JEE Main 4 April 2024 Shift 1 question paper with solutions

JEE Main 4 April 2024 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Continuity and Differentiability · Single correct

Let $f : \mathbb{R} \to \mathbb{R}$ be a function given by $$f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2}, & x 0 \end{cases}$$ where $\alpha, \beta \in \mathbb{R}$. If $f$ is continuous at $x = 0$, then $\alpha^2 + \beta^2$ is equal to :

  1. 3
  2. 12
  3. 48
  4. 6

Answer: (b)

Solution

Given $f(0^-) = \lim_{x \to 0^-} \frac{2 \sin^2 x}{x^2} = 2 = \alpha$. For $f(0^+)$, we have $f(0^+) = \lim_{x \to 0^+} \beta \times \sqrt{2} \frac{\sin \frac{x}{2}}{\frac{x}{2}} = \frac{\beta}{\sqrt{2}} = 2$. This implies $\beta = 2\sqrt{2}$. Therefore, $\alpha^2 + \beta^2 = 4 + 8 = 12$.

Question 2

Maths · Probability · Single correct

Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :

  1. $\frac{5}{18}$
  2. $\frac{5}{16}$
  3. $\frac{4}{17}$
  4. $\frac{7}{18}$

Answer: (a)

Solution

Given the bags A, B, and C with contents as follows: A has 7 red and 5 blue, B has 5 red and 7 blue, C has 6 red and 6 blue. The probability of selecting a blue ball, P(B), is calculated as: $$P(B) = \frac{1}{3} \cdot \frac{5}{12} + \frac{1}{3} \cdot \frac{7}{12} + \frac{1}{3} \cdot \frac{6}{12}$$ The required probability is: required probability = $$\frac{\frac{1}{3} \cdot \frac{5}{12}}{\frac{1}{3} \cdot \left[ \frac{5}{12} + \frac{7}{12} + \frac{6}{12} \right]} = \frac{5}{18}$$

Question 3

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

The vertices of a triangle are A(-1, 3), B(-2, 2) and C(3, -1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

  1. $x + y + \left(2 - \sqrt{2}\right) = 0$
  2. $-x + y - \left(2 - \sqrt{2}\right) = 0$
  3. $x + y - \left(2 - \sqrt{2}\right) = 0$
  4. $x - y - \left(2 + \sqrt{2}\right) = 0$

Answer: (c)

Solution

Equation of AC is $x + y = 2$. Equation of line parallel to AC is $x + y = d$. $$\left| \frac{d - 2}{\sqrt{2}} \right| = 1$$ $$d = 2 - \sqrt{2}$$ Equation of new required line is $x + y = 2 - \sqrt{2}$.

Question 4

Maths · Differential Equations · Single correct

If the solution $y = y(x)$ of the differential equation $(x^4 + 2x^3 + 3x^2 + 2x + 2) \, dy - (2x^2 + 2x + 3) \, dx = 0$ satisfies $y(-1) = -\frac{\pi}{4}$, then $y(0)$ is equal to:

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

Answer: (d)

Solution

Given $$\int dy = \int \frac{(2x^2 + 2x + 3)}{x^4 + 2x^3 + 3x^2 + 2x + 2} \, dx$$ we have $$y = \int \frac{(2x^2 + 2x + 3)}{(x^2 + 1)(x^2 + 2x + 2)} \, dx$$ which simplifies to $$y = \int \frac{dx}{x^2 + 2x + 2} + \int \frac{dx}{x^2 + 1}$$ resulting in $$y = \tan^{-1}(x + 1) + \tan^{-1} x + C$$ Given $$y(-1) = -\frac{\pi}{4}$$ we find $$-\frac{\pi}{4} = 0 - \frac{\pi}{4} + C \Rightarrow C = 0$$ Therefore, $$y = \tan^{-1}(x + 1) + \tan^{-1} x$$ Finally, $$y(0) = \tan^{-1} 1 = \frac{\pi}{4}$$

Question 5

Maths · Applications of Derivatives · Single correct

Let the sum of the maximum and the minimum values of the function $f(x) = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8}$ be $\frac{m}{n}$, where $\gcd(m, n) = 1$. Then $m + n$ is equal to:

  1. 195
  2. 201
  3. 217
  4. 182

Answer: (b)

Solution

Given $$y = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8}$$ $$x^2(2y - 2) + x(3y + 3) + 8y - 8 = 0$$ Use $D \geq 0$ $$(3y + 3)^2 - 4(2y - 2)(8y - 8) \geq 0$$ $$(11y - 5)(5y - 11) \leq 0$$ Therefore, $$y \in \left[ \frac{5}{11}, \frac{11}{5} \right]$$ $y = 1$ is also included.

Question 6

Maths · Applications of Integrals · Single correct

One of the points of intersection of the curves $y = 1 + 3x - 2x^2$ and $y = \frac{1}{x}$ is $\left( \frac{1}{2}, 2 \right)$. Let the area of the region enclosed by these curves be $\frac{1}{24} \left( l \sqrt{5} + m \right) - n \log_e \left( 1 + \sqrt{5} \right)$, where $l, m, n \in \mathbb{N}$. Then $l + m + n$ is equal to

  1. 29
  2. 31
  3. 30
  4. 32

Answer: (c)

Solution

The area $A$ is given by the integral: $$A = \int_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}} \left( 1 + 3x - 2x^2 - \frac{1}{x} \right) \, dx$$ Evaluating the integral, we have: $$A = \left[ x + \frac{3x^2}{2} - \frac{2x^3}{3} - \ln x \right]_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}}$$ Substituting the limits, we get: $$A = \left( 1 + \frac{\sqrt{5}}{2} + \frac{3}{2} \left( \frac{1+\sqrt{5}}{2} \right)^2 - \frac{2}{3} \left( \frac{1+\sqrt{5}}{2} \right)^3 - \ln \left( \frac{1+\sqrt{5}}{2} \right) \right)$$ $$- \left( \frac{1}{2} + \frac{3}{2} \left( \frac{1}{4} \right) - \frac{2}{3} \left( \frac{1}{8} \right) + \ln \left( \frac{1}{2} \right) \right)$$ Simplifying further, we have: $$A = 1 + \frac{\sqrt{5}}{2} + \frac{3}{8} + \frac{3}{4} \sqrt{5} + \frac{15}{8} - \frac{4}{3} - \frac{2}{3} \sqrt{5}$$ $$- \frac{1}{2} - \frac{3}{8} + \frac{1}{12} - \ln(1+\sqrt{5})$$ Combining terms, we get: $$A = \sqrt{5} \left( \frac{1}{2} + \frac{3}{4} - \frac{2}{3} \right) + \frac{15}{8} - \frac{4}{3} + \frac{1}{12} - \ln(1+\sqrt{5})$$ $$= \frac{14}{24} \sqrt{5} + \frac{15}{24} - \ln(1+\sqrt{5})$$

Question 7

Maths · Determinants · Single correct

If the system of equations $$x + (\sqrt{2} \sin \alpha) y + (\sqrt{2} \cos \alpha) z = 0$$ $$x + (\cos \alpha) y + (\sin \alpha) z = 0$$ $$x + (\sin \alpha) y - (\cos \alpha) z = 0$$ has a non-trivial solution, then $\alpha \in \left(0, \frac{\pi}{2}\right)$ is equal to:

  1. $\frac{11\pi}{24}$
  2. $\frac{5\pi}{24}$
  3. $\frac{7\pi}{24}$
  4. $\frac{3\pi}{4}$

Answer: (b)

Solution

The determinant of the matrix is set to zero: $$\begin{vmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \sin \alpha & -\cos \alpha \\ 1 & \cos \alpha & \sin \alpha \end{vmatrix} = 0$$ Expanding the determinant, we have: $$1 - \sqrt{2} \sin \alpha (\sin \alpha + \cos \alpha) + \sqrt{2} \cos \alpha (\cos \alpha - \sin \alpha) = 0$$ Simplifying, we get: $$1 + \sqrt{2} \cos 2\alpha - \sqrt{2} \sin 2\alpha = 0$$ This implies: $$\cos 2\alpha - \sin 2\alpha = -\frac{1}{\sqrt{2}}$$ Rewriting, we have: $$\cos \left(2\alpha + \frac{\pi}{4}\right) = -\frac{1}{2}$$ Solving for $2\alpha$, we get: $$2\alpha + \frac{\pi}{4} = 2n\pi \pm \frac{2\pi}{3}$$ Thus: $$\alpha + \frac{\pi}{8} = n\pi \pm \frac{\pi}{3}$$ For $n = 0$, we find: $$x = \frac{\pi}{3} - \frac{\pi}{8} = \frac{5\pi}{24}$$

Question 8

Maths · Permutations and Combinations · Single correct

There are 5 points $P_1, P_2, P_3, P_4, P_5$ on the side $AB$, excluding $A$ and $B$, of a triangle $ABC$. Similarly there are 6 points $P_6, P_7, \ldots, P_{11}$ on the side $BC$ and 7 points $P_{12}, P_{13}, \ldots, P_{18}$ on the side $CA$ of the triangle. The number of triangles, that can be formed using the points $P_1, P_2, \ldots, P_{18}$ as vertices, is:

  1. 776
  2. 796
  3. 751
  4. 771

Answer: (c)

Solution

The expression is given by: $$^{18}C_3 - ^{5}C_3 - ^{6}C_3 - ^{7}C_3$$ The result is: $$= 751$$

Question 9

Maths · Integrals · Single correct

Let $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \\ x - 2, & 0 < x \leq 2 \end{cases}$ and $h(x) = f(|x|) + |f(x)|$. Then $\int_{-2}^{2} h(x) \, dx$ is equal to :

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

Answer: (d)

Solution

Given the function $h(x)$ defined as follows: $$h(x) = \begin{cases} x - 2 + 2 - x = 0, & 0 \leq x \leq 2 \\ -x - 2 + 2 = -x, & -2 \leq x < 0 \end{cases}$$ We find that $$\int_0^2 h(x) \, dx = 0 and \int_{-2}^0 h(x) \, dx = 2$$

Question 10

Maths · Binomial Theorem · Single correct

The sum of all rational terms in the expansion of $\left(2^{\frac{1}{5}} + 5^{\frac{1}{3}}\right)^{15}$ is equal to:

  1. 3133
  2. 931
  3. 6131
  4. 633

Answer: (a)

Solution

$$T_{r+1} = {}^{15}C_r \left(5^{1/2}\right)^r \left(\frac{1}{2^5}\right)^{15-r}$$ $$= {}^{15}C_r \cdot 5^{\frac{r}{2}} \cdot 2^{-5(15-r)}$$ $R = 3\lambda,\ 15\mu$ $\Rightarrow r = 0,\ 15$ 2 rational terms $$\Rightarrow {}^{15}C_0 \cdot 2^5 + {}^{15}C_{15}(5)^8$$ $$= 8 + 3125 = 3133$$

Question 11

Maths · Vector Algebra · Single correct

Let a unit vector which makes an angle of $60^\circ$ with $2\hat{i} + 2\hat{j} - \hat{k}$ and angle $45^\circ$ with $\hat{i} - \hat{k}$ be $\vec{C}$. Then $\vec{C} + \left( -\frac{1}{2} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k} \right)$ is :

  1. $\frac{\sqrt{2}}{3} \hat{i} - \frac{1}{2} \hat{k}$
  2. $\left( \frac{1}{\sqrt{3}} + \frac{1}{2} \right) \hat{i} + \left( \frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{2}} \right) \hat{j} + \left( \frac{1}{\sqrt{3}} + \frac{\sqrt{2}}{3} \right) \hat{k}$
  3. $\frac{\sqrt{2}}{3} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{1}{2} \hat{k}$
  4. $-\frac{\sqrt{2}}{3} \hat{i} + \frac{\sqrt{2}}{3} \hat{j} + \left( \frac{1}{2} + \frac{2\sqrt{2}}{3} \right) \hat{k}$

Answer: (a)

Solution

Given $\vec{C} = C_1 \hat{i} + C_2 \hat{j} + C_3 \hat{k}$. $C_1^2 + C_2^2 + C_3^2 = 1$. $\vec{C} \cdot (2\hat{i} + 2\hat{j} - \hat{k}) = |\vec{C}| \sqrt{9} \cos 60^\circ$. $2C_1 + 2C_2 - C_3 = \frac{3}{2}$. $C_1 - C_3 = 1$. $C_1 + 2C_2 = \frac{1}{2}$. $C_1 = \frac{\sqrt{2}}{3} + \frac{1}{2}$. $C_2 = \frac{-1}{3\sqrt{2}}$. $C_3 = \frac{\sqrt{2}}{3} - \frac{1}{2}$.

Question 12

Maths · Sequences and Series · Single correct

Let the first three terms $2, p$ and $q$, with $q \neq 2$, of a G.P. be respectively the $7^{th}$, $8^{th}$ and $13^{th}$ terms of an A.P. If the $5^{th}$ term of the G.P. is the $n^{th}$ term of the A.P., then $n$ is equal to:

  1. 163
  2. 151
  3. 177
  4. 169

Answer: (a)

Solution

Given $p^2 = 2q$. $2 = a + 6d$ (i) $p = a + 7d$ (ii) $q = a + 12d$ (iii) $p - 2 = d$ ((ii) - (i)) $q - p = 5d$ ((iii) - (ii)) $q - p = 5d$ $q - p = 5(p - 2)$ $q = 6p - 10$ $p^2 = 2(6p - 10)$ $p^2 - 12p + 20 = 0$ $p = 10, 2$ $p = 10; q = 50$ $d = 8$ $a = -46$ $2, 10, 50, 250, 1250$ $a_4 = a + (n - 1)d$ $1250 = -46 + (n - 1)8$ $n = 163$

Question 13

Maths · Statistics · Single correct

Let $\alpha, \beta \in \mathbb{R}$. Let the mean and the variance of 6 observations $-3, 4, 7, -6, \alpha, \beta$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is:

  1. $\frac{13}{3}$
  2. $\frac{16}{3}$
  3. $\frac{11}{3}$
  4. $\frac{14}{3}$

Answer: (a)

Solution

Given $$\frac{\sum x_i}{6} = 2$$ and $$\frac{\sum x_i^2}{N} - \mu^2 = 23$$ $$\alpha + \beta = 10$$ $$\alpha^2 + \beta^2 = 52$$ Solving we get $$\alpha = 4, \beta = 6$$ $$\frac{\sum |x_i - \bar{x}|}{6} = \frac{5 + 2 + 5 + 8 + 2 + 4}{6} = \frac{13}{3}$$

Question 14

Maths · Complex Numbers and Quadratic Equations · Single correct

If 2 and 6 are the roots of the equation $ax^2 + bx + 1 = 0$, then the quadratic equation, whose roots are $\frac{1}{2a+b}$ and $\frac{1}{6a+b}$, is:

  1. $2x^2 + 11x + 12 = 0$
  2. $x^2 + 8x + 12 = 0$
  3. $4x^2 + 14x + 12 = 0$
  4. $x^2 + 10x + 16 = 0$

Answer: (b)

Solution

Sum $= 8 = -\frac{b}{a}$ Product $= 12 = \frac{1}{a} \implies a = \frac{1}{12}$ $b = -\frac{2}{3}$ $2a + b = \frac{2}{12} - \frac{2}{3} = -\frac{1}{2}$ $6a + b = \frac{6}{12} - \frac{2}{3} = -\frac{1}{6}$ sum $= -8$ $P = 12$ $x^2 + 8x + 12 = 0$

Question 15

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $\left(\bar{z}\right)^2 + |z| = 0, z \in \mathbb{C}$. Then $4 \left(\alpha^2 + \beta^2\right)$ is equal to:

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

Answer: (d)

Solution

Given $z = x + iy$ and $\bar{z} = x - iy$. Then $\bar{z}^2 = x^2 - y^2 - 2ixy$. This implies $$x^2 - y^2 - 2ixy + \sqrt{x^2 + y^2} = 0.$$ Therefore, $x = 0$ or $y = 0$. If $-y^2 + |y| = 0$, then $|y| = |y|^2$, which gives $y = 0, \pm 1$. Thus, $i, -i$ are roots. If $x^2 + |x| = 0$, then $x = 0$. This implies $\alpha = i - i = 0$ and $\beta = i(-i) = 1$. Therefore, $4(0 + 1) = 4$.

Question 16

Maths · Three Dimensional Geometry · Single correct

Let the point, on the line passing through the points $P(1, -2, 3)$ and $Q(5, -4, 7)$, farther from the origin and at distance of 9 units from the point $P$, be $(\alpha, \beta, \gamma)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

  1. 165
  2. 160
  3. 155
  4. 150

Answer: (c)

Solution

PQ line $$\frac{x-1}{4} = \frac{y+2}{-2} = \frac{z-3}{4}$$ Point: $$(4t+1, -2t-2, 4t+3)$$ Distance squared: $$16t^2 + 4t^2 + 16t^2 = 81$$ $$t = \pm \frac{3}{2}$$ Point: $$(7, -5, 9)$$ $$\alpha^2 + \beta^2 + \gamma^2 = 155$$ Option (1)

Question 17

Maths · Conic Sections · Single correct

A square is inscribed in the circle $x^2 + y^2 - 10x - 6y + 30 = 0$. One side of this square is parallel to $y = x + 3$. If $(x_i, y_i)$ are the vertices of the square, then $\Sigma (x_i^2 + y_i^2)$ is equal to:

  1. 148
  2. 152
  3. 160
  4. 156

Answer: (b)

Solution

Given the equations $y = x + c$ and $x + y + d = 0$. The conditions are: $$\left| \frac{5 - 3 + c}{\sqrt{2}} \right| = \sqrt{2}$$ $$\left| \frac{8 + d}{\sqrt{2}} \right| = \sqrt{2}$$ Solving these, we get: $$|c + 2| = 2$$ $$8 + d = \pm 2$$ Thus, $c = 0, -4$ and $d = -10, -6$. The points are $(5, 5), (3, 3), (7, 3), (5, 1)$. The sum is: $$\sum (x_i^2 + y_i^2) = 25 + 25 + 9 + 9 + 49 + 9 + 25 + 1 = 152$$

Question 18

Maths · Inverse Trigonometric Functions · Single correct

If the domain of the function $\sin^{-1}\left(\frac{3x-22}{2x-19}\right) + \log_e\left(\frac{3x^2-8x+5}{x^2-3x-10}\right)$ is $(\alpha, \beta]$, then $3\alpha + 10\beta$ is equal to:

  1. 100
  2. 95
  3. 97
  4. 98

Answer: (c)

Solution

Given the inequality $$-1 \leq \frac{3x - 22}{2x - 19} \leq 1$$ and $$\frac{3x^2 - 8x + 5}{x^2 - 3x - 10} > 0$$. The solution for $x$ is $$x \in \left(5, \frac{41}{5}\right]$$. Additionally, $$3\alpha + 10\beta = 97$$.

Question 19

Maths · Continuity and Differentiability · Single correct

Let $f(x) = x^5 + 2e^{x/4}$ for all $x \in \mathbb{R}$. Consider a function $g(x)$ such that $(g \circ f)(x) = x$ for all $x \in \mathbb{R}$. Then the value of $8g'(2)$ is:

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

Answer: (d)

Solution

Given $f(x) = 2$ when $x = 0$. Therefore, $g'(f(x)) f'(x) = 1$. $$g'(2) = \frac{1}{f'(0)}$$ Therefore, $f'(x) = 5x^4 + \frac{2}{4} e^{x/4}$. $g'(2) = 2$. Ans = 16 Option (1)

Question 20

Maths · Determinants · Single correct

Let $\alpha \in (0, \infty)$ and $A = \begin{bmatrix} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix}$. If $\det(adj(2A - A^T) \cdot adj(A - 2A^T)) = 2^8$, then $(\det(A))^2$ is equal to:

  1. 36
  2. 16
  3. 1
  4. 49

Answer: (b)

Solution

$adj(A-2A^T)(2A-A^T)=28$ $|A-2A^T||2A-A^T|=24$ $|A-2A^T|^2=\pm16$ $(A=2A^T)$ $=A^T=2A$ $|A-2A^T|=|A^T-2A|$ $\Rightarrow |A-2A^T|=16$ $|A-2A^T|=\pm4$ $\begin{bmatrix} 1&2&\alpha\\ 1&0&1\\ 0&1&2 \end{bmatrix}$ $-$ $\begin{bmatrix} 2&2&0\\ 4&0&2\\ 2\alpha&2&4 \end{bmatrix}$ $=$ $\begin{bmatrix} -1&0&\alpha\\ -3&0&-1\\ -2\alpha&-1&-2 \end{bmatrix}$ $|A-2A^T|$ $= \left| \begin{matrix} -1&0&\alpha\\ -3&0&-1\\ -2\alpha&-1&-2 \end{matrix} \right|$ $=1+3\alpha=4$ $=3\alpha-3$ $3\alpha=3$ $\alpha=1$ $|A|$ $= \left| \begin{matrix} 1&2&1\\ 1&0&1\\ 0&1&2 \end{matrix} \right|$ $=-1-3$ $=-4$ $|A|^2$ $=16$

Question 21

Maths · Limits and Derivatives · Numerical

If $\lim_{x \to 1} \frac{(5x+1)^{1/3}-(x+5)^{1/3}}{(2x+3)^{1/2}-(x+4)^{1/2}} = \frac{m\sqrt{5}}{n(2n)^{2/3}}$, where gcd$(m, n) = 1$, then $8m + 12n$ is equal to

Answer: 100

Solution

The limit is given by $$ \lim_{x \to 1} \frac{\frac{1}{3}(5x + 1)^{-2/3}5 - \frac{1}{3}(x + 5)^{-2/3}}{\frac{1}{2}(2x + 3)^{-1/2} \cdot 2 - \frac{1}{2}(x + 4)^{-1/2}} $$ This simplifies to $$ \frac{\frac{8}{3} \sqrt{5}}{6^{2/3}} m = 8 n = 3 $$ Given the equation $$ 8m + 12n = 100 $$

Question 22

Maths · Sets · Numerical

In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these subjects. Let $m$ and $n$ respectively be the least and the most number of students who studied all the three subjects. Then $m + n$ is equal to

Answer: 45

Solution

Given the Venn diagram, we have the following inequalities: $$125 \leq m + 90 - x \leq 130$$ $$85 \leq P + 70 - x \leq 95$$ $$75 \leq C + 80 - x \leq 90$$ The equation for the total is: $$m + P + C + 120 - 2x = 210$$ Solving these, we find: $$\Rightarrow 15 \leq x \leq 45 and 30 - x \geq 0$$ This implies: $$\Rightarrow 15 \leq x \leq 30$$ Finally, we have: $$30 + 15 = 45$$

Question 23

Maths · Differential Equations · Numerical

Let the solution $y = y(x)$ of the differential equation $\frac{dy}{dx} - y = 1 + 4 \sin x$ satisfy $y(\pi) = 1$. Then $y\left(\frac{\pi}{2}\right) + 10$ is equal to

Answer: 7

Solution

Given $y e^{-x} = \int \left( e^{-x} + 4 e^{-x} \sin x \right) \, dx$. $y e^{-x} = -e^{-x} - 2 \left( e^{-x} \sin x e^{-x} \cos x \right) + C$. $y = -1 - 2(\sin x + \cos x) + c e^x$. Therefore, $y(\pi) = 1 \Rightarrow c = 0$. $y'(\pi/2) = -1 - 2 = -3$. Ans = 10 - 3 = 7

Question 24

Maths · Three Dimensional Geometry · Numerical

If the shortest distance between the lines $\frac{x+2}{2} = \frac{y+3}{3} = \frac{z-5}{4}$ and $\frac{x-3}{1} = \frac{y-2}{-3} = \frac{z+4}{2}$ is $\frac{38}{3\sqrt{5}}k$, and $\int_0^k [x^2] \, dx = \alpha - \sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6\alpha^3$ is equal to .

Answer: 48

Solution

Given 38k/(3√5) = ((5i + 5j − 9k)/√5) · | i j k | | 2 3 4 | | 1 −3 2 | 38k/(3√5) = 19/√5 k = 19/√5 k = 3/2 ∫₀^(3/2) [x²] = ∫₀¹ 0 + ∫₁^√2 1 + ∫√2^(3/2) 2 = √2 − 1 + 2(3/2 − √2) = 2 − √2 α = 2 − √2 ⇒ 6α³ = 48

Question 25

Maths · Matrices · Fill in the blank

Let $A$ be a square matrix of order $2$ such that $|A|=2$ and the sum of its diagonal elements is $-3$. If the points $(x,y)$ satisfying \[ A^2+xA+yI=0 \] lie on a hyperbola, whose transverse axis is parallel to the $x$-axis, eccentricity is $e$ and the length of the latus rectum is $\ell$, then $e^4+\ell^4$ is equal to ____

Answer: 233

Solution

Question 26

Maths · Binomial Theorem · Numerical

Let $a = 1 + \frac{{^2C_2}}{{3!}} + \frac{{^3C_2}}{{4!}} + \frac{{^4C_2}}{{5!}} + \cdots,$ $b = 1 + \frac{{^1C_0 + ^1C_1}}{{1!}} + \frac{{^2C_0 + ^2C_1 + ^2C_2}}{{2!}} + \frac{{^3C_0 + ^3C_1 + ^3C_2 + ^3C_3}}{{3!}} + \cdots$ Then $\frac{{2b}}{{a^2}}$ is equal to

Answer: 8

Solution

Given $$f(x) = 1 + \frac{(1+x)}{1!} + \frac{(1+x)^2}{2!} + \frac{(1+x)^3}{3!} + \ldots$$ $$\frac{e^{(1+x)}}{1+x} = \frac{1}{1+x} + 1 + \frac{(1+x)}{2!} + \frac{(1+x)^2}{3!} + \frac{(1+x)^2}{4!}$$ Coefficient of $x^2$ in RHS: $$1 + \frac{2C_2}{3} + \frac{3C_2}{4} + \ldots = a$$ Coefficient of $x^2$ in LHS. $$e \left(1 + x + \frac{x^2}{2!}\right) \cdots \left(1 - x + \frac{x^2}{2!} \cdots \right)$$ is $e - e + \frac{e}{2!} = a$ $$b = 1 + \frac{2}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \ldots = e^2$$ $$\frac{2}{a^2} b = 8$$

Question 27

Maths · Matrices · Numerical

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$. Then the maximum value of $\det(A)$ is .

Answer: 27

Solution

Let $A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$. $$A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$ This implies: $$a_1 + a_2 + a_3 = 3 ...(1)$$ $$b_1 + b_2 + b_3 = 3 ...(2)$$ $$c_1 + c_2 + c_3 = 3 ...(3)$$ Now, $$|A| = (a_1b_2c_3 + a_2b_3c_1 + a_3b_1c_2) - (a_3b_2c_1 + a_2b_1c_3 + a_1b_3c_2)$$ Therefore, from above information, clearly $|A|_{\max} = 27$, when $a_1 = 3$, $b_2 = 3$, $c_3 = 3$.

Question 28

Maths · Conic Sections · Numerical

Let the length of the focal chord PQ of the parabola $y^2 = 12x$ be 15 units. If the distance of PQ from the origin is $p$, then $10p^2$ is equal to

Answer: 72

Solution

The length of the focal chord is given by $4a \csc^2 \theta = 15$. Thus, $12 \csc^2 \theta = 15$. We have $\sin^2 \theta = \frac{4}{5}$. Therefore, $\tan^2 \theta = 4$ and $\tan \theta = 2$. The equation is $\frac{y - 0}{x - 3} = 2$. This simplifies to $y = 2x - 6$. Rearranging gives $2x - y - 6 = 0$. We find $P = \frac{6}{\sqrt{5}}$. Finally, $10p^2 = 10 \cdot \frac{36}{5} = 72$.

Question 29

Maths · Vector Algebra · Numerical

Let ABC be a triangle of area $15\sqrt{2}$ and the vectors $\overrightarrow{AB} = \hat{i} + 2\hat{j} - 7\hat{k}$, $\overrightarrow{BC} = a\hat{i} + b\hat{j} + c\hat{k}$ and $\overrightarrow{AC} = 6\hat{i} + d\hat{j} - 2\hat{k}$, $d > 0$. Then the square of the length of the largest side of the triangle ABC is

Answer: 54

Solution

Given the triangle with vectors, we calculate the area using the determinant method. $$Area = \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{vmatrix} = 15 \sqrt{2}$$ Expanding the determinant, we have: $$(-4 + 7d) \hat{i} - \hat{j}(-2 + 42) + \hat{k}(d - 12)$$ Simplifying, we get: $$(7d - 4)^2 + (40)^2 + (d - 12)^2 = 1800$$ Expanding and simplifying: $$50d^2 - 80d - 40 = 0$$ Dividing by 10: $$5d^2 - 8d - 4 = 0$$ Factoring gives: $$5d(d - 2) + 2(d - 2) = 0$$ Thus, $d = 2$ or $d = -\frac{2}{5}$. Since $d > 0$, $d = 2$. Substituting back: $$(a + 1) \hat{i} + (b + 2) \hat{j} + (c - 7) \hat{k} = 6 \hat{i} + 2 \hat{j} - 2 \hat{k}$$ Solving the equations: $$a + 1 = 6, b + 2 = 2, c - 7 = -2$$ We find: $$a = 5, b = 0, c = 5$$ Calculating the magnitudes: $$|AB| = \sqrt{1 + 4 + 49} = \sqrt{54}$$ $$|BC| = \sqrt{25 + 25} = \sqrt{50}$$ $$|AC| = \sqrt{86 + 4 + 4} = \sqrt{44}$$ The answer is 54.

Question 30

Maths · Integrals · Fill in the blank

If $\displaystyle\int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x\cos x}\,dx = \frac{1}{a}\log_e\left(\frac{a}{3}\right) + \frac{\pi}{b\sqrt{3}}$, where $a, b \in \mathbb{N}$, then $a+b$ is equal to _____

Answer: 8

Solution

Given the integral $$\int_0^{\pi/2} \frac{\sin^2 x}{1 + \frac{1}{2} \sin 2x} \, dx = \int_0^{\pi/4} \frac{1 - \cos 2x}{2 + \sin 2x} \, dx$$ we have $$\int \frac{1}{2 + \sin 2x} - \int \frac{\cos 2x}{2 + \sin 2x}$$ which is $$(I_1) - (I_2).$$ For $$(I_1) = \int \frac{dx}{2 + \frac{2 \tan x}{1 + \tan^2 x}}$$ we have $$\int_0^{\pi/4} \frac{\sec^2 x \, dx}{2 \tan^2 x + 2 \tan x + 2}.$$ Let $$\tan x = t.$$ Then $$\frac{1}{2} \int_0^1 \frac{dt}{\left(t + \frac{1}{2}\right)^2 + \frac{3}{4}} = \frac{\pi}{6\sqrt{3}}.$$ For $$I_2 = \int_0^{\pi/4} \frac{\cos 2x}{2 + \sin 2x} \, dx = \frac{1}{2} \left(\ln \frac{3}{2}\right).$$ Therefore, $$I_1 - I_2 = \frac{1}{\sqrt{3}} \frac{\pi}{6} + \frac{1}{2} \ln \frac{2}{3}.$$ Thus, $$a = 2, \ b = 6.$$

Physics

Question 31

Physics · Moving Charges and Magnetism · Single correct

An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then :

  1. the electron will continue to move with uniform velocity along the axis of the solenoid.
  2. the electron will be accelerated along the axis.
  3. the electron path will be circular about the axis.
  4. the electron will experience a force at 45^$\circ$ to the axis and execute a helical path.

Answer: (a)

Solution

Since $\vec{v} \parallel \vec{B}$ so force on electron due to magnetic field is zero. So it will move along axis with uniform velocity.

Question 32

Physics · Electromagnetic Waves · Single correct

The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos \omega (t - z/c) \mathrm{NC}^{-1}$. The magnetic field induction of this wave is (in SI unit):

  1. $\vec{B} = \hat{k} \frac{40}{c} \cos \omega (t - z/c)$
  2. $\vec{B} = \hat{j} 40 \cos \omega (t - z/c)$
  3. $\vec{B} = \hat{i} \frac{40}{c} \cos \omega (t - z/c)$
  4. $\vec{B} = \hat{j} \frac{40}{c} \cos \omega (t - z/c)$

Answer: (d)

Solution

Given $\vec{E} = \hat{i} 40 \cos \omega \left( t - \frac{z}{c} \right)$. $\vec{E}$ is along $+x$ direction. $\vec{v}$ is along $+z$ direction. So direction of $\vec{B}$ will be along $+y$ and magnitude of $B$ will be $\frac{E}{c}$. So answer is $\frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right) \hat{j}$.

Question 33

Physics · Nuclei · Single correct

Which of the following nuclear fragments corresponding to nuclear fission between neutron $\left( ^1_0 \mathrm{n} \right)$ and uranium isotope $\left( ^{235}_{92} \mathrm{U} \right)$ is correct:

  1. $^{144}_{56} \mathrm{Ba} + ^{89}_{36} \mathrm{Kr} + 4 \, ^{1}_{0} \mathrm{n}$
  2. $^{144}_{56} \mathrm{Ba} + ^{89}_{36} \mathrm{Kr} + 3 \, ^{1}_{0} \mathrm{n}$
  3. $^{140}_{56} \mathrm{Xe} + ^{94}_{38} \mathrm{Sr} + 3 \, ^{1}_{0} \mathrm{n}$
  4. $^{153}_{51} \mathrm{Sb} + ^{99}_{41} \mathrm{Nb} + 3 \, ^{1}_{0} \mathrm{n}$

Answer: (b)

Solution

Balancing mass number and atomic number $$^{235}_{92} \mathrm{U} + ^{1}_{0} \mathrm{n} \rightarrow ^{144}_{56} \mathrm{Ba} + ^{89}_{36} \mathrm{Kr} + 3 \, ^{1}_{0} \mathrm{n}$$

Question 34

Physics · Mathematics in Physics · Single correct

In an experiment to measure focal length $(f)$ of convex lens, the least counts of the measuring scales for the position of object $(u)$ and for the position of image $(v)$ are $\Delta u$ and $\Delta v$, respectively. The error in the measurement of the focal length of the convex lens will be:

  1. $2f \left[ \frac{\Delta u}{u} + \frac{\Delta v}{v} \right]$
  2. $\frac{\Delta u}{u} + \frac{\Delta v}{v}$
  3. $f^2 \left[ \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2} \right]$
  4. $f \left[ \frac{\Delta u}{u} + \frac{\Delta v}{v} \right]$

Answer: (c)

Solution

Given $f^{-1} = v^{-1} - u^{-1}$. Differentiating, we have $-f^{-2} \, df = -v^{-2} \, dv - u^{-2} \, du$. Therefore, $$\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}.$$ Thus, $$df = f^2 \left[ \frac{dv}{v^2} + \frac{du}{u^2} \right].$$

Question 35

Physics · Mechanical Properties of Fluids · Single correct

Given below are two statements: Statement I : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation $P_1 - P_2 = \rho g \left( h_2 - h_1 \right)$. Statement II : In ventury tube shown $2gh = v_1^2 - v_2^2$ 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 correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.

Answer: (b)

Solution

Applying Bernoulli's equation $$P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$$ $\($[h_1 $\&$ h_2 are height of point from any reference level]$\)$ Given $V_1 = V_2 = 0$ (for statement-1) $$\therefore P_1 - P_2 = g (h_2 - h_1)$$ For statement-2 $$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$$ $$P_1 - P_2 = \rho g h$$ $$P_1 - P_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$$ $$\rho g h = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$$ $$2 g h = v_2^2 - v_1^2$$

Question 36

Physics · Thermal Properties of Matter · Single correct

The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $8\,\Omega$ and $10\,\Omega$ respectively. After inserting in a hot bath of temperature $400^\circ\mathrm{C}$, the resistance of platinum wire is:

  1. $10\,\Omega$
  2. $8\,\Omega$
  3. $16\,\Omega$
  4. $2\,\Omega$

Answer: (c)

Solution

Given $R_0 = 8 \, \Omega$, $R_{100} = 10 \, \Omega$. Therefore, $R_{100} = R_0 (1 + \alpha \Delta T)$. Also, $R_{400} = R_0 \left(1 + \alpha \Delta T^1 \right)$. Thus, $10 = 8(1 + \alpha \times 100) \Rightarrow 100 \alpha = \frac{1}{4}$. Therefore, $R_{400} = 8(1 + 400 \alpha) = 8(1 + 1) = 16 \, \Omega$.

Question 37

Physics · Gravitation · Single correct

A metal wire of uniform mass density having length $L$ and mass $M$ is bent to form a semicircular arc and a particle of mass $m$ is placed at the centre of the arc. The gravitational force on the particle by the wire is :

  1. $\frac{GmM\pi^2}{L^2}$
  2. $\frac{GMm\pi}{2L^2}$
  3. 0
  4. $\frac{2GmM\pi}{L^2}$

Answer: (d)

Solution

We have $R = \frac{L}{\pi}$. $$g_0 = \frac{2G \frac{M}{L}}{R} = \frac{2GM\pi}{L^2}$$ Therefore, $F_m = mg_0 = \frac{2GM\pi m}{L^2}$. Hence option (4)

Question 38

Physics · Thermal Properties of Matter · Single correct

On celcius scale the temperature of body increases by $40^\circ \mathrm{C}$. The increase in temperature on Fahrenheit scale is:

  1. $68^\circ \mathrm{F}$
  2. $75^\circ \mathrm{F}$
  3. $72^\circ \mathrm{F}$
  4. $70^\circ \mathrm{F}$

Answer: (c)

Solution

We know that per $1^\circ \mathrm{C}$ change is equivalent to $1.8^\circ$ change in $^\circ \mathrm{F}$. Therefore, $40^\circ$ change on Celsius scale will correspond to $72^\circ$ change on Fahrenheit scale. Hence option (3)

Question 39

Physics · Ray Optics and Optical Instruments · Single correct

An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is 25D. Focal length of each of the convex lens is :

  1. 20 $\mathrm{\ cm}$
  2. 50 $\mathrm{\ cm}$
  3. 500 $\mathrm{\ cm}$
  4. 25 $\mathrm{\ cm}$

Answer: (a)

Solution

We know that $P_{eq} = \Sigma P_i$. Given all lenses are identical, $5P = 25D$. Therefore, $P = 5D$. Thus, $\frac{1}{f} = 5 \Rightarrow f = \frac{1}{5} \, \mathrm{m} = 20 \, \mathrm{cm}$. Hence option (1).

Question 40

Physics · Dual Nature of Radiation and Matter · Single correct

Which figure shows the correct variation of applied potential difference (V) with photoelectric current (I) at two different intensities of light ($I_1 < I_2$) of same wavelengths :

Answer: (b)

Solution

Given lights are of same wavelength. Hence stopping potential will remain same. Since $I_2 > I_1$, hence saturation current corresponding to $I_2$ will be greater than that corresponding to $I_1$.

Question 41

Physics · Laws of Motion · Single correct

A wooden block, initially at rest on the ground, is pushed by a force which increases linearly with time $t$. Which of the following curve best describes acceleration of the block with time:

Answer: (d)

Solution

Given $F = ma \Rightarrow a = \frac{F}{m} = \frac{kt}{m}$. $a$ vs $t$ will be a straight line passing through the origin.

Question 42

Physics · Work, Energy and Power · Single correct

If a rubber ball falls from a height $h$ and rebounds upto the height of $h/2$. The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are :

  1. 50$\%$, $\sqrt{2gh}$
  2. 50$\%$, $\sqrt{gh}$
  3. 40$\%$, $\sqrt{2gh}$
  4. 50$\%$, $\sqrt{\frac{gh}{2}}$

Answer: (a)

Solution

Velocity just before collision is $\sqrt{2gh}$. Velocity just after collision is $\sqrt{2g \left( \frac{h}{2} \right)}$. Therefore, $\Delta KE = \frac{1}{2} m(2gh) - \frac{1}{2} mgh = \frac{1}{2} mgh$. Therefore, the percentage loss in energy is $$\frac{\Delta KE}{KE_i} \times 100 = \frac{\frac{1}{2} mgh}{\frac{1}{2} mg2h} \times 100 = 50\%$$

Question 43

Physics · Physical World, Units and Measurements · Single correct

The equation of stationary wave is : $$y = 2a \sin\left(\frac{2\pi nt}{\lambda}\right) \cos\left(\frac{2\pi x}{\lambda}\right).$$ Which of the following is NOT correct :

  1. The dimensions of $n/\lambda$ is $[T]$
  2. The dimensions of $n$ is $[LT^{-1}]$
  3. The dimensions of $x$ is $[L]$
  4. The dimensions of $nt$ is $[L]$

Answer: (a)

Solution

Comparing the given equation with standard equation of standing $$\frac{2\pi n}{\lambda} = \omega$$ and $$\frac{2\pi}{\lambda} = k$$ $$\left[ \frac{n}{\lambda} \right] = [\omega] = \mathrm{T}^{-1}$$ $$[nt] = [\lambda] = \mathrm{L}$$ $$[n] = [\lambda \omega] = \mathrm{LT}^{-1}$$ $$[x] = [\lambda] = \mathrm{L}$$

Question 44

Physics · Motion in a Straight Line · Single correct

A body travels $102.5\,\mathrm{m}$ in $n^{\text{th}}$ second and $115.0\,\mathrm{m}$ in $(n+2)^{\text{th}}$ second. The acceleration is:

  1. $9\,\mathrm{m/s^2}$
  2. $6.25\,\mathrm{m/s^2}$
  3. $12.5\,\mathrm{m/s^2}$
  4. $5\,\mathrm{m/s^2}$

Answer: (b)

Solution

Question 45

Physics · Current Electricity · Single correct

To measure the internal resistance of a battery, potentiometer is used. For $R = 10\Omega$, the balance point is observed at $l = 500 \, \mathrm{cm}$ and for $R = 1\Omega$ the balance point is observed at $l = 400 \, \mathrm{cm}$. The internal resistance of the battery is approximately :

  1. 0.2$\Omega$
  2. 0.3$\Omega$
  3. 0.4$\Omega$
  4. 0.1$\Omega$

Answer: (b)

Solution

Let potential gradient be $\lambda$. $$\therefore \; i \times 10 = \lambda \times 500 = \varepsilon - ir_s$$ $$\Rightarrow 500\lambda = \varepsilon - 50\lambda r_s$$ Also, $$i' \times 1 = \lambda \times 400 = \varepsilon - i'r_s$$ $$\Rightarrow 400\lambda = \varepsilon - 400\lambda r_5$$ $$\therefore 100\lambda = 350\lambda r_s \Rightarrow r_s = \frac{10}{35} \approx 0.3\Omega$$

Question 46

Physics · Electric Charges and Fields · Single correct

An infinitely long positively charged straight thread has a linear charge density $\lambda \mathrm{Cm}^{-1}$. An electron revolves along a circular path having axis along the length of the wire. The graph that correctly represents the variation of the kinetic energy of electron as a function of radius of circular path from the wire is:

Answer: (c)

Solution

Question 47

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

The value of net resistance of the network as shown in the given figure is :

  1. 6$\Omega$
  2. ($\frac{5}{2}$)$\Omega$
  3. ($\frac{15}{4}$)$\Omega$
  4. ($\frac{30}{11}$)$\Omega$

Answer: (a)

Solution

Diode 2 is in reverse bias. So current will not flow in the branch of the 2nd diode, so we can assume it to be a broken wire. Diode 1 is in forward bias, so it will behave like a conducting wire. So the new circuit will be $$R_{eq} = \frac{15 \times 10}{15 + 10} = \frac{15 \times 10}{25} = 6 \, \Omega$$

Question 48

Physics · Thermodynamics · Single correct

P-T diagram of an ideal gas having three different densities $\rho_1, \rho_2, \rho_3$ (in three different cases) is shown in the figure. Which of the following is correct:

  1. $\rho_1 > \rho_2$
  2. $\rho_2 < \rho_3$
  3. $\rho_1 = \rho_2 = \rho_3$
  4. $\rho_1 < \rho_2$

Answer: (a)

Solution

For ideal gas $$PV = nRT$$ $$PV = \frac{m}{M} RT$$ $$P = \left( \frac{M}{V} \right) \frac{RT}{M}$$ $$P = \frac{\rho RT}{M}$$ (Where $m$ is mass of gas and $M$ is molecular mass of gas) For same temperature $P_1 > P_2 > P_3$ So $\rho_1 > \rho_2 > \rho_3$

Question 49

Physics · Motion in a Plane · Single correct

The co-ordinates of a particle moving in $x - y$ plane are given by : $x = 2 + 4t, y = 3t + 8t^2$. The motion of the particle is :

  1. uniformly accelerated having motion along a parabolic path.
  2. uniform motion along a straight line.
  3. uniformly accelerated having motion along a straight line.
  4. non-uniformly accelerated.

Answer: (a)

Solution

Given $$x = 2 + 4t$$ $$\frac{dx}{dt} = v_x = 4$$ $$\frac{dv_x}{dt} = a_x = 0$$ $$y = 3t + 8t^2$$ $$\frac{dy}{dt} = v_y = 3 + 16t$$ $$\frac{dv_y}{dt} = a_y = 16$$ The motion will be uniformly accelerated motion. For path $$x = 2 + 4t$$ $$\frac{(x - 2)}{4} = t$$ Put this value of $t$ in equation of $y$ $$y = 3 \left(\frac{x-2}{4}\right) + 8 \left(\frac{x-2}{4}\right)^2$$ This is a quadratic equation so path will be parabola.

Question 50

Physics · Alternating Current · Single correct

In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to: \begin{enumerate} \item[(A)] pure inductor. \item[(B)] pure capacitor. \item[(C)] pure resistor. \item[(D)] combination of an inductor and capacitor. \end{enumerate} Choose the correct answer from the options given below:

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

Answer: (d)

Solution

This is possible when phase difference is $\frac{\pi}{2}$ between current and voltage so correct answer will be (4).

Question 51

Physics · Electric Charges and Fields · Numerical

An infinite plane sheet of charge having uniform surface charge density $+\sigma_s \, \mathrm{C/m^2}$ is placed on $x-y$ plane. Another infinitely long line charge having uniform linear charge density $+\lambda_e \, \mathrm{C/m}$ is placed at $z = 4 \, \mathrm{m}$ plane and parallel to $y$-axis. If the magnitude values $|\sigma_s| = 2 \, |\lambda_e|$ then at point $(0, 0, 2)$, the ratio of magnitudes of electric field values due to sheet charge to that of line charge is $\pi \sqrt{n} : 1$. The value of $n$ is _______.

Answer: 16

Solution

The ratio of $E_S$ to $E_\ell$ is given by $$\frac{E_S}{E_\ell} = \frac{\sigma}{2\epsilon_0} \times \frac{2\pi \epsilon_0 r}{\lambda}$$ Simplifying, we have $$= \frac{\pi \times \sigma r}{\lambda}$$ Further simplifying, $$= \frac{\pi \times 2 \lambda \times 2}{\lambda} = \frac{4\pi}{1}$$ Therefore, $n = 16$.

Question 52

Physics · Atoms · Numerical

A hydrogen atom changes its state from $n = 3$ to $n = 2$. Due to recoil, the percentage change in the wavelength of emitted light is approximately $1 \times 10^{-n}$. The value of $n$ is_____. [Given $\mathrm{Rhc}$ = 13.6 $\mathrm{eV}$, $\mathrm{hc}$ = 1242 $\mathrm{eVnm}$, $\mathrm{h}$ = 6.6 $\times$ $10^{-34}$ $\mathrm{Js}$ mass of the hydrogen atom = 1.6 $\times$ $10^{-27}$ $\mathrm{kg}$ ]

Answer: 7

Solution

Given $\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 1.9 \, \mathrm{eV}$. $\Delta E = \frac{hc}{\lambda}$ $\lambda = \frac{hc}{\Delta E}$ $P_i = P_t$ $0 = -mv + \frac{h}{\lambda'}$ Therefore, $v = \frac{h}{m\lambda'}$ $\Delta E = \frac{1}{2} mv^2 + \frac{hc}{\lambda'}$ $= \frac{1}{2} \left( \frac{h}{m\lambda'} \right)^2 + \frac{hc}{\lambda'}$ Now $\Delta E = \frac{h^2}{2m\lambda'^2} + \frac{hc}{\lambda'}$ $\lambda'^2 \Delta E - hc \lambda' - \frac{h^2}{2m} = 0$ $\lambda' = \frac{hc \pm \sqrt{h^2 c^2 + \frac{4 \Delta E h^2}{2m}}}{2 \Delta E}$ $\lambda' = \frac{hc \pm hc \sqrt{1 + \frac{2 \Delta E}{mc^2}}}{2 \Delta E}$ $\frac{\lambda'}{\lambda} = \frac{1 + \left(1 + \frac{2 \Delta E}{mc^2} \right)^{\frac{1}{2}}}{2} = 1 + 1 + \frac{\Delta E}{mc^2}$ $\frac{\lambda' - \lambda}{\lambda} = \frac{\Delta E}{2mc^2} = \frac{1.9 \times 1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27} \times 9 \times 10^{15}} = 10^{-9}$ Therefore, $\%$ change $\approx 10^{-7}$ Correct answer 7

Question 53

Physics · Moving Charges and Magnetism · Numerical

The magnetic field existing in a region is given by $\vec{B} = 0.2(1 + 2x)\hat{k}$ T. A square loop of edge 50 cm carrying 0.5 A current is placed in $x-y$ plane with its edges parallel to the $x-y$ axes, as shown in figure. The magnitude of the net magnetic force experienced by the loop is_____ mN.

Answer: 50

Solution

Force on segment parallel to x-axis will cancel each other. Hence $F_{net}$ will be due to portion parallel to y-axis. $$F = 0.5 \times 0.5 \times 6 \times 0.2 - 0.5 \times 0.5 \times 0.2 \times 5$$ $$= 0.5 \times 0.5 \times 0.2$$ $$= 0.25 \times 0.2$$ $$= 50 \times 10^{-3} \, \mathrm{N}$$ $$= 50 \, \mathrm{mN}$$

Question 54

Physics · Alternating Current · Numerical

A alternating current at any instant is given by $i = [6 + \sqrt{56} \sin(100 \pi t + \pi/3)] \, \mathrm{A}$. The $rms$ value of the current is ________ A.

Answer: 8

Solution

Given $$I_{DNs} = \sqrt{\frac{\int i^2 \, dt}{\int dt}}$$ We have $$I_{Dns} = \sqrt{\frac{(6)^2 + (\sqrt{56})^2}{2}}$$ Simplifying, $$= \sqrt{36 + 28}$$ $$= \sqrt{64}$$ $$= 8 \, \mathrm{A}$$

Question 55

Physics · Current Electricity · Numerical

Twelve wires each having resistance $2\,\Omega$ are joined to form a cube. A battery of $6\,\mathrm{V}$ emf is joined across point $a$ and $c$. The voltage difference between $e$ and $f$ is _____ $\mathrm{V}$.

Answer: 1

Solution

From symmetry, current through e-b and g-d is 0. Therefore, $R_{eq} = \frac{3}{4} \times R = \frac{3}{2} \, \Omega$. Therefore, current through battery $= \frac{6 \times 2}{3} = 4 \, A$. $i_2 = \frac{4}{8} \times 2 = 1 \, A$. Therefore, $\Delta V$ across e-f $= \frac{i_2}{2} \times R = \frac{1}{2} \times 2 = 1 \, V$.

Question 56

Physics · Mechanical Properties of Fluids · Numerical

A soap bubble is blown to a diameter of 7 cm. 36960 $\mathrm{erg}$ of work is done in blowing it further. If surface tension of soap solution is 40 $\mathrm{dyne/cm}$ then the new radius is ________ cm. Take ( $\pi$ = $\frac{22}{7}$ ) .

Answer: 7

Solution

Given $\omega = \Delta U = S \Delta A$. $$36960 \, \mathrm{erg} = \frac{40 \, \mathrm{dyne}}{\mathrm{cm}} 8 \pi [ (r)^2 - ( \frac{7}{2} )^2 ] \, \mathrm{cm}^2$$ $r = 7 \, \mathrm{cm}$

Question 57

Physics · Wave Optics · Numerical

Two wavelengths $\lambda_1$ and $\lambda_2$ are used in Young's double slit experiment. $\lambda_1 = 450 \, \mathrm{nm}$ and $\lambda_2 = 650 \, \mathrm{nm}$. The minimum order of fringe produced by $\lambda_2$ which overlaps with the fringe produced by $\lambda_1$ is $n$. The value of $n$ is _____.

Answer: 9

Solution

Given $n_2 \lambda_2 = n_1 \lambda_1$. $$\frac{n_2}{n_1} = \frac{\lambda_1}{\lambda_2} = \frac{450}{650} = \frac{9}{13}$$ Therefore, $n_2 = 9$.

Question 58

Physics · Mechanical Properties of Solids · Numerical

An elastic spring under tension of 3 N has a length $a$. Its length is $b$ under tension 2 N. For its length $(3a - 2b)$, the value of tension will be ______N.

Answer: 5

Solution

Given $$3 = K(a - \ell)$$ $$2 = K(b - \ell)$$ $$T = K(3a - 2b - \ell)$$ $$T = K \left[ 3(a - \ell) - 2(b - \ell) \right]$$ $$= K \left[ 3 \left( \frac{3}{K} \right) - 2 \left( \frac{2}{K} \right) \right]$$ $$= 9 - 4$$ $$= 5 \, \mathrm{N}$$

Question 59

Physics · Mathematics in Physics · Numerical

Two forces $\vec{F}_1$ and $\vec{F}_2$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between $\vec{F}_1$ and $\vec{F}_2$ is $\cos^{-1}\left(\frac{1}{n}\right)$. The value of $|n|$ is .

Answer: 6

Solution

Given $|\vec{F}_1| = F$ and $|\vec{F}_R| = |\vec{F}_2| = 3F$. $$F_R^2 = F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta$$ $$9F^2 = F^2 + 9F^2 + 6F^2 \cos \theta$$ Solving for $\cos \theta$, we get: $$\cos \theta = -\frac{1}{6}$$ Therefore, $$\theta = \cos^{-1}\left(-\frac{1}{6}\right)$$ Given $n = -6$, we have $$|n| = 6$$

Question 60

Physics · System of Particles and Rotational Motion · Numerical

A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed $v$. The sphere and the cylinder reaches upto maximum heights $h_1$ and $h_2$, respectively, above the initial level. The ratio $h_1 : h_2$ is $\frac{n}{10}$. The value of $n$ is _____.

Answer: 7

Solution

Gain in P.E. = Loss in K.E. $$mgh = \frac{1}{2} mv^2 \left( 1 + \frac{K^2}{R^2} \right)$$ $$h \propto 1 + \frac{K^2}{R^2}$$ $$\frac{h_1}{h_2} = \frac{1 + \frac{2}{5}}{1 + 1} = \frac{7}{5 \times 2} = \frac{7}{10}$$ $$n = 7$$

Chemistry

Question 61

Chemistry · Electrochemistry · Single correct

What pressure (bar) of $\mathrm{H}_2$ would be required to make emf of hydrogen electrode zero in pure water at $25^\circ \mathrm{C}$?

  1. $10^{-7}$
  2. $0.5$
  3. $1$
  4. $10^{-14}$

Answer: (d)

Solution

The reaction is given by $$2e^- + 2\mathrm{H}^+ (\mathrm{aq}) \rightarrow \mathrm{H}_2 (\mathrm{g})$$ The Nernst equation is $$E = E^\circ - \frac{0.059}{n} \log \frac{P_{\mathrm{H}_2}}{[\mathrm{H}^+]^2}$$ Substituting the values, we have $$0 = 0 - \frac{0.059}{2} \log \frac{P_{\mathrm{H}_2}}{(10^{-7})^2}$$ Simplifying, $$\log \frac{P_{\mathrm{H}_2}}{(10^{-7})^2} = 0$$ Therefore, $$\frac{P_{\mathrm{H}_2}}{10^{-14}} = 1$$ Thus, $$P_{\mathrm{H}_2} = 10^{-14} bar$$

Question 62

Chemistry · Co-ordination Compounds · Single correct

The correct sequence of ligands in the order of decreasing field strength is :

  1. $\mathrm{NCS}^{-} > \mathrm{EDTA}^{4-} > \mathrm{CN}^{-} > \mathrm{CO}$
  2. $\mathrm{CO} > \mathrm{H_2O} > \mathrm{F}^{-} > \mathrm{S}^{2-}$
  3. $\mathrm{S}^{2-} > \mathrm{-OH} > \mathrm{EDTA}^{4-} > \mathrm{CO}$
  4. $\mathrm{-OH} > \mathrm{F}^{-} > \mathrm{NH_3} > \mathrm{CN}^{-}$

Answer: (b)

Solution

According to spectrochemical series, ligand field strength is $\mathrm{CO} > \mathrm{H_2O} > \mathrm{F^-} > \mathrm{S^{2-}}$.

Question 63

Chemistry · Co-ordination Compounds · Single correct

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

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

Answer: (a)

Solution

The first reaction shows a +R effect with $\mathrm{NH_2}$ as an electron donating group. The second reaction shows a -R effect with $\mathrm{NO_2}$ as an electron withdrawing group. The third reaction shows a +E effect with $\mathrm{H^+}$ as an electron deficient species. The fourth reaction shows a -E effect with $\mathrm{CN}$ as an electron deficient species.

Question 64

Chemistry · Equilibrium · Single correct

What will be the decreasing order of basic strength of the following conjugate bases? $\mathrm{OH^-},\ \mathrm{RO^-},\ \mathrm{CH_3COO^-},\ \mathrm{Cl^-}$

  1. \quad $\mathrm{RO^-} > \mathrm{OH^-} > \mathrm{CH_3COO^-} > \mathrm{Cl^-}$
  2. \quad $\mathrm{Cl^-} > \mathrm{RO^-} > \mathrm{OH^-} > \mathrm{CH_3COO^-}$
  3. \quad $\mathrm{OH^-} > \mathrm{RO^-} > \mathrm{CH_3COO^-} > \mathrm{Cl^-}$
  4. \quad $\mathrm{Cl^-} > \mathrm{OH^-} > \mathrm{RO^-} > \mathrm{CH_3COO^-}$

Answer: (a)

Solution

Strong acids have weak conjugate bases. Acidic strength: $$H - Cl > CH_3COOH > H_2O > R - OH$$ Conjugate base strength: $$Cl^- < CH_3COO^- < OH^- < RO^-$$

Question 65

Chemistry · Analytical Chemistry · Single correct

In the precipitation of the iron group (III) in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to :

  1. increase concentration of $\mathrm{Cl}^-$ ions
  2. increase concentration of $\mathrm{NH}_4^+$ ions
  3. prevent interference by phosphate ions
  4. decrease concentration of $\mathrm{OH}^-$ ions

Answer: (d)

Solution

Given the reactions: $$\mathrm{NH_4OH} \rightleftharpoons \mathrm{NH_4^+} + \mathrm{OH^-}$$ $$\mathrm{NH_4Cl} \rightarrow \mathrm{NH_4^+} + \mathrm{Cl^-}$$ Due to the common ion effect of $\mathrm{NH_4^+}$, $[\mathrm{OH^-}]$ decreases to such an extent that only group-III cation can be precipitated, due to their very low $K_{sp}$ in the range of $10^{-38}$.

Question 66

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

Identify (B) and $(C)$ and how are (A) and $(C)$ related?

Answer: (c)

Solution

Compound (A) undergoes an elimination reaction (E2) with alcoholic NaOH to form compound (B). Compound (B) then reacts with HBr in ether through an electrophilic addition reaction to form compound (C). A and C are position isomers.

Question 67

Chemistry · Electrochemistry · Single correct

One of the commonly used electrode is calomel electrode. Under which of the following categories, calomel electrode comes?

  1. Oxidation - Reduction electrodes
  2. Metal ion - Metal electrodes
  3. Gas - Ion electrodes
  4. Metal - Insoluble Salt - Anion electrodes

Answer: (d)

Solution

Theory based

Question 68

Chemistry · Co-ordination Compounds · Single correct

Number of complexes from the following with even number of unpaired " d " electrons is $$[\mathrm{V(H_2O)_6}]^{3+}, \ [\mathrm{Cr(H_2O)_6}]^{2+}, \ [\mathrm{Fe(H_2O)_6}]^{3+}, \ [\mathrm{Ni(H_2O)_6}]^{3+}, \ [\mathrm{Cu(H_2O)_6}]^{2+}$$ [Given atomic numbers : $V = 23$, $Cr = 24$, $Fe = 26$, $Ni = 28$, $Cu = 29$]

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

Answer: (a)

Solution

$[\mathrm{V(H_2O)_6}]^{3+} \rightarrow d^2sp^3$ $_{23}\mathrm{V} : -[\mathrm{Ar}]\,3d^3\,4s^2$ $\mathrm{V}^{3+} : -[\mathrm{Ar}]\,3d^2,\; n=2$ (even number of unpaired $e^-$) --- $[\mathrm{Cr(H_2O)_6}]^{2+} \rightarrow sp^3d^2$ $_{24}\mathrm{Cr} : -[\mathrm{Ar}]\,3d^5\,4s^1$ $\mathrm{Cr}^{2+} : -[\mathrm{Ar}]\,3d^4,\; n=4$ (even number of unpaired $e^-$) \[ \begin{array}{c} e_g\\ \boxed{\uparrow}\boxed{\phantom{\uparrow}} \end{array} \] \[ \begin{array}{c} t_{2g}\\ \boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \end{array} \] --- $[\mathrm{Fe(H_2O)_6}]^{3+} \rightarrow sp^3d^2$ $\mathrm{Fe}^{3+} : -[\mathrm{Ar}]\,3d^5\,4s^0$ $n=5$ (odd number of unpaired $e^-$) --- $[\mathrm{Ni(H_2O)_6}]^{3+} \rightarrow sp^3d^2$ $\mathrm{Ni} : -[\mathrm{Ar}]\,3d^8\,4s^2$ $\mathrm{Ni}^{3+} : -[\mathrm{Ar}]\,3d^7,\; n=3$ (odd number of unpaired $e^-$) --- $[\mathrm{Cu(H_2O)_6}]^{2+} \rightarrow sp^3d^2$ $\mathrm{Cu} : -[\mathrm{Ar}]\,3d^9\,4s^0$ $n=1$ (odd number of unpaired $e^-$)

Question 69

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Which one of the following molecules has maximum dipole moment?

  1. $\mathrm{NF}_3$
  2. $\mathrm{CH}_4$
  3. $\mathrm{PF}_5$
  4. $\mathrm{NH}_3$

Answer: (d)

Solution

For $\mathrm{CH_4}$ and $\mathrm{PF_5}$, $\mu_{net} = 0$ (non polar). $$|\mu|_{\mathrm{NH_3}} > |\mu|_{\mathrm{NF_3}}$$ Vector addition of bond moment and lone pair moment is greater than vector subtraction of bond moment and lone pair moment.

Question 70

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Number of molecules/ions from the following in which the central atom is involved in $sp^3$ hybridization is _____ $\mathrm{NO_3^-}$, $\mathrm{BCl_3}$, $\mathrm{ClO_2^-}$, $\mathrm{ClO_3}$

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

Answer: (c)

Solution

Question 71

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

Which among the following is incorrect statement?

  1. Electromeric effect dominates over inductive effect
  2. The electromeric effect is, temporary effect
  3. Hydrogen ion ($H^+$) shows negative electromeric effect
  4. The organic compound shows electromeric effect in the presence of the reagent only.

Answer: (c)

Solution

Hydrogen ion ($\mathrm{H}^+$) shows positive electromeric effect.

Question 72

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Given below are two statements: Statements I: Acidity of $\alpha$-hydrogens of aldehydes and ketones is responsible for Aldol reaction. Statement II: Reaction between benzaldehyde and ethanal will NOT give Cross - Aldol product. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Answer: (a)

Solution

Aldehyde and ketones having acidic $\alpha$-hydrogen show aldol reaction. Benzaldehyde reacts with ethanal in the presence of a base to form a cross aldol product.

Question 73

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

Which of the following nitrogen containing compound does not give Lassaigne's test?

  1. Urea
  2. Phenyl hydrazine
  3. Glycene
  4. Hydrazine

Answer: (d)

Solution

Hydrazine ($\mathrm{NH_2 - NH_2}$) have no carbon so does not show Lassaigne's test.

Question 74

Chemistry · Biomolecules · Single correct

Which of the following is the correct structure of L-Glucose?

Answer: (a)

Solution

The structure of L-Glucose is shown with the following configuration: At the top, there is $\mathrm{CHO}$. The first hydroxyl group ($\mathrm{HO}$) is on the left. The second hydroxyl group ($\mathrm{HO}$) is also on the left. The third hydroxyl group ($\mathrm{HO}$) is on the left. The fourth hydroxyl group ($\mathrm{OH}$) is on the right. At the bottom, there is $\mathrm{CH_2OH}$.

Question 75

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

The element which shows only one oxidation state other than its elemental form is :

  1. Cobalt
  2. Titanium
  3. Nickel
  4. Scandium

Answer: (d)

Solution

Co, Ti, Ni can show $+2$, $+3$ and $+4$ oxidation states. But 'Sc' only shows $+3$ stable oxidation state.

Question 76

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Identify the product in the following reaction:

Answer: (a)

Solution

The reaction shown is a Clemmensen reduction. It involves the reduction of a carbonyl group to a methylene group using zinc amalgam (Zn-Hg) and hydrochloric acid (HCl).

Question 77

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

Number of elements from the following that CANNOT form compounds with valencies which match with their respective group valencies is _____. B, C, N, S, O, F, P, Al, Si

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

Answer: (b)

Solution

N, O, F can't extend their valencies up to their group number due to the non-availability of vacant 2 d like orbital.

Question 78

Chemistry · Solutions · Single correct

The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is : (Given : Molar Mass Na : 23 and Cl : 35.5 $\mathrm{g/mol}$ )

  1. 2
  2. 20
  3. 4
  4. 0.2

Answer: (d)

Solution

Molarity $M$ is given by the formula: $$M = \frac{n_{\mathrm{NaCl}}}{V_{\mathrm{sol}} (in L)}$$ Substituting the given values: $$M = \frac{5.85}{58.5}$$ $$M = \frac{1}{0.5} = 0.2 \, \mathrm{M}$$

Question 79

Chemistry · Haloalkanes and Haloarenes · Single correct

Identify the correct set of reagents or reaction conditions 'X' and 'Y' in the following set of transformation

  1. X = dil.aq. NaOH, 20°C, Y = Br₂/CHCl₃
  2. X = conc.alc. NaOH, 80°C, Y = Br₂/CHCl₃
  3. X = dil.aq. NaOH, 20°C, Y = HBr /acetic acid
  4. X = conc.alc. NaOH, 80°C, Y = HBr/ acetic acid

Answer: (d)

Solution

The reaction involves the conversion of $\mathrm{CH_3 - CH_2 - CH_2 - Br}$ to $\mathrm{CH_3 - CH = CH_2}$ using concentrated alcoholic $\mathrm{NaOH}$ (X). Then, $\mathrm{CH_3 - CH = CH_2}$ is converted to $\mathrm{CH_3 - CHBr - CH_3}$ using $\mathrm{HBr}$ in acetic acid (Y).

Question 80

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

The correct order of first ionization enthalpy values of the following elements is : (A) O (B) N (C) Be (D) F (E) B Choose the correct answer from the options given below :

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

Answer: (a)

Solution

The correct order of first ionization energy (IE) is: $$\mathrm{Li < B < Be < C < O < N < F < Ne}$$ This corresponds to the order: $$\mathrm{E < C < A < B < D}$$

Question 81

Chemistry · Thermodynamics · Numerical

The enthalpy of formation of ethane ($C_2H_6$) from ethylene by addition of hydrogen where the bond-energies of $\mathrm{C-H}$, $\mathrm{C-C}$, $\mathrm{C=C}$, $\mathrm{H-H}$ are $414 \, \mathrm{kJ}$, $347 \, \mathrm{kJ}$, $615 \, \mathrm{kJ}$ and $435 \, \mathrm{kJ}$ respectively is kJ

Answer: 125

Solution

For the reaction $\mathrm{C_2H_4 (g) + H_2 (g) \rightarrow C_2H_6 (g)}$, the change in enthalpy $\Delta H$ is calculated as follows. $\n$ $\n$ $\Delta$ H = \mathrm{BE(C=C) + 4BE(C-H) + BE(H-H)}$$ $\n$$$- \mathrm{BE(C-C) - 6BE(C-H)}$$ $\n$$\nSimplifying$, $\n$ $\n$ $\Delta$ H = \mathrm{BE(C=C) + BE(H-H) - BE(C-C)}$$ $\n$$$- 2\mathrm{BE(C-H)}$$ $\n$$\nSubstituting$ the bond energies, $\n$ $\n$= 615 + 435 - 347 - 2 $\times$ 414 $\n$= -125 \, \mathrm{kJ}

Question 82

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

The number of the correct reaction(s) among the following is

Answer: (a)

Solution

Option (A) involves the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl3, leading to the formation of a benzyl ketone, which is incorrect. Option (B) involves the reduction of benzoyl chloride to benzoic acid using hydrogen and Pd-BaSO4, which is incorrect. Option (C) involves the reaction of benzene with carbon monoxide and HCl in the presence of anhydrous AlCl3 and CuCl, leading to the formation of benzaldehyde, which is correct. Option (D) involves the hydrolysis of benzamide to aniline using $\mathrm{H_3O^+}$ and heat, which is incorrect.

Question 83

Chemistry · Amines · Numerical

X $\mathrm{g}$ of ethylamine is subjected to reaction with $\mathrm{NaNO_2/HCl}$ followed by water; evolved dinitrogen gas which occupied 2.24 $\mathrm{L}$ volume at STP. X is __ $\times 10^{-1}$ $\mathrm{g}$.

Answer: 45

Solution

Given: $\mathrm{N_2}$ evolved is $2.24 \, \mathrm{L}$ i.e. $0.1 \, mole$. i.e. $\mathrm{CH_3CH_2NH_2}$ (ethyl amine) will be $4.5 \, \mathrm{g}$ ($= 0.1 \, mole$). Hence the answer $= 45 \times 10^{-1} \, \mathrm{g}$.

Question 84

Chemistry · Structure of Atom · Numerical

The de-Broglie's wavelength of an electron in the 4th orbit is _______$\pi a_0 \cdot (a_0 = Bohr's radius)$

Answer: 8

Solution

Given $$2 \pi r_n = n \lambda_d$$ Substituting $$2 \pi a_0 \frac{n^2}{Z} = n \lambda_d$$ For $$2 \pi a_0 \frac{4^2}{1} = 4 \lambda_d$$ We find $$\lambda_d = 8 \pi a_0$$

Question 85

Chemistry · Redox Reactions · Numerical

Only 2 mL of $KMnO_4$ solution of unknown molarity is required to reach the end point of a titration of 20 $\mathrm{\, mL}$ of oxalic acid (2M) in acidic medium. The molarity of $KMnO_4$ solution should be____ $\mathrm{M.}$

Answer: 8

Solution

eq. ($\mathrm{KMnO_4}$) = eq. ($\mathrm{H_2C_2O_4}$) $$M \times 2 \times 5 = 2 \times 20 \times 2$$ Solving for $M$, we get: $$M = 8M$$

Question 86

Chemistry · The d-and f-Block Elements · Numerical

Consider the following reaction $$\mathrm{MnO_2 + KOH + O_2 \rightarrow A + H_2O}$$. Product ' A ' in neutral or acidic medium disproportionate to give products ' B ' and ' C ' along with water. The sum of spin-only magnetic moment values of B and C is ______$\mathrm{BM}$ . (nearest integer) (Given atomic number of Mn is 25)

Answer: 4

Solution

Given the reaction $\mathrm{MnO_2 + KOH + O_2 \rightarrow K_2MnO_4 + H_2O}$. In neutral or acidic solution, $\mathrm{K_2MnO_4}$ converts to $\mathrm{KMnO_4 + MnO_2}$. For $\mathrm{Mn^{+4}}$, the electronic configuration is $[\mathrm{Ar}] 3d^3$. The number of unpaired electrons $n = 3$, and the magnetic moment $\mu = \sqrt{3(3+2)} = 3.87 \, \mathrm{B.M.}$ The nearest integer is (4).

Question 87

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below. \[ A+B \xrightleftharpoons[\text{Step 3}]{\text{Step 1}} C \xrightarrow{\text{Step 2}} P \] Some details of the above reactions are listed below. \[ \begin{array}{ccc} \text{Step} & \text{Rate constant }(\mathrm{sec}^{-1}) & \text{Activation energy }(\mathrm{kJ\,mol}^{-1})\\ 1 & k_1 & 300\\ 2 & k_2 & 200\\ 3 & k_3 & E_{a3} \end{array} \] If the overall rate constant of the above transformation $(k)$ is given as \[ k=\frac{k_1k_2}{k_3} \] and the overall activation energy $(E_a)$ is $400\,\mathrm{kJ\,mol^{-1}}$, then the value of $E_{a3}$ is \[ \underline{\hspace{2cm}}\,\mathrm{kJ\,mol^{-1}} \] (nearest integer).

Answer: 100

Solution

Given $$K = \frac{K_1 K_2}{K_3}$$ $$Ae^{-\frac{E_3}{RT}} = \frac{A_1 e^{-\frac{E_{a1}}{RT}} A_2 e^{-\frac{E_{32}}{RT}}}{A_3 e^{-\frac{E_{k3}}{RT}}}$$ $$Ae^{-\frac{E_3}{RT}} = \frac{A_1 A_2}{A_3} e^{-\frac{(E_{a1} - E_{a2} - E_{33})}{RT}}$$ $$E_a = E_{a1} + E_{a2} - E_{a3}$$ $$400 = 300 + 200 - E_{a3}$$ $$E_{a3} = 100 \, \mathrm{kJ/mole}$$

Question 88

Chemistry · Solutions · Numerical

2.5 $\mathrm{g}$ of a non-volatile, non-electrolyte is dissolved in 100 $\mathrm{g}$ of water at $25^\circ$ $\mathrm{C}$. The solution showed a boiling point elevation by $2^\circ$ $\mathrm{C}$. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is _____ mm of Hg (nearest integer) [Given : Molal boiling point elevation constant of water ($K_b$) = 0.52 $\mathrm{K}$ $\cdot$ $\mathrm{kgmol}^{-1}$, 1 atm pressure = 760 mm of Hg, molar mass of water = 18 $\mathrm{g}$ $\cdot$ $\mathrm{mol}^{-1}$]

Answer: 707

Solution

Given $2 = 0.52 \times \mathrm{m}$. $$\mathrm{m} = \frac{2}{0.52}$$ According to the question, the solution is much diluted, so $$\frac{\Delta P}{P^\circ} = \frac{n_{solute}}{n_{solvent}}$$ $$\frac{\Delta P}{P^\circ} = \frac{\mathrm{m}}{1000} \times M_{solvent}$$ $$\Delta P = P^\circ \times \frac{\mathrm{m}}{1000} \times M_{solvent}$$ $$= 760 \times \frac{\frac{2}{0.52}}{1000} \times 18 = 52.615$$ $$P_5 = 760 - 52.615 = 707.385 \, mm of Hg$$

Question 89

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

The number of different chain isomers for $\mathrm{C_7H_{16}}$ is

Answer: 9

Solution

Question 90

Chemistry · Chemical Bonding and Molecular Structure · Numerical

Number of molecules/species from the following having one unpaired electron is ____$\mathrm{O_2}$, $\mathrm{O_2^{-1}}$, $\mathrm{NO}$, $\mathrm{CN^{-1}}$, $\mathrm{O_2^{2-}}$

Answer: 2

Solution

According to M.O.T. $\mathrm{O_2} \rightarrow$ no. of unpaired electrons $= 2$ $\mathrm{O_2^-} \rightarrow$ no. of unpaired electron $= 1$ $\mathrm{NO} \rightarrow$ no. of unpaired electron $= 1$ $\mathrm{CN^-} \rightarrow$ no. of unpaired electron $= 0$ $\mathrm{O_2^{2-}} \rightarrow$ no. of unpaired electron $= 0$