JEE Main 4 April 2026 Shift 1 question paper with solutions

JEE Main 4 April 2026 Shift 1: 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 · Inverse Trigonometric Functions · Single correct

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x) = \cos^{-1}\left(\frac{4x + 2[x]}{3}\right)$ is $[\alpha, \beta]$, then $12(\alpha + \beta)$ is equal to:

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

Answer: (a)

Solution

Given the inequality $$-1 \leq \frac{4x + 2[x]}{3} \leq 1$$ we can rewrite it as $$-3 - 4x \leq 2[x] \leq 3 - 4x.$$ From the graph, we observe the shaded region which corresponds to the solution set. The values of $x$ are within the interval $$x \in \left[ -\frac{1}{4}, \frac{3}{4} \right].$$ Additionally, we have the equation $$12(\alpha + \beta) = 6.$$

Question 2

Maths · Complex Numbers and Quadratic Equations · Single correct

If the set of all solutions of $|x^2 + x - 9| = |x| + |x^2 - 9|$ is $[\alpha, \beta] \cup [\gamma, \infty)$, then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to:

  1. 9
  2. 18
  3. 36
  4. 72

Answer: (b)

Solution

Given $|a| + |b| = |a + b|$. This implies $a \cdot b \geq 0$. Therefore, $x(x^2 - 9) \geq 0$. The solution is $x \in [-3, 0] \cup [3, \infty]$. Also, $\alpha^2 + \beta^2 + \gamma^2 = 18$.

Question 3

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $z$ be a complex number such that $|z+2| = |z-2|$ and $\arg \left( \frac{z+3}{z-i} \right) = \frac{\pi}{4}$. Then $|z|^2$ is equal to:

  1. 9
  2. 4
  3. 5
  4. 1

Answer: (a)

Solution

Given $|z + 2| = |z - 2|$. Therefore, $z = 0 + iy$ where $y \in \mathbb{R}$. $$\arg \left( \frac{3 + iy}{iy - i} \right) = \frac{\pi}{4}$$ This implies $$\arg \left( \frac{y}{y - 1} + \frac{3i}{1 - y} \right) = \frac{\pi}{4}$$ Thus, $$\frac{y}{y - 1} = \frac{3}{1 - y} > 0$$ Therefore, $z = -3i$. Finally, $|z|^2 = 9$.

Question 4

Maths · Relations and Functions · Single correct

The number of functions $f : \{1, 2, 3, 4\} \to \{a, b, c\}$, which are not onto, is:

  1. 48
  2. 45
  3. 51
  4. 35

Answer: (b)

Solution

Number of into functions is given by $$= \binom{3}{1} 2^4 - \binom{3}{2} 1 + 0$$ $$= 3 \times 16 - 3 \times 1$$ $$= 45$$

Question 5

Maths · Matrices · Single correct

Let $$ S = \left\{ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \{0, 1, 2, 3, 4\} and A^2 - 4A + 3I = 0 \right\} $$ be a set of $2 \times 2$ matrices. Then the number of matrices in $S$, for which the sum of the diagonal elements is equal to 4, is:

  1. 20
  2. 17
  3. 21
  4. 19

Answer: (d)

Solution

Given $a, b, c, d \in \{0, 1, 2, 3, 4\}$. We have $a + d = 4$ and $ad - bc = 3$. Case-I: $a = 0$, $d = 4$ $$0 - bc = 3$$ No solution. Case-II: $a = 1$, $d = 3$ $$3 - bc = 3$$ $$bc = 0 \Rightarrow 9 ways$$ Case-III: $a = 2$, $d = 2$ $$4 - bc = 3$$ $$bc = 1 \Rightarrow 1 ways$$ Case-IV: $a = 3$, $d = 1$ $$bc = 0 \Rightarrow 9 way$$ Case-V: $a = 4$, $d = 0 \Rightarrow No Solution$ Total 19 matrix

Question 6

Maths · Determinants · Single correct

Let $A = \begin{bmatrix} 1 & 1 & 2 \\ -2 & 0 & 1 \\ 1 & 3 & 5 \end{bmatrix}$. Then the sum of all elements of the matrix $adj(adj(2(adjA)^{-1}))$ is equal to:

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

Answer: (d)

Solution

Given $|A| = -4$. As $(adj A)^{-1} = \frac{A}{|A|} = \frac{-A}{4}$. B = $2(adj A)^{-1} = \frac{-A}{2}$. $|B| = \left| \frac{-A}{2} \right| = \frac{-1}{8} |A| = \frac{1}{2}$. $adj (adj B) = |B|^{n-2} B = |B| B = \frac{B}{2} = \frac{-A}{4}$. Sum of element $= \frac{-12}{4} = -3$.

Question 7

Maths · Sequences and Series · Single correct

The first term of an A.P. of 30 non-negative terms is $\frac{10}{3}$. If the sum of this A.P. is the cube of its last term, then its common difference is:

  1. $\frac{5}{87}$
  2. $\frac{25}{83}$
  3. $\frac{15}{29}$
  4. $\frac{5}{29}$

Answer: (a)

Solution

Given $a = \frac{10}{3}$, $S_{30} = (T_{30})^3$. $$\frac{30}{2} \left[ 2 \times \frac{10}{3} + 29d \right] = \left[ \frac{10}{3} + 29d \right]^3$$ $$15 \left( \frac{20}{3} + 29d \right) = \left( \frac{10}{3} + 29d \right)^3$$ $$15 \left( \frac{10}{3} + \frac{10}{3} + 29d \right) = \left( \frac{10}{3} + 29d \right)^3$$ By observation $\frac{10}{3} + 29d = 5$. $$29d = 5 - \frac{10}{3} = \frac{5}{3}$$ $$d = \frac{5}{87}$$

Question 8

Maths · Permutations and Combinations · Single correct

The number of ways, of forming a queue of 4 boys and 3 girls such that all the girls are not together, is:

  1. 5040
  2. 3050
  3. 3410
  4. 4320

Answer: (d)

Solution

No. of ways to arrange 7 students = $7!$ Cases where all girls are together $B_1 B_2 B_3 B_4 \boxed{G_1 G_2 G_3} = 5! \times 3!$ Ans. = $7! - 5!3!$ $$= 7! - 6!$$ $$= 6 \times 6!$$ $$= 6 \times 720$$ $$= 4320$$

Question 9

Maths · Binomial Theorem · Single correct

Let the smallest value of $k\in\mathbb{N}$, for which the coefficient of $x^3$ in $(1+x)^3+(1+x)^4+(1+x)^5+\cdots+(1+x)^{99}+(1+kx)^{100}$, $x\neq0$, is $\left(43n+\frac{101}{4}\right)\left({}^{100}C_3\right)$ for some $n\in\mathbb{N}$, be $p$.Then the value of $p$+n is

  1. 10
  2. 11
  3. 12
  4. 13

Answer: (b)

Solution

The expression $^3C_3 + ^4C_3 + ^5C_3 + ... + ^{99}C_3 + ^{100}C_3 K^3 = \left(43n + \frac{101}{4}\right)^{100} C_3$. This implies $^4C_4 + ^4C_3 + ^5C_3 + ... + ^{99}C_3 + ^{100}C_3 K^3 = \left(43n + \frac{101}{4}\right)^{100} C_3$. We have $^{100}C_4 + ^{100}C_3 K^3 = \left(43n + \frac{101}{4}\right)^{100} C_3$. Therefore, $$\frac{^{100}C_4}{^{100}C_3} + K^3 = 43n + \frac{101}{4}$$ Simplifying, $$\frac{97}{4} + K^3 = 43n + \frac{101}{4}$$ Thus, $K^3 = 43n + 1$. Given $n = 5$, we find $K^3 = 216$. Therefore, $K = 6 = P$. Finally, $P + n = 11$.

Question 10

Maths · Statistics · Single correct

Suppose that the mean and median of the non-negative numbers 21, 8, 17, a, 51, 103, b, 13, 67, (a > b), are 40 and 21, respectively. If the mean deviation about the median is 26, then 2a is equal to:

  1. 109
  2. 117
  3. 161
  4. 131

Answer: (d)

Solution

Given the numbers 8, 13, b, 17, 21, 51, a, 67, 103. The mean is given by $$40 = \frac{280 + a + b}{9}$$ which implies $$a + b = 80$$ The mean deviation about the median is $$\frac{13 + 8 + 4 + 30 + 46 + 82 + (a - 21) + (21 - b)}{9} = 26$$ Solving for $a$ and $b$, we have $$a - b = 51$$ Thus, $$2a = 131$$

Question 11

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

Let the line $L_1 : \ x + 3 = 0$ intersect the lines $L_2 : x - y = 0$ and $L_3 : 3x + y = 0$ at the points A and B, respectively. Let the bisector of the obtuse angle between the lines $L_2$ and $L_3$ intersect the line $L_1$ at the point C. Then $BC^2 : AC^2$ is equal to:

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

Answer: (a)

Solution

Given the diagram, we have the following calculation: $$\frac{BC^2}{AC^2} = \left(\frac{OB}{OA}\right)^2 = \frac{9 + 81}{9 + 9} = 5.$$

Question 12

Maths · Properties of Triangles · Single correct

Let the vertex A of a triangle ABC be (1, 2), and the mid-point of the side AB be (5, -1). If the centroid of this triangle is (3, 4) and its circumcenter is $(\alpha, \beta)$, then $21(\alpha + \beta)$ is equal to:

  1. 309
  2. 403
  3. 497
  4. 524

Answer: (c)

Solution

Equation of PM: $4x - 3y = 23$. Equation of PM: $5x - 9y = -25$. $P \left( \frac{94}{7}, \frac{215}{21} \right)$. $21(\alpha + \beta) = 497$.

Question 13

Maths · Conic Sections · Single correct

Suppose the two chords, drawn from the point (1, 2) on the circle $x^2 + y^2 + x - 3y = 0$ are bisected by the y-axis. If the other ends of these chords are R and S, and the mid point of the line segment RS is $(\alpha, \beta)$, then $6(\alpha + \beta)$ is equal to:

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

Answer: (b)

Solution

Chord PS bisected by y-axis at point M(0, $\lambda$). Therefore, S(-1, 2$\lambda$ - 2). Which satisfy equation of circle $$(-1)^2 + (2\lambda - 2)^2 - 1 - 3(2\lambda - 2) = 0$$ $$\Rightarrow \lambda = 1 \& \lambda = \frac{5}{2}$$ Therefore two such chords PS and PR will exist. When $\lambda$ = 1, $\Rightarrow$ S(-1, 0). When $\lambda$ = $\frac{5}{2}$, R(-1, 3). Therefore, midpoint of R and S will be $$\left(-1, \frac{3}{2}\right)$$ Therefore, $\alpha$ = -1, $\beta$ = $\frac{3}{2}$. Therefore, $$6(\alpha + \beta) = 3$$

Question 14

Maths · Three Dimensional Geometry · Single correct

A line with direction ratios 1, -1, 2 intersects the lines $\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3}$ and $\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4}$ at the points P and Q, respectively. If the length of the line segment PQ is $\alpha$, then $225\alpha^2$ is equal to:

  1. 1024
  2. 1014
  3. 1104
  4. 1204

Answer: (b)

Solution

P $(2\lambda, 3\lambda, 3\lambda - 1)$ and Q $(-\mu - 1, \mu + 2, 4\mu)$. D.R of PQ $2\lambda + \mu + 1, 3\lambda - \mu - 2, 3\lambda - 4\mu - 1$, $$\frac{2\lambda + \mu + 1}{1} = \frac{3\lambda - \mu - 2}{-1} = \frac{3\lambda - 4\mu - 1}{2}$$ Solving $\lambda = \frac{1}{5}$ and $\mu = \frac{-8}{15}$. $$\mathrm{P} \left( \frac{2}{5}, \frac{3}{5}, \frac{-2}{5} \right) \& \mathrm{Q} \left( \frac{-7}{15}, \frac{22}{15}, \frac{-32}{15} \right)$$ $$\mathrm{PQ}^2 = \left( \frac{13}{15} \right)^2 + \left( \frac{13}{15} \right)^2 + \left( \frac{26}{15} \right)^2$$ $$\alpha^2 = \frac{1014}{225} \therefore 225 \alpha^2 = 1014$$

Question 15

Maths · Three Dimensional Geometry · Single correct

The square of the distance of the point $(-2, -8, 6)$ from the line $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$ along the line $\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ is equal to:

  1. 3
  2. 6
  3. 8
  4. 12

Answer: (b)

Solution

The direction ratios (DR) or AB is $1, -1, 2$. Therefore, line AB is $$\frac{x+2}{1} = \frac{y+8}{-1} = \frac{z-6}{2} = \lambda$$ B is $(\lambda - 2, -\lambda - 8, 2\lambda + 6)$. B is $(\mu + 1, 2\mu + 1, -\mu)$. Solving point B is $(-3, -7, 4)$. $$AB = \sqrt{(-1)^2 + (1)^2 + (2)^2} = \sqrt{6}$$ $$AB^2 = 6$$

Question 16

Maths · Inverse Trigonometric Functions · Single correct

If $y = \tan^{-1} \left( \frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x} \right) + 2 \tan^{-1} \left( \frac{x}{1 + \sqrt{1 - x^2}} \right),$ then $\frac{dy}{dx}$ at $x = \frac{\sqrt{3}}{2}$ is equal to:

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

Answer: (c)

Solution

Given $$y = \tan^{-1}\left( \frac{\frac{3}{4} - \tan x}{1 + \frac{3}{4} \tan x} \right) + 2 \tan^{-1}\left( \frac{x}{1 + \sqrt{1-x^2}} \right)$$ For $$2 \tan^{-1}\left( \frac{x}{1 + \sqrt{1-x^2}} \right)$$ put $x = \sin \theta$ Therefore, $$2 \tan^{-1}\left( \frac{\sin \theta}{1 + \cos \theta} \right) = \theta = \sin^{-1} x$$ Hence $$y = \tan^{-1}\left( \frac{3}{4} \right) - \tan^{-1}(\tan x) + \sin^{-1} x$$ $$\frac{dy}{dx} = -1 + \frac{1}{\sqrt{1-x^2}}$$ $$x = \frac{\sqrt{3}}{2}$$ $$\frac{dy}{dx} = -1 + 2 = 1$$

Question 17

Maths · Integrals · Single correct

Let $f$ be a real polynomial of degree $n$ such that $f(x) = f'(x) f''(x)$, for all $x \in \mathbb{R}$. If $f(0) = 0$, then $36 \left( f'(2) + f''(2) + \int_0^2 f(x) \, dx \right)$ is equal to:

  1. 42
  2. 46
  3. 56
  4. 66

Answer: (c)

Solution

Let degree of polynomial be 'n'. $n = n - 1 + n - 2 \Rightarrow n = 3$ $f(x) = ax^3 + bx^2 + cx$ $(ax^3 + bx^2 + cx) = (3ax^2 + 2bx + c)(6ax + 2b)$ For $x^3$: $a = 18a^2 \Rightarrow a = \frac{1}{18}$ For $x^2$: $b = 6ab + 12ab$ For $x$: $c = 4b^2 + 6ac \Rightarrow \frac{2c}{3} = 4b^2 \Rightarrow b^2 = \frac{c}{6}$ For constant: $bc = 0 \Rightarrow b = c = 0$ $$\int_0^2 f(x) \, dx + f'(2) + f''(2)$$ $$\int_0^2 ax^3 \, dx + 12a + 12a = 4a + 24a = 28a = \frac{28}{18} = \frac{14}{9}$$ $$36 \left( \frac{14}{9} \right) = 56$$

Question 18

Maths · Applications of Integrals · Single correct

The area of the region $\{$(x,y): y $\leq \pi$ - |x|, y $\leq$ |x $\sin$ x|, y $\geq$ 0$\}$ is:

  1. 1 + $\frac{\pi^2}{8}$
  2. 2 + $\frac{\pi^2}{4}$
  3. $\frac{\pi^2}{8}$ - 1
  4. 4 + $\frac{\pi^2}{2}$

Answer: (b)

Solution

Area; $$2 \left( \int_{0}^{\pi/2} x \sin x \, dx + \int_{\pi/2}^{\pi} (\pi - x) \, dx \right)$$ $$= 2 \left( -x \cos x + \sin x \right)_{0}^{\pi/2} + 2 \left( \frac{(\pi - x)^2}{2} \right)_{\pi/2}^{\pi}$$ $$= 2 + \frac{\pi^2}{4}$$

Question 19

Maths · Integrals · Single correct

Let $\int_{-2}^{2}$ (|$\sin$ x| + [x $\sin$ x]) dx = 2(3 - $\cos$ 2) + $\beta$, where [$\cdot$] is the greatest integer function. Then $\beta \sin \left( \frac{\beta}{2} \right)$ equals.

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

Answer: (b)

Solution

Given $$\int_{-2}^{2} (|\sin x| + [x \sin x]) \, dx$$ $$\Rightarrow 2 \int_{0}^{2} (|\sin x| + [x \sin x]) \, dx$$ $$\Rightarrow 2 \left[ \int_{0}^{2} |\sin x| \, dx + \int_{0}^{2} [x \sin x] \, dx \right]$$ $$2 \left[ [-\cos x]_{0}^{\alpha} + \int_{0}^{\alpha} 0 \, dx + \int_{\alpha}^{2} 2 \, dx \right]$$ $$2 (3 - \cos 2 - 2 - \alpha)$$ $$2 (3 - \cos 2) - 2 \alpha$$ where $$\alpha$$ is the root of $$x \sin x = 1$$. Now comparing $$\beta = -2 \alpha$$ $$\alpha = -\frac{\beta}{2}$$ $$-\frac{\beta}{2} \sin \left( -\frac{\beta}{2} \right) = 1$$ $$\beta \sin \left( \frac{\beta}{2} \right) = 2$$

Question 20

Maths · Differential Equations · Single correct

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = (1 + x + x^2)(1 - y + y^2), y(0) = \frac{1}{2}$. Then $(2y(1) - 1)$ is equal to

  1. $\sqrt{3} \tan \left( \frac{11\sqrt{3}}{6} \right)$
  2. $\frac{\sqrt{3}}{2} \tan \left( \frac{11\sqrt{3}}{12} \right)$
  3. $\sqrt{3} \tan \left( \frac{11\sqrt{3}}{12} \right)$
  4. $\frac{\sqrt{3}}{2} \tan \left( \frac{11\sqrt{3}}{6} \right)$

Answer: (c)

Solution

Given $\frac{\mathrm{dy}}{y^2 - y + 1} = (x^2 + x + 1) \mathrm{dx}$ $$\frac{\mathrm{dy}}{\left( y - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = (x^2 + x + 1) \mathrm{dx}$$ $$\frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{y - \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) = \frac{x^3}{3} + \frac{x^2}{2} + x + C$$ $$\frac{2}{\sqrt{3}\sqrt{3}} = \tan^{-1} \left( \frac{2y - 1}{\sqrt{3}} \right) = \frac{x^3}{3} + \frac{x^2}{2} + x + C$$ $y(0) = \frac{1}{2} \Rightarrow 0 = 0 + 0 + 0 + C \Rightarrow C = 0$ $$\frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2y - 1}{\sqrt{3}} \right) = \frac{x^3}{3} + \frac{x^2}{2} + x$$ Put $x = 1$ $$\frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2y - 1}{\sqrt{3}} \right) = \frac{1}{3} + \frac{1}{2} + 1 = \frac{2 + 3 + 6}{6} = \frac{11}{6}$$ $$\tan^{-1} \left( \frac{2y - 1}{\sqrt{3}} \right) = \frac{11\sqrt{3}}{12}$$ $$\frac{2y - 1}{\sqrt{3}} = \tan \left( \frac{11\sqrt{3}}{12} \right)$$ $$\Rightarrow 2y(1) - 1 = \sqrt{3} \tan \left( \frac{11\sqrt{3}}{12} \right)$$

Question 21

Maths · Probability (Advanced) · Numerical

A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is $p$, then $96p$ is equal to $\ldots$.

Answer: 9

Solution

Given $x_1 x_2 x_3 x_4 x_5 x_6 x_7 x_8$. Exactly 4 heads in first six tosses. $(x_1 x_2 x_3 x_4 x_5 x_6)$ Exactly 3 heads in last 5 tosses. i.e. $(x_4 x_5 x_6 x_7 x_8)$ Common tosses $(x_4 x_5 x_6)$ Let $k$ be number of heads on $(x_4 x_5 x_6) k = 1$. Therefore, ways to overlap $= \binom{3}{1}$ $$\binom{3}{3} \times \binom{3}{1} \times \binom{2}{2} = 3 ways$$ $k = 2$ $$\binom{3}{2} \times \binom{3}{2} \times \binom{2}{1} = 18$$ $k = 3$ $$\binom{3}{1} \times 1 = 3$$ Total ways $24 \left( \frac{1}{2} \right)^8$ $$96 \times \frac{24}{28} = \frac{6 \times 24}{16} = 9$$

Question 22

Maths · Conic Sections · Numerical

Consider the parabola P : $y^2 = 4kx$ and the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let the line segment joining the points of intersection of P and E, be their latus rectums. If the eccentricity of E is $e$, then $e^2 + 2\sqrt{2}$ is equal to $\ldots$

Answer: 3

Solution

Focus of parabola $(k, 0)$ Focus of Ellipse $(ae, 0)$ $k = ae \ldots (1)$ Also $4k = 2 \frac{b^2}{a} \ldots (2)$ $$4ae = \frac{2b^2}{a}$$ $$\Rightarrow 4a^2 e = 2b^2$$ $$4a^2 e = 2a^2 (1 - e^2)$$ $$2e = 1 - e^2$$ $$e^2 + 2e + 1 = 2$$ $$(e + 1)^2 = 2$$ $$e + 1 = \sqrt{2} \therefore e = \sqrt{2} - 1$$ $$e^2 + 2\sqrt{2} = 3$$

Question 23

Maths · Trigonometric Functions · Numerical

If $A = \frac{\sin 3^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 27^\circ} + \frac{\sin 27^\circ}{\cos 81^\circ}$ and $B = \tan 81^\circ - \tan 3^\circ$, then $\frac{B}{A}$ is equal to $\ldots$.

Answer: 2

Solution

Consider $E = \frac{\sin \theta}{\cos 3\theta} \Rightarrow E = \frac{2 \sin \theta \cos \theta}{2 \cos 3\theta \cos \theta}$. Therefore, $E = \frac{\sin 2\theta}{2 \cos 3\theta \cos \theta} \Rightarrow E = \frac{\sin(30 - \theta)}{2 \cos 3\theta \cos \theta}$. Thus, $E = \frac{1}{2} [\tan 3\theta - \tan \theta]$. $A = \frac{1}{2} [\tan 9^\circ - \tan 3^\circ + \tan 27^\circ - \tan 9^\circ + \tan 81^\circ - \tan 27^\circ]$. Therefore, $A = \frac{1}{2} [\tan 81^\circ - \tan 3^\circ]$. Thus, $\frac{B}{A} = 2$.

Question 24

Maths · Vector Algebra · Numerical

Let $\vec{a}_k = (\tan \theta_k) \hat{i} + \hat{j}$ and $\vec{b}_k = \hat{i} - (\cos \theta_k) \hat{j}$, where $\theta_k = \frac{2^{k-1} \pi}{2^n + 1}$, for some $n \in \mathbb{N}$, $n > 5$. Then the value $$\frac{\sum_{k=1}^{n} |\vec{a}_k|^2}{\sum_{k=1}^{n} |\vec{b}_k|^2}$$ is $\ldots$.

Answer: 3

Solution

Given $|a_k|^2 = 1 + \tan^2 \theta = \sec^2 \theta_k$ and $|b_k|^2 = 1 + \cot^2 \theta_k = \csc^2 \theta_k$. Since, $\cot \theta - \tan \theta = 2 \cot 2 \theta$. Therefore, $-\csc^2 \theta - \sec^2 \theta = -4 \csc^2 2 \theta$. $\sec^2 \theta = 4 \csc^2 2 \theta - \csc^2 \theta$. $\Sigma \sec^2 \theta_k = 4 \Sigma \csc^2 2 \theta_k - \Sigma \csc^2 \theta_k$. Now, $\Sigma \csc^2 2 \theta_k = \Sigma \csc^2 \theta_k$ since $$\sum_{k=1}^{n} \cos \csc^2 \frac{2^k \pi}{2^n + 1} = \sum_{k=1}^{n} \cos \csc^2 \frac{2^{k-1} \pi}{2^n + 1}$$ because $\cos \csc \frac{2^n \pi}{2^n + 1} = \cos \csc \frac{\pi}{2^n + 1}$. Therefore, $\Sigma \sec^2 \theta_k = 3 \Sigma \csc^2 \theta_k$. $$\frac{\sum \sec^2 \theta_k}{\sum \csc^2 \theta_k} = 3$$

Question 25

Maths · Continuity and Differentiability · Numerical

The number of points, at which the function $f(x) = \max \{6x, 2 + 3x^2\} + |x - 1| \cos \left|x^2 - \frac{1}{4}\right|, x \in (-\pi, \pi)$, is not differentiable, is $\ldots$.

Answer: 3

Solution

Given $2 + 3x^2 = 6x$, we have $3x^2 - 6x + 2 = 0$. Solving for $x$, we find $$x = \frac{6 \pm \sqrt{12}}{6} = \frac{3 \pm \sqrt{2}}{3}.$$ The expression $|x - 1| \cos \left| x^2 - \frac{1}{4} \right|$ is differentiable at $x = 1$. The total number of points where $f(x)$ is non-differentiable is 3.

Physics

Question 26

Physics · Experimental Physics · Single correct

In a screw gauge when the circular scale is given five complete rotations it moves linearly by $2.5 \, \mathrm{mm}$. If the circular scale has 100 divisions, the least count of screw gauge is $\ldots$ mm.

  1. 1 $\times$ 10^{-2}
  2. 1 $\times$ 10^{-3}
  3. 5 $\times$ 10^{-2}
  4. 5 $\times$ 10^{-3}

Answer: (d)

Solution

Pitch $\Rightarrow \frac{2.5 \, \mathrm{mm}}{5} = 0.5 \, \mathrm{mm} = 5 \times 10^{-4} \, \mathrm{m}$ L.C. $\Rightarrow \frac{Pitch}{No. of divisions} = \frac{5 \times 10^{-4} \, \mathrm{m}}{100}$ $= 5 \times 10^{-6} \, \mathrm{m}$ $= 5 \times 10^{-3} \, \mathrm{mm}$

Question 27

Physics · Mechanical Properties of Solids · Single correct

The increase in the pressure required to decrease the volume ($\Delta V$) of water is $6.3 \times 10^7 \, \mathrm{N/m^2}$. The percentage decrease in the volume is $\ldots$. (Bulk modulus of water $= 2.1 \times 10^9 \, \mathrm{N/m^2}$.)

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

Answer: (b)

Solution

Bulk modulus $\beta = \frac{-\Delta P}{\left( \frac{\Delta V}{V} \right)}$ $$100 \times \frac{\Delta V}{V} = \frac{-\Delta P}{\beta} \times 100 = -\frac{6.3 \times 10^7}{2.1 \times 10^9} \times 100 = -3\%$$ Percentage decrease = 3$\%$

Question 28

Physics · Laws of Motion · Single correct

The time taken by a block of mass $m$ to slide down from the highest point to the lowest point on a rough inclined plane is 50$\%$ more compared to the time taken by the same block on identical inclined smooth plane. Both inclined planes are at $45^\circ$ with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is $\ldots$.

  1. 3/4
  2. 2/3
  3. 5/9
  4. 4/9

Answer: (c)

Solution

For rough incline plane, $a = g(\sin \theta - \mu \cos \theta)$. $$t_1 = \sqrt{\frac{2\ell}{g(\sin \theta - \mu \cos \theta)}}$$ For smooth incline plane, $$t_2 = \sqrt{\frac{2\ell}{g \sin \theta}} \& t_1 = 1.5 t_2$$ $$\sqrt{\frac{2\ell}{g(\sin \theta - \mu \cos \theta)}} = \frac{3}{2} \times \sqrt{\frac{2\ell}{g \sin \theta}}$$ $$(\sin \theta - \mu \cos \theta) = \frac{4}{9} \sin \theta$$ $$\frac{5}{9} \sin \theta = \mu \cos \theta \implies \frac{5}{9} \tan \theta$$ $$\mu = \frac{5}{9} \tan 45^\circ = \frac{5}{9}$$

Question 29

Physics · Nuclei · Single correct

Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass number 3, 4 and 10 are $5.6\,\mathrm{MeV}$, $7.4\,\mathrm{MeV}$ and $6.1\,\mathrm{MeV}$, respectively, then in the process, $\Delta Mc^2 = \ldots$ MeV.

  1. 6.9
  2. 7.9
  3. 2.2
  4. 4.3

Answer: (c)

Solution

Sol. $\Delta mc^2 = [2 \times 3 \times 5.6 + 4 \times (7.4) - 6.1 \times 10]$ $$= 2.2 \, \mathrm{MeV}$$

Question 30

Physics · System of Particles and Rotational Motion · Single correct

A solid sphere of mass M and radius R is divided into two unequal parts. The smaller part having mass M/8 is converted into a sphere of radius r and the larger part is converted into a circular disc of thickness t and radius 2R. If $I_1$ is moment of inertia of a sphere having radius r about an axis through its centre and $I_2$ is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia $I_2/I_1$ = $\ldots$.

  1. 35
  2. 70
  3. 140
  4. 210

Answer: (b)

Solution

The moment of inertia for the small sphere is given by $$I_1 = \frac{2}{5} \left( \frac{M}{8} \right) r^2$$ For the disc, the moment of inertia is $$I_2 = \frac{\frac{7M}{8} \times (2R)^2}{4}$$ The mass $M$ is given by $$M = \rho \times \frac{4\pi}{3} \times R^3$$ For the small sphere, $$\frac{M}{8} = \rho \times \frac{4\pi}{3} \times |r|^3$$ Thus, the radius $r$ is $$r = \frac{R}{2}$$ Now, calculating the ratio of the moments of inertia, $$\frac{I_1}{I_2} = \frac{\frac{2}{5} \left( \frac{M}{8} \right) \left( \frac{R}{2} \right)^2}{\frac{7M}{8 \times 4} \times (2R)^2} = \frac{2}{5} \times \frac{1}{4 \times 7} = \frac{1}{70}$$ Therefore, $$\frac{I_2}{I_1} = \frac{70}{1}$$

Question 31

Physics · Motion in a Plane · Single correct

The two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is 1 : x. The value of x is

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

Answer: (b)

Solution

Given the problem, we have: $$R_1 = \frac{u^2 \sin 2 \times 15^\circ}{g}$$ $$R_2 = \frac{u^2 \sin 2 \times 30^\circ}{g}$$ The ratio is given by: $$\frac{R_1}{R_2} = \frac{\sin 30^\circ}{\sin 60^\circ} = \frac{1}{\sqrt{3}} = \frac{1}{x}$$ Thus, we find: $$x = \sqrt{3}$$

Question 32

Physics · Dual Nature of Radiation and Matter · Single correct

The graph shows variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1$, $X_2$ and $X_3$. Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?

  1. $X_1$
  2. $X_2$
  3. $X_3$
  4. All the metals will give out photo electrons with same kinetic energies.

Answer: (a)

Solution

Given $h\nu = \phi + ev_0$. $$v_0 = \frac{h\nu}{e} - \frac{\phi}{e} \implies 0 \implies \frac{h\nu}{e} = \frac{\phi}{e}$$ $$\left\langle \phi_1 : \phi_2 : \phi_3 \right\rangle$$ $$\left\langle v_1 : v_2 : v_3 \right\rangle$$ $$\left\langle 1 : 1.5 : 2 \right\rangle$$ As metal (1) having minimum value of work function, so $x_1$ will having maximum kinetic energy.

Question 33

Physics · Wave Optics · Single correct

A slit of width $a$ is illuminated by light of wavelength $\lambda$. The linear separation between $1^{st}$ and $3^{rd}$ minima in the diffraction pattern produced on a screen placed at a distance $D$ from the slit system is $\ldots$.

  1. $\frac{D\lambda}{a}$
  2. $1.5 \frac{D\lambda}{a}$
  3. $2 \frac{D\lambda}{a}$
  4. $3 \frac{D\lambda}{a}$

Answer: (c)

Solution

Separation between $3^{rd}$ and $1^{st}$ minima is given by $$\frac{2D\lambda}{a}$$

Question 34

Physics · Mechanical Properties of Solids · Single correct

A string A of length 0.314 m of Young's modulus $2 \times 10^{10} \, \mathrm{N/m^2}$ is connected to another string B of length and Young's modulus both twice of those of A. This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass 0.8 kg. The net change in length of the combination is $\ldots$ mm. (radius of both the strings is 0.2 mm and acceleration due to gravity $= 10 \, \mathrm{m/s^2}$) (Mass of both strings is to be neglected as compared to the mass of load)

  1. 3
  2. 2
  3. 1.9
  4. 1

Answer: (b)

Solution

The total extension $\Delta \ell$ is given by $\Delta \ell_1 + \Delta \ell_2$. Therefore, $$\frac{mg \ell}{AY} + \frac{mg (2\ell)}{A (2Y)} = \frac{2mg \ell}{AY}$$ Calculating, $$= \frac{2 \times 0.8 \times 10 \times (0.314)}{3.14 \times (2 \times 10^{-4})^2 \times 2 \times 10^{10}} = 2 \, mm$$

Question 35

Physics · Kinetic Theory · Single correct

One gas of $n_1$ mole of molecules at temperature $T_1$, volume $V_1$, and pressure $P_1$, and another gas of $n_2$ mole of molecules at temperature $T_2$, volume $V_2$, and pressure $P_2$, are mixed resulting in pressure $P$ and volume $V$ of the mixture. The temperature of the mixture is $\ldots$

  1. $(T_1 + T_2)/2$
  2. $T_1 T_2 PV/(T_2 P_1 V_1 + T_1 P_2 V_2)$
  3. $(T_2 P_1 V_1 + T_1 P_2 V_2)(T_1 T_2 PV)$
  4. $|T_1 - T_2|/2$

Answer: (b)

Solution

Given $n = n_1 + n_2$. Therefore, $PV = nRT$ which implies $n = \frac{PV}{RT}$. $$\frac{P(V)}{T_{Final}} = \frac{P_1 V_1}{T_1} + \frac{P_2 V_2}{T_2}$$ $$\frac{PV}{R \, T_f} = \frac{P_1 V_1}{R \, T_1} + \frac{P_2 V_2}{R \, T_2}$$ $$T_f = \frac{PV}{\frac{P_1 V_1}{T_1} + \frac{P_2 V_2}{T_2}}$$

Question 36

Physics · Thermodynamics · Single correct

An ideal gas undergoes a process maintaining relation between pressure (P) and volume (V) as $$P = P_o \left( 1 + \left( \frac{V_o}{V} \right)^2 \right)^{-1},$$ where $P_o$ and $V_o$ are constants. If two samples A and B (two moles each) with initial volume $V_o$ and $3V_o$ respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, $T_B - T_A$ is $\ldots$ (R = gas constant)

  1. $\frac{9P_o V_o}{8R}$
  2. $\frac{11P_o V_o}{10R}$
  3. $\frac{7P_o V_o}{6R}$
  4. $\frac{13P_o V_o}{11R}$

Answer: (b)

Solution

For $V = V_0$, $P = \frac{P_0}{2}$. $$\frac{P_0}{2} \times V_0 = 2 \times R T_A$$ $$T_A = \frac{P_0 V_0}{4R}$$ For $V = 3V_0$, $P = \frac{9P_0}{10}$. $$\frac{9P_0}{10} \times 3V_0 = 2 \times R T_B$$ $$T_B = \frac{27P_0 V_0}{20R}$$ So, $T_B - T_A$ $$= \frac{11P_0 V_0}{10R}$$

Question 37

Physics · Current Electricity · Single correct

A voltmeter with internal resistance of $x \, \Omega$ can be used to measure upto $20 \, \mathrm{V}$. In order to increase its measuring range to $30 \, \mathrm{V}$, the required modification is to $\ldots$.

  1. connect resistor of $\frac{x}{2} \, \Omega$, in series with voltmeter.
  2. connect resistor of $\frac{x}{2} \, \Omega$, in parallel to voltmeter.
  3. connect resistor of $x \, \Omega$, in series with voltmeter.
  4. connect resistor of $2x \, \Omega$, in parallel to voltmeter.

Answer: (a)

Solution

Since maximum current through voltmeter remains same So, $\frac{20}{x} = \frac{30}{x + R}$ $20x + 20R = 30x$ $20R = 10x$ $R = \frac{x}{2}$ This extra resistance must be connected in series.

Question 38

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

Two 4 bits binary numbers, A = 1101 and B = 1010 are given in the inputs of a logic circuit shown in figure below. The output (Y) will be :

  1. Y = 1101
  2. Y = 0010
  3. Y = 0111
  4. Y = 1000

Answer: (a)

Solution

Given $Y = \overline{\overline{A \cdot B}} = A + \overline{B}$. $A = 1101$ $B = 1010$ $\overline{B} = 0101$ $A + \overline{B} = 1101$

Question 39

Physics · Ray Optics and Optical Instruments · Single correct

A rod of length $10\,\mathrm{\ cm}$ lies along the principle axis of a concave mirror of focal length $10\,\mathrm{\ cm}$ as shown in figure. The length of the image is $\ldots \mathrm{\ cm}$.

  1. 2.5
  2. 5
  3. 7.5
  4. 7

Answer: (b)

Solution

For A, $u = -20 \, \mathrm{cm}$, $f = -10 \, \mathrm{cm}$. $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \therefore v = \frac{u \times f}{u - f} = -20 \, \mathrm{cm}$$ For B, $u = -30 \, \mathrm{cm}$, $f = -10 \, \mathrm{cm}$. $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ $$v = \frac{u f}{u - f} = -15 \, \mathrm{cm}$$ Length of image $= 20 - 15 = 5 \, \mathrm{cm}$

Question 40

Physics · Electrostatic Potential and Capacitance · Single correct

A parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed $v$. If $x$ is the separation between the plates at any instant, then the time rate of change of electrostatic energy of the capacitor is proportional to $x^\alpha$, where $\alpha$ is $\ldots$.

  1. $-2$
  2. $1$
  3. $-1$
  4. $2$

Answer: (a)

Solution

Capacitance $= C = \frac{\varepsilon_0 A}{x}$ Energy stored $= U = \frac{1}{2} C v^2$ $$U = \frac{1}{2} \frac{\varepsilon_0 A}{x} v^2$$ $$\frac{dU}{dt} = \frac{1}{2} \varepsilon_0 A v^2 \left( -\frac{1}{x^2} \right) \frac{dx}{dt}$$ $$\frac{dU}{dt} \propto \frac{1}{x^2}$$

Question 41

Physics · Moving Charges and Magnetism · Single correct

An insulated wire is wound so that it forms a flat coil with $N = 200$ turns. The radius of the innermost turn is $r_1 = 3 \, \mathrm{cm}$, and of the outermost turn $r_2 = 6 \, \mathrm{cm}$. If $20 \, \mathrm{mA}$ current flow in it then the magnetic moment will be $\alpha \times 10^{-2} \, \mathrm{A} \cdot \mathrm{m}^2$. The value of $\alpha$ is $\ldots$

  1. 4.4
  2. 2.64
  3. 3.25
  4. 1.2

Answer: (b)

Solution

Given the differential element $dM = (dI) \pi r^2 = (IdN) \pi r^2$. The total magnetization $M$ is given by the integral $$ M = \int_{i}^{f} I \left( \frac{N}{r_{ex} - r_i} \right) dr \pi r^2 $$ which simplifies to $$ M = \frac{I \pi N}{r_{ex} - r_i} \left[ \frac{r^3}{3} \right]_{r_i}^{r_{ex}} $$ resulting in $$ M = \frac{I \pi N}{3(r_{ex} - r_{in})} (r_{ex}^3 - r_i^3) $$ Substituting the values, we have $$ M = \frac{20 \times 10^{-3} \times 3.14 \times 200 [6^3 - 3^3] 10^{-6}}{3[3] \times 10^{-2}} $$ which simplifies to $$ M = \frac{2373.84 \times 10^{-6}}{9 \times 10^{-2}} $$ giving $$ M = 263.76 \times 10^{-4} $$ and finally $$ M = 2.64 \times 10^{-2} \, \mathrm{Am^2} $$

Question 42

Physics · Current Electricity · Single correct

Consider a circuit consisting of a capacitor (20 $\mu \mathrm{F}$), resistor (100 $\Omega$) and two identical diodes as shown in figure. The resistance of diode under forward biasing condition is 10 $\Omega$. The time constant of the circuit is $\alpha \times 10^{-3}$ s. The value of $\alpha$ is $\ldots$

  1. 2.2
  2. 2.0
  3. 2.1
  4. 2.4

Answer: (a)

Solution

Given $\tau = RC$. $$\tau = (110) \times 20 \times 10^{-6} \, sec$$ $$\tau = 2200 \times 10^{-6} \, sec$$ $$\tau = 2.2 \times 10^{-3} \, sec$$

Question 43

Physics · Current Electricity · Single correct

The voltage and the current between A and B points shown in the circuit are $\ldots$.

  1. $24\,\mathrm{V}$, $12\,\mathrm{A}$
  2. $24\,\mathrm{V}$, $4\,\mathrm{A}$
  3. $18\,\mathrm{V}$, $12\,\mathrm{A}$
  4. $27\,\mathrm{V}$, $4\,\mathrm{A}$

Answer: (b)

Solution

This circuit can be simplified. Here: $$E_{eq} = \frac{\frac{27}{3} + \frac{27}{3} + \frac{27}{3} + \frac{27}{3}}{\frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3}}$$ $E_{eq} = 27 \, \mathrm{V}$ $r_{eq} = \frac{3}{4} \, \Omega$ $i_{AB} = \frac{27}{6 + \frac{3}{4}} = 4 \, \mathrm{A}$ $V_{AB} = 4 \times 6 = 24 \, \mathrm{Volt}$

Question 44

Physics · Ray Optics and Optical Instruments · Single correct

A telescope with objective diameter $R$ is used to observe a distant star emitting light of wavelength $500 \, \mathrm{nm}$, at a resolution of $5 \times 10^{-7}$ radian. The value of $R$ is $\ldots$ cm.

  1. 61
  2. 122
  3. 244
  4. 305

Answer: (b)

Solution

Given $\theta = 5 \times 10^{-7} radian$ and $\lambda = 500 \times 10^{-9} m$. Using the formula: $$\theta = \frac{1.22 \lambda}{R}$$ We find: $$R = \frac{1.22 \lambda}{\theta} = 1.22 m = 122 cm$$

Question 45

Physics · Wave Optics · Single correct

An unpolarized light is incident on the plane interface of air-dielectric medium shown in figure. If the incident angle is equal to Brewster angle, identify the expression representing reflected wave.

  1. $\left( E_x \hat{i} + E_y \hat{j} \right) \sin (kx - kz - \omega t)$
  2. $\left( E_x \hat{i} + E_z \hat{k} \right) \sin (kx + ky - \omega t)$
  3. $\left( E_x \hat{j} + E_y \hat{k} \right) \sin (ky + kz - \omega t)$
  4. $\left( E_x \hat{i} + E_y \hat{j} + E_z \hat{k} \right) \sin (kx + ky - kz - \omega t)$

Answer: (a)

Solution

Is the light propagating along the positive z-axis and it is reflected from the plane, then the correct phase for reflected light is $kx - kz - \omega t$ and for electric field as the light incident on Brewster angle so the reflected light should be perfectly plane polarized. That's why the polarization direction should be the y-axis, so the electric field of reflected light contains no z-component.

Question 46

Physics · Work, Energy and Power · Numerical

A $1\,\mathrm{kg}$ block subjected to two simultaneous forces (2$\hat{i}$ + 3$\hat{j}$ + 4$\hat{k}$) N and (3$\hat{i}$ - $\hat{j}$ - 2$\hat{k}$) N is moved a distance of $25\,\mathrm{m}$ along (3$\hat{i}$ - 4$\hat{j}$) direction. The work done in this process is $\ldots$ J.

Answer: 35

Solution

The net force is given by $$\vec{F}_{Net} = 5\hat{i} + 2\hat{j} + 2\hat{k}$$. The displacement is $$\vec{S} = 15\hat{i} - 20\hat{j}$$. The work done is $$W = \vec{F} \cdot \vec{S} = 75 - 40 = 35 \, J$$.

Question 47

Physics · Mechanical Properties of Fluids · Numerical

The surface tension of a soap solution is $3.5 \times 10^{-2} \, \mathrm{N/m}$. The work required to increase the radius of a soap bubble from $1 \, \mathrm{cm}$ to $2 \, \mathrm{cm}$ is $\alpha \times 10^{-6} \, \mathrm{J}$. The value of $\alpha$ is $\ldots$. $(\pi = 22/7)$

Answer: 264

Solution

W.D. = 2T $\left[ 4\pi R_2^2 - 4\pi R_1^2 \right]$ = 2 $\times$ 3.5 $\times$ 10^{-2} $\times$ 4 $\times \frac{22}{7} \left[ 2^2 - 1^2 \right] \times$ 10^{-4} = 2 $\times$ 132 $\times$ 10^{-6} = 264 $\times$ 10^{-6} $\mathrm{J}$ = $\alpha \times$ 10^{-6} $\mathrm{J} \alpha$ = 264

Question 48

Physics · Oscillations · Numerical

The velocity of a particle executing simple harmonic motion along x-axis is described as $v^2 = 50 - x^2$, where $x$ represents displacement. If the time period of motion is $\frac{x}{7} \, \mathrm{s}$, the value of $x$ is $\ldots$.

Answer: 44

Solution

Given $V^2 = \omega^2 A^2 - \omega^2 x^2$. Comparing $\omega = 1$. $$T = \frac{2\pi}{\omega} = \frac{2\pi}{1} = 2\pi = \frac{x}{7}$$ $x = 14\pi$. $x = 43.98 \approx 44$

Question 49

Physics · Work, Energy and Power · Numerical

A body of mass $2\,\mathrm{kg}$ begins to move under the influence of time dependent force $\vec{F} = (2t\hat{i} + 6t^2\hat{j}) \, \mathrm{N}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along x and y-axis respectively. The power produced by the force at $t = 2 \, \mathrm{s}$ is $\ldots$ W.

Answer: 200

Solution

Given $\vec{a} = \frac{\vec{F}}{m} = t \hat{i} + 3t^2 \hat{j}$. $$\vec{v} = \int_{0}^{2} \vec{a} \cdot dt = \left[ \frac{t^2}{2} \hat{i} + \frac{3t^3}{3} \hat{j} \right]_{0}^{2}$$ $$= (2 \hat{i} + 8 \hat{j}) \, \mathrm{m/s}$$ $$\vec{F}_{t=2} = (4 \hat{i} + 24 \hat{j}) \, \mathrm{N}$$ $$P = \vec{F} \cdot \vec{v} = 8 + 192 = 200 \, \mathrm{W}$$

Question 50

Physics · Alternating Current · Numerical

An inductor of $10\,\mathrm{mH}$, capacitor of 0.1 $\mu \mathrm{F}$ and a resistor of 100 $\Omega$ are connected in series across an a.c power supply $220\,\mathrm{V}$, $70\,\mathrm{Hz}$. The power factor of the given circuit is 0.5. The difference in the inductive reactance and capacitance reactance is $\sqrt{3} \alpha \Omega$. The value of $\alpha$ is $\ldots$

Answer: 100

Solution

Given $\cos \phi = 0.5 = \frac{1}{2}$. $\phi = 60^\circ$ $\tan \phi = \frac{|x_L - x_C|}{R}$ $\tan 60^\circ = \sqrt{3} = \frac{x_L - x_C}{R}$ $|x_L - x_L| = R \sqrt{3} = 100 \sqrt{3} = \sqrt{3} \alpha$ $\alpha = 100$

Chemistry

Question 51

Chemistry · Some Basic Concepts of Chemistry · Single correct

Number of moles and number of molecules in $1.4187\,\mathrm{L}$ of $\mathrm{SO}_2$ at STP respectively are

  1. 0.1266; $3.812 \times 10^{22}$
  2. 0.0633; $3.812 \times 10^{22}$
  3. 0.1266; $7.6238 \times 10^{22}$
  4. 0.0633; $7.6238 \times 10^{22}$

Answer: (b)

Solution

moles = $\frac{V lit}{22.7 lit} \Rightarrow \frac{1.4187 lit}{22.7 lit} \Rightarrow$ 0.0633 no. of molecules = moles $\times$ 6.023 $\times$ 10^{23} $\Rightarrow$ 0.0633 $\times$ 6.032 $\times$ 10^{3} = 3.812 $\times$ 10^{22}

Question 52

Chemistry · Structure of Atom · Single correct

What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

  1. 5:1
  2. 5:0.81
  3. 5:1.75
  4. 5:27

Answer: (b)

Solution

Given $\overline{V}_1 = R_H (z)^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] \Rightarrow 1st line of Balmer series$ $\overline{V}_2 = R_H (z)^2 \left[ \frac{1}{4^2} - \frac{1}{5^2} \right] \Rightarrow 1st line of Brackett series$ $\frac{\overline{V}_1}{\overline{V}_2} = \frac{500}{81} \Rightarrow 5 : 0.81$

Question 53

Chemistry · Structure of Atom · Single correct

Which of the following is correct set of 4 quantum number of $19^{th}$ electron in Chromium (Atomic number = 24) in accordance with Aufbau principle?

  1. n = 3, $\ell$ = 2, m = +2, s = +$\frac{1}{2}$
  2. n = 3, $\ell$ = 2, m = -2, s = +$\frac{1}{2}$
  3. n = 4, $\ell$ = 1, m = 0, s = +$\frac{1}{2}$
  4. n = 4, $\ell$ = 0, m = 0, s = +$\frac{1}{2}$

Answer: (d)

Solution

The electronic configuration of Cr is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \, 3d^5$. Therefore, the 19th electron belongs to the $4s$ subshell. The quantum numbers are $n = 4$, $\ell = 0$, $m = 0$, $m_s = +\frac{1}{2}$ or $-\frac{1}{2}$.

Question 54

Chemistry · Thermodynamics · Single correct

Given below are two statements: \ Statement I: For an ideal gas, heat capacity at constant volume is always greater than the heat capacity at constant pressure. \ Statement II: In a constant volume process, no work is produced and all the heat withdrawn goes into the chaotic motion and is reflected by a temperature increase of the ideal gas. \ In the light of the above statements, choose the correct answer from the options given below

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

Answer: (d)

Solution

From first law of thermodynamics $$\Delta U = q + w$$ constant volume $w = 0$ $$\Delta U = q$$ Also $C_p = C_v + R$ $C_p =$ heat capacity at constant pressure $C_v =$ Heat capacity at constant volume. $R =$ constant

Question 55

Chemistry · Equilibrium · Single correct

At T(K), the equilibrium constant of $$A_2(g) + B_2(g) \rightleftharpoons C(g)$$ is $2.7 \times 10^{-5}$. What is the equilibrium constant for $$\frac{1}{3}A_2(g) + \frac{1}{3}B_2(g) \rightleftharpoons \frac{1}{3}C(g)$$ at the same temperature?

  1. $(2.7 \times 10^{-5})^3$
  2. $6 \times 10^{-2}$
  3. $\sqrt{2.7 \times 10^{-5}}$
  4. $3 \times 10^{-2}$

Answer: (d)

Solution

Given $\mathrm{A_2(g)} + \mathrm{B_2(g)} \rightleftharpoons \mathrm{(g)}$, $K_{\mathrm{eq}} = 2.7 \times 10^{-5}$. $$\frac{1}{3} \mathrm{A_2(g)} + \frac{1}{3} \mathrm{B_e(g)} \rightleftharpoons \frac{1}{3} \mathrm{C(g)} K_{\mathrm{eq}}^1 = \left(K_{\mathrm{eq}}\right)^{1/3}$$ $$K_{\mathrm{eq}}^1 = \left(2.7 \times 10^{-5}\right)^{1/3}$$ $$= 3 \times 10^{-2}$$

Question 56

Chemistry · Redox Reactions · Single correct

In order to oxidise a mixture of 1 mole each of $\mathrm{FeC_2O_4}$, $\mathrm{Fe_2(C_2O_4)_3}$, $\mathrm{FeSO_4}$ and $\mathrm{Fe_2(SO_4)_3}$ in acidic medium, the number of moles of $\mathrm{KMnO_4}$ required is:

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

Answer: (b)

Solution

Meq of $\mathrm{KMnO_4}$ = meq of $\mathrm{FeC_2O_4}$ + meq of $\mathrm{Fe_2(C_2O_4)_3}$ + meq of $\mathrm{FeSO_4}$. moles $\times 5 = 1 \times 3 + 1 \times 6 + 1 \times 1$. moles = $\frac{10}{5}$ = 2.

Question 57

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

Consider the first order reaction $\mathrm{R} \rightarrow \mathrm{P}$. The fraction of molecules decomposed in the given first order reaction can be expressed as

  1. $1 - e^{k_1 t}$
  2. $1 + e^{k_1 t}$
  3. $1 + e^{-k_1 t}$
  4. $1 - e^{-k_1 t}$

Answer: (d)

Solution

First order of $R \mu^H$. $$a - x = a e^{-K_1 t}$$ $$\frac{x}{a} = 1 - e^{-K_1 t}$$

Question 58

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

A monoatomic anion $(\mathrm{A}^-)$ has 45 neutrons and 36 electrons. Atomic mass, group in the periodic table and physical state at room temperature of the element $(\mathrm{A})$ respectively are:

  1. 80, 17, liquid
  2. 81, 16, solid
  3. 80, 16, gas
  4. 81, 15, gas

Answer: (a)

Solution

X = $\mathrm{Br}$; Atomic mass = 80. Period 4. $\mathrm{Br_2}$ exists as liquid.

Question 59

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

Given below are two statements: \ Statement I: The covalency of oxygen is generally two but it can exceed upto four. The oxidation state of oxygen in $\mathrm{SO_2}$ is $-2$ and in $\mathrm{OF_2}$ it is $+2$. \ Statement II: The anomalous behaviour of oxygen when compared to the other elements of group 16 is due to its small size and high electronegativity. \ In the light of the above statements, choose the correct answer from the options given below:

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

Answer: (a)

Solution

Oxygen can have covalency 4. In $\mathrm{SO_2}$; O.S. of O $= -2$. In $\mathrm{OF_2}$; O.S. of O $= +2$. Anomalous behaviour of oxygen is due to high EN and small size.

Question 60

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

The correct statements among the following are, \ A. Mo(VI) and W(VI) are less stable than Cr(VI). \ B. Ce$^{4+}$ and Tb$^{4+}$ are oxidant while Eu$^{2+}$ and Yb$^{2+}$ are reductant. \ C. Cm and Am have seven unpaired electrons. \ D. Actinoid contraction is greater from element to element than lanthanoid contraction. \ Choose the correct answer from the options given below:

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

Answer: (c)

Solution

(A) Mo(VI) and W(VI) are more stable than Cr(VI). (B) In lanthanoids, the +3 oxidation state is more stable, hence +4 ions act as oxidising agents and +2 ions act as reducing agents. (C) Cm – [Rn] 5f^7 6d^1 7s^2 = 8 unpaired electrons. Am – [Rn] 5f^7 6s^2 = 7 unpaired electrons. (D) Actinoid contraction is more than lanthanoid contraction.

Question 61

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

Correct statements from the following are \ A. Potassium dichromate is an oxidising agent and it oxidises $\mathrm{FeSO_4}$ to $\mathrm{Fe_2(SO_4)_3}$ in acidic medium. \ B. Sodium dichromate can be used as primary standard in volumetric estimation. \ C. $\mathrm{CrO_4^{2-}}$ and $\mathrm{Cr_2O_7^{2-}}$ are interconvertible in aqueous solution by varying the pH of the solution. \ D. Cr–O–Cr bond angle in $\mathrm{Cr_2O_7^{2-}}$ is $126^\circ$. \ Choose the correct answer from the options given below:

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

Answer: (b)

Solution

(A) $\mathrm{MnO_4^- + Fe^{+2} + H^+ \longrightarrow Fe^{+3} + Mn^{+2} + H_2O}$ (B) $\mathrm{Na_2Cr_2O_7}$ generally not used as a primary standard in volumetric analysis because it is deliquescent. (C) In acidic medium $\mathrm{Cr_2O_7^{2-}}$ exists and it gets converted into $\mathrm{CrO_4^{2-}}$ in basic medium. $$\mathrm{2CrO_4^{2-} + 2H^+ \longrightarrow Cr_2O_7^{2-} + H_2O}$$ $$\mathrm{Cr_2O_7^{2-} + 2OH^- \longrightarrow 2CrO_4^{2-} + H_2O}$$ (D) $\mathrm{K_2Cr_2O_7}$ is a good oxidising agent and bond angle of $\mathrm{Cr-O-Cr}$ bond in $\mathrm{Cr_2O_7^{2-}}$ is $126^\circ$.

Question 62

Chemistry · Co-ordination Compounds · 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{Complex ion}} & \multicolumn{2}{c|}{\textbf{Calculated spin only}} \\ \multicolumn{2}{|c|}{} & \multicolumn{2}{c|}{\textbf{magnetic moment (BM)}} \\ \hline A. & $[\mathrm{Cr(H_2O)_6}]^{2+}$ & I. & $3.87$ \\ \hline B. & $[\mathrm{Co(H_2O)_6}]^{2+}$ & II. & $5.92$ \\ \hline C. & $[\mathrm{Cu(H_2O)_6}]^{2+}$ & III. & $4.90$ \\ \hline D. & $[\mathrm{Mn(H_2O)_6}]^{2+}$ & IV. & $1.73$ \\ \hline \end{tabular} Choose the correct answer from the options given below:

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

Answer: (d)

Solution

[$\mathrm{Cr(H_2O)_6}$]^{2+} $\rightarrow$ 4 unpaired e^- ; 4.90 BM(III) [$\mathrm{Co(H_2O)_6}$]^{2+} $\rightarrow$ 0 unpaired e^- ; 3.87 BM(I) [$\mathrm{Cu(H_2O)_6}$]^{2+} $\rightarrow$ 1 unpaired e^- ; 1.73 BM(IV) [$\mathrm{Mn(H_2O)_6}$]^{2+} $\rightarrow$ 5 unpaired e^- ; 5.92 BM(II)

Question 63

Chemistry · Amines · Single correct

Increasing order of electron withdrawing power of following functional groups is: \ a. $-\mathrm{CN}$ \ b. $-\mathrm{COOH}$ \ c. $-\mathrm{NO_2}$ \ d. $-\mathrm{I}$

  1. c < b < d < a
  2. c < a < b < d
  3. d < b < a < c
  4. a < b < c < d

Answer: (c)

Solution

Theory based

Question 64

Chemistry · Hydrocarbons · Single correct

An alkene (X) on ozonolysis followed by reduction gives following products. \ The alkene (X) is:

Answer: (d)

Solution

The reaction involves ozonolysis of the given alkene. The alkene is treated with $\mathrm{O_3}$ followed by reduction with $\mathrm{Zn/H_2O}$. This results in the cleavage of the double bonds and formation of carbonyl compounds. The products are $2 \, \mathrm{CH_2 = O}$ (formaldehyde), glyoxal, and biacetyl.

Question 65

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{Name of reaction}} & \multicolumn{2}{c|}{\textbf{Reagent or catalyst used}} \\ \hline A. & Finkelstein reaction & I. & $\mathrm{SbF_3}$ \\ \hline B. & Swarts reaction & II. & $\mathrm{Na,\ dry\ ether}$ \\ \hline C. & Sandmeyer's reaction & III. & $\mathrm{NaI}$ \\ \hline D. & Fitting reaction & IV. & $\mathrm{Cu_2Cl_2}$ \\ \hline \end{tabular} Choose the correct answer from the options given below:

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

Answer: (b)

Solution

Sol. Theory based

Question 66

Chemistry · Amines · Single correct

Amongst the following, the total number of compounds soluble in aqueous NaOH at room temperature is:

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

Answer: (a)

Solution

Compounds which are more acidic than $\mathrm{H_2O}$ can soluble in $\mathrm{NaOH}$.

Question 67

Chemistry · Amines · Single correct

Product C of the following reaction sequence will be

  1. 1-Bromo-4-nitrobenzene
  2. 1,3,5-Tribromo-2-nitrobenzene
  3. 4-Bromo-1-nitrobenzene
  4. 1,3,5-Tribromobenzene

Answer: (b)

Solution

The reaction starts with aniline, which is treated with $\mathrm{Br_2/H_2O}$ to form 2,4,6-tribromoaniline. This compound is then reacted with $\mathrm{NaNO_2/HCl}$ at $0{-}5^\circ \mathrm{C}$ to form the diazonium salt. The diazonium salt is then treated with $\mathrm{HBF_4}$ to form the tetrafluoroborate salt. Finally, the compound undergoes a Sandmeyer reaction with $\mathrm{NaNO_2}$ and copper at high temperature to form 1,3,5-tribromo-2-nitrobenzene.

Question 68

Chemistry · Biomolecules · Single correct

In the light of the above statements, choose the correct answer from the options given below

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

Answer: (d)

Solution

Maltose and lactose both have hemiacetal linkage. So both are reducing sugars.

Question 69

Chemistry · Biomolecules · Single correct

Match the LIST-I with LIST-II \begin{tabular}{|c|c|c|c|} \hline & \textbf{List-I} & &\textbf{List-II} \\ & \textbf{Name of amino acid} & &\textbf{One letter symbol/type} \\ \hline A & Arginine & I. & D / Non-essential \\ \hline B & Aspartic acid & II. & R / Essential \\ \hline C & Lysine & III. & E / Non-essential \\ \hline D & Glutamic acid & IV. & K / Essential \\ \hline \end{tabular} Choose the correct answer from the options given below :

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

Answer: (a)

Solution

Arginine (R) and Lysine (K) are essential amino acids. Aspartic acid (D) and Glutamic acid (E) are non-essential amino acids.

Question 70

Chemistry · Co-ordination Compounds · Single correct

Identify the colour of compound 'X' in the sequence of the reaction.

  1. Violet
  2. Green
  3. Red
  4. Colourless

Answer: (d)

Solution

Phenolphthalein indicator turns colourless in excess NaOH.

Question 71

Chemistry · Chemical Bonding and Molecular Structure · Numerical

According to Lewis theory, the total number of $\sigma$ bond-pairs and lone pair of electrons around the central atom of $\mathrm{XeO}_6^{4-}$ ion is $\ldots$.

Answer: 6

Solution

The structure shows xenon surrounded by six oxygen atoms. The sigma bonds are counted as $\sigma = 6$. There are no lone pairs on xenon, so lone pairs $= 0$.

Question 72

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

Consider the following sequence of reactions to give the major product (X) P g of the major product (X) formed is reacted with $\mathrm{NaHCO_3}$ solution to liberate a gas which occupied $11.2\,\mathrm{dm^3}$ at STP. P = $\ldots$ g. (Given molar mass in g $mol^{-1}$ H : 1, C : 12, O : 16, Cl : 35.5)

Answer: 78

Solution

Moles of P = $\frac{11.2}{22.4}$ = 0.5 mole M_{$\mathrm{C_7H_5O_2Cl}$} = 156.5 gm W_{$\mathrm{P(C_7H_5O_2Cl)}$} = $\frac{156.5}{2}$ = 78.25 $\approx$ 78 grams

Question 73

Chemistry · Analytical Chemistry · Numerical

$2.0\,\mathrm{g}$ of a bromo hydrocarbon (X) was subjected to Carius analysis, gave $3.36\,\mathrm{g}$ of AgBr. The percentage of carbon in the compound (X) is 26.7$\%$. Total number of carbon atoms in the empirical formula for compound (X) is $\ldots$ . (Given molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ H : 1, C : 12, Br : 80, Ag : 108)

Answer: 5

Solution

Given $n_{Br} = \frac{3.36}{(108 + 80)}$. $r$ is the number of Br atoms in the organic compound. $n_{organic compound} = \frac{3.36}{(108 + 80) \times x} = \frac{2}{M}$. $M = \frac{2 \times 188 \times x}{3.36}$. $M = \frac{1}{2} \times x$. $(x is integer)$. $W_c = \frac{112 \times x \times 26.7}{100} = 30 \times x$. $n_C = \frac{30 \times x}{12} = 2.5 \times x$. Take $x = 2, n_C = 5$. (Empirical formula is $\mathrm{C_5H_4Br}$)

Question 74

Chemistry · Equilibrium · Numerical

The pH a solution obtained by mixing $5\,\mathrm{mL}$ of $0.1\,\mathrm{M}\ \mathrm{NH_4OH}$ solution with $250\,\mathrm{mL}$ of $0.1\,\mathrm{M}\ \mathrm{NH_4Cl}$ solution is $\ldots \times 10^{-2}$. (Nearest integer) Given : $\mathrm{pK_b} (\mathrm{NH_4OH}) = 4.74$ $\log 2 = 0.30$ $\log 3 = 0.48$ $\log 5 = 0.70$

Answer: 756

Solution

On mixing final volume = $255\,\mathrm{ml}$. $$[\mathrm{NH_4OH}] = \frac{5 \times 0.1}{255}, [\mathrm{NH_4Cl}] = \frac{250 \times 0.1}{255}$$ $$\mathrm{p}^{\mathrm{OH}} = \mathrm{p}k_b + \log \frac{[\mathrm{NH_4Cl}]}{\mathrm{NH_4OH}}$$ $$\mathrm{p}^{\mathrm{OH}} = 4.74 + \log \frac{250 \times 0.1}{5 \times 0.1}$$ $$\mathrm{pH} = 14 - \mathrm{p}^{\mathrm{OH}} = 14 - 6.44$$

Question 75

Chemistry · Solutions · Numerical

A non-volatile, non-electrolyte solid solute when dissolved in $40\,\mathrm{g}$ of a solvent, the vapour pressure of the solvent decreased from $760\,\mathrm{mm}\ \mathrm{Hg}$ to $750\,\mathrm{mm}\ \mathrm{Hg}$. If the same solution boils at $320\,\mathrm{K}$, then the number of moles of the solvent present in the solution is $\ldots$ (Nearest integer) $\newline$ [Given : boiling point of the pure solvent = $319.5\,\mathrm{K}$, K_b of the solvent = $0.3\,\mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$]

Answer: 5

Solution

Given $\frac{P^\circ - P_s}{P_s} = i \cdot molality \times \frac{(M.solvent)}{1000}$. $\Delta T_b = i \cdot K_b \cdot molality \Rightarrow molality = \frac{0.5}{0.3}$. (Molecular Mass) = $\frac{600}{75}$ g Moles = $\frac{40}{600/75}$ = 5