JEE Main 30 January 2024 Shift 2 question paper with solutions

JEE Main 30 January 2024 Shift 2: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Determinants · Single correct

Consider the system of linear equations $$x + y + z = 5, \; x + 2y + \lambda^2 z = 9$$ $$x + 3y + \lambda z = \mu, \; where \; \lambda, \mu \in \mathbb{R}.$$ Then, which of the following statement is NOT correct?

  1. System has infinite number of solution if $\lambda = 1$ and $\mu = 13$
  2. System is inconsistent if $\lambda = 1$ and $\mu \neq 13$
  3. System is consistent if $\lambda \neq 1$ and $\mu = 13$
  4. System has unique solution if $\lambda \neq 1$ and $\mu \neq 13$

Answer: (d)

Solution

Given the determinant equation: $$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{vmatrix} = 0$$ This simplifies to: $$2\lambda^2 - \lambda - 1 = 0$$ Solving for $\lambda$, we get: $$\lambda = 1, -\frac{1}{2}$$ Now consider the determinant: $$\begin{vmatrix} 1 & 1 & 5 \\ 2 & \lambda^2 & 9 \\ 3 & \lambda & \mu \end{vmatrix} = 0 \Rightarrow \mu = 13$$ For an infinite solution, $\lambda = 1$ and $\mu = 13$. For a unique solution, $\lambda \neq 1$. For no solution, $\lambda = 1$ and $\mu \neq 13$. If $\lambda \neq 1$ and $\mu \neq 13$. Considering the case when $\lambda = -\frac{1}{2}$ and $\mu \neq 13$, this will generate no solution case.

Question 2

Maths · Trigonometric Functions · Single correct

For $\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$, let $3 \sin(\alpha + \beta) = 2 \sin(\alpha - \beta)$ and a real number $k$ be such that $\tan \alpha = k \tan \beta$. Then the value of $k$ is equal to:

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

Answer: (b)

Solution

Given the equation $3 \sin \alpha \cos \beta + 3 \sin \beta \cos \alpha = 2 \sin \alpha \cos \beta - 2 \sin \beta \cos \alpha$. Simplifying, we have: $$5 \sin \beta \cos \alpha = - \sin \alpha \cos \beta$$ Dividing both sides by $\cos \alpha \cos \beta$, we get: $$\tan \beta = -\frac{1}{5} \tan \alpha$$ Therefore, we can write: $$\tan \alpha = -5 \tan \beta$$

Question 3

Maths · Conic Sections · Single correct

Let $A(\alpha, 0)$ and $B(0, \beta)$ be the points on the line $5x + 7y = 50$. Let the point $P$ divide the line segment $AB$ internally in the ratio $7 : 3$. Let $3x - 25 = 0$ be a directrix of the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the corresponding focus be $S$. If from $S$, the perpendicular on the x-axis passes through $P$, then the length of the latus rectum of $E$ is equal to

  1. $\frac{25}{3}$
  2. $\frac{32}{9}$
  3. $\frac{25}{9}$
  4. $\frac{32}{5}$

Answer: (d)

Solution

Given points: A = (10, 0) B = $\left$(0, $\frac{50}{7}$$\right$) P = (3, 5) The equation of the line is given by x = $\frac{25}{3}$ Calculations: ae = 3 $\frac{a}{e}$ = $\frac{25}{3}$ a = 5 b = 4 Length of LR = $\frac{2b^2}{a}$ = $\frac{32}{5}$

Question 4

Maths · Vector Algebra · Single correct

Let $\vec{a}$ = $\hat{i}$ + $\alpha$ $\hat{j}$ + $\beta$ $\hat{k}$, $\alpha$, $\beta$ $\in$ $\mathbb{R}$. Let a vector $\vec{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$ and |$\vec{b}|^2$ = 6, If $\vec{a}$ $\cdot$ $\vec{b}$ = 3$\sqrt{2}$, then the value of ($\alpha^2$ + $\beta^2$) $|\vec{a}\times\vec{b}|^2$ is equal to

  1. 90
  2. 75
  3. 95
  4. 85

Answer: (a)

Solution

Given $|\vec{b}|^2 = 6$; $|\vec{a}| \, |\vec{b}| \cos \theta = 3 \sqrt{2}$. $|\vec{a}|^2 \, |\vec{b}|^2 \cos^2 \theta = 18$. $|\vec{a}|^2 = 6$. Also $1 + \alpha^2 + \beta^2 = 6$. $\alpha^2 + \beta^2 = 5$. To find $$(\alpha^2 + \beta^2) \, |\vec{a}|^2 \, |\vec{b}|^2 \sin^2 \theta$$ $$= (5)(6)(6) \left( \frac{1}{2} \right)$$ $$= 90$$

Question 5

Maths · Applications of Derivatives · Single correct

Let $f(x) = (x + 3)^2(x - 2)^3$, $x \in [-4, 4]$. If $M$ and $m$ are the maximum and minimum values of $f$, respectively in $[-4, 4]$, then the value of $M - m$ is:

  1. 600
  2. 392
  3. 608
  4. 108

Answer: (c)

Solution

Given $f'(x) = (x+3)^2 \cdot 3(x-2)^2 + (x-2)^3 2(x+3)$. This simplifies to $5(x+3)(x-2)^2(x+1)$. We have $f'(x) = 0$ at $x = -3, -1, 2$. Calculating $f(-4) = -216$. For $f(-3) = 0$ and $f(4) = 49 \times 8 = 392$. The maximum $M = 392$ and the minimum $m = -216$. The difference $M - m = 392 + 216 = 608$. Ans = '3'

Question 6

Maths · Sequences and Series · Single correct

Let $a$ and $b$ be two distinct positive real numbers. Let $11^{th}$ term of a GP, whose first term is $a$ and third term is $b$, is equal to $p^{th}$ term of another GP, whose first term is $a$ and fifth term is $b$. Then $p$ is equal to

  1. 20
  2. 25
  3. 21
  4. 24

Answer: (c)

Solution

$1^{\text{st}}$ GP $\Rightarrow t_1 = a,\quad t_3 = b = ar^2 \Rightarrow r^2 = \dfrac{b}{a}$ $$t_{11} = ar^{10} = a(r^2)^5 = a\cdot\left(\frac{b}{a}\right)^5$$ $2^{\text{nd}}$ GP $\Rightarrow T_1 = a,\quad T_5 = ar^4 = b$ $$\Rightarrow r^4 = \left(\frac{b}{a}\right) \Rightarrow r = \left(\frac{b}{a}\right)^{1/4}$$ $$T_p = ar^{p-1} = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$$ $$t_{11} = T_p \Rightarrow a\left(\frac{b}{a}\right)^5 = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$$ $$\Rightarrow 5 = \frac{p-1}{4} \Rightarrow p = 21$$

Question 7

Maths · Conic Sections · Single correct

If $x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines $x + 2y + 7 = 0$ and $2x - y + 8 = 0$, then the value of $g + c + h - f$ equals

  1. 14
  2. 6
  3. 8
  4. 29

Answer: (a)

Solution

Cocus of point $P(x, y)$ whose distance from $x + 2y + 7 = 0$ and $2x - y + 8 = 0$ are equal is $$\frac{x + 2y + 7}{\sqrt{5}} = \pm \frac{2x - y + 8}{\sqrt{5}}$$ $$(x + 2y + 7)^2 - (2x - y + 8)^2 = 0$$ Combined equation of lines $$(x - 3y + 1)(3x + y + 15) = 0$$ $$3x^2 - 3y^2 - 8xy + 18x - 44y + 15 = 0$$ $$x^2 - y^2 - \frac{8}{3}xy + 6x - \frac{44}{3}y + 5 = 0$$ $$x^2 - y^2 + 2hxy + 2gx2 + 2fy + c = 0$$ $$h = \frac{4}{3}, \ g = 3, \ f = -\frac{22}{3}, \ c = 5$$ $$g + c + h - f = 3 + 5 - \frac{4}{3} + \frac{22}{3} = 8 + 6 = 14$$

Question 8

Maths · Vector Algebra · Single correct

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}| = 1$ and $|\vec{b} \times \vec{a}| = 2$. Then $|(\vec{b} \times \vec{a}) - \vec{b}|^2$ is equal to

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

Answer: (b)

Solution

Given $|\vec{b}| = 1$ and $|\vec{b} \times \vec{a}| = 2$. $\left(\vec{b} \times \vec{a}\right) \cdot \vec{b} = \vec{b} \cdot \left(\vec{b} \times \vec{a}\right) = 0$ $|\left(\vec{b} \times \vec{a}\right) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2$ $= 4 + 1 = 5$

Question 9

Maths · Applications of Integrals · Single correct

Let $y = f(x)$ be a thrice differentiable function in $(-5, 5)$. Let the tangents to the curve $y = f(x)$ at $(1, f(1))$ and $(3, f(3))$ make angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$, respectively with positive $x$-axis. If $$27 \int_{1}^{3} \left( (f'(t))^2 + 1 \right) f''(t) dt = \alpha + \beta \sqrt{3}$$ where $\alpha$, $\beta$ are integers, then the value of $\alpha + \beta$ equals

  1. -14
  2. 26
  3. -16
  4. 36

Answer: (b)

Solution

Given $y = f(x)$, it follows that $\frac{dy}{dx} = f'(x)$. $$\left( \frac{dy}{dx} \right)_{(1,(1))} = f'(1) = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \Rightarrow f'(1) = \frac{1}{\sqrt{3}}$$ $$\left( \frac{dy}{dx} \right)_{(3,r(3))} = f'(3) = \tan \frac{\pi}{4} = 1 \Rightarrow f'(3) = 1$$ $$27 \int_{1}^{3} \left( (f'(t))^2 + 1 \right) f''(t) dt = \alpha + \beta \sqrt{3}$$ Let $I = \int_{1}^{3} \left( (f'(t))^2 + 1 \right) f''(t) dt$. Given $f(t) = z \Rightarrow f'(t) dt = dz$. We have $z = f(3) = 1$ and $z = f(1) = \frac{1}{\sqrt{3}}$. Then, $$I = \int_{1/\sqrt{3}}^{1} \left( z^2 + 1 \right) dz = \left( \frac{z^3}{3} + z \right)_{1/\sqrt{5}}^{1}$$ This evaluates to: $$= \left( \frac{1}{3} + 1 \right) - \left( \frac{1}{3} \cdot \frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}} \right)$$ $$= \frac{4}{3} - \frac{10}{9\sqrt{3}} = \frac{4}{3} - \frac{10}{27\sqrt{3}}$$ Thus, $$\alpha + \beta \sqrt{3} = 27 \left( \frac{4}{3} - \frac{10}{27\sqrt{3}} \right) = 36 - 10 \sqrt{3}$$ Therefore, $\alpha = 36$, $\beta = -10$. Finally, $\alpha + \beta = 36 - 10 = 26$.

Question 10

Maths · Conic Sections · Single correct

Let P be a point on the hyperbola H : $\frac{x^2}{9} - \frac{y^2}{4} = 1$, in the first quadrant such that the area of triangle formed by P and the two foci of H is $2\sqrt{13}$. Then, the square of the distance of P from the origin is

  1. 18
  2. 26
  3. 22
  4. 20

Answer: (c)

Solution

Given $\($ $\frac{x^2}{9}$ - $\frac{y^2}{4}$ = 1 $\)$ with $\($ a^2 = 9, b^2 = 4 $\)$. $\($ b^2 = a^2 (e^2 - 1) $\Rightarrow$ e^2 = 1 + $\frac{b^2}{a^2}$ $\)$ $\($ e^2 = 1 + $\frac{4}{9}$ = $\frac{13}{9}$ $\)$ $\($ e = $\frac{\sqrt{13}}{3}$ $\Rightarrow$ s_1 s_2 = 2ae = 2 $\times$ 3 $\times$ $\frac{\sqrt{13}}{3}$ = 2$\sqrt{13}$ $\)$ Area of $\($ $\Delta$ PS_1 S_2 = $\frac{1}{2}$ $\times$ $\beta$ $\times$ s_1 s_2 = 2$\sqrt{13}$ $\)$ $\($ $\Rightarrow$ $\frac{1}{2}$ $\times$ $\beta$ $\times$ (2$\sqrt{13}$) = 2$\sqrt{13}$ $\Rightarrow$ $\beta$ = 2 $\)$ $\($ $\frac{\alpha^2}{9}$ - $\frac{\beta^2}{4}$ = 1 $\Rightarrow$ $\frac{\alpha^2}{9}$ - 1 = 1 $\Rightarrow$ $\alpha$^2 = 18 $\Rightarrow$ $\alpha$ = 3$\sqrt{2}$ $\)$ Distance of $\($ P $\)$ from origin $\($ = $\sqrt{\alpha^2 + \beta^2}$ $\)$ $\($ = $\sqrt{18 + 4}$ = $\sqrt{22}$ $\)$

Question 11

Maths · Probability · Single correct

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

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

Answer: (c)

Solution

\begin{tabular}{|c|c|c|} \hline & A & B \\ \hline $E_1:$ A is selected & 3W & 3W \\ \cline{2-3} & 7R & 2R \\ \hline \end{tabular} Given $E_1$: A is selected, $E_2$: B is selected, $E$: white ball is drawn. The probability $P(E_1/E)$ is given by: $$P(E_1/E) = \frac{P(E) \cdot P(E/E_1)}{P(E_1) \cdot P(E/E_1) + P(E_2) \cdot P(E/E_2)} = \frac{\frac{1}{2} \times \frac{3}{10}}{\frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{3}{5}}$$ Simplifying, we get: $$= \frac{3}{3+6} = \frac{1}{3}$$

Question 12

Maths · Integrals · Single correct

Let $f : \mathbb{R} \to \mathbb{R}$ be defined $f(x) = ae^{2x} + be^x + cx$. If $f(0) = -1$, $f'(\log_e 2) = 21$ and $$\int_0^{\log_4 4} (f(x) - cx) \, dx = \frac{39}{2}$$, then the value of $|a + b + c|$ equals:

  1. 16
  2. 10
  3. 12
  4. 8

Answer: (d)

Solution

Given $f(x) = ae^{2x} + be^x + cx$ and $f(0) = -1$. Therefore, $a + b = -1$. The derivative is $f'(x) = 2ae^{2x} + be^x + c$ and $f'(\ln 2) = 21$. Thus, $8a + 2b + c = 21$. The integral $$\int_0^{\ln 4} \left(ae^{2x} + be^x\right) \, dx = \frac{39}{2}$$ gives $$\left[ \frac{ae^{2x}}{2} + be^x \right]_0^{\ln 4} = \frac{39}{2} \Rightarrow 8a + 4b - \frac{a}{2} - b = \frac{39}{2}$$ leading to $15a + 6b = 39$. Solving $15a - 6a - 6 = 39$ gives $9a = 45 \Rightarrow a = 5$. Then $b = -6$. For $c$, $c = 21 - 40 + 12 = -7$. Finally, $a + b + c = -8$ and $|a + b + c| = 8$.

Question 13

Maths · Three Dimensional Geometry · Single correct

Let $L_1 : \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda (\hat{i} - \hat{j} + 2\hat{k}), \lambda \in \mathbb{R}$, $L_2 : \vec{r} = (\hat{j} - \hat{k}) + \mu (3\hat{i} + \hat{j} + p\hat{k}), \mu \in \mathbb{R}$ and $L_3 : \vec{r} = \delta (\ell \hat{i} + m \hat{j} + n \hat{k}) \delta \in \mathbb{R}$ Be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then the point which lies on $L_3$ is

  1. $(-1, 7, 4)$
  2. $(-1, -7, 4)$
  3. $(1, 7, -4)$
  4. $(1, -7, 4)$

Answer: (a)

Solution

Given $L_1 \perp L_2$. $$3 - 1 + 2P = 0$$ $$P = -1$$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & -1 \end{vmatrix} = -\hat{i} + 7\hat{j} + 4\hat{k}$$ Therefore, $(-\delta, 7\delta, 4\delta)$ will lie on $L_3$. For $\delta = 1$ the point will be $(-1, 7, 4)$.

Question 14

Maths · Continuity and Differentiability · Single correct

Let $a$ and $b$ be real constants such that the function $f$ defined by $f(x) = \begin{cases} x^2 + 3x + a, & x \leq 1 \\ bx + 2, & x > 1 \end{cases}$ be differentiable on $\mathbb{R}$. Then, the value of $\int_{-2}^{2} f(x) \, dx$ equals

  1. $\($ $\frac{15}{6}$ $\)$
  2. $\($ $\frac{19}{6}$ $\)$
  3. 21
  4. 17

Answer: (d)

Solution

Given that $f$ is continuous. $f'(x) = 2x + 3$, $k 1$, we have $a = b - 2$. Given that $f$ is differentiable. Therefore, $b = 5$ and $a = 3$. Now, evaluate the integral: $$\int_{-2}^{1} (x^2 + 3x + 3) \, dx + \int_{1}^{2} (5x + 2) \, dx$$ $$= \left[ \frac{x^3}{3} + \frac{3x^2}{2} + 3x \right]_{-2}^{1} + \left[ \frac{5x^2}{2} + 2x \right]_{1}^{2}$$ $$= \left( \frac{1}{3} + \frac{3}{2} + 3 \right) - \left( \frac{-8}{3} + 6 - 6 \right) + \left( 10 + 4 - \frac{5}{2} - 2 \right)$$ $$= 6 + \frac{3}{2} + 12 - \frac{5}{2} = 17$$

Question 15

Maths · Continuity and Differentiability · Single correct

Let $f : \mathbb{R} - \{0\} \to \mathbb{R}$ be a function satisfying $f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$ for all $x, y$, $f(y) \neq 0$. If $f'(1) = 2024$, then

  1. $x f'(x) - 2024 f(x) = 0$
  2. $x f'(x) + 2024 f(x) = 0$
  3. $x f'(x) + f(x) = 2024$
  4. $x f'(x) - 2023 f(x) = 0$

Answer: (a)

Solution

Given $f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$, $f'(1) = 2024$, and $f(1) = 1$. Partially differentiating with respect to $x$: $$f'\left(\frac{x}{y}\right) \cdot \frac{1}{y} = \frac{1}{f(y)} f'(x)$$ Let $y \to x$: $$f'(1) \cdot \frac{1}{x} = \frac{f'(x)}{f(x)}$$ $$2024 f(x) = x f'(x) \Rightarrow x f'(x) - 2024 f(x) = 0$$

Question 16

Maths · Complex Numbers and Quadratic Equations · Single correct

If $z$ is a complex number, then the number of common roots of the equation $z^{1985} + z^{100} + 1 = 0$ and $z^3 + 2z^2 + 2z + 1 = 0$, is equal to:

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

Answer: (b)

Solution

Given $z^{1985} + z^{100} + 1 = 0$ and $z^3 + 2z^2 + 2z + 1 = 0$. $$(z + 1)(z^2 - z + 1) + 2z(z + 1) = 0$$ $$(z + 1)(z^2 + z + 1) = 0$$ This implies $z = -1$, $z = w, w^2$. Now putting $z = -1$ does not satisfy. Now put $z = w$ $$w^{1985} + w^{100} + 1$$ $$w^2 + w + 1 = 0$$ Also, $z = w^2$ $$w^{3970} + w^{200} + 1$$ $$w + w^2 + 1 = 0$$ There are two common roots.

Question 17

Maths · Binomial Theorem · Single correct

Suppose $2 - p$, $p$, $2 - \alpha$, $\alpha$ are the coefficient of four consecutive terms in the expansion of $(1 + x)^n$. Then the value of $p^2 - \alpha^2 + 6\alpha + 2p$ equals

  1. 4
  2. 10
  3. 8
  4. Data Inconsistent

Answer: (a)

Solution

No solution. Refer to answer key.

Question 18

Maths · Inverse Trigonometric Functions · Single correct

If the domain of the function $f(x) = \log_e \left( \frac{2x+3}{4x^2+x-3} \right) + \cos^{-1} \left( \frac{2x-1}{x+2} \right)$ is $(\alpha, \beta]$, then the value of $5\beta - 4\alpha$ is equal to

  1. 10
  2. 12
  3. 11
  4. 9

Answer: (b)

Solution

Given $\($ $\frac{2x + 3}{4x^2 + x - 3}$ > 0 $\)$ and $\($ -1 $\leq$ $\frac{2x - 1}{x + 2}$ $\leq$ 1 $\)$. $\($ $\frac{2 \times 3 + 3}{(4x - 3)(x + 1)}$ > 0 $\)$. $\($ $\frac{3x + 1}{x + 2}$ $\geq$ 0 $\)$ and $\($ $\frac{x - 3}{x + 2}$ $\leq$ 0 $\)$. $\($ (-$\infty$/2 + -1 $\)$. $\($ (-2, 3] $\ldots$ $\left$[ -$\frac{1}{3}$, $\infty$ $\right$) $\ldots$ $\ldots$ (1) $\)$. $\($ $\left$[ -$\frac{1}{3}$, 3 $\right$] $\ldots$ $\ldots$ (3) (1) $\cap$ (2) $\cap$ (3) $\)$. $\($ $\left$( $\frac{3}{4}$, 3 $\right$] $\)$. $\($ $\alpha$ = $\frac{3}{4}$ $\)$ and $\($ $\beta$ = 3 $\)$. $\($ 5 $\beta$ - 4 $\alpha$ = 15 - 3 = 12 $\)$.

Question 19

Maths · Integrals · Single correct

Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined $f(x) = \frac{x}{(1+x^4)^{1/4}}$ and $g(x) = f(f(f(f(x))))$ then $18 \int_0^{\sqrt{2\sqrt{5}}} x^2 g(x) \, dx$

  1. 33
  2. 36
  3. 42
  4. 39

Answer: (d)

Solution

Given $f(x) = \frac{x}{(1 + x^4)^{1/4}}$. Then $f \circ f(x) = \frac{f(x)}{(1 + f(x)^4)^{1/4}} = \frac{\frac{x}{(1+x^4)^{1/4}}}{\left(1 + \frac{x^4}{1+x^4}\right)^{1/4}} = \frac{x}{(1 + 2x^4)^{1/4}}$. Therefore, $f(f(f(f(x)))) = \frac{x}{(1 + 4x^4)^{1/4}}$. Now, consider $18 \int_0^{\sqrt{2\sqrt{5}}} \frac{x^3}{(1 + 4x^4)^{1/4}} \, dx$. Let $1 + 4x^4 = t^4$. Then $16x^3 \, dx = 4t^3 \, dt$. Thus, $\frac{18}{4} \int_1^3 \frac{t^3}{t} \, dt = \frac{9}{2} \left( \frac{t^3}{3} \right)_1^3 = \frac{3}{2} [26] = 39$.

Question 20

Maths · Matrices · Single correct

Let $R = \begin{pmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{pmatrix}$ be a non-zero $3 \times 3$ matrix, where $x \sin \theta = y \sin \left( \theta + \frac{2\pi}{3} \right) = z \sin\left(\theta+\frac{4\pi}{3}\right) \neq 0, \qquad \theta\in(0,2\pi)$. For a square matrix $M$, let trace $(M)$ denote the sum of all the diagonal entries of $M$. Then, among the statements: (I) Trace $(R) = 0$ (II) If trace $(adj(adj(R))) = 0$, then $R$ has exactly one non-zero entry.

  1. Both (I) and (II) are true
  2. Neither (I) nor (II) is true
  3. Only (II) is true
  4. Only (I) is true

Answer: (b)

Solution

1. $x \sin \theta = y \sin \left( \theta + \frac{2\pi}{3} \right) = z \sin \left( \theta + \frac{4\pi}{3} \right) \neq 0$ implies $x, y, z \neq 0$. Also, $$\sin \theta + \sin \left( \theta + \frac{2\pi}{3} \right) + \sin \left( \theta + \frac{4\pi}{3} \right) = 0 \forall \theta \in \mathbb{R}$$ implies $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$$ implies $$xy + yz + zx = 0$$ (i) $Trace(R) = x + y + z$. If $x + y + z = 0$ and $xy + yz + zx = 0$, then $x = y = z = 0$. Statement (i) is False. (ii) $Adj(Adj(R)) = |R|R$. $Trace (Adj(Adj(R))) = xyz(x + y + z) \neq 0$. Statement (ii) is also False.

Question 21

Maths · Applications of Integrals · Numerical

Let $Y = Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the line $$Y - y = Y'(x)(X - x)$$ and the co-ordinate axes, where $(x, y)$ is any point on the curve, is always $$\frac{-y^2}{2Y'(x)} + 1, \; Y'(x) \neq 0.$$ If $Y(1) = 1$, then $12Y(2)$ equals .

Answer: 20

Solution

Given $$A = \frac{1}{2} \left( \frac{-y}{Y'(x)} + x \right) \left( y - xY/x \right) = \frac{-y^2}{2Y'(x)} + 1$$ The points are $$(0, y - xY'(x))$$ and $$\left( \frac{-y}{Y'(x)} + x, 0 \right)$$ The expression $$(-y + xY'(x))(y - xY'(x)) = -y^2 + 2Y'(x)$$ Simplifies to $$-y^2 + xyY'(x) + xyY'(x) - x^2[Y'(x)]^2$$ This gives $$2xy - y^2y'(x) = 2Y'(x)$$ The derivative is $$\frac{dy}{dx} = \frac{2xy - 2}{x^2}$$ Thus, $$\frac{dy}{dx} - \frac{2}{x}y = \frac{-2}{x^2}$$ The integrating factor is $$I.F. = e^{-2 \ln x} = \frac{1}{x^2}$$ The solution is $$y \cdot \frac{1}{x^2} = \frac{2}{3}x^{-3} + c$$ Substitute $x = 1$, $y = 1$: $$1 = \frac{2}{3} + c \Rightarrow c = \frac{1}{3}$$ Thus, $$Y = \frac{2}{3} \cdot \frac{1}{x} + \frac{1}{3}x^2$$ Finally, $$\Rightarrow 12Y(2) = \frac{5}{3} \times 12 = 20$$

Question 22

Maths · Three Dimensional Geometry · Numerical

Let a line passing through the point $(-1, 2, 3)$ intersect the lines $L_1 : \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2}$ at $M(\alpha, \beta, \gamma)$ and $L_2 : \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4}$ at $N(a, b, c)$. Then the value of $\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$ equals .

Answer: 196

Solution

Given the points M and N, we have the equations for M as $M(3\lambda + 1, 2\lambda + 2, -2\lambda - 1)$ and for N as $N(-3\mu - 2, -2\mu + 2, 4\mu + 1)$. The conditions are $\alpha + \beta + \gamma = 3\lambda + 2$ and $a + b + c = -\mu + 1$. The ratios are given by: $$\frac{3\lambda + 2}{-3\mu - 1} = \frac{2\lambda}{-2\mu} = \frac{-2\lambda - 4}{4\mu - 2}$$ Solving these, we get: $$3\lambda + 2\mu = 3\lambda + \lambda$$ $$2\mu = \lambda$$ $$2\lambda - \lambda = \lambda\mu + 2\mu$$ $$\lambda\mu = \lambda + 2\mu$$ Thus, $\lambda\mu = 2\lambda$ and $\mu = 2$ (since $\lambda \neq 0$). Therefore, $\lambda = 4$. Substituting back, we find: $$\alpha + \beta + \gamma = 14$$ $$a + b + c = -1$$ Finally, we have: $$\frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2} = 196$$

Question 23

Maths · Conic Sections · Numerical

Consider two circles $C_1 : x^2 + y^2 = 25$ and $C_2 : (x - \alpha)^2 + y^2 = 16$, where $\alpha \in (5, 9)$. Let the angle between the two radii (one to each circle) drawn from one of the intersection points of $C_1$ and $C_2$ be $\sin^{-1}\left(\frac{\sqrt{63}}{8}\right)$. If the length of common chord of $C_1$ and $C_2$ is $\beta$, then the value of $(\alpha \beta)^2$ equals

Answer: 1575

Solution

Given the circles $C_1: x^2 + y^2 = 25$ and $C_2: (x - \alpha)^2 + y^2 = 16$ with $5 < \alpha < 9$. The angle $\theta$ is given by $$\theta = \sin^{-1}\left(\frac{\sqrt{63}}{8}\right)$$ Thus, $$\sin \theta = \frac{\sqrt{63}}{8}$$ The area of $\triangle OAP$ is $$\frac{1}{2} \times \alpha \left(\frac{\beta}{2}\right) = \frac{1}{2} \times 5 \times 4 \sin \theta$$ This implies $$\alpha \beta = 40 \times \frac{\sqrt{63}}{8}$$ Therefore, $$\alpha \beta = 5 \times \sqrt{63}$$ Finally, $$(\alpha \beta)^2 = 25 \times 63 = 1575$$

Question 24

Maths · Binomial Theorem · Numerical

Let $\alpha=\sum_{k=0}^{n}\left(\frac{\left({}^{n}C_{k}\right)^2}{k+1}\right)$ and $\beta=\sum_{k=0}^{n-1}\left(\frac{{}^{n}C_{k}\,{}^{n}C_{k+1}}{k+2}\right)$ If $5\alpha=6\beta$, then $n$ equals

Answer: 10

Solution

$$\alpha = \sum_{k=0}^{n} \frac{{}^nC_k \cdot {}^nC_k}{k+1} \cdot \frac{n+1}{n+1}$$ $$= \frac{1}{n+1} \sum_{k=0}^{n} {}^{n+1}C_{k+1} \cdot {}^nC_{n-k}$$ $$\alpha = \frac{1}{n+1} \cdot {}^{2n+1}C_{n+1}$$ $$\beta = \sum_{k=0}^{n-1} {}^nC_k \cdot \frac{{}^nC_{k+1}}{k+2} \cdot \frac{n+1}{n+1}$$ $$= \frac{1}{n+1} \sum_{k=0}^{n-1} {}^nC_{n-k} \cdot {}^{n+1}C_{k+2}$$ $$= \frac{1}{n+1} \cdot {}^{2n+1}C_{n+2}$$ $$\frac{\beta}{\alpha} = \frac{{}^{2n+1}C_{n+2}}{{}^{2n+1}C_{n+1}}$$ $$\frac{\beta}{\alpha} = \frac{2n+1-(n+2)+1}{n+2} = \frac{5}{6}$$ $n = 10$

Question 25

Maths · Sequences and Series · Numerical

Let $S_n$ be the sum to n-terms of an arithmetic progression 3, 7, 11, $\ldots$ $\ldots$ . If $40 < \left( \frac{6}{n(n+1)} \sum_{k=1}^{n} s_k \right) < 42$, then $n$ equals ________.

Answer: 9

Solution

Given $S_n = 3 + 7 + 11 + \ldots n terms$. $$= \frac{n}{2} (6 + (n-1)4) = 3n + 2n^2 - 2n$$ $$= 2n^2 + n$$ $$\sum_{k=1}^{n} S_k = 2 \sum_{k=1}^{n} K^2 + \sum_{k=1}^{n} K$$ $$= 2 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$$ $$= n(n+1) \left[ \frac{2n+1}{3} + \frac{1}{2} \right]$$ $$= \frac{n(n+1)(4n+5)}{6}$$ Therefore, $$40 < \frac{6}{n(n+1)} \sum_{k=1}^{n} S_k < 42$$ $$40 < 4n + 5 < 42$$ $$35 < 4n < 37$$ $$n = 9$$

Question 26

Maths · Permutations and Combinations · Numerical

In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections: A, B and C. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C has 6 questions, then the total number of ways a student can select 15 questions is

Answer: 11376

Solution

If 4 questions from each section are selected, remaining 3 questions can be selected either in (1, 1, 1) or (3, 0, 0) or (2, 1, 0). Therefore, Total ways ${}^{8}C_{5}\cdot{}^{6}C_{5}\cdot{}^{6}C_{5}$ $+{}^{8}C_{6}\cdot{}^{6}C_{4}\cdot{}^{2}C_{0}$ $+{}^{8}C_{5}\cdot{}^{6}C_{6}\cdot{}^{6}C_{4}$ $+{}^{8}C_{4}\cdot{}^{6}C_{6}\cdot{}^{6}C_{5}$ $+{}^{8}C_{2}$ $\cdot{}^{8}C_{7}$ $\cdot{}^{6}C_{4}$ $\cdot{}^{6}C_{4}$ $=56\cdot6\cdot6$ $+28\cdot6\cdot15\cdot2$ $+56\cdot15\cdot2$ $+70\cdot6\cdot2$ $+8\cdot15\cdot15$ $$= 2016 + 5040 + 1680 + 840 + 1800 = 11376$$

Question 27

Maths · Relations and Functions · Numerical

The number of symmetric relations defined on the set {$1, 2, 3, 4$\} which are not reflexive is $\ldots$

Answer: 960

Solution

Question 28

Maths · Complex Numbers and Quadratic Equations · Numerical

The number of real solutions of the equation $x \left( x^2 + 3|x| + 5|x - 1| + 6|x - 2| \right) = 0$ is_____.

Answer: 1

Solution

Given $x = 0$ and $x^2 + 3|x| + 5|x - 1| + 6|x - 2| = 0$. Here all terms are positive except at $x = 0$. So there is no value of $x$ that satisfies this equation. Only solution $x = 0$. Number of solutions is 1.

Question 29

Maths · Applications of Integrals · Numerical

The area of the region enclosed by the parabola $(y - 2)^2 = x - 1$, the line $x - 2y + 4 = 0$ and the positive coordinate axes is ________.

Answer: 5

Solution

Solving the equations $(y - 2)^2 = x - 1$ and $x - 2y + 4 = 0$. $x = 2(y - 2)$ $$\frac{x^2}{4} = x - 1$$ $$x^2 - 4x + 4 = 0$$ $$(x - 2)^2 = 0$$ $x = 2$ Exclose area (w.r.t. y-axis) = $\int_0^3 x \, dy$ - Area of $\Delta$. $$= \int_0^3 ((y - 2)^2 + 1) \, dy - \frac{1}{2} \times 1 \times 2$$ $$= \int_0^3 (y^2 - 4y + 5) \, dy - 1$$ $$= \left[ \frac{y^3}{3} - 2y^2 + 5y \right]_0^3 - 1$$ $$= 9 - 18 + 15 - 1 = 5$$

Question 30

Maths · Statistics · Fill in the blank

The variance $\sigma^2$ of the data \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline $x_i$ & $0$ & $1$ & $5$ & $6$ & $10$ & $12$ & $17$ \\ \hline $f_i$ & $3$ & $2$ & $3$ & $2$ & $6$ & $3$ & $3$ \\ \hline \end{tabular} Is ___

Answer: 29

Solution

\begin{tabular}{|c|c|c|c|} \hline $x_i$ & $f_i$ & $f_i x_i$ & $f_i x_i^2$ \\ \hline $0$ & $3$ & $0$ & $0$ \\ \hline $1$ & $2$ & $2$ & $2$ \\ \hline $5$ & $3$ & $15$ & $75$ \\ \hline $6$ & $2$ & $12$ & $72$ \\ \hline $10$ & $6$ & $60$ & $600$ \\ \hline $12$ & $3$ & $36$ & $432$ \\ \hline $17$ & $3$ & $51$ & $867$ \\ \hline & $\sum f_i = 22$ & & $\sum f_i x_i^2 = 2048$ \\ \hline \end{tabular} $\therefore\ \sum f_i x_i = 176$ $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{176}{22} = 8$ $\sigma^2 = \dfrac{1}{N}\sum f_i x_i^2 - (\bar{x})^2$ $= \dfrac{1}{22} \times 2048 - (8)^2$ $= 93.090964 - 64$ $= 29.0909$

Physics

Question 31

Physics · Experimental Physics · Single correct

If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \, \mathrm{mm}$, the Vernier constant of travelling microscope is:

  1. $0.1 \, \mathrm{mm}$
  2. $0.1 \, \mathrm{cm}$
  3. $0.01 \, \mathrm{cm}$
  4. $0.01 \, \mathrm{mm}$

Answer: (d)

Solution

Given the equation $50V + S = 49S + S$. Simplifying, we have $S = 50(S - V)$. This leads to $0.5 = 50(S - V)$. Solving for $S - V$, we get $$S - V = \frac{0.5}{50} = \frac{1}{100} = 0.01 \, mm.$$

Question 32

Physics · Laws of Motion · Single correct

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of $60^\circ$ by a force of 10 $\,$ $\mathrm{N}$ parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is: [g = 10 $\,$ $\mathrm{m/s^2}$]

  1. 5 $\sqrt{3}$ $\,$ $\mathrm{J}$
  2. 5 $\,$ $\mathrm{J}$
  3. 5 $\times$ 10^3 $\,$ $\mathrm{J}$
  4. 10 $\,$ $\mathrm{J}$

Answer: (b)

Solution

Work done against frictional force $$= \mu N \times 10$$ $$= 0.1 \times 5 \times 10 = 5 \, \mathrm{J}$$

Question 33

Physics · Dual Nature of Radiation and Matter · Single correct

For the photoelectric effect, the maximum kinetic energy $(E_k)$ of the photoelectrons is plotted against the frequency $(\nu)$ of the incident photons as shown in figure. The slope of the graph gives

  1. Ratio of Planck's constant to electric charge
  2. Work function of the metal
  3. Charge of electron
  4. Planck's constant

Answer: (d)

Solution

Kinetic energy is given by the equation $$\mathrm{K.E.} = hf - \phi$$ where $hf$ is the energy of the photon and $\phi$ is the work function. The equation for the tangent of the angle $\theta$ is $$\tan \theta = h$$ where $h$ is the height.

Question 34

Physics · Thermal Properties of Matter · Single correct

A block of ice at $-10^\circ \mathrm{C}$ is slowly heated and converted to steam at $100^\circ \mathrm{C}$. Which of the following curves represent the phenomenon qualitatively:

Answer: (d)

Solution

Most of the students will think no option is correct but we will give the benefit of doubt to NTA since no coordinates are mentioned. In transition we should have two horizontal lines and they assumed the starting point is $-10$.

Question 35

Physics · Nuclei · Single correct

In a nuclear fission reaction of an isotope of mass M, three similar daughter nuclei of same mass are formed. The speed of a daughter nuclei in terms of mass defect $\Delta M$ will be:

  1. $\sqrt{\frac{2c \Delta M}{M}}$
  2. $\frac{\Delta M c^2}{3}$
  3. $c \sqrt{\frac{2 \Delta M}{M}}$
  4. $c \sqrt{\frac{3 \Delta M}{M}}$

Answer: (c)

Solution

(X) $\rightarrow$ (Y) + (Z) + (P) $M \frac{M}{3} \frac{M}{3} \frac{M}{3}$ $$\Delta M c^2 = \frac{1}{2} \frac{M}{3} V^2 + \frac{1}{2} \frac{M}{3} V^2 + \frac{1}{2} V^2$$ $$V = c \sqrt{\frac{2 \Delta M}{M}}$$

Question 36

Physics · Thermodynamics · Single correct

Choose the correct statement for processes A $\&$ B shown in figure.

  1. $PV^\gamma = k$ for process B and $PV = k$ for process A.
  2. $PV = k$ for process B and A.
  3. $\frac{T^\gamma}{P^{\gamma-1}} = k$ for process A and $PV = k$ for process B.
  4. None of these

Answer: (a)

Solution

Steeper curve (B) is adiabatic. Adiabatic implies $PV^v = const.$ Or $P \left( \frac{T}{P} \right)^v = const.$ $$\frac{T^v}{P^{v-1}} = const.$$ Curve (A) is isothermal. $T = const.$ $PV = const.$

Question 37

Physics · Atoms · Single correct

An electron revolving in $n^{th}$ Bohr orbit has magnetic moment $\mu_n$. If $\mu_n \alpha n^x$, the value of $x$ is:

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

Answer: (b)

Solution

Magnetic moment = $i \pi r^2$ $$\mu = \frac{evr}{2}$$ $$\mu \alpha \left( \frac{1}{n} \right) n^2$$ $$\mu \alpha n$$ $$x = 1$$

Question 38

Physics · Alternating Current · Single correct

An alternating voltage $V(t) = 220 \sin 100 \pi t$ volt is applied to a purely resistive load of $50 \Omega$. The time taken for the current to rise from half of the peak value to the peak value is:

  1. $5\,\mathrm{ms}$
  2. $3.3\,\mathrm{ms}$
  3. $7.2\,\mathrm{ms}$
  4. $2.2\,\mathrm{ms}$

Answer: (b)

Solution

Rising half to peak $t = \frac{T}{6}$ $$t = \frac{2\pi}{6\omega} = \frac{\pi}{3\omega} = \frac{\pi}{300\pi} = \frac{1}{300} = 3.33 \, \mathrm{ms}$$

Question 39

Physics · Laws of Motion · Single correct

A block of mass $m$ is placed on a surface having vertical cross section given by $y = x^2/4$. If coefficient of friction is $0.5$, the maximum height above the ground at which block can be placed without slipping is:

  1. 1/4 m
  2. 1/2 m
  3. 1/6 m
  4. 1/3 m

Answer: (a)

Solution

Given $\($ $\frac{dy}{dx}$ = $\tan$ $\theta$ = $\frac{x}{2}$ = $\mu$ = $\frac{1}{2}$ $\)$. At $\($ x = 1, y = $\frac{1}{4}$ $\)$.

Question 40

Physics · System of Particles and Rotational Motion · Single correct

If the total energy transferred to a surface in time $t$ is $6.48 \times 10^5 \, \mathrm{J}$, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

  1. $2.46 \times 10^{-3} \, \mathrm{kg \, m/s}$
  2. $2.16 \times 10^{-3} \, \mathrm{kg \, m/s}$
  3. $1.58 \times 10^{-3} \, \mathrm{kg \, m/s}$
  4. $4.32 \times 10^{-3} \, \mathrm{kg \, m/s}$

Answer: (b)

Solution

Given $p = \frac{E}{C} = \frac{6.48 \times 10^5}{3 \times 10^8} = 2.16 \times 10^{-3}$.

Question 41

Physics · Wave Optics · Single correct

A beam of unpolarised light of intensity $I_0$ is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of $45^\circ$ relative to that of A. The intensity of emergent light is :

  1. $I_0/4$
  2. $I_0$
  3. $I_0/2$
  4. $I_0/8$

Answer: (a)

Solution

Intensity of emergent light $$= \frac{I_0}{2} \cos^2 45^\circ = \frac{I_0}{4}$$

Question 42

Physics · Gravitation · Single correct

Escape velocity of a body from earth is $11.2 \, \mathrm{km/s}$. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the plate is:

  1. $11.2 \, \mathrm{km/s}$
  2. $8.4 \, \mathrm{km/s}$
  3. $4.2 \, \mathrm{km/s}$
  4. $7.9 \, \mathrm{km/s}$

Answer: (d)

Solution

Given $R_P = \frac{R_E}{3}$, $M_P = \frac{M_E}{6}$. $V_e = \sqrt{\frac{2GM_e}{R_e}} \ldots (i)$ $V_P = \sqrt{\frac{2GM_P}{R_P}} \ldots (ii)$ $\frac{V_e}{V_P} = \sqrt{2}$ $V_P = \frac{V_e}{\sqrt{2}} = \frac{11.2}{\sqrt{2}} = 7.9 \, \mathrm{km/sec}$

Question 43

Physics · Electric Charges and Fields · Single correct

A particle of charge ' -q ' and mass ' m ' moves in a circle of radius ' r ' around an infinitely long line charge of linear density ' +$\lambda$ '. Then time period will be given as: (Consider k as Coulomb's constant)

  1. $T^2 = \frac{4\pi^2 \, m}{2k\lambda q} \, r^3$
  2. $T = 2\pi r \sqrt{\frac{m}{2k\lambda q}}$
  3. $T = \frac{1}{2\pi} \sqrt{\frac{m}{2k\lambda q}}$
  4. $T = \frac{1}{2\pi} \sqrt{\frac{2k\lambda q}{m}}$

Answer: (b)

Solution

Given $\($ $\frac{2k \lambda q}{r}$ = m $\omega$^2 r $\)$. $\($ $\omega$^2 = $\frac{2k \lambda q}{mr^2}$ $\)$. $\($ $\left$( $\frac{2\pi}{T}$ $\right$)^2 = $\frac{2k \lambda q}{mr^2}$ $\)$. $\($ T = 2$\pi$ r $\sqrt{\frac{m}{2k \lambda q}}$ $\)$.

Question 44

Physics · Physical World, Units and Measurements · Single correct

If mass is written as $m = k \, c^P \, G^{-1/2} \, h^{1/2}$ then the value of $P$ will be: (Constants have their usual meaning with $k$ a dimensionless constant)

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

Answer: (a)

Solution

Question 45

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

In the given circuit, the voltage across load resistance ($R_L$) is:

  1. 8.75 V
  2. 9.00 V
  3. 8.50 V
  4. 14.00 V

Answer: (a)

Solution

The current $i$ is calculated as follows: $$i = \frac{14}{4} = 3.5 \, mA$$ The voltage across $R_L$ is given by: $$V_L = i R_L = 3.5 \times 2.5 \, volt$$ $$= 8.75 \, volt$$

Question 46

Physics · Kinetic Theory · Single correct

If three moles of monoatomic gas $\left( \gamma = \frac{5}{3} \right)$ is mixed with two moles of a diatomic gas $\left( \gamma = \frac{7}{5} \right)$, the value of adiabatic exponent $\gamma$ for the mixture is:

  1. 1.75
  2. 1.40
  3. 1.52
  4. 1.35

Answer: (c)

Solution

Given $f_1 = 3$, $f_2 = 5$ and $n_1 = 3$, $n_2 = 2$. The formula for $f_{mixture}$ is: $$f_{mixture} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{9 + 10}{5} = \frac{19}{5}$$ The formula for $\gamma_{mixture}$ is: $$\gamma_{mixture} = 1 + \frac{2 \times 5}{19} = \frac{29}{19} = 1.52$$

Question 47

Physics · Laws of Motion · Single correct

Three blocks A, B and C are pulled on a horizontal smooth surface by a force of 80 N as shown in figure The tensions $T_1$ and $T_2$ in the string are respectively:

  1. 40 N, 64 N
  2. 60 N, 80 N
  3. 88 N, 96 N
  4. 80 N, 100 N

Answer: (a)

Solution

Given $a_A = a_B = a_C = \frac{F}{5+3+2} = \frac{80}{10} = 8 \, \mathrm{m/s^2}$. For block A: $$T_1 = 5 \times 8 = 40$$ For block B: $$T_2 - T_1 = 3 \times 8 \Rightarrow T_2 = 64$$

Question 48

Physics · Current Electricity · Single correct

When a potential difference $V$ is applied across a wire of resistance $R$, it dissipates energy at a rate $W$. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the same supply, the energy dissipation rate will become:

  1. 1/4 $W$
  2. 1/2 $W$
  3. 2 $W$
  4. 4 $W$

Answer: (d)

Solution

Given $$\frac{v^2}{R} = W$$ ... (i) $$\frac{v^2}{\frac{1}{2} \left( \frac{R}{2} \right)} = W'$$ ... (ii) From (i) and (ii), we get $$W' = 4 \, W$$

Question 49

Physics · Electric Charges and Fields · Single correct

Match List-I with List-II. \begin{tabular}{|c|l|c|l|} \hline & \textbf{List-I} & & \textbf{List-II} \\ \hline A. & Gauss's law of magnetism & I. & $\displaystyle \oint \vec{E}\cdot d\vec{a}=\frac{1}{\varepsilon_0}\int \rho\,dV$ \\ \hline B. & Faraday's law of electromagnetic induction & II. & $\displaystyle \oint \vec{B}\cdot d\vec{a}=0$ \\ \hline C. & Ampere's law & III. & $\displaystyle \oint \vec{E}\cdot d\vec{l}=-\frac{d}{dt}\int \vec{B}\cdot d\vec{a}$ \\ \hline D. & Gauss's law of electrostatics & IV. & $\displaystyle \oint \vec{B}\cdot d\vec{l}=-\mu_0 I$ \\ \hline \end{tabular} Choose the correct answer from the options given below:

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

Answer: (d)

Solution

Question 50

Physics · Motion in a Plane · Single correct

Projectiles A and B are thrown at angles of $45^\circ$ and $60^\circ$ with vertical respectively from top of a $400 \, \mathrm{m}$ high tower. If their ranges and times of flight are same, the ratio of their speeds of projection $v_A : v_B$ is :

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

Answer: (d)

Solution

For $u_A$ and $u_B$, time of flight and range cannot be the same. So above options are incorrect.

Question 51

Physics · Alternating Current · Numerical

A power transmission line feeds input power at $2.3 \, \mathrm{kV}$ to a step down transformer with its primary winding having 3000 turns. The output power is delivered at $230 \, \mathrm{V}$ by the transformer. The current in the primary of the transformer is $5 \, \mathrm{A}$ and its efficiency is $90\%$. The winding of transformer is made of copper. The output current of transformer is_____ $\mathrm{A}$.

Answer: 45

Solution

Given $\($ P_i = 2300 $\times$ 5 watt $\)$. $\($ P_0 = 2300 $\times$ 5 $\times$ 0.9 = 230 $\times$ I_2 $\)$. $\($ I_2 = 45 A $\)$.

Question 52

Physics · Mechanical Properties of Fluids · Numerical

A big drop is formed by coalescing 1000 small identical drops of water. If $E_1$ be the total surface energy of 1000 small drops of water and $E_2$ be the surface energy of single big drop of water, the $E_1 : E_2$ is $x : 1$ where $x =$

Answer: 10

Solution

Given $\rho \left( \frac{4}{3} \pi r^3 \right) 1000 = \frac{4}{3} \pi R^3 \rho$. Let $R = 10r$. Then, $E_1 = 1000 \times 4 \pi r^2 \times S$ and $E_2 = 4 \pi (10r)^2 S$. Therefore, $\[$ $\frac{E_1}{E_2}$ = $\frac{10}{1}$, $\ $x = 10 $\]$

Question 53

Physics · System of Particles and Rotational Motion · Numerical

Two discs of moment of inertia $I_1 = 4 \, \mathrm{kg} \, \mathrm{m}^2$ and $I_2 = 2 \, \mathrm{kg} \, \mathrm{m}^2$ about their central axes $\&$ normal to their planes, rotating with angular speeds $10 \, \mathrm{rad/s}$ $\&$ $4 \, \mathrm{rad/s}$ respectively are brought into contact face to face with their axe of rotation coincident. The loss in kinetic energy of the system in the process is_____ J.

Answer: 24

Solution

Given $I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_0$ (C.O.A.M.) gives $\omega_0 = 8 \, \mathrm{rad/s}$. $$E_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 = 216 \, \mathrm{J}$$ $$E_2 = \frac{1}{2} (I_1 + I_2) \omega_0^2 = 192 \, \mathrm{J}$$ Therefore, $\Delta E = 24 \, \mathrm{J}$

Question 54

Physics · Ray Optics and Optical Instruments · Numerical

In an experiment to measure the focal length (f) of a convex lens, the magnitude of object distance (x) and the image distance (y) are measured with reference to the focal point of the lens. The $y-x$ plot is shown in figure. The focal length of the lens is_____cm.

Answer: 20

Solution

Given $\frac{1}{f+20}-\frac{1}{-(f+20)}=\frac{1}{f}$. $\frac{2}{f+20}=\frac{1}{f}$ gives $f=20\,\mathrm{cm}$ Or $x_1x_2=f^2$ gives $f=20\,\mathrm{cm}$

Question 55

Physics · Mathematics in Physics · Numerical

A vector has magnitude same as that of $\vec{A} = 3\hat{j} + 4\hat{j}$ and is parallel to $\vec{B} = 4\hat{i} + 3\hat{j}$. The $x$ and $y$ components of this vector in first quadrant are $x$ and 3 respectively where $x =$

Answer: 4

Solution

Given $\overline{\mathbf{N}} = \overline{\mathbf{A}} |\mathbf{\hat{B}}| = \frac{5(4\mathbf{\hat{i}} + 3\mathbf{\hat{j}})}{5} = 4\mathbf{\hat{i}} + 3\mathbf{\hat{j}}$. Therefore, $x = 4$.

Question 56

Physics · Moving Charges and Magnetism · Numerical

The current of 5 A flows in a square loop of sides 1 m is placed in air. The magnetic field at the centre of the loop is $X \sqrt{2} \times 10^{-7} \, \mathrm{T}$. The value of $X$ is_____

Answer: 40

Solution

Given $$B = 4 \times \frac{\mu_0 i}{4 \pi (1/2)} \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right)$$ Simplifying, $$= 4 \times 10^{-7} \times 5 \times 2 \times \sqrt{2}$$ Finally, $$= 40 \sqrt{2} \times 10^{-7} \, \mathrm{T}$$

Question 57

Physics · Electric Charges and Fields · Numerical

Two identical charged spheres are suspended by string of equal lengths. The string make an angle of $37^\circ$ with each other. When suspended in a liquid of density $0.7 \, \mathrm{g/cm^3}$, the angle remains same. If density of material of the sphere is $1.4 \, \mathrm{g/cm^3}$, the dielectric constant of the liquid is_____ ( $tan 37^\circ$ = $\frac{3}{4}$ ).

Answer: 2

Solution

T $\cos$ $\theta$ = mg $\newline$ T $\sin$ $\theta$ = F_e $\newline$ $\tan$ $\theta$ = $\frac{F_e}{mg}$ $\newline$ $\tan$ $\theta$ = $\frac{F_e}{\rho_B V g}$ $\ldots$ (i) $\newline$ $\tan$ $\theta$ = $\frac{F_e}{\frac{k}{(\rho_B - \rho_L) V g}}$ $\ldots$ (ii) $\newline$ From Eq. (i) $\&$ (ii) $\newline$ $\rho$_B V g = ($\rho$_B - $\rho$_L) k V g $\newline$ 1.4 = 0.7k $\newline$ k = 2

Question 58

Physics · Oscillations · Numerical

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be _____ s. [ take $g = \pi^2 \, \mathrm{ms^{-2}}$]

Answer: 8

Solution

Acceleration due to gravity $g' = \frac{g}{4}$ $$T = 2\pi \sqrt{\frac{4\ell}{g}}$$ $$T = 2\pi \sqrt{\frac{4 \times 4}{g}}$$ $$T = 2\pi \frac{4}{\pi} = 8 \, \mathrm{s}$$

Question 59

Physics · Waves · Numerical

A point source is emitting sound waves of Power $16 \times 10^{-8} \, \mathrm{W}$ at the origin. The ratio of intensity (magnitude only) at two points located at a distances of $2 \, \mathrm{m}$ and $4 \, \mathrm{m}$ from the origin respectively will be _____ $\mathrm{Wm}^{-2}$.

Answer: 4

Solution

Intensity = Power/Area Ratio of intensity will be inverse square ratio of distances $$Ratio = \frac{4^2}{2^2} = 4$$

Question 60

Physics · Current Electricity · Fill in the blank

Two resistances of $100\,\Omega$ and $200\,\Omega$ are connected in series with a battery of $4\,\mathrm{V}$ and negligible internal resistance. A voltmeter is used to measure voltage across $100\,\Omega$ resistance, which gives reading as $1\,\mathrm{V}$. The resistance of voltmeter must be ___ $\Omega$.

Answer: 200

Solution

Given the circuit, we have the equation for the resistors in parallel: $$\frac{R_v \cdot 100}{R_v + 100} = \frac{200}{3}$$ Solving for $R_v$, we get: $$3R_v = 2R_v + 200$$ $$R_v = 200$$

Chemistry

Question 61

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

Which among the following purification methods is based on the principle of "Solubility" in two different solvents?

  1. Column Chromatography
  2. Sublimation
  3. Distillation
  4. Differential Extraction

Answer: (d)

Solution

Different layers are formed which can be separated in funnel. (Theory based).

Question 62

Chemistry · Alcohols, Phenols and Ethers · Single correct

Salicylaldehyde is synthesized from phenol, when reacted with

  1. CO_2, NaOH
  2. CCl_4, NaOH
  3. HCCl_3, NaOH
  4. $\mathrm{CHCl_3,\ NaOH}$

Answer: (d)

Solution

Question 63

Chemistry · Haloalkanes and Haloarenes · Single correct

Given below are two statements:\ Statement-I: High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow $S_N2$ mechanism.\ Statement-II: A secondary alkyl halide when treated with a large excess of ethanol follows $S_N1$ mechanism.\ In the light of the above statements, choose the most appropriate from the questions 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: (d)

Solution

Statement-I: Rate of $S_N \propto [\mathrm{R} - \mathrm{X}][\mathrm{Nu}]$. $S_N2$ reaction is favoured by high concentration of nucleophile (Nu) and less crowding in the substrate molecule. Statement-II: Solvolysis follows $S_N1$ path. Both are correct Statements.

Question 64

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

m-chlorobenzaldehyde on treatment with 50$\%$ KOH solution yields

Answer: (b)

Solution

Meta-chlorobenzaldehyde will undergo Cannizzaro reaction with 50$\%$ $\mathrm{KOH}$ to give m-chlorobenzoate ion and m-chlorobenzyl alcohol.

Question 65

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

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.\ Assertion A: $\mathrm{H}_2\mathrm{Te}$ is more acidic than $\mathrm{H}_2\mathrm{S}$.\ Reason R: Bond dissociation enthalpy of $\mathrm{H}_2\mathrm{Te}$ is lower than $\mathrm{H}_2\mathrm{S}$.\ In the light of the above statements. Choose the most appropriate from the options given below.

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

Answer: (b)

Solution

Due to lower bond dissociation enthalpy of $\mathrm{H_2Te}$, it ionizes to give $\mathrm{H^+}$ more easily than $\mathrm{H_2S}$.

Question 66

Chemistry · Hydrocarbons · Single correct

Product A and B formed in the following set of reactions are:

Answer: (b)

Solution

The reaction involves the conversion of a cyclohexene derivative to an alcohol. The first step is the hydroboration of the alkene using $\mathrm{B_2H_6}$, followed by oxidation with $\mathrm{H_2O_2}$ and $\mathrm{NaOH}$ to form the alcohol (B). The second step involves the acid-catalyzed hydration of the alkene to form the alcohol (A).

Question 67

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

IUPAC name of following compound is

  1. 2-Aminopentanenitrile
  2. 2-Aminobutanenitrile
  3. 3-Aminobutanenitrile
  4. 3-Aminopropanenitrile

Answer: (c)

Solution

The compound shown is 3-Aminobutanenitrile. The structure indicates the presence of an amino group ($\mathrm{NH_2}$) attached to the third carbon of a butanenitrile chain.

Question 68

Chemistry · Amines · Single correct

The products A and B formed in the following reaction scheme are respectively

Answer: (c)

Solution

The reaction sequence involves the following steps: 1. Nitration of benzene using concentrated $\mathrm{HNO_3}$ and $\mathrm{H_2SO_4}$ to form nitrobenzene ($\mathrm{NO_2}$ group attached to benzene). 2. Reduction of nitrobenzene using $\mathrm{Sn/HCl}$ to form aniline ($\mathrm{NH_2}$ group attached to benzene). 3. Diazotization of aniline using $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ at $0{-}5^\circ \mathrm{C}$ to form a diazonium salt ($\mathrm{N \equiv NCl^-}$). 4. Coupling of the diazonium salt with phenol to form an azo compound ($\mathrm{HO{-}N=N{-}}$ benzene).

Question 69

Chemistry · Co-ordination Compounds · Single correct

The molecule/ion with square pyramidal shape is:

  1. $[Ni(CN)_4]^{2-}$
  2. $PCl_5$
  3. $BrF_5$
  4. $PF_5$

Answer: (c)

Solution

The molecular geometry of $\mathrm{BrF_5}$ is square pyramidal.

Question 70

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

The orange colour of $\mathrm{K_2Cr_2O_7}$ and purple colour of $\mathrm{KMnO_4}$ is due to

  1. Charge transfer transition in both.
  2. $d \rightarrow d$ transition in $\mathrm{KMnO_4}$ and charge transfer transitions in $\mathrm{K_2Cr_2O_7}$.
  3. $d \rightarrow d$ transition in $\mathrm{K_2Cr_2O_7}$ and charge transfer transitions in $\mathrm{KMnO_4}$.
  4. $d \rightarrow d$ transition in both.

Answer: (a)

Solution

For $\mathrm{K_2Cr_2O_7} \rightarrow \mathrm{Cr^{+6}}$, there is no d-d transition. For $\mathrm{KMnO_4} \rightarrow \mathrm{Mn^{7+}}$, there is no d-d transition. This is due to charge transfer.

Question 71

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

Alkaline oxidative fusion of $\mathrm{MnO_2}$ gives "A" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are:

  1. $\mathrm{Mn_2O_7}$ and $\mathrm{MnO_4^-}$
  2. $\mathrm{MnO_4^{2-}}$ and $\mathrm{MnO_4^-}$
  3. $\mathrm{Mn_2O_3}$ and $\mathrm{MnO_4^{2-}}$
  4. $\mathrm{MnO_4^{2-}}$ and $\mathrm{Mn_2O_7}$

Answer: (b)

Solution

Alkaline oxidative fusion of $\mathrm{MnO_2}$: $$2\mathrm{MnO_2} + 4\mathrm{OH}^- + \mathrm{O_2} \rightarrow 2\mathrm{MnO_4^{2-}} + 2\mathrm{H_2O}$$ Electrolytic oxidation of $\mathrm{MnO_4^{2-}}$ in alkaline medium. $$\mathrm{MnO_4^{2-}} \rightarrow \mathrm{MnO_4^-} + e^-$$

Question 72

Chemistry · Solutions · Single correct

If a substance ' A ' dissolves in solution of a mixture of ' B ' and ' C ' with their respective number of moles as $n_A, n_B$ and $n_C$, mole fraction of C in the solution is:

  1. $\frac{n_C}{n_A \times n_B \times n_C}$
  2. $\frac{n_C}{n_A + n_B + n_C}$
  3. $\frac{n_C}{n_A - n_B - n_C}$
  4. $\frac{n_B}{n_A + n_B}$

Answer: (b)

Solution

Mole fraction of $C = \frac{n_C}{n_A + n_B + n_C}$

Question 73

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

Given below are two statements:\ Statement-I: Along the period, the chemical reactivity of the element gradually increases from group 1 to group 18.\ Statement-II: The nature of oxides formed by group 1 element is basic while that of group 17 elements is acidic.\ In the light above statements, choose the most appropriate from the questions given below:

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

Answer: (c)

Solution

Chemical reactivity of elements decreases along the period therefore statement I is false. Group 1 elements form basic nature oxides while group 17 elements form acidic oxides therefore statement II is true.

Question 74

Chemistry · Co-ordination Compounds · Single correct

The coordination geometry around the manganese in decacarbonyl dimanganese(0)

  1. Octahedral
  2. Trigonal bipyramidal
  3. Square pyramidal
  4. Square planar

Answer: (a)

Solution

The compound $\mathrm{Mn_2(CO)_{10}}$ has an octahedral geometry around each $\mathrm{Mn}$ atom.

Question 75

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Given below are two statements:\ Statement-I: Since fluorine is more electronegative than nitrogen, the net dipole moment of $\mathrm{NF}_3$ is greater than $\mathrm{NH}_3$.\ Statement-II: In $\mathrm{NH}_3$, the orbital dipole due to lone pair and the dipole moment of N-H bonds are in opposite direction, but in $\mathrm{NF}_3$ the orbital dipole due to lone pair and dipole moments of N-F bonds are in same direction.\ In the light of the above statements. Choose the most appropriate from the options given below.

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

Answer: (b)

Solution

Question 76

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

The correct stability order of carbocations is

  1. \quad (CH_3)_3C^{+}>CH_3-CH_2^{+}>(CH_3)_2CH^{+}>CH_3^{+}
  2. \quad CH_3^{+}>(CH_3)_2CH^{+}>CH_3-CH_2^{+}>(CH_3)_3C^{+}
  3. \quad (CH_3)_3C^{+}>(CH_3)_2CH^{+}>CH_3-CH_2^{+}>CH_3^{+}
  4. \quad CH_3^{+}>CH_3-CH_2^{+}>CH_3-CH(CH_3)^{+}>(CH_3)_3C^{+}

Answer: (c)

Solution

More number of hyperconjugable hydrogens, more stable is the carbocations.

Question 77

Chemistry · Solutions · Single correct

The solution from the following with highest depression in freezing point/lowest freezing point is

  1. 180 g of acetic acid dissolved in water
  2. 180 g of acetic acid dissolved in benzene
  3. 180 g of benzoic acid dissolved in benzene
  4. 180 g of glucose dissolved in water

Answer: (a)

Solution

Given $\Delta T_f$ is maximum when $i \times m$ is maximum. 1) $m_1 = \frac{180}{60} = 3$, $i = 1 + \alpha$ Hence $$\Delta T_f = (1 + \alpha) \cdot k_f = 3 \times 1.86 = 5.58^\circ \mathrm{C} (\alpha < 1)$$ 2) $m_2 = \frac{180}{60} = 3$, $i = 0.5$, $\Delta T_f = \frac{3}{2} \times k_f' = 7.68^\circ \mathrm{C}$ 3) $m_3 = \frac{180}{122} = 1.48$, $i = 0.5$, $\Delta T_f = \frac{1.48}{2} \times k_f' = 3.8^\circ \mathrm{C}$ 4) $m_4 = \frac{180}{180} = 1$, $i = 1$, $\Delta T_f = 1 \cdot k_f' = 1.86^\circ \mathrm{C}$ As per NCERT, $k_f' \,(\mathrm{H_2O}) = 1.86\mathrm{k} \cdot \mathrm{kgmol}^{-1}$ $k_f'\,(\mathrm{Benzene}) = 5.12\mathrm{k} \cdot \mathrm{kgmol}^{-1}$

Question 78

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

A and B formed in the following reactions are: $$\mathrm{CrO_2Cl_2 + 4NaOH \rightarrow A + 2NaCl + 2H_2O}$$ $$\mathrm{A + 2HCl + 2H_2O_2 \rightarrow B + 3H_2O}$$

  1. $A = \mathrm{Na_2CrO_4}$, $B = \mathrm{CrO_5}$
  2. $A = \mathrm{Na_2Cr_2O_4}$, $B = \mathrm{CrO_4}$
  3. $A = \mathrm{Na_2Cr_2O_7}$, $B = \mathrm{CrO_3}$
  4. $A = \mathrm{Na_2Cr_2O_7}$, $B = \mathrm{CrO_5}$

Answer: (a)

Solution

The balanced chemical equations are given as follows. (A) $\mathrm{CrO_2Cl_2} + 4\mathrm{NaOH} \rightarrow \mathrm{Na_2CrO_4} + 2\mathrm{NaCl} + 2\mathrm{H_2O}$ (B) $\mathrm{Na_2CrO_4} + 2\mathrm{H_2O_2} + 2\mathrm{HCl} \rightarrow \mathrm{CrO_5} + 2\mathrm{NaCl} + 3\mathrm{H_2O}$ In equation (B), $2\mathrm{NaCl}$ is missing from the balanced equation.

Question 79

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

Choose the correct statements about the hydrides of group 15 elements. A. The stability of the hydrides decreases in the order $\mathrm{NH}_3 > \mathrm{PH}_3 > \mathrm{AsH}_3 > \mathrm{SbH}_3 > \mathrm{BiH}_3$ B. The reducing ability of the hydrides increases in the order $\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3 < \mathrm{BiH}_3$ C. Among the hydrides, $\mathrm{NH}_3$ is strong reducing agent while $\mathrm{BiH}_3$ is mild reducing agent. D. The basicity of the hydrides increases in the order $\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3 < \mathrm{BiH}_3$\ Choose the most appropriate from the option given below:

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

Answer: (c)

Solution

On moving down the group, bond strength of $\mathrm{M} - \mathrm{H}$ bond decreases, which reduces the thermal stability but increases reducing nature of hydrides, hence A and B are correct statements.

Question 80

Chemistry · Electrochemistry · Single correct

Reduction potential of ions are given below: $\begin{array}{ccc} \mathrm{ClO_4^-} & \mathrm{IO_4^-} & \mathrm{BrO_4^-} \\ E^\circ = 1.19\,\mathrm{V} & E^\circ = 1.65\,\mathrm{V} & E^\circ = 1.74\,\mathrm{V} \end{array}$ $ The correct order of their oxidising power is:

  1. $\mathrm{ClO_4^- > IO_4^- > BrO_4^-}$
  2. $\mathrm{BrO_4^- > IO_4^- > ClO_4^-}$
  3. $\mathrm{BrO_4^- > ClO_4^- > IO_4^-}$
  4. $\mathrm{IO_4^- > BrO_4^- > ClO_4^-}$

Answer: (b)

Solution

Higher the value of $\oplus$ve SRP (Std. reduction potential) more is tendency to undergo reduction, so better is oxidising power of reactant. Hence, ox. Power: $\mathrm{BrO_4^- > IO_4^- > ClO_4^-}$

Question 81

Chemistry · Co-ordination Compounds · Numerical

Number of complexes which show optical isomerism among the following is $\underline{\phantom{123456789}}$ cis $-[\mathrm{Cr(ox)_2Cl_2}]^{3-}$, $[\mathrm{Co(en)_3}]^{3+}$,\ cis $-[\mathrm{Pt(en)_2Cl_2}]^{2+}$, cis $-[\mathrm{Co(en)_2Cl_2}]^{+}$,\ trans $-[\mathrm{Pt(en)_2Cl_2}]^{2+}$, trans $-[\mathrm{Cr(ox)_2Cl_2}]^{3-}$

Answer: 4

Solution

cis - [$\mathrm{Cr(ox)_2Cl_2}$]^{3-} $\rightarrow$ can show optical isomerism (no POS $\&$ COS) [$\mathrm{Co(en)_3}$]^{3+} $\rightarrow$ can show (no POS $\&$ COS) cis - [$\mathrm{Pt(en)_2Cl_2}$]^{2+} $\rightarrow$ can show (no POS $\&$ COS) cis - [$\mathrm{Co(en)_2Cl_2}$]^{+} $\rightarrow$ can show (no POS $\&$ COS) trans - [$\mathrm{Pt(en)_2Cl_2}$]^{2+} $\rightarrow$ can't show (contains POS $\&$ COS) trans - [$\mathrm{Cr(ox)_2Cl_2}$]^{3-} $\rightarrow$ can't show (contains POS $\&$ COS)

Question 82

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

$\mathrm{NO_2}$ required for a reaction is produced by decomposition of $\mathrm{N_2O_5}$ in $\mathrm{CCl_4}$ as by equation $$2\,\mathrm{N_2O_{5(g)}} \rightarrow 4\,\mathrm{NO_{2(g)}} + \mathrm{O_{2(g)}}$$ The initial concentration of $\mathrm{N_2O_5}$ is $3\,\mathrm{mol\,L^{-1}}$ and it is $2.75\,\mathrm{mol\,L^{-1}}$ after $30$ minutes. The rate of formation of $\mathrm{NO_2}$ is $x \times 10^{-3}\,\mathrm{mol\,L^{-1}\,min^{-1}}$, value of $x$ is ______

Answer: 17

Solution

Rate of reaction (ROR) $$= -\frac{1}{2} \frac{\Delta [\mathrm{N_2O_5}]}{\Delta t} = \frac{1}{4} \frac{[\mathrm{NO_2}]}{\Delta t} = \frac{\Delta [\mathrm{O_2}]}{\Delta t}$$ $$ROR = -\frac{1}{2} \frac{\Delta [\mathrm{N_2O_5}]}{\Delta t} = -\frac{1}{2} \frac{(2.75 - 3)}{30} \, \mathrm{mol \, L^{-1} \, min^{-1}}$$ $$ROR = -\frac{1}{2} \frac{(-0.25)}{30} \, \mathrm{mol \, L^{-1} \, min^{-1}}$$ $$ROR = \frac{1}{240} \, \mathrm{mol \, L^{-1} \, min^{-1}}$$ Rate of formation of $\mathrm{NO_2}$ $$= \frac{\Delta [\mathrm{NO_2}]}{\Delta t} = 4 \times ROR$$ $$= \frac{4}{240} = 16.66 \times 10^{-3} \, \mathrm{mol \, L^{-1} \, min^{-1}} \simeq 17 \times 10^{-3}.$$

Question 83

Chemistry · Thermodynamics · Numerical

Two reactions are given below: $$2\mathrm{Fe}_{(s)} + \frac{3}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{Fe}_2\mathrm{O}_{3(s)}, \Delta H^\circ = -822 \, \mathrm{kJ/mol}$$ $$\mathrm{C}_{(s)} + \frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}, \Delta H^\circ = -110 \, \mathrm{kJ/mol}$$ Then enthalpy change for following reaction $$3\mathrm{C}_{(s)} + \mathrm{Fe}_2\mathrm{O}_{3(s)} \rightarrow 2\mathrm{Fe}_{(s)} + 3\mathrm{CO}_{(g)}$$

Answer: 492

Solution

Given the reactions: $$2\mathrm{Fe}_{(s)} + \frac{3}{2}\mathrm{O}_2{(g)} \rightarrow \mathrm{Fe}_2\mathrm{O}_3{(s)}, \Delta H^\circ = -822 \, \mathrm{kJ/mol} \ldots (1)$$ $$\mathrm{C}_{(s)} + \frac{1}{2}\mathrm{O}_2{(g)} \rightarrow \mathrm{CO}_{(g)}, \Delta H^\circ = -110 \, \mathrm{kJ/mol} \ldots (2)$$ For the reaction: $$3\mathrm{C}_{(s)} + \mathrm{Fe}_2\mathrm{O}_3{(s)} \rightarrow 2\mathrm{Fe}_{(s)} + 3\mathrm{CO}_{(g)}, \Delta H_3 = ?$$ Using the equation: $$(3) = 3 \times (2) - (1)$$ Calculate $\Delta H_3$: $$\Delta H_3 = 3 \times \Delta H_2 - \Delta H_1$$ $$= 3(-110) + 822$$ $$= 492 \, \mathrm{kJ/mole}$$

Question 84

Chemistry · Biomolecules · Single correct

The total number of correct statements, regarding the nucleic acids is

  1. RNA is regarded as the reserve of genetic information.
  2. DNA molecule self-duplicates during cell division
  3. DNA synthesizes proteins in the cell.
  4. The message for the synthesis of particular proteins is present in DNA
  5. Identical DNA strands are transferred to daughter cells.

Answer: (c)

Solution

A. RNA is regarded as the reserve of genetic information. (False) B. DNA molecule self-duplicates during cell division. (True) C. DNA synthesizes proteins in the cell. (False) D. The message for the synthesis of particular proteins is present in DNA. (True) E. Identical DNA strands are transferred to daughter cells. (True)

Question 85

Chemistry · Equilibrium · Numerical

The pH of an aqueous solution containing 1M benzoic acid ($pK_a = 4.20$) and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is _____ mL.

Answer: 100

Solution

Given 1M benzoic acid and 1M sodium benzoate. Milimole for benzoic acid is $V_a \times 1$ and for sodium benzoate is $V_s \times 1$. The pH is given as $4.5$. Using the equation for pH: $$pH = pKa + \log \left( \frac{[salt]}{[acid]} \right)$$ Substitute the values: $$4.5 = 4.2 + \log \left( \frac{V_s}{V_a} \right)$$ This simplifies to: $$\frac{V_s}{V_a} = 2 \ldots (1)$$ Also, given: $$V_s + V_a = 300 \ldots (2)$$ Solving equations (1) and (2), we find: $$V_a = 100 \, ml$$

Question 86

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

Number of geometrical isomers possible for the given structure is/are _____

Answer: 4

Solution

3 stereocenters, symmetrical. Total Geometrical isomers $\rightarrow 4$. EE, ZZ, EZ (two isomers)

Question 87

Chemistry · Redox Reactions · Numerical

Total number of species from the following which can undergo disproportionation reaction . $\mathrm{H_2O_2, ClO_3^-, P_4, Cl_2, Ag,}$$Cu^{+1}$, $F_2,$ $NO_2,$ $K^+$

Answer: 6

Solution

Intermediate oxidation state of element can undergo disproportionation. $\mathrm{H_2O_2}$, $\mathrm{ClO_3^-}$, $\mathrm{P_4}$, $\mathrm{Cl_2}$, $\mathrm{Cu^{+1}}$, $\mathrm{NO_2}$

Question 88

Chemistry · Analytical Chemistry · Numerical

Number of metal ions characterized by flame test among the following is . $\mathrm{Sr}^{2+}, \mathrm{Ba}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Cu}^{2+}, \mathrm{Zn}^{2+}, \mathrm{Co}^{2+}, \mathrm{Fe}^{2+}$

Answer: 4

Solution

All the following metal ions will respond to flame test. $\mathrm{Sr^{2+}}$, $\mathrm{Ba^{2+}}$, $\mathrm{Ca^{2+}}$, $\mathrm{Cu^{2+}}$

Question 89

Chemistry · Haloalkanes and Haloarenes · Numerical

2-chlorobutane + $\mathrm{Cl}_2 \rightarrow \mathrm{C}_4\mathrm{H}_8\mathrm{Cl}_2$ (isomers) Total number of optically active isomers shown by $\mathrm{C}_4\mathrm{H}_8\mathrm{Cl}_2$, obtained in the above reaction is _______.

Answer: 6

Solution

The reaction of the given compound with $\mathrm{Cl_2/h\nu}$ produces three isomers. One of these isomers is optically inactive.

Question 90

Chemistry · Structure of Atom · Numerical

Number of spectral lines obtained in $\mathrm{He}^+$ spectra, when an electron makes transition from fifth excited state to first excited state will be

Answer: 10

Solution

5th excited state implies $n_1 = 6$. 1st excited state implies $n_2 = 2$. The change in $n$ is $\Delta n = n_1 - n_2 = 6 - 2 = 4$. The maximum number of spectral lines is given by $$\frac{\Delta n (\Delta n + 1)}{2} = \frac{4(4+1)}{2} = 10.$$