is :-
1-Chloroethene
1-Chloroethyne
1,2-Dichloroethane
1,1-Dichloroethane
Answer: (d)
Solution
1, 1-Dichloroethane is Ethylidene chloride.
Question 75
Chemistry · Haloalkanes and Haloarenes · Single correct
The major product in the reaction
t-Butyl ethyl ether
2,2-Dimethyl butane
2-Methyl pent-1-ene
2-Methyl prop-1-ene
Answer: (d)
Solution
We have been given a bulky base, hence elimination will take place and not substitution.
Question 76
Chemistry · Alcohols, Phenols and Ethers · Single correct
The intermediate X, in the reaction is :
Answer: (c)
Solution
It's a classic Reimer-Tiemann reaction. Will be the intermediate formed.
Question 77
Chemistry · Alcohols, Phenols and Ethers · Single correct
In the following reaction: The compounds A and B respectively are :-
Answer: (c)
Solution
Given reaction is cumene-Peroxide method for the preparation of phenol. In this reaction, cumene hydroperoxide is converted to phenol and acetone in the presence of an acid catalyst.
Question 78
Chemistry · Amines · Single correct
The reaction of \begin{array}{c} R-C-NH_2\\ \| \ \\\ \ O \ \end{array} with bromine and $KOH$ gives $RNH_2$ as the end product. Which one of the following is the intermediate product formed in this reaction?
R−NH−Br
R−N=C=O
Answer: (c)
Solution
Question 79
Chemistry · Chemistry in Everyday Life · Single correct
Using very little soap while washing clothes, does not serve the purpose of cleaning of clothes because
soap particles remain floating in water as ions
the hydrophobic part of soap is not able to take away grease
the micelles are not formed due to concentration of soap, below its CMC value
colloidal structure of soap in water is completely disturbed.
Answer: (c)
Solution
Micelle formation only takes place above CMC.
Question 80
Chemistry · Chemistry in Everyday Life · Single correct
Which one of the following is an example of artificial sweetner?
Bithional
Alitame
Salvarsan
Lactose
Answer: (b)
Solution
Alitame is a second generation dipeptide sweetener that is 200 times sweeter than sucrose.
Question 81
Chemistry · Some Basic Concepts of Chemistry · Numerical
The number of N atoms is 681 g of $C_7H_5N_3O_6$ is $x \times 10^{21}$. The value of $x$ is _____ ($N_A = 6.02 \times 10^{23} \, \mathrm{mol}^{-1}$) (Nearest Integer)
Answer: 5418
Solution
M.M. of $\mathrm{C_7H_5N_3O_6}$ is $84 + 5 + 42 + 96 = 227$. $$n_{\mathrm{C_7H_5N_3O_6}} = \frac{681}{227} = 3$$ $$n_{\mathrm{N}} = \frac{681}{227} \times 3 = 9 \, \mathrm{mol}$$ The number of $\mathrm{N}$ atoms $= 9 \times 6.02 \times 10^{23}$. $$= 5418 \times 10^{21}$$ Therefore, the answer is 5418.
Question 82
Chemistry · The Solid State · Numerical
The distance between $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ ions in solid $\mathrm{NaCl}$ of density $43.1\,\mathrm{g\,cm^{-3}}$ is \_\_\_\_ $\times 10^{-10}\,\mathrm{m}$. (Nearest Integer) (Given: $N_A = 6.02 \times 10^{23}\,\mathrm{mol^{-1}}$)
Answer: 1
Solution
Unit cell formula – $\mathrm{Na_4Cl_4}$ Mass per unit cell = $\frac{Z \times \mathrm{M.M.}}{N_A}$ g $$= \frac{4 \times 58.5}{N_A} g$$ $$d_{unit cell} = \frac{m}{V} = \frac{m}{a^3}$$ $$\Rightarrow \frac{4 \times 58.5}{N_A \cdot a^3} = 43.1$$ $$\Rightarrow a^3 = 9.02 \times 10^{-24} cm^3$$ $$\Rightarrow a = 2.08 \times 10^{-8} cm$$ $$\Rightarrow a = 2.08 \times 10^{-10} m$$ Also $a = 2 \left( r_{\mathrm{Na^+}} + r_{\mathrm{Cl^-}} \right)$ $$\Rightarrow r_{\mathrm{Na^+}} + r_{\mathrm{Cl^-}} = 1.04 \times 10^{-10} m$$ Therefore, the answer is 1.
Question 83
Chemistry · Structure of Atom · Numerical
The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is $x \times 10^{-8}$ m. The value of $x$ is ___________. (Nearest Integer) (Given : Energy of the electron in the first shell of the hydrogen atom is $-2.2 \times 10^{-18}$ J; $h = 6.63 \times 10^{-34}$ Js and $c = 3 \times 10^{8}$ ms$^{-1}$)
Answer: 4
Solution
We can not calculate I.E. of lithium atom.
Question 84
Chemistry · Thermodynamics · Numerical
The standard entropy change for the reaction $$4\mathrm{Fe}(s) + 3\mathrm{O}_2(g) \rightarrow 2\mathrm{Fe}_2\mathrm{O}_3(s)$$ is $-550 \, \mathrm{JK}^{-1}$ at $298 \, \mathrm{K}$. [Given: The standard enthalpy change for the reaction is $-165 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$]. The temperature in K at which the reaction attains equilibrium is _________. (Nearest Integer)
Answer: 300
Solution
At equilibrium, $\Delta G = \Delta H - T \Delta S = 0$. $$\Rightarrow -165 \times 10^3 - T \times (-505) = 0$$ $$\Rightarrow T = 300 \, \mathrm{K}$$ The answer is 300.
Question 85
Chemistry · Solutions · Numerical
1 L aqueous solution of $\mathrm{H_2SO_4}$ contains $0.02$ m mol $\mathrm{H_2SO_4}$. 50$\%$ of this solution is diluted with deionized water to give 1 L solution (A). In solution (A), $0.01$ m mol of $\mathrm{H_2SO_4}$ are added. Total m mols of $\mathrm{H_2SO_4}$ in the final solution is $\times 10^3$ m mols.
Answer: 15
Solution
Given $n_{\mathrm{H_2SO_4}}$ in solution A is 50$\%$ of the original solution. $$= 0.01 \, \mathrm{mmol}.$$ $n_{\mathrm{H_2SO_4}}$ in the final solution is $0.01 + 0.01$. $$= 0.02 \, \mathrm{mmol}$$ $$= 0.00002 \times 10^3 \, \mathrm{mmol}$$ The answer is 0.
Question 86
Chemistry · Equilibrium · Numerical
The standard free energy change ($\Delta G^\circ$) for 50$\%$ dissociation of $\mathrm{N_2O_4}$ into $\mathrm{NO_2}$ at $27^\circ\mathrm{C}$ and $1 \, \mathrm{atm}$ pressure is $-x \, \mathrm{J \, mol^{-1}}$. The value of $x$ is _________. (Nearest Integer) [Given : $R = 8.31 \, \mathrm{J \, K}^{-1} \, \mathrm{mol}^{-1}$, $\log 1.33 = 0.1239$, $\ln 10 = 2.3$]
In a cell, the following reactions take place $$\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-} E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}} = 0.77 \, \mathrm{V}$$ $$\mathrm{2I^- \rightarrow I_2 + 2e^-} E^\circ_{\mathrm{I_2/I^-}} = 0.54 \, \mathrm{V}$$ The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \, \mathrm{V}$ at 298 K. The value of $x$ is _________ (Nearest Integer)
Answer: 23
Solution
The reaction is given by: $$\mathrm{Fe^{+3} + I^- \rightarrow I_2 + Fe^{+2}}$$ The cell potential is calculated as: $$E^0_{Cell} = E^0_{cathode} - E^0_{anode}$$ Substituting the values: $$= 0.77 - 0.54$$ $$= 0.23$$ $$= 23 \times 10^{-2} \, \mathrm{V}$$
Question 88
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For a given chemical reaction $$\gamma_1\mathrm{A} + \gamma_2\mathrm{B} \rightarrow \gamma_3\mathrm{C} + \gamma_4\mathrm{D}$$ Concentration of C changes from $10\ \mathrm{mmol\ dm^{-3}}$ to $20\ \mathrm{mmol\ dm^{-3}}$ in $10$ seconds. Rate of appearance of D is $1.5$ times the rate of disappearance of B which is twice the rate of disappearance of A. The rate of appearance of D has been experimentally determined to be $9\ \mathrm{mmol\ dm^{-3}\ s^{-1}}$. Therefore the rate of reaction is \_\_\_\_\_ $\mathrm{mmol\ dm^{-3}\ s^{-1}}$. (Nearest Integer)
If $\mathrm{[Cu(H_2O)_4]^{2+}}$ absorbs a light of wavelength $600 \, \mathrm{nm}$ for d–d transition, then the value of octahedral crystal field splitting energy for $\mathrm{[Cu(H_2O)_6]^{2+}}$ will be $\times 10^{-21} \, \mathrm{J}$. (Nearest Integer) (Given: $h = 6.63 \times 10^{-34} \, \mathrm{Js}$ and $c = 3.08 \times 10^8 \, \mathrm{ms}^{-1}$)
Number of grams of bromine that will completely react with $5.0 \, \mathrm{g}$ of pent-1-ene is _______ $\times 10^{-2} \, \mathrm{g}$. (Atomic mass of Br = 80 g/mol) [Nearest Integer]
Answer: 1136
Solution
Moles of $\mathrm{Br_2}$ $=$ moles of $\mathrm{C_5H_{10}}$. Therefore, $$\frac{w}{160} = \frac{5}{70}$$ Solving for $w$, we have $$w = \frac{5 \times 160}{70}\,\mathrm{g}$$ $$= 11.428\,\mathrm{g}$$ $$= 1142.8 \times 10^{-2}\,\mathrm{g} \approx 1143 \times 10^{-2}\,\mathrm{g}$$