JEE Main 4 April 2026 Shift 2 question paper with solutions

JEE Main 4 April 2026 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 · Relations and Functions · Single correct

For the function $f : [1, \infty) \rightarrow [1, \infty)$ defined by $f(x) = (x-1)^4 + 1$, among the two statements: (I) The set $S = \{ x \in [1, \infty) : f(x) = f^{-1}(x) \}$ contains exactly two elements, and (II) The set $S = \{ x \in [1, \infty) : f(x) = f^{-1}(x+1) \}$ is an empty set,

  1. only (I) is TRUE
  2. only (II) is TRUE
  3. both (I) and (II) are TRUE
  4. neither (I) nor (II) is TRUE

Answer: (a)

Solution

Given $f(x) = (x - 1)^4 + 1$. The derivative is $f'(x) = 4(x - 1)^3$; hence $f'(x) \geq 0$. Therefore, $f(x)$ is increasing. Thus, $(x - 1)^4 + 1 = x$. Solving $(x - 1) [(x - 1)^3 - 1] = 0$, we find $x = 1, x = 2$ are two solutions. Now, $f^{-1}(x) = (x - 1)^{1/4} + 1$. For $f^{-1}(x + 1) = x^{1/4} + 1$, we have $f^{-1}(x + 1) = f(x)$. Therefore, $x^{1/4} + 1 = (x - 1)^4 + 1$. Solving $x = (x - 1)^{16}$, the above equation has one solution using the graph.

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $S = \{ z \in \mathbb{C} : z^2 + 4z + 16 = 0 \}$. Then \[ \sum_{z\in S}\left|z+\sqrt{3}i\right|^2\] is equal to:

  1. 42
  2. 23
  3. 27
  4. 38

Answer: (d)

Solution

Given $z^2 + 4z + 16 = 0$. Therefore, $$(z + 2)^2 = -12$$ which implies $$z = -2 \pm 2\sqrt{3}i$$ Thus, $$S = \{-2 + 2\sqrt{3}i, -2 - 2\sqrt{3}i\}$$ Therefore, $$\sum_{z \in S} \left| z + \sqrt{3}i \right|^2$$ is calculated as follows: $$= \left|-2 + 2\sqrt{3}i + \sqrt{3}i\right|^2 + \left|-2 - 2\sqrt{3}i + \sqrt{3}i\right|^2$$ $$= \left|-2 + 3\sqrt{3}i\right|^2 + \left|-2 - \sqrt{3}i\right|^2$$ $$= 38$$

Question 3

Maths · Determinants · Single correct

If the system of equations : $$x + y + z = 5$$ $$x + 2y + 3z = 9$$ $$x + 3y + \lambda z = \mu$$ has infinitely many solutions, then the value of $\lambda + \mu$ is :

  1. 16
  2. 18
  3. 19
  4. 21

Answer: (b)

Solution

Given $D = \lambda - 5$. $$D_1 = \lambda + \mu - 18$$ $$D_2 = 4\lambda - 2\mu + 6$$ $$D_3 = \mu - 13$$ Here $x = \frac{D_1}{D}$, $y = \frac{D_2}{D}$, $z = \frac{D_3}{D}$. For infinitely many solutions, $D = D_1 = D_2 = D_3 = 0$. $$\Rightarrow \lambda = 5; \mu = 13$$ Therefore, $\lambda + \mu = 18$.

Question 4

Maths · Complex Numbers and Quadratic Equations · Single correct

If $\alpha = 1$ and $\beta = 1 + i \sqrt{2}$, where $i = \sqrt{-1}$ are two roots of the equation $x^3 + ax^2 + bx + c = 0$, $a, b, c \in \mathbb{R}$, then $$\int_{-1}^{1} \left( x^3 + ax^2 + bx + c \right) dx$$ is equal to:

  1. $-2$
  2. $-4$
  3. $-8$
  4. $-10$

Answer: (c)

Question 5

Maths · Complex Numbers and Quadratic Equations · Single correct

If the quadratic equation $(\lambda + 2)x^2 - 3\lambda x + 4\lambda = 0$, $\lambda \neq -2$, has two positive roots, then the number of possible integral values of $\lambda$ is:

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

Answer: (b)

Solution

Given $f(x) = (\lambda + 2)x^2 - 3\lambda x + 4\lambda$. C-1: $af(0) > 0$ $$(\lambda + 2)4\lambda > 0$$ $$\Rightarrow \lambda 0$$ C-2: $-\frac{b}{2a} > 0$ $$\frac{3\lambda}{2(\lambda + 2)} > 0$$ $$\Rightarrow \lambda 0$$ C-3: $D \geq 0$ $$(-3\lambda)^2 - 4(\lambda + 2) \times 4\lambda \geq 0$$ $$\lambda(7\lambda + 32) \leq 0$$ $$\Rightarrow \lambda \in \left[\frac{-32}{7}, 0\right]$$ Intersection of C-1, C-2 and C-3: $$\Rightarrow \lambda \in \left[\frac{-32}{7}, -2\right)$$ $$\lambda \in [-4.57, -2)$$ $$\Rightarrow \lambda = -4, -3$$ Therefore, the number of values of $\lambda = 2$.

Question 6

Maths · Determinants · Single correct

Let $A=\begin{bmatrix}1&2&7\\4&-2&8\\3&8&-7\end{bmatrix}$ and $\det(A-\alpha I)=0$, where $\alpha$ is a real number. If the largest possible value of $\alpha$ is $p$, then the circle $(x-p)^2+(y-2p)^2=320$, intersects the co-ordinate axes at

  1. 1 point
  2. 2 points
  3. 3 points
  4. 4 points

Answer: (c)

Solution

The determinant of the matrix is given by: $$\begin{vmatrix} 1 - \alpha & 2 & 7 \\ 4 & -2 - \alpha & 8 \\ 3 & 8 & -7 - \alpha \end{vmatrix} = 0$$ This simplifies to: $$(1 - \alpha) [(\alpha + 2)(\alpha + 7) - 64] - 2[-28 - 4\alpha - 24] + 7[32 + 6 + 3\alpha] = 0$$ Simplifying further, we get: $$\alpha^3 + 8\alpha^2 - 88\alpha - 320 = 0$$ Factoring gives: $$(\alpha - 8)(\alpha^2 + 16\alpha + 40) = 0$$ Thus, $\alpha = 8$ or $\alpha = -8 \pm 2\sqrt{6}$. So $p = -8$. The equation becomes: $$(x - 8)^2 + (y - 16)^2 = 320 = 8^2 + 16^2$$ Substituting $y = 0$ gives $x = 16, 0$. Substituting $x = 0$ gives $y = 32, 0$. Thus, the points of intersection are $(16, 0), (0, 32), (0, 0)$. Therefore, there are 3 points of intersection.

Question 7

Maths · Sequences and Series · Single correct

Let $\alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \infty$ and $\beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \infty$. Then the value of $(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_{5}(\beta)}$ is equal to:

  1. 4
  2. 5
  3. 8
  4. 25

Answer: (c)

Solution

Given $\alpha = \frac{1/4}{1 - \frac{1}{2}} = \frac{1}{2}$ $\beta = \frac{1/3}{1 - \frac{1}{3}} = \frac{1}{2}$ $(0.2)^{\log_{\sqrt{5}} \alpha} = \alpha^{\log_{\sqrt{5}} 5} = \left( \frac{1}{2} \right)^{-2} = 4$ $(0.04)^{\log_5 \beta} = 5^{-2 \log_5 \beta} = \left( \frac{1}{2} \right)^{-2} = 4$ $\Rightarrow 4 + 4 = 8$

Question 8

Maths · Statistics · Single correct

For 10 observations $x_1, x_2, \ldots, x_{10}$, if $$\sum_{i=1}^{10} (x_i + 2)^2 = 180$$ and $$\sum_{i=1}^{10} (x_i - 1)^2 = 90$$, then their standard deviation is:

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

Answer: (d)

Solution

Given $$\sum_{i=1}^{10} (x_i + 2)^2 = 180$$ We expand to get: $$\sum_{i=1}^{10} x_i^2 + 4 \sum_{i=1}^{10} x_i + \sum_{i=1}^{10} 4 = 180$$ Simplifying, we have: $$\sum_{i=1}^{10} x_i^2 + 4 \sum_{i=1}^{10} x_i = 180 - 40 (1)$$ Also, given $$\sum_{i=1}^{10} (x_i - 1)^2 = 90$$ We expand to get: $$\sum_{i=1}^{10} x_i^2 - 2 \sum_{i=1}^{10} x_i + \sum_{i=1}^{10} 1 = 90$$ Simplifying, we have: $$\sum_{i=1}^{10} x_i^2 - 2 \sum_{i=1}^{10} x_i = 90 - 10 (2)$$ From equations (1) and (2), we have: $$\sum_{i=1}^{10} x_i^2 + 4 \sum_{i=1}^{10} x_i = 140$$ $$\sum_{i=1}^{10} x_i^2 - 2 \sum_{i=1}^{10} x_i = 80$$ Adding these equations, we get: $$\sum_{i=1}^{10} x_i^2 = 100 and \sum_{i=1}^{10} x_i = 10$$ The variance is given by: $$\sigma^2 = \frac{\sum_{i=1}^{10} x_i^2}{N} - \left(\frac{\sum_{i=1}^{10} x_i}{N}\right)^2$$ Substituting the values, we have: $$= \frac{100}{10} - \left(\frac{10}{10}\right)^2$$ Thus, $$\sigma^2 = 10 - 1 = 9$$ Therefore, $$\sigma = 3$$

Question 9

Maths · Binomial Theorem · Single correct

In the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}$, $x > 0$, if the term independent of $x$ is $(221)k$, then $k$ is equal to:

  1. 84
  2. 78
  3. 168
  4. 198

Answer: (a)

Solution

General term in above expression is $$T_{r+1} = \binom{18}{r} (9x)^{18-r} \left( \frac{-1}{3\sqrt{x}} \right)^r$$ $$= \left( \frac{-1}{3} \right)^r \binom{18}{r} 9^{18-r} \cdot x^{\frac{18-3r}{2}}$$ For term independent of $x$, $18 - \frac{3r}{2} = 0 \implies r = 12$ Therefore, coefficient of term independent of $x$ $$= \left( \frac{-1}{3} \right)^{12} \binom{18}{12} 9^{18-12}$$ $$= 18564$$ $$= 221 \ k \ (given)$$ $$\implies k = 84$$

Question 10

Maths · Conic Sections · Single correct

Let $P(3\cos\alpha,2\sin\alpha)$, $\alpha\neq0$, be a point on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$, $Q$ be a point on the circle $x^2+y^2-14x-14y+82=0$ and $R$ be a point on the line $x+y=5$ such that the centroid of the triangle $PQR$ is $\left(2+\cos\alpha,3+\frac{2}{3}\sin\alpha\right)$. Then the sum of the ordinates of all possible points $R$ is:

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

Answer: (d)

Solution

Centroid: $$\left( \frac{3 \cos \alpha + 7 + 4 \cos \theta + 5 - y}{3}, \frac{2 \sin \alpha + 7 + 4 \sin \theta + y}{3} \right)$$ $$= \left( \cos \alpha + 2, \frac{2}{3} \sin \alpha + 3 \right)$$ On comparison $$\cos \alpha = \frac{y - 6}{4} \& \sin \alpha = \frac{2 - y}{4}$$ $$\sin^2 \alpha + \cos^2 \alpha = 1 \Rightarrow (y - 6)^2 + (y - 2)^2 = 16$$ $$y^2 - 8y + 12 = 0$$ $\Rightarrow$ Sum of ordinates = 8

Question 11

Maths · Conic Sections · Single correct

Let $\mathrm{H}: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be a hyperbola such that the distance between its foci is $6$ and the distance between its directrices is $\frac{8}{3}$. If the line $x = \alpha$ intersects the hyperbola H at the points A and B such that the area of the triangle AOB is $4\sqrt{15}$, where O is the origin, then $\alpha^2$ equals

  1. 12
  2. 16
  3. 24
  4. 25

Answer: (b)

Solution

Given: $$2ae = 6 \Rightarrow ae = 3$$ $$\frac{2a}{e} = \frac{8}{3} \Rightarrow \frac{a}{e} = \frac{4}{3}$$ $$a^2 = 4 e = \frac{3}{2} b^2 = 5$$ $$\frac{x^2}{4} - \frac{y^2}{5} = 1$$ solve with $x = \alpha$ $$y = \pm \sqrt{\frac{5(\alpha^2 - 4)}{4}}$$ $$A \left( \alpha, \sqrt{\frac{5(\alpha^2 - 4)}{4}} \right) B \left( \alpha, -\sqrt{\frac{5(\alpha^2 - 4)}{4}} \right)$$ Area of $\triangle OAB = \frac{1}{2} \left| \alpha \right| \sqrt{5(\alpha^2 - 4)} = 4\sqrt{15}$ $$\Rightarrow 16 \times 15 = \frac{1}{4} \alpha^2 (5)(\alpha^2 - 4)$$ $$\Rightarrow \alpha^2 = 16$$

Question 12

Maths · Applications of Derivatives · Single correct

$\max_{0\leq x\leq\pi}\left(16\sin\left(\frac{x}{2}\right)\cos^3\left(\frac{x}{2}\right)\right)$ is equal to:

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

Answer: (b)

Solution

Given $E = 16 \sin \frac{x}{2} \cos^3 \frac{x}{2}$. $E = 4 \sin x \left[1 + \cos x\right]$ Differentiating, $\frac{dE}{dx} = 4 \left[ \cos x + \cos 2x \right]$. Setting the derivative to zero: $$8 \cos \frac{3x}{2} \cos \frac{x}{2} = 0$$ This implies: $$\cos \frac{3x}{2} = 0 or \cos \frac{x}{2} = 0$$ Thus, $x = \left\{ \frac{\pi}{3}, \pi \right\}$ are critical points of the function. Evaluating $E$ at these points: $E(0) = 0$ $E(\pi) = 0$ $E\left(\frac{\pi}{3}\right) = 3\sqrt{3}$ Therefore, the maximum value of $E = 3\sqrt{3}$.

Question 13

Maths · Three Dimensional Geometry · Single correct

The shortest distance between the lines $$\vec{r} = \left( \frac{1}{3} \hat{i} + 2 \hat{j} + \frac{8}{3} \hat{k} \right) + \lambda \left( 2 \hat{i} - 5 \hat{j} + 6 \hat{k} \right)$$ and $$\vec{r} = \left( -\frac{2}{3} \hat{i} - \frac{1}{3} \hat{k} \right) + \mu \left( \hat{j} - \hat{k} \right), \lambda, \mu \in \mathbb{R},$$ is :

  1. $\sqrt{5}$
  2. 3
  3. 2$\sqrt{3}$
  4. $\sqrt{15}$

Answer: (b)

Solution

Given $\vec{a}_1 = \frac{1}{3} \hat{i} + 2 \hat{j} + \frac{8}{3} \hat{k}$. $\vec{a}_2 = -\frac{2}{3} \hat{i} - \frac{1}{3} \hat{k}$. $\vec{b}_1 = 2 \hat{i} - 5 \hat{j} + 6 \hat{k}$, $\vec{b}_2 = \hat{j} - \hat{k}$. $$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -5 & 6 \\ 0 & 1 & -1 \end{vmatrix} = -\hat{i} + 2 \hat{j} + 2 \hat{k}$$ $\vec{a}_1 - \vec{a}_2 = \hat{i} + 2 \hat{j} + 3 \hat{k}$. The scalar distance (S.D) is given by: $$S.D = \frac{|(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$ Substituting the values: $$S.D = \frac{|(\hat{i} + 2 \hat{j} + 3 \hat{k}) \cdot (-\hat{i} + 2 \hat{j} + 2 \hat{k})|}{3}$$ $$= \frac{9}{3} = 3$$

Question 14

Maths · Three Dimensional Geometry · Single correct

If $\left(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2}\right)$ is the image of ($\alpha$, 2$\alpha$, 1) in the line $\frac{x-2}{3}$ = $\frac{y-1}{2}$ = $\frac{z}{1}$, then the possible value(s) of $\alpha$ is(are) :

  1. Only 3
  2. Only 3 and -1
  3. Only 3, $\frac{1}{4}$ and -1
  4. Only 3 and $\frac{1}{4}$

Answer: (a)

Solution

P is the midpoint of AB and lies on the given line. Therefore, $$ P \left( \frac{3\alpha + 1}{2}, \frac{\alpha^2 - \alpha}{2}, \frac{\alpha + 1}{4} \right) $$ Now put coordinates of P in the line $$ \frac{\frac{3\alpha + 1}{2} - 2}{3} = \frac{\frac{\alpha^2 - \alpha}{2} - 1}{2} = \frac{\alpha + 1}{4} $$ This implies $$ \frac{\alpha - 1}{2} = \frac{\alpha + 1}{4} = \frac{\alpha^2 - \alpha - 2}{4} $$ Solving gives $$ \alpha = 3 $$ Now for $\alpha = 3$, clearly AB is perpendicular to $3\hat{i} + 2\hat{j} + \hat{k}$. For $\alpha = 3$, point B is the image of point A.

Question 15

Maths · Vector Algebra · Single correct

Let $\hat{u}$ and $\hat{v}$ be unit vectors inclined at an acute angle such that $|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}$. If $\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v})$. Then $\lambda$ is equal to:

  1. $\frac{4}{3} (\vec{A} \cdot \hat{u}) - \frac{2}{3} (\vec{A} \cdot \hat{v})$
  2. $\frac{2}{3} (\vec{A} \cdot \hat{u}) - \frac{1}{3} (\vec{A} \cdot \hat{v})$
  3. $\frac{4}{3} (\vec{A} \cdot \hat{u}) + \frac{2}{3} (\vec{A} \cdot \hat{v})$
  4. $(\vec{A} \cdot \hat{u}) - \frac{1}{2} (\vec{A} \cdot \hat{v})$

Answer: (a)

Solution

Given $|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}$. $|\hat{u}| |\hat{v}| \sin \theta = \frac{\sqrt{3}}{2}$. Therefore, $\theta = \frac{\pi}{3}$. And $\hat{u} \cdot \hat{v} = |\hat{u}| |\hat{v}| \cos \frac{\pi}{3} = \frac{1}{2}$. $$\vec{A} = \lambda \hat{u} + \hat{v} + \hat{u} \times \hat{v} ...(1)$$ Dot with $\hat{u}$ $$\vec{A} \cdot \hat{u} = \lambda (1) + \hat{u} \cdot \hat{v} + \hat{u} \cdot (\hat{u} \times \hat{v})$$ $$\vec{A} \cdot \hat{u} = \lambda + \frac{1}{2}$$ $$\Rightarrow 2 \vec{A} \cdot \hat{u} = 2 \lambda + 1 ...(2)$$ Dot equation (1) with $\hat{v}$ $$\vec{A} \cdot \hat{v} = \lambda (\hat{u} \cdot \hat{v}) + \hat{v} \cdot \hat{v} + \hat{v} \cdot (\hat{u} \times \hat{v})$$ $$\vec{A} \cdot \hat{v} = \frac{\lambda}{2} + 1$$ $$\vec{A} \cdot \hat{v} = 1 ...(3)$$ From (2) and (3) $$2 \vec{A} \cdot \hat{u} - 2 \lambda = \vec{A} \cdot \hat{v} - \frac{\lambda}{2}$$ Therefore, $\lambda = \frac{4}{3} \vec{A} \cdot \hat{u} - \frac{2}{3} \vec{A} \cdot \hat{v}$.

Question 16

Maths · Relations and Functions · Single correct

Let for some $\alpha \in \mathbb{R}$, $f : \mathbb{R} \to \mathbb{R}$ be a function satisfying $f(x+y) = f(x) + 2y^2 + y + \alpha xy$ for all $x,y \in \mathbb{R}$. If $f(0) = -1$ and $f(1) = 2$, then the value of $$\sum_{n=1}^{5} (\alpha + f(n))$$ is:

  1. 110
  2. 140
  3. 150
  4. 170

Answer: (b)

Solution

Given $f(x+y) = f(x) + 2y^2 + y + \alpha xy$ ...(1) Put $x = 0$ in eq.(1) $$\Rightarrow f(y) = -1 + 2y^2 + y$$ Now put $x = y = 1$ in eq. (1) $$\Rightarrow f(2) = f(1) + 3 + \alpha$$ $$\Rightarrow 9 = 2 + 3 + \alpha \Rightarrow \alpha = 4$$ Now $$\sum_{n=1}^{5} (\alpha + f(n)) = \sum_{n=1}^{5} (2y^2 + y + 3)$$ $$= \frac{2 \times 5 \times 6 \times 11}{6} + \frac{5 \times 6}{2} + 3 \times 5 = 140$$

Question 17

Maths · Permutations and Combinations · Single correct

Let $A = \{(a, b, c) : a, b, c$ are non-negative integers and $a + b + 2c = 22\}$. The $n(A)$ is equal to :

  1. 121
  2. 124
  3. 144
  4. 169

Answer: (c)

Solution

Given $c = 0$ implies $a + b = 22$ which leads to $$\binom{22 + 2 - 1}{2 - 1} = \binom{23}{1}$$ $c = 1$ implies $a + b = 20$ which leads to $$\binom{20 + 2 - 1}{2 - 1} = \binom{21}{1}$$ $\vdots$ $c = 10$ implies $a + b = 2$ which leads to $$\binom{2 + 2 - 1}{2 - 1} = \binom{3}{1}$$ $c = 11$ implies $a + b = 0$ which leads to $$\binom{0 + 2 - 1}{2 - 1} = \binom{1}{1}$$ Number of non-negative integral solutions $$\binom{1}{1} + \binom{3}{1} + \binom{5}{1} + \ldots + \binom{23}{1}$$ $$1 + 3 + 5 + \ldots + 23$$ $$= 144$$

Question 18

Maths · Applications of Integrals · Single correct

The area of the region bounded by the curves $x + 3y^2 = 0$ and $x + 4y^2 = 1$ is equal to:

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

Answer: (c)

Solution

Given $x = -3y^2$ and $x = 1 - 4y^2$. $$A = \int_{-1}^{1} \left( (1 - 4y^2) + 3y^2 \right) \, dy$$ $$A = \int_{-1}^{1} (1 - y^2) \, dy = 2 \int_{0}^{1} (1 - y^2) \, dy$$ $$A = \frac{4}{3}$$

Question 19

Maths · Differential Equations · Single correct

Let $y = y(x)$ be the solution of the differential equation: $$\frac{dy}{dx} + \left( \frac{6x^2 + (3x^2 + 2x^3 + 4)e^{-2x}}{(x^3 + 2)(2 + e^{-2x})} \right) y = 2 + e^{-2x},$$ $x \in (-1, 2)$, satisfying $y(0) = \frac{3}{2}$. If $y(1) = \alpha(2 + e^{-2})$, than $\alpha$ is equal to:

  1. $\frac{13}{8}$
  2. $\frac{6}{13}$
  3. $\frac{12}{13}$
  4. $\frac{13}{12}$

Answer: (d)

Solution

I.F. = e^{$\int \frac{6x^2 + e^{-2x}(3x^2 + 2x^3 + 4)}{(x^3 + 2)(2 + e^{-2x})}$ dx} = e^{$\int \frac{6x^2 + e^{-2x}(3x^2 - 2x^3 - 4) + (4x^3 + 8)e^{-2x}}{(x^3 + 2)(2 + e^{-2x})}$ dx}. Let (x^3 + 2)(2 + e^{-2x}) = t. (6x^2 + e^{-2x}(3x^2 - 2x^3 - 4))dx = dt. I.F. = e^{$\ln$|(x^3 + 2)(2 + e^{-2x})| - $\int \frac{4e^{-2x}dx}{2 + e^{-2x}}$} = e^{$\ln$|(x^3 + 2)(2 + e^{-2x})| - 2$\ln$|2 + e^{-2x}|} = e^{$\ln \left|\frac{x^3 + 2}{2 + e^{-2x}}\right|$} = $\frac{x^3 + 2}{2 + e^{-2x}}$. $\Rightarrow$ y $\cdot \left(\frac{x^3 + 2}{2 + e^{-2x}}\right)$ = $\int \left(\frac{x^3 + 2}{2 + e^{-2x}}\right)$(e^{-2x} + 2)dx + C. y $\cdot \left(\frac{x^3 + 2}{2 + e^{-2x}}\right)$ = $\frac{x^4}{4}$ + 2x + C at y(0) = $\frac{3}{2} \Rightarrow$ C = 1. So y $\cdot \left(\frac{x^3 + 2}{2 + e^{-2x}}\right)$ = $\frac{x^4}{4}$ + 2x + 1 at x = 1. y $\cdot \left(\frac{3}{2 + e^{-2}}\right)$ = $\frac{1}{4}$ + 2 + 1. $\Rightarrow$ y = $\frac{13}{12}$(e^{-2} + 2). So $\alpha$ = $\frac{13}{12}$.

Question 20

Maths · Integrals · Single correct

The integral $\int_{0}^{1} \cot^{-1} \left( 1 + x + x^2 \right) dx$ is equal to:

  1. $2 \tan^{-1} 2 + \frac{1}{2} \log_e \left( \frac{5}{4} \right) + \frac{\pi}{2}$
  2. $2 \tan^{-1} 2 + \frac{1}{2} \log_e \left( \frac{5}{4} \right) - \frac{\pi}{2}$
  3. $2 \tan^{-1} 2 - \frac{1}{2} \log_e \left( \frac{5}{4} \right) + \frac{\pi}{2}$
  4. $2 \tan^{-1} 2 - \frac{1}{2} \log_e \left( \frac{5}{4} \right) - \frac{\pi}{2}$

Answer: (d)

Solution

The solution is given by the integral $$ \int_0^1 \tan^{-1} \left( \frac{1}{1 + x + x^2} \right) \, dx $$ which can be rewritten as $$ \int_0^1 \tan^{-1} \left( \frac{(x+1) - x}{1 + x(x+1)} \right) \, dx $$ This simplifies to $$ \int_0^1 (\tan^{-1}(x+1) - \tan^{-1} x) \, dx $$ Evaluating the integral, we have $$ \left( x \tan^{-1}(x+1) - \frac{1}{2} \ln |1 + (1+x)^2| + \tan^{-1}(x+1) \right) \bigg|_0^1 $$ minus $$ \left( x \tan^{-1} x - \frac{1}{2} \ln |1 + x^2| \right) \bigg|_0^1 $$ This results in $$ \left( 2 \tan^{-1} 2 - \frac{\pi}{4} - \frac{1}{2} \ln \frac{5}{2} \right) - \left( \frac{\pi}{4} - \frac{1}{2} \ln 2 \right) $$ Simplifying further, we get $$ 2 \tan^{-1} 2 - \frac{1}{2} \ln \left( \frac{5}{4} \right) - \frac{\pi}{2} $$

Question 21

Maths · Probability · Numerical

From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to $\frac{a}{b}$, where $a, b \in \mathbb{N}$ and $\gcd(a,b) = 1$, then $a + b$ is equal to $\ldots$

Answer: 944

Solution

Total numbers of ways of selecting 3 numbers = $$^{31}C_3 = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 4495$$ a, b, c are in A.P. Therefore, either a and c both odd numbers or both even numbers $$= ^{16}C_2 + ^{15}C_2$$ $$= 120 + 105$$ $$= 225$$ Probability = $$\frac{225}{4495} = \frac{45}{899}$$ $$= \frac{a}{b} \Rightarrow a + b = 944$$

Question 22

Maths · Continuity and Differentiability · Numerical

Let $f(x) = \begin{cases} e^{x-1}, & x < 0 \\ x^2 - 5x + 6, & x \geq 0 \end{cases}$ and $g(x) = f(|x|) + |f(x)|$. If the number of points where $g$ is not continuous and is not differentiable are $\alpha$ and $\beta$ respectively, then $\alpha + \beta$ is equal to $\ldots$

Answer: 4

Solution

Graph of $f(x)$ Now graph of $f(|x|)$ $f(|x|)$ is continuous function and it is non diff. at $x = 0$ Graph of $|f(x)|$ $|f(x)|$ is discontinuous at $x = 0$ $|f(x)|$ is non-diff. at $x = 0, 2, 3$ $g(x) = f(|x|) + |f(x)|$ $g(x)$ will be discontinuous at $x = 0$ $g(x)$ will be non diff at $x = 0, 2, 3$ $\alpha = 1$, $\beta = 3$ $\alpha + \beta = 1 + 3 = 4$

Question 23

Maths · Properties of Triangles · Numerical

Let A, B be points on the two half-lines $x - \sqrt{3} |y| = \alpha$, $\alpha > 0$ at a distance of $\alpha$ from their point of intersection P. The line segment AB meets the angle bisector of the given half-lines at the point Q. If $PQ = \frac{9}{2}$ and R is the radius of the circumcircle of $\triangle PAB$, then $\frac{\alpha^2}{R}$ is equal to $\ldots$.

Answer: 9

Solution

Given that $\triangle PAB$ is equilateral. $$PQ = \frac{\sqrt{3} \alpha}{2} = \frac{9}{2} \therefore \alpha = \frac{9}{\sqrt{3}}$$ $$\alpha = 3 \sqrt{3}$$ $$\frac{\alpha}{2r} = \cos 30^\circ$$ $$\alpha = \frac{\sqrt{3}}{2} \cdot 2R$$ $$\alpha = \sqrt{3} R$$ $$R = 3$$ $$\frac{\alpha^2}{R} = \frac{(3 \sqrt{3})^2}{3} = 9$$

Question 24

Maths · Conic Sections · Numerical

Let A, B and C be the vertices of a variable right angled triangle inscribed in the parabola $y^2 = 16x$. Let the vertex B containing the right angle be $(4, 8)$ and the locus of the centroid of $\triangle ABC$ be a conic $C_0$. Then three times the length of latus rectum of $C_0$ is $\ldots$.

Answer: 16

Solution

Variable point A and C are shown in the figure. Therefore, $m_{AB} \cdot m_{BC} = -1$. Thus, $t_1 + t_2 + t_1 t_2 = -5 \ldots (1)$. Suppose the locus of the centroid of $\Delta ABC$ is $(h, k)$. Therefore, $$3h = 4 + 4t_1^2 + 4t_2^2 and 3k = 8 + 8t_1 + 8t_2.$$ By eliminating $t_1$ and $t_2$ using equation (1), we also get $$h = \frac{9}{48} k^2 + \frac{40}{3}.$$ Therefore, the locus of the centroid of the parabola is $$x = \frac{9}{48} y^2 + \frac{40}{3}.$$ Thus, $\ell(\mathrm{L.R.}) = \frac{48}{9}$. Therefore, $3\ell(\mathrm{L.R.}) = \frac{48}{3} = 16$.

Question 25

Maths · Integrals · Numerical

Let $f$ be a twice differentiable function such that $$f(x) = \int_{0}^{x} \tan(t-x) \, dt - \int_{0}^{x} f(t) \tan t \, dt, x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$$ Then $f''\left( \frac{\pi}{6} \right) + 12 f'\left( -\frac{\pi}{6} \right) + f\left( \frac{\pi}{6} \right)$ is equal to $\ldots$.

Answer: 5

Solution

Given $f(x) = \int_0^x \tan(x - t - x) \, dt - \int_0^x f(t) \tan t \, dt$. $f(x) = -\int_0^x \tan t \, dt - \int_0^x f(t) \tan t \, dt$. $f'(x) = -\tan x - f(x) \tan x$. $$\frac{dy}{y+1} = -\tan x \, dx$$ $y + 1 = \cos x \cdot c$. $f(0) = 0 \implies 1 = c$. $y = \cos x - 1 = f(x)$. $$f\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} - 1$$ $f'(x) = -\sin x$. $$f'\left(-\frac{\pi}{6}\right) = \frac{1}{2}$$ $f''(x) = -\cos x$. $$f''\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$ $$f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right) = 5$$

Physics

Question 26

Physics · Dual Nature of Radiation and Matter · Single correct

Match the $\textbf{LIST-I}$ with $\textbf{LIST-II}$ Choose the $\textbf{correct}$ answer from the options given below:

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

Answer: (a)

Solution

(i) $[h] = \left[ \frac{E}{v} \right] = \frac{ML^2 T^{-2}}{T^{-1}} = ML^2 T^{-1}$ (ii) Stopping potential: $E = q V_0$ $$[V_0] = [v_0] = \left[ \frac{E}{q} \right] = \frac{ML^2 T^{-2}}{AT} = ML^2 T^{-3} A^{-1}$$ (iii) Work function ($\phi$) $\rightarrow$ Energy form $$[\phi] = ML^2 T^{-2}$$ (iv) Threshold freq. $[v_0] = T^{-1}$

Question 27

Physics · Motion in a Straight Line · Single correct

Two cars A and B are moving in the same direction along a straight line with speeds 100 km/h and 80 km/h, respectively such that car A is moving ahead of car B. A person in car B throws a stone with a speed $v$ so that it hits the car A with a speed of $5 \, \mathrm{m/s}$. The value of $v$ is $\ldots$ km/h.

  1. 18
  2. 28
  3. 38
  4. 48

Answer: (c)

Solution

Given $V = \vec{V}_{SB}$. $V = \vec{V}_S - \vec{V}_B$. $V_B + V = \vec{V}_S$. $$\vec{V}_S = V + 80$$ $\vec{V}_{SA} = 5 \, \mathrm{m/s} = 18 \, \mathrm{km/hr}$. $\vec{V}_S - \vec{V}_A = \vec{V}_{SA}$. $V + 80 - 100 = 18 \, \mathrm{km/hr}$. $V - 20 = 18$. $V = 38 \, \mathrm{km/hr}$.

Question 28

Physics · Laws of Motion · Single correct

At $t = 0$, a body of mass $100 \, \mathrm{g}$ starts moving under the influence of a force $(5\hat{i} + 10\hat{j}) \, \mathrm{N}$. After $2 \, \mathrm{s}$ its position is $(2x\hat{i} + 5y\hat{j}) \, \mathrm{m}$. The ratio $x : y$ is $\ldots$.

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

Answer: (d)

Solution

Given $\vec{a} = \frac{\vec{F}}{m} = \frac{5\hat{i} + 10\hat{j}}{0.1} = (50\hat{i} + 100\hat{j})$. $\vec{s} = \vec{u}t + \frac{1}{2} \vec{a} t^2$ $\vec{s} = 0 + \frac{1}{2} [50\hat{i} + 100\hat{j}][2]^2$ $\vec{s} = 100\hat{i} + 200\hat{j} = \vec{r} = 2x\hat{i} + 5y\hat{j}$ $2x = 100 5y = 200$ $x = 50 y = 100$ $\frac{x}{y} = \frac{5}{4}$

Question 29

Physics · Motion in a Plane · Single correct

If x and y coordinates of a projectile as a function of time (t) are given as $24t$ and $43.6t - 4.9t^2$, respectively, then the angle (in degrees) made by the projectile with horizontal when $t = 2 \, \mathrm{s}$ is $\ldots$.

  1. 60
  2. 45
  3. 30
  4. 75

Answer: (b)

Solution

Given $x = 24t$ and $y = 43.6t - 4.9t^2$. The velocity in the $x$ direction is $v_x = \frac{dx}{dt} = 24$. The velocity in the $y$ direction is $v_y = \frac{dy}{dt} = 43.6 - 9.8t$. The tangent of the angle $\theta$ is given by $\tan \theta = \frac{v_y}{v_x} = \frac{43.6 - 9.8t}{24}$. At $t = 2$, $$\frac{43.6 - 19.6}{24}$$ which gives $\tan \theta = 1$. Therefore, $\theta = 45^\circ$.

Question 30

Physics · Gravitation · Single correct

The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $\frac{g}{9}$, where $g$ is acceleration due to gravity on earth's surface, is $\ldots$

  1. $\sqrt{3}R$
  2. $2\sqrt{2}R$
  3. $2R$
  4. $\frac{4}{9}R$

Answer: (c)

Solution

Given $g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$. $$\frac{g}{9} = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$$ Solving for $h$, we have: $$1 + \frac{h}{R} = 3$$ Therefore, $h = 2R$.

Question 31

Physics · Mechanical Properties of Solids · Single correct

A metal string A is suspended from a rigid support and its free end is attached to a block of mass M. Second block having mass 2M is suspended at the bottom of the first block using a string B. The area of cross sections of strings A and B are same. The ratio of lengths of strings of A to B is 2 and the ratio of their Young's moduli $(Y_A/Y_B)$ is 0.5. The ratio of elongations in A to B is $\ldots$.

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

Answer: (d)

Solution

The elongation $\Delta \ell$ is given by $$\Delta \ell = \frac{T \ell}{YA}$$ For the first part, $$\Delta \ell_1 = \frac{3mg \ell}{YA}$$ For the second part, $$\Delta \ell_2 = \frac{2mg(\ell/2)}{2YA} = \frac{mg \ell}{2YA}$$ The ratio of elongations is $$\frac{\Delta \ell_1}{\Delta \ell_2} = 6$$

Question 32

Physics · Mechanical Properties of Fluids · Single correct

A water spray gun is attached to a hose of cross sectional area $30 \, \mathrm{cm}^2$. The gun comprises of 10 perforations each of cross sectional area of $15 \, \mathrm{mm}^2$. If the water flows in the hose with the speed of $50 \, \mathrm{cm/s}$, calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)

  1. 100 m/s
  2. 10 m/s
  3. 1000 m/s
  4. 15 $\times$ 10^2 $\mathrm{m/s}$

Answer: (b)

Solution

Using equation of continuity $$AV = 10av$$ $$v = \frac{AV}{10a} = \frac{30 \times 50}{10 \times 0.15} = 1000 \, \mathrm{cm/s} = 10 \, \mathrm{m/s}$$

Question 33

Physics · Kinetic Theory · Single correct

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: If the average kinetic energy of $\mathrm{H_2}$ and $\mathrm{O_2}$ molecules, kept in two different sized containers are same, then their temperatures will be same. Reason R: The r.m.s. speed of $\mathrm{H_2}$ and $\mathrm{O_2}$ molecules are same at same temperature.

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

Answer: (c)

Solution

Avg. kinetic energy of molecule will be $\frac{f}{2} KT$. But $v_{rms} = \sqrt{\frac{3RT}{M}}$. Molar mass $m$ different so $v_{rms}$ is different.

Question 34

Physics · Thermal Properties of Matter · Single correct

The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $T_1$ to $T_2$, resulting in increase of its length by $\Delta L_1$. The temperature is further increased from $T_2$ to $T_3$ such that the increase in its length is $\Delta L_2$. Given $T_3 + T_1 = 2T_2$ and $T_2 - T_1 = \Delta T$, the value of $\Delta L_2$ is $\ldots$.

  1. $\Delta L_1 [1 + 2\alpha^2 (\Delta T)^2]$
  2. $\Delta L_1 [1 + \alpha^2 (\Delta T)^2]$
  3. $\Delta L_1 [1 + 2\alpha \Delta T]$
  4. $\Delta L_1 [1 + \alpha \Delta T]$

Answer: (d)

Solution

Given $\Delta L_1 = L_0 \, \alpha \, (T_2 - T_1)$. So final length at $T_2 = L_0 + \Delta L_1$. $\Delta L_2 = (L_0 + \Delta L_1) \, \alpha \, (T_3 - T_2)$ and $T_3 - T_2 = T_2 - T_1 = \Delta T$. $\Delta L_2 = L_0 \, \alpha \, \Delta T + \Delta L_1 \, \alpha \, \Delta T$. $= \Delta L_1 + \Delta L_1 \, \alpha \, \Delta T$. $= \Delta L_1 \, (1 + \alpha \Delta T)$.

Question 35

Physics · Oscillations · Single correct

A uniform disc of radius R and mass M is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is _______. (g is acceleration due to gravity)

  1. 2$\pi \sqrt{\frac{5R}{4g}}$
  2. 2$\pi \sqrt{\frac{2R}{3g}}$
  3. 2$\pi \sqrt{\frac{3R}{2g}}$
  4. 2$\pi \sqrt{\frac{3R}{g}}$

Answer: (a)

Solution

The moment of inertia of the disc is given by $$I_{disc} = \frac{mR^2}{4} + mR^2 = \frac{5mR^2}{4}$$ Therefore, for small displacement $$\tau = -mgR\theta$$ Thus, $$\alpha = -\frac{mgR}{\frac{5mR^2}{4g}} \theta$$ Therefore, $$\omega = \sqrt{\frac{4g}{5R}}$$ Thus, $$T = 2\pi \sqrt{\frac{5R}{4g}}$$

Question 36

Physics · Oscillations · Single correct

A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field $\vec{E}_1 = E_0 \hat{x}$. If another electric field $\vec{E}_2 = 2E_0 (\hat{y} + \hat{z})$ is introduced to the system, what will be the percentage change in the frequency of the oscillation (approximate)?

  1. 73$\%$
  2. 63$\%$
  3. 83$\%$
  4. 53$\%$

Answer: (a)

Solution

Initial freq. $f_1 = \frac{1}{2\pi} \sqrt{\frac{P(E_0)}{I}}$ Final EF $= E_0 \hat{i} + 2E_0 (\hat{j} + \hat{k})$ Magnitude $= 3E_0$ Final Freq. $f_2 = \frac{1}{2\pi} \sqrt{\frac{P(3E_0)}{I}}$ $\%$ change $= \frac{\Delta f}{f_1} = (\sqrt{3} - 1) \times 180 \approx 73$

Question 37

Physics · Electrostatic Potential and Capacitance · Single correct

From the circuit given below, the capacitance between terminals A and B shown in the circuit is _____ $\mu \mathrm{F}$. (take $C_1 = C_2 = C_3 = 1 \, \mu \mathrm{F}$ and $C_4 = 2 \, \mu \mathrm{F}$.)

  1. 2
  2. 7/2
  3. 7/3
  4. 5/2

Answer: (b)

Solution

The equivalent capacitance between points A and B is calculated. The circuit is rearranged to simplify the calculation. The equivalent capacitance is given by: $$C_{AB} = 2 \, \mu\mathrm{F}$$

Question 38

Physics · Electrostatic Potential and Capacitance · Single correct

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: In electrostatics, a conductor does not store any net charge inside. Reason R: Inside the capacitor (with no dielectric medium), the free charge carriers, if placed between the plates of capacitor, experience force and drift. Choose the correct answer from the options given below.

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

Answer: (b)

Solution

Theoretical

Question 39

Physics · Magnetism and Matter · Single correct

A solenoid has a core made of material with relative permeability 400. The magnetic field produced in the interior of solenoid is $1.0\,\mathrm{T}$. The magnetic intensity is SI units is $\alpha \times 10^5$. The value of $\alpha$ is $\ldots$ (Free space permeability $\mu_0 = 4\pi \times 10^{-7}$ SI units).

  1. $\frac{25}{\pi}$
  2. $\frac{1}{16\pi}$
  3. $\frac{1}{\pi}$
  4. $\frac{1}{4\pi}$

Answer: (b)

Solution

Given $\mathbf{H} = \frac{\mathbf{B}}{\mu} = \frac{\mathbf{B}}{\mu_0 \mu_r} = \frac{1}{400 \times 4 \pi \times 10^{-7}}$ $$\mathbf{H} = \frac{1}{16 \pi} \times 10^5$$

Question 40

Physics · Electromagnetic Waves · Single correct

A magnetic field vector in an electromagnetic wave is represented by, $\vec{B} = B_0 \sin \left( 2 \pi vt - \frac{2 \pi x}{\lambda} \right) \hat{j}$. Its associated electric field vector is $\ldots$.

  1. $\vec{E} = -v \lambda B_0 \sin \left( 2 \pi vt - \frac{2 \pi x}{\lambda} \right) \hat{k}$
  2. $\vec{E} = -v \lambda B_0 \sin \left( 2 \pi vt - \frac{2 \pi x}{\lambda} \right) \hat{i}$
  3. $\vec{E} = v \lambda B_0 \sin \left( 2 \pi vt - \frac{2 \pi x}{\lambda} \right) \hat{k}$
  4. $\vec{E} = v \lambda B_0 \sin \left( 2 \pi vt - \frac{2 \pi x}{\lambda} \right) \hat{i}$

Answer: (a)

Solution

Given $\hat{E} \times \hat{B} = \hat{C}$. $\hat{B} = \hat{j}$ $\hat{C} = \hat{i}$ Therefore, $\hat{E} = -\hat{k}$. $$\frac{E_0}{B_0} = C = v \lambda$$ $$E_0 = B \times v \lambda$$

Question 41

Physics · Ray Optics and Optical Instruments · Single correct

A convex lens is made from glass material having refractive index of 1.4 with same radius of curvature on both sides. The ratio of its focal length and radius of curvature is $\ldots$.

  1. 0.5
  2. 2.5
  3. 0.8
  4. 1.25

Answer: (d)

Solution

Given $\frac{1}{f} = (1.4 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$. $\frac{1}{f} = \frac{0.8}{R}$ $\frac{f}{R} = 1.25$

Question 42

Physics · Wave Optics · Single correct

An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at $30^\circ$ and $90^\circ$, respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at $60^\circ$ with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is $\ldots$.

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

Answer: (c)

Solution

The initial intensity is $I_0$. After passing through the first polarizer at $30^\circ$, the intensity becomes $\frac{I_0}{2}$. After the second polarizer at $90^\circ$, the intensity is $\frac{I_0}{2} \cos^2(60^\circ) = \frac{I_0}{8}$. For the second setup, the initial intensity is $I_0$. After the first polarizer at $30^\circ$, the intensity is $\frac{I_0}{2}$. After the second polarizer at $60^\circ$, the intensity becomes $\frac{3}{8} I_0$. Finally, after the third polarizer at $90^\circ$, the intensity is $\frac{3I_0}{8} \times \frac{3}{4} = \frac{9}{32} I_0$. The ratio of the final intensities is given by: $$\frac{\frac{9I_0}{32}}{\frac{I_0}{8}} \times \frac{8}{I_0} = \frac{9}{4}$$

Question 43

Physics · Atoms · Single correct

In Rutherford's alpha-particle scattering experiment, only a few alpha particles rebound back because A.The size of gold nucleus is very small as compared to the size of gold atom. B. Alpha particle and gold nucleus have equal charge. C.The impact parameter is minimum for a few alpha particles. D. A few alpha particles have very high kinetic energy. E. Only a few alpha particles undergo head-on collision with the nuclei.

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

Answer: (i)

Solution

A, C, E only. Theoretical.

Question 44

Physics · Dual Nature of Radiation and Matter · Single correct

The de Broglie wavelength associated with an electron accelerated through a potential difference $V$ is $\lambda_e$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_p$. If their corresponding masses are $m_e$ and $m_p$, respectively, then the ratio of their de Broglie wavelengths $\left( \frac{\lambda_e}{\lambda_p} \right)$ is $\ldots$.

  1. $\sqrt{\frac{m_p}{m_e}}$
  2. $\sqrt{\frac{m_e}{m_p}}$
  3. $\frac{m_p}{m_e}$
  4. $\left( \frac{m_p}{m_e} \right)^2$

Answer: (a)

Solution

Given $$\lambda = \frac{h}{\sqrt{2mqV}} \propto \frac{1}{\sqrt{m}}$$. So $$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$$ $$\sqrt{\frac{m_e}{m_p}}$$

Question 45

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

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : A diode under reverse-biased condition provides very small current which is nearly independent of voltage until a critical limit at which the current increases drastically. Reason R : Below the critical voltage limit, only majority charge carriers flow which increases drastically above critical voltage. Choose the correct answer from the options given below.

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

Answer: (c)

Solution

Theoretical

Question 46

Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Numerical

A diode has Zener voltage of $10 \, \mathrm{V}$ and maximum power dissipation of $0.5 \, \mathrm{W}$, then the minimum resistance to be used in series with this diode for safety when it is connected to a $25 \, \mathrm{V}$ power supply is $\ldots \, \Omega$.

Answer: 300

Solution

Maximum current from diode $$i = \frac{P}{V} = \frac{0.5}{10} = 0.05 \, \mathrm{A}$$ Now let resistance is $R$ $$V_B = i (R + R_{diode})$$ $$25 = 0.05 \times R + 10$$ $$R = \frac{15}{0.05} = 300 \, \Omega$$

Question 47

Physics · Motion in a Plane · Fill in the blank

A gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is $6.4 \, \mathrm{m}$. The speed of the bullets from the gun is $\ldots \mathrm{m/s}$. (take $g = 10 \, \mathrm{m/s^2}$)

Answer: 8

Solution

Given $R_{max} = \frac{v^2}{g} = 6.4$. $v = \sqrt{64} = 8 \, \mathrm{m/s}$

Question 48

Physics · Magnetism and Matter · Numerical

Two identical small bar magnets each of dipole moment $3\sqrt{5} \, \mathrm{J/T}$ are placed at a center to center separation of $10 \, \mathrm{cm}$, with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point $P$ midway between the magnets is $\alpha \times 10^{-3} \, \mathrm{T}$. The value of $\alpha$ is $\ldots$. $(\mu_0 = 4\pi \times 10^{-7} \, \mathrm{Tm/A})$

Answer: 12

Solution

The net magnetic field $B_{net}$ is given by the equation: $$B_{net} = \sqrt{B_1^2 + B_2^2} = \frac{\mu_0}{4\pi} \frac{M}{(r)^3} \sqrt{5}$$ The diagram shows $B_1$ and $B_2$ as perpendicular vectors. Calculating further: $$= 10^{-7} \times 3\sqrt{5} \times \sqrt{5} \times \frac{8}{10^{-3}}$$ $$= 120 \times 10^{-4} = 12 \, mT$$

Question 49

Physics · Moving Charges and Magnetism · Numerical

A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of 30^$\circ$ with the direction of the magnetic field. The torque acting on the coil is $\alpha \times$ 10^{-4} $\mathrm{N} \cdot \mathrm{m}$. The value of $\alpha$ is $\ldots$. ($\pi$ = 3.14)

Answer: 314

Solution

The torque $\tau$ is given by $\tau = |\vec{M} \times \vec{B}|$. This can be expressed as $\tau = |N I \vec{A} \times \vec{B}|$. Substituting the values, we have $$\tau = 125 \times 1 \times \pi \left( \frac{2}{100} \right)^2 \times 0.4 \times \sin 30^\circ$$ $$= 100 \pi \times 10^{-4} \, \mathrm{N\cdot m}$$ So $\alpha = 100 \pi = 314$.

Question 50

Physics · Wave Optics · Numerical

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe shifts to the position of 7^{th} bright fringe, obtained with both slits uncovered. If the light source wavelength is $450\,\mathrm{nm}$, the thickness of mica sheet is $\alpha \times$ 10^{-9} $\mathrm{m}$. The value of $\alpha$ is$\ldots$.

Answer: 5625

Solution

Given $\Delta y = (\mu - 1) t \frac{D}{d} = 7 \frac{\lambda D}{d}$. $(\mu - 1) t = 7 \lambda$. $(1.56 - 1) t = 7 \times 450 \, \mathrm{m}$. $t = 5625 \, \mathrm{nm}$.

Chemistry

Question 51

Chemistry · Some Basic Concepts of Chemistry · Single correct

The correct order of total number of atoms of present in (A) 2 moles of cyclohexane (B) 684 g of sucrose (C) 90.8 L of dihydrogen at STP

  1. C > A > B
  2. C > B > A
  3. B > C > A
  4. B > A > C

Answer: (c)

Solution

(A) Mole of atoms in $\mathrm{C_6H_{12}}$ $$= 2 \times 18$$ $$= 36$$ (B) Moles of atoms of sucrose $$= \frac{604}{342} \times 45$$ (C) $n_{\mathrm{H_2}} = \frac{90.8}{22.7} = 4$ moles Mole of atoms $= 4 \times 2 = 8$ (B) $>$ (A) $>$ (C)

Question 52

Chemistry · Structure of Atom · Single correct

The species having identical radii according to the Bohr's theory are: A. H (first orbit) B. $\mathrm{He}^{+}$ (first orbit) C. $\mathrm{He}^{+}$ (second orbit) D. $\mathrm{Li}^{2+}$ (first orbit) E. $\mathrm{Be}^{3+}$ (Second orbit) Choose the correct answer from the options given below:

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

Answer: (b)

Solution

(A) $\rightarrow$ same (B) $\rightarrow$ same (C) $\rightarrow$ same (D) $\rightarrow$ r = $\frac{90 \times (1)^2}{3}$ = $\frac{90}{3}$ (E) $\rightarrow$ same

Question 53

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Which of the following pictorial diagram most correctly represents the $\pi^*$ ($\pi$ - antibonding) molecular orbital between two atoms if the internuclear axis is taken to be in the z-direction ($\overset{\text{z-axis}}{\longrightarrow}$) ?

Answer: (c)

Solution

Destructive interference between $p_y$ orbitals leads to the formation of an antibonding molecular orbital (ABMO). The resulting orbital is denoted as $\pi^*_g(p_y)$.

Question 54

Chemistry · Solutions · Single correct

At $27^\circ\mathrm{C}$, $0.1\,\mathrm{M}$, $1\,\mathrm{L}$ $\mathrm{K_4[Fe(CN)_6]}$ aqueous solution and $0.1\,\mathrm{M}$, $1\,\mathrm{L}$ $\mathrm{FeCl_3}$ aqueous solution are placed in a container separated by a semi permeable membrane AB. Assume complete dissociation of both the solutes. Which of the following statement is \textit{correct}?

  1. Blue color is formed on both sides.
  2. Ionic solutes in aqueous solution can pass through semi-permeable membrane.
  3. Solution on side 'y' is hypotonic.
  4. To cause the reverse flow of solvent during osmosis, external pressure (any value) should be applied to side 'x'.

Answer: (c)

Solution

The diagram shows two compartments X and Y, each containing a solution of concentration $0.1 \, \mathrm{M}$ and volume $1 \, \mathrm{litre}$.

Question 55

Chemistry · Equilibrium · Single correct

$20\,\mathrm{mL}$ of a solution of acetic acid required $28.4\,\mathrm{mL}$ of $0.1\,\mathrm{M}$ NaOH for its neutralization. A solution (X) was prepared by mixing $20\,\mathrm{mL}$ of the above acetic acid and $14.2\,\mathrm{mL}$ of $0.1\,\mathrm{M}$ NaOH solution. What is the pH of the solution (X)? ($pK_a$ value of acetic acid is 4.75).

  1. 7.0
  2. 4.45
  3. 3.5
  4. 4.82

Answer: (b)

Solution

In experiment 1, $28.4 \, \mathrm{ml}$ of NaOH is required for complete neutralisation, therefore for $14.2 \, \mathrm{ml}$, half equivalence point will be achieved to form acidic buffer.

Question 56

Chemistry · Haloalkanes and Haloarenes · Single correct

Match the LIST-I with LIST-II \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{\textbf{List-I}} & \multicolumn{2}{c|}{\textbf{List-II}} \\ \multicolumn{2}{|c|}{\textbf{Reaction}} & \multicolumn{2}{c|}{\textbf{Mechanism}} \\ \hline A. & Williamson Synthesis & I. & Electrophilic addition \\ \hline B. & Friedel Craft Reaction & II. & Free radical substitution \\ \hline C. & Bromination of vinyl benzene & III. & Nucleophilic substitution \\ \hline D. & Chlorination of toluene in light & IV. & Electrophilic substitution \\ \hline \end{tabular} Choose the correct answer from the options given below:

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

Answer: (c)

Solution

Theory based

Question 58

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

The electronegativity of a group 13 element 'E' is same as that of Ge (on Pauling scale and upto one decimal point). The CORRECT statements about $\mathrm{E}^{3+}$ are A. It can act as a reducing agent. B. It can act as an oxidizing agent. C. $\mathrm{E}^{3+}$ is more stable than $\mathrm{E}^{+}$. D. The standard electrode potential value for $\mathrm{E}^{3+}/\mathrm{E}$ is positive. Choose the correct answer from the options given below

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

Answer: (c)

Solution

EN of 'Tl' is 1.8 that is same as 'Ge'. 'E' = Tl. $\mathrm{Tl}^{+3}$ acts as oxidizing agent. Stability order: $\mathrm{Tl}^{+3} < \mathrm{Tl}^{+1}$ (due to inert pair effect). The standard electrode potential value $\mathrm{E}^\circ(\mathrm{Tl}^{3+}/\mathrm{Tl}) = +1.26 \, \mathrm{V}$.

Question 59

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

Pairs of elements with the same number of electrons in their respective 4f orbital are [Atomic number, Eu-63, Gd-64, Dy-66, Ho-67, Tm-69, Yb-70, Lu-71, Hf-72] A. (Eu and Gd) B. (Dy and Ho) C. (Yb and Hf) D. (Lu and Tm) Choose the correct answer from the options given below:

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

Answer: (d)

Solution

Gd – $[\mathrm{Xe}]4f^7 5d^1 6s^2$ Eu – $[\mathrm{Xe}]4f^7 6s^2$ Dy – $[\mathrm{Xe}]4f^9 6s^2$ Ho – $[\mathrm{Xe}]4f^{11} 6s^2$ Hf – $[\mathrm{Xe}]4f^{14} 6s^2 5d^2$ Yb – $[\mathrm{Xe}]4f^{14} 6s^2$ Lu – $[\mathrm{Xe}]4f^{14} 5d^1 6s^2$ Tm – $[\mathrm{Xe}]4f^{13} 6s^2$

Question 60

Chemistry · Co-ordination Compounds · Single correct

Consider the metal complexes $[Ni(en)_3]^{2+}$ (A), $[NiCl_4]^{2-}$ (B) and $[Ni(NH_3)_6]^{2+}$ (C). Choose the CORRECT option by considering the number of unpaired electron present in (A), (B) and (C) respectively and the order of frequency of absorption.

  1. 2, 2, 2 and (A) > (C) > (B)
  2. 0, 2, 0 and (A) > (C) > (B)
  3. 2, 2, 0 and (B) > (C) > (A)
  4. 2, 2, 2 and (C) > (A) > (B)

Answer: (a)

Solution

(A) $\mathrm{Ni^{2+} \Rightarrow 3d^8}$, SFL, $t_{2g}^{2,2,2} e_g^{1,1} \Rightarrow$ unpaired $e^- = 2$ (B) $\mathrm{Ni^{2+} \Rightarrow 3d^8}$, WFL, tetrahedral, $e^{2,2} t_2^{2,1,1} \Rightarrow$ unpaired $e^- = 2$ (C) $\mathrm{Ni^{2+} \Rightarrow 3d^8}$, SFL, $t_{2g}^{2,2,2} e_g^{1,1} \Rightarrow$ unpaired $e^- = 2$ As $\Delta_o > \Delta_t$ and Ligand strength: en $> \mathrm{NH_3} > \mathrm{Cl^-}$ Strength of ligand increases, $\Delta$ increases, hence more is the frequency required of absorption.

Question 61

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Consider the following molecules/species: The correct order of carbon – oxygen double bond length is:

  1. x > y > z
  2. y > z > x
  3. z > x > y
  4. x > z > y

Answer: (d)

Solution

C–O bond length is maximum in compound (b) because RS2 of this compound has aromatic character.

Question 62

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

Consider $|x|$ is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with $|x|$ number of unpaired electrons from the following are: A. $\mathrm{Sc^{3+}}$ \qquad B. $\mathrm{Zn^{2+}}$ C. $\mathrm{V^{2+}}$ \qquad D. $\mathrm{Fe^{2+}}$ E. $\mathrm{Co^{2+}}$ Choose the correct answer from the options given below:

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

Answer: (c)

Solution

Oxide of Mn in its highest O.S. = $\mathrm{Mn_2O_7} [\mathrm{Mn}^{+7}]$ Fluoride of Mn in its highest O.S. = $\mathrm{MnF_4} [\mathrm{Mn}^{+4}]$ Therefore, the difference in O.S. is $|x| = 3$. $\mathrm{Sc}^{+3} \Rightarrow [\mathrm{Ar}]$; 0 unpaired $e^-$. $\mathrm{Zn}^{2+} \Rightarrow [\mathrm{Ar}] \, 3d^{10}$; 0 unpaired electrons. $\mathrm{Fe}^{+2} \Rightarrow [\mathrm{Ar}] \, 3d^6$; 4 unpaired electrons. $\mathrm{Co}^{+2} \Rightarrow [\mathrm{Ar}] \, 3d^7$; 3 unpaired electrons. $\mathrm{V}^{+2} \Rightarrow [\mathrm{Ar}] \, 3d^3$; 3 unpaired electrons.

Question 63

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

Consider the given graph showing variation of reactant concentration with time. Three different reactions were started with identical initial concentration of reactants. Which of the following statement is correct?

  1. The order of all the three reactions is same.
  2. The rate constant of reaction 3 is larger than the rate constant of reaction 2 if the order of reaction is same for both.
  3. The SI unit of rate constant of reaction 1 is $\mathrm{s}^{-1}$.
  4. Thermal decomposition of HI on gold surface is an example of reaction 2.

Answer: (b)

Solution

Curve I$\textsuperscript{st}$ represents zero order kinetics. $$[R]_t = [R]_0 - Kt$$ Therefore, $K_3 > K_2$. Reaction-1 is zero order having unit of rate constant same as rate of reaction i.e., molarity/second.

Question 64

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Compound (X) is subjected to the sequence of reactions as shown above. Molar mass of the major product (Y) formed is $\ldots$ g mol$^{-1}$ (Given molar mass in g mol$^{-1}$ C : 12, H : 1, O : 16)

  1. 90
  2. 118
  3. 160
  4. 125

Answer: (b)

Solution

The reaction starts with the addition of $\mathrm{Br_2}$ to the alkene, forming a dibromo compound. This undergoes an elimination reaction (E2) with $\mathrm{NaNH_2}$ to form a terminal alkyne, $\mathrm{Ph{-}C\equiv CH}$. The alkyne is deprotonated by $\mathrm{NH_2^-}$ to form $\mathrm{Ph{-}C\equiv C^\Theta}$. This undergoes an $\mathrm{S_N2}$ reaction with $\mathrm{CH_3{-}I}$ to form $\mathrm{Ph{-}C\equiv C{-}CH_3}$. Finally, the alkyne is reduced using $\mathrm{Na/liquid\ NH_3}$ to form the alkene $\mathrm{Ph{-}CH{=}CH{-}CH_3}$, which is compound $P$ with the formula $\mathrm{C_9H_{10}}$. The molar mass of $Y$ is $118 \, \mathrm{gm}$.

Question 65

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

The following structures are

  1. enantiomer
  2. Identical molecules
  3. Diastereomers
  4. meso compound

Answer: (b)

Solution

No chiral center (2 methyl groups)

Question 66

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

The descending order of acidity among the following compounds is Choose the correct answer from the option given below options

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

Answer: (d)

Solution

Theory based. Benzoic acid is more acidic than phenol. Electron withdrawing group increases acidity.

Question 67

Chemistry · Amines · Single correct

The strongest conjugate acid will result from

Answer: (c)

Solution

The reaction of the given compounds with $\mathrm{H^+}$ results in the formation of conjugate acids. The stability order of the conjugate acids is as follows: $$\begin{array}{c} \mathrm{\oplus NH_3} \, \mathrm{OCH_3} \\ \end{array} > \begin{array}{c} \mathrm{\oplus NH_3} \, \mathrm{CH_3} \\ \end{array} > \begin{array}{c} \mathrm{\oplus NH_3} \\ \end{array} > \begin{array}{c} \mathrm{\oplus NH_3} \, \mathrm{NO_2} \\ \end{array}$$ For stronger conjugate acid, we need weaker base.

Question 68

Chemistry · Biomolecules · Single correct

A D-aldotetrose on oxidation with concentrated $\mathrm{HNO}_3$ resulted in optically inactive dicarboxylic acid. The structure of the D-aldotetrose is :

Answer: (c)

Solution

D-Aldotetrose reacts with concentrated $\mathrm{HNO_3}$ to form an optically inactive dicarboxylic acid.

Question 69

Chemistry · Analytical Chemistry · Single correct

Among $\mathrm{Fe}^{3+}$, $\mathrm{Pb}^{2+}$, $\mathrm{Cu}^{2+}$ and $\mathrm{Mn}^{2+}$, identify the one that gets precipitated out while passing $\mathrm{H}_2\mathrm{S}$ in presence of $\mathrm{NH}_4\mathrm{OH}$ as group reagent. The highest possible oxidation state of the corresponding metal is

  1. +3
  2. +4
  3. +2
  4. +7

Answer: (d)

Solution

Group IV reagent: $\mathrm{NH_4OH} + \mathrm{H_2S}$ $\mathrm{Mn^{+2}}$ will be precipitated out and hence highest possible oxidation state of $\mathrm{Mn}$ is $+7$

Question 70

Chemistry · Analytical Chemistry · Single correct

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

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

Answer: (a)

Solution

Theory based. Tollen's test is used for test for aldehydes. Heinsberg's reagent is used to test for amines. Lucas test is used to test for alcohols. Phthalein dye test is used for test of phenols.

Question 71

Chemistry · Thermodynamics · Numerical

If 3.365 g of ethanol (l) burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The $|\Delta H_f^\circ|$ of ethanol at 298.15 K is $\ldots \times 10^2$ kJ mol$^{-1}$. (Nearest integer) Given: Standard enthalpy of combustion of graphite $= -393.5$ kJ mol$^{-1}$ Standard enthalpy of formation of water (l) $= -285.8$ kJ mol$^{-1}$ Molar mass in gmol$^{-1}$ of C, H, O are 12, 1 and 16 respectively.

Answer: 282

Solution

The change in internal energy for combustion, $\Delta U^\circ_{combustion}$, is given by: $$\Delta U^\circ_{combustion} = \frac{-q}{moles} = \frac{-99.472}{3.365/46} \Rightarrow 1359.8 \, KJ/mole$$ The reaction is: $$\mathrm{C_2H_5OH(\ell) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(\ell)}; \Delta H^\circ_r = \Delta H^\circ_C$$ The enthalpy change is: $$\Delta H^\circ_C = \Delta U^\circ + \Delta n_g RT$$ Calculating: $$= -1359.8 + \frac{(-1) \times 8.314 \times 298.15}{1000}$$ $$= -1362.27 \, KJ/mole$$ Setting up the equation: $$-1362.27 = [2 \times (-393.5) + 3 \times (-285.8)] - \Delta H^\circ_{f \, \mathrm{C_2H_5OH(\ell)}}$$ Solving for $\Delta H^\circ_{f \, \mathrm{C_2H_5OH(\ell)}}$: $$-1362.27 = -787 - 857.4 - \Delta H^\circ_{f \, [\mathrm{C_2H_5OH(\ell)}]}$$ Therefore: $$\Delta H^\circ_{f \, [\mathrm{C_2H_5OH(\ell)}]} = -2.82 \times 10^2 \, KJ/mole$$ Approximately: $$\approx -3 \times 10^2$$

Question 72

Chemistry · Equilibrium · Numerical

For the following reaction at $50^{\circ}C$ and $2 \, \mathrm{atm}$ pressure, $$2\mathrm{N}_2\mathrm{O}_5\,(\mathrm{g}) \rightleftharpoons 2\mathrm{N}_2\mathrm{O}_4\,(\mathrm{g}) + \mathrm{O}_2\,(\mathrm{g})$$ $\mathrm{N}_2\mathrm{O}_5$ is 50$\%$ dissociated The magnitude of standard free energy change at this temperature is $x$. $x = \ldots \, \mathrm{J} \, \mathrm{mol}^{-1}$ [Nearest integer] Given : $R = 8.314 \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\ln 10 = 2.303$, $^{\circ}C + 273 = \mathrm{K}$

Answer: 2474

Solution

The reaction is given as $2\mathrm{N_2O_5 (g)} \rightleftharpoons 2\mathrm{N_2O_4 (g)} + \mathrm{O_2 (g)}$. At time $t=0$, the concentrations are $1$, $-$, and $-$. At equilibrium $t_{eqm.}$, the concentrations are $1-2x$, $2x$, and $x$. Solving for $x$, we have $1-0.5$, $0.5$, and $0.25$. The concentrations become $0.5$, $0.5$, and $0.25$. The equilibrium concentrations are $\frac{0.5}{1.25} \times 2$, $\frac{0.5}{1.25} \times 2$, and $\frac{0.25}{1.25} \times 2$. The equilibrium constant $K_p$ is given by $$K_p = \frac{P_{\mathrm{N_2O_4}}^2 \cdot P_{\mathrm{O_2}}}{P_{\mathrm{N_2O_5}}^2} = 0.4.$$ The change in Gibbs free energy $\Delta G^\circ$ is calculated as $$\Delta G^\circ = -2.303 \, RT \, \log K_p$$ $$= -2.303 \times 8.314 \times 323 \, \log \left(\frac{4}{10}\right)$$ $$= 2473.81 \, \mathrm{J/mole}.$$

Question 73

Chemistry · Electrochemistry · Numerical

An electrochemical cell. consist of the following two redox couples, $\mathrm{M}^{x+}(\mathrm{aq})/ \mathrm{M}(\mathrm{s}) [E^\circ_{red} = +0.15\, \mathrm{V}]$ and $\mathrm{Fe}^{3+}(\mathrm{aq}) / \mathrm{Fe}(\mathrm{s}) [E^\circ_{red} = -0.036\, \mathrm{V}]$ The cell EMF $(E_{cell})$ is recorded to be $0.2057\, \mathrm{V}$. If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$, then the value of $x$ is $\ldots$ (Nearest integer) [Given : M is a p-block metal and $\frac{2.303RT}{F} = 0.059\, \mathrm{V}$]

Answer: 2

Solution

Given $E_{cell} = E^\circ_{cell} - \frac{0.059}{3x} \log 10^{-2}$. $$0.2057 = 0.186 - \frac{0.059}{3x}(-2)$$ $$3x = \frac{0.059 \times 2}{0.0197}$$ $$x = 1.99 \approx 2$$

Question 74

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

For a first order reaction A $\rightarrow$ B \begin{tabular}{|c|c|} \hline $t/\mathrm{min}$ & $[\mathrm{A}]/\mathrm{M}$ \\ \hline $0$ & $0.6500$ \\ \hline $x$ & $0.0650$ \\ \hline $20$ & $0.00065$ \\ \hline \end{tabular} x = $\ldots$ min. (Nearest integer)

Answer: 7

Solution

Given $K = \frac{1}{t} \ln \left[ \frac{A_0}{A_t} \right]$. $$K = \frac{1}{20} \ln \left[ \frac{0.6500}{0.00065} \right] \Rightarrow \frac{3 \ln 10}{20} \, \mathrm{min}^{-1}$$ Also, $x = \frac{1}{K} \ln \left[ \frac{A_0}{A_x} \right] \Rightarrow \frac{20}{3 \ln 10} \ln \left( \frac{0.6500}{0.0650} \right)$ $$= \frac{20}{3} \, \mathrm{min} \Rightarrow 6.7 \, \mathrm{min}$$

Question 75

Chemistry · Analytical Chemistry · Numerical

In sulphur estimation, $2.0 \times 10^{-3} \, mol$ of an organic compound (X) (molar mass $76 \, \mathrm{g \, mol^{-1}}$) gave $0.4813 \, \mathrm{g}$ of barium sulphate (molar mass $233 \, \mathrm{g \, mol^{-1}}$). The percentage of sulphur in the compound (X) is $\ldots \times 10^{-1}\%$ (Nearest integer)

Answer: 435

Solution

% of S = $\frac{Mass of S}{Mass of BaSO_4} \times \frac{m}{w} \times$ 100 m = gm of BaSO_4 ppt w = gm of Organic compound w = 2 $\times$ 10^{-3} $\times$ 76 = 0.152 gm % of S = $\frac{32}{233} \times \frac{0.4813}{0.152} \times$ 100 = 43.487 In form of 10^{-1} = 43.487 $\times$ 10^{-1} = 434.87 $\approx$ 435