JEE Main 2 April 2025 Shift 2 question paper with solutions

JEE Main 2 April 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 · Three Dimensional Geometry · Single correct

If the image of the point P(1, 0, 3) in the line joining the points A(4, 7, 1) and B(3, 5, 3) is Q($\alpha$, $\beta$, $\gamma$), then $\($ $\alpha$ + $\beta$ + $\gamma$ $\)$ is equal to

  1. $\frac{47}{3}$
  2. $\frac{46}{3}$
  3. 18
  4. 13

Answer: (b)

Solution

Given points $P(1, 0, 3)$, $A(4, 7, 1)$, and $B(3, 5, 3)$. The line $AB$ is given by $$\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 3}{-2} = \lambda.$$ Let the foot of the perpendicular from $P$ on $AB$ be $R$. Then $$R \equiv (\lambda + 3, 2\lambda + 5, -2\lambda + 3).$$ Solving $$(\lambda + 3 - 1)(1) + (2\lambda + 5 - 0)(2) + (-2\lambda + 3 - 3)(-2) = 0,$$ we get $$\lambda + 2 + 4\lambda + 10 + 4\lambda = 0.$$ Thus, $$\lambda = -\frac{4}{3}.$$ Therefore, $$R \equiv \left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right).$$ Now, $$Q \equiv \left(\frac{10}{3}, \frac{14}{3}, \frac{34}{3}\right) \equiv \left(\frac{7}{3}, \frac{14}{3}, \frac{25}{3}\right).$$ Finally, $$\alpha + \beta + \gamma = \frac{7 + 14 + 25}{3} = \frac{46}{3}.$$

Question 2

Maths · Differential Equations · Single correct

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function, If $10 \int_1^x f(t) \, dt = 5x f(x) - x^5 - 9$ for all $x \geq 1$, then the value of $f(3)$ is:

  1. 18
  2. 32
  3. 22
  4. 26

Answer: (b)

Solution

Given $10 \frac{d}{dx} \int_1^x f(t) dt = \frac{d}{dx} (5xf(x) - x^5 - 9)$. This implies $10f(x) = 5f(x) + 5xf'(x) - 5x^4$. Therefore, $f(x) + x^4 = xf'(x)$. This can be rewritten as $y + x^4 = x \frac{dy}{dx}$. Rearranging gives $\frac{dy}{dx} + y \left( -\frac{1}{x} \right) = x^3$. Multiplying by the integrating factor $e^{-\frac{1}{x} dx}$, we have $ye^{-\frac{1}{x} dx} = \int x^3 e^{-\frac{1}{x} dx} + c$. Simplifying, $ye^{-\ln |x|} = \int x^3 e^{-\ln |x|} + c$. Thus, $\frac{y}{|x|} = \int \frac{x^3}{|x|} + c$. This simplifies to $\frac{y}{x} = \int x^2 + c$. Therefore, $\frac{y}{x} = \frac{x^3}{3} + c$. Putting $x = 1$ in the given equation, $0 = 5f(1) - 1 - 9 \Rightarrow f(1) = 2$. Solving $2 = \frac{1}{3} + c \Rightarrow c = \frac{5}{3}$. Finally, $f(3) = \frac{27}{3} + \frac{5}{3} = 32$.

Question 3

Maths · Sequences and Series · Single correct

The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $21/2$. Then the number of terms which are integers in the A.P. is:

  1. 4
  2. 10
  3. 6
  4. 8

Answer: (a)

Solution

Given $$a_2 + a_4 + \ldots + a_n = 30 \ldots (1)$$ $$(1) a_1 + a_3 + \ldots + a_{n-1} = 24 \ldots (2)$$ Subtracting (2) from (1): $$(a_2 - a_1) + (a_4 - a_3) \ldots (a_n - a_{n-1}) = 6$$ This implies $$\frac{n}{2} d = 6 \Rightarrow nd = 12$$ Also, $$a_n - a_1 = (n-1)d = \frac{21}{2}$$ Thus, $$\Rightarrow nd - d = \frac{21}{2} \Rightarrow 12 - \frac{21}{2} = d$$ Therefore, $$d = \frac{3}{2}, \; n = 8$$ The sum of odd terms is $$\frac{4}{2} [2a + (4-1)3] = 24$$ This implies $$a = \frac{3}{2}$$ The arithmetic progression is $$\frac{3}{2}, 3, \frac{9}{2}, 6, \frac{15}{2}, 9, \frac{21}{2}, 12$$ The number of integer terms is 4.

Question 4

Maths · Sets · Single correct

Let $A = \{1, 2, 3, \ldots, 10\}$ and $R$ be a relation on $A$ such that $R = \{(a, b) : a = 2b + 1\}$. Let $(a_1, a_2), (a_2, a_3), (a_3, a_4), \ldots, (a_k, a_{k+1})$ be a sequence of $k$ elements of $R$ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer $k$, for which such a sequence exists, is equal to:

  1. 6
  2. 7
  3. 5
  4. 8

Answer: (c)

Solution

Given $a = 2b + 1$ and $2b = a - 1$. Let $R = \{(3, 1), (5, 2), \ldots, (99, 49)\}$. Let $(2m + 1, m), (2\lambda - 1, \lambda)$ are such ordered pairs. According to the condition $m = 2\lambda - 1 \Rightarrow m =$ odd number. Therefore, the first element of ordered pair $(a, b)$ is $a = 2(2\lambda - 1) + 1 = 4\lambda - 1$. Hence $a \in \{3, 7, \ldots, 99\}$ which implies $\lambda \in \{1, 2, \ldots, 25\}$. Thus, the set of sequence is $$\left\{(4\lambda - 1, 2\lambda - 1), (2\lambda - 1, \lambda - 1), \left(\lambda - 1, \frac{\lambda - 2}{2}\right), \ldots \right\}$$ The second element of each ordered pair is $$\frac{\lambda - 2^{r-2}}{2^{r-2}}$$ For maximum number of ordered pairs in such sequence, $$\frac{\lambda - 2^{r-2}}{2^{r-2}} = 1 or 2; 1 \leq \lambda \leq 25$$ $$\lambda = 2^{r-1} or \lambda = 3 \cdot 2^{r-2}$$ Case-I: $\lambda = 2r - 1$ $$\lambda = 2, 2^2, 2^3, 2^4$$ $$r = 2, 3, 4, 5$$ Hence maximum value of $r$ is 5 when $\lambda = 16$. Case-II: $\lambda = 3 \cdot 2^{r-2}$ $$\lambda = 3, 6, 12, 24$$ $$r = 2, 3, 4, 5$$ Hence maximum value of $r$ is 5 when $\lambda = 24$.

Question 5

Maths · Conic Sections · Single correct

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :

  1. $\frac{4}{\sqrt{17}}$
  2. $\frac{\sqrt{3}}{16}$
  3. $\frac{3}{\sqrt{19}}$
  4. $\frac{\sqrt{5}}{7}$

Answer: (a)

Solution

Given $2b = \frac{1}{4}(2ae)$. We have $\frac{b}{a} = \frac{e}{4}$. Then $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting, $e = \sqrt{1 - \frac{e^2}{16}}$. This gives $e^2 \left(1 + \frac{1}{16}\right) = 1$. Solving for $e$, we find $e = \frac{4}{\sqrt{17}}$.

Question 6

Maths · Three Dimensional Geometry · Single correct

The line $L_1$ is parallel to the vector $\vec{a} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ and passes through the point $(7, 6, 2)$ and the line $L_2$ is parallel to the vector $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is:

  1. $\frac{23}{\sqrt{38}}$
  2. $\frac{21}{\sqrt{57}}$
  3. $\frac{23}{\sqrt{57}}$
  4. $\frac{21}{\sqrt{38}}$

Answer: (a)

Solution

Given the lines $$L_1 : (7\hat{i} + 6\hat{j} + 2\hat{k}) + \lambda (-3\hat{i} + 2\hat{j} + 4\hat{k})$$ $$L_2 : (5\hat{i} + 3\hat{j} + 4\hat{k}) + \lambda (2\hat{i} + \hat{j} + 3\hat{k})$$ The distance between skew lines is given by $$(2\hat{i} + 3\hat{j} - 2\hat{k}) \cdot (2\hat{i} + 17\hat{j} - 7\hat{k})$$ This simplifies to $$= \frac{69}{\sqrt{342}} = \frac{69}{3\sqrt{38}} = \frac{23}{\sqrt{38}}$$

Question 7

Maths · Integrals · Single correct

Let (a, b) be the point of intersection of the curve $x^2 = 2y$ and the straight line $y - 2x - 6 = 0$ in the second quadrant. Then the integral $I = \int_a^b \frac{9x^2}{1+5^x} \, dx$ is equal to:

  1. 24
  2. 27
  3. 18
  4. 21

Answer: (a)

Solution

Given $x^2 = 2y$ and $y = 2x + 6$. (1) $x^2 = 4x + 12$ Solving $x^2 - 4x - 12 = 0$ gives $x = 6$, $y = 18$ and if $x = -2$, $y = 2$. Therefore, the points are $(6, 18)$ and $(-2, 2)$. Here $(6, 18)$ is rejected because $(a, b)$ lies in the second quadrant. Thus, $a = -2$ and $b = 2$. Therefore, $$I = \int_{-2}^{2} \frac{9x^2}{1 + 5^x} \, dx = \int_{-2}^{2} \frac{9 \cdot 5^x \cdot x^2}{1 + 5^x} \, dx$$ Thus, $$2I = \int_{-2}^{2} 9x^2 \, dx = 18 \int_{0}^{2} x^2 \, dx = 18 \left( \frac{x^3}{3} \right)_0^2$$ Calculating gives $2I = 48$. Therefore, $I = 24$.

Question 8

Maths · Determinants · Single correct

If the system of equation $$2x + \lambda y + 3z = 5$$ $$3x + 2y - z = 7$$ $$4x + 5y + \mu z = 9$$ has infinitely many solutions, then $\left( \lambda^2 + \mu^2 \right)$ is equal to :

  1. \quad 22
  2. \quad 18
  3. \quad 26
  4. \quad 30

Answer: (c)

Solution

Given $$\Delta = 0 \implies \begin{vmatrix} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{vmatrix} = 0$$ This implies $$2(2\mu + 5) + \lambda(-4 - 3\mu) + 3(7) = 0$$ Simplifying, we get $$4\mu - 3\lambda\mu - 4\lambda + 31 = 0 \ldots (1)$$ Now, $$\Delta_3 = 0 \implies \begin{vmatrix} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{vmatrix} = 0$$ This implies $$2(-17) + \lambda(1) + 5(7) = 0$$ Solving, we find $$\lambda = -1$$ From equation (1), $$4\mu + 3\mu + 4 + 31 = 0 \implies \mu = -5$$ Therefore, $$\lambda^2 + \mu^2 = 26$$

Question 9

Maths · Trigonometric Functions · Single correct

If $\theta\in\left[-\frac{7\pi}{6},\frac{4\pi}{3}\right]$, then the number of solutions of $\sqrt{3}\operatorname{cosec}^2\theta-2(\sqrt{3}-1)\operatorname{cosec}\theta-4=0$ is equal to

  1. 6
  2. 8
  3. 10
  4. 7

Answer: (a)

Solution

Given $$\csc \theta = \frac{2(\sqrt{3} - 1) \pm \sqrt{4(3 + 1 - 2\sqrt{3}) + 16\sqrt{3}}}{2\sqrt{3}}$$ Simplifying, we have $$= \frac{2(\sqrt{3} - 1) \pm \sqrt{16 + 8\sqrt{3}}}{2\sqrt{3}}$$ Further simplifying, $$= \frac{2(\sqrt{3} - 1) \pm (2 + 2\sqrt{3})}{2\sqrt{3}}$$ Thus, $$\csc \theta = 2 or \frac{-2}{\sqrt{3}}$$ Therefore, $$\sin \theta = \frac{1}{2} or \frac{-\sqrt{3}}{2}$$ Since $$\sin \theta = \frac{1}{2}$$ has 3 solutions and also $$\sin \theta = \frac{-\sqrt{3}}{2}$$ has 3 solutions in $$\left[ -\frac{7\pi}{6}, \frac{4\pi}{3} \right]$$

Question 10

Maths · Probability · Single correct

"Given three indentical bags each containing 10 balls, whose colours are as follows: \[ \begin{array}{c|ccc} & \text{Red} & \text{Blue} & \text{Green} \\ \hline \text{Bag I} & 3 & 2 & 5 \\ \text{Bag II} & 4 & 3 & 3 \\ \text{Bag III} & 5 & 1 & 4 \end{array} \] A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is $p$ and if the ball is Green, the probability that it is from bag III is $q$, then the value of ( $\frac{1}{p}$ + $\frac{1}{q}$) is:"

  1. 6
  2. 9
  3. 7
  4. 8

Answer: (c)

Solution

Let $\($ p = P $\left$( B_I $\mid$ R $\right$) $\)$. Then, $\($ p = $\frac{1}{3}$ $\left$( $\frac{3}{10}$ $\right$) = $\frac{1}{4}$ $\)$. Let $\($ q = $\left$( B_{III} $\mid$ G $\right$) $\)$. Then, $\($ q = $\frac{1}{3}$ $\left$( $\frac{5}{10}$ + $\frac{3}{10}$ + $\frac{4}{10}$ $\right$) = $\frac{1}{3}$ $\)$. Therefore, $\($ $\frac{1}{p}$ + $\frac{1}{q}$ = 7 $\)$.

Question 11

Maths · Statistics · Single correct

If the mean and the variance of 6, 4, $a$, 8, $b$, 12, 10, 13 are 9 and 9.25 respectively, then $a + b + ab$ is equal to:

  1. 105
  2. 103
  3. 100
  4. 106

Answer: (b)

Solution

Given mean $= 9$. Therefore, $53 + a + b = 72$. This implies $a + b = 19$. Therefore, $\sigma^2 = \frac{37}{4}$ and $\overline{(X)}^2 + \sigma^2 = \frac{\sum x_1^2}{N}$. Therefore, $81 + \frac{37}{4} = \frac{529 + a^2 + b^2}{8}$. This implies $648 + 74 = 529 + a^2 + b^2$. Therefore, $a^2 + b^2 = 193$. Since $a + b = 19$, it follows that $a^2 + b^2 + 2ab = 361$. Therefore, $2ab = 168$. This implies $ab = 84$. Therefore, $a + b + ab = 103$.

Question 12

Maths · Relations and Functions · Single correct

If the domain of the function $f(x) = \frac{1}{\sqrt{10 + 3x - x^2}} + \frac{1}{\sqrt{x + |x|}}$ is $(a, b)$, then $(1 + a)^2 + b^2$ is equal to :

  1. 26
  2. 29
  3. 25
  4. 30

Answer: (a)

Solution

Given $x + |x| > 0$ which implies $x \in (0, \infty)$ ...(1) (1) $10 + 3x - x^2 > 0$ This implies $x^2 - 3x - 10 < 0$ Thus, $x \in (-2, 5)$ ...(2) From (1) and (2), $x \in (0, 5)$ Therefore, $a = 0$ and $b = 5$ Thus, $(1 + a^2) + b^2 = 1 + 25 = 26$

Question 13

Maths · Integrals · Single correct

$4\displaystyle\int_0^1\left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right)\,dx-3\log_e(\sqrt{3})$ is equal to:

  1. 2 + \sqrt{2} + \log_e(1+\sqrt{2})
  2. 2 - \sqrt{2} - \log_e(1+\sqrt{2})
  3. 2 + \sqrt{2} - \log_e(1+\sqrt{2})
  4. 2 - \sqrt{2} - \log_e(1+\sqrt{2})

Answer: (b)

Solution

Given $$4 \int_0^1 \frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}} \, dx - 3 \ln \sqrt{3}$$ This is equal to $$4 \int_0^1 \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{(3 + x^2) - (1 - x^2)} \, dx - \frac{3}{2} \ln 3$$ This simplifies to $$2 \left[ \left\{ \frac{x}{2} \sqrt{3 + x^2} + \frac{3}{2} \ln \left( x + \sqrt{3 + x^2} \right) \right\}_0^1 \right.$$ Subtract $$\left. \left\{ \frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \ln \left( x + \sqrt{1 + x^2} \right) \right\}_0^1 \right] - \frac{3}{2} \ln 3$$ This evaluates to $$= 2 \left[ \left\{ \frac{1}{2} \sqrt{4} + \frac{3}{2} \ln (1 + \sqrt{4}) \right\} - \left\{ 0 + \frac{3}{2} \ln \sqrt{3} \right\} \right.$$ Subtract $$\left. \left\{ \frac{1}{2} \sqrt{2} + \frac{1}{2} \ln (1 + \sqrt{2}) \right\} + \left\{ 0 + \frac{1}{2} (0) \right\} \right] - \frac{3}{2} \ln 3$$ This simplifies to $$= 2 \left[ 1 + \frac{3}{2} \ln 3 - \frac{3}{4} \ln 3 - \frac{1}{\sqrt{2}} - \frac{1}{2} \ln (1 + \sqrt{2}) \right] - \frac{3}{2} \ln 3$$ Finally, $$= 2 + 3 \ln 3 - \frac{3}{2} \ln 3 - \sqrt{2} - \ln (1 + \sqrt{2}) - \frac{3}{2} \ln 3$$ This results in $$= 2 - \sqrt{2} - \ln (1 + \sqrt{2})$$

Question 14

Maths · Limits and Derivatives · Single correct

If $\lim_{x \to 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4}$ is finite, then $(a + b)$ is equal to:

  1. $\frac{1}{2}$
  2. 0
  3. $\frac{3}{4}$
  4. $-1$

Answer: (a)

Solution

Given $\lim_{x \to 0} \frac{\cos 2x + a \cos 4x - b}{x^4} = finite$. $$L = \left\{ 1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{4 \cdot \cdots} \right\} + a \left\{ 1 - \frac{(4x)^2}{2} + \frac{(4x)^4}{4 \cdot \cdots} \right\} - b$$ $$L = \frac{x^4}{x^4} \left( (1 + a - b) - x^2 (2 + 8a) + x^4 \left( \frac{2}{3} + \frac{32}{3} a \right) + x^6 (\cdots) \right)$$ Therefore, $1 + a - b = 0$ and $2 + 8a = 0 \Rightarrow a = -\frac{1}{4}$. $$b = a + 1$$ $$= -\frac{1}{4} + 1 = \frac{3}{4}$$ Thus, $a + b = -\frac{1}{4} + \frac{3}{4} = \frac{1}{2}$.

Question 15

Maths · Binomial Theorem · Single correct

If $\sum_{r=0}^{10} \left( \frac{10^{r+1} - 1}{10^r} \right) \cdot {}^{11}C_{r+1} = \frac{\alpha^{11} - 11 \cdot 11^{11}}{10^{10}}$, then $\alpha$ is equal to:

  1. 15
  2. 11
  3. 24
  4. 20

Answer: (d)

Solution

Given $$\sum_{r=0}^{10} \left( \frac{1 \cdot r^{r-1} - 1}{10^r} \right) \binom{11}{r+1}$$ This simplifies to $$\sum_{r=0}^{10} \left( 10 - \frac{1}{10^r} \right) \binom{11}{r+1}$$ Which is equal to $$10 \sum_{r=0}^{10} \binom{11}{r+1} - 10 \sum_{r=0}^{10} \binom{11}{r+1} \left( \frac{1}{10} \right)^{r+1}$$ This becomes $$10 \left[ \binom{11}{1} + \binom{11}{2} + \ldots + \binom{11}{11} \right]$$ Minus $$10 \left[ \binom{11}{1} \left( \frac{1}{10} \right)^1 + \binom{11}{2} \left( \frac{1}{10} \right)^2 + \ldots + \binom{11}{11} \left( \frac{1}{10} \right)^{11} \right]$$ This simplifies to $$10 \left[ 2^{11} - 1 \right] - 10 \left[ \left( 1 + \frac{1}{10} \right)^{11} - 1 \right]$$ Which equals $$10(2)^{11} - 10 - \frac{11^{11}}{10^{10}} + 10$$ This results in $$(20)^{11} - 11^{11}$$ Finally, $$10^{10}$$ Therefore, $\alpha = 20$

Question 16

Maths · Permutations and Combinations · Single correct

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is :

  1. 5880
  2. 960
  3. 840
  4. 5760

Answer: (d)

Solution

= Total - [ (All in R_1 and R_3) + (All in R_2 and R_3) + (All in R_1 and R_2) ] = {^8C_5} $\cdot$ 5 - $\{$ 5 + [5 + {^6C_5} $\cdot$ 5] $\}$ = 5(56 - 1 - 1 - 6) = 120(48) = 5760

Question 17

Maths · Conic Sections · Single correct

Let the point $P$ of the focal chord $PQ$ of the parabola $y^2 = 16x$ be $(1, -4)$. If the focus of the parabola divides the chord $PQ$ in the ratio $m : n$, $\mathrm{gcd}(m, n) = 1$, then $m^2 + n^2$ is equal to:

  1. 17
  2. 10
  3. 37
  4. 26

Answer: (a)

Solution

Given $y^2 = 16x$; $a = 4$ with focus $S \equiv (4, 0)$. $2at_1 = -4$ $$\Rightarrow 2(4)t_1 = -4$$ $$\Rightarrow t_1 = -\frac{1}{2}$$ Therefore, $t_1 t_2 = -1$ $$\Rightarrow t_2 = 2$$ Thus, $Q(at_2^2, 2at_2) = (16, 16)$. Let $S$ divides $PQ$ internally in $\lambda : 1$ ratio. $$16\lambda - 4 = 0$$ Therefore, $\lambda + 1 = \frac{1}{4} = \frac{m}{n}$. Thus, $m^2 + n^2 = 1 + 16 = 17$.

Question 18

Maths · Vector Algebra · Single correct

Let $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$, $\vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$ and a vector $\vec{c}$ be such that $(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k}$ and $\vec{a} \cdot \vec{c} = 3$. If $\vec{b} \times \vec{c} = \vec{d}$, then $|\vec{a} \cdot \vec{d}|$ is equal to:

  1. 18
  2. 12
  3. 9
  4. 15

Answer: (d)

Solution

Given $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$, $\vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$. $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 3 & 2 & 5 \end{vmatrix}$$ $$= -17\hat{i} - 7\hat{j} + 13\hat{k}$$ $$(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k}$$ $$\Rightarrow (\vec{a} \times \vec{b}) - (\vec{c} \times \vec{b}) = -18\hat{i} - 3\hat{j} + 12\hat{k}$$ $$\Rightarrow \vec{b} \times \vec{c} = (-18\hat{i} - 3\hat{j} + 12\hat{k}) - (\vec{a} \times \vec{b})$$ $$= (-18\hat{i} - 3\hat{j} + 12\hat{k}) - (-17\hat{i} - 7\hat{j} + 13\hat{k})$$ $$\vec{b} \times \vec{c} = -\hat{i} + 4\hat{j} - \hat{k}$$ Therefore, $\vec{a} \cdot \vec{d} = \vec{a} \cdot (\vec{b} \times \vec{c}) = (2\hat{i} - 3\hat{j} + \hat{k}) \cdot (-\hat{i} + 4\hat{j} - \hat{k})$ $$= -2 - 12 - 1 = -15$$ Thus, $|\vec{a} \cdot \vec{d}| = 15$

Question 19

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

Let the area of the triangle formed by the straight line $L:\; x+by+c=0$ with the coordinate axes be $48$ square units. If the perpendicular drawn from the origin to the line $L$ makes an angle of $45^\circ$ with the positive $x$-axis, then the value of $b^2+c^2$ is:

  1. 90
  2. 93
  3. 97
  4. 83

Answer: (c)

Solution

The area of the triangle is given by $\frac{1}{2} \left| \frac{c^2}{b} \right| = 48$. Solving for $\left| \frac{c^2}{b} \right| = 96$, we have: $$-c = -\frac{c}{b}$$ This implies $b = 1$ and therefore $c^2 = 96$. Thus, $b^2 + c^2 = 97$.

Question 20

Maths · Matrices · Single correct

Let A be a 3 $\times$ 3 real matrix such that $A^2(A - 2I) - 4(A - I) = O$, where I and O are the identity and null matrices, respectively. If $A^5 = \alpha A^2 + \beta A + \gamma I$, where $\alpha, \beta$ and $\gamma$ are real constants, then $\alpha + \beta + \gamma$ is equal to:

  1. 12
  2. 20
  3. 76
  4. 4

Answer: (a)

Solution

Given $A^3 - 2A^2 - 4A + 4I = 0$. $A^3 = 2A^2 + 4A - 4I$ $A^4 = 2A^3 + 4A^2 - 4A$ $= 2 \left(2A^2 + 4A - 4I\right) + 4A^2 - 4A$ $A^4 = 8A^2 + 4A - 8I$ $A^5 = 8A^3 + 4A^2 - 8A$ $= 8 \left(2A^2 + 4A - 4I\right) + 4A^2 - 8A$ $A^5 = 20A^2 + 24A - 32I$ Therefore, $\alpha = 20$, $\beta = 24$, $\gamma = -32$ Thus, $\alpha + \beta + \gamma = 12$

Question 21

Maths · Differential Equations · Numerical

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x$ such that $y(0) = \frac{5}{4}$. Then $12 \left( y \left( \frac{\pi}{4} \right) - e^{-2} \right)$ is equal to .

Answer: 21

Solution

\[ \text{I.F.} = e^{\int 2\sec^2 x\,dx} = e^{2\tan x}. \] Hence, the solution of the differential equation is $$y \cdot e^{2 \tan x} = \int e^{2 \tan x} (2 \sec^2 x + 3 \tan x \cdot \sec^2 x) \, dx$$ $$y \cdot e^{2 \tan x} = \int e^{2 \tan x} \cdot (2 \sec^2 x) \, dx + \int e^{2 \tan x} \cdot (3 \tan x \cdot \sec^2 x) \, dx$$ $$y \cdot e^{2 \tan x} = e^{2 \tan x} \cdot 2 \tan x - \int e^{2 \tan x} \cdot 2 \sec^2 x \times 2 \tan x \, dx + \int e^{2 \tan x} \cdot 3 \tan x \cdot \sec^2 x \, dx$$ $$y \cdot e^{2 \tan x} = 2 \tan x \cdot e^{2 \tan x} - \int e^{2 \tan x} \cdot \tan x \sec^2 x \, dx$$ $$y \cdot e^{2 \tan x} = 2 \tan x \cdot e^{2 \tan x} - \frac{\tan x \cdot e^{2 \tan x}}{2} + \frac{e^{2 \tan x}}{4} + C$$ $$y = 2 \tan x - \frac{\tan x}{2} + \frac{1}{4} + Ce^{-2 \tan x}$$ $$x = 0, y = \frac{5}{4}$$ $$c = 1$$ $$y \left( \frac{\pi}{4} \right) = \frac{7}{4} + e^{-2}$$ Then $$12 \left( y \left( \frac{\pi}{4} \right) - e^{-2} \right) = 12 \left( \frac{7}{4} \right) = 21$$

Question 22

Maths · Sequences and Series · Numerical

If the sum of the first 10 terms of the series $\frac{4.1}{1+4.1^4}$ + $\frac{4.2}{1+4.2^4}$ + $\frac{4.3}{1+4.3^4}$ + $\ldots$ is $\frac{m}{n}$ , where $\gcd$(m, n) = 1, then m + n is equal to

Answer: 441

Solution

Given $T_r = \frac{4 \cdot r}{1 + 4 \cdot r^4}$. $$T_r = \frac{4 \cdot r}{(2r^2 + 2r + 1)(2r^2 - 2r + 1)}$$ $$T_r = \frac{(2r^2 + 2r + 1) - (2r^2 - 2r + 1)}{(2r^2 + 2r + 1)(2r^2 - 2r + 1)}$$ $$T_r = \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1}$$ For $T_1$: $$T_1 = \frac{1}{1} - \frac{1}{5}$$ $$T_1 = \frac{1}{1} - \frac{1}{5}$$ For $T_2$: $$T_2 = \frac{1}{5} - \frac{1}{13}$$ For $T_{10}$: $$T_{10} = \frac{1}{181} - \frac{1}{221}$$ For $S_{10}$: $$S_{10} = 1 - \frac{1}{221} = \frac{220}{221} = \frac{m}{n}$$ Therefore, $m + n = 441$.

Question 23

Maths · Inverse Trigonometric Functions · Fill in the blank

If $y = \cos\left( \frac{\pi}{3} + \cos^{-1} \frac{x}{2} \right)$, then $(x-y)^2 + 3y^2$ is equal to .

Answer: 3

Solution

Given $$y = \cos \left( \cos^{-1} \frac{1}{2} + \cos^{-1} \frac{x}{2} \right)$$ We have $$y = \frac{1}{2} \times \frac{x}{2} - \sqrt{1 - \frac{1}{4}} \sqrt{1 - \frac{x^2}{4}}$$ Simplifying, $$4y = x - \sqrt{3} \sqrt{4 - x^2}$$ Rearranging, $$3 \left( 4 - x^2 \right) = x^2 + 16y^2 - 8xy$$ This gives $$12 - 3x^2 = x^2 + 16y^2 - 8xy$$ Rearranging terms, $$4x^2 + 16y^2 - 8xy = 12$$ Simplifying further, $$x^2 + 4y^2 - 2xy = 3$$ Finally, $$x^2 + y^2 - 2xy - 3y^2 = 3$$ This simplifies to $$(x - y)^2 + 3y^2 = 3$$

Question 24

Maths · Straight Lines and Pair of Straight Lines · Numerical

Let A(4, -2), B(1, 1) and C(9, -3) be the vertices of a triangle $ABC$. Then the maximum area of the parallelogram $AFDE$, formed with vertices D, E and F on the sides BC, CA and AB of the triangle $ABC$ respectively, is ______.

Answer: 3

Solution

Area of $\triangle ABC = \frac{1}{2} \begin{vmatrix} 4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1 \end{vmatrix}$ $$= 6 square units$$ Maximum area of $AFDE = \frac{1}{2} \times 6 = 3 sq. units$

Question 25

Maths · Complex Numbers and Quadratic Equations · Numerical

If the set of all $a \in \mathbb{R} - \{1\}$, for which the roots of the equation $(1-a)x^2 + 2(a-3)x + 9 = 0$ are positive is $(-\infty, -\alpha] \cup [\beta, \gamma)$, then $2\alpha + \beta + \gamma$ is equal to .

Answer: 7

Solution

Both the roots are positive. Given $D \geq 0$ $$4(a-3)^2 - 4 \times 9(1-a) \geq 0$$ $$a^2 - 6a + 9 - 9 + 9a \geq 0$$ $$a^2 + 3a \geq 0$$ $$a(a+3) \geq 0$$ $$a \in (-\infty, -3] \cup [0, \infty) ...(i)$$ $$\frac{b}{2a} > 0$$ $$2(a-3) > 0$$ $$2(a-1) > 0$$ $$a \in (-\infty, 1) \cup (3, \infty) ...(ii)$$ $f(0) = 9 > 0$ Equation (i) $\cap$ (ii) $$a \in (-\infty, -3] \cup [0, 1)$$ $$2\alpha + \beta + \gamma - 6 + 0 + 1 = 7$$

Physics

Question 26

Physics · Electric Charges and Fields · Single correct

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field. Reason (R): In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (d)

Solution

A: Since polar dielectrics are randomly oriented $\vec{P}_{net} = \vec{0}$. R: If $\vec{E}$ is absent, polar dielectrics remain polar and are randomly oriented.

Question 27

Physics · Moving Charges and Magnetism · Single correct

In a moving coil galvanometer, two moving coils $M_1$ and $M_2$ have the following particulars : $R_1 = 5\,\Omega$, $N_1 = 15$, $A_1 = 3.6 \times 10^{-3} \, \mathrm{m}^2$, $B_1 = 0.25 \, \mathrm{T}$ $R_2 = 7\,\Omega$, $N_2 = 21$, $A_2 = 1.8 \times 10^{-3} \, \mathrm{m}^2$, $B_2 = 0.50 \, \mathrm{T}$ Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_1$ and $M_2$ ?

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

Answer: (a)

Solution

Voltage sensitivity $= \frac{\theta}{V} = \frac{NAB}{cR}$ $$Ratio = \left( \frac{N_1 A_1 B_1}{N_2 A_2 B_2} \right) \frac{R_2}{R_1} = \frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5} \times \frac{7}{5} = \frac{1}{1}$$

Question 28

Physics · System of Particles and Rotational Motion · Single correct

The moment of inertia of a circular ring of mass $M$ and diameter $r$ about a tangential axis lying in the plane of the ring is:

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

Answer: (b)

Solution

Diameter is given as $R$. Therefore, Radius $= R/2$. $$I_{tangent} = \frac{3}{2} m \left( \frac{R}{2} \right)^2 = \frac{3}{8} m R^2$$

Question 29

Physics · Mechanical Properties of Fluids · Single correct

Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is :

  1. $4 \pi r^2 \, T \left[ 2 - 2^{\frac{2}{3}} \right]$
  2. $4 \pi r^2 \, T \left[ 2 - 2^{\frac{1}{3}} \right]$
  3. $4 \pi r^2 \, T [1 + \sqrt{2}]$
  4. $4 \pi r^2 T [\sqrt{2} - 1]$

Answer: (a)

Solution

Given $2 \times \frac{4}{3} \pi R^3 = \frac{4}{3} \pi r^3 \Rightarrow r = 2^{1/3} R$. $U_i = 2 \times 4 \pi R^2 T$. $U_f = 4 \pi R^2 T = 4 \pi R^2 T 2^{2/3}$. Therefore, Heat lost $= u_i - u_f = 4 \pi R^2 T \left[ 2 - 2^{2/3} \right]$.

Question 30

Physics · Dual Nature of Radiation and Matter · Single correct

An electron with mass $m$ with an initial velocity $(t = 0)$, $\vec{v} = v_0\hat{i}$ $(v_0 > 0)$ enters a magnetic field $\vec{B} = B_0\hat{j}$. If the initial de-Broglie wavelength at $t = 0$ is $\lambda_0$, then its value after time $t$ would be:

  1. $\frac{\lambda_0}{\sqrt{1 - \frac{e^2 B_0^2 t^2}{m^2}}}$
  2. $\frac{\lambda_0}{\sqrt{1 + \frac{e^2 B_0^2 t^2}{m^2}}}$
  3. $\lambda_0 \sqrt{1 + \frac{e^2 B_0^2 t^2}{m^2}}$
  4. $\lambda_0$

Answer: (d)

Solution

Magnetic field does not work. Therefore, speed will not change, so De-Broglie wavelength remains the same.

Question 31

Physics · Waves · Single correct

A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the x-direction in 0.3 sec. The crest P is at x = 0 at t = 0 sec and maximum displacement of the wave is 2 cm. Which equation correctly represents this wave?

  1. y = 2 $\cos$(0.83x - 3.35t) $\mathrm{cm}$
  2. y = 2 $\sin$(0.83x - 3.5t) $\mathrm{cm}$
  3. y = 2 $\cos$(3.35x - 0.83t) $\mathrm{cm}$
  4. y = 2 $\cos$(0.13x - 0.5t) $\mathrm{cm}$

Answer: (a)

Solution

Given $v = \frac{distance}{time}$. $v = \frac{12}{0.3} = 4 \, \mathrm{cm/s}$ $k = \frac{2\pi}{\lambda} = \frac{2\pi}{7.5} = \frac{4\pi}{15} = 0.83$ $v = \frac{\omega}{k} \Rightarrow \omega = vk = 4 \times \frac{4\pi}{15} = 3.35$ So $y = A \cos(kx - \omega t)$

Question 32

Physics · Physical World, Units and Measurements · Single correct

Given a charge $q$, current $I$ and permeability of vacuum $\mu_0$. Which of the following quantity has the dimension of momentum?

  1. $qI/\mu_0$
  2. $q\mu_0I$
  3. $q^2\mu_0I$
  4. $q\mu_0/I$

Answer: (b)

Solution

Q = AT I = A $\mu$_0 = MLT^{-2} A^{-2} P = Q^x $\mu$_0^y I^z = [AT]^x [MLT^{-2} A^{-2}]^y [A]^z MLT^{-1} = M^y L^{x-2y} A^{-2y+z+x} Now; y = 1 x - 2y = -1 -2y + z = 0 $\therefore$ x = y = z = 1

Question 33

Physics · Magnetism and Matter · Single correct

A solenoid having area $A$ and length $l$ is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is:

  1. $\frac{B^2 A l}{\mu_0}$
  2. $\frac{B^2 A l}{2 \mu_0}$
  3. $B^2 A l$
  4. $\frac{B^2 A l}{4 \mu_0}$

Answer: (d)

Solution

Given $\($ $\frac{U}{V}$ = $\frac{B^2}{2 \mu_r \mu_0}$ $\)$ which implies $\($ U = $\frac{B^2}{4 \mu_0}$ $\)$ and $\($ V = $\frac{B^2}{4 \mu_0}$ A $\ell$ $\)$.

Question 34

Physics · Electric Charges and Fields · Single correct

Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is $V$. The potential difference between the points $A$ and $B$ (shown in the figure) is:

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

Answer: (b)

Solution

Using $\Delta V = E(\Delta d)$ $V = E(10)$ $$V_{AB} = E \cdot 4 = \frac{V}{10} \times 4 = \frac{2V}{5}$$

Question 35

Physics · Kinetic Theory · Single correct

Identify the characteristics of an adiabatic process in a monoatomic gas. (A) Internal energy is constant. (B) Work done in the process is equal to the change in internal energy. (C) The product of temperature and volume is a constant. (D) The product of pressure and volume is a constant. (E) The work done to change the temperature from $T_1$ to $T_2$ is proportional to $(T_2 - T_1)$ Choose the correct answer from the options given below:

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

Answer: (c)

Solution

Given $Q = \Delta U + W = 0 \implies -\Delta U = W$. $WD = -nC_v \Delta T \implies |WD| = nC_v \Delta T \propto T_2 - T_1$. Therefore, B $\&$ E [Only possibility]

Question 36

Physics · Atoms · Single correct

Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of $\mathrm{Li}^{++}$ ion in its ground state is given by $\frac{1}{X} a_0$, where $X =$. (Where $a_0$ is the first Bohr's radius.)

  1. 2
  2. 1
  3. 4
  4. 9

Answer: (c)

Solution

Given $r = r_0 \frac{n^2}{z}$ and $z = 3$ for $\mathrm{Li}^{+2}$ and $n = 1$. Therefore, $r = r_0 \frac{1^2}{3} = \frac{r_0}{3}$. Thus, $x = 3$.

Question 37

Physics · Nuclei · Single correct

Energy released when two deuterons $\left( {}_1^2\mathrm{H} \right)$ fuse to form a helium nucleus $\left( {}_2^4\mathrm{He} \right)$ is: (Given: Binding energy per nucleon of ${}_1^2\mathrm{H} = 1.1 \, \mathrm{MeV}$ and binding energy per nucleon of ${}_2^4\mathrm{He} = 7.0 \, \mathrm{MeV}$)

  1. 8.1 MeV
  2. 5.9 MeV
  3. 23.6 MeV
  4. 26.8 MeV

Answer: (c)

Solution

The reaction is given by $$^{2}_{1}\mathrm{H} + ^{2}_{1}\mathrm{H} \rightarrow ^{4}_{1}\mathrm{He}$$ with binding energies $1.1 \, \mathrm{MeV}$ for $^{2}_{1}\mathrm{H}$ and $7.0 \, \mathrm{MeV}$ for $^{4}_{1}\mathrm{He}$. The binding energy difference $E_B$ is calculated as follows: $$E_B = \mathrm{BE}_{reactant} - \mathrm{BE}_{product}$$ Substituting the values, we have: $$= 1.1 \times 2 + 1.1 \times 2 - 7 \times 4 = -23.6 \, \mathrm{MeV}$$ Thus, the energy released $Q$ is: $$Q = 23.6 \, \mathrm{MeV}$$

Question 38

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

In the digital circuit shown in the figure, for the given inputs the P and Q values are:

  1. $P = 1, Q = 1$
  2. $P = 0, Q = 0$
  3. $P = 0, Q = 1$
  4. $P = 1, Q = 0$

Answer: (b)

Solution

The circuit consists of logic gates with inputs and outputs. The first AND gate receives inputs 1 and 1, producing an output of 1. This output is fed into the next AND gate along with another input of 0, resulting in an output of 0. The output of this gate is connected to the input of the OR gate, which also receives another input of 0, resulting in an output of 0 for P. The lower part of the circuit has an AND gate receiving inputs 1 and 1, producing an output of 1. This output is inverted by a NOT gate, resulting in 0, which is fed into an OR gate along with another input of 0, resulting in an output of 0 for Q.

Question 39

Physics · Ray Optics and Optical Instruments · Single correct

Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 $\mathrm{cm}$ , at same distance of 18 $\mathrm{cm}$ from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is :

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

Answer: (a)

Solution

Using $m = \frac{f}{u-f}$, $$m_1 = \frac{6}{18-6} = \frac{1}{2}$$ $$m_2 = \frac{6}{18+6} = \frac{1}{4} \therefore \frac{m_2}{m_1} = \frac{1}{2}$$

Question 40

Physics · Motion in a Plane · Single correct

A sportsman runs around a circular track of radius $r$ such that he traverses the path $ABAB$. The distance travelled and displacement, respectively, are

  1. $2r, 3\pi r$
  2. $3\pi r, \pi r$
  3. $\pi r, 3r$
  4. $3\pi r, 2r$

Answer: (d)

Solution

Displacement is $2r$. Distance is $2\pi r + \pi r = 3\pi r$.

Question 41

Physics · Laws of Motion · Single correct

A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions $T_1$ and $T_2$, respectively, are (in N):

  1. 5, 5$\sqrt{3}$
  2. 5$\sqrt{3}$, 5
  3. 5$\sqrt{3}$, 5$\sqrt{3}$
  4. 5, 5

Answer: (b)

Solution

Given the forces in the diagram, we have: $$T_1 = mg \cos 30^\circ$$ $$T_2 = mg \sin 30^\circ$$

Question 42

Physics · Ray Optics and Optical Instruments · Single correct

A bi-convex lens has radius of curvature of both the surfaces same as $1/6 \, \mathrm{cm}$. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides $(R_1 \neq R_2)$, without any change in lens power then possible combination of $R_1$ and $R_2$ is:

  1. $\frac{1}{3} \, \mathrm{cm}$ and $\frac{1}{3} \, \mathrm{cm}$
  2. $\frac{1}{5} \, \mathrm{cm}$ and $\frac{1}{7} \, \mathrm{cm}$
  3. $\frac{1}{3} \, \mathrm{cm}$ and $\frac{1}{7} \, \mathrm{cm}$
  4. $\frac{1}{6} \, \mathrm{cm}$ and $\frac{1}{9} \, \mathrm{cm}$

Answer: (b)

Solution

This will happen when $$\frac{1}{f_1} = \frac{1}{f_2}$$ $$ (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{-R_2} \right) = (\mu - 1) \left( \frac{2}{R} \right) $$ $$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} $$

Question 43

Physics · Physical World, Units and Measurements · Single correct

If $\mu_0$ and $\varepsilon_0$ are the permeability and permittivity of free space, respectively, then the dimension of $\left( \frac{1}{\mu_0 \varepsilon_0} \right)$ is:

  1. $L/T^2$
  2. $L^2/T^2$
  3. $T^2/L$
  4. $T^2/L^2$

Answer: (b)

Solution

Given $C = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$. This implies $\frac{1}{\mu_0 \varepsilon_0} = C^2 = L^2 T^{-2}$.

Question 44

Physics · Thermal Properties of Matter · Single correct

Match List-I with List-II. List-I (A) Heat capacity of body (B) Specific heat capacity of body (C) Latent heat (D) Thermal conductivity List-II (I) $\mathrm{Jkg^{-1}}$ (II) $\mathrm{JK^{-1}}$ (III) $\mathrm{Jkg^{-1}K^{-1}}$ (IV) $\mathrm{Jm^{-1}K^{-1}s^{-1}}$ Choose the correct answer from the options given below :

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

Answer: (d)

Solution

Given $C' = \frac{\Delta Q}{\Delta T} = \mathrm{JK}^{-1}$. $S = \frac{\Delta Q}{m \Delta T} = \mathrm{Jkg}^{-1} \mathrm{K}^{-1}$. $L = \frac{\Delta Q}{m} = \mathrm{Jkg}^{-1}$. $\Delta Q = \frac{KA \Delta T}{L} \Rightarrow K = \frac{\Delta Q (L)}{A \Delta T} = \mathrm{Jm}^{-1} \mathrm{K}^{-1} \mathrm{s}^{-1}$.

Question 45

Physics · Electric Charges and Fields · Single correct

Consider a circular loop that is uniformly charged and has a radius $a\sqrt{2}$. Find the position along the positive $z$-axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in $xy$-plane at the origin:

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

Answer: (c)

Solution

Given $E = \frac{KQr}{(x^2 + R^2)^{3/2}}$. Differentiating, $\frac{dE}{dx} = 0$. Therefore, $x = \frac{R}{\sqrt{2}} = \frac{\sqrt{2}a}{\sqrt{2}} = a$.

Question 46

Physics · System of Particles and Rotational Motion · Numerical

A wheel of radius $0.2 \, \mathrm{m}$ rotates freely about its center when a string that is wrapped over its rim is pulled by force of $10 \, \mathrm{N}$ as shown in figure. The established torque produces an angular acceleration of $2 \, \mathrm{rad/s^2}$. Moment of inertia of the wheel is ______$\mathrm{kgm^2}$. (Acceleration due to gravity $= 10 \, \mathrm{m/s^2}$)

Answer: 1

Solution

Given $FR = I \alpha$. Therefore, $$I = \frac{FR}{\alpha} = \frac{10 \times 0.2}{2} = 1 \, \mathrm{kg} - \mathrm{m}^2$$

Question 47

Physics · Kinetic Theory · Numerical

The internal energy of air in $4 \, \mathrm{m} \times 4 \, \mathrm{m} \times 3 \, \mathrm{m}$ sized room at $1$ atmospheric pressure will be _____ $\times10^6$J (Consider air as diatomic molecule)

Answer: 12

Solution

To find the internal energy of gas in the room. $$U = nC_v T = n \frac{5RT}{2}$$ $$= \frac{5}{2} PV = \frac{5}{2} \times 10^5 \times 48 = 12 \times 10^6 \, \mathrm{J}$$

Question 48

Physics · Ray Optics and Optical Instruments · Numerical

A ray of light suffers minimum deviation when incident on a prism having angle of the prism equal to $60^\circ$. The refractive index of the prism material is $\sqrt{2}$. The angle of incidence (in degrees) is ______.

Answer: 45

Solution

Given $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$, since $A = 60^\circ$. Therefore, $\delta_m = 30^\circ$. $\delta_m = 2i - A$ [as $i = e$] Thus, $i = 45^\circ$.

Question 49

Physics · Mechanical Properties of Solids · Fill in the blank

The length of a light string is $1.4\ \mathrm{m}$ when the tension on it is $5\ \mathrm{N}.$ If the tension increases to $7\ \mathrm{N},$ the length of the string is $1.56\ \mathrm{m}.$ The original length of the string is $\underline{\hspace{2cm}}\ \mathrm{m}.$

Answer: 1

Solution

Given $T = K(\ell - \ell_0)$. From (1), $$5 = K(1.4 - \ell_0)$$ $$7 = K(1.56 - \ell_0)$$ Dividing the equations, $$\frac{5}{7} = \frac{1.4 - \ell_0}{1.56 - \ell_0}$$ Therefore, $\ell_0 = 1 \, \mathrm{m}$.

Question 50

Physics · Gravitation · Numerical

A satellite of mass $1000\,\mathrm{kg}$ is launched to revolve around the earth in an orbit at a height of $270\,\mathrm{km}$ from the earth's surface. Kinetic energy of the satellite in this orbit is \[ \underline{\hspace{2cm}}\times 10^{10}\,\mathrm{J}. \] (Mass of earth $=6\times10^{24}\,\mathrm{kg}$, Radius of earth $=6.4\times10^6\,\mathrm{m}$, Gravitational constant $=6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$.)

Answer: 3

Solution

The kinetic energy is given by $KE = \frac{1}{2} mv^2$. This can be expressed as $\frac{1}{2} m \frac{GM_e}{r} = \frac{GM_em}{2r} = \frac{GM_em}{2(R_E + h)}$. Substituting the values, we have: $$= \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 6.4 \times 10^6}{2(6.4 \times 10^6 + 2.7 \times 10^5)} = 3 \times 10^{10} \, \mathrm{J}$$

Chemistry

Question 51

Chemistry · Amines · Single correct

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid ($273\,\mathrm{K}$) and acidified with acetic acid, the mass ($\mathrm{g}$) of $0.1$ mole of product formed is: (Given molar mass in $\mathrm{g\,mol}^{-1}$: $\mathrm{H}$: $1$, $\mathrm{C}$: $12$, $\mathrm{N}$: $14$, $\mathrm{O}$: $16$, $\mathrm{S}$: $32$)

  1. 343
  2. 330
  3. 33
  4. 66

Answer: (c)

Solution

Sulphanilic acid reacts with acetic acid to form an acetylated product. This product then reacts with nitrous acid to form a diazonium salt. The diazonium salt couples with a naphthylamine derivative to form a red-azo dye. 0.1 mole of red-azo dye (Molar Mass = 327 $\mathrm{gm/mol}$) will have 32.7 gm mass. Nearly 33 gm.

Question 52

Chemistry · Co-ordination Compounds · Single correct

The d-orbital electronic configuration of the complex among $[\mathrm{Co(en)}_3]^{3+}$, $[\mathrm{CoF}_6]^{3-}$, $[\mathrm{Mn(H_2O)}_6]^{2+}$ and $[\mathrm{Zn(H_2O)}_6]^{2+}$ that has the highest CFSE is:

  1. $t_{2g}^6 e_g^0$
  2. $t_{2g}^6 e_g^4$
  3. $t_{2g}^3 e_g^2$
  4. $t_{2g}^4 e_g^2$

Answer: (a)

Solution

In $[\mathrm{Co(en)}_3]^{+3}$ S.F.L. is present and hence highest value of CFSE. In rest all complexes WFL is present hence CFSE will be low. $$[\mathrm{Co(en)}_3]^{+3} \Rightarrow \mathrm{Co}^{+3} and en (SFL)$$ $$\mathrm{Co}^{+3} \Rightarrow [\mathrm{Ar}]3d^6$$

Question 53

Chemistry · Hydrocarbons · Single correct

Given below are two statements: Statement (I): Neopentane forms only one monosubstituted derivative. Statement (II): Melting point of neopentane is higher than n-pentane In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Statement I is correct but Statement II is incorrect
  2. Both Statement I and Statement II are correct
  3. Both Statement I and Statement II are incorrect
  4. Statement I is incorrect but Statement II is correct

Answer: (b)

Solution

Both Statement-I and Statement-II are correct. The reaction is: $$\mathrm{CH_3-C(CH_3)_2-CH_3 + X_2 \xrightarrow{h\nu} CH_3-C(CH_3)_2-CH_2-X}$$ where $X = \mathrm{Cl, Br}$. The melting point of neo-pentane is greater than that of n-pentane.

Question 54

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Which among the following molecules is (a) involved in $sp^3 d$ hybridization, (b) has different bond lengths and $(c)$ has lone pair of electrons on the central atom?

  1. PF_5
  2. XeF_4
  3. SF_4
  4. XeF_2

Answer: (c)

Solution

Question 55

Chemistry · Analytical Chemistry · Single correct

Formation of Na$_4$[Fe(CN)$_5$NOS], a purple coloured complex formed by addition of sodium nitroprusside in sodium carbonate extract of salt indicates the presence of :

  1. Sodium ion
  2. Sulphate ion
  3. Sulphide ion
  4. Sulphite ion

Answer: (c)

Solution

$Na_2S+Na_2[Fe(CN)_5NO]\longrightarrow Na_4[Fe(CN)_5NOS]$ \[ \begin{array}{ccc} \text{Sodium sulphide} & \text{Sodium nitroprusside} & \text{Violet Colour} \end{array} \]

Question 56

Chemistry · Chemical Bonding and Molecular Structure · Single correct

In 3, 3-dimethylhex-1-ene-4-yne, there are _______ $sp^3$, _______ $sp^2$ and _______ $sp$ hybridised carbon atoms respectively:

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

Answer: (a)

Solution

Question 57

Chemistry · Structure of Atom · Single correct

Which of the following statements are true? (A) The subsidiary quantum number $l$ describes the shape of the orbital occupied by the electron. (B) is the boundary surface diagram of the $2p_x$ orbital. (C) The $+$ and $-$ signs in the wave function of the $2p_x$ orbital refer to charge. (D) The wave function of $2p_x$ orbital is zero everywhere in the xy plane.

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

Answer: (d)

Solution

(A) Azimuthal quantum number ($\ell$) indicates the shape of orbital occupied by the electron. (B) $p_x$ orbital. (C) The $+$ and $-$ sign in the wave function of $2p_x$ orbital refer to the sign (Phase) of the wave function, not the charge. (D) The wave function of $2p_x$ orbital will be zero in $yz$ plane (Nodal plane).

Question 58

Chemistry · Co-ordination Compounds · Single correct

The type of hybridization and the magnetic property of $[\mathrm{MnCl}_6]^{3-}$ are:

  1. $d^2 sp^3$, paramagnetic with four unpaired electrons
  2. $sp^3 d^2$, paramagnetic with four unpaired electrons
  3. $d^2 sp^3$, paramagnetic with two unpaired electrons
  4. $sp^3 d^2$, paramagnetic with two unpaired electrons

Answer: (b)

Solution

The complex $[\mathrm{MnCl}_6]^{3-}$ contains $\mathrm{Mn}^{+3}$. The electronic configuration of $\mathrm{Mn}^{+3}$ is $[\mathrm{Ar}] 3d^4$. The ligand is $\mathrm{Cl}^-$, which is a weak field ligand (WFL). The $3d^4$ configuration leads to the following electron arrangement. Hybridisation is $sp^3 d^2$. There are 4 unpaired electrons.

Question 59

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Multiple correct

Consider the following reactions. From these reactions which reaction will give carboxylic acid as a major product? (A) R - $\mathrm{C} \equiv \mathrm{N}$ $\xrightarrow[\text{mild condition}]{(i)\ \mathrm{H^+/H_2O}}$ (B) R - $\mathrm{MgX}$ $\xrightarrow[(ii)\ \mathrm{H_3O^+}]{(i)\ \mathrm{CO_2}}$ (C) R - $\mathrm{C} \equiv \mathrm{N}$ $\xrightarrow[(ii)\ \mathrm{H_3O^+}]{(i)\ \mathrm{SnCl_2/HCl}}$ (D) R - $\mathrm{CH_2OH}$ $\xrightarrow{\mathrm{PCC}}$ (E)

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

Answer: (d)

Solution

Under mild condition amide is formed because this reaction is typically slow if further more heat will supplied then it gets convert in to -COOH.

Question 60

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

Electronic configuration of four elements A, B, C and D are given below: (A) $1s^2 2s^2 2p^3$ (B) $1s^2 2s^2 2p^4$ $(C) 1s^2 2s^2 2p^5$ (D) $1s^2 2s^2 2p^2$ Which of the following is the correct order of increasing electronegativity (Pauling's scale)?

  1. A < D < B < C
  2. A < C < B < D
  3. A < B < C < D
  4. D < A < B < C

Answer: (d)

Solution

N: $1s^2 2s^2 2p^3$ (Electronegativity = 3) O: $1s^2 2s^2 2p^4$ (Electronegativity = 3.5) F: $1s^2 2s^2 2p^5$ (Electronegativity = 4) C: $1s^2 2s^2 2p^2$ (Electronegativity = 2.55) Correct order = C > B > A > D

Question 61

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

Match List-I with List-II \begin{tabular}{|c|p{4.8cm}|c|p{4.8cm}|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Purification technique)} & \multicolumn{2}{c|}{(Mixture of organic compounds)} \\ \hline (A) & Distillation (simple) & (I) & Diesel + Petrol \\ \hline (B) & Fractional distillation & (II) & Aniline + Water \\ \hline (C) & Distillation under reduced pressure & (III) & Chloroform + Aniline \\ \hline (D) & Steam distillation & (IV) & Glycerol + Spent-lye \\ \hline \end{tabular} Choose the correct answer from the options given below :

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

Answer: (d)

Solution

\begin{tabular}{|c|p{4.8cm}|c|p{4.8cm}|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Purification technique)} & \multicolumn{2}{c|}{(Mixture of organic compounds)} \\ \hline (A) & Distillation (simple) & (III) & Chloroform + Aniline \\ \hline (B) & Fractional distillation & (I) & Diesel + Petrol \\ \hline (C) & Distillation under reduced pressure & (IV) & Glycerol + Spent-lye \\ \hline (D) & Steam distillation & (II) & Aniline + Water \\ \hline \end{tabular}

Question 62

Chemistry · Solutions · Single correct

' x ' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from $1^\circ \mathrm{C}$ to $25^\circ \mathrm{C}$. Which out of the following plots, is best suited for the change in the molarity (M) of the solution with respect to temperature? [Consider the solubility of NaCl remains unchanged over the temperature range]

Answer: (b)

Solution

Water at $1^\circ \mathrm{C}$ decreases in volume to water at $4^\circ \mathrm{C}$. Then, the volume increases due to thermal expansion to water at $25^\circ \mathrm{C}$. Molarity is given by $$Molarity = \frac{n_{solute}}{(Volume of solution) \ell}$$ Hence, the molarity from $1^\circ \mathrm{C}$ to $4^\circ \mathrm{C}$ increases, and from $4^\circ \mathrm{C}$ to $25^\circ \mathrm{C}$ decreases.

Question 63

Chemistry · Thermodynamics · Single correct

Arrange the following in order of magnitude of work done by the system / on the system at constant temperature: (a) $|w_{reversible}|$ for expansion in infinite stage. (b) $|w_{irreversible}|$ for expansion in single stage. (c) $|w_{reversible}|$ for compression in infinite stage. (d) $|w_{irreversible}|$ for compression in single stage. Choose the correct answer from the options given below:

  1. a > b > c > d
  2. d > c = a > b
  3. c = a > d > b
  4. a > c > b > d

Answer: (b)

Solution

For isothermal process: $$|W_{reversible expansion}| = |W_{reversible compression}|$$ $$= -nRT \ln \frac{V_f}{V_i}$$ $$|W_{irreversible expansion}| c = a > b$$ $$|W_{irreversible expansion}| = -P_{ext} (V_f - V_i)$$ Graphical representation: We can compare work by area of PV graph.

Question 64

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

Reactant A converts to product D through the given mechanism (with the net evolution of heat): A $\rightarrow$ B slow ; $\Delta$ H = +ve B $\rightarrow$ C fast ; $\Delta$ H = -ve C $\rightarrow$ D fast ; $\Delta$ H = -ve Which of the following represents the above reaction mechanism?

Answer: (a)

Solution

A to B is slow. $\($ $\Delta$ H = +ve $\)$, ($E_{a_1}$ $\to$ High). B to C is fast. $\($ $\Delta$ H = -ve $\)$, ($E_{a_2}$ $\to$ Low). C to D. $\($ $\Delta$ D = -ve $\)$, ($E_{a_3}$ $\to$ Low).

Question 65

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

The nature of oxide $(TeO_2)$ and hydride $(TeH_2)$ formed by $Te$, respectively are:

  1. Oxidising and acidic
  2. Reducing and basic
  3. Reducing and acidic
  4. Oxidising and basic

Answer: (a)

Solution

TeO$_2$ is oxidizing in nature because it can be reduced from $+4$ oxidation state to lower oxidation state. TeH$_2$ due to less bond dissociation energy easily breaks and hence acidic in nature.

Question 66

Chemistry · Haloalkanes and Haloarenes · Single correct

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

  1. $(A)\text{-}(III),\ (B)\text{-}(II),$ $(C)\text{-}(IV),\ (D)\text{-}(I)$
  2. $(A)\text{-}(III),\ (B)\text{-}(IV),$ $(C)\text{-}(II),\ (D)\text{-}(I)$
  3. $(A)\text{-}(IV),\ (B)\text{-}(III),$ $(C)\text{-}(I),\ (D)\text{-}(II)$
  4. $(A)\text{-}(IV),\ (B)\text{-}(I),$ $(C)\text{-}(II),\ (D)\text{-}(III)$

Answer: (b)

Solution

The reactions are matched with their respective names as follows: (A) The reaction $2 C_6H_5X + 2Na \xrightarrow{Dry Ether} C_6H_5-C_6H_5$ corresponds to the Fittig reaction. (B) The reaction $ArN_2^+X^- \xrightarrow{Cu/HCl} ArCl + N_2 \uparrow + CuX$ corresponds to the Gatterman reaction. (C) The reaction $C_2H_5Br + NaI \xrightarrow{Dry Acetone} C_2H_5I + NaBr$ corresponds to the Finkelstein reaction. (D) The reaction $CH_3C(OH)(CH_3)CH_3 \xrightarrow{HCl/ZnCl_2} CH_3C(Cl)(CH_3)CH_3$ corresponds to the Lucas reaction.

Question 67

Chemistry · Equilibrium · Single correct

Consider the following chemical equilibrium of the gas phase reaction at a constant temperature: $$\mathrm{A(g) \rightleftharpoons B(g) + C(g)}$$ If $p$ being the total pressure, $K_p$ is the pressure equilibrium constant and $\alpha$ is the degree of dissociation, then which of the following is true at equilibrium?

  1. If $p$ value is extremely high compared to $K_p$, $\alpha \approx 1$
  2. When $p$ increases $\alpha$ decreases
  3. If $K_p$ value is extremely high compared to $p$, $\alpha$ becomes much less than unity
  4. When $p$ increases $\alpha$ increases

Answer: (b)

Solution

a moles of $A(g)$ taken initially and at time Now moles fraction of $A(g)$, $B(g)$ and $C(g)$ are $$X_A = \frac{a - a\alpha}{a + a\alpha} = \frac{1 - \alpha}{1 + \alpha}$$ $$X_B = \frac{a\alpha}{a + a\alpha} = \frac{\alpha}{1 + \alpha}$$ $$X_C = \frac{a\alpha}{a + a\alpha} = \frac{\alpha}{1 + \alpha}$$ Now if $P$ is total pressure then partial pressure of $A(g)$, $B(g)$ and $C(g)$ are $$P_A = \left( \frac{1 - \alpha}{1 + \alpha} \right) P$$ $$P_B = \left( \frac{\alpha}{1 + \alpha} \right) P$$ $$P_C = \left( \frac{\alpha}{1 + \alpha} \right) P$$ $$K_P = \left( \frac{\alpha}{1 + \alpha} \right) P \left( \frac{\alpha}{1 + \alpha} \right) P \left( \frac{1 - \alpha}{1 + \alpha} \right) P$$ $$K_P = \frac{\alpha^2 P}{1 - \alpha^2}$$ As $K_P$ is only function of temperature. So as $P \uparrow \alpha \downarrow$$

Question 68

Chemistry · Thermodynamics · Single correct

Which of the following graphs correctly represents the variation of thermodynamic properties of Haber’s process?

Answer: (a)

Question 69

Chemistry · Biomolecules · Single correct

A tetrapeptide "x" on complete hydrolysis produced glycine (Gly), alanine (Ala), valine (Val), leucine (Leu) in equimolar proportion each. The number of tetrapeptides (sequences) possible involving each of these amino acids is

  1. 16
  2. 32
  3. 8
  4. 24

Answer: (d)

Solution

The number of tetrapeptides (sequences) possible involving each of these amino acids (glycine, alanine, valine, leucine); it has three (3) peptides linkage. The number of permutations in which they can be arranged is $$= 4 \times 3 \times 2 \times 1$$ $$= 24$$

Question 70

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

In Dumas' method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound (Aqueous tension at 300 K = 15 $\mathrm{mmHg}$) is

  1. 1.257
  2. 20.87
  3. 18.67
  4. 12.57

Answer: (d)

Solution

Pressure of $\mathrm{N_2}$ gas $= (715 - 15)$ $= 700 \, \mathrm{mmHg}$ $$n_{\mathrm{N_2}} = \frac{PV}{RT}$$ $$n_{\mathrm{N_2}} = \frac{700 \times 60 \times 10^{-3}}{760 \times 0.0821 \times 300}$$ $$= 2.24 \times 10^{-3} \, \mathrm{mol}$$ Mass of $\mathrm{N_2} = 2.24 \times 10^{-3} \times 28 \, \mathrm{g} = 0.06272 \, \mathrm{g}$ $$\% \, \mathrm{N_2} = \frac{0.06272}{0.5} \times 100 \simeq 12.57$$

Question 71

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

For the reaction A $\rightarrow$ B the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to 2.5 $\mathrm{g \, L^{-1}}$ (if the initial concentration of A was 50 $\mathrm{g \, L^{-1}}$) is _______ (Nearest integer) Given : $\log$ 2 = 0.3010

Answer: 43

Solution

As it is difficult to predict order using data provided in graph. For specific time interval 0 – 5 sec, 5 – 10 sec and 10 – 15 sec, order comes to be zero, but graph is not a straight line. Assuming 1st order kinetics $$K = \frac{1}{t} \ln \frac{A_0}{A_t}$$ $$K = \frac{1}{10} \ln \frac{40}{20}$$ Time required to reduce to 2.5 g/L $$K = \frac{1}{t} \ln \frac{50}{2.5}$$ $$\frac{1}{10} \ln 2 = \frac{1}{t} \ln 20$$ $$t = \frac{1.3010 \times 10}{0.3010} = 43.3 sec$$

Question 72

Chemistry · Electrochemistry · Numerical

0.2$\%$ (w/v) solution of NaOH is measured to have resistivity $870.0 \, \mathrm{m\Omega \, m}$. The molar conductivity of the solution will be ______ $\times 10^2 \, \mathrm{mS \, dm^2 \, mol^{-1}}$. (Nearest integer)

Answer: 23

Solution

Given: Concentration of NaOH = 0.2$\%$ (w/v). Therefore, 0.2 g of NaOH in 100 ml of solution. Molarity of NaOH solution is given by $$Molarity = \frac{moles of solute}{V_{ml}} \times 1000$$ $$= \frac{0.2/40}{100} \times 1000 = \frac{0.2}{40 \times 100} \times 1000 = \frac{2}{40} \, M$$ Given resistivity of solution = 870 m ohm m = 870 $\times$ 10^{-3} $\,$ ohm m $$= 870 \times 10^{-3} \times 100 \, ohm dm$$ $$= 870 \times 10^{-2} \, ohm dm$$ $$= 8.7 \, ohm dm$$ Now conductivity $$K = \frac{1}{\rho} = \frac{1}{8.7} \, ohm^{-1} dm^{-1}$$ Now molar conductivity of solution is $$\lambda_m = \frac{K}{M} = \frac{1}{8.7} \times \frac{40}{\frac{2}{40}} = 2.29 \, S dm^2 mol^{-1}$$ $$2.29 \times 10^3 \, m S dm^2 mol^{-1}$$ $$= 22.9 \times 10^2 \, m S dm^2 mol^{-1}$$ $$= 23 \times 10^2 \, m S dm^2 mol^{-1}$$

Question 73

Chemistry · Haloalkanes and Haloarenes · Numerical

Consider the above sequence of reactions. 151 g of 2-bromopentane is made to react. Yield of major product P is 80$\%$ whereas Q is 100$\%$. Mass of product Q obtained is $\,$ $\mathrm{g}$. (Given molar mass in $\mathrm{gmol^{-1}}$ H : 1, C : 12, O : 16, Br : 80)

Answer: 184

Solution

1 mole of 2-Bromopentane reacts with alcoholic KOH to give 0.8 mole of P (major product) with 80% yield. P is then reacted with $\mathrm{Br_2}$ to give Q. The molecular mass of Q is $230 \, \mathrm{g \, mol^{-1}}$. The mass of Q is calculated as follows: $$Mass of Q = 0.8 \times 230$$ $$= 184 \, \mathrm{g}$$

Question 74

Chemistry · Solutions · Numerical

When $1 \, \mathrm{g}$ each of compounds AB and $\mathrm{AB_2}$ are dissolved in $15 \, \mathrm{g}$ of water separately, they increased the boiling point of water by $2.7 \, \mathrm{K}$ and $1.5 \, \mathrm{K}$ respectively. The atomic mass of A (in amu) is ____ $\times 10^{-1}$ (Nearest integer) (Given: Molal boiling point elevation constant is $0.5 \, \mathrm{K \, kg \, mol^{-1}}$)

Answer: 25

Solution

For AB, $\Delta T_b = 2.7 \, \mathrm{K}$ $2.7 = 1 \times 0.5 \times m$ $m = \frac{27}{5}$ Let molar mass of AB $= x$ So $\frac{1}{x} \times 1000 = \frac{27}{5}$ $x = 12.34$ For $\mathrm{AB_2}$, $\Delta T_b = 1.5 \, \mathrm{K}$ $1.5 = 1 \times 0.5 \times m$ $m = 3$ Let molar mass of $\mathrm{AB_2} = y$ So $\frac{1}{y} \times 1000 = 3$ $\frac{1000}{45}$ $y = 22.22$ Now let $a$ and $b$ be atomic masses of A and B respectively, then $a + b = 12.34$ ...(i) $a + 2b = 22.22$ ...(ii) $b = 22.22 - 12.34 = 9.88$ Now $a = 12.34 - 9.88 = 2.46$ $= 24.6 \times 10^{-1} = 25 \times 10^{-1}$

Question 75

Chemistry · Co-ordination Compounds · Numerical

The spin-only magnetic moment value of $\mathrm{M}^{n+-}$ ion formed among Ni, ZnMn and Cu that has the least enthalpy of atomisation is ________. (in nearest integer) Here $n$ is equal to the number of diamagnetic complexes among $\mathrm{K}_2 [\mathrm{NiCl}_4]$, $[\mathrm{Zn(H_2O)}_6] \mathrm{Cl}_2$, $\mathrm{K}_3 [\mathrm{Mn(CN)}_6]$ and $[\mathrm{Cu(PPh}_3)_3\mathrm{I}]$

Answer: 0

Solution

$K_2$ [$\mathrm{NiCl_4}$] $\Rightarrow$ $sp^3$, Paramagnetic [$\mathrm{Zn(H_2O)_6}$] $\mathrm{Cl_2}$ $\Rightarrow$ $sp^3 d^2$, Diamagnetic $K_3$ [$\mathrm{Mn(CN)_6}$] $\Rightarrow$ $d^2 sp^3$, Paramagnetic [$\mathrm{Cu(PPh_3)_3I}$] $\Rightarrow$ $sp^3$, Diamagnetic Hence the value of $n$ is 2 Least value of enthalpy of atomisation among Ni, Zn, Mn and Cu is of Zn $\mathrm{Zn}^{+2}$ : -[$\mathrm{Ar}]3d^{10}$ $\mu$ = 0