JEE Main 29 January 2024 Shift 1 question paper with solutions

JEE Main 29 January 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.

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Maths

Question 1

Maths · Sequences and Series · Single correct

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

  1. 7
  2. 4
  3. 5
  4. 6

Answer: (d)

Solution

Given the series $a + ar + ar^2 + ar^3 + \ldots + ar^{63}$. This can be expressed as $7 \left( a + ar^2 + ar^4 + \ldots + ar^{62} \right)$. Therefore, we have: $$\frac{a \left( 1 - r^{64} \right)}{1 - r} = \frac{7a \left( 1 - r^{64} \right)}{1 - r^2}$$ Solving for $r$, we find $r = 6$.

Question 2

Maths · Sequences and Series · Single correct

In an A.P., the sixth terms $a_6 = 2$. If the $a_1 a_4 a_5$ is the greatest, then the common difference of the A.P., is equal to

  1. $\frac{3}{2}$
  2. $\frac{8}{5}$
  3. $\frac{2}{3}$
  4. $\frac{5}{8}$

Answer: (b)

Solution

\[ a_6 = 2 \Rightarrow a + 5d = 2 \] \[ a_1a_4a_5 = a(a+3d)(a+4d) \] \[ = (2-5d)(2-2d)(2-d) \] \[ f(d)=8-32d+34d^2-20d+30d^2-10d^3 \] \[ f'(d)=-2(5d-8)(3d-2) \] \[ d=\frac{8}{5} \]

Question 3

Maths · Relations and Functions · Single correct

If $f(x) = \begin{cases} 2 + 2x, & -1 \leq x < 0 \\ 1 - \frac{x}{3}, & 0 \leq x \leq 3 \end{cases}; g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases},$ then range of $(f \circ g(x))$ is

  1. (0, 1]
  2. [0, 3)
  3. [0, 1]
  4. [0, 1)

Answer: (c)

Solution

Given $$f(g(x)) = \begin{cases} 2 + 2g(x), & -1 \leq g(x) < 0 .....(1) \\ 1 - \frac{g(x)}{3}, & 0 \leq g(x) \leq 3 .....(2) \end{cases}$$ By (1) $x \in \emptyset$ And by (2) $x \in [-3, 0]$ and $x \in [0, 1]$ Range of $f(g(x))$ is $[0, 1]$

Question 4

Maths · Probability · Single correct

A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is

  1. $\frac{5}{6}$
  2. $\frac{1}{6}$
  3. $\frac{5}{11}$
  4. $\frac{6}{11}$

Answer: (c)

Solution

Required probability = $$\frac{5}{6} \times \frac{1}{6} + \left(\frac{5}{6}\right)^3 \times \frac{1}{6} + \left(\frac{5}{6}\right)^5 \times \frac{1}{6} + \ldots$$ $$= \frac{1}{6} \times \frac{\frac{5}{6}}{1 - \frac{25}{36}} = \frac{5}{11}$$

Question 5

Maths · Complex Numbers and Quadratic Equations · Single correct

If $z = \frac{1}{2} - 2i$, is such that $|z + 1| = \alpha z + \beta (1 + i)$, $i = \sqrt{-1}$ and $\alpha, \beta \in \mathbb{R}$, then $\alpha + \beta$ is equal to

  1. -4
  2. 3
  3. 2
  4. -1

Answer: (b)

Solution

Given $z = \frac{1}{2} - 2i$. The equation $|z + 1| = \alpha z + \beta (1 + i)$ is given. Substituting $z = \frac{1}{2} - 2i$, we have: $$\left| \frac{3}{2} - 2i \right| = \frac{\alpha}{2} - 2\alpha i + \beta + \beta i$$ This can be rewritten as: $$\left| \frac{3}{2} - 2i \right| = \left( \frac{\alpha}{2} + \beta \right) + (\beta - 2\alpha)i$$ We have $\beta = 2\alpha$ and $\frac{\alpha}{2} + \beta = \sqrt{\frac{9}{4} + 4}$. Solving these, we find $\alpha + \beta = 3$.

Question 6

Maths · Limits and Derivatives · Single correct

$\lim\limits_{x\to\frac{\pi}{2}} \left( \dfrac{1}{\left(x-\frac{\pi}{2}\right)^2} \int_{x^3}^{\left(\frac{\pi}{2}\right)^3} \cos\!\left(\dfrac{1}{t^3}\right)\,dt \right)$ is equal to

  1. $\frac{3\pi}{8}$
  2. $\frac{3\pi^2}{4}$
  3. $\frac{3\pi^2}{8}$
  4. $\frac{3\pi}{4}$

Answer: (c)

Solution

Using L'hopital rule $$= \lim_{x \to \frac{\pi}{2}} \frac{0 - \cos x \times 3x^2}{2 \left( x - \frac{\pi}{2} \right)}$$ $$= \lim_{x \to \frac{\pi}{2}} \frac{\sin \left( x - \frac{\pi}{2} \right)}{2 \left( x - \frac{\pi}{2} \right)} \times \frac{3\pi^2}{4}$$ $$= \frac{3\pi^2}{8}$$

Question 7

Maths · Properties of Triangles · Single correct

In a $\triangle ABC$, suppose $y = x$ is the equation of the bisector of the angle $B$ and the equation of the side $AC$ is $2x - y = 2$. If $2AB = BC$ and the point $A$ and $B$ are respectively $(4, 6)$ and $(\alpha, \beta)$, then $\alpha + 2\beta$ is equal to

  1. 42
  2. 39
  3. 48
  4. 45

Answer: (a)

Solution

Given the ratio $AD : DC = 1 : 2$. We have the equation $$\frac{4 - \alpha}{6 - \alpha} = \frac{10}{8}.$$ Solving for $\alpha$, we find $\alpha = \beta$. Therefore, $\alpha = 14$ and $\beta = 14$.

Question 8

Maths · Vector Algebra · Single correct

Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear if $\vec{a} + 5\vec{b}$ is collinear with $\vec{c}$, $\vec{b} + 6\vec{c}$ is collinear with $\vec{a}$ and $\vec{a} + \alpha \vec{b} + \beta \vec{c} = \vec{0}$, then $\alpha + \beta$ is equal to

  1. 35
  2. 30
  3. -30
  4. -25

Answer: (a)

Solution

Given $\vec{a} + 5 \vec{b} = \lambda \vec{c}$ and $\vec{b} + 6 \vec{c} = \mu \vec{a}$. Eliminating $\vec{a}$, we have $$\lambda \vec{c} - 5 \vec{b} = \frac{6}{\mu} \vec{c} + \frac{1}{\mu} \vec{b}.$$ Therefore, $\mu = -\frac{1}{5}$ and $\lambda = -30$. Thus, $\alpha = 5$ and $\beta = 30$.

Question 9

Maths · Properties of Triangles · Single correct

Let $( 5, \frac{a}{4})$, be the circumcenter of a triangle with vertices A(a, -2), B(a, 6) and C $(\frac{a}{4}, -2)$. Let $\alpha$ denote the circumradius, $\beta$ denote the area and $\gamma$ denote the perimeter of the triangle. Then $\alpha$ + $\beta$ + $\gamma$ is

  1. 60
  2. 53
  3. 62
  4. 30

Answer: (b)

Solution

Given points $A(a, -2)$, $B(a, 6)$, $C\left(\frac{a}{4}, -2\right)$, $O\left(5, \frac{a}{4}\right)$. $AO = BO$ $$(a - 5)^2 + \left(\frac{a}{4} + 2\right)^2 = (a - 5)^2 + \left(\frac{a}{4} - 6\right)^2$$ Solving gives $a = 8$. Thus, $AB = 8$, $AC = 6$, $BC = 10$. The angles are $\alpha = 5$, $\beta = 24$, $\gamma = 24$.

Question 10

Maths · Integrals · Single correct

For $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, if $y(x) = \int \frac{\csc x + \sin x}{x \sec x + \tan x \sin^2 x} \, dx$ and $\lim_{x \to \left( \frac{\pi}{2} \right)^-} y(x) = 0$ then $y\left( \frac{\pi}{4} \right)$ is equal to

  1. $\tan^{-1} \left( \frac{1}{\sqrt{2}} \right)$
  2. $\frac{1}{2} \tan^{-1} \left( \frac{1}{\sqrt{2}} \right)$
  3. $-\frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{1}{\sqrt{2}} \right)$
  4. $\frac{1}{\sqrt{2}} \tan^{-1} \left( -\frac{1}{2} \right)$

Answer: (c)

Solution

Given $$y(x) = \int \frac{(1 + \sin^2 x) \cos x}{1 + \sin^4 x} \, dx$$ Put $\sin x = t$ $$= \int \frac{1 + t^2}{t^4 + 1} \, dt = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t - \frac{1}{t}}{\sqrt{2}} \right) + C$$ For $x = \frac{\pi}{2}, t = 1$ therefore $C = 0$ $$y\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{-1}{\sqrt{2}} \right)$$

Question 11

Maths · Trigonometric Functions · Single correct

If $\alpha$, $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$ is the solution of $4 \cos \theta + 5 \sin \theta = 1$, then the value of $\tan \alpha$ is

  1. $\frac{10 - \sqrt{10}}{6}$
  2. $\frac{10 - \sqrt{10}}{12}$
  3. $\frac{\sqrt{10} - 10}{12}$
  4. $\frac{\sqrt{10} - 10}{6}$

Answer: (c)

Solution

Given $4 + 5 \tan \theta = \sec \theta$. Squaring: $$24 \tan^2 \theta + 40 \tan \theta + 15 = 0$$ $$\tan \theta = \frac{-10 \pm \sqrt{10}}{12}$$ and $\tan \theta = -\left(\frac{10 + \sqrt{10}}{12}\right)$ is Rejected. (3) is correct.

Question 12

Maths · Differential Equations · Single correct

A function $y = f(x)$ satisfies $$f(x) \sin 2x + \sin x - (1 + \cos^2 x) f'(x) = 0$$ with condition $f(0) = 0$. Then $f\left(\frac{\pi}{2}\right)$ is equal to

  1. 1
  2. 0
  3. -1
  4. 2

Answer: (a)

Solution

Given $\dfrac{dy}{dx}-\left(\dfrac{\sin 2x}{1+\cos^2x}\right)y=\sin x$ I.F. $=1+\cos^2x$ $y(1+\cos^2x)=\int \sin x\,dx$ $=-\cos x+C$ $x=0,\ C=1$ $y\left(\dfrac{\pi}{2}\right)=1$

Question 13

Maths · Three Dimensional Geometry · Single correct

Let O be the origin and the position vector of A and B be $2\hat{i} + 2\hat{j} + \hat{k}$ and $2\hat{i} + 4\hat{j} + 4\hat{k}$ respectively. If the internal bisector of $\angle AOB$ meets the line $AB$ at $C$, then the length of $OC$ is

  1. $\frac{2}{3}\sqrt{31}$
  2. $\frac{2}{3}\sqrt{34}$
  3. $\frac{3}{4}\sqrt{34}$
  4. $\frac{3}{2}\sqrt{31}$

Answer: (b)

Solution

The length of $OC$ is calculated as follows: $$\frac{\sqrt{136}}{3} = \frac{2\sqrt{34}}{3}.$$

Question 14

Maths · Relations and Functions · Single correct

Consider the function $f : \left[ \frac{1}{2}, 1 \right] \to \mathbb{R}$ defined by $f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$. Consider the statements (I) The curve $y = f(x)$ intersects the $x$-axis exactly at one point (II) The curve $y = f(x)$ intersects the $x$-axis at $x = \cos \frac{\pi}{12}$ Then

  1. Only (II) is correct
  2. Both (I) and (II) are incorrect
  3. Only (I) is correct
  4. Both (I) and (II) are correct

Answer: (d)

Solution

Given $f'(x) = 12 \sqrt{2} x^2 - 3 \sqrt{2} \geq 0$ for $\left[ \frac{1}{2}, 1 \right]$. $f\left( \frac{1}{2} \right) 0 \Rightarrow (A)$ is correct. $f(x) = \sqrt{2} \left( 4x^3 - 3x \right) - 1 = 0$. Let $\cos \alpha = x$, $\cos 3\alpha = \cos \frac{\pi}{4} \Rightarrow \alpha = \frac{\pi}{12}$. $x = \cos \frac{\pi}{12}$. $(4)$ is correct.

Question 15

Maths · Matrices · Single correct

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$ and $|2A|^3 = 2^{21}$ where $\alpha, \beta \in \mathbb{Z}$, Then a value of $\alpha$ is

  1. 3
  2. 5
  3. 17
  4. 9

Answer: (b)

Solution

$|A|=\alpha^2-\beta^2$ $|2A|^3=2^{6^1}\Rightarrow |A|=2^4$ $\alpha^2-\beta^2=16$ $(\alpha+\beta)(\alpha-\beta)=16$ $\Rightarrow \alpha=4\ \text{or}\ 5$

Question 16

Maths · Three Dimensional Geometry · Single correct

Let $PQR$ be a triangle with $R(-1, 4, 2)$. Suppose $M(2, 1, 2)$ is the mid point of $PQ$. The distance of the centroid of $\triangle PQR$ from the point of intersection of the line $\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$ and $\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$ is

  1. 69
  2. 9
  3. $\sqrt{69}$
  4. $\sqrt{99}$

Answer: (c)

Solution

Centroid G divides MR in 1 : 2. G(1, 2, 2). Point of intersection A of given lines is (2, -6, 0). AG = $\sqrt{69}$.

Question 17

Maths · Relations and Functions · Single correct

Let R be a relation on $\mathbb{Z} \times \mathbb{Z}$ defined by $(a, b)R(c, d)$ if and only if $ad - bc$ is divisible by 5. Then R is

  1. Reflexive and symmetric but not transitive
  2. Reflexive but neither symmetric not transitive
  3. Reflexive, symmetric and transitive
  4. Reflexive and transitive but not symmetric

Answer: (a)

Solution

Question 18

Maths · Integrals · Single correct

If the value of the integral $$\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \left( \frac{x^2 \cos x}{1+\pi^x} + \frac{1+\sin^2 x}{1+e^{\sin x^{2033}}} \right) dx = \frac{\pi}{4} (\pi + a) - 2,$$ then the value of $a$ is

  1. 3
  2. $-\frac{3}{2}$
  3. 2
  4. $\frac{3}{2}$

Answer: (a)

Solution

Given $$I = \int_{-\pi/2}^{\pi/2} \left( \frac{x^2 \cos x}{1 + \pi^x} + \frac{1 + \sin^2 x}{1 + e^{\sin x^{2023}}} \right) \, dx$$ $$I = \int_{-\pi/2}^{\pi/2} \left( \frac{x^2 \cos x}{1 + \pi^{-x}} + \frac{1 + \sin^2 x}{1 + e^{\sin(-x)^{2023}}} \right) \, dx$$ On adding, we get $$2I = \int_{-\pi/2}^{\pi/2} \left( x^2 \cos x + 1 + \sin^2 x \right) \, dx$$ On solving $$I = \frac{\pi^2}{4} + \frac{3\pi}{4} - 2$$ a = 3

Question 19

Maths · Continuity and Differentiability · Single correct

Suppose $$f(x) = \frac{(2^x + 2^{-x}) \tan x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3}$$ Then the value of $f'(0)$ is equal to

  1. $\pi$
  2. $0$
  3. $\sqrt{\pi}$
  4. $\frac{\pi}{2}$

Answer: (c)

Solution

Given $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$. $$= \lim_{h \to 0} \frac{\left(2^h + 2^{-h}\right) \tan h \sqrt{\tan^{-1}(h^2 - h + 1)} - 0}{(7h^2 + 3h + 1)^3 h}$$ $$= \sqrt{\pi}$$

Question 20

Maths · Matrices · Single correct

Let A be a square matrix such that $AA^T = I$. Then $\frac{1}{2} A \left[ \left( A + A^T \right)^2 + \left( A - A^T \right)^2 \right]$ is equal to

  1. $A^2 + I$
  2. $A^3 + I$
  3. $A^2 + A^T$
  4. $A^3 + A^T$

Answer: (d)

Solution

Given $AA^\top = I = A^\top A$. On solving the given expression, we get $$\frac{1}{2} A \left[ A^2 + \left(A^\top\right)^2 + 2 \, A \, A^\top + A^2 + \left(A^\top\right)^2 - 2 \, A \, A \, A^\top \right]$$ $$= A \left[ A^2 + \left(A^\top\right)^2 \right] = A^3 + A^\top$$

Question 21

Maths · Conic Sections · Numerical

Equation of two diameters of a circle are $2x - 3y = 5$ and $3x - 4y = 7$. The line joining the points $\left(-\frac{22}{7}, -4\right)$ and $\left(-\frac{1}{7}, 3\right)$ intersects the circle at only one point $P(\alpha, \beta)$. Then $17\beta - \alpha$ is equal to

Answer: 2

Solution

Centre of circle is (1, -1). Equation of AB is $7x - 3y + 10 = 0$ (i). Equation of CP is $3x + 7y + 4 = 0$ (ii). Solving (i) and (ii): $$\alpha = \frac{-41}{29}, \beta = \frac{1}{29} \therefore 17\beta - \alpha = 2$$

Question 22

Maths · Permutations and Combinations · Numerical

All the letters of the word "GTWENTY" are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word "GTWENTY" IS

Answer: 553

Solution

Words starting with E = 360 Words starting with GE = 60 Words starting with GN = 60 Words starting with GTE = 24 Words starting with GTN = 24 Words starting with GTT = 24 GTWENTY = 1 Total = 553

Question 23

Maths · Complex Numbers and Quadratic Equations · Numerical

Let $\alpha, \beta$ be the roots of the equation $x^2 - x + 2 = 0$ with $\mathrm{Im}(\alpha) > \mathrm{Im}(\beta)$. Then $\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$ is equal to

Answer: 13

Solution

Given $\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$. $$= \alpha^4(\alpha - 2) + \alpha^4 - 5\alpha^2 + (\beta - 2)^2$$ $$= \alpha^5 - \alpha^4 - 5\alpha^2 + \beta^2 - 4\beta + 4$$ $$= \alpha^3(\alpha - 2) - \alpha^4 - 5\alpha^2 + \beta - 2 - 4\beta + 4$$ $$= -2\alpha^3 - 5\alpha^2 - 3\beta + 2$$ $$= -2\alpha(\alpha - 2) - 5\alpha^2 - 3\beta + 2$$ $$= -7\alpha^2 + 4\alpha - 3\beta + 2$$ $$= -7(\alpha - 2) + 4\alpha - 3\beta + 2$$ $$= -3\alpha - 3\beta + 16 = -3(1) + 16 = 13$$

Question 24

Maths · Applications of Derivatives · Numerical

Let $f(x) = 2^x - x^2$, $x \in \mathbb{R}$. If $m$ and $n$ are respectively the number of points at which the curves $y = f(x)$ and $y = f'(x)$ intersects the $x$-axis, then the value of $m + n$ is

Answer: 5

Solution

Given $m = 3$. The derivative $f'(x) = 2^x \ln 2 - 2x = 0$. Solving gives $2^x \ln 2 = 2x$. Therefore, $n = 2$. Thus, $m + n = 5$.

Question 25

Maths · Conic Sections · Numerical

If the points of intersection of two distinct conics $x^2 + y^2 = 4b$ and $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ lie on the curve $y^2 = 3x^2$, then $3\sqrt{3}$ times the area of the rectangle formed by the intersection points is

Answer: 432

Solution

Putting $y^2 = 3x^2$ in both the conics. We get $x^2 = b$ and $\frac{b}{16} + \frac{3}{b} = 1$. Therefore, $b = 4, 12$ ( $b = 4$ is rejected because curves coincide). Thus, $b = 12$. Hence points of intersection are $(\pm \sqrt{12}, \pm 6)$ which implies area of rectangle $= 432$.

Question 26

Maths · Differential Equations · Numerical

If the solution curve $y = y(x)$ of the differential equation $(1 + y^2)\left(1 + \log_e x\right) dx + x dy = 0, x > 0$ passes through the point $(1, 1)$ and $y(e) = \frac{\alpha - \tan\left(\frac{3}{2}\right)}{\beta + \tan\left(\frac{3}{2}\right)}$, then $\alpha + 2\beta$ is

Answer: 3

Solution

Given $\($ $\int$ $\left$( $\frac{1}{x}$ + $\frac{\ln x}{x}$ $\right$) dx + $\int$ $\frac{dy}{1+y^2}$ = 0 $\)$ $\($ $\ln$ x + $\frac{(\ln x)^2}{2}$ + $\tan$^{-1} y = C $\)$ Put $\($ x = y = 1 $\)$ $\($ $\therefore$ C = $\frac{\pi}{4}$ $\)$ $\($ $\Rightarrow$ $\ln$ x + $\frac{(\ln x)^2}{2}$ + $\tan$^{-1} y = $\frac{\pi}{4}$ $\)$ Put $\($ x = e $\)$ $\($ $\Rightarrow$ y = $\tan$ $\left$( $\frac{\pi}{4}$ - $\frac{3}{2}$ $\right$) = $\frac{1 - \tan \frac{3}{2}}{1 + \tan \frac{3}{2}}$ $\)$ $\($ $\therefore$ $\alpha$ = 1, $\beta$ = 1 $\)$ $\($ $\Rightarrow$ $\alpha$ + 2$\beta$ = 3 $\)$

Question 27

Maths · Statistics · Numerical

If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $\alpha$, $\beta$, 60 where $\alpha > \beta$ are 56 and 66.2 respectively, then $\alpha^2 + \beta^2$ is equal to

Answer: 6344

Solution

Given $\bar{x} = 56$ and $\sigma^2 = 66.2$. Therefore, $$\frac{\alpha^2 + \beta^2 + 25678}{10} - (56)^2 = 66.2$$ Thus, $\alpha^2 + \beta^2 = 6344$.

Question 28

Maths · Applications of Integrals · Numerical

The area (in sq. units) of the part of circle $x^2 + y^2 = 169$ which is below the line $5x - y = 13$ is $$\frac{\pi \alpha}{2 \beta} - \frac{65}{2} + \frac{\alpha}{\beta} \sin^{-1} \left( \frac{12}{13} \right)$$ where $\alpha, \beta$ are coprime numbers. Then $\alpha + \beta$ is equal to

Answer: 171

Solution

Area = $\int$_{-13}^{12} $\sqrt{169 - y^2}$ $\,$ dy - $\frac{1}{2}$ $\times$ 25 $\times$ 5 = $\frac{\pi}{2}$ $\times$ $\frac{169}{2}$ - $\frac{65}{2}$ + $\frac{169}{2}$ $\sin$^{-1} $\frac{12}{13}$ $\therefore$ $\alpha$ + $\beta$ = 171

Question 29

Maths · Binomial Theorem · Numerical

If $\frac{{^{11}C_1}}{2} + \frac{{^{11}C_2}}{3} + \ldots + \frac{{^{11}C_9}}{10} = \frac{n}{m}$ with $\gcd(n, m) = 1$, then $n + m$ is equal to

Answer: 2041

Solution

Given $$\sum_{r=1}^{9} \frac{{^{11}C_r}}{{r+1}}$$ This is equal to $$\frac{1}{12} \sum_{r=1}^{9} {^{12}C_{r+1}}$$ Which simplifies to $$\frac{1}{12} \left[ 2^{12} - 2^6 \right] = \frac{2035}{6}$$ Therefore, $m + n = 2041$.

Question 30

Maths · Three Dimensional Geometry · Numerical

A line with direction ratios 2, 1, 2 meets the lines $x = y + 2 = z$ and $x + 2 = 2y = 2z$ respectively at the point $P$ and $Q$. if the length of the perpendicular from the point $(1, 2, 12)$ to the line $PQ$ is $l$, then $l^2$ is

Answer: 65

Solution

Let $\mathrm{P}(t, t-2, t)$ and $\mathrm{Q}(2s-2, s, s)$. D.R's of $\mathrm{PQ}$ are $2, 1, 2$. $$\frac{2s-2-t}{2} = \frac{s-t+2}{1} = \frac{s-t}{2}$$ implies $t = 6$ and $s = 2$. Therefore, $\mathrm{P}(6, 4, 6)$ and $\mathrm{Q}(2, 2, 2)$. $$\mathrm{PQ} : \frac{x-2}{2} = \frac{y-2}{1} = \frac{z-2}{2} = \lambda$$ Let $\mathrm{F}(2\lambda + 2, \lambda + 2, 2\lambda + 2)$. $\mathrm{A}(1, 2, 12)$. $$\overrightarrow{\mathrm{AF}} \cdot \overrightarrow{\mathrm{PQ}} = 0$$ Therefore, $\lambda = 2$. So $\mathrm{F}(6, 4, 6)$ and $\mathrm{AF} = \sqrt{65}$.

Physics

Question 31

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

In the given circuit, the breakdown voltage of the Zener diode is 3.0 $\mathrm{V}$. What is the value of $I_2$?

  1. 3.3 $\mathrm{mA}$
  2. 5.5 $\mathrm{mA}$
  3. 10 $\mathrm{mA}$
  4. 7 $\mathrm{mA}$

Answer: (b)

Solution

Given $V_Z = 3 \, \mathrm{V}$. Let potential at $B = 0 \, \mathrm{V}$. Potential at $E (V_E) = 10 \, \mathrm{V}$. $V_C = V_A = 3 \, \mathrm{V}$. $I_Z + I_1 = I$. $$I = \frac{10 - 3}{1000} = \frac{7}{1000} \, \mathrm{A}$$ $$I_1 = \frac{3}{2000} \, \mathrm{A}$$ Therefore $I_Z = \frac{7 - 1.5}{1000} = 5.5 \, \mathrm{mA}$.

Question 32

Physics · Current Electricity · Single correct

The electric current through a wire varies with time as $I = I_0 + \beta \, t$. where $I_0 = 20 \, \mathrm{A}$ and $\beta = 3 \, \mathrm{A/s}$. The amount of electric charge crossed through a section of the wire in 20 s is :

  1. 80C
  2. 1000C
  3. 800C
  4. 1600C

Answer: (b)

Solution

Given that Current $I = I_0 + \beta t$ $I_0 = 20 \, \mathrm{A}$ $\beta = 3 \, \mathrm{A/s}$ $I = 20 + 3t$ $$\frac{dq}{dt} = 20 + 3t$$ $$\int_0^q dq = \int_0^{20} (20 + 3t) \, dt$$ $$q = \int_0^{20} 20 \, dt + \int_0^{20} 3t \, dt$$ $$q = \left[ 20t + \frac{3t^2}{2} \right]_0^{20} = 1000 \, \mathrm{C}$$

Question 33

Physics · Mechanical Properties of Fluids · Single correct

Given below are two statements: Statement I : If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water. Statement II : If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in cold water. In the light of the above statements, choose the most appropriate from the options given below

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is false but Statement II is true

Answer: (c)

Solution

Surface tension will be less as temperature increases $$h = \frac{2T \cos \theta}{\rho gr}$$ Height of capillary rise will be smaller in hot water and larger in cold water.

Question 34

Physics · Ray Optics and Optical Instruments · Single correct

A convex mirror of radius of curvature 30 cm forms an image that is half the size of the object. The object distance is :

  1. -15 $\mathrm{cm}$
  2. 45 $\mathrm{cm}$
  3. -45 $\mathrm{cm}$
  4. 15 $\mathrm{cm}$

Answer: (a)

Solution

Given $R = 30 \, \mathrm{cm}$. $f = R/2 = +15 \, \mathrm{cm}$. Magnification $(m) = \pm \frac{1}{2}$. For convex mirror, virtual image is formed for real object. Therefore, $m$ is $+\mathrm{ve}$. $$\frac{1}{2} = \frac{f}{f - u}$$ $$u = -15 \, \mathrm{cm}$$

Question 35

Physics · Electric Charges and Fields · Single correct

Two charges of $5Q$ and $-2Q$ are situated at the points $(3a, 0)$ and $(-5a, 0)$ respectively. The electric flux through a sphere of radius '4a' having center at origin is:

  1. $\frac{2Q}{\varepsilon_0}$
  2. $\frac{5Q}{\varepsilon_0}$
  3. $\frac{7Q}{\varepsilon_0}$
  4. $\frac{3Q}{\varepsilon_0}$

Answer: (b)

Solution

The $5Q$ charge is inside the spherical region. The flux through the sphere is given by $$\frac{5Q}{\varepsilon_0}.$$

Question 36

Physics · Motion in a Straight Line · Single correct

A body starts moving from rest with constant acceleration covers displacement $S_1$ in first $(p - 1)$ seconds and $S_2$ in first $p$ seconds. The displacement $S_1 + S_2$ will be made in time:

  1. $(2p + 1)\,\mathrm{s}$
  2. $\sqrt{(2p^2 - 2p + 1)}\,\mathrm{s}$
  3. $(2p - 1)\,\mathrm{s}$
  4. $(2p^2 - 2p + 1)\,\mathrm{s}$

Answer: (b)

Solution

Given $S_1$ in first $(p-1)$ sec and $S_2$ in first $p$ sec. $S_1 = \frac{1}{2} a (p-1)^2$ $S_2 = \frac{1}{2} a (p)^2$ $S_1 + S_2 = \frac{1}{2} a t^2$ $(p-1)^2 + p^2 = t^2$ $t = \sqrt{2p^2 + 1 - 2p}$

Question 37

Physics · Work, Energy and Power · Single correct

The potential energy function (in $J$) of a particle in a region of space is given as $U = (2x^2 + 3y^3 + 2z)$. Here $x$, $y$ and $z$ are in meter. The magnitude of $x$ - component of force (in $N$) acting on the particle at point $P(1, 2, 3)$ m is :

  1. 2
  2. 6
  3. 4
  4. 8

Answer: (c)

Solution

Given $U = 2x^2 + 3y^3 + 2z$ $$F_x = -\frac{\partial U}{\partial x} = -4x$$ At $x = 1$ magnitude of $F_x$ is $4 \mathrm{N}$

Question 38

Physics · Mathematics in Physics · Single correct

The resistance $R = \frac{V}{I}$ where $V = (200 \pm 5) \, \mathrm{V}$ and $I = (20 \pm 0.2) \, \mathrm{A}$, the percentage error in the measurement of $R$ is:

  1. 3.5%
  2. 7%
  3. 3%
  4. 5.5%

Answer: (a)

Solution

Given $R = \frac{V}{I}$. According to error analysis, $$\frac{dR}{R} = \frac{dV}{V} + \frac{dI}{I}$$ $$\frac{dR}{R} = \frac{5}{200} + \frac{0.2}{20}$$ $$\frac{dR}{R} = \frac{7}{200}$$ The percentage error is $$\frac{dR}{R} \times 100 = \frac{7}{200} \times 100 = 3.5\%$$

Question 39

Physics · Work, Energy and Power · Single correct

A block of mass $100 \, \mathrm{kg}$ slides over a distance of $10 \, \mathrm{m}$ on a horizontal surface. If the co-efficient of friction between the surfaces is $0.4$, then the work done against friction (in J) is:

  1. 4200
  2. 3900
  3. 4000
  4. 4500

Answer: (c)

Solution

Given $m = 100 \, \mathrm{kg}$, $s = 10 \, \mathrm{m}$, $\mu = 0.4$. As $f = \mu mg = 0.4 \times 100 \times 10 = 400 \, \mathrm{N}$. Now $W = f \cdot s = 400 \times 10 = 4000 \, \mathrm{J}$.

Question 40

Physics · Electromagnetic Waves · Single correct

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

  1. A-IV, B-I, C-III, D-II
  2. A-II, B-III, C-I, D-IV
  3. A-IV, B-III, C-I, D-II
  4. A-I, B-II, C-III, D-IV

Answer: (c)

Solution

Ampere-Maxwell law $$\oint \vec{B} \cdot \vec{dl} = \mu_0 i_c + \mu_0 \varepsilon_0 \frac{d \phi_E}{dt}$$ Faraday law $$\oint \vec{E} \cdot \vec{d} = -\frac{d \phi_B}{dt}$$ Gauss' law for electricity $$\oint \vec{E} \cdot \vec{dA} = \frac{Q}{\varepsilon_0}$$ Gauss' law for magnetism $$\oint \vec{B} \cdot \vec{dA} = 0$$

Question 41

Physics · Motion in a Plane · Single correct

If the radius of curvature of the path of two particles of same mass are in the ratio 3:4, then in order to have constant centripetal force, their velocities will be in the ratio of:

  1. $\sqrt{3} : 2$
  2. $1 : \sqrt{3}$
  3. $\sqrt{3} : 1$
  4. $2 : \sqrt{3}$

Answer: (a)

Solution

Given $m_1 = m_2$ and $\frac{r_1}{r_2} = \frac{3}{4}$. As centripetal force $F = \frac{mv^2}{r}$. In order to have constant (same in this question) centripetal force $F_1 = F_2$. $$\frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2}$$ $$\Rightarrow \frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}} = \frac{\sqrt{3}}{2}$$

Question 42

Physics · Current Electricity · Single correct

A galvanometer having coil resistance $10 \, \Omega$ shows a full scale deflection for a current of $3 \, \mathrm{mA}$. For it to measure a current of $8 \, \mathrm{A}$, the value of the shunt should be:

  1. $3 \times 10^{-3} \, \Omega$
  2. $4.85 \times 10^{-3} \, \Omega$
  3. $3.75 \times 10^{-3} \, \Omega$
  4. $2.75 \times 10^{-3} \, \Omega$

Answer: (c)

Solution

Given $G = 10 \, \Omega$. $I_g = 3 \, \mathrm{mA}$. $I = 8 \, \mathrm{A}$. In case of conversion of galvanometer into ammeter. We have $I_g G = (I - I_g) S$. $$S = \frac{I_g G}{I - I_g}$$ $$S = \frac{(3 \times 10^{-3}) \times 10}{8 - 0.003} = 3.75 \times 10^{-3} \, \Omega$$

Question 43

Physics · Work, Energy and Power · Single correct

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 25$\%$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

  1. $\frac{1}{1}$
  2. $\frac{1}{8}$
  3. $\frac{8}{1}$
  4. $\frac{1}{4}$

Answer: (b)

Solution

For photon $$E_p = \frac{hc}{\lambda_p} \Rightarrow \lambda_p = \frac{hc}{E_p}$$ For electron $$\lambda_e = \frac{h}{m_e v_e} = \frac{h v_e}{2 K_e}$$ Given $v_c = 0.25c$ $$\lambda_e = \frac{h \times 0.25c}{2 K_e} = \frac{hc}{8 K_e}$$ Also $\lambda_p = \lambda_e$ $$\frac{hc}{E_p} = \frac{hc}{8 K_e}$$ $$\frac{K_e}{E_p} = \frac{1}{8}$$

Question 44

Physics · Current Electricity · Single correct

The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of 24\,$\Omega$ is applied. The resistance of galvanometer coil will be :

  1. 12\,$\Omega$
  2. 96\,$\Omega$
  3. 48\,$\Omega$
  4. 100\,$\Omega$

Answer: (b)

Solution

Let $x = current/division$. Initially, $I_g = 25x$. After applying shunt: $I_g = 5x$ and $I - I_g = 20x$ with a $24 \, \Omega$ resistor. Now $5x \times G = 20x \times 24$. Solving for $G$: $$G = 4 \times 24$$ $$G = 96 \, \Omega$$

Question 45

Physics · Ray Optics and Optical Instruments · Single correct

A biconvex lens of refractive index 1.5 has a focal length of 20 $\mathrm{\ cm}$ in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:

  1. -16 $\mathrm{\ cm}$
  2. -160 $\mathrm{\ cm}$
  3. +160 $\mathrm{\ cm}$
  4. +16 $\mathrm{\ cm}$

Answer: (b)

Solution

Given $\mu_1 = 1.5$, $\mu_m = 1.6$, $f_a = 20 \, \mathrm{cm}$. As $$\frac{f_m}{f_a} = \frac{(\mu_1 - 1) \mu_m}{(\mu_1 - \mu_m)}$$ $$\frac{f_m}{20} = \frac{(1.5 - 1) 1.6}{(1.5 - 1.6)}$$ $$f_m = -160 \, \mathrm{cm}$$

Question 46

Physics · Thermodynamics · Single correct

A thermodynamic system is taken from an original state $A$ to an intermediate state $B$ by a linear process as shown in the figure. It's volume is then reduced to the original value from $B$ to $C$ by an isobaric process. The total work done by the gas from $A$ to $B$ and $B$ to $C$ would be :

  1. 33800 J
  2. 2200 J
  3. 600 J
  4. 800 J

Answer: (d)

Solution

Work done AB = $\frac{1}{2}$ (8000 + 6000) $\mathrm{Dyne/cm^2}$ $\times$ $$4 \, \mathrm{m^3} = (6000 \, \mathrm{Dyne/cm^2}) \times 4 \, \mathrm{m^3}$$ Work done BC = - (4000 $\mathrm{Dyne/cm^2}$) $\times$ 4 $\mathrm{m^3}$ Total work done = 2000 $\mathrm{Dyne/cm^2}$ $\times$ 4 $\mathrm{m^3}$ $$= 2 \times 10^3 \times \frac{1}{10^5} \frac{\mathrm{N}}{\mathrm{cm^2}} \times 4 \, \mathrm{m^3}$$ $$= 2 \times 10^{-2} \times \frac{\mathrm{N}}{10^{-4} \, \mathrm{m^2}} \times 4 \, \mathrm{m^3}$$ $$= 2 \times 10^2 \times 4 \, \mathrm{Nm} = 800 \, \mathrm{J}$$

Question 47

Physics · Gravitation · Single correct

At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as R.)

  1. $\sqrt{5}R - R$
  2. $\frac{\sqrt{3}R - R}{2}$
  3. $\frac{R}{2}$
  4. $\frac{\sqrt{5}R - R}{2}$

Answer: (d)

Solution

Given $\($ g_p = $\frac{gR^2}{(R+h)^2}$ $\)$ and $\($ g_q = g $\left$( 1 - $\frac{h}{R}$ $\right$) $\)$. Equating $\($ g_p = g_q $\)$, we have: $$ \frac{g}{\left( 1 + \frac{h}{R} \right)^2} = g \left( 1 - \frac{h}{R} \right) $$ $$ \left( 1 - \frac{h^2}{R^2} \right) \left( 1 + \frac{h}{R} \right) = 1 $$ Take $\($ $\frac{h}{R}$ = x $\)$. So, $$ x^3 - x + x^2 = 0 $$ $$ x = \frac{\sqrt{5} - 1}{2} $$ $$ h = \frac{R}{2} (\sqrt{5} - 1) $$

Question 48

Physics · Alternating Current · Single correct

A capacitor of capacitance $100\mu F$ is charged to a potential of $12 \, \mathrm{V}$ and connected to a $6.4 \, \mathrm{mH}$ inductor to produce oscillations. The maximum current in the circuit would be:

  1. 3.2 $\mathrm{A}$
  2. 1.5 $\mathrm{A}$
  3. 2.0 $\mathrm{A}$
  4. 1.2 $\mathrm{A}$

Answer: (b)

Solution

By energy conservation $$\frac{1}{2} CV^2 = \frac{1}{2} LI_{max}^2$$ $$I_{max} = \sqrt{\frac{C}{L}} V$$ $$= \sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12$$ $$= \frac{12}{8} = \frac{3}{2} = 1.5 \, \mathrm{A}$$

Question 49

Physics · Nuclei · Single correct

The explosive in a Hydrogen bomb is a mixture of $_1H^2$, $_1H^3$ and $_3Li^6$ in some condensed form. The chain reaction is given by $$_5Li^6 + _0n^1 \rightarrow _2He^4 + _1H^3$$ $$_1H^2 + _1H^3 \rightarrow _2He^4 + _0n^1$$ During the explosion the energy released is approximately [Given : M(Li) = 6.01690 amu. M($_1H^2$) = 2.01471 amu. M($_2He^4$) = 4.00388 amu, and 1 amu = 931.5 MeV]

  1. 28.12 MeV
  2. 12.64 MeV
  3. 16.48 MeV
  4. 22.22 MeV

Answer: (d)

Solution

The reactions are as follows: $$^3\mathrm{Li}^6 + ^0\mathrm{n}^1 \rightarrow ^2\mathrm{He}^4 + ^1\mathrm{H}^3$$ $$^2\mathrm{H}^2 + ^1\mathrm{H}^3 \rightarrow ^2\mathrm{He}^4 + ^0\mathrm{n}^1$$ Combining these, we get: $$^3\mathrm{Li}^6 + ^1\mathrm{H}^2 \rightarrow 2 \left( ^2\mathrm{He}^4 \right)$$ Energy released in process: $$Q = \Delta mc^2$$ $$Q = \left[ \mathrm{M(Li)} + \mathrm{M} \left( ^2\mathrm{H}^2 \right) - 2 \times \mathrm{M} \left( ^2\mathrm{He}^4 \right) \right] \times 931.5 \mathrm{MeV}$$ $$Q = \left[ 6.01690 + 2.01471 - 2 \times 4.00388 \right] \times 931.5 \mathrm{MeV}$$ $$Q = 22.216 \mathrm{MeV}$$ $$Q = 22.22 \mathrm{MeV}$$

Question 50

Physics · Kinetic Theory · Single correct

Two vessels A and B are of the same size and are at same temperature. A contains 1 g of hydrogen and B contains 1 g of oxygen. $P_A$ and $P_B$ are the pressures of the gases in A and B respectively, then $\frac{P_A}{P_B}$ is :

  1. 16
  2. 8
  3. 4
  4. 32

Answer: (a)

Solution

Given $V_A = V_B$ and $T_A = T_B$, we have $$\frac{P_A V_A}{P_B V_B} = \frac{n_A R T_A}{n_B R T_B}.$$ This simplifies to $$\frac{P_A}{P_B} = \frac{n_A}{n_B}.$$ Substituting the given values, $$\frac{P_A}{P_B} = \frac{1/2}{1/32} = 16.$$

Question 51

Physics · Atoms · Numerical

When a hydrogen atom going from $n = 2$ to $n = 1$ emits a photon, its recoil speed is $\frac{x}{5}$ m/s. Where $x =$ _______. (Use : mass of hydrogen atom $= 1.6 \times 10^{-27} \, \mathrm{kg}$)

Answer: 17

Solution

Given $\Delta E = 10.2 \, \mathrm{eV}$. Recoil speed $(v)$ is given by $$v = \frac{\Delta E}{mc}$$ Substituting the values, $$v = \frac{10.2 \, \mathrm{eV}}{1.6 \times 10^{-27} \times 3 \times 10^8}$$ $$= \frac{10.2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27} \times 3 \times 10^8}$$ $$v = 3.4 \, \mathrm{m/s} = \frac{17}{5} \, \mathrm{m/s}$$ Therefore, $x = 17$.

Question 52

Physics · Motion in a Plane · Numerical

A ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1 \, \mathrm{m}$ high and $0.1 \, \mathrm{m}$ wide. The minimum velocity $u$ with which that ball just hits the step 5 of the stairway will be $\sqrt{x} \, \mathrm{ms}^{-1}$ where x = _______ [use $g = 10 \, \mathrm{m/s}^2$].

Answer: 2

Solution

The ball needs to just cross 4 steps to just hit the 5th step. Therefore, horizontal range (R) = 0.4 m. $$R = u \cdot t$$ Similarly, in the vertical direction, $$h = \frac{1}{2} g t^2$$ $$0.4 = \frac{1}{2} g t^2$$ $$0.4 = \frac{1}{2} g \left( \frac{0.4}{u} \right)^2$$ $$u^2 = 2$$ $$u = \sqrt{2} \, \mathrm{m/s}$$ Therefore, $$x = 2$$

Question 53

Physics · Electromagnetic Induction · Numerical

A square loop of side 10 cm and resistance 0.7 $\Omega$ is placed vertically in east-west plane. A uniform magnetic field of 0.20 T is set up across the plane in north east direction. The magnetic field is decreased to zero in 1 s at a steady rate. Then, magnitude of induced emf is $\sqrt{x} \times 10^{-3}$ V. The value of $x$ is _____

Answer: 2

Solution

Given $\vec{A} = (0.1)^{2} \hat{j}$ and $\vec{B} = \frac{0.2}{\sqrt{2}} \hat{i} + \frac{0.2}{\sqrt{2}} \hat{j}$. The magnitude of induced emf is given by $$c = \frac{\Delta \phi}{\Delta t} = \frac{\vec{B} \cdot \vec{A} - 0}{1} = \sqrt{2} \times 10^{-3} \, \mathrm{V}$$

Question 54

Physics · System of Particles and Rotational Motion · Numerical

A cylinder is rolling down on an inclined plane of inclination $60^{\circ}$. Its acceleration during rolling down will be $\left( \frac{x}{\sqrt{3}} \right)$ m/s$^2$, where $x$ = _____ (use $g = 10$ m/s$^2$).

Answer: 10

Solution

For rolling motion, $a = \frac{g \sin \theta}{1 + \frac{I_{cm}}{MR^2}}$ $$a = \frac{g \sin \theta}{1 + \frac{1}{2}}$$ $$= \frac{2 \times 10 \times \frac{\sqrt{3}}{2}}{3}$$ $$= \frac{10}{\sqrt{3}}$$ Therefore $x = 10$

Question 55

Physics · Magnetism and Matter · Numerical

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm from its center is $1.5 \times 10^{-5} \, \mathrm{Tm}$.The magnetic moment of the dipole is $\mathrm{A\,m}^2$. (Given: $\frac{\mu_0}{4\pi} = 10^{-7}\,\mathrm{T\,m\,A}^{-1}$)

Answer: 6

Solution

Given $V = \frac{\mu_0}{4\pi} \frac{M}{r^2}$. Therefore, $$1.5 \times 10^{-5} = 10^{-7} \times \frac{M}{(20 \times 10^{-2})^2}$$ This implies $$M = \frac{1.5 \times 10^{-5} \times 20 \times 20 \times 10^{-4}}{10^{-7}}$$ Thus, $$M = 1.5 \times 4 = 6$$

Question 56

Physics · Wave Optics · Numerical

In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at $P$. If $D = 0.2 \, \mathrm{m}$, the minimum distance between the slits $S_1$ and $S_2$ is ______ mm.

Answer: 0.2

Solution

Path difference for minima at $P$ $$2 \sqrt{D^2 + d^2} - 2D = \frac{\lambda}{2}$$ Therefore, $$\sqrt{D^2 + d^2} - D = \frac{\lambda}{4}$$ Thus, $$\sqrt{D^2 + d^2} = \frac{\lambda}{4} + D$$ Therefore, $$D^2 + d^2 = D^2 + \frac{\lambda^2}{16} + \frac{D \lambda}{2}$$ Thus, $$d^2 = \frac{D \lambda}{2} + \frac{\lambda^2}{16}$$ Substituting values, $$d^2 = \frac{0.2 \times 400 \times 10^{-9}}{2} + \frac{4 \times 10^{-14}}{4}$$ Therefore, $$d^2 \approx 400 \times 10^{-10}$$ Thus, $$d = 20 \times 10^{-5}$$ Therefore, $$d = 0.20 \, \mathrm{mm}$$

Question 57

Physics · Current Electricity · Numerical

A 16 $\Omega$ wire is bend to form a square loop. A 9 V battery with internal resistance 1$\Omega$ is connected across one of its sides. If a 4$\mu$F capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $\frac{x}{2}$ $\mu$J. where x = .

Answer: 81

Solution

Given the circuit, we calculate the current $I$ using the formula $I = \frac{V}{R_{eq}}$. Thus, $$I = \frac{9}{1 + \frac{12 \times 4}{12 + 4}} = \frac{9}{4}$$ Next, we find $I_1$: $$I_1 = \frac{9}{4} \times \frac{4}{16} = \frac{9}{16}$$ The voltage difference $V_A - V_B$ is given by: $$V_A - V_B = I_1 \times 8 = \frac{9}{16} \times 8 = \frac{9}{2} \, V$$ The energy $U$ is calculated as: $$\therefore \, U = \frac{1}{2} \times 4 \times \frac{81}{4} \, \mu J$$ Simplifying, we get: $$\therefore \, U = \frac{81}{2} \, \mu J$$ Therefore, $$\therefore \, x = 81$$

Question 58

Physics · Oscillations · Fill in the blank

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\frac{x}{8}$, where x = ____ .

Answer: 9

Solution

Let total energy $= E = \frac{1}{2} K A^2$. $$U = \frac{1}{2} K \left( \frac{A}{3} \right)^2 = \frac{K A^2}{2 \times 9} = \frac{E}{9}$$ $$KE = E - \frac{E}{9} = \frac{8E}{9}$$ Ratio $\frac{Total}{KE} = \frac{E}{\frac{8E}{9}} = \frac{9}{8}$ $x = 9$

Question 59

Physics · Electric Charges and Fields · Numerical

An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having surface charge density $+\sigma$. The electron at $t = 0$ is at a distance of 1 m from S and has a speed of 1 m/s. The maximum value of $\sigma$ if the electron strikes S at $t = 1$ s is $\alpha \left[ \frac{m \varepsilon_0}{e} \right] \frac{C}{m^2}$ the value of $\alpha$ is

Answer: 8

Solution

Given $u = 1 \, \mathrm{m/s}$; $a = -\frac{\sigma e}{2 \varepsilon_0 \, \mathrm{m}}$. $t = 1 \, \mathrm{s}$ $S = -1 \, \mathrm{m}$ Using $S = ut + \frac{1}{2} at^2$ $$-1 = 1 \times 1 - \frac{1}{2} \times \frac{\sigma e}{2 \varepsilon_0 \, \mathrm{m}} \times (1)^2$$ Therefore, $\sigma = 8 \frac{\varepsilon_0 \, \mathrm{m}}{e}$ Thus, $\alpha = 8$

Question 60

Physics · Mechanical Properties of Fluids · Numerical

In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are $70 \, \mathrm{ms^{-1}}$ and $65 \, \mathrm{ms^{-1}}$ respectively. If the wing area is $2 \, \mathrm{m^2}$ the lift of the wing is N. (Given density of air = $1.2 \, \mathrm{kg \, m^{-3}}$)

Answer: 810

Solution

Given the equation for force: $$F = \frac{1}{2} \rho \left( v_1^2 - v_2^2 \right) A$$ Substitute the values: $$F = \frac{1}{2} \times 1.2 \times \left( 70^2 - 65^2 \right) \times 2$$ The calculated force is: $$= 810 \, \mathrm{N}$$

Chemistry

Question 61

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

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: The first ionisation enthalpy decreases across a period. Reason R: The increasing nuclear charge outweighs the shielding across the period. In the light of the above statements, choose the most appropriate from the options given below:

  1. Both A and R are true and R is the correct explanation of A
  2. A is true but R is false
  3. A is false but R is true
  4. Both A and R are true but R is NOT the correct explanation of A

Answer: (c)

Solution

First ionisation energy increases along the period. Along the period $Z$ increases which outweighs the shielding effect.

Question 62

Chemistry · Biomolecules · Single correct

Match List I with List II \begin{tabular}{|l|l|} \hline \textbf{LIST-I} & \textbf{LIST-II} \\ \textbf{(Substances)} & \textbf{(Element Present)} \\ \hline A. Ziegler catalyst & I. Rhodium \\ \hline B. Blood Pigment & II. Cobalt \\ \hline C. Wilkinson catalyst & III. Iron \\ \hline D. Vitamin B$_{12}$ & IV. Titanium \\ \hline \end{tabular} Choose the correct answer from the options given below:

  1. A-II, B-IV, C-I, D-III
  2. A-II, B-III, C-IV, D-I
  3. A-III, B-II, C-IV, D-I
  4. A-IV, B-III, C-I, D-II

Answer: (d)

Solution

Q5 Ziegler catalyst $\rightarrow$ Titanium Blood pigment $\rightarrow$ Iron Wilkinson catalyst $\rightarrow$ Rhodium Vitamin B_{12} $\rightarrow$ Cobalt

Question 63

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

In chromyl chloride test for confirmation of $\mathrm{Cl}^-$ ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10$\%$ $\mathrm{H}_2\mathrm{O}_2$ turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is

  1. +6
  2. +5
  3. +10
  4. +3

Answer: (a)

Solution

In a basic medium, $\mathrm{Cl^-} + \mathrm{K_2Cr_2O_7} + \mathrm{H_2SO_4} \rightarrow \mathrm{CrO_2Cl_2}$ (yellow solution) $\rightarrow \mathrm{CrO_4^{2-}} + \mathrm{Cl^-}$ (yellow solution). The reaction sequence is as follows: 1. Acidification 2. Amyl alcohol 3. 10$\%$ $\mathrm{H_2O_2}$ This leads to the formation of $\mathrm{CrO_5}$ (blue compound).

Question 64

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

The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is

  1. electromeric energy
  2. resonance energy
  3. ionization energy
  4. hyperconjugation energy

Answer: (b)

Solution

The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is known as resonance energy.

Question 65

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

Given below are two statements: Statement I: The electronegativity of group 14 elements from Si to Pb gradually decreases. Statement II: Group 14 contains non-metallic, metallic, as well as metalloid elements. In the light of the above statements, choose the most appropriate from the options given below:

  1. Statement I is false but Statement II is true
  2. Statement I is true but Statement II is false
  3. Both Statement I and Statement II are true
  4. Both Statement I and Statement II are false

Answer: (a)

Solution

The electronegativity values for elements from Si to Pb are almost same. So Statement I is false.

Question 66

Chemistry · Structure of Atom · Single correct

The correct set of four quantum numbers for the valence electron of rubidium atom $(Z = 37)$ is:

  1. $5, 0, 0, +\frac{1}{2}$
  2. $5, 0, 1, +\frac{1}{2}$
  3. $5, 1, 0, +\frac{1}{2}$
  4. $5, 1, 1, +\frac{1}{2}$

Answer: (a)

Solution

Rb = [$\mathrm{Kr}$] 5s^1 n = 5 l = 0 m = 0 s = +1/2 or -1/2

Question 67

Chemistry · Hydrocarbons · Single correct

The major product (P) in the following reaction is

Answer: (d)

Solution

The reaction involves the addition of concentrated HBr to the given compound. The alkene group undergoes electrophilic addition with HBr, leading to the formation of a carbocation intermediate. The bromide ion then attacks the carbocation, resulting in the formation of the product with a bromine atom added to the alkene carbon. The excess HBr ensures complete conversion, leading to the final product.

Question 68

Chemistry · Hydrocarbons · Single correct

The arenium ion which is not involved in the bromination of Aniline is

Answer: (c)

Solution

Since $-\mathrm{NH_2}$ group is o/p directing hence arenium ion will not be formed by attack at meta position i.e. $Nc1cccc(Br)c1$. Hence Answer is (3)

Question 69

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

Appearance of blood red colour, on treatment of the sodium fusion extract of an organic compound with $\mathrm{FeSO_4}$ in presence of concentrated $\mathrm{H_2SO_4}$ indicates the presence of element/s

  1. Br
  2. N
  3. N and S
  4. S

Answer: (c)

Solution

The reaction starts with $\mathrm{Fe^{2+}}$ which is converted to $\mathrm{Fe^{3+}}$ in the presence of $\mathrm{H^+}$ and concentrated $\mathrm{H_2SO_4}$. Then, $\mathrm{Fe^{3+}}$ reacts with $\mathrm{SCN^-}$ to form $\mathrm{Fe(SCN)_3}$, which has a blood red color. The appearance of blood red color indicates the presence of both nitrogen and sulphur.

Question 70

Chemistry · Haloalkanes and Haloarenes · Single correct

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Aryl halides cannot be prepared by replacement of hydroxyl group of phenol by halogen atom. Reason R : Phenols react with halogen acids violently. In the light of the above statements, choose the most appropriate from the options given below:

  1. Both A and R are true but R is NOT the correct explanation of A
  2. A is false but R is true
  3. A is true but R is false
  4. Both A and R are true and R is the correct explanation of A

Answer: (c)

Solution

Assertion (A): Given statement is correct because in phenol hydroxyl group cannot be replaced by halogen atom. Reason (R): Given reason is false. Hence Assertion (A) is correct but Reason (R) is false.

Question 71

Chemistry · Hydrocarbons · Single correct

Identify product A and product B:

  1. A:
  2. A:
  3. A:
  4. A:

Answer: (d)

Solution

The reaction of the given alkene with $\mathrm{Cl_2}$ can proceed via two different mechanisms. Under the influence of $h\nu$, a free radical mechanism occurs, leading to the formation of Product A. In the presence of $\mathrm{CCl_4}$, an electrophilic addition reaction takes place on the alkene, resulting in Product B. Hence, the correct answer is option (4).

Question 72

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

Identify the incorrect pair from the following:

  1. Fluorspar- $BF_3$
  2. Cryolite- $Na_3AlF_6$
  3. Fluoroapatite- $3Ca_3(PO_4)_2$ $\cdot$ $CaF_2$
  4. Carnallite- $KCl$ $\cdot$ $MgCl_2$ $\cdot$ $6H_2O$

Answer: (a)

Solution

Fluorspar is $\mathrm{CaF_2}$

Question 73

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

The interaction between $\pi$ bond and lone pair of electrons present on an adjacent atom is responsible for

  1. Hyperconjugation
  2. Inductive effect
  3. Electromeric effect
  4. Resonance effect

Answer: (d)

Solution

It is a type of conjugation responsible for resonance.

Question 74

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

$KMnO_4$ decomposes on heating at $513 \, \mathrm{K}$ to form $O_2$ along with

  1. $MnO_2$ & $K_2O_2$
  2. $K_2MnO_4$ & $Mn$
  3. $Mn$ & $KO_2$
  4. $K_2MnO_4$ & $MnO_2$

Answer: (d)

Solution

The reaction is given by the equation: $$\mathrm{KMnO_4} \xrightarrow{\Delta} \mathrm{K_2MnO_4} + \mathrm{MnO_2} + \mathrm{O_2}.$$

Question 75

Chemistry · Co-ordination Compounds · Single correct

In which one of the following metal carbonyls, CO forms a bridge between metal atoms?

  1. $\mathrm{[Co_2(CO)_8]}$
  2. $\mathrm{[Mn_2(CO)_{10}]}$
  3. $\mathrm{[Os_3(CO)_{12}]}$
  4. $\mathrm{[Ru_3(CO)_{12}]}$

Answer: (a)

Solution

Question 76

Chemistry · Biomolecules · Single correct

Type of amino acids obtained by hydrolysis of proteins is:

  1. $\beta$
  2. $\alpha$
  3. $\delta$
  4. $\gamma$

Answer: (b)

Solution

Proteins are natural polymers composed of $\alpha$-amino acids which are connected by peptide linkages. Hence proteins upon acidic hydrolysis produce $\alpha$-amino acids.

Question 77

Chemistry · Hydrocarbons · Single correct

The final product A formed in the following multistep reaction sequence is

Answer: (a)

Solution

The reaction sequence involves the following steps: 1. The alkene undergoes hydration in the presence of $\mathrm{H_2O}$ and $\mathrm{H^+}$ to form an alcohol. 2. The alcohol is oxidized using $\mathrm{CrO_3}$ to form a ketone. 3. The ketone undergoes Wolff-Kishner reduction with $\mathrm{NH_2NH_2}$ and $\mathrm{KOH}$ under heat to form an alkane.

Question 78

Chemistry · Thermodynamics · Single correct

Which of the following is not correct?

  1. $\Delta G$ is negative for a spontaneous reaction
  2. $\Delta G$ is positive for a spontaneous reaction
  3. $\Delta G$ is zero for a reversible reaction
  4. $\Delta G$ is positive for a non-spontaneous reaction

Answer: (b)

Solution

$(\Delta G)_{P,T} = (+)\,\text{ve}$ for non-spontaneous process

Question 79

Chemistry · Redox Reactions · Single correct

Chlorine undergoes disproportionation in alkaline medium as shown below: $a\ \mathrm{Cl_2(g)}$ $+\,b\ \mathrm{OH^{-}(aq)}$$\rightarrow$ $c\ \mathrm{ClO^{-}(aq)}$ $+\,d\ \mathrm{Cl^{-}(aq)}$ $+\,e\ \mathrm{_2O(l)}$ The values of $a,\ b,\ c$ and $d$ in the balanced redox reaction are respectively:

  1. 1, 2, 1 and 1
  2. 2, 2, 1 and 3
  3. 3, 4, 4 and 2
  4. 2, 4, 1 and 3

Answer: (a)

Solution

The reaction involves the reduction and oxidation of chlorine. The oxidation states change as follows: $$\mathrm{Cl_2} \rightarrow \mathrm{Cl^-} + \mathrm{ClO^-}$$ The balanced equation is: $$\mathrm{Cl_2} + 2\mathrm{\overline{O}H} \rightarrow \mathrm{Cl^-} + \mathrm{ClO^-} + \mathrm{H_2O}$$

Question 80

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

In alkaline medium. $\mathrm{MnO_4^-}$ oxidises $\mathrm{I^-}$ to

  1. $\mathrm{IO_4^-}$
  2. $\mathrm{IO^-}$
  3. $\mathrm{I_2}$
  4. $\mathrm{IO_3^-}$

Answer: (d)

Solution

In an alkaline medium, the reaction is given by: $$2\mathrm{MnO_4^-} + \mathrm{H_2O} + \mathrm{I^-} \rightarrow 2\mathrm{MnO_2} + 2\mathrm{OH^-} + \mathrm{IO_3^-}$$

Question 81

Chemistry · Chemical Bonding and Molecular Structure · Numerical

Number of compounds with one lone pair of electrons on central atom amongst following is _______ $\mathrm{O}_3$, $\mathrm{H}_2\mathrm{O}$, $\mathrm{SF}_4$, $\mathrm{ClF}_3$, $\mathrm{NH}_3$, $\mathrm{BrF}_5$, $\mathrm{XeF}_4$

Answer: 4

Solution

Question 82

Chemistry · Electrochemistry · Numerical

The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A for 15 minutes is _______ $\times 10^{-4}$ g. (Atomic mass of zinc = 65.4 amu)

Answer: 46

Solution

The reaction is given by: $\mathrm{Zn}^{2+} + 2e^- \rightarrow \mathrm{Zn}$. The formula for weight is: $W = Z \times i \times t$. Substituting the values: $W = \frac{65.4}{2 \times 96500} \times 0.015 \times 15 \times 60$ $= 45.75 \times 10^{-4}\,\mathrm{gm}$

Question 83

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

For a reaction taking place in three steps at same temperature, overall rate constant $K = \frac{K_1 \, K_2}{K_3}$. If $E_{a_1}$, $E_{a_2}$ and $E_{a_3}$ are 40, 50 and 60 $\,$ $\mathrm{kJ/mol}$$ respectively, the overall $E_a$ is \, \mathrm{kJ/mol}.

Answer: 30

Solution

Given $$K = \frac{K_1 \cdot K_2}{K_3} = \frac{A_1 \cdot A_2}{A_3} \cdot e^{-\frac{(E_{a_1} + E_{a_2} - E_{a_3})}{RT}}$$ $$A \cdot e^{-E_a/RT} = \frac{A_1 A_2}{A_3} \cdot e^{-\frac{(E_{a_1} + E_{a_2} - E_{a_3})}{RT}}$$ $$E_a = E_{a_1} + E_{a_2} - E_{a_3} = 40 + 50 - 60 = 30 \, \mathrm{kJ/mole}$$

Question 84

Chemistry · Equilibrium · Fill in the blank

For the reaction $\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$, $K_p = 0.492\,\mathrm{atm}$ at $300\,\mathrm{K}$. $K_c$ for the reaction at same temperature is $\underline{\hspace{1cm}} \times 10^{-2}$. (Given: $R = 0.082\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}}$)

Answer: 2

Solution

Given $K_P = K_C \cdot (RT)^{\Delta n_g}$. $\Delta n_g = 1$ Therefore, $$K_C = \frac{K_P}{RT} = \frac{0.492}{0.082 \times 300} = 2 \times 10^{-2}$$

Question 85

Chemistry · Solutions · Numerical

A solution of $\mathrm{H_2SO_4}$ is $31.4\%$ $\mathrm{H_2SO_4}$ by mass and has a density of $1.25 \, \mathrm{g/mL}$. The molarity of the $\mathrm{H_2SO_4}$ solution is _________ M (nearest integer). [Given molar mass of $\mathrm{H_2SO_4} = 98 \, \mathrm{g \, mol}^{-1}$]

Answer: 4

Solution

Given $M = \frac{n_{solute}}{V} \times 1000$. $M = \frac{\left(\frac{31.4}{98}\right)}{\left(\frac{100}{1.25}\right)} \times 1000$ $= 4.005 \approx 4$

Question 86

Chemistry · Solutions · Numerical

The osmotic pressure of a dilute solution is $7 \times 10^5 \, \mathrm{Pa}$ at $273 \, \mathrm{K}$. Osmotic pressure of the same solution at $283 \, \mathrm{K}$ is ______ $\times 10^4 \, \mathrm{Nm}^{-2}$.

Answer: 73

Solution

Given $\pi = CRT$. Therefore, $\frac{\pi_1}{\pi_2} = \frac{T_1}{T_2}$ Thus, $\pi_2 = \frac{\pi_1 \, T_2}{T_1} = \frac{7 \times 10^5 \times 283}{273}$ $= 72.56 \times 10^4 \, \mathrm{Nm^{-2}}$

Question 87

Chemistry · Biomolecules · Numerical

Number of compounds among the following which contain sulphur as heteroatom is ______ Furan, Thiophene, Pyridine, Pyrrole, Cysteine, Tyrosine

Answer: 2

Solution

The image shows the structures of thiophene and cysteine. Thiophene is a five-membered aromatic ring containing sulfur. Cysteine is an amino acid with a thiol group (SH) and a carboxyl group (COOH).

Question 88

Chemistry · Chemical Bonding and Molecular Structure · Numerical

The number of species from the following which are paramagnetic and have bond order equal to one is $\underline{\hspace{2cm}}$. $H_2$, $He_2^+$, $O_2^+$, $N_2^{2-}$, $O_2^{2-}$, $F_2$, $Ne_2^+$, $B_2$

Answer: 1

Solution

\begin{tabular}{|c|c|c|} \hline & Magnetic behaviour & Bond order \\ \hline H$_2$ & Diamagnetic & 1 \\ \hline H$_2^+$ & Paramagnetic & 0.5 \\ \hline O$_2^+$ & Paramagnetic & 2.5 \\ \hline N$_2^+$ & Paramagnetic & 2 \\ \hline O$_2^{2-}$ & Diamagnetic & 1 \\ \hline F$_2$ & Diamagnetic & 1 \\ \hline Ne$_2^+$ & Paramagnetic & 0.5 \\ \hline B$_2$ & Paramagnetic & 1 \\ \hline \end{tabular}

Question 89

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

From the compounds given below, number of compounds which give positive Fehling's test is _______ Benzaldehyde, Acetaldehyde, Acetone, Acetophenone, Methanal, 4-nitrobenzaldehyde, cyclohexane carbaldehyde.

Answer: 3

Solution

Acetaldehyde ($\mathrm{CH_3CHO}$), Methanal ($\mathrm{HCHO}$), and cyclohexane carbaldehyde.

Question 90

Chemistry · Hydrocarbons · Numerical

Consider the given reaction. The total number of oxygen atoms present per molecule of the product $(P)$ is

Answer: 1

Solution

Hence total number of oxygen atom present per molecule is 1.