JEE Main 8 April 2024 Shift 1 question paper with solutions

JEE Main 8 April 2024 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.

Maths

Question 1

Maths · Integrals · Single correct

The value of $k \in \mathbb{N}$ for which the integral $I_n = \int_0^1 (1-x^k)^n \, dx$, $n \in \mathbb{N}$, satisfies $147I_{20} = 148I_{21}$ is

Answer: (d)

Solution

Given $$I_n = \int_0^1 (1-x^k)^n \cdot 1 \, dx$$ We have $$I_n = (1-x^k)^n \cdot x - nk \int_0^1 (1-x^k)^{n-1} \cdot x^{k-1} \, dx$$ This simplifies to $$I_n = nk \int_0^1 \left[ (1-x^k)^n - (1-x^k)^{n-1} \right] \, dx$$ Thus, $$I_n = nk I_n - nk I_n$$ We find $$\frac{I_n}{I_{n-1}} = \frac{nk}{nk+1}$$ For $$\frac{I_{21}}{I_{20}} = \frac{21k}{1+21k}$$ This gives $$= \frac{147}{148} \Rightarrow k = 7$$

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

The sum of all the solutions of the equation $(8)^{2x} - 16 \cdot (8)^x + 48 = 0$ is:

$1 + \log_8(6)$
$1 + \log_6(8)$
$\log_8(6)$
$\log_8(4)$

Answer: (a)

Solution

Given $(8)^{2x} - 16 \cdot (8)^x + 48 = 0$. Put $8^x = t$. Then, $$t^2 - 16 + 48 = 0$$ which implies $t = 4$ or $t = 12$. Thus, $8^x = 4$ and $8^x = 12$. This gives $x = \log_8 4$ and $x = \log_8 12$. The sum of the solution is $\log_8 4 + \log_8 12$. This equals $\log_8 48 = \log_8 (6.8)$. Therefore, the sum is $1 + \log_8 6$.

Question 3

Maths · Conic Sections · Single correct

Let the circles $C_1 : (x - \alpha)^2 + (y - \beta)^2 = r_1^2$ and $C_2 : (x - 8)^2 + \left(y - \frac{15}{2}\right)^2 = r_2^2$ touch each other externally at the point $(6, 6)$. If the point $(6, 6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2 : 1$, then $(\alpha + \beta) + 4\left(r_1^2 + r_2^2\right)$ equals

Answer: (b)

Solution

Given the points $C_1(\alpha, \beta)$ and $C_2(8, \frac{15}{2})$, we have the equation: $$\frac{16 + \alpha}{3} = 6 and \frac{15 + \beta}{3} = 6$$ This implies $(\alpha, \beta) \equiv (2, 3)$. Also, $C_1 C_2 = r_1 + r_2$. Thus, $$\sqrt{(2 - 8)^2 + \left(3 - \frac{15}{2}\right)^2} = 2r_2 + r_2$$ This simplifies to $r_2 = \frac{5}{2}$, which implies $r_1 = 2r_2 = 5$. Therefore, $$\alpha + \beta + 4\left(r_1^2 + r_2^2\right)$$ Calculating further, $$= 5 + 4\left(\frac{25}{4} + 25\right) = 130$$

Question 4

Maths · Three Dimensional Geometry · Single correct

Let $P(x, y, z)$ be a point in the first octant, whose projection in the $xy$-plane is the point $Q$. Let $OP = \gamma$; the angle between $OQ$ and the positive $x$-axis be $\theta$; and the angle between $OP$ and the positive $z$-axis be $\phi$, where $O$ is the origin. Then the distance of $P$ from the $x$-axis is

$\gamma \sqrt{1 - \sin^2 \phi \cos^2 \theta}$
$\gamma \sqrt{1 - \sin^2 \theta \cos^2 \phi}$
$\gamma \sqrt{1 + \cos^2 \phi \sin^2 \theta}$
$\gamma \sqrt{1 + \cos^2 \theta \sin^2 \phi}$

Answer: (a)

Solution

Given points $P(x, y, z)$ and $Q(x, y, 0)$, with $x^2 + y^2 + z^2 = \gamma^2$. The vector $\overline{OQ} = x \hat{i} + y \hat{j}$. The cosine of $\theta$ is given by $$\cos \theta = \frac{x}{\sqrt{x^2 + y^2}}.$$ The cosine of $\phi$ is given by $$\cos \phi = \frac{z}{\sqrt{x^2 + y^2 + z^2}}.$$ Therefore, $$\Rightarrow \sin^2 \phi = \frac{x^2 + y^2}{x^2 + y^2 + z^2}.$$ The distance of $P$ from the x-axis is $\sqrt{y^2 + z^2}$. Thus, $$\Rightarrow \sqrt{\gamma^2 - x^2} \Rightarrow \gamma \sqrt{1 - \frac{x^2}{\gamma^2}}$$ $$= \gamma \sqrt{1 - \cos^2 \theta \sin^2 \phi}.$$

Question 5

Maths · Applications of Derivatives · Single correct

The number of critical points of the function $f(x) = (x - 2)^{2/3}(2x + 1)$ is

Answer: (b)

Solution

Given $f(x) = (x - 2)^{2/3}(2x + 1)$. Differentiating, we have: $$f'(x) = \frac{2}{3}(x - 2)^{-1/3}(2x + 1) + (x - 2)^{2/3}$$ Simplifying further: $$f'(x) = 2 \times \frac{(2x + 1) + (x - 2)}{3(x - 2)^{1/3}}$$ Setting the derivative to zero: $$\frac{3x - 1}{(x - 2)^{1/3}} = 0$$ The critical points are $x = \frac{1}{3}$ and $x = 2$.

Question 6

Maths · Differential Equations · Single correct

Let $f(x)$ be a positive function such that the area bounded by $y=f(x),$ $y=0$ from $x=0$ to $x=a>0$ is $e^{-a}+4a^{2}+a-1.$ Then the differential equation, whose general solution is $y=c_{1}f(x)+c_{2},$ where $c_{1}$ and $c_{2}$ are arbitrary constants, is

(8e^x - 1) $\frac{d^2y}{dx^2}$ + $\frac{dy}{dx}$ = 0
(8e^x - 1) $\frac{d^2y}{dx^2}$ - $\frac{dy}{dx}$ = 0
(8e^x + 1) $\frac{d^2y}{dx^2}$ - $\frac{dy}{dx}$ = 0
(8e^x + 1) $\frac{d^2y}{dx^2}$ + $\frac{dy}{dx}$ = 0

Answer: (d)

Solution

Given $$$ $\int$_0^a f(x) dx = e^{-a} + 4a^2 + a - 1 $$$. $$$ f(a) = -e^{-a} + 8a + 1 $$$ $$$ f(x) = -e^{-x} + 8x + 1 $$$ Now $$$ y = C_1 f(x) + C_2 $$$ $$$ $\frac{dy}{dx}$ = C_1 f'(x) = C_1 $\left$( e^{-x} + 8 $\right$) $$$ ...(1) $$$ $\frac{d^2y}{dx^2}$ = -C_1 e^{-x} $\Rightarrow$ -e^x $\frac{d^2y}{dx^2}$ $$$ Put in equation (1) $$$ $\frac{dy}{dx}$ = -e^x $\frac{d^2y}{dx^2}$ $\left$( e^{-x} + 8 $\right$) $$$ $$$ (8e^x + 1) $\frac{d^2y}{dx^2}$ + $\frac{dy}{dx}$ = 0 $$$

Question 7

Maths · Applications of Derivatives · Single correct

Let $f(x) = 4 \cos^3 x + 3 \sqrt{3} \cos^2 x - 10$. The number of points of local maxima of $f$ in interval $(0, 2\pi)$ is

Answer: (d)

Solution

Given $f(x) = 4 \cos^3(x) + 3\sqrt{3} \cos^2(x) - 10$; $x \in (0, 2\pi)$. Therefore, $f'(x) = 12 \cos^2 x [-\sin(x)] + 3\sqrt{3}(2 \cos(x))[-\sin(x)]$. This simplifies to $f'(x) = -6 \sin(x) \cos(x) [2 \cos(x) + \sqrt{3}]$. The local maxima occur at $x = \frac{5\pi}{6}, \frac{7\pi}{6}$.

Question 8

Maths · Matrices · Single correct

Let $A = \begin{pmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{pmatrix}$. If $A^3 = 4A^2 - A - 21I$, where $I$ is the identity matrix of order $3 \times 3$, then $2a + 3b$ is equal to

-9
-13
-10
-12

Answer: (b)

Solution

Given $A^3 - 4A^2 + A + 21I = 0$. $\mathrm{tr}(A) = 4 = 5 + 6 \Rightarrow b = -1$ $|A| = -21$ $-16 + a = -21 \Rightarrow a = -5$ $2a + 3b = -13$

Question 9

Maths · Three Dimensional Geometry · Single correct

If the shortest distance between the lines $$L_1 : \vec{r} = (2 + \lambda)\hat{i} + (1 - 3\lambda)\hat{j} + (3 + 4\lambda)\hat{k}, \lambda \in \mathbb{R}$$ $$L_2 : \vec{r} = 2(1 + \mu)\hat{i} + 3(1 + \mu)\hat{j} + (5 + \mu)\hat{k}, \mu \in \mathbb{R}$$ is $\frac{m}{\sqrt{n}}$, where $\mathrm{gcd}(m, n) = 1$, then the value of $m + n$ equals

Answer: (d)

Solution

Shortest distance (CD) is given by $$\frac{\left| \overrightarrow{AB} \cdot \overrightarrow{p} \times \overrightarrow{q} \right|}{\left| \overrightarrow{p} \times \overrightarrow{q} \right|}$$ where $$\overrightarrow{p} = \hat{i} - 3\hat{j} + 4\hat{k}$$ $$\overrightarrow{q} = 2\hat{i} + 3\hat{j} + \hat{k}$$ The cross product is $$\overrightarrow{p} \times \overrightarrow{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 2 & 3 & 1 \end{vmatrix} = -15\hat{i} + 7\hat{k}$$ The magnitude is $$\left| \overrightarrow{p} \times \overrightarrow{q} \right| = \sqrt{(-15)^2 + 7^2} = \sqrt{355}$$ The vector $$$ $\overrightarrow{AB}$ $$$ is $$\overrightarrow{AB} = (2\hat{i} + 3\hat{j} + 5\hat{k}) - (2\hat{i} + \hat{j} + 3\hat{k}) = 0\hat{i} + 2\hat{j} + 2\hat{k}$$ The dot product is $$\overrightarrow{AB} \cdot (\overrightarrow{p} \times \overrightarrow{q}) = (0\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-15\hat{i} + 7\hat{k}) = 0 + 14 + 18 = 32$$ Thus, the shortest distance is $$\frac{32}{\sqrt{355}}$$ Therefore, $$$ m + n = 32 + 355 = 387 $$$.

Question 10

Maths · Probability · Single correct

Let the sum of two positive integers be 24. If the probability, that their product is not less than $\frac{3}{4}$ times their greatest possible product, is $\frac{m}{n}$, where $\gcd(m, n) = 1$, then $n - m$ equals

Answer: (a)

Solution

Given $x + y = 24$, $x, y \in \mathbb{N}$. $AM > GM \Rightarrow xy \leq 144$ $xy \geq 108$ Favorable pairs of $(x,y)$ are $(13,11),(12,12),(14,10),(15,9),(16,8),$ $(17,7),(18,6),(6,18),(7,17),(8,16),(9,15),$ $(10,14),(11,13)$ i.e. 13 cases Total choices for $x + y = 24$ is 23 Probability $= \frac{13}{23} = \frac{m}{n}$ $n - m = 10$

Question 11

Maths · Trigonometric Functions · Single correct

If $\sin x = -\frac{3}{5}$, where $\pi < x < \frac{3\pi}{2}$, then $80 \left( \tan^2 x - \cos x \right)$ is equal to

Answer: (b)

Solution

Given $\sin x = -\frac{3}{5}$, $\pi < x < \frac{3\pi}{2}$. $\tan x = \frac{3}{4}$, $\cos x = -\frac{4}{5}$. Calculate $80 \left( \tan^2 x - \cos x \right)$. $$= 80 \left( \frac{9}{16} + \frac{4}{5} \right) = 45 + 64 = 109$$

Question 12

Maths · Integrals · Single correct

Let $I(x) = \int \frac{6}{\sin^2 x (1 - \cot x)^2} \, dx$. If $I(0) = 3$, then $I\left(\frac{\pi}{12}\right)$ is equal to

2$\sqrt{3}$
$\sqrt{3}$
3$\sqrt{3}$
6$\sqrt{3}$

Answer: (c)

Solution

Given $I(x) = \int \frac{6dx}{\sin^2 x (1 - \cot x)^2} = \int \frac{6 \csc^2 x dx}{(1 - \cot x)^2}$. Put $1 - \cot x = t$. Then $\csc^2 x dx = dt$. Thus, $I = \int \frac{6dt}{t^2} = \frac{-6}{t} + c$. Therefore, $I(x) = \frac{-6}{1 - \cot x} + c$, where $c = 3$. So, $I(x) = 3 - \frac{6}{1 - \cot x}$, and $I\left(\frac{\pi}{12}\right) = 3 - \frac{6}{1 - (2 + \sqrt{3})}$. Calculating $I\left(\frac{\pi}{12}\right) = 3 + \frac{6}{\sqrt{3} + 1} = 3 + \frac{6(\sqrt{3} - 1)}{2} = 3\sqrt{3}$.

Question 13

Maths · Straight Lines and Pair of Straight Lines · Single correct

The equations of two sides AB and AC of a triangle ABC are $4x + y = 14$ and $3x - 2y = 5$, respectively. The point $\left(2, -\frac{4}{3}\right)$ divides the third side BC internally in the ratio $2 : 1$. The equation of the side BC is

$x + 3y + 2 = 0$
$x - 6y - 10 = 0$
$x - 3y - 6 = 0$
$x + 6y + 6 = 0$

Answer: (a)

Solution

Given the triangle with points A, B, and C, and the equations of the lines, we start by solving for the coordinates. The equation for point P is given by: $$ \frac{2x_2 + x_1}{3} = 2, \frac{2 \left( \frac{3x_2 - 5}{2} \right) + (14 - 4x_1)}{3} = \frac{-4}{3} $$ Solving these equations, we have: $$ 2x_2 + x_1 = 6, 3x_2 - 4x_1 = -13 $$ Solving for $x_2$ and $x_1$, we find: $$ x_2 = 1, x_1 = 4 $$ Thus, the coordinates are $C(1, -1)$ and $B(4, -2)$. The slope $m$ of line BC is: $$ m = \frac{-1}{3} $$ The equation of line BC is: $$ y + 1 = \frac{-1}{3}(x - 1) $$ Simplifying, we get: $$ 3y + 3 = -x + 1 $$ $$ x + 3y + 2 = 0 $$

Question 14

Maths · Relations and Functions · Single correct

Let [t] be the greatest integer less than or equal to t. Let A be the set of all prime factors of 2310 and $f : A \to \mathbb{Z}$ be the function $f(x) = \left\lfloor \log_2 \left( x^2 + \left\lfloor \frac{x^3}{5} \right\rfloor \right) \right\rfloor$. The number of one-to-one functions from A to the range of f is

Answer: (d)

Solution

Given $N = 2310 = 231 \times 10$. This can be factored as $3 \times 11 \times 7 \times 2 \times 5$. Let $A = \{2, 3, 5, 7, 11\}$. The function is defined as: $$f(x) = \left[ \log_2 \left( x^2 + \left\lfloor \frac{x^3}{5} \right\rfloor \right) \right]$$ Calculating for each element in $A$: $f(2) = \left[ \log_2(5) \right] = 2$ $f(3) = \left[ \log_2(14) \right] = 3$ $f(5) = \left[ \log_2(25 + 25) \right] = 5$ $f(7) = \left[ \log_2(117) \right] = 6$ $f(11) = \left[ \log_2 387 \right] = 8$ The range of $f$ is $B = \{2, 3, 5, 6, 8\}$. The number of one-one functions is $5! = 120$.

Question 15

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $z$ be a complex number such that $|z + 2| = 1$ and $\mathrm{Im}\left(\frac{z+1}{z+2}\right) = \frac{1}{5}$. Then the value of $|\mathrm{Re}(\overline{z+2})|$ is

$\frac{2\sqrt{6}}{5}$
$\frac{24}{5}$
$\frac{1+\sqrt{6}}{5}$
$\frac{\sqrt{6}}{5}$

Answer: (a)

Solution

Given $|z + 2| = 1$, $\mathrm{Im}\left(\frac{z+1}{z+2}\right) = \frac{1}{5}$. Let $z + 2 = \cos \theta + i \sin \theta$. Then, $$\frac{1}{z+2} = \cos \theta - i \sin \theta$$ which implies $$\frac{z+1}{z+2} = 1 - \frac{1}{z+2} = 1 - (\cos \theta - i \sin \theta)$$ $$= (1 - \cos \theta) + i \sin \theta$$ $$\mathrm{Im}\left(\frac{z+1}{z+2}\right) = \sin \theta, \sin \theta = \frac{1}{5}$$ $$\cos \theta = \pm \sqrt{1 - \frac{1}{25}} = \pm \frac{2\sqrt{6}}{5}$$ Therefore, $$|\mathrm{Re}(\overline{z+2})| = \frac{2\sqrt{6}}{5}$$

Question 16

Maths · Sequences and Series · Single correct

If the set $R = \{(a, b) : a + 5b = 42, a, b \in \mathbb{N}\}$ has $m$ elements and $\sum_{n=1}^{m} \left(1 - i^{n!}\right) = x + iy$, where $i = \sqrt{-1}$, then the value of $m + x + y$ is

Answer: (a)

Solution

Given $a + 5b = 42$, $a, b \in \mathbb{N}$. $a = 42 - 5b$, for $b = 1$, $a = 37$. $b = 2$, $a = 32$. $b = 3$, $a = 27$. $\vdots$ $b = 8$, $a = 2$. $R$ has "8" elements $\Rightarrow m = 8$. $$\sum_{n=1}^{8} \left(1 - i^{n!}\right) = x + iy$$ For $n \geq 4$, $i^{n!} = 1$. $\Rightarrow (1 - i) + \left(1 - i^{2!}\right) + \left(1 - i^{3!}\right)$ $= 1 - i + 2 + 1 + 1$ $= 5 - i = x + iy$. $m + x + y = 8 + 5 - 1 = 12$

Question 17

Maths · Applications of Derivatives · Single correct

For the function $f(x) = (\cos x) - x + 1, x \in \mathbb{R}$, between the following two statements (S1) $f(x) = 0$ for only one value of $x$ in $[0, \pi]$. (S2) $f(x)$ is decreasing in $\left[0, \frac{\pi}{2}\right]$ and increasing in $\left[\frac{\pi}{2}, \pi\right]$.

Both (S1) and (S2) are correct.
Both (S1) and (S2) are incorrect.
Only (S2) is correct.
Only (S1) is correct.

Answer: (d)

Solution

Given $f'(x) = \cos x - x + 1$. $f'(x) = -\sin x - 1$. $f$ is decreasing $\forall x \in \mathbb{R}$. $f(x) = 0$. $f(0) = 2$, $f(\pi) = -\pi$. $f$ is strictly decreasing in $[0, \pi]$ and $f(0) \cdot f(\pi) < 0$. Therefore, only one solution of $f(x) = 0$. S1 is correct and S2 is incorrect.

Question 18

Maths · Vector Algebra · Single correct

The set of all $\alpha$, for which the vectors $\vec{a} = \alpha t \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = t \hat{i} - 2 \alpha t \hat{k}$ are inclined at an obtuse angle for all $t \in \mathbb{R}$, is

( -$\frac{4}{3}$, 1)
[0, 1)
( -$\frac{4}{3}$, 0]
(-2, 0]

Answer: (c)

Solution

Given $\vec{a} = \alpha \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = t \hat{i} - 2 \hat{j} - 2 \alpha \hat{k}$. So $\vec{a} \cdot \vec{b} < 0$, $\forall t \in \mathbb{R}$. $$\alpha t^2 - 12 + 6 \alpha t < 0$$ $$\alpha t^2 + 6 \alpha t - 12 < 0, \forall t \in \mathbb{R}$$ $\alpha < 0$, and $D < 0$. $$36 \alpha^2 + 48 \alpha < 0$$ $$12 \alpha (3 \alpha + 4) < 0$$ $$\frac{-4}{3} < \alpha < 0$$ Also for $\alpha = 0$, $\vec{a} \cdot \vec{b} < 0$. Hence $\alpha \in \left( \frac{-4}{3}, 0 \right]$

Question 19

Maths · Differential Equations · Single correct

Let $y = y(x)$ be the solution of the differential equation $$(1 + y^2) e^{\tan x} dx + \cos^2 x \left(1 + e^{2 \tan x}\right) dy = 0, y(0) = 1.$$ Then $y\left(\frac{\pi}{4}\right)$ is equal to

$\frac{2}{e}$
$\frac{2}{e^2}$
$\frac{1}{e}$
$\frac{1}{e^2}$

Answer: (c)

Solution

(1 + y^2) e^{$\tan$ x} dx + $\cos$^2 x $\left$(1 + e^{2 $\tan$ x}$\right$) dy = 0 $\int$ $\frac{\sec^2 x e^{\tan x}}{1 + e^{2 \tan x}}$ dx + $\int$ $\frac{dy}{1 + y^2}$ = C $\Rightarrow$ $\tan$^{-1} $\left$(e^{$\tan$ x}$\right$) + $\tan$^{-1} y = C for x = 0, y = 1, $\tan$^{-1} (1) + $\tan$^{-1} 1 = C C = $\frac{\pi}{2}$ $\tan$^{-1} $\left$(e^{$\tan$ x}$\right$) + $\tan$^{-1} y = $\frac{\pi}{2}$ Put x = $\pi$, $\tan$^{-1} e + $\tan$^{-1} y = $\frac{\pi}{2}$ $\tan$^{-1} y = $\cot$^{-1} e y = $\frac{1}{e}$

Question 20

Maths · Conic Sections · Single correct

Let $H : -\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha, 6), \alpha > 0$ lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2 + \beta$ is equal to

Answer: (b)

Solution

Given the hyperbola: $$\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1, \ e = \sqrt{3}$$ Calculate the eccentricity: $$e = \sqrt{1 + \frac{a^2}{b^2}} = \sqrt{3} \Rightarrow \frac{a^2}{b^2} = 2$$ Thus, $$a^2 = 2b^2$$ The length of the latus rectum is: $$\frac{2a^2}{b} = 4\sqrt{3}$$ Given $$a = \sqrt{6}$$ Point $$P(\alpha, 6)$$ lies on $$\frac{y^2}{3} - \frac{x^2}{6} = 1$$ Substitute to find $$\alpha^2$$: $$12 - \frac{\alpha^2}{6} = 1 \Rightarrow \alpha^2 = 66$$ The foci are: $$(0, \pm be) = (0, 3) \& (0, -3)$$ Let $$d_1 \& d_2$$ be the focal distances of $$P(\alpha, 6)$$ Calculate $$d_1$$ and $$d_2$$: $$d_1 = \sqrt{\alpha^2 + (6 + be)^2}, \ d_2 = \sqrt{\alpha^2 + (6 - be)^2}$$ Substitute values: $$d_1 = \sqrt{66 + 81}, \ d_2 = \sqrt{66 + 9}$$ Calculate $$\beta$$: $$\beta = d_1 d_2 = \sqrt{147 \times 75} = 105$$ Finally, $$\alpha^2 + \beta = 66 + 105 = 171$$

Question 21

Maths · Matrices · Numerical

Let $A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$. If the sum of the diagonal elements of $A^{13}$ is $3^n$, then $n$ is equal to ________

Answer: 7

Solution

Given $$A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$$ Calculate $$A^2 = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -3 \\ 3 & 0 \end{bmatrix}$$ Then $$A^3 = \begin{bmatrix} 3 & -3 \\ 3 & 0 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -6 \\ 6 & -3 \end{bmatrix}$$ Next $$A^4 = \begin{bmatrix} 3 & -6 \\ 6 & -3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -9 \\ 9 & -9 \end{bmatrix}$$ Then $$A^5 = \begin{bmatrix} 0 & -9 \\ 9 & -9 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -9 & -9 \\ 9 & -18 \end{bmatrix}$$ Next $$A^6 = \begin{bmatrix} -9 & -9 \\ 9 & -18 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -27 & 0 \\ 0 & -27 \end{bmatrix}$$ Finally $$A^7 = \begin{bmatrix} -27 & 0 \\ 0 & -27 \end{bmatrix} \begin{bmatrix} -54 & 27 \\ -27 & -27 \end{bmatrix} = \begin{bmatrix} 3^6 \times 2 & -27^2 \\ 27^2 & 3^6 \end{bmatrix}$$ Since $$3^7 = 3^n \Rightarrow n = 7$$

Question 22

Maths · Properties of Triangles · Numerical

If the orthocentre of the triangle formed by the lines $2x + 3y - 1 = 0$, $x + 2y - 1 = 0$ and $ax + by - 1 = 0$, is the centroid of another triangle, whose circumcentre and orthocentre respectively are $(3, 4)$ and $(-6, -8)$, then the value of $|a - b|$ is

Answer: 16

Solution

Given the equations $2x + 3y - 1 = 0$, $x + 2y - 1 = 0$, and $ax + by - 1 = 0$. The points are $H(-6, -8)$, $G(6, 6)$, and $O(3, 4)$. The intersection point of the lines $2x + 3y - 1 = 0$ and $x + 2y - 1 = 0$ is calculated as follows: $$\left( \frac{6 - 6}{3}, \frac{8 - 8}{3} \right) = (0, 0)$$ For the line $ax + by - 1 = 0$, we have: $$\begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} -a \\ b \end{pmatrix} = -1$$ This implies $-a = b$. Therefore, $ax - ay - 1 = 0$. Rewriting, we get: $$ax - a \left( 1 - \frac{2x}{3} \right) - 1$$ Solving for $x$: $$x \left( a + \frac{2a}{3} \right) = \frac{a}{3}$$ Thus, $$x = \frac{a + 3}{5a}$$ Substituting into $2 \left( \frac{a + 3}{5a} \right) + 3y - 1 = 0$, we find: $$y = \frac{1 - \frac{2a + 6}{5a}}{3} = \frac{3a - 6}{3 \times 5a}$$ This simplifies to: $$y = \frac{a - 2}{5a}$$ Equating: $$\frac{a - 2}{5a} = 2 \Rightarrow a - 2 = 2a + 6$$ Solving gives $a = -8$ and $b = 8$. Therefore, the equation becomes $-8x + 8y - 1 = 0$. Finally, $|a - b| = 16$.

Question 23

Maths · Probability · Numerical

Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables $X$ and $Y$ respectively denote the number of blue and yellow balls. If $\bar{X}$ and $\bar{Y}$ are the means of $X$ and $Y$ respectively, then $7\bar{X} + 4\bar{Y}$ is equal to

Answer: 17

Solution

\begin{tabular}{|c|c|c|c|c|c|c|} \hline Blue balls & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline Prob & $\dfrac{{}^{5}C_{0}\cdot{}^{4}C_{3}}{{}^{9}C_{3}}$ & $\dfrac{{}^{5}C_{1}\cdot{}^{4}C_{2}}{{}^{9}C_{3}}$ & $\dfrac{{}^{5}C_{2}\cdot{}^{4}C_{1}}{{}^{9}C_{3}}$ & $\dfrac{{}^{5}C_{3}\cdot{}^{4}C_{0}}{{}^{9}C_{3}}$ & 0 & 0 \\ \hline \end{tabular} $7x=\dfrac{30+2\times40+3\times10}{84}\times7$ $=\dfrac{30+80+30}{84}\times7$ $=\dfrac{140}{84}\times7$ $=\dfrac{140}{12}$ $=\dfrac{70}{6}$ $=\dfrac{35}{3}$ \begin{tabular}{|c|c|c|c|c|c|} \hline yellow & 0 & 1 & 2 & 3 & 4 \\ \hline & ${}^{5}C_{2}\cdot{}^{4}C_{1}$ & ${}^{5}C_{1}\cdot{}^{4}C_{2}$ & ${}^{5}C_{0}\cdot{}^{4}C_{3}$ & 0 & 0 \\ \hline \end{tabular} $4y=\dfrac{40+60+12}{84}\times4$ $=\dfrac{112}{21}$ $=\dfrac{16}{3}$

Question 24

Maths · Permutations and Combinations · Numerical

The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to

Answer: 36

Solution

Total number of three-digit numbers not divisible by 3 will be formed by using the digits (4, 5, 7) (3, 4, 7) (2, 5, 7) (2, 4, 7) (2, 4, 5) (2, 3, 5) Number of ways = 6 $\times$ 3! = 36

Question 25

Maths · Sequences and Series · Fill in the blank

Let the positive integers be written in the form : If the $k^{th}$ row contains exactly $k$ numbers for every natural number $k$, then the row in which the number 5310 will be, is _______

Answer: 103

Solution

Given $$S = 1 + 2 + 4 + 7 + \ldots + T_n$$ $$S = 1 + 2 + 4 + \ldots$$ $$T_n = 1 + 1 + 2 + 3 + \ldots + (T_n - T_{n-1})$$ We have $$T_n = 1 + \left(\frac{n-1}{2}\right)[2 + (n-2) \times 1]$$ Simplifying, $$T_n = 1 + 1 + \frac{n(n-1)}{2}$$ For $n = 100$, $$T_n = 1 + \frac{100 \times 99}{2} = 4950 + 1$$ For $n = 101$, $$T_n = 1 + \frac{101 \times 100}{2} = 5050 + 1 = 5051$$ For $n = 102$, $$T_n = 1 + \frac{102 \times 101}{2} = 5151 + 1 = 5152$$ For $n = 103$, $$T_n = 1 + \frac{103 \times 102}{2} = 5254$$ For $n = 104$, $$T_n = 1 + \frac{104 \times 103}{2} = 5357$$

Question 26

Maths · Trigonometric Functions · Numerical

If the range of $f(\theta) = \frac{\sin^4 \theta + 3 \cos^2 \theta}{\sin^4 \theta + \cos^2 \theta}$, $\theta \in \mathbb{R}$ is $[\alpha, \beta]$, then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{\alpha}{\beta}$, is equal to

Answer: 96

Solution

Given $$f(\theta) = \frac{\sin^4 \theta + 3 \cos^2 \theta}{\sin^4 \theta + \cos^2 \theta}$$ We can rewrite it as $$f(\theta) = 1 + \frac{2 \cos^2 \theta}{\sin^4 \theta + \cos^2 \theta}$$ Simplifying further, $$f(\theta) = \frac{2 \cos^2 \theta}{\cos^4 \theta - \cos^2 \theta + 1} + 1$$ This becomes $$f(\theta) = \frac{2}{\cos^2 \theta + \sec^2 \theta - 1} + 1$$ The minimum value of $f(\theta)$ is 1. The maximum value of $f(\theta)$ is 3. Therefore, $$S = \frac{64}{1 - 1/3} = 96$$

Question 27

Maths · Binomial Theorem · Numerical

Let $\alpha = \sum_{r=0}^{n} (4r^2+2r+1) {}^nC_r$ and $\beta = (\sum_{r=0}^{n} \frac{{}^nC_r}{r+1}) + \frac{1}{n+1}$. If $140 < \frac{2\alpha}{\beta} < 281$, then the value of $n$ is \_\_\_\_\_\_.

Answer: 5

Solution

Given $\alpha = \displaystyle\sum_{r=0}^{n} (4r^2 + 2r + 1) \cdot \binom{n}{r}$. $$\alpha = 4\sum_{r=0}^{n} r^2 \cdot \frac{n}{r} \cdot \binom{n-1}{r-1} + 2\sum_{r=0}^{n} r \cdot \frac{n}{r} \cdot \binom{n-1}{r-1} + \sum_{r=0}^{n} \binom{n}{r}$$ $$= 4n\sum_{r=0}^{n} r \cdot \binom{n-1}{r-1} + 2n\sum_{r=0}^{n} \binom{n-1}{r-1} + \sum_{r=0}^{n} \binom{n}{r}$$ $$\alpha = 4n(n-1) \cdot 2^{n-2} + 4n \cdot 2^{n-1} + 2n \cdot 2^{n-1} + 2^n$$ $$\alpha = 2^{n-2}\left[4n(n-1) + 8n + 4n + 4\right]$$ $$\alpha = 2^{n-2}\left[4n^2 + 8n + 4\right]$$ $$\alpha = 2^n(n+1)^2$$ $$\beta = \sum_{r=0}^{n} \frac{\binom{n}{r}}{r+1} = \frac{1}{n+1}\sum_{r=0}^{n} \binom{n+1}{r+1}$$ $$= \frac{1}{n+1}\left(\binom{n+1}{1} + \binom{n+1}{2} + \ldots + \binom{n+1}{n+1}\right)$$ $$= \frac{2^{n+1} - 1}{n+1}$$ $$\frac{2\alpha}{\beta} = \frac{2^{n+1}(n+1)^2}{\dfrac{2^{n+1}-1}{n+1}} \approx (n+1)^3$$ $$140 < (n+1)^3 < 281$$ $n = 4 \Rightarrow (n+1)^3 = 125$ $n = 5 \Rightarrow (n+1)^3 = 216$ $n = 6 \Rightarrow (n+1)^3 = 343$ Therefore, $n = 5$.

Question 28

Maths · Vector Algebra · Numerical

Let $\vec{a}$ = 9$\hat{i}$ - 13$\hat{j}$ + 25$\hat{k}$, $\vec{b}$ = 3$\hat{i}$ + 7$\hat{j}$ - 13$\hat{k}$ and $\vec{c}$ = 17$\hat{i}$ - 2$\hat{j}$ + $\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r}$ $\times$ $\vec{a}$ = ($\vec{b}$ + $\vec{c}$) $\times$ $\vec{a}$ and $\vec{r}$ $\cdot$ ($\vec{b}$ - $\vec{c}$) = 0, then $\frac{|593\vec{r} + 67\vec{a}|^2}{(593)^2}$ is equal to

Answer: 569

Solution

Given $\vec{a}=9\hat{i}-13\hat{j}+25\hat{k}$ and $\vec{b}=3\hat{i}+7\hat{j}-13\hat{k}$. Also, $\vec{c}=17\hat{i}-2\hat{j}+\hat{k}$. Then $\vec{b}+\vec{c}=20\hat{i}+5\hat{j}-12\hat{k}$ and $\vec{b}-\vec{c}=-14\hat{i}+9\hat{j}-14\hat{k}$. $(\vec{r}-(\vec{b}+\vec{c}))\times\vec{a}=0$ implies $\vec{r}-(\vec{b}+\vec{c})=\lambda\vec{a}$. Thus, $\vec{r}=\lambda\vec{a}+\vec{b}+\vec{c}$. But $\vec{r}\cdot(\vec{b}-\vec{c})=0$. This implies $(\lambda\vec{a}+\vec{b}+\vec{c})\cdot(\vec{b}-\vec{c})=0$. Expanding, $\lambda\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{b}+\vec{c}\cdot\vec{b}-\lambda\vec{a}\cdot\vec{c}-\vec{b}\cdot\vec{c}-\vec{c}\cdot\vec{c}=0$. Solving for $\lambda$, $\lambda=\frac{\vec{c}\cdot\vec{c}-\vec{b}\cdot\vec{c}}{\vec{a}\cdot\vec{b}-\vec{a}\cdot\vec{c}}=\frac{294-227}{-389-204}=\frac{-67}{593}$. Therefore, $\vec{r}=\vec{b}+\vec{c}-\frac{67}{593}\vec{a}$. Thus, $593\vec{r}+67\vec{a}=593(\vec{b}+\vec{c})$. Finally, $|\vec{b}+\vec{c}|^2=569$.

Question 29

Maths · Applications of Integrals · Numerical

Let the area of the region enclosed by the curve $y = \min\{\sin x, \cos x\}$ and the $x$ axis between $x = -\pi$ to $x = \pi$ be $A$. Then $A^2$ is equal to

Answer: 16

Solution

Given $y = \min \{ \sin x, \cos x \}$. The $x$-axis is from $x = -\pi$ to $x = \pi$. $$\int_0^{\pi/4} \sin x = (\cos x)^0_{\pi/4} = 1 - \frac{1}{\sqrt{2}}$$ $$\int_{-\pi}^{-3\pi/4} (\sin x - \cos x) = (-\cos x - \sin x)^{-3\pi/4}_{-\pi}$$ $$= (\cos x + \sin x)^{-\pi}_{-3\pi/4}$$ $$= (-1 + 0) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right)$$ $$= -1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$$ $$\int_{\pi/4}^{\pi/2} \cos x \, dx = (\sin x)^{\pi/2}_{\pi/4} = 1 - \frac{1}{\sqrt{2}}$$ $A = 4$ $A^2 = 16$

Question 30

Maths · Limits and Derivatives · Numerical

The value of $\lim_{x \to 0} 2 \left( \frac{1 - \cos x \sqrt[]{\cos 2x} \sqrt[3]{\cos 3x} \ldots \sqrt[10]{\cos 10x}}{x^2} \right)$ is

Answer: 55

Solution

Given the limit expression: $$\lim_{x \to 0} 2 \left( \frac{1 - \left( 1 - \frac{x^2}{2!} \right) \left( 1 - \frac{4x^2}{2!} \right) \left( 1 - \frac{9x^2}{2!} \right) \cdots \left( 1 - \frac{100x^2}{2!} \right)}{x^2} \right)$$ By expansion, $$\lim_{x \to 0} \frac{2 \left( 1 - \left( 1 - \frac{x^2}{2} \right) \right) \left( 1 - \frac{1}{2} \cdot \frac{4x^2}{2} \right) \left( 1 - \frac{1}{3} \cdot \frac{9x^2}{2} \right) \cdots \left( 1 - \frac{1}{10} \cdot \frac{100x^2}{2} \right) \right)}{x^2}$$ Simplifying further, $$\lim_{x \to 0} 2 \left( \frac{1 - \left( 1 - \frac{x^2}{2} \right) \left( 1 - \frac{2x^2}{2} \right) \left( 1 - \frac{3x^2}{2} \right) \cdots \left( 1 - \frac{10x^2}{2} \right)}{x^2} \right)$$ This becomes, $$\lim_{x \to 0} \frac{2 \left( 1 - 1 + x^2 \left( \frac{1}{2} + \frac{2}{2} + \frac{3}{2} + \cdots + \frac{10}{2} \right) \right)}{x^2}$$ Which simplifies to, $$2 \left( \frac{1}{2} + \frac{2}{2} + \frac{3}{2} + \cdots + \frac{10}{2} \right)$$ The sum is, $$1 + 2 + \cdots + 10 = \frac{10 \times 11}{2} = 55$$

Physics

Question 31

Physics · System of Particles and Rotational Motion · Single correct

Three bodies A, B and C have equal kinetic energies and their masses are 400 $\mathrm{g}$, 1.2 $\mathrm{kg}$ and 1.6 $\mathrm{kg}$ respectively. The ratio of their linear momenta is:

$\sqrt{2} : \sqrt{3} : 1$
1 : $\sqrt{3}$ : $2$
1 : $\sqrt{3}$ : $\sqrt{2}$
$\sqrt{3}$ : $\sqrt{2}$ : $1$

Answer: (b)

Solution

$KE=\frac{P^2}{2m}$ $P\propto\sqrt{m}$ Hence, $P_A:P_B:P_C$ $=\sqrt{400}:\sqrt{1200}:\sqrt{1600}$ $=1:\sqrt3:2$

Question 32

Physics · Dual Nature of Radiation and Matter · Single correct

Average force exerted on a non-reflecting surface at normal incidence is $2.4 \times 10^{-4} \, \mathrm{N}$. If $360 \, \mathrm{W/cm^2}$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:

$0.2 \, \mathrm{m^2}$
$20 \, \mathrm{m^2}$
$0.1 \, \mathrm{m^2}$
$0.02 \, \mathrm{m^2}$

Answer: (d)

Solution

Pressure $= \frac{I}{C} = \frac{F}{A}$ $$\Rightarrow \frac{360}{10^{-4} \times 3 \times 10^8} = \frac{2.4 \times 10^{-4}}{A}$$ $$\Rightarrow A = 2 \times 10^{-2} \, \mathrm{m^2} = 0.02 \, \mathrm{m^2}$$

Question 33

Physics · Dual Nature of Radiation and Matter · Single correct

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume $h=6.63 \times 10^{-34} \, \mathrm{J \, s}$, $m_e = 9.0 \times 10^{-31} \, \mathrm{kg}$ and $m_p = 1836$ times $m_e$)

1 : $\sqrt{1836}$
1 : $\frac{1}{1836}$
1 : $\frac{1}{\sqrt{1836}}$
1 : 1836

Answer: (d)

Solution

Given $\lambda$ is same for both. $P = \frac{h}{\lambda}$ is same for both. $P = \sqrt{2mK}$. Hence, $K \propto \frac{1}{m}$. Therefore, $$$ $\frac{\mathrm{KE}_p}{\mathrm{KE}_e}$ = $\frac{m_e}{m_p}$ = $\frac{1}{1836}$ $$$.

Question 34

Physics · Kinetic Theory · Single correct

A mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature (27^{$\circ$} $\mathrm{C}$). The ratio of specific heat of gases at constant volume respectively is:

$\frac{7}{5}$
$\frac{3}{5}$
$\frac{5}{3}$
$\frac{3}{2}$

Answer: (b)

Solution

The ratio of specific heats for monatomic and diatomic gases is given by: $$\frac{(C_v)_{mono}}{(C_v)_{dia}} = \frac{\frac{3}{2} R}{\frac{5}{2} R} = \frac{3}{5}$$

Question 35

Physics · Mathematics in Physics · Single correct

In an expression $a \times 10^b$;

$b$ is order of magnitude for $a \geq 5$
$b$ is order of magnitude for $a \leq 5$
$a$ is order of magnitude for $b \leq 5$
$b$ is order of magnitude for $5 < a \leq 10$

Answer: (b)

Solution

Given $a \times 10^b$. If $a \leq 5$, the order is $b$. If $a > 5$, the order is $b + 1$.

Question 36

Physics · Current Electricity · Single correct

In the given circuit, the terminal potential difference of the cell is :

2 $\mathrm{V}$
3 $\mathrm{V}$
4 $\mathrm{V}$
1.5 $\mathrm{V}$

Answer: (a)

Solution

The current $i$ is calculated as $$i = \frac{3}{1 + 2} = 1 \, \mathrm{A}.$$ The voltage $v$ is given by $$v = E - ir = 3 - 1 \times 1 = 2 \, \mathrm{V}.$$

Question 37

Physics · Nuclei · Single correct

Binding energy of a certain nucleus is $18 \times 10^8 \, \mathrm{J}$. How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus:

$10 \, \mu\mathrm{g}$
$20 \, \mu\mathrm{g}$
$0.2 \, \mu\mathrm{g}$
$2 \, \mu\mathrm{g}$

Answer: (b)

Solution

Given $\Delta mc^2 = 18 \times 10^8$. $\Delta m \times 9 \times 10^{16} = 18 \times 10^8$. $\Delta m = 2 \times 10^{-8} \, \mathrm{kg} = 20 \, \mu \mathrm{g}$.

Question 38

Physics · Magnetism and Matter · Single correct

Paramagnetic substances: A. align themselves along the directions of external magnetic field. B. attract strongly towards external magnetic field. C. has susceptibility little more than zero. D. move from a region of strong magnetic field to weak magnetic field. Choose the most appropriate answer from the options given below:

A, B, C Only
A, B, C, D
A, C Only
B, D Only

Answer: (c)

Solution

A, C only

Question 39

Physics · Motion in a Plane · Single correct

A clock has 75 cm, 60 cm long second hand and minute hand respectively. In 30 minutes duration the tip of second hand will travel $x$ distance more than the tip of minute hand. The value of $x$ in meter is nearly (Take $\pi = 3.14$):

139.4
140.5
220.0
118.9

Answer: (a)

Solution

Question 40

Physics · Mathematics in Physics · Single correct

Young's modulus is determined by the equation given by $Y = 49000 \frac{m}{l} \, \mathrm{dyn/cm^2}$ where $M$ is the mass and $l$ is the extension of wire used in the experiment. Now error in Young modules $(Y)$ is estimated by taking data from $M - l$ plot in graph paper. The smallest scale divisions are $5 \, \mathrm{g}$ and $0.02 \, \mathrm{cm}$ along load axis and extension axis respectively. If the value of $M$ and $l$ are $500 \, \mathrm{g}$ and $2 \, \mathrm{cm}$ respectively then percentage error of $Y$ is :

0.5%
2%
0.02%
0.2%

Answer: (b)

Solution

Given $$\frac{\Delta Y}{Y} = \frac{\Delta m}{m} + \frac{\Delta \ell}{\ell}$$ we have $$\frac{5}{500} + \frac{0.02}{2} = 0.01 + 0.01$$ Therefore, $$\frac{\Delta Y}{Y} = 0.02 \Rightarrow \% \frac{\Delta Y}{Y} = 2\%$$

Question 41

Physics · Thermodynamics · Single correct

Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P-V diagram. The relation between the ratio $\frac{V_a}{V_d}$ and the ratio $\frac{V_b}{V_c}$ is:

$\frac{V_a}{V_d} \neq \frac{V_b}{V_c}$
$\frac{V_a}{V_d} = \frac{V_b}{V_c}$
$\frac{V_a}{V_d} = \left(\frac{V_b}{V_c}\right)^{-1}$
$\frac{V_a}{V_d} = \left(\frac{V_b}{V_c}\right)^2$

Answer: (b)

Solution

For adiabatic process, $TV^{\gamma-1} = constant$. $$T_a \cdot V_a^{\gamma-1} = T_d \cdot V_d^{\gamma-1}$$ $$\left( \frac{V_a}{V_d} \right)^{\gamma-1} = \frac{T_d}{T_a}$$ $$T_b \cdot V_b^{\gamma-1} = T_c \cdot V_c^{\gamma-1}$$ $$\left( \frac{V_b}{V_c} \right)^{\gamma-1} = \frac{T_c}{T_b}$$ $$\frac{V_a}{V_d} = \frac{V_b}{V_c} \left( \therefore T_d = T_c T_a = T_b \right)$$

Question 42

Physics · Gravitation · Single correct

Two planets $A$ and $B$ having masses $m_1$ and $m_2$ move around the sun in circular orbits of $r_1$ and $r_2$ radii respectively. If angular momentum of $A$ is $L$ and that of $B$ is $3L$, the ratio of time period $\left( \frac{T_A}{T_B} \right)$ is:

$\left(\dfrac{r_2}{r_1}\right)^{\frac{3}{2}}$
$\dfrac{1}{27}\left(\dfrac{m_2}{m_1}\right)^3$
$27\left(\dfrac{m_1}{m_2}\right)^3$
$\left(\dfrac{r_1}{r_2}\right)^3$

Answer: (b)

Solution

Given the equations: $$\frac{\pi r_1^2}{T_A} = \frac{L}{2 \, m_1} \ldots \ldots \ldots (1)$$ $$\frac{\pi r_2^2}{T_B} = \frac{3 \, L}{2 \, m_2} \ldots \ldots \ldots (2)$$ Therefore, $$\frac{T_A}{T_B} = 3 \cdot \frac{m_1}{m_2} \cdot \left( \frac{r_1}{r_2} \right)^2$$ $$\left( \frac{T_A}{T_B} \right)^2 = \left( \frac{r_1}{r_2} \right)^3 \implies \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{T_A}{T_B} \right)^{\frac{4}{3}}$$ $$\implies \frac{1}{27} \cdot \left( \frac{m_2}{m_1} \right)^3 = \left( \frac{T_A}{T_B} \right)$$

Question 43

Physics · Alternating Current · Single correct

A LCR circuit is at resonance for a capacitor $C$, inductance $L$ and resistance $R$. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:

Zero
same
halved
double

Answer: (d)

Solution

In resonance $Z = R$ $$I = \frac{V}{R}$$ $R \rightarrow$ halved $\Rightarrow I \rightarrow 2I$ $I$ becomes doubled.

Question 44

Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct

The output Y of following circuit for given inputs is :

$\overline{A}$ $\cdot B$($\overline{A} + B$)
$0$
$\overline{A} \cdot B$
$A \cdot B$

Answer: (b)

Solution

By truth table \begin{tabular}{|c|c|c|} \hline $A$ & $B$ & $Y$ \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 0 \\ \hline 1 & 1 & 0 \\ \hline \end{tabular}

Question 45

Physics · Electric Charges and Fields · Single correct

Two charged conducting spheres of radii $a$ and $b$ are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:

$\frac{a}{b}$
$\sqrt{ab}$
$\frac{b}{a}$
$ab$

Answer: (a)

Solution

Potential at surface will be same $$\frac{K q_1}{a} = \frac{K q_2}{b}$$ $$\frac{q_1}{q_2} = \frac{a}{b}$$

Question 46

Physics · Mechanical Properties of Fluids · Single correct

Correct Bernoulli's equation is (symbols have their usual meaning) :

$P + mgh + \frac{1}{2} mv^2 = constant$
$P + \rho gh + \frac{1}{2} \rho v^2 = constant$
$P + \rho gh + \rho v^2 = constant$
$P + \frac{1}{2} \rho gh + \frac{1}{2} \rho v^2 = constant$

Answer: (b)

Solution

The equation given is Bernoulli's equation: $$P + \rho gh + \frac{1}{2} \rho V^2 = constant$$

Question 47

Physics · System of Particles and Rotational Motion · Single correct

A player caught a cricket ball of mass 150 $\mathrm{g}$ moving at a speed of 20 $\mathrm{m/s}$. If the catching process is completed in 0.1 $\mathrm{s}$, the magnitude of force exerted by the ball on the hand of the player is:

3 $\mathrm{N}$
300 $\mathrm{N}$
150 $\mathrm{N}$
30 $\mathrm{N}$

Answer: (d)

Solution

Given the change in momentum $\Delta P$ and the change in time $\Delta t$, the force $F$ is calculated as follows: $$F = \frac{\Delta P}{\Delta t} = \frac{mv - 0}{0.1}$$ $$= \frac{150 \times 10^{-3} \times 20}{0.1} = 30 \, \mathrm{N}$$

Question 48

Physics · Work, Energy and Power · Single correct

A stationary particle breaks into two parts of masses $m_A$ and $m_B$ which move with velocities $v_A$ and $v_B$ respectively. The ratio of their kinetic energies $(K_B : K_A)$ is :

$v_B : v_A$
$m_B : m_A$
$m_B v_B : m_A v_A$
$1 : 1$

Answer: (a)

Solution

Initial momentum is zero. Hence $|P_A| = |P_B|$. Therefore, $m_A v_B = m_B v_B$. $$\frac{(\mathrm{KE})_A}{(\mathrm{KE})_B} = \frac{\frac{1}{2} m_A v_A^2}{\frac{1}{2} m_B v_B^2} = \frac{v_A}{v_B}$$ $$\frac{(\mathrm{KE})_B}{(\mathrm{KE})_A} = \frac{v_B}{v_A}$$

Question 49

Physics · Ray Optics and Optical Instruments · Single correct

Critical angle of incidence for a pair of optical media is $45^\circ$. The refractive indices of first and second media are in the ratio:

1 : $\sqrt{2}$
$\sqrt{2}$ : 1
2 : 1
1 : 2

Answer: (b)

Solution

Given $\sin\theta_c=\frac{\mu_R}{\mu_d}=\frac{\mu_2}{\mu_1}$. $\sin45^\circ=\frac{\mu_2}{\mu_1}$. Therefore, $\frac{1}{\sqrt{2}}=\frac{\mu_2}{\mu_1}$. This implies $\frac{\mu_1}{\mu_2}=\frac{\sqrt{2}}{1}$.

Question 50

Physics · Experimental Physics · Single correct

The diameter of a sphere is measured using a vernier caliper whose 9 divisions of main scale are equal to 10 divisions of vernier scale. The shortest division on the main scale is equal to 1 mm. The main scale reading is 2 cm and second division of vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 g, the density of the sphere is:

$2.0\,\mathrm{g/cm^3}$ \\
$1.7\,\mathrm{g/cm^3}$ \\
$2.2\,\mathrm{g/cm^3}$ \\
$2.5\,\mathrm{g/cm^3}$

Answer: (a)

Solution

Given $9 MSD = 10 VSD$. Mass $= 8.635 \, g$. $LC = 1 MSD - 1 VSD$. $LC = 1 MSD - \frac{9}{10} MSD$. $LC = \frac{1}{10} MSD$. $LC = 0.01 \, cm$. Reading of diameter $= MSR + LC \times VSR$. $= 2 \, cm + (0.01) \times (2)$. $= 2.02 \, cm$. Volume of sphere $= \frac{4}{3} \pi \left( \frac{d}{2} \right)^3 = \frac{4}{3} \pi \left( \frac{2.02}{2} \right)^3$. $= 4.32 \, cm^3$. Density $= \frac{mass}{volume} = \frac{8.635}{4.32} = 1.998 \sim 2.00 \, g$. SECTION -B

Question 51

Physics · System of Particles and Rotational Motion · Numerical

A uniform thin metal plate of mass 10 kg with dimensions is shown. The ratio of x and y coordinates of center of mass of plate in $\frac{n}{9}$. The value of $n$ is

Answer: 15

Solution

Given $m_1 = \sigma \times 5 = 10 \, \mathrm{Kg}$ and $m_2 = \sigma \times 1 = 2 \, \mathrm{Kg}$. Also, $m_3 = \sigma \times 6 = 12 \, \mathrm{Kg}$. Using the equation for center of mass: $$m_1 x_1 + m_2 x_2 = m_3 x_3$$ Substituting the values: $$10 x_1 + 2 (1.5) = 12 (1.5) \implies x_1 = 1.5 \, \mathrm{cm}$$ Similarly, for the $y$-coordinates: $$m_1 y_1 + m_2 y_2 = m_3 y_3$$ Substituting the values: $$10 y_1 + 2 (1.5) = 12 \times 1 \implies y_1 = 0.9 \, \mathrm{cm}$$ The ratio is: $$\frac{x_1}{y_1} = \frac{1.5}{0.9} = \frac{15}{9}$$ Thus, $n = 15$.

Question 52

Physics · Moving Charges and Magnetism · Numerical

An electron with kinetic energy \[ 5\,\mathrm{eV} \] enters a region of uniform magnetic field of \[ 3\,\mu\mathrm{T} \] perpendicular to its direction. An electric field \[ E \] is applied perpendicular to the direction of velocity and magnetic field. The value of \[ E, \] so that electron moves along the same path, is \[ \underline{\hspace{2cm}}\ \mathrm{NC^{-1}}. \] \[ \text{(Given, mass of electron }=9\times10^{-31}\ \mathrm{kg}, \text{ electric charge }=1.6\times10^{-19}\ \mathrm{C}) \]

Answer: 4

Solution

For the given condition of moving undeflected, net force should be zero. $qE = qVB$ $E = VB$ $$E = \sqrt{\frac{2 \times \mathrm{KE}}{m}} \times B$$ $$= \sqrt{\frac{2 \times 5 \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}} \times 3 \times 10^{-6}$$ $$= 4 \, \mathrm{N/C}$$

Question 53

Physics · Electromagnetic Induction · Numerical

A square loop PQRS having 10 turns, area $3.6 \times 10^{-3} \, \mathrm{m}^2$ and resistance $100\, \Omega$ is slowly and uniformly being pulled out of a uniform magnetic field of magnitude $B = 0.5 \, \mathrm{T}$ as shown. Work done in pulling the loop out of the field in $1.0 \, \mathrm{s}$ is ________ $\times 10^{-6}$ $\mathrm{J}$.

Answer: 3

Solution

Given $\epsilon = NB\ell v$. The current $i = \frac{\epsilon}{R} = \frac{NB\ell v}{R}$. The force $F = N(i\ell B) = \frac{N^2 B^2 \ell^2 v}{R}$. The work done $W = F \times \ell = \frac{N^2 B^2 \ell^3}{R} \left( \frac{\ell}{t} \right)$. The area $A = \ell^2$. Calculating work done: $$W = \frac{(10 \times 10)(0.5)^2 \times (3.6 \times 10^{-3})^2}{100 \times 1}$$ Finally, $W = 3.24 \times 10^{-6} \, \mathrm{J}$.

Question 54

Physics · Current Electricity · Numerical

Resistance of a wire at 0°C, 100°C and $t$°C is found to be $10\,\Omega$, $10.2\,\Omega$ and $10.95\,\Omega$ respectively. The temperature $t$ in Kelvin scale is

Answer: 748

Solution

Given $R = R_0(1 + \alpha \Delta T)$. The equation $\frac{\Delta R}{R_0} = \alpha \Delta T$ is used. Case-I From $0^\circ \mathrm{C} \rightarrow 100^\circ \mathrm{C}$: $$\frac{10.2 - 10}{10} = \alpha (100 - 0) \ldots(1)$$ Case-II From $0^\circ \mathrm{C} \rightarrow t^\circ \mathrm{C}$: $$\frac{10.95 - 10}{10} = \alpha (t - 0) \ldots(2)$$ Solving for $t$: $$\Rightarrow \frac{t}{100} = \frac{0.95}{0.2} = 475^\circ \mathrm{C}$$ Thus, $t = 475 + 273 = 748 \, \mathrm{K}$.

Question 55

Physics · Electric Charges and Fields · Numerical

An electric field, $\vec{E} = \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}}$ passes through the surface of $4 \, \mathrm{m}^2$ area having unit vector $\hat{n} = \left( \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \right)$. The electric flux for that surface is Vm.

Answer: 12

Solution

The flux $\phi$ is given by the dot product of $\vec{E}$ and $\vec{A}$. $$\phi = \vec{E} \cdot \vec{A}$$ Substituting the given vectors, we have: $$= \left( \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}} \right) \cdot 4 \left( \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \right)$$ Simplifying the expression: $$= \frac{4}{6} \times (4 + 6 + 8) = 12 \, \mathrm{Vm}$$

Question 56

Physics · Mechanical Properties of Fluids · Numerical

A liquid column of height 0.04 cm balances excess pressure of a soap bubble of certain radius. If density of liquid is $8 \times 10^3 \, \mathrm{kg \, m^{-3}}$ and surface tension of soap solution is $0.28 \, \mathrm{Nm^{-1}}$, then diameter of the soap bubble is _____ cm. (if $g = 10 \, \mathrm{m \, s^{-2}}$)

Answer: 7

Solution

$\rho g h = \dfrac{4S}{R}$ $\Rightarrow R = \dfrac{4 \times 0.28}{8 \times 10^3 \times 10 \times 4 \times 10^{-4}}$ $\Rightarrow \dfrac{0.28}{8}\,\text{m} = \dfrac{28}{8}\,\text{cm}$ $\Rightarrow R = 3.5\,\text{cm}$ $\text{Diameter} = 7\,\text{cm}$

Question 57

Physics · Waves · Numerical

A closed and an open organ pipe have same lengths. If the ratio of frequencies of their seventh overtones is $\left( \frac{a-1}{a} \right)$ then the value of $a$ is

Answer: 16

Solution

For closed organ pipe $$f_c = (2n + 1) \frac{v}{4\ell} = \frac{15v}{4\ell}$$ For open organ pipe $$f_o = (n + 1) \frac{v}{2\ell} = \frac{8v}{2\ell}$$ $$\frac{f_c}{f_o} = \frac{15}{16} = \frac{a - 1}{a}$$ Therefore, $a = 16$

Question 58

Physics · Mathematics in Physics · Numerical

Three vectors $\overrightarrow{OP}$, $\overrightarrow{OQ}$ and $\overrightarrow{OR}$ each of magnitude $A$ are acting as shown in figure. The resultant of the three vectors is $A\sqrt{x}$. The value of $x$ is _______.

Answer: 3

Solution

The resultant vector $\vec{R}$ is given by: $$\vec{R} = \left( A + \frac{A}{\sqrt{2}} \right) \hat{i} + \left( A - \frac{A}{\sqrt{2}} \right) \hat{j}$$ The magnitude of $\vec{R}$ is: $$|\vec{R}| = \sqrt{\left( A + \frac{A}{\sqrt{2}} \right)^2 + \left( A - \frac{A}{\sqrt{2}} \right)^2} = \sqrt{3}A$$

Question 59

Physics · Wave Optics · Numerical

A parallel beam of monochromatic light of wavelength $600 \, \mathrm{nm}$ passes through single slit of $0.4 \, \mathrm{mm}$ width. Angular divergence corresponding to second order minima would be ____ $\times 10^{-3} \, \mathrm{rad}$.

Answer: 6

Solution

Given $\sin \theta \simeq \theta \simeq \frac{2 \lambda}{b}$ $$= \frac{2 \times 600 \times 10^{-9}}{4 \times 10^{-4}} = 3 \times 10^{-3} \, \mathrm{rad}$$ Total divergence $= (3 + 3) \times 10^{-3} = 6 \times 10^{-3} \, \mathrm{rad}$

Question 60

Physics · Nuclei · Numerical

In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \, \mathrm{m}$. If target nucleus has atomic number $80$, then maximum velocity of $\alpha$ - particle is _______ $\times 10^5 \, \mathrm{m/s}$ approximately. ( $\frac{1}{4 \pi \varepsilon_0}$ = 9 $\times$ $10^9$ $\mathrm{SI \, unit}$, mass of $\alpha$ particle = 6.72 $\times$ $10^{-27}$ $\mathrm{kg}$ )

Answer: 156

Solution

Given $$v = \sqrt{\frac{4KZe^2}{mr_{\min}}}$$ Substitute the values: $$= \sqrt{\frac{4 \times 9 \times 10^9 \times 80}{6.72 \times 10^{-27} \times 4.5 \times 10^{-14} \times 1.6 \times 10^{-19}}}$$ Calculate: $$= 9.759 \times 10^{25} \times 1.6 \times 10^{-19}$$ Simplify: $$= 156 \times 10^5 \, \mathrm{m/s}$$

Chemistry

Question 61

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Given below are two statements: Statements I: IUPAC name of Compound A is 4-chloro-1,3-dinitrobenzene. Statements II: IUPAC name of Compound B is 4-ethyl-2-methylaniline. In the light of the above statements, choose the most appropriate answer from the options given below:

Both Statement I and Statement II are incorrect.
Both Statement I and Statement II are correct.
Statement I is correct but Statement II is incorrect.
Statement I is incorrect but Statement II is correct.

Answer: (d)

Solution

Statement I: The IUPAC name is 1-chloro-2, 4-dinitrobenzene. Therefore, statement I is incorrect. Statement II: The IUPAC name is 4-ethyl-2-methylaniline. Therefore, statement II is correct.

Question 62

Chemistry · Haloalkanes and Haloarenes · Single correct

Which among the following compounds will undergo fastest $S_N2$ reaction.

Answer: (a)

Solution

The fastest $\mathrm{S_N^2}$ reaction is given by the primary (1°) alkyl halide. Rate of $\mathrm{S_N^2}$ is Me-x > 1°-x > 2°-x > 3°-x.

Question 63

Chemistry · Some Basic Concepts of Chemistry · Single correct

Combustion of glucose $(C_6H_{12}O_6)$ produces $CO_2$ and water. The amount of oxygen (in g) required for the complete combustion of 900 g of glucose is: [Molar mass of glucose in $\mathrm{g/mol}^{-1}$ = 180]

Answer: (c)

Solution

The reaction is given by $$\mathrm{C_6H_{12}O_{6(s)} + 6O_{2(g)} \rightarrow 6CO_{2(g)} + 6H_2O_{(l)}}$$ Calculating the moles of glucose: $$\frac{900}{180} = 5 \, mol$$ Thus, 30 mol of $\mathrm{O_2}$ is required. The mass of $\mathrm{O_2}$ required is calculated as follows: $$30 \times 32 = 960 \, gm$$

Question 64

Chemistry · Alcohols, Phenols and Ethers · Single correct

Identify the major products A and B respectively in the following set of reactions.

Answer: (d)

Solution

The reaction starts with the compound having a hydroxyl group and a methyl group attached to a cyclohexane ring. It undergoes acetylation with $\mathrm{CH_3COCl}$ in the presence of pyridine to form compound (B), which has an acetoxy group and a methyl group attached to the cyclohexane ring. Compound (B) is then subjected to an $\mathrm{E_1}$ reaction using concentrated $\mathrm{H_2SO_4}$ and heat ($\Delta$) to form compound (A), which has a methyl group attached to the cyclohexane ring and water as a byproduct.

Question 65

Chemistry · The p-Block Elements (Group-13 and 14) · Single correct

Give below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A: The stability order of +1 oxidation state of Ga, In and Tl is Ga < In < Tl. Reason R: The inert pair effect stabilizes the lower oxidation state down the group. In the light of the above statements, choose the correct answer from the options given below:

A is true but R is false.
A is false but R is true.
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is NOT the correct explanation of A.

Answer: (c)

Solution

The relative stability of $+1$ oxidation state progressively increases for heavier elements due to inert pair effect. Therefore, the stability of $\mathrm{Al}^{+1} < \mathrm{Ga}^{+1} < \mathrm{In}^{+1} < \mathrm{Tl}^{+1}$.

Question 66

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Match List I with List II Choose the correct answer from the options given below:

A-III, B-II, C-IV, D-I
A-III, B-IV, C-I, D-II
A-III, B-I, C-II, D-IV
A-III, B-I, C-IV, D-II

Answer: (b)

Solution

Cobalt nitrate test $\mathrm{MCO_3 \rightarrow MO}$ $\mathrm{\xrightarrow[+\Delta]{Co(NO_3)_2}}CoO\cdot MO$ Flame test $\mathrm{MCO_3 \rightarrow MCl_2 \rightarrow M^{2+}}$ Borax Bead test $\mathrm{MSO_4 {Na_2B_4O_7} M(BO_2)_2}$$MBO_2$ $\rightarrow{M}$ Charcoal cavity test $\mathrm{MSO_4 \xrightarrow[\Delta]{Na_2CO_3}}$ $\mathrm{MCO_3 \rightarrow MO \rightarrow M}$

Question 67

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Match List I with List II \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Molecule)} & \multicolumn{2}{c|}{(Shape)} \\ \hline A. & NH$_3$ & I. & Square pyramid \\ \hline B. & BrF$_5$ & II. & Tetrahedral \\ \hline C. & PCl$_5$ & III. & Trigonal pyramidal \\ \hline D. & CH$_4$ & IV. & Trigonal bipyramidal \\ \hline \end{tabular} Choose the correct answer from the options given below:

A-II, B-IV, C-I, D-III
A-III, B-I, C-IV, D-II
A-IV, B-III, C-I, D-II
A-III, B-IV, C-I, D-II

Answer: (b)

Solution

Question 68

Chemistry · Equilibrium · Single correct

For the given hypothetical reactions, the equilibrium constants are as follows : $$X \rightleftharpoons Y; \; K_1 = 1.0$$ $$Y \rightleftharpoons Z; \; K_2 = 2.0$$ $$Z \rightleftharpoons W; \; K_3 = 4.0$$ The equilibrium constant for the reaction $$X \rightleftharpoons W$$ is

6.0
12.0
7.0
8.0

Answer: (d)

Solution

Given the reactions: $$X \rightleftharpoons Y k_1 = 1$$ $$Y \rightleftharpoons Z k_2 = 2$$ $$Z \rightleftharpoons \omega k_3 = 4$$ For the overall reaction: $$X \rightleftharpoons \omega$$ The equilibrium constant is given by: $$k = k_1 \cdot k_2 \cdot k_3$$ Substituting the values: $$k = 1 \times 2 \times 4$$ Therefore, the equilibrium constant is: $$k = 8$$

Question 69

Chemistry · Redox Reactions · Single correct

Thiosulphate reacts differently with iodine and bromine in the reactions given below: $$2 \mathrm{S}_2\mathrm{O}_3^{2-} + \mathrm{I}_2 \rightarrow \mathrm{S}_4\mathrm{O}_6^{2-} + 2\mathrm{I}^-$$ $$\mathrm{S}_2\mathrm{O}_3^{2-} + 5\mathrm{Br}_2 + 5\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{SO}_4^{2-} + 4\mathrm{Br}^- + 10\mathrm{H}^+$$ Which of the following statement justifies the above dual behaviour of thiosulphate?

Bromine is a stronger oxidant than iodine
Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reaction
Bromine is a weaker oxidant than iodine
Bromine undergoes oxidation and iodine undergoes reduction in these reactions

Answer: (a)

Solution

In the reaction of $\mathrm{S_2O_3^{2-}}$ with $\mathrm{I_2}$, oxidation state of sulphur changes to $+2$ to $+2.5$. In the reaction of $\mathrm{S_2O_3^{2-}}$ with $\mathrm{Br_2}$, oxidation state of sulphur changes from $+2$ to $+6$. Therefore, both $\mathrm{I_2}$ and $\mathrm{Br_2}$ are oxidant (oxidising agent) and $\mathrm{Br_2}$ is stronger oxidant than $\mathrm{I_2}$.

Question 70

Chemistry · Co-ordination Compounds · Single correct

An octahedral complex with the formula $CoCl_3 \cdot n{NH}_3$ upon reaction with excess of $\mathrm{AgNO}_3$ solution gives 2 moles of $\mathrm{AgCl}$. Consider the oxidation state of Co in the complex is 'x'. The value of "x + n" is ______

Answer: (b)

Solution

Given $[\mathrm{Co(NH_3)_5Cl}]\mathrm{Cl_2} + excess \mathrm{AgNO_3} \rightarrow 2\mathrm{AgCl}$ (2 moles). $x + 0 - 1 - 2 = 0$ $x = +3$ $n = 5$ Therefore, $x + n = 8$

Question 71

Chemistry · Biomolecules · Single correct

The incorrect statement regarding the given structure is

can be oxidized to a dicarboxylic acid with $\mathrm{Br}_2$ water
will coexist in equilibrium with 2 other cyclic structure
despite the presence of - CHO does not give Schiff's test
has 4 asymmetric carbon atom

Answer: (a)

Solution

Statement 1 is incorrect (monocarboxylic acid). Statement 2 is correct. Statement 3: c.c. is 4 (correct). Statement 4: $\alpha$-D-glucose converts to $\beta$-D-glucose (correct).

Question 72

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

In the given compound, the number of $2^\circ$ carbon atom /s is _______.

Four
Two
One
Three

Answer: (c)

Solution

Only one $2^\circ$ carbon is present in this compound.

Question 73

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Which of the following are aromatic?

A and C only
B and D only
C and D only
A and B only

Answer: (b)

Solution

Option A is non-aromatic. Option B is aromatic. Option C is non-aromatic. Option D is aromatic.

Question 74

Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct

Among the following halogens $\mathrm{F_2}$, $\mathrm{Cl_2}$, $\mathrm{Br_2}$ and $\mathrm{I_2}$ Which can undergo disproportionation reactions?

$\mathrm{F_2}$, $\mathrm{Cl_2}$ and $\mathrm{Br_2}$
$\mathrm{F_2}$ and $\mathrm{Cl_2}$
Only $\mathrm{I_2}$
$\mathrm{Cl_2}$, $\mathrm{Br_2}$ and $\mathrm{I_2}$

Answer: (d)

Solution

$F_2$ do not disproportionate because fluorine does not exist in a positive oxidation state. However, $Cl_2$, $Br_2$, and $I_2$ undergo disproportionation.

Question 75

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Given below are two statements: Statement I: $N(\mathrm{CH}_3)_3$ and $P(\mathrm{CH}_3)_3$ can act as ligands to form transition metal complexes. Statement II: As N and P are from same group, the nature of bonding of $N(\mathrm{CH}_3)_3$ and $P(\mathrm{CH}_3)_3$ is always same with transition metals. In the light of the above statements, choose the most appropriate answer from the options given below:

Statement I is correct but Statement II is incorrect.
Statement I is incorrect but Statement II is correct.
Both Statement I and Statement II are correct.
Both Statement I and Statement II are incorrect.

Answer: (a)

Solution

$N(\mathrm{CH}_3)_3$ and $P(\mathrm{CH}_3)_3$ both are Lewis base and acts as ligand. However, $P(\mathrm{CH}_3)_3$ has a $\pi$-acceptor character.

Question 76

Chemistry · Classification of Elements and Periodicity in Properties · Single correct

Match List I with List II \begin{tabular}{|c|c|c|l|} \hline \multicolumn{2}{|c|}{List - I (Elements)} & \multicolumn{2}{c|}{List - II (Properties in their respective groups)} \\ \hline A. & Cl, S & I. & Elements with highest electronegativity \\ \hline B. & Ge, As & II. & Elements with largest atomic size \\ \hline C. & Fr, Ra & III. & Elements which show properties of both metals and non-metal \\ \hline D. & F, O & IV. & Elements with highest negative electron gain enthalpy \\ \hline \end{tabular} Choose the correct answer from the options given below:

A-II, B-I, C-IV, D-III
A-III, B-II, C-I, D-IV
A-IV, B-III, C-II, D-I
A-II, B-III, C-IV, D-I

Answer: (c)

Solution

Elements with highest electronegativity $\rightarrow$ F, O Elements with largest atomic size $\rightarrow$ Fr, Ra Elements which shows properties of both metal and non-metals i.e. metalloids $\rightarrow$ Ge, As Elements with highest negative electron gain enthalpy $\rightarrow$ Cl, S

Question 77

Chemistry · The d-and f-Block Elements · Multiple correct

Iron (III) catalyses the reaction between iodide and persulphate ions, in which A. $\mathrm{Fe}^{3+}$ oxidises the iodide ion B. $\mathrm{Fe}^{3+}$ oxidises the persulphate ion C. $\mathrm{Fe}^{2+}$ reduces the iodide ion D. $\mathrm{Fe}^{2+}$ reduces the persulphate ion Choose the most appropriate answer from the options given below:

B only
A only
B and C only
A and D only

Answer: (d)

Solution

Given the reactions: $$ 2\mathrm{Fe}^{3+} + 2\mathrm{I}^{-} \rightarrow 2\mathrm{Fe}^{2+} + \mathrm{I}_2 $$ $$ 2\mathrm{Fe}^{2+} + \mathrm{S}_2\mathrm{O}_8^{2-} \rightarrow 2\mathrm{Fe}^{3+} + 2\mathrm{SO}_4^{2-} $$ $\mathrm{Fe}^{3+}$ oxidises $\mathrm{I}^{-}$ to $\mathrm{I}_2$ and converts itself into $\mathrm{Fe}^{2+}$. This $\mathrm{Fe}^{2+}$ reduces $\mathrm{S}_2\mathrm{O}_8^{2-}$ to $\mathrm{SO}_4^{2-}$ and converts itself into $\mathrm{Fe}^{3+}$.

Question 78

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Match List I with List II \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Molecule)} & \multicolumn{2}{c|}{(Shape)} \\ \hline A. & Fe$_4$[Fe(CN)$_6$]$_3\cdot x$H$_2$O & I. & Violet \\ \hline B. & [Fe(CN)$_5$NOS]$^{4-}$ & II. & Blood Red \\ \hline C. & [Fe(SCN)]$^{2+}$ & III. & Prussian Blue \\ \hline D. & (NH$_4$)$_3$PO$_4\cdot12$MoO$_3$ & IV. & Yellow \\ \hline \end{tabular} Choose the correct answer from the options given below:

A-III, B-I, C-II, D-IV
A-I, B-II, C-III, D-IV
A-IV, B-I, C-II, D-III
A-II, B-III, C-IV, D-I

Answer: (a)

Solution

Fe_4[Fe(CN)_6]_3 $\cdot$ xH_2O $\rightarrow$ Prussian Blue [Fe(CN)_5NOS]^{4-} $\rightarrow$ Violet [Fe(SCN)]^{2+} $\rightarrow$ Blood Red (NH_4)_3PO_4 $\cdot$ 12MoO_3 $\rightarrow$ Yellow

Question 79

Chemistry · Co-ordination Compounds · Single correct

Number of Complexes with even number of electrons in $t_{2g}$ orbitals is - $[\mathrm{Fe(H_2O)_6}]^{2+}$, $[\mathrm{Co(H_2O)_6}]^{2+}$, $[\mathrm{Co(H_2O)_6}]^{3+}$, $[\mathrm{Cu(H_2O)_6}]^{2+}$, $[\mathrm{Cr(H_2O)_6}]^{2+}$

Answer: (b)

Solution

$[\mathrm{Fe(H_2O)_6}]^{2+}$ $\mathrm{Fe}^{2+} \rightarrow d^6$ Electron in $t_{2g}=4$ (even) --- $[\mathrm{Co(H_2O)_6}]^{2+}$ $\mathrm{Co}^{2+} \rightarrow d^7$ Electron in $t_{2g}=5$ (odd) --- $[\mathrm{Co(H_2O)_6}]^{3+}$ $\mathrm{Co}^{3+} \rightarrow d^6$ Electron in $t_{2g}=6$ (even) --- $[\mathrm{Cu(H_2O)_6}]^{2+}$ $\mathrm{Cu}^{2+} \rightarrow d^9$ Electron in $t_{2g}=6$ (even) --- $[\mathrm{Cr(H_2O)_6}]^{2+}$ $\mathrm{Cr}^{2+} \rightarrow d^4$ Electron in $t_{2g}=3$ (odd)

Question 80

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Identify the product $(P)$ in the following reaction:

Answer: (a)

Solution

Question 81

Chemistry · Structure of Atom · Numerical

A hypothetical electromagnetic wave is show below. The frequency of the wave is x $\times$ $10^{19}$ Hz. x = (nearest integer)

Answer: 5

Solution

Given $\lambda = 1.5 \times 4 \, \mathrm{pm}$, which equals $6 \times 10^{-12} \, \mathrm{meter}$. Using the equation $\lambda v = C$, we have: $$6 \times 10^{-12} \times v = 3 \times 10^8$$ Solving for $v$, we get: $$v = 5 \times 10^{19} \, \mathrm{Hz}$$

Question 82

Chemistry · Thermodynamics · Numerical

Consider the figure provided. 1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at $18^\circ \mathrm{C}$. If the piston is moved to position B, keeping the temperature unchanged, then '$x$' L atm work is done in this reversible process. x = L atm. (nearest integer) [Given : Absolute temperature = ${}^\circ$ C + 273.15, R = 0.08206 L atm $mol^{-1}$ $K^{-1}$]

Answer: 55

Solution

\[ w=-nRT\ln\left(\frac{V_2}{V_1}\right) \] \[ =-1\times 0.08206 \times 291 \times 15 \ln\left(\frac{100}{10}\right) \] \[ =-55.0128 \] \[ \text{Work done by system} \approx 55\ \text{atm L} \]

Question 83

Chemistry · Amines · Numerical

Number of amine compounds from the following giving solids which are soluble in NaOH upon reaction with Hinsberg's reagent is _______

Answer: 5

Solution

Primary amines give an ionic solid upon reaction with Hinsberg reagent which is soluble in $\mathrm{NaOH}$. The primary amine in the given options is the one with the $\mathrm{NH_2}$ group attached directly to the benzene ring.

Question 84

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical

The number of optical isomers in following compound is:

Answer: 32

Solution

Total chiral centre = 5 No. of optical isomers = $2^5$ = 32.

Question 85

Chemistry · The d-and f-Block Elements · Fill in the blank

The 'spin only' magnetic moment value of $\mathrm{MO}_4^{2-}$ is ______ BM. (Where M is a metal having least metallic radii. among Sc, Ti, V, Cr, Mn and Zn ). (Given atomic number: Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25 and Zn = 30)

Answer: 0

Solution

Metal having least metallic radii among Sc, Ti, V, Cr, Mn & Zn is Cr. Spin only magnetic moment of $\mathrm{CrO_4^{2-}}$. Here $\mathrm{Cr^{+6}}$ is in $d^0$ configuration (diamagnetic).

Question 86

Chemistry · Chemical Bonding and Molecular Structure · Numerical

Number of molecules from the following which are exceptions to octet rule is $CO_2, NO_2, H_2SO_4, BF_3, CH_4, SiF_4, ClO_2, PCl_5, BeF_2, C_2H_6, CHCl_3, CBr_4$

Answer: 6

Solution

Question 87

Chemistry · Amines · Numerical

If 279 g of aniline is reacted with one equivalent of benzenediazonium chloride, the maximum amount of aniline yellow formed will be ______ g. (nearest integer) (consider complete conversion).

Answer: 591

Solution

Question 88

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

Consider the following reaction $$A + B \rightarrow C$$ The time taken for A to become $1/4^{th}$ of its initial concentration is twice the time taken to become $1/2$ of the same. Also, when the change of concentration of $B$ is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is ______.

Answer: 1

Solution

For first order reaction, 75$\%$ life = 2 $\times$ 50$\%$ life. So order with respect to A will be first order. So order with respect to B will be zero. Overall order of reaction = 1 + 0 = 1.

Question 89

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

Major product $B$ of the following reaction has ______ $\pi$-bond.

Answer: 5

Solution

The major product B is formed by the oxidation of the ethyl group to a carboxylic acid group using $\mathrm{KMnO_4}$ and $\mathrm{KOH}$. The subsequent nitration with $\mathrm{HNO_3}$ and $\mathrm{H_2SO_4}$ introduces a nitro group. The total number of $\pi$ bonds in B are 5.

Question 90

Chemistry · Solutions · Numerical

A solution containing 10 g of an electrolyte $\mathrm{AB_2}$ in 100 g of water boils at $100.52^\circ \mathrm{C}$. The degree of ionization of the electrolyte $(\alpha)$ is _______ $\times 10^{-1}$. (nearest integer) [Given : Molar mass of $\mathrm{AB_2} = 200 \, \mathrm{g} \, \mathrm{mol}^{-1}$, $K_b$ (molal boiling point elevation const. of water) $= 0.52 \, \mathrm{K} \, \mathrm{kg} \, \mathrm{mol}^{-1}$, boiling point of water $= 100^\circ \mathrm{C}$; $\mathrm{AB_2}$ ionises as $\mathrm{AB_2} \rightarrow \mathrm{A^{2+}} + 2 \mathrm{B^-}$]

Answer: 5

Solution

The reaction is given by $$$ $\mathrm{AB_2}$ $\rightarrow$ $\mathrm{A^{+2}}$ + 2 $\mathrm{B^{\ominus}}$ $$$. The van't Hoff factor $$$ i $$$ is calculated as follows: $$$ i = 1 + (3 - 1) $\alpha$ $$$. Simplifying, we get $$$ i = 1 + 2$\alpha$ $$$. The boiling point elevation is given by $$$ $\Delta$ T_b = k_b i m $$$. Substituting the values, we have: $$0.52 = 0.52(1 + 2\alpha) \frac{\frac{10}{200}}{\frac{100}{1000}}$$ Simplifying further, $$1 = (1 + 2\alpha) \frac{10}{20}$$ Solving for $$$ $\alpha$ $$$, we get: $$2 = 1 + 2\alpha$$ Thus, $$$ $\alpha$ = 0.5 $$$. The answer is $$$ $\alpha$ = 5 $\times$ 10^{-1} $$$.