JEE Main 4 April 2025 Shift 2 question paper with solutions

JEE Main 4 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 · Applications of Derivatives · Single correct

Let $a > 0$. If the function $f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$ attains its local maximum and minimum values at the points $x_1$ and $x_2$ respectively such that $x_1 x_2 = 54$, then $a + x_1 + x_2$ is equal to :-

  1. 15
  2. 18
  3. 24
  4. 13

Answer: (b)

Solution

Given $f'(x) = 18x^2 - 90ax + 108a^2 = 0$. Let $x = 2a$ and $x = 3a$. Then $x_1 = 2a$ and $x_2 = 3a$. We have $x_1 x_2 = 54$. Therefore, $6a^2 = 54$. Solving for $a$, we get $a = 3$. Now, $a + x_1 + x_2 = 3 + 2 \times 3 + 3 \times 3 = 18$. Thus, the correct option is option (2).

Question 2

Maths · Applications of Derivatives · Single correct

Let $f$ be a differentiable function on $\mathbb{R}$ such that $f(2)=1$ and $f'(2)=4$. Let \[ \lim_{x \to 0} \left(f(2+x)\right)^{\frac{3}{x}} = e^{\alpha}. \] Then, the number of times the curve \[ y = 4x^3 - 4x^2 - 4(\alpha - 7)x - \alpha \] meets the $x$-axis is:

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

Answer: (a)

Solution

Given $\lim_{x \to 0} \left( f(2+x) \right)^3 = x$. We have $\lim_{e^x \to 0} \frac{x}{f(2+x) - 1} = 3$. Then $e^{3f'(2)} = (e)^{12} = (e)^a \Rightarrow a = 12$. The equation is $y = 4x^3 - 4x^2 - 4(a - 7)x - a$. Substituting $a = 12$, we get $y = 4x^3 - 4x^2 - 20x - 12$. The roots are $x = -1, -1, 3$. Therefore, the correct option is option (1).

Question 3

Maths · Inverse Trigonometric Functions · Single correct

The sum of the infinite series \[ \cot^{-1}\left(\frac{7}{4}\right) + \cot^{-1}\left(\frac{19}{4}\right) + \cot^{-1}\left(\frac{39}{4}\right) + \cot^{-1}\left(\frac{67}{4}\right) + \ldots \text{ is:} \]

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

Answer: (d)

Solution

Given $T_n = \tan^{-1} \left( \frac{4}{4n^2 + 3} \right)$. $T_n = \tan^{-1} \left( \left( n + \frac{1}{2} \right) - \left( n - \frac{1}{2} \right) \right)$. $T_n = \tan^{-1} \left( \frac{n + \frac{1}{2} - (n - \frac{1}{2})}{1 + \left( n + \frac{1}{2} \right) \left( n - \frac{1}{2} \right)} \right)$. $T_n = \tan^{-1} \left( n + \frac{1}{2} \right) - \tan^{-1} \left( n - \frac{1}{2} \right)$. $T_1 + T_2 + \ldots + T_n = \tan^{-1} \left( n + \frac{1}{2} \right) - \tan^{-1} \left( \frac{1}{2} \right)$. $S_{\infty} = \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{2} \right)$. option (4)

Question 4

Maths · Sets · Single correct

Let A = $\{$-3, -2, -1, 0, 1, 2, 3$\}$ and R be a relation on A defined by $xRy$ if and only if $2x - y \in \{0, 1\}$. Let $l$ be the number of elements in R. Let $m$ and $n$ be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then $l + mn$ is equal to :-

  1. 18
  2. 17
  3. 15
  4. 16

Answer: (b)

Solution

Given the equations $2x - y = 0$ and $2x - y = 1$. The solutions are: For $2x - y = 0$: $\{$(0, 0), (-1, -2), (1, 2)$\}$ For $2x - y = 1$: $\{$(0, -1), (1, 1), (2, 3), (-1, -3)$\}$ Total: $(0, 0), (-1, -2), (1, 2), (0, -1), (1, 1), (2, 3), (-1, -3)$ Reflexive: $m = 5$, $\ell = 7$ Symmetric: $n = 5$, $\ell + m + n = 17$ Option (2)

Question 5

Maths · Complex Numbers and Quadratic Equations · Single correct

Let the product of $\omega_1 = (8 + i) \sin \theta + (7 + 4i) \cos \theta$ and $\omega_2 = (1 + 8i) \sin \theta + (4 + 7i) \cos \theta$ be $\alpha + i \beta$, $i = \sqrt{-1}$. Let $p$ and $q$ be the maximum and the minimum values of $\alpha + \beta$ respectively.

  1. 140
  2. 130
  3. 160
  4. 150

Answer: (b)

Solution

Given $\omega_1 = (8 \sin \theta + 7 \cos \theta) + i (\sin \theta + 4 \cos \theta)$ and $\omega_2 = (\sin \theta + 4 \cos \theta) + i (8 \sin \theta + 7 \cos \theta)$. Then, $$\omega_1 \omega_2 = 8 \sin^2 \theta + 7 \sin \theta \cos \theta + 3 \sin \theta \cos \theta + 28 \cos^2 \theta - 8 \sin^2 \theta - 32 \sin \theta \cos \theta - 7 \sin \theta \cos \theta - 28 \cos^2 \theta + i (\sin^2 \theta + 4 \sin \theta \cos \theta + 4 \sin \theta \cos \theta + 16 \cos^2 \theta + 64 \sin^2 \theta + 56 \sin \theta \cos \theta + 56 \sin \theta \cos \theta + 49 \cos^2 \theta).$$ Simplifying, $$\omega_1 \omega_2 = 0 + i (65 \sin^2 \theta + 120 \sin \theta \cos \theta + 65 \cos^2 \theta).$$ Therefore, $$\alpha + \beta = 65 + 60 \sin 2q.$$ The maximum value is $$\alpha + \beta |_{\max} = 125$$ and the minimum value is $$\alpha + \beta |_{\min} = 5.$$ Thus, $$Ans. = 125 + 5 = 130.$$ option (2)

Question 6

Maths · Three Dimensional Geometry · Single correct

Let the values of $p$, for which the shortest distance between the lines $\frac{x+1}{3} = \frac{y}{4} = \frac{z}{5}$ and $\vec{r} = (p\hat{i} + 2\hat{j} + \hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 4\hat{k})$ is $\frac{1}{\sqrt{6}}$, be $a$, $b$, $(a < b)$. Then the length of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is :-

  1. 9
  2. $\frac{3}{2}$
  3. $\frac{2}{3}$
  4. 18

Answer: (c)

Solution

The shortest distance is given by $$\frac{|(\bar{a} - \bar{b}) \cdot (\bar{p} \times \bar{q})|}{|\bar{p} \times \bar{q}|}$$ where $$\bar{a} = -\hat{i} + 0\hat{j} + 0\hat{k}$$ $$\bar{b} = p\hat{i} + 2\hat{j} + \hat{k}$$ $$\bar{p} = 3\hat{i} + 4\hat{j} + 5\hat{k} - \bar{b} = (-1 - p)\hat{i} - 2\hat{j} - \hat{k}$$ $$\bar{q} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$ $$\frac{1}{16} = | -1 - p + 4 - 1|$$ $$\sqrt{6}$$ $$|-p + 2| = 1$$ $$p = 3$$ $$\frac{x^2}{1^2} + \frac{y^2}{3^3} = 1$$ $$L \cdot R = \frac{2a^2}{b} = \frac{2 \times 1}{3} = \frac{2}{3}$$ option (3)

Question 7

Maths · Conic Sections · Single correct

The axis of a parabola is the line $y = x$ and its vertex and focus are in the first quadrant at distances $\sqrt{2}$ and $2\sqrt{2}$ units from the origin, respectively. If the point $(1, k)$ lies on the parabola, then a possible value of $k$ is :-

  1. 4
  2. 9
  3. 3
  4. 8

Answer: (b)

Solution

Directrix $x + y = 0$. PS = PM $$\sqrt{(1 - 2)^2 + (K - 2)^2} = \frac{(1 + K)}{\sqrt{2}}$$ $$2K^2 + 8 - 8K + 2 = K^2 + 1 + 2K$$ $$K^2 - 10K + 9 = 0$$ $$K = 9$$ option (2)

Question 8

Maths · Relations and Functions · Single correct

Let the domains of the functions $$f(x) = \log_4 \log_3 \log_7 \left( 8 - \log_2 (x^2 + 4x + 5) \right)$$ and $$g(x) = \sin^{-1} \left( \frac{7x + 10}{x^2 + 4x + 5} \right)$$ be $(\alpha, \beta)$ and $[\gamma, \delta]$, respectively. Then $\alpha^2 + \beta^2 + \gamma^2 + \delta^2$ is equal to:

  1. 15
  2. 13
  3. 16
  4. 14

Answer: (a)

Solution

Given $\log_3 \left( \log_7 \left( 8 - \log_2 (x^2 + 4x + 5) \right) \right) > 0$. $\log_2 (x^2 + 4x + 5) < 1$ $x^2 + 4x + 3 < 0$ $\Rightarrow x \in (-3, -1)$ $-1 \leq \frac{7x + 10}{x - 2} \leq 1$ $\Rightarrow x \in [-2, -1]$ $\alpha = -3, \beta = -1, \gamma = -2, \delta = -1$ $\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 15$ option (1)

Question 9

Maths · Conic Sections · Single correct

A line passing through the point $A(-2, 0)$, touches the parabola $P: y^2 = x - 2$ at the point $B$ in the first quadrant. The area, of the region bounded by the line $AB$, parabola $P$ and the $x$-axis, is :-

  1. $\frac{7}{3}$
  2. 2
  3. $\frac{8}{3}$
  4. 3

Answer: (c)

Solution

Given $y = m(x + 2)$. $$y^2 = x - 2$$ $$(m(n + 2))^2 = n - 2$$ $$m^2 x^2 + (4m^2 - 1)x + (4m^2 + 2) = 0$$ $$D = 0$$ $$(4m^2 - 1)^2 - 4m^2(4m^2 + 2) = 0$$ $$m = \frac{1}{4}$$ $$y = \frac{1}{4}(n + 2)$$ and point of tangency $(6, 2)$. Area $A = \int_0^2 \left((y^2 + 2) - (4y - 2)\right) \, dy$ $$A = \frac{8}{3}$$ option (3)

Question 10

Maths · Conic Sections · Single correct

Let the sum of the focal distances of the point P(4, 3) on the hyperbola H : $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $8 \sqrt{\frac{5}{3}}$. If for $H$, the length of the latus rectum is $l$ and the product of the focal distances of the point P is $m$, then $9l^2 + 6m$ is equal to :-

  1. 184
  2. 186
  3. 185
  4. 187

Answer: (c)

Solution

Given $ex + a + ex - a = 8 \sqrt{\frac{5}{3}}$. $2ex = 8 \sqrt{\frac{5}{3}}$. $2e \times 4 = 8 \sqrt{\frac{5}{3}}$. $e = \sqrt{\frac{5}{3}}$. $b^2 = a^2 \left( \left( \sqrt{\frac{5}{3}} \right)^2 - 1 \right)$. $b^2 = \frac{2}{3} a^2$. $\frac{16}{a^2} - \frac{9}{b^2} = 1$ and $b^2 = \frac{2}{3} a^2$. $\Rightarrow a^2 = \frac{5}{2}, \; b^2 = \frac{5}{3}$. Now, $\ell = \frac{2 \; b^2}{a}$. $\ell^2 = \frac{4 \; b^4}{a^2}$. $9 \ell^2 = 36 \times \frac{25}{9 \times 5 \times 2}$. $9 \ell^2 = 40$. $m = (ex + a)(ex - a)$. $m = e^2 x^2 - a^2$. $= \frac{5}{3} \times 16 - 2 = 6$. $= 6 \; m = 145$. $9 \ell^2 + 6 \; m = 145$. $40 + 145 = 185$. option (3)

Question 11

Maths · Matrices · Single correct

Let the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$ satisfy $A^n = A^{n-2} + A^2 - I$ for $n \geq 3$. Then the sum of all the elements of $A^{50}$ is :-

  1. 53
  2. 52
  3. 39
  4. 44

Answer: (a)

Solution

Given $A^{50} = A^{48} + A^2 - I$. $$= A^{46} + 2 \left( A^2 - I \right)$$ $$= A^{44} + 3 \left( A^2 - I \right)$$ $$= A^2 + 24 \left( A^2 - I \right)$$ $$= 25 A^2 - 24I$$ $$= 25 \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} - 24 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$= \begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$$ Sum = 53 option (1)

Question 12

Maths · Sequences and Series · Single correct

If the sum of the first 20 terms of the series \[\frac{4 \cdot 1}{4 + 3 \cdot 1^2 + 1^4} + \frac{4 \cdot 2}{4 + 3 \cdot 2^2 + 2^4} + \frac{4 \cdot 3}{4 + 3 \cdot 3^2 + 3^4} + \frac{4 \cdot 4}{4 + 3 \cdot 4^2 + 4^4} + \cdots \] is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $m + n$ is equal to :-

  1. 423
  2. 420
  3. 421
  4. 422

Answer: (c)

Solution

Given the expression $$\sum_{r=1}^{20} \frac{4r}{4 + 3r^2 + r^4}$$ we can rewrite it as $$\sum_{r=1}^{20} \frac{4r}{(r^2 + r + 2)(r^2 - r + 2)}$$ which simplifies to $$2 \sum_{r=1}^{20} \left( \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right).$$ This further simplifies to $$2 \left( \frac{1}{2} - \frac{1}{4} \right)$$ which equals $$1 - \frac{1}{4} = \frac{3}{4}.$$ Continuing, we have $$\frac{1}{4} - \frac{1}{8} = \frac{1}{8},$$ $$\frac{1}{8} - \frac{1}{14} = \frac{3}{56},$$ $$\frac{1}{14} - \frac{1}{382} = \frac{368}{5348},$$ $$\frac{1}{382} - \frac{1}{422} = \frac{40}{161044}.$$ Therefore, the final result is $$2 \left( \frac{1}{2} - \frac{1}{422} \right) = 420.$$ Thus, the answer is 422 - 210 = 211. Therefore, the correct option is option (3).

Question 13

Maths · Binomial Theorem · Single correct

If $1^2 \cdot \left(^{15}C_1\right) + 2^2 \cdot \left(^{15}C_2\right) + 3^2 \cdot \left(^{15}C_3\right) + \ldots + 15^2 \cdot \left(^{15}C_{15}\right) = 2^m \cdot 3^n \cdot 5^k$, where $m, n, k \in \mathbb{N}$, then $m + n + k$ is equal to:

  1. 19
  2. 21
  3. 18
  4. 20

Answer: (a)

Solution

Given $$\sum_{r=1}^{15} r^2 \binom{15}{r} \Rightarrow 15 \sum_{r=1}^{15} r^{14} \binom{r-1}{r-1}$$ We have $$15 \sum_{r=1}^{15} (r-1+1)^{14} \binom{r-1}{r-1}$$ This becomes $$15 \cdot 14 \sum_{r=1}^{15} \binom{13}{r-2} + 15 \sum_{r=1}^{15} \binom{14}{r-1}$$ Simplifying further, $$15 \cdot 14 \cdot 2^{13} + 15 \cdot 2^{14}$$ This results in $$3^1 \cdot 2^{13} (70 + 10)$$ Which simplifies to $$3^1 \cdot 2^{13} \cdot 80$$ Finally, $$3^1 \cdot 5^1 \cdot 2^{17}$$ Thus, we have $$m = 17, n = 1, k = 1$$ option (1)

Question 14

Maths · Conic Sections · Single correct

Let for two distinct values of $p$ the lines $y = x + p$ touch the ellipse $E: \frac{x^2}{4^2} + \frac{y^2}{3^2} = 1$ at the points $A$ and $B$. Let the line $y$ = $x$ intersect $E$ at the points $C$ and $D$. Then the area of the quadrilateral $ABCD$ is equal to:

  1. 36
  2. 24
  3. 48
  4. 20

Answer: (b)

Solution

Point of contact are $$\left( \frac{\pm a^2 m}{\sqrt{a^2 m^2 + b^2}}, \frac{\pm b^2}{\sqrt{a^2 m^2 + b^2}} \right)$$ A $$\left( \frac{-16}{5}, \frac{9}{5} \right)$$ B $$\left( \frac{16}{5}, \frac{-9}{5} \right)$$ Point D is $$\left( \frac{12}{5}, \frac{12}{5} \right)$$ Area of ABD $$= \frac{1}{2} \begin{vmatrix} \frac{-16}{5} & \frac{9}{5} & 1 \\ \frac{16}{5} & \frac{-9}{5} & 1 \\ \frac{12}{5} & \frac{12}{5} & 1 \end{vmatrix}$$ $$= 12$$ Area of ABCD is $$= 24$$ option (2)

Question 15

Maths · Sequences and Series · Single correct

Consider two sets $A$ and $B$, each containing three numbers in A.P. Let the sum and the product of the elements of $A$ be 36 and $p$ respectively and the sum and the product of the elements of $B$ be 36 and $q$ respectively. Let $d$ and $D$ be the common differences of A.P's in $A$ and $B$ respectively such that $D = d + 3, d > 0$. If $\frac{p+q}{p-q} = \frac{19}{5}$, then $p - q$ is equal to

  1. 600
  2. 450
  3. 630
  4. 540

Answer: (d)

Solution

Let $A(a-d, a, a+d)$ and $B(b-D, b, b+D)$. Given $a = 12$ and $b = 12$. $p = 12(144 - d^2)$ $q = 12(144 - D^2)$ $\[$ $\frac{p+q}{p-q}$ = $\frac{19}{5}$ $\]$ $\[$ $\frac{p}{q}$ = $\frac{24}{14}$ = $\frac{12}{7}$ $\]$ $\[$ 144 - d^2 = $\frac{12}{7}$ $\]$ $\[$ 144 - (d^2 + 6d + 9) = $\frac{12}{7}$ $\]$ $\[$ 1008 - 7d^2 = -12d^2 - 72d + 1620 $\]$ $\[$ 5d^2 + 72d - 612 = 0 $\]$ $d = 6$ $D = 9$ $p - q = 12(D^2 - d^3)$ $\[$ = 12(81 - 36) $\]$ $\[$ = 12(45) $\]$ $\[$ = 540 $\]$ option (4)

Question 16

Maths · Differential Equations · Single correct

If a curve $y = y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation $$\left(7x^4 \cot y - e^x \csc y\right) \frac{dx}{dy} = x^5, x \geq 1,$$ then at $x = 2$, the value of $\cos y$ is:

  1. $\frac{2e^2-e}{64}$
  2. $\frac{2e^2+e}{64}$
  3. $\frac{2e^2-e}{128}$
  4. $\frac{2e^2+e}{128}$

Answer: (c)

Solution

\[ \frac{dy}{dx} = \frac{7\cot y}{x} - \frac{e^x \csc y}{x^5} \] \[ \frac{dy}{dx} = \frac{7\cot y}{\sin y \cdot x} - \frac{e^x}{\sin y \cdot x^5} \] \[ \sin y \frac{dy}{dx} - \cos y \cdot \frac{7}{x} = \frac{-e^x}{x^5} \] \text{let } -\cos y = t \[ \sin y \frac{dy}{dx} = \frac{dt}{dx} \] \[ \frac{dt}{dx} + \frac{7t}{x} = \frac{-e^x}{x^5} \] \[ \text{I.F.} = x^7 \] \[ t \cdot x^7 = -\int x^2 e^x\,dx \] \[ \cos y\, x^7 = x^2 e^x - 2\int x e^x\,dx \] \[ \cos y\, x^7 = x^2 e^x - 2xe^x + 2e^x + c \] \[ x = 1,\ y = \frac{\pi}{2},\ c = -e \] \[ \cos y = \frac{2e^2 - e}{128} \] \text{option (3)}

Question 17

Maths · Conic Sections · Single correct

The centre of a circle C is at the centre of the ellipse E : $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a > b$. Let C pass through the foci $F_1$ and $F_2$ of E such that the circle C and the ellipse E intersect at four points. Let P be one of these four points. If the area of the triangle $PF_1 F_2$ is 30 and the length of the major axis of E is 17, then the distance between the foci of E is :

  1. 26
  2. 13
  3. 12
  4. $\frac{13}{2}$

Answer: (b)

Solution

Given the equations: $\($ $\frac{1}{2}$ $\cdot$ $\mathrm{PF}$_1 $\cdot$ $\mathrm{PF}$_2 = 30 $\)$, $\($ $\mathrm{PF}$_1 + $\mathrm{PF}$_2 = 17 $\)$, $\($ $\mathrm{PF}$_1 = 12 $\)$, $\($ $\mathrm{PF}$_2 = 5 $\)$, and $\($ $\mathrm{F}$_1 $\mathrm{F}$_2 = 13 $\)$. The correct option is (2).

Question 18

Maths · Relations and Functions · Single correct

Let $f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and $2g(x) - 3g\left(\frac{1}{2}\right) = x, x > 0$. If $\alpha = \int_1^2 f(x) \, dx$, and $\beta = \int_1^2 g(x) \, dx$, then the value of $9\alpha + \beta$ is:

  1. 1
  2. 0
  3. 10
  4. 11

Answer: (d)

Solution

Given $f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$. $f\left(\frac{1}{x}\right) + 2f(x) = \frac{1}{x^2} + 5$. $f(x) = \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3}$. $\alpha = \int_1^2 \left(\frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3}\right) \, dx$. $$\left(-\frac{2}{3x} - \frac{x^3}{9} + \frac{5x}{3}\right)_1^2$$ $$= -\frac{1}{3} - \frac{8}{9} + \frac{10}{3} + \frac{2}{3} + \frac{1}{9} - \frac{5}{3}$$ $\alpha = 2 - \frac{7}{9} = \frac{11}{9}$. $2g(x) - 3g\left(\frac{1}{2}\right) = x$. $g\left(\frac{1}{2}\right) = -\frac{1}{2}$. $g(x) = \frac{x}{2} - \frac{3}{4}$. $\beta = \int_1^2 \left(\frac{x}{2} - \frac{3}{4}\right) \, dx$. $$\left(\frac{x^2}{4} - \frac{3x}{4}\right)_1^2 = 1 - \frac{3}{2} - \frac{1}{4} + \frac{3}{4} = 0$$ $9\alpha + \beta = 11$. option (4)

Question 19

Maths · Three Dimensional Geometry · Single correct

Let A be the point of intersection of the lines $L_1 : \frac{x-7}{1} = \frac{y-5}{0} = \frac{z-3}{-1}$ and $L_2 : \frac{x-1}{3} = \frac{y+3}{4} = \frac{z+7}{5}$. Let $B$ and $C$ be the point on the lines $L_1$ and $L_2$ respectively such that $AB = AC = \sqrt{15}$. Then the square of the area of the triangle $ABC$ is:

  1. 54
  2. 63
  3. 57
  4. 60

Answer: (a)

Solution

Given $\($ $\cos$ $\theta$ = $\left$| $\frac{3 + 0 - 5}{\sqrt{2 \sqrt{50}}}$ $\right$| $\)$. $\($ $\sin$ $\theta$ = $\frac{2}{10}$ = $\frac{1}{5}$ $\)$. $\($ $\sin$ $\theta$ = $\frac{\sqrt{24}}{5}$ $\)$. Area $\($ = $\frac{1}{2}$ ab $\sin$ $\theta$ $\)$. $\($ $\frac{1}{2}$ $\sqrt{15}$ $\sqrt{15}$ $\frac{\sqrt{24}}{5}$ $\)$. Square of area $\($ $\frac{15 \cdot 15 \cdot 24}{4.25}$ $\)$.

Question 20

Maths · Statistics · Single correct

Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is:

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

Answer: (a)

Solution

Given $24 + a + b = 4$. $$8$$ $a + b = 8$. $$2 = \frac{4 + 1 + 1 + 0 + 1 + 9 + (a - 4)^2 + (b - 4)^2}{8}$$ $16 = 48 + a^2 + b^2 - 8a - 8b$. $a^2 + b^2 = 32$. $32 = 2ab$. $ab = 16$. $a = 4b = 4$. $mode = 4$. $$mean deviation = \frac{2 + 1 + 1 + 0 + 1 + 3 + 0 + 0}{8} = 1$$ Option (1)

Question 21

Maths · Complex Numbers and Quadratic Equations · Numerical

If $\alpha$ is a root of the equation $x^2 + x + 1 = 0$ and $\sum_{k=1}^{n} \left( \alpha^k + \frac{1}{\alpha^k} \right)^2 = 20$, then $n$ is equal to

Answer: 11

Solution

Given $\alpha = \omega$. Therefore, $$\left( \omega^k + \frac{1}{\omega^k} \right)^2 = \omega^{2k} + \frac{1}{\omega^{2k}} + 2$$ $$= \omega^{2k} + \omega^k + 2 \therefore \omega^{3k} = 1$$ Thus, $$\sum_{k=1}^{n} (\omega^{2k} + \omega^k + 2) = 20$$ This implies $$(\omega^2 + \omega^4 + \omega^6 + \ldots + \omega^{2n}) + (\omega + \omega^2 + \omega^3 + \ldots + \omega^n) + 2n = 20$$ Now if $n = 3m$, $m \in I$. Then $0 + 0 + 2n = 20 \Rightarrow n = 10$ (not satisfy). If $n = 3m + 1$, then $$\omega^2 + \omega + 2n = 20$$ $$-1 + 2n = 20 \Rightarrow n = \frac{21}{2} (not possible)$$ If $n = 3m + 2$, $$(\omega^8 + \omega^{10}) + (\omega^4 + \omega^5) + 2n = 20$$ This implies $$(\omega^2 + \omega) + (\omega + \omega^2) + 2n = 20$$ $$2n = 22$$ $$n = 11$$ satisfies $n = 3m + 2$. Therefore, $n = 11$.

Question 22

Maths · Integrals · Numerical

If $\int \frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^9}\,dx = \frac{1}{m}\left(\left(\sqrt{1+x^2}+x\right)^n\left(n\sqrt{1+x^2}-x\right)\right)+C$, where $C$ is the constant of integration and $m,n\in\mathbb{N}$, then $m+n$ is equal to

Answer: 379

Solution

Rationalise $$\Rightarrow \int \left( \sqrt{1+x^2+x} \right)^{10} \times \left( \frac{\sqrt{1+x^2+x}}{\sqrt{1+x^2-x}} \right)^9 dx$$ $$\Rightarrow \int_1^{\sqrt{1+x^2+x}} \left( \sqrt{1+x^2+x} \right)^{19} dx$$ Put $\sqrt{1+x^2+x} = t$ $$\left( \frac{x}{\sqrt{1+x^2}+1} \right) dx = dt$$ $$dx = dt \sqrt{1+x^2}$$ Now as $\sqrt{1+x^2+x} = t$ $$so \sqrt{1+x^2-x} = \frac{1}{t}$$ $$\therefore \sqrt{1+x^2} = \frac{1}{2} \left( t + \frac{1}{t} \right)$$ Thus $I = \int t^{19} \cdot dt \cdot \frac{1}{t} \cdot \frac{1}{2} \left( t + \frac{1}{t} \right)$ $$\Rightarrow \frac{1}{2} \int \left( t^{19} + t^{17} \right) dt$$ $$= \frac{1}{2} \left[ \frac{t^{20}}{20} + \frac{t^{18}}{18} \right] + C$$ $$= \frac{1}{360} \left[ 9t + \frac{10}{t} \right] + C$$ $$= \frac{1}{360} \left[ 9 \left( t + \frac{1}{t} \right) + 1 \right] + C$$ $$\Rightarrow \left( \sqrt{1+x^2+x} \right)^{19} \left[ 9 \left( 2\sqrt{1+x^2} \right) + \left( \sqrt{1+x^2-x} \right) \right] + C$$ $$\Rightarrow \left( \sqrt{1+x^2+x} \right)^{19} \frac{1}{360} \left[ 19\sqrt{1+x^2-x} \right] + C$$ $$\therefore m = 360, n = 19$$ $$\therefore m+n = 379$$

Question 23

Maths · Probability · Numerical

A card from a pack of 52 cards is lost. From the remaining 51 cards, $n$ cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$, the $n$ is equal to

Answer: 2

Solution

n cards are drawn and are found all spade, thus remaining spades = $13 - x$, remaining total cards = $52 - x$. Now given that $P(lost card is spade) = \frac{11}{50}$. i.e. $$\frac{{^{13-n}C_1}}{{^{52-n}C_1}} = \frac{11}{50}$$ $$50(13-n) = 11(52-n)$$ $$39n = 78$$ $$n = 2$$

Question 24

Maths · Sets · Numerical

Let m and n, (m < n) be two 2-digit numbers. Then the total numbers of pairs (m, n), such that $\mathrm{gcd}(m, n) = 6$, is

Answer: 64

Solution

Let $m = 6a$, $n = 6b$. Given $m < n \Rightarrow a < b$ where $a$ and $b$ are co-prime numbers. Also, since $m$ and $n$ are 2-digit numbers, $10 \leq m \leq 99$ and $10 \leq n \leq 99$. Thus, $2 \leq a \leq 16$ and $2 \leq b \leq 16$ (since $a$ is an integer). Now, $2 \leq a < b \leq 16$ and $a$ and $b$ are co-prime. Therefore, if $a = 2$, $b = 3, 5, 7, 9, 11, 13, 15$; $a = 3$, $b = 4, 5, 7, 8, 10, 11, 13, 14, 16$; $a = 4$, $b = 5, 7, 9, 11, 13, 15$; $a = 5$, $b = 6, 7, 8, 9, 11, 12, 13, 14, 16$; $a = 6$, $b = 7, 11, 13$; $a = 7$, $b = 8, 9, 10, 11, 12, 13, 15, 16$; $a = 8$, $b = 9, 11, 13, 15$; $a = 9$, $b = 10, 11, 13, 14, 16$; $a = 10$, $b = 11, 13$; $a = 11$, $b = 12, 13, 14, 15, 16$; $a = 12$, $b = 13$; $a = 13$, $b = 14, 15, 16$; $a = 14$, $b = 15$; $a = 15$, $b = 16$. There are 64 ordered pairs.

Question 25

Maths · Vector Algebra · Fill in the blank

Let the three sides of a triangle $ABC$ be given by the vectors $2\hat{i} - \hat{j} + \hat{k}$, $\hat{i} - 3\hat{j} - 5\hat{k}$ and $3\hat{i} - 4\hat{j} - 4\hat{k}$. Let $G$ be the centroid of the triangle $ABC$. Then $6 \left( |\overrightarrow{AG}|^2 + |\overrightarrow{BG}|^2 + |\overrightarrow{CG}|^2 \right)$ is equal to

Answer: 164

Solution

By given data $\overrightarrow{AB} + \overrightarrow{AC} = \overrightarrow{CB}$. Let pv of $\overrightarrow{A}$ are $\overrightarrow{O}$ then $\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A}$. i.e. pv of $\overrightarrow{B} = -2\hat{i} - \hat{j} + \hat{k}$. $\overrightarrow{CA} = \overrightarrow{A} - \overrightarrow{C}$. i.e. pv of $\overrightarrow{C} = -(\hat{i} - 3\hat{j} - 5\hat{k})$. Now pv of centroid $\overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3} = \frac{\overrightarrow{0} + (2, -1, 1) + (-1, 3, 5)}{3}$. $\overrightarrow{G} = \frac{1}{3}(\hat{i} + 2\hat{j} + 6\hat{k})$. Now $\overrightarrow{AG} = \frac{1}{3}(\hat{i} + 2\hat{j} + 6\hat{k})$. $\Rightarrow |\overrightarrow{AG}|^2 = \frac{1}{9} \times 41$. $\overrightarrow{BG} = \left(\frac{1}{3} - 2\right)\hat{i} + \left(\frac{2}{3} + 1\right)\hat{j} + (2 - 1)\hat{k}$. $\Rightarrow |\overrightarrow{BG}|^2 = \frac{59}{9}$. $\overrightarrow{CG} = \left(\frac{1}{3} + 1\right)\hat{i} + \left(\frac{2}{3} - 3\right)\hat{j} + (2 - 5)\hat{k}$. $\Rightarrow |\overrightarrow{CG}|^2 = \frac{146}{9}$. Now $6 \left[|\overrightarrow{AG}|^2 + |\overrightarrow{BG}|^2 + |\overrightarrow{CG}|^2\right] = 6 \times \left[\frac{41}{9} + \frac{59}{9} + \frac{146}{9}\right] = 6 \times \frac{246}{9} = 164$.

Physics

Question 26

Physics · Nuclei · Single correct

A radioactive material P first decays into Q and then Q decays to non-radioactive material R. Which of the following figure represents time dependent mass of P, Q and R?

Answer: (b)

Solution

Question 27

Physics · Current Electricity · Single correct

There are ' n ' number of identical electric bulbs, each is designed to draw a power $p$ independently from the mains supply. They are now joined in series across the main supply. The total power drawn by the combination is:

  1. $np$
  2. $\frac{p}{n^2}$
  3. $\frac{p}{n}$
  4. $p$

Answer: (c)

Solution

Given $R_s = R_1 + R_2 + R_3 + \ldots + R_n$. $$V^2 = \frac{V^2}{P} + \frac{V^2}{P} + \ldots + \frac{V^2}{P_n}$$ Therefore, $P_s = \frac{P}{n}$.

Question 28

Physics · Thermal Properties of Matter · Single correct

Consider a rectangular sheet of solid material of length $\ell = 9 \, \mathrm{cm}$ and width $d = 4 \, \mathrm{cm}$. The coefficient of linear expansion is $\alpha = 3.1 \times 10^{-5} \, \mathrm{K}^{-1}$ at room temperature and one atmospheric pressure. The mass of sheet $m = 0.1 \, \mathrm{kg}$ and the specific heat capacity $C_v = 900 \, \mathrm{J} \, \mathrm{kg}^{-1} \, \mathrm{K}^{-1}$. If the amount of heat supplied to the material is $8.1 \times 10^2 \, \mathrm{J}$ then change in area of the rectangular sheet is :-

  1. $2.0 \times 10^{-6} \, \mathrm{m}^2$
  2. $3.0 \times 10^{-7} \, \mathrm{m}^2$
  3. $6.0 \times 10^{-7} \, \mathrm{m}^2$
  4. $4.0 \times 10^{-7} \, \mathrm{m}^2$

Answer: (a)

Solution

Given $\Delta Q = ms \Delta T$. $8.1 \times 10^2 = 0.1 \times 900 \times \Delta T$. $\Delta A = A_0 2 \alpha \Delta T = 2.0 \times 10^{-6} \, \mathrm{m^2}$.

Question 29

Physics · Atoms · Single correct

Given below are two statements: Statement (I): The dimensions of Planck's constant and angular momentum are same. Statement (II): In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (c)

Solution

Given $E = hf$. The dimensional formula is $ML^2 T^{-2} = [h] \times [T^{-1}]$. Therefore, $[h] = [ML^2 T^{-1}]$. The angular momentum $L = [MVR] = [ML^2 T^{-1}]$. Thus, $L = \frac{nh}{2\pi}$. Therefore, $L$ is an integral multiple of $\frac{h}{2\pi}$.

Question 30

Physics · Mechanical Properties of Solids · Single correct

A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of $10^5 \, \mathrm{N}$ at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement $\theta$ of the rod axis from its original position would be: (shear moduli, $G = 10^{10} \, \mathrm{N/m^2}$)

  1. $1/160\pi$
  2. $1/4\pi$
  3. $1/40\pi$
  4. $1/2\pi$

Answer: (a)

Solution

Shear moduli is given by the formula $\frac{\sigma_{shear}}{\theta}$. We have: $$10^{10} = \frac{10^5}{\pi \times 16 \times 10^{-4}} \times \frac{1}{\theta}$$ Solving for $\theta$ gives: $$\theta = \frac{1}{160\pi} Radian$$

Question 31

Physics · Current Electricity · Single correct

From the combination of resistors with resistance values $R_1 = R_2 = R_3 = 5\,\Omega$ and $R_4 = 10\,\Omega$, which of the following combination is the best circuit to get an equivalent resistance of $6\,\Omega$?

Answer: (a)

Solution

The reciprocal of the parallel resistance $R_P$ is given by: $$\frac{1}{R_P} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{1}{6}.$$

Question 32

Physics · Electric Charges and Fields · Single correct

A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to 'O' is 'E' is magnitude. What would be the magnitude of electric field at 'O' due to arc ABC?

  1. 2E
  2. $\sqrt{2}$E
  3. E/2
  4. Zero

Answer: (b)

Solution

The diagrams show two different configurations of electric fields. In the first diagram, the electric field $E$ is divided into components along the horizontal and vertical axes. The horizontal component is $\frac{E}{\sqrt{2}}$ and the vertical component is also $\frac{E}{\sqrt{2}}$. In the second diagram, the electric field is shown as $\sqrt{2}E$ along the horizontal axis and $\sqrt{2}E$ along the vertical axis. The points A, B, and C are marked on the circles, and the positive charges are distributed along the arcs.

Question 33

Physics · Kinetic Theory · Single correct

There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large vessel contains the gas at 8 $\,$ $\mathrm{kPa}$ at 1000 $\,$ $\mathrm{K}$ while the smaller vessel contains the gas at 7 $\,$ $\mathrm{kPa}$ at 500 $\,$ $\mathrm{K}$. If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 $\,$ $\mathrm{K}$, at steady state the pressure in the vessels will be (in kPa).

  1. 4.4
  2. 6
  3. 24
  4. 18

Answer: (b)

Solution

Number of masses will remain constant $$n_1 + n_2 = n_f$$ $$P_1 V_1 + P_2 V_2 = P_f V_f$$ $$RT_1 + RT_2 = RT_f$$ $$8 \times 2 \, V + 7 \times V = P_f (3 \, V)$$ $$R \times 1000 + R \times 500 = R \times 600$$ $$16 + 14 = P_f$$ $$1000 + 1000 = R \times 600$$ $$30 = P_f$$ $$1000 = 200$$ $$P_f = 6 \, kPa$$

Question 34

Physics · Gravitation · Single correct

An object is kept at rest at a distance of $3 \, R$ above the earth's surface where $R$ is earth's radius. The minimum speed with which it must be projected so that it does not return to earth is: (Assume $M =$ mass of earth, $G =$ Universal gravitational constant)

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

Answer: (a)

Solution

Given the equation for energy conservation: $$P_P + k_P = P_o + k_0$$ Substituting the values: $$-\frac{GMm}{4R} + \frac{1}{2} m V_P^2 = 0$$ Solving for $V_P$: $$V_P = \sqrt{\frac{GM}{2R}}$$ Choice 1 is correct.

Question 35

Physics · Electrostatic Potential and Capacitance · Single correct

Three parallel plate capacitors $C_1$, $C_2$ and $C_3$ each of capacitance $5\mu \mathrm{F}$ are connected as shown in figure. The effective capacitance between points $A$ and $B$, when the space between the parallel plates of $C_1$ capacitor is filled with a dielectric medium having dielectric constant of 4, is :

  1. $22.5\mu \mathrm{F}$
  2. $7.5\mu \mathrm{F}$
  3. $9\mu \mathrm{F}$
  4. $30\mu \mathrm{F}$

Answer: (c)

Solution

After dielectric $C_1 = 4C$ $C_1 = 4 \times 5 = 20 \, \mu \mathrm{F}$ $C_2 = C_3 = 5 \, \mu \mathrm{F}$ $C_1$ and $C_2$ are in series which is parallel to $C_3$. So $$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} + C_3 \Rightarrow \frac{20 \times 5}{20 + 5} + 5$$ $$= 4 + 5 = 9 \, \mu \mathrm{F}$$ Correct Option (3)

Question 36

Physics · Motion in a Straight Line · Single correct

The displacement x versus time graph is shown below. (A) The average velocity during $0$ to $3\,\mathrm{s}$ is $10\,\mathrm{m/s}$ (B) The average velocity during $3$ to $5\,\mathrm{s}$ is $0\,\mathrm{m/s}$ (C) The instantaneous velocity at $t = 2\,\mathrm{s}$ is $5\,\mathrm{m/s}$ (D) The average velocity during $5$ to $7\,\mathrm{s}$ and instantaneous velocity at $t = 6.5\,\mathrm{s}$ are equal (E) The average velocity from $t = 0$ to $t = 9\,\mathrm{s}$ is zero Choose the correct answer from the options given below:

  1. The average velocity during 0 to 3 s is 10 $\mathrm{m/s}$
  2. The average velocity during 3 to 5 s is 0 $\mathrm{m/s}$
  3. The instantaneous velocity at t = 2 $\mathrm{s}$ is 5 $\mathrm{m/s}$
  4. The average velocity during 5 to 7 s and instantaneous velocity at t = 6.5 $\mathrm{s}$ are equal

Answer: (a), (b), (c), (d), (e)

Solution

The average velocity $\langle \vec{v} \rangle$ is given by the change in displacement $\Delta \vec{s}$ over the change in time $\Delta t$, which is $\frac{S_f - S_i}{t_f - t_i}$. The instantaneous velocity $\vec{v}$ is the derivative of displacement with respect to time, $\frac{ds}{dt}$, which is the slope. (A) From 0 to 3 seconds, $\langle \vec{v} \rangle = \frac{5 - 0}{3 - 0} = \frac{5}{3} m/s$. (B) From 0 to 5 seconds, $\langle \vec{v} \rangle = \frac{5 - 5}{5 - 2} = 0$. (C) At $t = 2$, the slope is $\vec{v} = 5 m/s$. (D) From 5 to 7 seconds, $\langle \vec{v} \rangle = \frac{0 - 5}{2} = -2.5 m/s$. At $t = 6.5$ seconds, $\vec{v} = 10$. (E) From $t = 0$ to $t = 9$, $\langle \vec{v} \rangle = 0$.

Question 37

Physics · System of Particles and Rotational Motion · Single correct

A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is 8 m/s. The speed of the particle on the rim of the wheel at the same level as the centre of wheel, will be:

  1. 4$\sqrt{2}$ $\,$ $\mathrm{m/s}$
  2. 8 $\,$ $\mathrm{m/s}$
  3. 4 $\,$ $\mathrm{m/s}$
  4. 8$\sqrt{2}$ $\,$ $\mathrm{m/s}$

Answer: (a)

Solution

If $V_B = 2V$. Point A is the instantaneous center of rotation. Given $V_B = 8 \, \mathrm{m/s}$. $V = 4 \, \mathrm{m/s}$. $V_P = \sqrt{2}v \Rightarrow V_P = 4\sqrt{2} \, \mathrm{m/s}$. Correct (1).

Question 38

Physics · Experimental Physics · Single correct

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm):

  1. 0.001
  2. 0.002
  3. 0.0005
  4. 0.0025

Answer: (b)

Solution

Given $300 msd = 15 cm$. 1. Calculate $1 msd$: $$1 msd = \frac{15}{300} cm = 0.05 cm$$ 2. Given $25 vsd = 24 msd$, find $1 vsd$: $$1 vsd = \frac{24}{25} msd$$ 3. Calculate the least count (LC): $$LC = 1 msd - 1 vsd$$ Substitute the values: $$LC = 1 msd - \frac{24}{25} msd = \frac{1}{25} msd$$ 4. Substitute $1 msd = 0.05 cm$: $$LC = \frac{1}{25} \times 0.05 = 0.002 cm$$ The correct option is (2).

Question 39

Physics · Work, Energy and Power · Single correct

A block of mass $25 \, \mathrm{kg}$ is pulled along a horizontal surface by a force at an angle $45^\circ$ with the horizontal. The friction coefficient between the block and the surface is $0.25$. The displacement of $5 \, \mathrm{m}$ of the block is:

  1. $970 \, \mathrm{J}$
  2. $735 \, \mathrm{J}$
  3. $245 \, \mathrm{J}$
  4. $490 \, \mathrm{J}$

Answer: (c)

Solution

Block travels with uniform velocity. So $a = 0 \Rightarrow F \cos 45^\circ = friction$. $$\frac{F}{\sqrt{2}} = \mu \left[ mg - \frac{F}{\sqrt{2}} \right]$$ $$\frac{F}{\sqrt{2}} = 0.25 \left[ 25 \times 9.8 - \frac{F}{\sqrt{2}} \right]$$ $$\Rightarrow 1.25 \frac{F}{\sqrt{2}} = 61.25$$ $$F = \frac{61.25 \times \sqrt{2}}{1.25} = 49 \sqrt{2}$$ $$W_{ext} = FS \cos 45^\circ$$ $$= 49 \sqrt{2} \times 5 \times \frac{1}{\sqrt{2}} = 245 \, J$$

Question 40

Physics · Wave Optics · Single correct

Two polarisers $P_1$ and $P_2$ are placed in such a way that the intensity of the transmitted light will be zero. A third polariser $P_3$ is inserted in between $P_1$ and $P_2$, at the particular angle between $P_2$ and $P_3$. The transmitted intensity of the light passing through all three polarisers is maximum. The angle between the polarisers $P_2$ and $P_3$ is:

  1. $\frac{\pi}{4}$
  2. $\frac{\pi}{6}$
  3. $\frac{\pi}{8}$
  4. $\frac{\pi}{3}$

Answer: (a)

Solution

Through $P_2$, $I_1 = I_0 \sin^2 \left( \frac{\pi}{2} - \theta \right)$. $I_1 = I_0 \cos^2 \theta$. Through $P_3$, $I_{net} = (I_0 \cos^2 \theta) \sin^2 \theta$. $I_{nct} = \frac{I_0}{4} [\sin(2\theta)]^2$ for max $I_{net}$, $\theta = 45^\circ$. So angle between $P_2$ and $P_3 = \frac{\pi}{4}$. Correct Ans. (1)

Question 41

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

Consider a n-type semiconductor in which $n_e$ and $n_h$ are number of electrons and holes, respectively. $(A)$ Holes are minority carriers $(B)$ The dopant is a pentavalent atom $(C)$ $n_e n_h \neq n_i^2$ (where $n_i$ is number of electrons or holes in semiconductor when it is intrinsic form) $(D)$ $n_e n_h \geq n_i^2$ $(E)$ The holes are not generated due to the donors Choose the correct answer from the options given below:

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

Answer: (c)

Solution

$(A)$ In n type semiconductor holes are minority carriers and $e^-$ are majority carriers. $(B)$ Dopant are pentavalent atom. $(C)$ $n_e \cdot n_h = n_i^2$ for intrinsic semiconductor. $(E)$ In n type semiconductor primary source of holes generation are thermal excitation.

Question 42

Physics · Kinetic Theory · Single correct

Match List-I with List-II. \begin{tabular}{|l|l|} \hline \textbf{LIST-I} & \textbf{LIST-II} \\ \hline A. Isobaric & I. $\Delta Q = \Delta W$ \\ \hline B. Isochoric & II. $\Delta Q = \Delta U$ \\ \hline C. Adiabatic & III. $\Delta Q = 0$ \\ \hline D. Isothermal & IV. $\Delta Q = \Delta U + P\Delta V$ \\ \hline \end{tabular} $\Delta Q = \text{Heat supplied}$ $\Delta W = \text{Work done by the system}$ $\Delta U = \text{Change in internal energy}$ $P = \text{Pressure of the system}$ $\Delta V = \text{Change in volume of the system}$ Choose the correct answer from the options given below

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

Answer: (a)

Solution

(A) Isobaric ($P = C$) $$\Delta Q = \Delta U + P \Delta V$$ (B) Isochoric ($V = C$) $$\Delta Q = \Delta U$$ (C) Adiabatic ($\Delta Q = 0$) $$\Delta Q = 0$$ (D) Isothermal ($\Delta U = 0$) $$\Delta Q = \Delta W$$

Question 43

Physics · Waves · Single correct

Displacement of a wave is expressed as $x(t) = 5 \cos \left( 628t + \frac{\pi}{2} \right) \, \mathrm{m}$. The wavelength of the wave when its velocity is $300 \, \mathrm{m/s}$ is:

  1. 5 m
  2. 3 m
  3. 0.5 m
  4. 0.33 m

Answer: (b)

Solution

Given $x(t) = 5 \cos \left[ 628t + \frac{\pi}{2} \right] \, \mathrm{m}$. The velocity $v_{\omega} = 300 \, \mathrm{m/s}$. The wave velocity $v_w$ is given by $$v_w = \frac{\omega}{K}$$ Substituting the given values: $$300 = \frac{628}{K} \implies K = \frac{628}{300}$$ The wavelength $\lambda$ is given by $$2\pi = \frac{628}{\lambda}$$ Substituting the value of $K$: $$\frac{2 \times 3.14 \times 300}{628}$$ Thus, $$\lambda = 2 \, \mathrm{m}$$

Question 44

Physics · Ray Optics and Optical Instruments · Single correct

A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm. A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is :

  1. 45 cm
  2. 7.5 cm
  3. 22.5 cm
  4. 15 cm

Answer: (b)

Solution

For Convex mirror $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{v} - \frac{1}{30} = \frac{1}{30}$$ $$\frac{1}{v} = \frac{2}{30} = \frac{1}{15} \Rightarrow v = 15 \, cm$$ Image formed by convex mirror is at 45 cm from object so plane mirror should be placed midway at 22.5 cm from object so that both of their images may coincide. Therefore distance between both mirrors $$= 30 - 22.5 = 7.5 \, cm$$ Correct Answer : Option 2

Question 45

Physics · Physical World, Units and Measurements · Single correct

In an electromagnetic system, a quantity defined as the ratio of electric dipole moment and magnetic dipole moment has dimension of $\left[ M^P L^0 T^R A^S \right]$. The value of P and Q are:

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

Answer: (d)

Solution

Electric dipole moment $\left( \vec{P} \right) = q \times 2\ell$ Magnetic dipole moment $\left( \vec{M} \right) = IA$ $$\begin{bmatrix} P \\ M \end{bmatrix} = \begin{bmatrix} LTA \\ L^2 A \end{bmatrix} = L^{-1} \; T = M^0 \; L^{-1} \; T^1 \; A^0$$ After comparing values of P and Q are $0, -1$. Correct Answer: Option 4

Question 46

Physics · Moving Charges and Magnetism · Fill in the blank

A particle of charge 1.6 $\mu C$ and mass 16 $\mu g$ is present in a strong magnetic field of 6.28 $\mathrm{T}$. The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is ________ S. ($\pi = 3.14$)

Answer: 0

Solution

Angle between $\vec{V}$ of charge and $\vec{B}$ is $90^\circ$. Motion will be uniform circular motion. Time period is given by $$T = \frac{2\pi m}{qB} = \frac{2\pi \times 16 \times 10^{-9} \, \mathrm{kg}}{1.6 \times 10^{-6} \times 6.28}$$ $$T = 0.01 \, \mathrm{seconds}$$ NTA Answer is 10 Correct Answer is 0 (nearest integer)

Question 47

Physics · System of Particles and Rotational Motion · Fill in the blank

A solid sphere with uniform density and radius $R$ is rotating initially with constant angular velocity ($\omega_1$) about its diameter. After some time during the rotation its starts loosing mass at a uniform rate, with no change in its shape. The angular velocity of the sphere when its radius become $R/2$ is $x \omega_1$. The value of $x$ is ________

Answer: 32

Solution

When sphere is of radius $R$, its mass is $M$, when radius is reduced to $\frac{R}{2}$, mass will be reduced to $\frac{M}{8}$. Now by conservation of angular momentum $$(\tau_{ext} = 0)$$ $$L_1 = L_2$$ $$I_1 \omega_1 = I_2 \omega_2$$ $$\left( \frac{2}{5} MR^2 \right) \omega_1 = \left( \frac{2}{5} \left( \frac{M}{8} \right) \left( \frac{R}{2} \right)^2 \right) \omega_2$$ $$\omega_2 = 32 \omega_1$$ value of $x$ is 32 Answer is 32

Question 48

Physics · Electromagnetic Waves · Numerical

If an optical medium possesses a relative permeability of $\frac{10}{\pi}$ and relative permittivity of $\frac{1}{0.0885}$, then the velocity of light is greater in vacuum than that in this medium by ________ times. ($\mu_0 = 4\pi \times 10^{-7} \mathrm{H/m}, \epsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}$ $c = 3 \times 10^8 \mathrm{m/s}$)

Answer: 6

Solution

Since velocity of light in terms of $\mu$ and $\epsilon$ is $$V = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_0 \mu_r}} \times \frac{1}{\sqrt{\epsilon_0 \epsilon_r}}$$ $$= \frac{1}{\sqrt{\mu_r \epsilon_r}} \times \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$ $$= \frac{C}{\sqrt{\mu_r \epsilon_r}} = \frac{C}{\sqrt{\frac{10}{\pi}} \times \frac{1}{0.0885}}$$ $$= \frac{C}{\sqrt{36}} = \frac{C}{6}$$ $$V = \frac{C}{6}$$ $$C = 6V$$ Velocity of light in vacuum is greater by 6 times the velocity of light in medium. Answer is 6

Question 49

Physics · Wave Optics · Numerical

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is $x \times 10^{-3}$ cm, where $x$-value is

Answer: 15

Solution

Width of 20 maxima of double slit = width of central maxima of single slit $$20 \lambda D = \frac{2 \lambda D}{a}$$ $$\frac{d}{10} = \frac{1}{a}$$ $$a = \frac{d}{10} = \frac{1.5 \times 10^{-1}}{10} \, \mathrm{cm} = 15 \times 10^{-3} \, \mathrm{cm}$$ Value of $x$ is 15 Answer is 15

Question 50

Physics · Alternating Current · Fill in the blank

An inductor of self inductance 1 H connected in series with a resistor of $100\pi \mathrm{ohm}$ and an ac supply of $100\pi$ volt, 50 Hz. Maximum current flowing in the circuit is _______ A.

Answer: 1

Solution

Impedance of circuit $$Z = \sqrt{R^2 + (X_L)^2} = \sqrt{R^2 + (\omega L)^2}$$ $$= \sqrt{(100\pi)^2 + (2\pi \times 50 \times 1)^2}$$ $$= \sqrt{(100\pi)^2 + (100\pi)^2}$$ $$= \sqrt{2} \times 100\pi$$ $$I_{rms} = \frac{V}{Z} = \frac{100\pi}{\sqrt{2} \times 100\pi} = \frac{1}{\sqrt{2}}$$ $$I_{max} = \sqrt{2} I_{rms} = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1 Ampere$$

Chemistry

Question 51

Chemistry · Amines · Single correct

The correct order of basicity for the following molecules is:

  1. P > Q > R
  2. R > P > Q
  3. Q > P > R
  4. R > Q > P

Answer: (d)

Solution

According to Bredt's rule, it is a localized lone pair. Cross conjugation.

Question 52

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

The incorrect relationship in the following pairs in relation to ionisation enthalpies is:

  1. $\mathrm{Mn}^{+} < \mathrm{Cr}^{+}$
  2. $\mathrm{Mn}^{+} < \mathrm{Mn}^{2+}$
  3. $\mathrm{Fe}^{2+} < \mathrm{Fe}^{3+}$
  4. $\mathrm{Mn}^{2+} < \mathrm{Fe}^{2+}$

Answer: (d)

Solution

$Mn^{2+}$ : [$\mathrm{Ar}$] $3d^5$ Half filled stability More 1 E than $Fe^{2+}$ $Fe^{2+}$ : [$\mathrm{Ar}$] $3d^6$

Question 53

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Which among the following compounds give yellow solid when reacted with NaOI/NaOH?$ Choose the correct answer from the options given below:

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

Answer: (b)

Solution

Question 54

Chemistry · Biomolecules · Single correct

A dipeptide, "x" on complete hydrolysis gives "y" and "z". "y" on treatment with aq. $\mathrm{HNO_2}$ produces lactic acid. On the other hand "z" on heating gives the following cyclic molecule. Based on the information given, the dipeptide X is:

  1. valine-glycine
  2. alanine-glycine
  3. valine-leucine
  4. alanine-alanine

Answer: (b)

Solution

The compound (x) is alanine-glycine, represented as $\mathrm{NH_2{-}CH{-}C(=O){-}NH{-}CH_2{-}C(=O){-}OH}$. Upon hydrolysis, it yields compound (y) and compound (z). Compound (y) is $\mathrm{NH_2{-}CH{-}C(=O){-}OH}$, and compound (z) is $\mathrm{NH_2{-}CH_2{-}C(=O){-}OH}$. When compound (y) is treated with aqueous $\mathrm{HNO_2}$, it forms lactic acid, represented as $\mathrm{HO{-}CH{-}C(=O){-}OH}$. When compound (z) is heated, it forms a cyclic compound with the structure shown in the image.

Question 55

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

In which pairs, the first ion is more stable than the second?

  1. (B) & (D) only
  2. (A) & (B) only
  3. (B) & $(C)$ only
  4. (A) & $(C)$ only

Answer: (b)

Solution

Question 56

Chemistry · Haloalkanes and Haloarenes · Single correct

Given below are two statements: Statement (I): Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction. Statement (II): In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the $\beta$-carbon. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (b)

Solution

Question 57

Chemistry · Solutions · Single correct

Given below are two statements: Statement (I): Molal depression constant $K_f$ is given by $\frac{M_1RT_f}{\Delta S_{fus}}$, where symbols have their usual meaning. Statement (II): $K_f$ for benzene is less than the $K_f$ for water. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (d)

Solution

Statement-I Molar depression constant $k_f = \frac{M_1 R T_f^2}{\Delta H_{fus}}$ $$k_f = M_1 R T_f \left[ \frac{\Delta H_{fus}}{T_f} \right]$$ $$k_f = M_1 R T_f$$ $$k_f = \Delta S_{fus}$$ Hence statement-I is correct but $k_f$ for benzene $= 5.12 \, ^\circ \mathrm{C_{molal}}$ $k_f$ for water $= 1.86 \, ^\circ \mathrm{C_{molal}}$ Hence statement-II is incorrect

Question 58

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

The IUPAC name of the following compound is -

  1. 4-Hydroxyhept-1-en-6-yne
  2. 4- Hydroxyhept-6-en-1-yne
  3. Hept-6-en-1-yn-4-ol
  4. Hept-1-en-6-yn-4-ol

Answer: (d)

Solution

Question 59

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

Match List-I with List-II \begin{tabular}{|c|p{5.3cm}|c|p{4.8cm}|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{(Separation of)} & \multicolumn{2}{c|}{(Separation Technique)} \\ \hline (A) & Aniline from aniline-water mixture & (I) & Simple distillation \\ \hline (B) & Glycerol from spent-lye in soap industry & (II) & Fractional distillation \\ \hline (C) & Different fractions of crude oil in petroleum industry & (III) & Distillation at reduced pressure \\ \hline (D) & Chloroform-Aniline mixture & (IV) & Steam distillation \\ \hline \end{tabular} Choose the correct answer from the options given below :

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

Answer: (a)

Solution

(A) Aniline $-$ $\mathrm{H_2O}$ : Steam Distillation (1) (B) Glycerol from spent-lye in soap industry $-$ Distillation under reduced pressure $(C)$ Different fraction of crude oil in petroleum industry $-$ Fractional distillation (D) $\mathrm{CHCl_3}$$-$ Aniline $-$ Simple distillation

Question 60

Chemistry · Alcohols, Phenols and Ethers · Single correct

A toxic compound " A " when reacted with NaCN in aqueous acidic medium yields an edible cooking component and food preservative ' B '. " B " is converted to "C" by diborane and can be used as an additive to petrol to reduce emission. "C" upon reaction with oleum at 140°C yields an inhalable anesthetic "D". Identify "A", "B", "C" and "D", respectively.

  1. Methanol; formaldehyde; methyl chloride; chloroform
  2. Ethanol; acetonitrile; ethylamine; ethylene
  3. Methanol; acetic acid; ethanol; diethyl ether
  4. Acetaldehyde; 2-hydroxypropanoic acid; propanoic acid; dipropyl ether

Answer: (c)

Solution

Methanol is converted to acetic acid, then to ethanol, and finally to diethylether. Starting with methanol $\mathrm{CH_3OH_2^+}$ (A), it reacts with $\mathrm{Na^+CN^-}$ in an acidic medium to form $\mathrm{CH_3CN}$. This undergoes hydrolysis with $\mathrm{H_3O^+}$ to form acetic acid $\mathrm{CH_3C(=O)OH}$ (B). Acetic acid is reduced by $\mathrm{B_2H_6}$ to form ethanol $\mathrm{CH_3CH_2OH}$ (C). Ethanol reacts with oleum $\mathrm{(H_2SO_4 + SO_3)}$ to form diethylether $\mathrm{C_2H_5OC_2H_5}$ (D), which is an inhalable anesthetic.

Question 61

Chemistry · Co-ordination Compounds · Single correct

The correct order of $[\mathrm{FeF}_6]^{3-}$, $[\mathrm{CoF}_6]^{3-}$, $[\mathrm{Ni(CO)}_4]$ and $[\mathrm{Ni(CN)}_4]^{2-}$ complex species based on the number of unpaired electrons present is:

  1. $[\mathrm{FeF}_6]^{3-} > [\mathrm{CoF}_6]^{3-} > [\mathrm{Ni(CN)}_4]^{2-} > [\mathrm{Ni(CO)}_4]$
  2. $[\mathrm{Ni(CN)}_4]^{2-} > [\mathrm{FeF}_6]^{3-} > [\mathrm{CoF}_6]^{3-} > [\mathrm{Ni(CO)}_4]$
  3. $[\mathrm{CoF}_6]^{3-} > [\mathrm{FeF}_6]^{3-} > [\mathrm{Ni(CO)}_4] > [\mathrm{Ni(CN)}_4]^{2-}$
  4. $[\mathrm{FeF}_6]^{3-} > [\mathrm{CoF}_6]^{3-} > [\mathrm{Ni(CN)}_4]^{2-} = [\mathrm{Ni(CO)}_4]$

Answer: (d)

Solution

Question 62

Chemistry · Thermodynamics · Single correct

Consider the given data: (a) $\mathrm{HCl(g)} + 10\mathrm{H_2O(l)} \rightarrow \mathrm{HCl} \cdot 10\mathrm{H_2O}$ $\Delta H = -69.01 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$ (b) $\mathrm{HCl(g)} + 40\mathrm{H_2O(l)} \rightarrow \mathrm{HCl} \cdot 40\mathrm{H_2O}$ $\Delta H = -72.79 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$ Choose the correct statement:

  1. Dissolution of gas in water is an endothermic process
  2. The heat of solution depends on the amount of solvent.
  3. The heat of dilution for the $\mathrm{HCl}$ ($\mathrm{HCl} \cdot 10\mathrm{H_2O}$ to $\mathrm{HCl} \cdot 40\mathrm{H_2O}$) is 3.78 $\mathrm{kJ} \, \mathrm{mol}^{-1}$.
  4. The heat of formation of $\mathrm{HCl}$ solution is represented by both (a) and (b)

Answer: (b)

Solution

From the given information, $\Delta H$ is negative so it means dissolution of gas $\mathrm{HCl(g)}$ is exothermic. $$\mathrm{HCl(g)} + 10\mathrm{H_2O(l)} \rightarrow \mathrm{HCl} \cdot 10\mathrm{H_2O} ...(1)$$ $$\Delta H_1 = -69.01 \, \mathrm{\frac{kJ}{mol}}$$ $$\mathrm{HCl(g)} + 40\mathrm{H_2O(l)} \rightarrow \mathrm{HCl} \cdot 40\mathrm{H_2O} ...(2)$$ $$\Delta H_2 = -72.79 \, \mathrm{\frac{kJ}{mol}}$$ Hence heat of solution depends upon amount of solvent. By equation (2) - equation (1): $$\mathrm{HCl} \cdot 10\mathrm{H_2O} + 30\mathrm{H_2O(l)} \rightarrow \mathrm{HCl} \cdot 40\mathrm{H_2O}$$ So Heat of dilution $= -72.79 - (-69.01)$ $$= -3.78 \, \mathrm{\frac{kJ}{mol}}$$ Hence option (3) is incorrect. For heat of formation reactant should be in elemental form hence option (4) is incorrect.

Question 63

Chemistry · Structure of Atom · Single correct

Consider the ground state of chromium atom ($Z = 24$). How many electrons are with Azimuthal quantum number $l = 1$ and $l = 2$ respectively?

  1. 12 and 4
  2. 16 and 4
  3. 12 and 5
  4. 16 and 5

Answer: (c)

Solution

Cr: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$ $\ell = 1$ $ \ell = 1$ $ \ell = 2$ electrons having $\ell = 1 \Rightarrow 12$ electrons having $\ell = 2 \Rightarrow 5$

Question 64

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

Given below are two statements: Statement (I): The first ionisation enthalpy of group 14 elements is higher than the corresponding elements of group 13. Statement (II): Melting points and boiling points of group 13 elements are in general much higher than those of the corresponding elements of group 14. 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. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are incorrect
  4. Both Statement I and Statement II are correct

Answer: (a)

Solution

Statement 1 is correct since left to right $1E$ increases in general in periodic table. Statement 2 is incorrect since M.P. of group 14 elements is more than group 13 elements.

Question 65

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

Consider the following plots of log of rate constant k (log k) vs $\frac{1}{T}$ for three different reactions. The correct order of activation energies of these reactions is

  1. E_{a_2} > E_{a_1} > E_{a_3}
  2. E_{a_1} > E_{a_3} > E_{a_2}
  3. E_{a_1} > E_{a_2} > E_{a_3}
  4. E_{a_3} > E_{a_2} > E_{a_1}

Answer: (a)

Solution

Given $K = A e^{-\frac{E_a}{RT}}$. $$\log k = \log A - \frac{E_a}{2.303RT}$$ For graph between $\log k$ with $\frac{1}{T}$ $$|Slope of curve| = \frac{E_a}{2.303R}$$ From given graph Magnitude of slope $\Rightarrow (2) > (1) > (3)$ Hence $E_{a2} > E_{a1} > E_{a3}$

Question 66

Chemistry · Co-ordination Compounds · Single correct

'X' is the number of electrons in $t_{2g}$ orbitals of the most stable complex ion among $[\mathrm{Fe(NH_3)_6}]^{3+}$, $[\mathrm{Fe(Cl_6)}]^{3-}$, $[\mathrm{Fe(C_2O_4)_3}]^{3-}$ and $[\mathrm{Fe(H_2O)_6}]^{3+}$. The nature of oxide of vanadium of the type $\mathrm{V_2O_x}$ is:

  1. Acidic
  2. Neutral
  3. Basic
  4. Amphoteric

Answer: (d)

Solution

Most stable is $[\mathrm{Fe(C_2O_4)_3}]^{3-}$ due to Chelation effect. $\mathrm{V_2O_5}$ is amphoteric.

Question 67

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

The elements of Group 13 with highest and lowest first ionisation enthalpies are respectively:

  1. B & Ga
  2. B & Tl
  3. Tl & B
  4. B & In

Answer: (d)

Solution

IE order is given as $\mathrm{B} > \mathrm{Tl} > \mathrm{Ga} > \mathrm{Al} > \mathrm{In}$.

Question 68

Chemistry · Hydrocarbons · Single correct

Consider the following molecule (X). The structure of X is

Answer: (b)

Solution

The compound (X) reacts with $\mathrm{H}^+$ to form a tertiary carbocation. This carbocation is then attacked by $\mathrm{Br}^\Theta$ to form the major product.

Question 69

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Given below are two statements: Statement (I): for $\mathrm{ClF}_3$, all three possible structures may be drawn as follows. Statement (II): Structure III is most stable, as the orbitals having the lone pairs are axial, where the $\ell_p - bp$ repulsion is minimum. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (b)

Solution

Statement 1 is correct. Statement 2 is incorrect since in $sp^3d$ hybridization; lone pair cannot occupy axial position.

Question 70

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

Half life of zero order reaction A $\rightarrow$ product is 1 hour, when initial concentration of reaction is 2.0 $\mathrm{mol \, L^{-1}}$. The time required to decrease concentration of A from 0.50 to 0.25 $\mathrm{mol \, L^{-1}}$ is:

  1. 0.5 hour
  2. 4 hour
  3. 15 min
  4. 60 min

Answer: (c)

Solution

For zero order reaction, half life is given by $\frac{A_0}{2k}$. Given $60 \, min = \frac{2}{2k}$, we find $k = \frac{1}{60} \, M/min$. Now, $A_t = A_0 - kt$. Rearranging gives $t = \frac{A_0 - A_t}{k}$. Substituting the values, $t = \frac{0.5 - 0.25}{1/60}$. This simplifies to $0.25 \times 60$. Therefore, $t = 15 \, min$.

Question 71

Chemistry · Solutions · Numerical

Sea water, which can be considered as a $6$ molar $(6 \, \mathrm{M})$ solution of NaCl, has a density of $2 \, \mathrm{g} \, \mathrm{mL}^{-1}$. The concentration of dissolved oxygen $(\mathrm{O}_2)$ in sea water is $5.8 \, \mathrm{ppm}$. Then the concentration of dissolved oxygen $(\mathrm{O}_2)$ in sea water, is $x \times 10^{-4} \, \mathrm{m}$. $x$ = ______. (Nearest integer) Given: Molar mass of NaCl is $58.5 \, \mathrm{g} \, \mathrm{mol}^{-1}$ Molar mass of $\mathrm{O}_2$ is $32 \, \mathrm{g} \, \mathrm{mol}^{-1}$

Answer: 2

Solution

Sea water is 6 Molar in NaCl, so 1000 ml of sea water contains 6 mol of NaCl. mass of solution = Volume $\times$ density $$= 1000 \times 2$$ mass of solution = 2000 $\,$ $\mathrm{g}$ $$ppm = \frac{mass of \mathrm{O_2}}{2000} \times 10^6$$ mass of $\mathrm{O_2} = 5.8 \times 2 \times 10^{-3}$ $$= 1.16 \times 10^{-2} \, \mathrm{g}$$ molality for $\mathrm{O_2} = \frac{1.16 \times 10^{-2}/32}{(2000 - 6 \times 58.5)} \times 1000$ $$= \frac{1.16 \times 10}{32 \times 1649}$$ $$= 0.000219$$ $$= 2.19 \times 10^{-4}$$ Correct answer $\Rightarrow 2$

Question 72

Chemistry · Some Basic Concepts of Chemistry · Numerical

The amount of calcium oxide produced on heating 150 $\,$ $\mathrm{kg}$ limestone (75$\%$ pure) is _____ kg. (Nearest integer) Given : Molar mass (in $\mathrm{g/mol}^{-1}$) of $Ca-40, O-16, C-12$

Answer: 63

Solution

The reaction is given by $\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}$. The mass of $\mathrm{CaCO_3}$ is calculated as follows: $$mass of \mathrm{CaCO_3} = \frac{150 \times 75}{100} = 112.5 \, \mathrm{kg}$$ which is equal to $112500 \, \mathrm{g}$. The number of moles of $\mathrm{CaCO_3}$ is $1125$. Therefore, the moles of $\mathrm{CaO}$ are also $1125$. The mass of $\mathrm{CaO}$ is calculated as: $$mass of \mathrm{CaO} = \frac{1125 \times 56}{1000} = 63 \, \mathrm{kg}$$ The correct answer is $63$.

Question 73

Chemistry · Chemical Bonding and Molecular Structure · Numerical

A metal complex with a formula $\mathrm{MCl}_4 \cdot 3\mathrm{NH}_3$ is involved in $\mathrm{sp}^3 \mathrm{d}^2$ hybridisation. It upon reaction with excess of $\mathrm{AgNO}_3$ solution gives ' $x$ ' moles of $\mathrm{AgCl}$. Consider ' $x$ ' is equal to the number of lone pairs of electron present in central atom of $\mathrm{BrF}_5$. Then the number of geometrical isomers exhibited by the complex is .

Answer: 2

Solution

1 lone pair hence 1 mole AgCl. Complex is $[\mathrm{M(NH_3)_3Cl_3}] \mathrm{Cl}$. It shows 2 geometrical isomers ($\mathrm{Ma_3b_3}$ type) facial (fac) and meridional (Mer).

Question 74

Chemistry · Electrochemistry · Numerical

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be $185 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$ and the ionic conductance of hydroxyl and chloride ions are $170$ and $70 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$, respectively. If molar conductance of $0.02 \, \mathrm{M}$ solution of ammonium hydroxide is $85.5 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$, its degree of dissociation is given by $x \times 10^{-1}$. The value of $x$ is ______ (Nearest integer)

Answer: 3

Solution

Given $\lambda_m^\circ$ of $\mathrm{NH_4Cl} = 185$. $$(\lambda_m^\circ)_{\mathrm{NH_4^+}} + (\lambda_m^\circ)_{\mathrm{Cl^-}} = 185$$ $$(\lambda_m^\circ)_{\mathrm{NH_4^+}} = 185 - 70 = 115 \, \mathrm{Scm^2 \, mol^{-1}}$$ $$(\lambda_m^\circ)_{\mathrm{NH_4OH}} = (\lambda_m^\circ)_{\mathrm{NH_4^+}} + (\lambda_m^\circ)_{\mathrm{OH^-}}$$ $$= 115 + 170$$ $$(\lambda_m^\circ)_{\mathrm{NH_4OH}} = 285 \, \mathrm{Scm^2 \, mol^{-1}}$$ The degree of dissociation is given by: $$\frac{(\lambda_m)_{\mathrm{NH_4OH}}}{(\lambda_m^\circ)_{\mathrm{NH_4OH}}}$$ $$= \frac{85.5}{285}$$ $$= 0.3$$ $$= 3 \times 10^{-1}$$

Question 75

Chemistry · Equilibrium · Numerical

x mg of Mg(OH)$_2$ (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of x is _____ mg. (Nearest integer) (Given: Mg(OH)$_2$ is assumed to dissociate completely in H$_2$O)

Answer: 3

Solution

pH = 10 pOH = 4 $$[OH^-] = 10^{-4}$$ Number of moles of OH$^-$ = $10^{-4}$ Number of moles of Mg(OH)$_2$ = $\frac{10^{-4}}{2} = 5 \times 10^{-5}$ Mass of Mg(OH)$_2$ = $5 \times 10^{-5} \times 58 \times 10^3 \, mg$ = 2.9