JEE Main 29 January 2025 Shift 2 question paper with solutions

JEE Main 29 January 2025 Shift 2: all 75 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 · Complex Numbers and Quadratic Equations · Single correct

If the set of all $a \in \mathbb{R}$, for which the equation $2x^2 + (a - 5)x + 15 = 3a$ has no real root, is the interval $(\alpha, \beta)$, and $X = \{x \in \mathbb{Z} : \alpha < x < \beta\}$, then $\sum_{x \in X} x^2$ is equal to:

  1. 2109
  2. 2129
  3. 2119
  4. 2139

Answer: (d)

Solution

Given $ (a - 5)^2 - 8(15 - 3a) < 0 $. Simplifying, we have: $$ a^2 + 14a + 25 - 120 < 0 $$ $$ a^2 + 14a - 95 < 0 $$ Factoring gives: $$ (a + 19)(a - 5) < 0 $$ Thus, $ a \in (-19, 5) $. Therefore, $ -19 < x < 5 $. The sum is: $$ \sum_{x \in X} x^2 = (1^2 + 2^2 + \ldots + 4^2) + (1^2 + 2^2 + \ldots + 18^2) $$ Calculating each part: $$ = \frac{4 \times 5 \times 9}{6} + \frac{18 \times 19 \times 37}{6} $$ $$ = 30 + 2109 $$ $$ = 2139 $$

Question 2

Maths · Trigonometric Functions · Single correct

If $\sin x + \sin^2 x = 1$, $x \in \left(0, \frac{\pi}{2}\right)$, then $(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^8 x + \tan^8 x) + (\cos^6 x + \tan^6 x)$ is equal to:

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

Answer: (d)

Solution

Given $\sin x + \sin^2 x = 1$. This implies $\sin x = \cos^2 x$ and $\tan x = \cos x$. Therefore, the given expression is $$2 \cos^{12} x + 6 \left[ \cos^{10} x + \cos^8 x \right] + 2 \cos^6 x$$ which simplifies to $$2 \left[ \sin^6 x + 3 \sin^5 x + 3 \sin^4 x + \sin^3 x \right]$$ This further simplifies to $$2 \sin^3 x \left[ (\sin x + 1)^3 \right]$$ which is $$2 \left[ \sin^2 x + \sin x \right]^3$$ Finally, this equals $$2$$

Question 3

Maths · Applications of Integrals · Single correct

Let the area enclosed between the curves $|y| = 1 - x^2$ and $x^2 + y^2 = 1$ be $\alpha$. If $9\alpha = \beta \pi + \gamma$; $\beta, \gamma$ are integers, then the value of $|\beta - \gamma|$ equals.

  1. 27
  2. 33
  3. 15
  4. 18

Answer: (b)

Solution

Required area $= \pi - 4 \int_0^1 (1 - x^2) \, dx$ $$= \pi - 4 \left[ x - \frac{x^3}{3} \right]_0^1$$ $$= \pi - 4 \times \frac{2}{3} = \pi - \frac{8}{3}$$ Therefore, $\alpha = \pi - \frac{8}{3}$. $9\alpha = 9\pi - 24 \rightarrow \beta = 9, \gamma = -24$ $|\beta - \gamma| = |9 + 24| = 33$

Question 4

Maths · Relations and Functions · Single correct

If the domain of the function $\log_5 (18x - x^2 - 77)$ is $(\alpha, \beta)$ and the domain of the function $\log_{(x-1)} \left( \frac{2x^2 + 3x - 2}{x^2 - 3x - 4} \right)$ is $(\gamma, \delta)$, then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

  1. 195
  2. 179
  3. 186
  4. 174

Answer: (c)

Solution

Given $f_1(x) = \log_5(18x - x^2 - 77)$. Therefore, $$18x - x^2 - 77 > 0$$ $$x^2 - 18x + 77 1, x - 1 \neq 1, \frac{2x^2 + 3x - 2}{x^2 - 3x - 4} > 0$$ $$x > 1, x \neq 2, \frac{(2x - 1)(x + 2)}{(x - 4)(x + 1)} > 0$$ From the number line, $x > 1, x \neq 2$. Therefore, $x \in (4, \infty)$. Thus, $\gamma = 4$. Therefore, $$\alpha^2 + \beta^2 + \gamma^2 = 49 + 121 + 16$$ $$= 186$$

Question 5

Maths · Continuity and Differentiability · Single correct

Let the function $f(x) = (x^2 + 1) |x^2 - ax + 2| + \cos |x|$ be not differentiable at the two points $x = \alpha = 2$ and $x = \beta$. Then the distance of the point $(\alpha, \beta)$ from the line $12x + 5y + 10 = 0$ is equal to :

  1. 5
  2. 4
  3. 3
  4. 2

Answer: (c)

Solution

Given $f(x) = (x^2 + 1) \left| x^2 - ax + 2 \right| + \cos |x|$. Notice that $\cos(-x) = \cos x = \cos |x|$ which means $\cos |x|$ is differentiable everywhere in $x \in \mathbb{R}$. Therefore, $f(x)$ can be non-differentiable where $|x^2 - ax + 2| = 0$. This implies $x^2 - ax + 2 = 0$. Solving $4 - 2a + 2 = 0$ gives $a = 3$. Thus, $(x^2 - 3x + 2) = 0$ implies $x = 1, 2$. Let $x = \alpha = 2$ and $x = \beta = 1$. The distance of $(\alpha, \beta)$ from the line $12x + 5y + 10 = 0$ is calculated as follows: $$\frac{|2(12) + 5(1) + 10|}{13} = \frac{39}{13} = 3.$$

Question 6

Maths · Three Dimensional Geometry · Single correct

Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-2}$ and $\frac{x-3}{1} = \frac{y-2}{3} = \frac{z+2}{4}$. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:

  1. $\sqrt{10}$
  2. $2\sqrt{3}$
  3. 2
  4. 3

Answer: (d)

Solution

Vector parallel to $L$ is given by the determinant: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix} = 10\hat{i} - 10\hat{j} + 5\hat{k}$$ This simplifies to: $$= 5(2\hat{i} - 2\hat{j} + \hat{k})$$ Equation of $L'$ is: $$\frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-3}{1} = \lambda (say)$$ Let $Q(2\lambda + 2, -2\lambda - 1, \lambda + 3)$. Solving $2\lambda + 2 = 0$ gives $\lambda = -1$. Thus, $Q(0, 1, 2)$. The distance $d(P, Q) = 3$.

Question 7

Maths · Sets · Single correct

Let $S = \mathbb{N} \cup \{0\}$. Define a relation $R$ from $S$ to $\mathbb{R}$ by : $$R = \left\{ (x, y) : \log_e y = x \log_e \left( \frac{2}{5} \right), x \in S, y \in \mathbb{R} \right\}$$ Then, the sum of all the elements in the range of $R$ is equal to :

  1. $\frac{10}{9}$
  2. $\frac{3}{2}$
  3. $\frac{5}{2}$
  4. $\frac{5}{3}$

Answer: (d)

Solution

Given the set $S = \{0, 1, 2, 3, \ldots\}$. We have $\log_y y = \log_e \left( \frac{2}{5} \right)$ which implies $y = \left( \frac{2}{5} \right)^x$. The required sum is given by: $$Sum = 1 + \left( \frac{2}{5} \right)^1 + \left( \frac{2}{5} \right)^2 + \left( \frac{2}{5} \right)^3 + \ldots = \frac{1}{1 - \frac{2}{5}} = \frac{5}{3}$$

Question 8

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

Let the line $x + y = 1$ meet the axes of $x$ and $y$ at $A$ and $B$, respectively. A right angled triangle $AMN$ is inscribed in the triangle $OAB$, where $O$ is the origin and the points $M$ and $N$ lie on the lines $OB$ and $AB$, respectively. If the area of the triangle $AMN$ is $\frac{4}{9}$ of the area of the triangle $OAB$ and $AN : NB = \lambda : 1$, then the sum of all possible value(s) of $\lambda$:

  1. 2
  2. $\frac{5}{2}$
  3. $\frac{1}{2}$
  4. $\frac{13}{6}$

Answer: (a)

Solution

Area of $\triangle AOB = \frac{1}{2}$. Area of $\triangle AMN = \frac{4}{9} \times \frac{1}{2} = \frac{2}{9}$. Equation of $AB$ is $x + y = 1$. $OA = 1$, $AM = \sec(45^\circ - \theta)$. $AN = \sec(45^\circ - \theta) \cos \theta$. $MN = \sec(45^\circ - \theta) \sin \theta$. $Ar(\triangle AMN) = \frac{1}{2} \times \sec^2(45^\circ - \theta) \sin \theta \cdot \cos \theta = \frac{2}{9}$. $\Rightarrow \tan \theta = 2, \frac{1}{2}$. $\tan \theta = 2$ is rejected. $AN = \lambda$. $NB = 1 = \cot \theta = 2$.

Question 9

Maths · Conic Sections · Single correct

If $\alpha x + \beta y = 109$ is the equation of the chord of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$, whose mid point is $\left( \frac{5}{2}, \frac{1}{2} \right)$, then $\alpha + \beta$ is equal to :

  1. 58
  2. 46
  3. 37
  4. 72

Answer: (a)

Solution

Equation of chord $T = S_1$ $$\frac{5}{2} \left( \frac{x}{9} \right) + \frac{1}{2} \left( \frac{y}{4} \right) = \frac{25}{36} + \frac{1}{16}$$ $$\Rightarrow \frac{5x}{18} + \frac{y}{8} = \frac{100 + 9}{144} = \frac{109}{144}$$ $$\Rightarrow 40x + 18y = 109$$ $$\Rightarrow \alpha = 40, \beta = 18$$ $$\Rightarrow \alpha + \beta = 58$$

Question 10

Maths · Permutations and Combinations · Single correct

If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at $440^{th}$ position in this arrangement, is :

  1. PRNAUK
  2. PRKANU
  3. PRKAUN
  4. PRNAKU

Answer: (c)

Solution

A ............... $5! = 120$ K ............... $5! = 120$ N ............... $5! = 120$ PA ............. $4! = 24$ PK ............. $4! = 24$ PN ............. $4! = 24$ PRA ........... $3! = 6$ $$\begin{array}{c} \boxed{P} \boxed{R} \boxed{K} \boxed{A} \boxed{N} \boxed{U} = 1 \\ \boxed{P} \boxed{R} \boxed{K} \boxed{A} \boxed{U} \boxed{N} = 1 \end{array}$$ Total $= 440$ $\Rightarrow 440^{th}$ word is $P \ R \ K \ A \ U \ N$

Question 11

Maths · Determinants · Single correct

Let $\alpha, \beta (\alpha \neq \beta)$ be the values of $m$, for which the equations $x + y + z = 1; x + 2y + 4z = m$ and $x + 4y + 10z = m^2$ have infinitely many solutions. Then the value of $\sum_{n=1}^{10} \left( n^\alpha + n^\beta \right)$ is equal to:

  1. 3080
  2. 560
  3. 3410
  4. 440

Answer: (d)

Solution

Given the determinant $$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{vmatrix} = 1(20 - 16) - 1(10 - 4) + 1(4 - 2)$$ which simplifies to $$= 4 - 6 + 2 = 0$$ For infinite solutions, $$\Delta_x = \Delta_y = \Delta_z = 0$$ The equation $$m^2 - 3x + 2 = 0$$ gives $$m = 1, 2$$ Therefore, $$\alpha = 1, \beta = 2$$ Thus, $$\sum_{n=1}^{10} \left( n^\alpha + n^\beta \right) = \sum_{n=1}^{10} n^1 + \sum_{n=1}^{10} n^2$$ This equals $$\frac{10(11)}{2} + \frac{10(11)(21)}{6}$$ which simplifies to $$= 55 + 385$$ resulting in $$= 440$$

Question 12

Maths · Matrices · Single correct

Let $A = [a_{ij}]$ be a matrix of order $3 \times 3$, with $a_{ij} = (\sqrt{2})^{i+j}$. If the sum of all the elements in the third row of $A^2$ is $\alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}$, then $\alpha + \beta$ is equal to:

  1. 280
  2. 224
  3. 210
  4. 168

Answer: (b)

Solution

Given $$A = \begin{bmatrix} (\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \\ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \\ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6 \end{bmatrix}$$ Simplifying, we have $$A = \begin{bmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{bmatrix}$$ Calculating $A^2$, $$A^2 = 2^2 \begin{bmatrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2\sqrt{2} \\ 2 & 2\sqrt{2} & 4 \end{bmatrix} \begin{bmatrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2\sqrt{2} \\ 2 & 2\sqrt{2} & 4 \end{bmatrix}$$ This results in $$= 4 \begin{bmatrix} (2 + 4 + 8) & (2\sqrt{2} + 4\sqrt{2} + 8\sqrt{2}) & (4 + 8 + 16) \end{bmatrix}$$ Sum of elements of 3rd row $$= 4(14 + 14\sqrt{2} + 28)$$ $$= 4(42 + 14\sqrt{2})$$ $$= 168 + 56\sqrt{2}$$ Let $\alpha + \beta\sqrt{2}$ be the sum, then $$\therefore \alpha + \beta = 168 + 56 = 224$$

Question 13

Maths · Three Dimensional Geometry · Single correct

Let P be the foot of the perpendicular from the point (1, 2, 2) on the line L: $\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2}$. Let the line $\vec{r} = (-\hat{i} + \hat{j} - 2\hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k}), \lambda \in \mathbb{R}$, intersect the line L at Q. Then $2(PQ)^2$ is equal to:

  1. 25
  2. 19
  3. 29
  4. 27

Answer: (d)

Solution

General point on line $L$: $\($ $\frac{x-1}{1}$ = $\frac{y+1}{-1}$ = $\frac{z-2}{2}$ $\)$ is $\($($\lambda$ + 1, -$\lambda$ - 1, 2$\lambda$ + 2)$\)$. DR's of $PM$ are $\($($\lambda$, -$\lambda$ - 3, 2$\lambda$)$\)$. $PM \perp L$. $\[$ $\Rightarrow$ $\lambda$ + (-1)(-$\lambda$ - 3) + 2(2$\lambda$) = 0 $\]$ $\[$ $\Rightarrow$ 6$\lambda$ + 3 = 0 $\]$ $\[$ P $\left$( $\frac{1}{2}$, -1, 1 $\right$) $\]$ Let another line $L'$: $\($ $\frac{x+1}{2}$ = $\frac{y-1}{-1}$ = $\frac{z+2}{1}$ $\)$. General point on line $L'$ is $\($($\mu$ - 1, -$\mu$ + 1, $\mu$ - 2)$\)$. Point of intersection of line $L$ and $L'$ is $\[$ $\lambda$ + 1 = $\mu$ - 1 $\Rightarrow$ $\mu$ - $\lambda$ = 2 $\ldots$ (1) $\]$ $\[$ 2$\lambda$ + 2 = $\mu$ - 2 $\Rightarrow$ 2$\lambda$ = $\mu$ - 4 $\]$ $\[$ $\Rightarrow$ $\lambda$ = -2 and $\mu$ = 0 $\]$ $\[$ Q(-1, 1, -2) $\]$ $\[$ 2(PQ)^2 = 2 $\left$( $\left$( $\frac{1}{2}$ + 1 $\right$)^2 + $\left$( -$\frac{1}{2}$ - 1 $\right$)^2 + (1 + 2)^2 $\right$) $\]$ $\[$ = 2 $\left$( $\frac{9}{4}$ + $\frac{9}{4}$ + 9 $\right$) $\]$ $\[$ = 27 $\]$

Question 14

Maths · Conic Sections · Single correct

Let a circle $C$ pass through the points $(4, 2)$ and $(0, 2)$, and its centre lie on $3x + 2y + 2 = 0$. Then the length of the chord, of the circle $C$, whose mid-point is $(1, 2)$, is :

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

Answer: (c)

Solution

Let the centre be $$\left(-2a, \frac{6a-2}{2}\right) \equiv (-2a, 3a-1)$$. Centre is equal distance from $(4, 2)$ and $(0, 2)$. Therefore, $$\sqrt{(4 + 2a)^2 + (3a - 3)^2} = \sqrt{(-2a - 0)^2 + (3a - 3)^2}$$ $$\Rightarrow (2a + 4)^2 + 9(a - 1)^2 = 4a^2 + 9(a - 1)^2$$ $$\Rightarrow 4a^2 + 16 + 16a = 4a^2 \Rightarrow a = -1$$ $$\Rightarrow centre \equiv (2, -4) \Rightarrow Radius = \sqrt{40}$$ $$\Rightarrow AM^2 = (\sqrt{40})^2 - (\sqrt{37})^2$$ $$\Rightarrow 2AM = AB = 2\sqrt{3}$$

Question 15

Maths · Determinants · Single correct

Let $A = \begin{bmatrix} a_{ij} \end{bmatrix}$ be a $2 \times 2$ matrix such that $a_{ij} \in \{0,1\}$ for all $i$ and $j$. Let the random variable $X$ denote the possible values of the determinant of the matrix $A$. Then, the variance of $X$ is:

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

Answer: (c)

Solution

Given the values of $x$ and $P(x)$: $$x: 0 1 -1$$ $$P(x): \frac{10}{16} \frac{3}{16} \frac{3}{16}$$ The variance is calculated as: $$Var(x) = E(x^2) - [E(x)]^2$$ Substituting the values, we have: $$= \sum_{i=1}^{3} x_i^2 P(x_i) - (\mu)^2$$ $$= 1 \times \frac{3}{16} + 1 \times \frac{3}{16} [\mu = 0]$$ $$= \frac{6}{16} = \frac{3}{8}$$

Question 16

Maths · Probability · Single correct

Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains $n$ white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is $29/45$, then $n$ is equal to:

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

Answer: (a)

Solution

Bag 1 contains $4w, 5B$. Bag 2 contains $nw, 3B$. (I) Transferred ball is white $P(w) = \frac{n+1}{n+4} \cdot \frac{4}{9}$ (II) Transferred ball is black $P(w) = \frac{5}{9} \cdot \frac{n}{n+4}$ $$\frac{4n + 4}{9n + 36} + \frac{5n}{9n + 36} = \frac{29}{45}$$ $$\frac{9n + 4}{9n + 36} = \frac{29}{45} \Rightarrow n = 6$$

Question 17

Maths · Binomial Theorem · Single correct

The remainder, when $7^{103}$ is divided by 23, is equal to:

  1. 6
  2. 17
  3. 9
  4. 14

Answer: (d)

Solution

Given $7^{103} = 7 \left(7^{102}\right) = 7(343)^{34} = 7(345 - 2)^{34}$. $7^{103} = 23 \, K_1 + 7 \cdot 2^{34}$. Now $7 \cdot 2^{34} = 7 \cdot 2^2 \cdot 2^{32}$ $$= 28 \cdot (256)^4$$ $$= 28(253 + 3)^4$$ Therefore, $28 \times 81 \Rightarrow (23 + 5)(69 + 12)$ $$23 \, K_2 + 60$$ Therefore, Remainder $= 14$

Question 18

Maths · Integrals · Single correct

Let $f(x) = \int_{0}^{t} t \left(t^2 - 9t + 20\right) \, dt$, $1 \leq x \leq 5$. If the range of $f$ is $[\alpha, \beta]$, then $4(\alpha + \beta)$ equals:

  1. 253
  2. 154
  3. 125
  4. 157

Answer: (d)

Solution

Given $f'(x) = x(x^2 - 9x + 20)$, $x \in (1, 5)$. $$f'(x) = (x-4)x(x-5)$$ From the sign chart, we have: $f'(x) > 0$ for all $x \in (1, 4)$ $f'(x) < 0$ for all $x \in (4, 5)$ Thus, $f(x)$ is increasing in $(1, 4)$ and $f(x)$ is decreasing in $(4, 5)$. Critical points to check are $x = 1, 4, 5$. $$f(x) = \int_0^x (t^3 - 9t^2 + 20t) \, dt$$ $$= \frac{t^4}{4} - 3t^3 + 10t^2 \bigg|_0^x = \frac{x^4}{4} - 3x^3 + 10x^2$$ Calculating at critical points: $$f(1) = \frac{1}{4} - 3 + 10 = \frac{29}{4}$$ $$f(4) = 4^3 - 3 \cdot 4^3 + 10 \cdot 4^2 = -2 \cdot 4^3 + 10 \cdot 4^2 = 32$$ $$f(5) = \frac{5^4}{4} - 3 \cdot 5^3 + 10 \cdot 2.5 = \frac{5^4}{4} - 125 = \frac{125}{4}$$ Range is given by: $$\left[ \frac{29}{4}, 32 \right] \Rightarrow 4(\alpha + \beta) = 128 + 29$$ Thus, the range is $157$.

Question 19

Maths · Vector Algebra · Single correct

Let $\hat{a}$ be a unit vector perpendicular to the vectors $\vec{b}=\hat{i}-2\hat{j}+3\hat{k}$ and $\vec{c}=2\hat{i}+3\hat{j}-\hat{k}$, and makes an angle of $\cos^{-1}\left(-\frac{1}{3}\right)$ with the vector $\hat{i}+\hat{j}+\hat{k}$. If $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with the vector $\hat{i}+\alpha\hat{j}+\hat{k}$, then the value of $\alpha$ is equal to

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

Answer: (b)

Solution

Given the matrix: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix}$$ We have $\vec{a} = \hat{i}(-7) + 7\hat{j} + 7\hat{k}$. The unit vector $\hat{a}$ is given by: $$\hat{a} = \pm \frac{(-7\hat{i} + 7\hat{j} + 7\hat{k})}{\sqrt{7^2 + 7^2 + 7^2}} = \pm \frac{(-\hat{i} + \hat{j} + \hat{k})}{\sqrt{3}}$$ Now, $\cos \theta = \pm \frac{(-1 + 1 + 1)}{\sqrt{3} \cdot \sqrt{3}} = \pm \frac{1}{3}$. Thus, $$\cos^{-1}\left(\frac{-1}{3}\right) \Rightarrow \hat{a} = \frac{-(\hat{i} + \hat{j} + \hat{k})}{\sqrt{3}}$$ Therefore, $$\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$$ For $\cos \frac{\pi}{3} = \frac{1 - \alpha - 1}{\sqrt{3} \cdot \sqrt{\alpha^2 + 2}}$, we have: $$\frac{1}{2} = \frac{-\alpha}{\sqrt{3} \cdot \sqrt{\alpha^2 + 2}} \Rightarrow \alpha < 0$$ Solving, $$3(\alpha^2 + 2) = 4\alpha^2$$ $$6 = \alpha^2$$ $$\alpha = \pm \sqrt{6}$$ Clearly, $\alpha = -\sqrt{6}$.

Question 20

Maths · Differential Equations · Single correct

If for the solution curve $y = f(x)$ of the differential equation $\frac{dy}{dx} + (\tan x) y = \frac{2 + \sec x}{(1 + 2 \sec x)^2}$, $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, $f\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{10}$, then $f\left( \frac{\pi}{4} \right)$ is equal to:

  1. $\frac{\sqrt{3} + 1}{10(4 + \sqrt{3})}$
  2. $\frac{5 - \sqrt{3}}{2\sqrt{2}}$
  3. $\frac{9\sqrt{3} + 3}{10(4 + \sqrt{3})}$
  4. $\frac{4 - \sqrt{2}}{14}$

Answer: (d)

Solution

Given $e^{f \tan x} = e^{\ln(\sec x)} = \sec x$. Therefore, $y \cdot \sec x = \int \left\{ \frac{2 + \sec x}{(1 + 2 \sec x)^2} \right\} \sec x \, dx$. Let $\cos x = \frac{1 - t^2}{1 + t^2}$. Then, $$\int \frac{2 \cos x + 1}{(\cos x + 2)^2} \, dx = \int \frac{2 \left( \frac{1 - t^2}{1 + t^2} \right) + 1}{\left( \frac{1 - t^2}{1 + t^2} + 2 \right)^2} \, 2 \, dt$$ $$= \int \frac{2 - 2t^2 + 1 + t^2}{(1 - t^2 + 2 + 2t^2)^2} \, 2 \, dt = 2 \int \frac{3 - t^2}{(t^2 + 3)^2} \, dt$$ Let $t + \frac{3}{t} = u$. Then, $\left(1 - \frac{3}{t^2}\right) \, dt = du$. Therefore, $$= -2 \int \frac{du}{u^2}$$ $$y \cdot (\sec x) = \frac{2}{u} + c$$ $$y \cdot \sec x = \frac{2}{t + \frac{3}{t}} + c \ldots (I)$$ At $x = \frac{\pi}{3}$, $t = \tan \frac{x}{2} = \frac{1}{\sqrt{3}}$. $$2 \cdot \frac{\sqrt{3}}{10} = \frac{2}{\frac{1}{\sqrt{3}} + \sqrt{3}} + c$$ $$\frac{\sqrt{3}}{10} = \frac{2 \sqrt{3}}{10} + c \Rightarrow c = 0$$ At $x = \frac{\pi}{4}$, $t = \tan \frac{x}{2} = \sqrt{2} - 1$. Therefore, $y \cdot \sqrt{2} = \frac{\sqrt{2} - 1 + \frac{3}{\sqrt{2} - 1}}{2(\sqrt{2} - 1)} = \frac{1}{\sqrt{2}} \times \frac{2 \sqrt{2} - 1}{7}$ $= \frac{4 - \sqrt{2}}{14}$

Question 21

Maths · Integrals · Numerical

If $\displaystyle 24\int_{0}^{\pi}\left(\sin\left|4x-\frac{\pi}{12}\right|+\left[2\sin x\right]\right)\,dx=2\pi+\alpha,$ where $[\cdot]$ denotes the greatest integer function, then $\alpha$ is equal to _______.

Answer: 12

Solution

Let $I = 24 \int_0^{\frac{\pi}{2}} \left( \sin \left( 4x - \frac{\pi}{2} \right) + [2 \sin x] \right) \, dx \ldots (i)$ Now $\left| 4x - \frac{\pi}{12} \right| = \begin{cases} -4x + \frac{\pi}{12} ; & x < \frac{\pi}{48} \\ 4x - \frac{\pi}{12} ; & x \geq \frac{\pi}{48} \end{cases}$ Therefore, from (i) $$I = 24 \int_0^{\frac{\pi}{48}} -\sin \left( 4x - \frac{\pi}{12} \right) \, dx + \int_{\frac{\pi}{48}}^{\frac{\pi}{4}} \sin \left( 4x - \frac{\pi}{12} \right)$$ $$+ \int_0^{\frac{\pi}{6}} [2 \sin x] \, dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} [2 \sin x] \, dx$$ $$I = 24 \left[ \left( \frac{1 - \cos \frac{\pi}{12}}{4} \right) - \left( \frac{-\cos \frac{\pi}{12} - 1}{4} \right) \right] + \frac{\pi}{4} - \frac{\pi}{6}$$ $$I = 24 \left( \frac{1}{2} \right) + \frac{\pi}{4} - \frac{\pi}{6}$$ $$I = 2\pi + 12 = 2\pi + \alpha (from above)$$ Therefore, $\alpha = 12$

Question 22

Maths · Limits and Derivatives · Numerical

If $\lim_{t \to 0} \left( \int_{0}^{1} (3x + 5)^t \, dx \right)^{\frac{1}{t}} = \frac{\alpha}{5e} \left( \frac{8}{5} \right)^{\frac{2}{3}}$, then $\alpha$ is equal to

Answer: 64

Solution

$\dfrac{\alpha}{5e} = \exp\left(\lim_{t \to 0} \dfrac{1}{t}\left(\int_0^1 (3x+5)^t dx - 1\right)\right)$ $= \exp\left(\lim_{t \to 0} \dfrac{1}{t}\left(\dfrac{(3x+5)^{t+1}}{3(t+1)}\Bigg|_0^1 - 1\right)\right)$ $= \exp\left(\lim_{t \to 0} \dfrac{1}{t}\left(\dfrac{8^{t+1} - 5^{t+1}}{3(t+1)} - 1\right)\right)$ $= \exp\left(\lim_{t \to 0} \dfrac{1}{t}\left(\dfrac{8^{t+1} - 5^{t+1} - 3t - 3}{3(t+1)}\right)\right)$ $= \exp\left(\lim_{t \to 0}\left(\dfrac{8^{t+1} \cdot \ln 8 - 5^{t+1} \ln 5 - 3}{3(t+1)}\right)\right)$ $= \exp\left(\dfrac{\ln 8^8 - \ln 5^5 - 3}{5}\right)$ $= \left(\dfrac{8}{5}\right)^{2/3} \dfrac{\alpha}{5e} = \exp\left(\dfrac{\ln\left(\dfrac{8^8}{5^5}\right)}{5} - 1\right)$ $\Rightarrow \left(\dfrac{8}{5}\right)^{2/3} \dfrac{\alpha}{5} = \left(\dfrac{8^8}{5^5}\right)^{1/3} = \left(\dfrac{8^6 \cdot 8^2}{5^3 \cdot 5^2}\right)^{1/3} = \dfrac{64}{5}\left(\dfrac{8}{5}\right)^{2/3}$ $\Rightarrow \alpha = 64$

Question 23

Maths · Sequences and Series · Numerical

Let $a_1, a_2, \ldots, a_{2024}$ be an Arithmetic Progression such that $a_1 + (a_5 + a_{10} + a_{15} + \ldots + a_{2020}) + a_{2024} = 2233$. Then $a_1 + a_2 + a_3 + \ldots + a_{2024}$ is equal to

Answer: 11132

Solution

As $a_1 + a_5 + a_{10} + \ldots + a_{2020} + a_{2024} = 2233$ ...(1) We know in arithmetic progression. Sum of terms equidistant from ends is equal. Therefore, from (1) $$a_1 + a_{2024} = a_5 + a_{2020} = a_{10} + a_{2015} = \ldots$$ $$\underbrace{\phantom{a_1 + a_{2024} = a_5 + a_{2020} = a_{10} + a_{2015} = \ldots}}_{203 pairs}$$ $$\Rightarrow 203 \left(a_1 + a_{2024}\right) = 2233$$ $$\Rightarrow a_1 + a_{2024} = 11$$ $$\sum_{i=1}^{2024} a_i = S_{2024} = \frac{2024}{2} [a_1 + a_{2024}]$$ Now $$= 1012(11)$$ $$= 11132$$

Question 24

Maths · Complex Numbers and Quadratic Equations · Numerical

Let integers $a, b \in [-3, 3]$ be such that $a + b \neq 0$. Then the number of all possible ordered pairs $(a, b)$, for which $\left| \frac{x-a}{x+b} \right| = 1$ and $$\begin{vmatrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix}$$ $= 1, z \in \mathbb{C}$, where $\omega$ and $\omega^2$ are the roots of $x^2 + x + 1 = 0$, is equal to .

Answer: 10

Solution

Given $a, b \in I$, $-3 \leq a, b \leq 3$, $a + b \neq 0$. $$|z - a| = |z + b|$$ $$\begin{vmatrix} z + 1 & \omega & \omega^2 \\ \omega & z + \omega^2 & 1 \\ \omega^2 & 1 & z + \omega \end{vmatrix} = 1$$ $$\Rightarrow \begin{vmatrix} z & \omega & z + \omega^2 & 1 \\ \omega^2 & 1 & z + \omega \\ 1 & 1 & 1 \end{vmatrix} = 1$$ $$\Rightarrow z \begin{vmatrix} \omega & z + \omega^2 & 1 \\ \omega^2 & 1 & z + \omega \\ 1 & 0 & 0 \end{vmatrix} = 1$$ $$\Rightarrow z \begin{vmatrix} \omega & z + \omega^2 - \omega & 1 - \omega \\ \omega^2 & 1 - \omega^2 & z + \omega - \omega^2 \end{vmatrix} = 1$$ $$\Rightarrow z^3 = 1$$ $$\Rightarrow z = \omega, \omega^2, 1$$ Now $$|1 - a| = |1 + b|$$ $$\Rightarrow 10 pairs$$

Question 25

Maths · Conic Sections · Numerical

Let $y^2 = 12x$ be the parabola and $S$ be its focus. Let $PQ$ be a focal chord of the parabola such that $(SP)(SQ) = \frac{147}{4}$. Let $C$ be the circle described taking $PQ$ as a diameter. If the equation of a circle $C$ is $64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta$, then $\beta - \alpha$ is equal to _______.

Answer: 1328

Solution

Given $y^2 = 12x$, $a = 3$, $SP \times SQ = \frac{147}{4}$. Let $P \left(3t^2, 6t\right)$ and $t_1 t_2 = -1$ (ends of focal chord). So, $Q \left(\frac{3}{t^2}, \frac{-6}{t}\right)$. $S(3, 0)$. $SP \times SQ = PM_1 \times QM_2$ (dist. from directrix). $$= \left(3 + 3t^2\right) \left(3 + \frac{3}{t^2}\right) = \frac{147}{4}$$ $$\Rightarrow (1 + t^2)^2 = \frac{49}{12}$$ $$t^2 = \frac{3}{4} \cdot \frac{4}{3}$$ $$t = \pm \frac{\sqrt{3}}{2}, \pm \frac{2}{\sqrt{3}}$$ considering $t = -\frac{\sqrt{3}}{2}$. $P \left(\frac{9}{4}, -3\sqrt{3}\right)$ and $Q(4, 4\sqrt{3})$. Hence, diametric circle: $$(x - 4)\left(x - \frac{9}{4}\right) + (y + 3\sqrt{3})(y - 4\sqrt{3}) = 0$$ $$\Rightarrow x^2 + y^2 - \frac{25}{4}x - \sqrt{3}y - 27 = 0$$ $$\Rightarrow \alpha = 400, \beta = 1728$$ $$\beta - \alpha = 1328$$

Physics

Question 26

Physics · Current Electricity · Single correct

The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have

  1. high thermal conductivity and high electrical conductivity
  2. low thermal conductivity and low electrical conductivity
  3. high thermal conductivity and low electrical conductivity
  4. low thermal conductivity and high electrical conductivity

Answer: (d)

Solution

Material should have low thermal conductivity and high electrical conductivity.

Question 27

Physics · Kinetic Theory · Single correct

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process. Reason (R) : In isothermal process, $PV = constant$, while in adiabatic process $PV^\gamma = constant$. Here $\gamma$ is the ratio of specific heats, $P$ is the pressure and $V$ is the volume of the ideal gas. In the light of the above statements, choose the correct answer from the options given below :

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

Answer: (a)

Solution

The graph shows the relationship between pressure $P$ and volume $V$ for adiabatic and isothermal processes. In an adiabatic process, pressure increases more rapidly with compression compared to an isothermal process. Therefore, the slope of the $P$-$V$ curve, which is $\left( \frac{dP}{dV} \right)$, is greater for an adiabatic process than for an isothermal process: $$\left( \frac{dP}{dV} \right)_{Adiabatic} > \left( \frac{dP}{dV} \right)_{Isothermal}.$$

Question 28

Physics · Electric Charges and Fields · Single correct

An electric dipole is placed at a distance of 2 cm from an infinite plane sheet having positive charge density $\sigma_0$. Choose the correct option from the following.

  1. Potential energy and torque both are maximum.
  2. Torque on dipole is zero and net force is directed away from the sheet.
  3. Torque on dipole is zero and net force acts towards the sheet.
  4. Potential energy of dipole is minimum and torque is zero.

Answer: (d)

Solution

Electric field due to sheet $E = \frac{\sigma}{2\varepsilon_0}$ and torque on dipole $\vec{\tau} = \vec{P} \times \vec{E}$. Here $\vec{\tau} = 0$ and $U = -\vec{P} \cdot \vec{E}$ should be minimum.

Question 29

Physics · Dual Nature of Radiation and Matter · Single correct

In an experiment with photoelectric effect, the stopping potential,

  1. increases with increase in the intensity of the incident light
  2. decreases with increase in the intensity of the incident light
  3. increases with increase in the wavelength of the incident light
  4. is ($\frac{1}{e}$) times the maximum kinetic energy of the emitted photoelectrons

Answer: (d)

Solution

From Einstein photoelectric equation $$\frac{hc}{\lambda} = \phi + eV_S$$ Maximum K.E. = $$(K)_{max} = eV_S$$ So, $$V_S = \frac{(K)_{max}}{e}$$

Question 30

Physics · Electric Charges and Fields · Single correct

A point charge causes an electric flux of $-2 \times 10^4 \, \mathrm{Nm}^2 \mathrm{C}^{-1}$ to pass through a spherical Gaussian surface of $8.0 \, \mathrm{cm}$ radius, centred on the charge. The value of the point charge is : (Given $\epsilon_0 = 8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{N}^{-1} \mathrm{m}^{-2}$)

  1. $15.7 \times 10^{-8} \, \mathrm{C}$
  2. $17.7 \times 10^{-8} \, \mathrm{C}$
  3. $-15.7 \times 10^{-8} \, \mathrm{C}$
  4. $-17.7 \times 10^{-8} \, \mathrm{C}$

Answer: (d)

Solution

Flux $(\phi) = \frac{\theta_{inc}}{\epsilon_0}$ $$\theta_{inc} = \epsilon_0 \phi$$ $$= -17.7 \times 10^{-8} \, \mathrm{C}$$

Question 31

Physics · Kinetic Theory · Single correct

A poly-atomic molecule ( $C_V = 3R$, $C_P = 4R$, where $R$ is gas constant) goes from phase space point $A$ ($P_A = 10^5$ Pa, $V_A = 4 \times 10^{-6}$ m$^3$) to point $B$ ($P_B = 5 \times 10^4$ Pa, $V_B = 6 \times 10^{-6}$ m$^3$) to point $C$ ($P_C = 10^4$ Pa, $V_C = 8 \times 10^{-6}$ m$^3$). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is:

  1. 500R($\ln$ 3 + $\ln$ 4)
  2. 450R($\ln$ 4 - $\ln$ 3)
  3. 500R $\ln$ 2
  4. 400R $\ln$ 4

Answer: (b)

Solution

Given $\Delta Q_{AB} = 0$ adiabatic. $\Delta Q_{BC} = \Delta W_{BC}$. $$= nRT \ln \left( \frac{V_C}{V_B} \right) = 450R \ln \left( \frac{8 \times 10^{-6}}{6 \times 10^{-6}} \right)$$ $$= 450R \ln \left( \frac{4}{3} \right) = 450R (\ln 4 - \ln 3)$$ Therefore, $\Delta Q = \Delta Q_{AB} + \Delta Q_{BC}$. $$\Delta Q = 450R (\ln 4 - \ln 3)$$

Question 32

Physics · Ray Optics and Optical Instruments · Single correct

Two identical symmetric double convex lenses of focal length $f$ are cut into two equal parts $L_1$, $L_2$ by $AB$ plane and $L_3$, $L_4$ by $XY$ plane as shown in figure respectively. The ratio of focal lengths of lenses $L_1$ and $L_3$ is

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

Answer: (b)

Solution

Given $f_1 = f$, $f_2 = f$, and $f_3 = 2f$, so $\($ $\frac{f_1}{f_3}$ = 1 : 2 $\)$.

Question 33

Physics · Electromagnetic Waves · Single correct

A plane electromagnetic wave propagates along the $+x$ direction in free space. The components of the electric field, $\vec{E}$ and magnetic field, $\vec{B}$ vectors associated with the wave in Cartesian frame are

  1. $E_x$, $B_y$
  2. $E_y$, $B_z$
  3. $E_z$, $B_y$
  4. $E_y$, $B_x$

Answer: (c)

Solution

Direction of propagation is given by $$\vec{E} \times \vec{B}$$.

Question 34

Physics · Ray Optics and Optical Instruments · Single correct

Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object $O$ is placed midway, between $P$ and $B$. The separation between the images of $O$, formed by each refracting surface is :

  1. 0.214 $R$
  2. 0.411 $R$
  3. 0.124 $R$
  4. 0.114 $R$

Answer: (d)

Solution

For B $$\frac{\mu_2}{V} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$ $$\frac{1.5}{V} + \frac{1}{\frac{R}{2}} = \frac{0.5}{-R}$$ $$\frac{1.5}{V} = -\frac{1}{2R} - \frac{2}{R}$$ $$\frac{1.5}{V} = -\frac{5}{2R} \Rightarrow V_B = -0.6R$$ For A $$\frac{1.5}{V} + \frac{2}{3R} = \frac{0.5}{-R}$$ $$\frac{1.5}{V} = -\frac{1}{2R} - \frac{2}{3R}$$ $$\frac{1.5}{V} = -\frac{7}{6R}$$ $$V_A = -\frac{9}{7}R$$ Distance between images $$= 2R - \left(0.6R + \frac{9}{7}R\right) = 0.114R$$

Question 35

Physics · Oscillations · Single correct

Two bodies A and B of equal mass are suspended from two massless springs of spring constant $k_1$ and $k_2$, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is

  1. $\frac{k_1}{k_2}$
  2. $\sqrt{\frac{k_1}{k_2}}$
  3. $\sqrt{\frac{k_2}{k_1}}$
  4. $\frac{k_2}{k_1}$

Answer: (b)

Solution

Given $V_1 = A_1 \omega_1$ and $V_2 = A_2 \omega_2$. Also, $A_1 = A_2$. Therefore, $$\frac{V_1}{V_2} = \frac{\omega_1}{\omega_2} = \frac{\sqrt{K_1/m}}{\sqrt{K_2/m}}$$ Simplifying, $$\frac{V_1}{V_2} = \sqrt{\frac{K_1}{K_2}}$$

Question 36

Physics · Work, Energy and Power · Single correct

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Three identical spheres of same mass undergo one dimensional motion as shown in figure with initial velocities $v_A=5\,\mathrm{m/s}$, $v_B=2\,\mathrm{m/s}$, $v_C=4\,\mathrm{m/s}$. If we wait sufficiently long for elastic collision to happen, then $v_A=4\,\mathrm{m/s}$, $v_B=2\,\mathrm{m/s}$, $v_C=5\,\mathrm{m/s}$ will be the final velocities. Reason (R): In an elastic collision between identical masses, two objects exchange their velocities. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (a)

Solution

For A and B. Before collision: A has velocity $5 \, \mathrm{m/s}$ and B has velocity $2 \, \mathrm{m/s}$. After collision: A has velocity $2 \, \mathrm{m/s}$ and B has velocity $5 \, \mathrm{m/s}$. For B and C. Before collision: B has velocity $5 \, \mathrm{m/s}$ and C has velocity $4 \, \mathrm{m/s}$. After collision: B has velocity $4 \, \mathrm{m/s}$ and C has velocity $5 \, \mathrm{m/s}$. Final velocity: $V_A = 2 \, \mathrm{m/s}$, $V_B = 4 \, \mathrm{m/s}$, $V_C = 5 \, \mathrm{m/s}$. Therefore, velocity exchange for two identical mass in elastic collision.

Question 37

Physics · Work, Energy and Power · Single correct

A sand dropper drops sand of mass $m(t)$ on a conveyer belt at a rate proportional to the square root of speed $(v)$ of the belt, i.e. $\frac{dm}{dt} \propto \sqrt{v}$. If $P$ is the power delivered to run the belt at constant speed then which of the following relationship is true?

  1. $P \propto \sqrt{v}$
  2. $P \propto v$
  3. $P^2 \propto v^5$
  4. $P^2 \propto v^3$

Answer: (c)

Solution

Power $= \vec{F} \cdot \vec{V}$ and $F = \frac{dp}{dt}$ = Rate of change of linear momentum $F = V \cdot \frac{dm}{dt} = K_1 V^{\frac{3}{2}}, K$ is constant Power $(P) = \left(K V^{\frac{3}{2}}\right) \cdot (V)$ $$= K V^{\frac{5}{2}}$$ So, $P^2 \propto V^5$

Question 38

Physics · Ray Optics and Optical Instruments · Single correct

A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to

  1. 24 cm
  2. 96 cm
  3. 36 cm
  4. 72 cm

Answer: (b)

Solution

Given $\($ $\frac{1}{8}$ = $\left$( $\frac{\mu_\ell}{\mu_s - 1}$ $\right$) $\left$[ $\frac{1}{R_1}$ - $\frac{1}{R_2}$ $\right$] $\)$ $\($ $\frac{1}{24}$ = (1.5 - 1) $\left$[ $\frac{2}{R}$ $\right$] $\ldots$ (i) $\)$ $\($ $\frac{1}{f'}$ = $\left$( $\frac{1.5}{1.33}$ - 1 $\right$) $\left$( $\frac{2}{R}$ $\right$) $\)$ $\($ $\frac{1}{f'}$ = $\left$( $\frac{1.5 \times 3}{4}$ - 1 $\right$) $\frac{2}{R}$ $\ldots$ (ii) $\)$ (i) divided by (ii) $\($ $\frac{f'}{24}$ = 4 $\)$ $\($ f' = 96 $\,$ $\mathrm{cm}$ $\)$

Question 39

Physics · Electrostatic Potential and Capacitance · Single correct

A capacitor, $C_1 = 6 \mu \mathrm{F}$ is charged to a potential difference of $V_0 = 5 \, \mathrm{V}$ using a $5 \, \mathrm{V}$ battery. The battery is removed and another capacitor, $C_2 = 12 \mu \mathrm{F}$ is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges ($q_1$ and $q_2$) on the capacitors $C_1$ and $C_2$ when equilibrium condition is reached.

  1. $q_1 = 10 \mu \mathrm{C}, q_2 = 20 \mu \mathrm{C}$
  2. $q_1 = 30 \mu \mathrm{C}, q_2 = 15 \mu \mathrm{C}$
  3. $q_1 = 20 \mu \mathrm{C}, q_2 = 10 \mu \mathrm{C}$
  4. $q_1 = 15 \mu \mathrm{C}, q_2 = 30 \mu \mathrm{C}$

Answer: (a)

Solution

Given $q_1' = 6 \times 5 = 30 \, \mu \mathrm{C}$. Finally, $$6 \, V_C + 12 \, V_C = 30 + 0$$ $$18 \, V_C = 30$$ $$V_C = \frac{30}{18} = \frac{5}{3} \, Volt$$ Thus, $$q_1 = \frac{6 \times 5}{3} = 10 \, \mu \mathrm{C}$$ $$q_2 = \frac{12 \times 5}{3} = 20 \, \mu \mathrm{C}$$

Question 40

Physics · System of Particles and Rotational Motion · Single correct

Three equal masses $m$ are kept at vertices $(A, B, C)$ of an equilateral triangle of side $a$ in free space. At $t = 0$, they are given an initial velocity $\vec{V}_A = V_0 \vec{AC}$, $\vec{V}_B = V_0 \vec{BA}$ and $\vec{V}_C = V_0 \vec{CB}$. Here, $\vec{AC}$, $\vec{CB}$ and $\vec{BA}$ are unit vectors along the edges of the triangle. If the three masses interact gravitationally, then the magnitude of the net angular momentum of the system at the point of collision is:

  1. 3amV_0
  2. $\frac{3}{2}$ a $\,$ mV_0
  3. $\frac{\sqrt{3}}{2}$ a $\,$ mV_0
  4. $\frac{1}{2}$ a $\,$ mV_0

Answer: (c)

Solution

Given $d = \frac{a}{2\sqrt{3}}$. Angular momentum of one mass about point $O$ is $L = mvd$. Therefore, $$L = mv_0 \cdot \frac{a}{2\sqrt{3}}.$$ Net angular momentum about point $O$ is $$L_{net} = 3 \cdot 2 \cdot \frac{\sqrt{3}mv_0 a}{2}.$$

Question 41

Physics · Physical World, Units and Measurements · Single correct

Match List - I with List - II \vspace{0.9cm} \[ \begin{array}{clcl} \textbf{List-I} & & \textbf{List-II} & \\ (A) & \text{Young's Modulus} & (I) & M L^{-1} T^{-1}\\ (B) & \text{Torque} & (II) & M L^{-1} T^{-2}\\ (C) & \text{Coefficient of Viscosity} & (III) & M^{-1} L^3 T^{-2}\\ (D) & \text{Gravitational Constant} & (IV) & M L^2 T^{-2} \end{array} \] \vspace{0.3cm} Choose the correct answer from the options given below:

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

Answer: (c)

Solution

(A) $[Y] = \frac{F}{A\left(\frac{\Delta \ell}{\ell}\right)} \Rightarrow \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$ (B) Torque ($\vec{\tau}$) $= \vec{r} \times \vec{F}$ ($\vec{\tau}$) $= L \times MLT^{-2} = ML^2T^{-2}$ (IV) (C) Coefficient of viscosity $\Rightarrow F = \eta A \frac{dV}{dt}$ $\eta \rightarrow \mathrm{Pa} \cdot \mathrm{sec}$ $[\eta] = \frac{MLT^{-2}}{L^2} \times T = ML^{-1}T^{-1}$ (D) Gravitational constant (G) $F = \frac{GM_1M_2}{r^2}$ $[G] = \frac{F \cdot r^2}{m_1 \ m_2} = \frac{MLT^{-2} \times L^2}{M^2} = M^{-1}L^3T^{-2}$

Question 42

Physics · Magnetism and Matter · Single correct

Match List - I with List - II \[ \begin{array}{clcl} \textbf{List-I} & & \textbf{List-II} & \\ (A) & \text{Magnetic induction} & (I) & \text{Ampere meter}^2\\ (B) & \text{Magnetic intensity} & (II) & \text{Weber}\\ (C) & \text{Magnetic flux} & (III) & \text{Gauss}\\ (D) & \text{Magnetic moment} & (IV) & \text{Ampere/meter} \end{array} \]

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

Answer: (d)

Solution

Q6. (A) Magnetic induction $\rightarrow$ Gauss (III) (B) Magnetic intensity $$\left( \mathbf{H} = \frac{\mathbf{B}}{\mu} \right) \rightarrow Ampere / meter (IV)$$ (C) Magnetic flux $\rightarrow$ Weber (Wb) (II) (D) Magnetic moment $\rightarrow$ Ampere-meter^2 $$\left( \vec{\mathbf{M}} = i \vec{\mathbf{A}} \right)$$

Question 43

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

The truth table for the circuit given below is:

  1. \begin{tabular}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ \hline 1 & 0 & 0 \\ \hline 1 & 1 & 0 \\ \hline 0 & 1 & 1 \\ \hline \end{tabular}
  2. \begin{tabular}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ \hline 1 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 0 & 1 & 1 \\ \hline \end{tabular}
  3. \begin{tabular}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ \hline 1 & 0 & 1 \\ \hline 0 & 1 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{tabular}
  4. \begin{tabular}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{tabular}

Answer: (d)

Solution

Question 44

Physics · Thermal Properties of Matter · Single correct

A cup of coffee cools from $90^\circ\mathrm{C}$ to $80^\circ\mathrm{C}$ in $t$ minutes when the room temperature is $20^\circ\mathrm{C}$. The time taken by the similar cup of coffee to cool from $80^\circ\mathrm{C}$ to $60^\circ\mathrm{C}$ at the same room temperature is:

  1. $\frac{13}{10}$ t
  2. $\frac{10}{13}$ t
  3. $\frac{5}{13}$ t
  4. $\frac{13}{5}$ t

Answer: (d)

Solution

By using average form of Newton's law of cooling $$\frac{90 - 80}{t} = k \left( \frac{90 + 80}{2} - 20 \right)$$ $$\frac{80 - 60}{t'} = k \left( \frac{80 + 60}{2} - 20 \right)$$ Dividing (i) by (ii) $$\frac{10 \times t'}{t \times 20} = \frac{65}{50}$$ $$t' = \frac{65}{50} \times 2t = \frac{65}{25} t = \frac{13}{5} t$$

Question 45

Physics · Atoms · Single correct

The number of spectral lines emitted by atomic hydrogen that is in the $4^{th}$ energy level, is

  1. 3
  2. 1
  3. 6
  4. 0

Answer: (c)

Solution

Total possible transition = 6

Question 46

Physics · Moving Charges and Magnetism · Numerical

The magnetic field inside a 200 turns solenoid of radius 10 cm is $2.9 \times 10^{-4}$ Tesla. If the solenoid carries a current of 0.29 A, then the length of the solenoid is ________ $\pi$ cm.

Answer: 8

Solution

Assuming long solenoid $$B = \mu_0 \left( \frac{N}{\ell} \right) i$$ $$\ell = \frac{\mu_0 N i}{B} = \frac{(4 \pi \times 10^{-7}) (200) (0.29)}{2.9 \times 10^{-4}} \ \mathrm{m}$$ $$= 8 \pi \ \mathrm{cm}$$

Question 47

Physics · Electrostatic Potential and Capacitance · Numerical

A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A. If the rate of change of potential difference between the plates is $7 \times 10^8 \, \mathrm{V/s}$ then the integer value of the distance between the parallel plates is (Take, $\epsilon_0 = 9 \times 10^{-12} \, \mathrm{F/m}$, $\pi = \frac{22}{7}$) ________ $\mu \mathrm{m}$.

Answer: 1320

Solution

Given $Q = cV$. $$V = \frac{Q}{c} = \frac{it}{\left( \frac{\varepsilon_0 A}{d} \right)}$$ $$d = \frac{\varepsilon_0 \pi r^2}{i} \left( \frac{v}{t} \right)$$ Putting values $$d = \frac{9 \times 10^{-12} \times \frac{22}{7} \times (0.1)^2 \times (7 \times 10^8)}{0.15}$$ $$= 1320 \, \mu \mathrm{m}$$

Question 48

Physics · Physical World, Units and Measurements · Numerical

A physical quantity $Q$ is related to four observables $a, b, c, d$ as follows: $$Q = \frac{ab^4}{cd}$$ where, $a = (60 \pm 3) \, \mathrm{Pa}$; $b = (20 \pm 0.1) \, \mathrm{m}$; $c = (40 \pm 0.2) \, \mathrm{Nsm^{-2}}$ and $d = (50 \pm 0.1) \, \mathrm{m}$, then the percentage error in $Q$ is $\frac{x}{1000}$, where x = ___.

Answer: 7700

Solution

Given $Q = \frac{ab^4}{cd}$. Therefore, $$\frac{\Delta Q}{Q} \times 100 = \left[ \frac{\Delta a}{a} + 4 \frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d} \right] \times 100$$ Thus, $$\frac{x}{1000} = \left[ \frac{3}{60} + 4 \left( \frac{0.1}{20} \right) + \frac{0.1}{50} \right] \times 100$$ Therefore, $x = 7700$.

Question 49

Physics · Gravitation · Numerical

Two planets, A and B are orbiting a common star in circular orbits of radii $R_A$ and $R_B$, respectively, with $R_B = 2R_A$. The planet B is $4\sqrt{2}$ times more massive than planet A. The ratio $\left( \frac{L_B}{L_A} \right)$ of angular momentum $(L_B)$ of planet B to that of planet A $(L_A)$ is closest to integer ______.

Answer: 8

Solution

Given $L = mv_0 R = m \sqrt{\frac{GM}{R}} R = m \sqrt{GMR}$. Here $M$ is the mass of the star. $$\frac{L_B}{L_A} = \frac{m_B}{m_A} \sqrt{\frac{R_B}{R_A}}$$ $$= 4 \sqrt{2} \sqrt{\frac{2}{1}}$$ $$\frac{L_B}{L_A} = 8$$

Question 50

Physics · Motion in a Straight Line · Numerical

Two cars $P$ and $Q$ are moving on a road in the same direction. Acceleration of car $P$ increases linearly with time whereas car $Q$ moves with a constant acceleration. Both cars cross each other at time $t = 0$, for the first time. The maximum possible number of crossing(s) (including the crossing at $t = 0$) is

Answer: 3

Solution

Given $a_P = kt$, $k$ is constant and $a_Q = a$, $a$ is constant. $a_{QP} = a_Q - a_P = a - kt$. As initial velocities are not mentioned in the question, we will have to assume two cases. Case-I: $u_{QP}$ and $a_{QP}$ in the same direction. Total number of crossings $= 2$. Case-II: $u_{QP}$ and $a_{QP}$ in opposite direction. Total number of crossings $= 3$.

Chemistry

Question 51

Chemistry · Co-ordination Compounds · Single correct

The calculated spin-only magnetic moments of $K_3 [\mathrm{Fe(OH)}_6]$ and $K_4 [\mathrm{Fe(OH)}_6]$ respectively are:

  1. 3.87 and 4.90 B.M.
  2. 4.90 and 5.92 B.M.
  3. 4.90 and 4.90 B.M.
  4. 5.92 and 4.90 B.M.

Answer: (d)

Solution

For $\mathrm{K_3[Fe(OH)_6]}$, $\mathrm{Fe^{3+} \Rightarrow 3\, d^5}$. $\mathrm{Fe^{3+}}$ with $\mathrm{OH^-}$ (WFL) is $t_{2g}^3 e_g^2$. The number of unpaired electrons $(n) = 5$. The spin-only magnetic moment $\mu_{spin only} = 5.92\, \mathrm{BM}$. For $\mathrm{K_4[Fe(OH)_6]}$, $\mathrm{Fe^{2+} \Rightarrow OH^-}$ WFL. $\mathrm{Fe^{2+} \Rightarrow 3\, d^6 = t_{2g}^4 e_g^2}$. $n = 4$. The spin-only magnetic moment $\mu_{spin only} = 4.90\, \mathrm{BM}$.

Question 52

Chemistry · Structure of Atom · Single correct

For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot for a constant n ? [E: Energy of the stationary state, Z : atomic number, n = principal quantum number]

Answer: (d)

Solution

The energy level $E_n$ is given by the formula $$E_n = -13.6 \frac{z^2}{n^2}$$ where $z$ is the atomic number and $n$ is the principal quantum number. This implies that $$E_n \propto -z^2.$$ The equation of the parabola is given by $$y = kx^2.$$

Question 53

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

Given below are two statements: Statement (I): In partition chromatography, stationary phase is thin film of liquid present in the inert support. Statement (II): In paper chromatography, the material of paper acts as a stationary phase. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (a)

Solution

Statement I is true. In partition chromatography, stationary phase is thin liquid film present in the inert support. Statement II is false. Because stationary phase in paper chromatography is water.

Question 54

Chemistry · Biomolecules · Single correct

Identify the essential amino acids from below: (A) Valine (B) Proline (C) Lysine (D) Threonine (E) Tyrosine Choose the correct answer from the options given below:

  1. , (C) and (E) only
  2. , (C) and (D) only
  3. , (D) and (E) only
  4. , (C) and (E) only

Answer: (b)

Solution

Valine, Lysine and Threonine are essential amino acids.

Question 55

Chemistry · Hydrocarbons · Single correct

Which among the following halides will generate the most stable carbocation in the nucleophilic substitution reaction?

Answer: (c)

Solution

Question 56

Chemistry · Equilibrium · Single correct

Consider the equilibrium CO($\,$ $\mathrm{g}$) + 3$\mathrm{H}$_2($\,$ $\mathrm{g}$) $\rightleftharpoons$ $\mathrm{CH}$_4($\,$ $\mathrm{g}$) + $\mathrm{H}$_2$\mathrm{O}$($\,$ $\mathrm{g}$) If the pressure applied over the system increases by two fold at constant temperature then

  1. , (B) and (C) only
  2. and (B) only
  3. , (B) and (D) only
  4. and (D) only

Answer: (b)

Solution

Given the reaction: $$\mathrm{CO(g) + 3H_2 \rightleftharpoons CH_4(g) + H_2O(g)}$$ $$\Delta n_g = -2$$ If pressure of system increases then according to Le-Chatelier's principle, the reaction will move in the forward direction. Concentration of reactant and products both increases but concentration of product increases more.

Question 57

Chemistry · Solutions · Single correct

Given below are two statements: Statement (I): NaCl is added to the ice at $0^\circ \mathrm{C}$, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at $0^\circ \mathrm{C}$, there is a depression in freezing point. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (d)

Solution

A mixture of salt and ice is known as freezing mixture. Freezing mixture decreases freezing point of ice. Both statements are true.

Question 58

Chemistry · Hydrocarbons · Single correct

Given below are two statements : Statement (I) : On nitration of m-xylene with $\mathrm{HNO_3}$, $\mathrm{H_2SO_4}$ followed by oxidation, 4-nitrobenzene-1,3-dicarboxylic acid is obtained as the major product. Statement (II) : $- \mathrm{CH_3}$ group is o/p-directing while $- \mathrm{NO_2}$ group is m-directing group. In the light of the above statements, choose the correct answer from the options given below :

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

Answer: (d)

Solution

Question 59

Chemistry · Redox Reactions · Multiple correct

0.1 $\,$ $\mathrm{M}$ solution of $\mathrm{KI}$ reacts with excess of $\mathrm{H_2SO_4}$ and $\mathrm{KIO_3}$ solutions. According to equation 5$\mathrm{I^-}$ + $\mathrm{IO_3^-}$ + 6$\mathrm{H^+}$ $\rightarrow$ 3$\mathrm{I_2}$ + 3$\mathrm{H_2O}$ Identify the correct statements:

  1. 200 mL of KI solution reacts with 0.004 mol of KIO$_3$
  2. 200 mL of KI solution reacts with 0.006 mol of H$_2$SO$_4$
  3. 0.5 L of KI solution produced 0.005 mol of I$_2$
  4. Equivalent weight of KIO$_3$ is equal to $\left( \frac{Molecular weight}{5} \right)$

Answer: (a)

Solution

Given $E_{\mathrm{KIO_3}} = \frac{Molecular weight}{n_f}$. $n_{fff} = 5$. $E_{\mathrm{KIO_3}} = \frac{Molecular weight}{5}$. (D) is correct. meq of KI $= 0.1 \times 200 = 20$. meq of KIO$_3 = 4 \times 5 = 20$. (A) is correct.

Question 60

Chemistry · General Principles and Processes of Isolation of Elements · Single correct

Match List-I with List-II: \begin{tabular}{|c|l|c|l|} \hline & \textbf{List-I} & & \textbf{List-II} \\ & \textbf{Applications} & & \textbf{Batteries/Cell} \\ \hline (A) & Transistors & (I) & Anode -- Zn/Hg; Cathode -- HgO + C \\ \hline (B) & Hearing aids & (II) & Hydrogen fuel cell \\ \hline (C) & Inverters & (III) & Anode -- Zn; Cathode -- Carbon \\ \hline (D) & Apollo space ship & (IV) & Anode -- Pb; Cathode -- PbO$_2$ \\ \hline \end{tabular} Choose the correct answer from the options given below:

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

Answer: (c)

Solution

In transistor, anode is Zn and cathode is carbon. In hearing aids, mercury battery are used. In invertors, lead storage battery is used. In apollo space ship, hydrogen fuel cell was used.

Question 61

Chemistry · Electrochemistry · Single correct

O$_2$ gas will be evolved as a product of electrolysis of: (A) an aqueous solution of AgNO$_3$ using silver electrodes. (B) an aqueous solution of AgNO$_3$ using platinum electrodes. (C) a dilute solution of H$_2$SO$_4$ using platinum electrodes. (D) a high concentration solution of H$_2$SO$_4$ using platinum electrodes. Choose the correct answer from the options given below:

  1. and (C) only
  2. and (C) only
  3. and (D) only
  4. and (D) only

Answer: (b)

Solution

When an aqueous solution of $\mathrm{AgNO_3}$ is electrolysed using Pt electrodes. Cathode: $\mathrm{Ag^+ (aq) + e^- \rightarrow Ag(s)}$ Anode: $2\mathrm{H_2O(l)} \rightleftharpoons 4\mathrm{H^+ (aq)} + \mathrm{O_2(g)} + 4e^-$ When dilute $\mathrm{H_2SO_4}$ is electrolysed using Pt electrodes. Anode: $2\mathrm{H_2O(l)} \rightarrow \mathrm{O_2(g)} + 4\mathrm{H^+ (aq)} + 4e^-$ Cathode: $2\mathrm{H^+ (aq)} + 2e^- \rightarrow \mathrm{H_2(g)}$

Question 62

Chemistry · Co-ordination Compounds · Single correct

Identify the homoleptic complexes with odd number of d electrons in the central metal: (A) $[\mathrm{FeO}_4]^{2-}$ (B) $[\mathrm{Fe(CN)}_6]^{3-}$ (C) $[\mathrm{Fe(CN)}_5\mathrm{NO}]^{2-}$ (D) $[\mathrm{CoCl}_4]^{2-}$ (E) $[\mathrm{Co(H}_2\mathrm{O)}_3\mathrm{F}_3]$ Choose the correct answer from the options given below:

  1. , (B) and (D) only
  2. and (E) only
  3. and (D) only
  4. , (C) and (E) only

Answer: (c)

Solution

(A) $[\mathrm{FeO}_4]^{2-} \Rightarrow \mathrm{Fe}^{6+} = 3\, d^2$ (B) $[\mathrm{Fe(CN)}_8]^{3-} \Rightarrow \mathrm{Fe}^{3+} = 3\, d^5$ (C) $[\mathrm{Fe(CN)}_5\mathrm{NO}]^{2-} \Rightarrow \mathrm{Fe}^{2+} = 3\, d^6$ (D) $[\mathrm{CoCl}_4]^{2-} \Rightarrow \mathrm{Co}^{2+} = 3\, d^7$ (E) $[\mathrm{Co(H_2O)}_3\, \mathrm{F}_3] \Rightarrow \mathrm{Co}^{3+} = 3\, d^6$ (B) and (D) are homoleptic complex having odd no. of d electrons.

Question 63

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Total number of sigma ($\sigma$) ____ and pi($\pi$) _____ bonds respectively present in hex-1-en-4-yne are:

  1. 3 and 13
  2. 11 and 3
  3. 13 and 3
  4. 14 and 3

Answer: (c)

Solution

The number of $\sigma$ bonds is $13$ and the number of $\pi$ bonds is $3$.

Question 64

Chemistry · Thermodynamics · Single correct

If $C$ (diamond) $\rightarrow C$ (graphite) $+ X \ \mathrm{kJ \ mol}^{-1}$ $C$ (diamond) $+ O_2$ (g) $\rightarrow CO_2$ (g) $+ Y \ \mathrm{kJ \ mol}^{-1}$ $C$ (graphite) $+ O_2$ (g) $\rightarrow CO_2$ (g) $+ Z \ \mathrm{kJ \ mol}^{-1}$ at constant temperature. Then

  1. X = -Y + Z
  2. -X = Y + Z
  3. X = Y + Z
  4. X = Y - Z

Answer: (d)

Solution

The given reactions are: $$\mathrm{C(diamond) + O_2(g) \rightarrow CO_2(g)}; \Delta H_1 = -Y \, \mathrm{kJ \, mol^{-1}}$$ $$\mathrm{CO_2(g) \rightarrow C(graphite) + O_2(g)}; \Delta H_2 = Z \, \mathrm{kJ \, mol^{-1}}$$ By adding these reactions, we get: $$\mathrm{C(diamond) \rightarrow C(graphite)}; \Delta H_3 = -Y + Z$$ Thus, $$-X = -Y + Z$$ Therefore, $$X = Y - Z$$

Question 65

Chemistry · Structure of Atom · Single correct

Given below are two statements: Statement (I): It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle. Statement (II): If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron, then the uncertainty in the measurement of velocity is $\geq \sqrt{\frac{h}{\pi}} \times \frac{1}{2 \, m}$. In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (b)

Solution

According to Heisenberg's uncertainty principle, it is impossible to determine simultaneously the exact position and momentum of a particle like an electron. If $$\Delta p = \Delta x$$ $$\Delta p \cdot \Delta x \geq \frac{h}{4\pi}$$ $$(\Delta p)^2 \geq \frac{h}{4\pi}$$ $$\Delta p \geq \sqrt{\frac{h}{\pi}} \times \frac{1}{2}$$ $$m \Delta v \geq \sqrt{\frac{h}{\pi}} \times \frac{1}{2}$$ $$\Delta v \geq \sqrt{\frac{h}{\pi}} \times \frac{1}{2m}$$

Question 66

Chemistry · Hydrocarbons · Single correct

Which one of the following reaction sequences will give an azo dye?

Answer: (a)

Solution

Question 67

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

Drug X becomes ineffective after $50\%$ decomposition. The original concentration of the drug in a bottle was $16\,\mathrm{mg\,mL^{-1}}$, which becomes $4\,\mathrm{mg\,mL^{-1}}$ in $12$ months. Assuming that the decomposition of the drug follows first-order kinetics, the expiry time of the drug is $\underline{\hspace{1cm}}$ months.

  1. 2
  2. 6
  3. 12
  4. 3

Answer: (b)

Solution

The reaction is first order. The initial concentration of the drug is $16 \, \mathrm{mg/mL}$. The concentration of the drug after $12$ months is $4 \, \mathrm{mg/mL}$. The half-life of the drug is $6$ months. The drug becomes ineffective after $50\%$ decomposition. The expiry time of the drug is $6$ months.

Question 68

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

The type of oxide formed by the element among Li, Na, Be, Mg, B and Al that has the least atomic radius is :

  1. A_2O
  2. A_2O_3
  3. AO_2
  4. AO

Answer: (b)

Solution

Among given atoms, Boron has least atomic radius. Oxide of Boron is $\mathrm{B_2O_3}$.

Question 69

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

First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family:

  1. $1402\,\mathrm{kJ\,mol^{-1}}$
  2. $834\,\mathrm{kJ\,mol^{-1}}$
  3. $1012\,\mathrm{kJ\,mol^{-1}}$
  4. $947\,\mathrm{kJ\,mol^{-1}}$

Answer: (c)

Solution

Q11. The main component of apatite family is phosphorus. (3) Order of $\mathrm{IE}_1$ of group 15 elements $\mathrm{N} > \mathrm{P} > \mathrm{As} > \mathrm{Sb}$. IE of phosphorus $= 1012 \, \mathrm{kJ \, mol^{-1}}$.

Question 70

Chemistry · Alcohols, Phenols and Ethers · Single correct

Which one of the following, with HBr will give a phenol?

Answer: (c)

Solution

The reaction involves the treatment of anisole with HBr. The first step is the formation of a methyl bromide ion and a phenol ion. The reaction proceeds via an $S_N2$ mechanism where the bromide ion attacks the methyl group, resulting in the formation of phenol and methyl bromide.

Question 71

Chemistry · Co-ordination Compounds · Numerical

Consider the following low-spin complexes $\mathrm{K_3[Co(NO_2)_6]}$, $\mathrm{K_4[Fe(CN)_6]}$, $\mathrm{K_3[Fe(CN)_6]}$, $\mathrm{Cu_2[Fe(CN)_6]}$ and $\mathrm{Zn_2[Fe(CN)_6]}$ The sum of the spin-only magnetic moment values of complexes having yellow colour is. $\_$$\_$$\_$$\_$ B.M. (answer in nearest integer)

Answer: 0

Solution

For $\mathrm{K_3[Co(NO_2)_6]}$, the color is Yellow, which implies $\mathrm{Co^{3+}} = 3d^6 \Rightarrow t_{2g}^6 e_g^0$. For $\mathrm{K_4[Fe(CN)_6]}$, the color is Yellow, which implies $\mathrm{Fe^{2+}} = 3d^6 \Rightarrow t_{2g}^6 e_g^0$. For $\mathrm{K_3[Fe(CN)_6]}$, the color is Bright Red, which implies $\mathrm{Fe^{3+}} = 3d^5 \Rightarrow t_{2g}^5 e_g^0$. For $\mathrm{Cu_2[Fe(CN)_6]}$, the color is Chocolate brown. For $\mathrm{Zn_2[Fe(CN)_6]}$, the color is White. Spin only magnetic moment of complex having Yellow colour is zero.

Question 72

Chemistry · Hydrocarbons · Numerical

Isomeric hydrocarbons $\rightarrow$ negative Baeyer's test (Molecular formula $\mathrm{C}_9\mathrm{H}_{12}$) The total number of isomers from above with four different non-aliphatic substitution sites is -

Answer: 2

Solution

Degree of unsaturation = C + 1 - $\frac{H}{2}$ = 9 + 1 - 6 = 4. Benzene shows negative Baeyer's test. Both compounds have four different non-aliphatic substitution sites.

Question 73

Chemistry · Alcohols, Phenols and Ethers · Numerical

In the Claisen-Schmidt reaction to prepare, dibenzalacetone from $5.3 \, \mathrm{g}$ of benzaldehyde, a total of $3.51 \, \mathrm{g}$ of product was obtained. The percentage yield in this reaction was ______ $\%$.

Answer: 60

Solution

Benzaldehyde is given as 5.3 gm. The moles of benzaldehyde are calculated as $$\frac{5.3}{106} = \frac{1}{20} Mol$$. The product, dibenzalacetone, has a mass of 3.51 gm. The moles of dibenzalacetone are calculated as $$\frac{3.51}{234} = 0.015 Mol$$ (Actual). The theoretical yield is $$\frac{1}{40} Mol$$. The percentage yield is calculated as $$\frac{0.015}{1/40} \times 100$$ which gives $$60\%$$.

Question 74

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

In the sulphur estimation, 0.20 $\,$ $\mathrm{g}$ of a pure organic compound gave 0.40 $\,$ $\mathrm{g}$ of barium sulphate. The percentage of sulphur in the compound is $\times$ 10^{-1}$\%$. (Molar mass: O = 16, S = 32, Ba = 137 in gmol^{-1})

Answer: 275

Solution

Organic Compound $\rightarrow \mathrm{BaSO_4}$ $0.20 \, \mathrm{gm}$ $0.40 \, \mathrm{gm}$ $\frac{0.40}{233} \, \mathrm{mol} \, (\mathrm{BaSO_4})$ $\frac{0.40}{233} \, \mathrm{mol} \, (\mathrm{Sulphur})$ $\frac{0.40}{233} \times 32 \, \mathrm{gm} \, (\mathrm{sulphur})$ $$\% \, \mathrm{S} = \frac{0.40 \times 32}{233 \times 0.20} \times 100 = 27.5\% or 275 \times 10^{-1}\%$$

Question 75

Chemistry · Chemical Bonding and Molecular Structure · Fill in the blank

Total number of non bonded electrons present in $\mathrm{NO_2^-}$ ion based on Lewis theory is

Answer: 3

Solution

Number of non-bonding electrons = 12