JEE Main 9 April 2024 Shift 1 question paper with solutions

JEE Main 9 April 2024 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Three Dimensional Geometry · Single correct

Let the line L intersect the lines $x - 2 = -y = z - 1, 2(x + 1) = 2(y - 1) = z + 1$ and be parallel to the line $$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$$. Then which of the following points lies on L?

  1. $(-\\frac{1}{3}, 1, -1)$
  2. $(-\\frac{1}{3}, -1, 1)$
  3. $(-\\frac{1}{3}, 1, 1)$
  4. $(-\\frac{1}{3}, -1, -1)$

Answer: (a)

Solution

For line $L_1$: $\frac{x-2}{1}=\frac{y-1}{-1}=\frac{z-1}{2}=\lambda$. For line $L_2$: $\frac{x+1}{\frac{1}{2}}=\frac{y-1}{1}=\frac{z+1}{1}=\mu$. The direction ratios of line $MN$ will be $ $ and it will be proportional to $ $. $\frac{3+\lambda-\frac{\mu}{2}}{3}=\frac{-1-\lambda-\frac{\mu}{2}}{1}=\frac{2+\lambda-\mu}{2}$ Thus, $4\lambda+\mu=-6$ $4+3\lambda=0$ Therefore, $\lambda=-\frac{4}{3}$ and $\mu=-\frac{2}{3}$. The coordinate of $M$ will be $\left(\frac{2}{3},\frac{4}{3},-\frac{1}{3}\right)$ and the equation of the required line will be: $\frac{x-\frac{2}{3}}{3}=\frac{y-\frac{4}{3}}{1}=\frac{z+\frac{1}{3}}{2}=k$ So any point on this line will be $\left(\frac{2}{3}+3k,\frac{4}{3}+k,-\frac{1}{3}+2k\right)$. Thus, $\frac{2}{3}+3k=-\frac{1}{3}$ implies $k=-\frac{1}{3}$. Therefore, the point lies on the line for $k=-\frac{1}{3}$ is $\left(-\frac{1}{3},1,-1\right)$.

Question 2

Maths · Applications of Integrals · Single correct

The parabola $y^2 = 4x$ divides the area of the circle $x^2 + y^2 = 5$ in two parts. The area of the smaller part is equal to:

  1. $\frac{1}{3} + 5 \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$
  2. $\frac{1}{3} + \sqrt{5} \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$
  3. $\frac{2}{3} + 5 \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$
  4. $\frac{2}{3} + \sqrt{5} \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$

Answer: (c)

Solution

Given $y^2 = 4x$ and $x^2 + y^2 = 5$. The area of the shaded region as shown in the figure will be $$A_1 = \int_0^1 \sqrt{4x} \, dx + \int_1^{\sqrt{5}} \sqrt{5 - x^2} \, dx$$ $$= \frac{4}{3} \left[ x^{3/2} \right]_0^1 + \left[ \frac{x}{2} \sqrt{5 - x^2} + \frac{5}{2} \sin^{-1} \frac{x}{\sqrt{5}} \right]_1^{\sqrt{5}}$$ $$= \frac{1}{3} + \frac{5\pi}{4} - \frac{5}{2} \sin^{-1} \left( \frac{1}{\sqrt{5}} \right)$$ Therefore, the required area $= 2A_1$ $$= \frac{2}{3} + \frac{5\pi}{2} - 5 \sin^{-1} \left( \frac{1}{\sqrt{5}} \right)$$ $$= \frac{2}{3} + 5 \left( \frac{\pi}{2} - \sin^{-1} \frac{1}{\sqrt{5}} \right)$$ $$= \frac{2}{3} + 5 \cos^{-1} \frac{1}{\sqrt{5}}$$ $$= \frac{2}{3} + 5 \sin^{-1} \left( \frac{2}{\sqrt{5}} \right)$$

Question 3

Maths · Differential Equations · Single correct

The solution curve, of the differential equation $2y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx}$, passing through the point $(0, 1)$ is a conic, whose vertex lies on the line:

  1. $2x + 3y = 9$
  2. $2x + 3y = -9$
  3. $2x + 3y = -6$
  4. $2x + 3y = 6$

Answer: (a)

Solution

(2y - 5) $\frac{dy}{dx}$ = -3 (2y - 5) dy = -3 dx 2 $\cdot$ $\frac{y^2}{2}$ - 5y = -3x + $\lambda$ Since the curve passes through (0, 1), $\Rightarrow$ $\lambda$ = -4 Therefore, the curve will be $\left$(y - $\frac{5}{2}$$\right$)^2 = -3 $\left$(x - $\frac{3}{4}$$\right$) The vertex of the parabola will be $\left$($\frac{3}{4}$, $\frac{5}{2}$$\right$) Thus, 2x + 3y = 9

Question 4

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

A ray of light coming from the point $P(1, 2)$ gets reflected from the point $Q$ on the $x$-axis and then passes through the point $R(4, 3)$. If the point $S(h, k)$ is such that $PQRS$ is a parallelogram, then $hk^2$ is equal to:

  1. 70
  2. 80
  3. 60
  4. 90

Answer: (a)

Solution

Image of $P$ with respect to the $x$-axis will be $P'(1, -2)$. The equation of the line joining $P'R$ will be $$y - 3 = \frac{5}{3}(x - 4)$$ The above line will meet the $x$-axis at $Q$ where $$y = 0 \Rightarrow x = \frac{11}{5}$$ Therefore, $Q \left( \frac{11}{5}, 0 \right)$. Thus, $PQRS$ is a parallelogram so their diagonals will bisect each other. $$\Rightarrow \frac{4 + 1}{2} = \frac{11}{5} + \frac{h}{2} \& \frac{2 + 3}{2} = \frac{k + 0}{2}$$ $$\Rightarrow h = \frac{14}{5} \& k = 5$$ Therefore, $$hk^2 = \frac{14}{5} \times 5^2 = 70$$

Question 5

Maths · Determinants · Single correct

Let $\lambda, \mu \in \mathbb{R}$. If the system of equations $$3x + 5y + \lambda z = 3$$ $$7x + 11y - 9z = 2$$ $$97x + 155y - 189z = \mu$$ has infinitely many solutions, then $\mu + 2\lambda$ is equal to:

  1. 24
  2. 25
  3. 22
  4. 27

Answer: (b)

Solution

Given the equations: $$3x + 5y + \lambda z = 3$$ $$7x + 11y - 9z = 2$$ $$97x + 155y - 189z = \mu$$ $$93x + 155y + 31\lambda z = 93$$ $$97x + 155y - 189z = \mu$$ Subtracting: $$-4x + (31\lambda + 189)z = 93 - \mu$$ Adding: $$1085x + 1705y - 1395z = 310$$ $$1067x + 1705y - 2079z = 11\mu$$ Subtracting: $$18x + 684z = 310 - 11\mu$$ Subtracting: $$-36x + 9(31\lambda + 189)z = 9(93 - \mu)$$ Adding: $$36x + 1368z = 2(310 - 11\mu)$$ Combining: $$(279\lambda + 3069)z = 1457 - 31\mu$$ For infinite solutions: $$\lambda = \frac{-3069}{279} = \frac{-341}{31}$$ $$\mu = \frac{1457}{31}$$ $$\mu + 2\lambda = \frac{1457 - 682}{31} = \frac{775}{31} = 25$$

Question 6

Maths · Binomial Theorem · Single correct

The coefficient of $x^{70}$ in $x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\ldots+x^{54}(1+x)^{46}$ is ${}^{99}C_p-{}^{46}C_q$. Then a possible value of $p+q$ is:

  1. 55
  2. 83
  3. 61
  4. 68

Answer: (b)

Solution

Given $$x^2(1+x)^{98} + x^3(1+x)^{97} + x^4(1+x)^{96} + \ldots$$ $$x^{54}(1+x)^{46}$$ Coefficient of $x^{70}$: $$^{98}C_{68} + ^{97}C_{67} + ^{96}C_{66} + \ldots$$ $$^{47}C_{17} + ^{46}C_{16}$$ $$= ^{46}C_{30} + ^{47}C_{30} + \ldots \ ^{98}C_{30} - ^{46}C_{31}$$ $$= (^{46}C_{31} + ^{46}C_{30}) + ^{47}C_{30} + \ldots \ ^{98}C_{30} - ^{46}C_{31}$$ $$= ^{47}C_{31} + ^{47}C_{30} + \ldots \ ^{98}C_{30} - ^{46}C_{31}$$ $$\ldots$$ $$= ^{99}C_{31} - ^{46}C_{31} = ^{99}C_{p} - ^{46}C_{q}$$ Possible values of $(p+q)$ are 62, 83, 99, 46. Therefore, $p+q = 83$

Question 7

Maths · Integrals · Single correct

Let \[ \int \frac{2-\tan x}{3+\tan x}\,dx = \frac12\left(\alpha x+\log_e\left|\beta\sin x+\gamma\cos x\right|\right)+C, \] where $C$ is the constant of integration. Then \[ \alpha+\frac{\gamma}{\beta} \] is equal to:

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

Answer: (b)

Solution

The integral $$ \int \frac{2 - \tan x}{3 + \tan x} \, dx = \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} \, dx $$ We have: $$ 2 \cos x - \sin x = A(3 \cos x + \sin x) + B(\cos x - 3 \sin x) $$ Solving for $A$ and $B$: $$ 3A + B = 2 $$ $$ A - 3B = -1 $$ Thus, $$ A = \frac{1}{2}, B = \frac{1}{2} $$ Therefore: $$ \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} \, dx $$ This becomes: $$ = \frac{x}{2} + \frac{1}{2} \ln |3 \cos x + \sin x| + C $$ Which simplifies to: $$ = \frac{1}{2} \left( x + \ln |3 \cos x + \sin x| \right) + C $$ And further: $$ = \frac{1}{2} \left( \alpha x + \ln |\beta \sin x + \gamma \cos x| \right) + C $$ Where $\alpha = 1$, $\beta = 1$, $\gamma = 3$ Thus: $$ \therefore \alpha + \frac{\gamma}{\beta} = 1 + \frac{3}{1} = 4 $$

Question 8

Maths · Applications of Derivatives · Single correct

A variable line $L$ passes through the point $(3, 5)$ and intersects the positive coordinate axes at the points $A$ and $B$. The minimum area of the triangle $OAB$, where $O$ is the origin, is :

  1. 30
  2. 25
  3. 40
  4. 35

Answer: (a)

Solution

Given $\($ $\frac{x}{a}$ + $\frac{y}{b}$ = 1 $\)$. $\($ $\frac{3}{a}$ + $\frac{5}{b}$ = 1 $\Rightarrow$ b = $\frac{5a}{a-3}$, a > 3 $\)$. $\($ A = $\frac{1}{2}$ ab = $\frac{1}{2}$ a $\frac{5a}{a-3}$ = $\frac{5}{2}$ $\cdot$ $\frac{a^2}{a-3}$ $\)$. $\($ = $\frac{5}{2}$ $\left$( $\frac{a^2 - 9 + 9}{a-3}$ $\right$) $\)$. $\($ = $\frac{5}{2}$ $\left$( a + 3 + $\frac{9}{a-3}$ $\right$) $\)$. $\($ = $\frac{5}{2}$ $\left$( a - 3 + $\frac{9}{a-3}$ + 6 $\right$) $\geq$ 30 $\)$.

Question 9

Maths · Trigonometric Functions · Single correct

Let $\left| \cos \theta \cos(60 - \theta) \cos(60 + \theta) \right| \leq \frac{1}{8}, \theta \in [0, 2\pi]$. Then, the sum of all $\theta \in [0, 2\pi]$, where $\cos 3\theta$ attains its maximum value, is:

  1. 15$\pi$
  2. 18$\pi$
  3. 6$\pi$
  4. 9$\pi$

Answer: (c)

Solution

We know that $$\left( \cos \theta \right) \left( \cos(60^\circ - \theta) \right) \left( \cos(60^\circ + \theta) \right) = \frac{1}{4} \cos 3\theta$$ So equation reduces to $$\left| \frac{1}{4} \cos 3\theta \right| \leq \frac{1}{8}$$ $$\Rightarrow \left| \cos 3\theta \right| \leq \frac{1}{2}$$ $$\Rightarrow -\frac{1}{2} \leq \cos 3\theta \leq \frac{1}{2}$$ $$\Rightarrow maximum value of \cos 3\theta = \frac{1}{2}, here$$ $$\Rightarrow 3\theta = 2n\pi \pm \frac{\pi}{3}$$ $$\theta = \frac{2n\pi}{3} \pm \frac{\pi}{9}$$ As $\theta \in [0, 2\pi]$ possible values are $$\theta = \left\{ \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{11\pi}{9}, \frac{13\pi}{9}, \frac{17\pi}{9} \right\}$$ Whose sum is $$\frac{\pi}{9} + \frac{5\pi}{9} + \frac{7\pi}{9} + \frac{11\pi}{9} + \frac{13\pi}{9} + \frac{17\pi}{9} = \frac{54\pi}{9} = 6\pi$$

Question 10

Maths · Vector Algebra · Single correct

Let $\overrightarrow{OA} = 2\overrightarrow{a}$, $\overrightarrow{OB} = 6\overrightarrow{a} + 5\overrightarrow{b}$ and $\overrightarrow{OC} = 3\overrightarrow{b}$, where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $OABC$ is equal to:

  1. 32
  2. 40
  3. 38
  4. 35

Answer: (d)

Solution

Area of parallelogram having sides $\overrightarrow{OA} \& \overrightarrow{OC} = |\overrightarrow{OA} \times \overrightarrow{OC}| = |2\, \overrightarrow{a} \times 3\, \overrightarrow{b}| = 15$. $6|\overrightarrow{a} \times \overrightarrow{b}| = 15$ $\Rightarrow |\overrightarrow{a} \times \overrightarrow{b}| = \frac{5}{2} \ldots (1)$ Area of quadrilateral $$OABC = \frac{1}{2} \left| \overrightarrow{d_1} \times \overrightarrow{d_2} \right|$$ $$= \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{OB}| = \frac{1}{2} |(3\, \overrightarrow{b} - 2\, \overrightarrow{a}) \times (6\, \overrightarrow{a} + 5\, \overrightarrow{b})|$$ $$= \frac{1}{2} |18\, \overrightarrow{b} \times \overrightarrow{a} - 10\, \overrightarrow{a} \times \overrightarrow{b}| = |14\, \overrightarrow{a} \times \overrightarrow{b}|$$ $$= 14 \times \frac{5}{2} = 35$$

Question 11

Maths · Inverse Trigonometric Functions · Single correct

If the domain of the function $f(x) = \sin^{-1} \left( \frac{x-1}{2x+3} \right)$ is $\mathbb{R} - (\alpha, \beta)$, then $12\alpha\beta$ is equal to :

  1. 32
  2. 40
  3. 24
  4. 36

Answer: (a)

Solution

Domain of $f(x) = \sin^{-1}\left(\frac{x-1}{2x+3}\right)$ is $2x + 3 \neq 0$ and $x \neq -\frac{3}{2}$ and $$\left|\frac{x-1}{2x+3}\right| \leq 1$$ $$|x-1| \leq |2x+3|$$ For $|2x+3| \geq |x-1|$ $$x \in (-\infty, -4] \cup \left(-\frac{2}{3}, \infty\right)$$ $$\alpha = -4 and \beta = -\frac{2}{3} : 12\alpha\beta = 32$$

Question 12

Maths · Sequences and Series · Single correct

If the sum of the series $\frac{1}{1 \cdot (1+d)} + \frac{1}{(1+d)(1+2d)} + \cdots + \frac{1}{(1+9d)(1+10d)}$ is equal to 5, then $50d$ is equal to:

  1. 10
  2. 5
  3. 15
  4. 20

Answer: (b)

Solution

The given series is $\dfrac{1}{1\cdot(1+d)} + \dfrac{1}{(1+d)(1+2d)} + \ldots$. We need to find the sum of this series which is equal to $5$. Starting with the first term: $$\frac{1}{(1+9d)(1+10d)} = 5$$ Rewriting the series: $$\frac{1}{d}\left[\frac{(1+d)-1}{1\cdot(1+d)} + \frac{(1+2d)-(1+d)}{(1+d)(1+2d)}\right] + \ldots$$ Simplifying further: $$\frac{(1+10d)-(1+9d)}{(1+9d)(1+10d)} = 5$$ This becomes: $$\frac{1}{d}\left[\left(1 - \frac{1}{1+d}\right) + \left(\frac{1}{1+d} - \frac{1}{1+2d}\right) + \ldots\right]$$ Continuing the simplification: $$\frac{1}{d}\left[1 - \frac{1}{1+10d}\right] = 5$$ Solving for $d$: $$\frac{10d}{1+10d} = 5d$$ Finally: $$50d = 5$$

Question 13

Maths · Continuity and Differentiability · Single correct

Let $f(x) = ax^3 + bx^2 + cx + 41$ be such that $f(1) = 40$, $f'(1) = 2$ and $f''(1) = 4$. Then $a^2 + b^2 + c^2$ is equal to:

  1. 73
  2. 62
  3. 51
  4. 54

Answer: (c)

Solution

Given $f(x) = ax^3 + bx^2 + cx + 41$. The derivative is $f'(x) = 3ax^2 + 2bx + c$. Thus, $f'(1) = 3a + 2b + c = 2 \ldots (1)$. The second derivative is $f''(x) = 6ax + 2b$. Thus, $f''(1) = 6a + 2b = 4$. From this, $3a + b = 2 \ldots (2)$. Subtracting (2) from (1), we get $b + c = 0 \ldots (3)$. Given $f(1) = 40$, we have $a + b + c + 41 = 40$. Using (3), $a + 41 = 40$. By (2), $-3 + b = 2 \Rightarrow b = 5$ and $c = -5$. Finally, $a^2 + b^2 + c^2 = 1 + 25 + 25 = 51$.

Question 14

Maths · Conic Sections · Single correct

Let a circle passing through $(2, 0)$ have its centre at the point $(h, k)$. Let $(x_c, y_c)$ be the point of intersection of the lines $3x + 5y = 1$ and $(2 + c)x + 5c^2y = 1$. If $h = \lim_{c \to 1} x_c$ and $k = \lim_{c \to 1} y_c$, then the equation of the circle is:

  1. $25x^2 + 25y^2 - 2x + 2y - 60 = 0$
  2. $5x^2 + 5y^2 - 4x + 2y - 12 = 0$
  3. $5x^2 + 5y^2 - 4x - 2y - 12 = 0$
  4. $25x^2 + 25y^2 - 20x + 2y - 60 = 0$

Answer: (d)

Solution

Given $$(2+c)x + 5c^2 \left(1 - \frac{3x}{5}\right) = 1$$ We have $$x = \frac{1-c^2}{2+c-3c^2}, y = \frac{1-3x}{5} = \frac{c-1}{5(2+c-3c^2)}$$ Calculate $$h = \lim_{c \to 1} \frac{(1-c)(1+c)}{(1-c)(2+3c)} = \frac{2}{5}$$ And $$K = \lim_{c \to 1} \frac{c-1}{-5(c-1)(3c+2)} = -\frac{1}{25}$$ The centre is $$\left(\frac{2}{25}, -\frac{1}{25}\right)$$ The radius is $$r = \sqrt{\left(2 - \frac{2}{5}\right)^2 + \left(0 - \left(-\frac{1}{25}\right)\right)^2} = \sqrt{\frac{64}{25} + \frac{1}{625}}$$ Simplifying gives $$r = \frac{\sqrt{161}}{25}$$ The equation of the circle is $$\left(x - \frac{2}{5}\right)^2 + \left(y + \frac{1}{25}\right)^2 = \frac{161}{125}$$ Thus, $$\Rightarrow 25x^2 + 25y^2 - 20x + 2y - 60 = 0$$

Question 15

Maths · Three Dimensional Geometry · Single correct

The shortest distance between the lines $\frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5}$ and $\frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1}$ is:

  1. $\frac{178}{\sqrt{563}}$
  2. $\frac{187}{\sqrt{563}}$
  3. $\frac{185}{\sqrt{563}}$
  4. $\frac{179}{\sqrt{563}}$

Answer: (b)

Solution

Given $\vec{p} = 4\hat{i} - 11\hat{j} + 5\hat{k}$ and $\vec{q} = 3\hat{i} - 6\hat{j} + \hat{k}$. The vector $\vec{n}$ is given by the cross product $\vec{p} \times \vec{q}$. $$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{vmatrix} = 19\hat{i} + 11\hat{j} + 9\hat{k}$$ The scalar distance (S.d.) is the projection of $\overrightarrow{AB}$ on $\vec{n}$. $$S.d. = \frac{\left| \overrightarrow{AB} \cdot \vec{n} \right|}{\left| \vec{n} \right|}$$ Calculating $\overrightarrow{AB} = (2\hat{i} + 16\hat{j} - 3\hat{k})$ and $$\overrightarrow{AB} \cdot \vec{n} = (2\hat{i} + 16\hat{j} - 3\hat{k}) \cdot (19\hat{i} + 11\hat{j} + 9\hat{k}) = 38 + 176 - 27$$ Thus, $$S.d. = \frac{187}{\sqrt{361 + 121 + 81}} = \frac{187}{\sqrt{563}}$$

Question 16

Maths · Statistics · Single correct

The frequency distribution of the age of students in a class of $40$ students is given below. \begin{tabular}{|c|c|c|c|c|c|c|} \hline Age & 15 & 16 & 17 & 18 & 19 & 20 \\ \hline No. of Students & 5 & 8 & 5 & 12 & $x$ & $y$ \\ \hline \end{tabular} If the mean deviation about the median is $1.25$, then $4x+5y$ is equal to:

  1. 46
  2. 43
  3. 44
  4. 47

Answer: (c)

Solution

$x + y = 10 \hfill \ldots(1)$ $\text{Median} = 18 = M$ $\text{M.D.} = \dfrac{\sum f_i\,|x_i - M|}{\sum f_i}$ $1.25 = \dfrac{36 + x + 2y}{40}$ $x + 2y = 14 \hfill \ldots(2)$ By (1) \& (2) $x = 6,\ y = 4$ $\Rightarrow 4x + 5y = 24 + 20 = 44$ \begin{tabular}{|c|c|c|c|} \hline Age$(x_i)$ & $f$ & $|x_i - M|$ & $f_i\,|x_i - M|$ \\ \hline 15 & 5 & 3 & 15 \\ \hline 16 & 8 & 2 & 16 \\ \hline 17 & 5 & 1 & 5 \\ \hline 18 & 12 & 0 & 0 \\ \hline 19 & $x$ & 1 & $x$ \\ \hline 20 & $y$ & 2 & $2y$ \\ \hline \end{tabular}

Question 17

Maths · Differential Equations · Single correct

The solution of the differential equation $(x^2 + y^2) \, \mathrm{dx} - 5xy \, \mathrm{dy} = 0, y(1) = 0$, is :

  1. $|x^2 - 2y^2|^6 = x$
  2. $|x^2 - 4y^2|^6 = x$
  3. $|x^2 - 4y^2|^5 = x^2$
  4. $|x^2 - 2y^2|^5 = x^2$

Answer: (c)

Solution

Given $(x^2+y^2)\,dx=5xy\,dy$. This implies $\dfrac{dy}{dx}=\dfrac{x^2+y^2}{5xy}$. Put $y=Vx$. This gives $V+x\dfrac{dV}{dx}=\dfrac{1+V^2}{5V}$. Thus, $x\dfrac{dV}{dx}=\dfrac{1-4V^2}{5V}$. Integrating, $\int \dfrac{V}{1-4V^2}\,dV=\int \dfrac{dx}{5x}$. Let $1-4V^2=t$. Then $-8V\,dV=dt$. This implies $\int \dfrac{dt}{(-8)t}=\int \dfrac{dx}{5x}$. Thus, $-\dfrac{1}{8}\ln|t|=\dfrac{1}{5}\ln|x|+\ln C$. Therefore, $-5\ln|t|=8\ln|x|+\ln K$. This gives $\ln x^8+\ln|t|^5+\ln K=0$. Thus, $x^8|t|^5=C$. Therefore, $x^8|1-4V^2|^5=C$. This implies $x^8\left|\dfrac{x^2-4y^2}{x^2}\right|^5=C$. Thus, $|x^2-4y^2|^5=Cx^2$. Given $y(1)=0$. Then $|1|^5=C\Rightarrow C=1$. Therefore, $|x^2-4y^2|^5=x^2$.

Question 18

Maths · Vector Algebra · Single correct

Let three vectors $\vec{a} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k}$, $\vec{b} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$, $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$ form a triangle such that $\vec{c} = \vec{a} - \vec{b}$ and the area of the triangle is $5 \sqrt{6}$. If $\alpha$ is a positive real number, then $|\vec{c}|^2$ is equal to:

  1. 16
  2. 14
  3. 12
  4. 10

Answer: (b)

Solution

Given $\vec{c} = \vec{a} - \vec{b}$, we have $(x, y, z) = (\alpha - 5, 1, -2)$. Therefore, $x = \alpha - 5$, $y = 1$, $z = -2$...(1). The area of $\Delta = 5 \sqrt{6}$ (given). $$\frac{1}{2} \left| \vec{a} \times \vec{c} \right| = 5 \sqrt{6}$$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ x & 1 & -2 \end{vmatrix} = 10 \sqrt{6}$$ This implies: $$\left| -10 \hat{i} - \hat{j} (-2\alpha - 2x) + \hat{k} (\alpha - 4x) \right| = 10 \sqrt{6}$$ $$\Rightarrow (2\alpha + 2x - 10)^2 + (\alpha - 4x + 20)^2 = 500$$ $$\Rightarrow (4\alpha - 10)^2 + (20 - 3\alpha)^2 = 500$$ $$\Rightarrow 25\alpha^2 - 80\alpha - 120\alpha = 0$$ $$\Rightarrow \alpha (25\alpha - 200) = 0$$ $$\Rightarrow \alpha = 8 (given \alpha is positive number)$$ Therefore, $x = \alpha - 5 = 3$. Now, $|\vec{c}|^2 = x^2 + y^2 + z^2 = 9 + 1 + 4 = 14$.

Question 19

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $\alpha, \beta$ be the roots of the equation $x^2 + 2\sqrt{2}x - 1 = 0$. The quadratic equation, whose roots are $\alpha^4 + \beta^4$ and $\frac{1}{10} \left( \alpha^6 + \beta^6 \right)$, is:

  1. $x^2 - 190x + 9466 = 0$
  2. $x^2 - 180x + 9506 = 0$
  3. $x^2 - 195x + 9506 = 0$
  4. $x^2 - 195x + 9466 = 0$

Answer: (c)

Solution

Given the equation $x^2 + 2\sqrt{2}x - 1 = 0$. We have $\alpha + \beta = -2\sqrt{2}$ and $\alpha \beta = -1$. Now, $\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2 \beta^2$. This equals $((\alpha + \beta)^2 - 2\alpha \beta)^2 - 2(\alpha \beta)^2$. Substituting the values, we get $(8 + 2)^2 - 2(-1)^2 = 100 - 2 = 98$. Next, $\alpha^6 + \beta^6 = (\alpha^3 + \beta^3)^2 - 2\alpha^3 \beta^3$. This equals $((\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)^2 - 2(\alpha \beta)^3$. Substituting the values, we get $(-2\sqrt{2}(8 + 3))^2 + 2 = (8)(121) + 2 = 970$. Then, $\frac{1}{10}(\alpha^6 + \beta^6) = 97$. The equation becomes $x^2 - (98 + 97)x + (98)(97) = 0$. Thus, $\Rightarrow x^2 - 195x + 9506 = 0$.

Question 20

Maths · Conic Sections · Single correct

Let $f(x) = x^2 + 9$, $g(x) = \frac{x}{x-9}$ and $a = f \circ g(10)$, $b = g \circ f(3)$. If $e$ and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^2}{a} + \frac{y^2}{b} = 1$, then $8e^2 + l^2$ is equal to.

  1. 8
  2. 16
  3. 6
  4. 12

Answer: (a)

Solution

Given $f(x) = x^2 + 9$ and $g(x) = \frac{x}{x-9}$. $a = f(g(10)) = f\left(\frac{10}{10-9}\right)$ $= f(10) = 109$ $b = g(f(3)) = g(9 + 9)$ $= g(18) = \frac{18}{9} = 2$ $E : \frac{x^2}{109} + \frac{y^2}{2} = 1$ $e^2 = 1 - \frac{2}{109} = \frac{107}{109}$ $\ell = \frac{2(2)}{\sqrt{109}} = \frac{4}{\sqrt{109}}$ $8e^2 + \ell^2 = \frac{8(107)}{109} + \frac{16}{109}$ $= 8$

Question 21

Maths · Probability · Numerical

Let a, b and c denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that $ax^2 + bx + c = 0$ has all real roots is $\frac{m}{n}$, $\mathrm{gcd}(m, n) = 1$, then m + n is equal to

Answer: 19

Solution

Given $a, b, c \in \{1, 2, 3, 4\}$. The equation $ax^2 + bx + c = 0$ has all real roots. This implies $D \geq 0$, which means $b^2 - 4ac \geq 0$. Let $b = 1$, then $1 - 4ac \geq 0$ (Not feasible). For $b = 2$, $4 - 4ac \geq 0$, which implies $1 \geq ac$. Therefore, $a = 1, c = 1$. For $b = 3$, $9 - 4ac \geq 0$, which implies $\frac{9}{4} \geq ac$. Therefore, $a = 1, c = 1$, $a = 1, c = 2$, $a = 2, c = 1$. For $b = 4$, $16 - 4ac \geq 0$, which implies $4 \geq ac$. Therefore, $a = 1, c = 1$, $a = 1, c = 2$, $a = 2, c = 1$, $a = 1, c = 3$, $a = 3, c = 1$, $a = 1, c = 4$, $a = 4, c = 1$, $a = 2, c = 2$. The probability is $$\frac{12}{(4)(4)(4)} = \frac{3}{16} = \frac{m}{m}$$ where $m + n = 19$.

Question 22

Maths · Complex Numbers and Quadratic Equations · Numerical

The sum of the square of the modulus of the elements in the set $$ \{ z = a + ib : a, b \in \mathbb{Z}, z \in \mathbb{C}, |z - 1| \leq 1, |z - 5| \leq |z - 5i| \} $$ is

Answer: 9

Solution

Given $|z - 1| \leq 1$ implies $|(x - 1) + iy| \leq 1$. This leads to $\sqrt{(x - 1)^2 + y^2} \leq 1$ which implies $(x - 1)^2 + y^2 \leq 1$. Also $|z - 5| \leq |z - 5i|$ gives $(x - 5)^2 + y^2 \leq x^2 + (y - 5)^2$. Simplifying, $-10x \leq -10y$ implies $x \geq y$ (2). Solving (1) and (2), we have $(x - 1)^2 + x^2 = 1$ leading to $2x^2 - 2x = 0$. This implies $x(x - 1) = 0$ so $x = 0$ or $x = 1$. Therefore, $y = 0$ or $y = 1$. Given $x, y \in I$, the points $(0, 0), (1, 0), (2, 0), (1, 1), (1, -1)$ are used to find $|z_1|^2 + |z_2|^2 + |z_3|^2 + |z_4|^2 + |z_5|^2 = 0 + 1 + 4 + 1 + 1 + 1 + 1 = 9$.

Question 23

Maths · Applications of Derivatives · Numerical

Let the set of all positive values of $\lambda$, for which the point of local minimum of the function $\left(1 + x \left(\lambda^2 - x^2\right)\right)$ satisfies $\frac{x^2 + x + 2}{x^2 + 5x + 6} < 0$, be $(\alpha, \beta)$. Then $\alpha^2 + \beta^2$ is equal to

Answer: 39

Solution

Given $\($ $\frac{x^2 + x + 2}{x^2 + 5x + 6}$ $\lambda$ > 2$\sqrt{3}$ $\)$. $\($ $\alpha$ = 2$\sqrt{3}$, $\beta$ = 3$\sqrt{3}$ $\)$. $\($ $\alpha$^2 + $\beta$^2 = 12 + 27 = 39 $\)$.

Question 24

Maths · Limits and Derivatives · Numerical

Let $\lim_{n\to\infty}$( $\frac{n}{\sqrt{n^4+1}}$ - $\frac{2n}{(n^2+1)\sqrt{n^4+1}}$ + $\frac{n}{\sqrt{n^4+16}}$ - $\frac{8n}{(n^2+4)\sqrt{n^4+16}}$ + $\cdots$ + $\frac{n}{\sqrt{n^4+n^4}}$ - $\frac{2n \cdot n^2}{(n^2+n^2)\sqrt{n^4+n^4}})$ be $\frac{\pi}{k}$, using only the principal values of the inverse trigonometric functions. Then $k^2$ is equal to

Answer: 32

Solution

The given series is $$\sum_{r=1}^{\infty} \frac{n}{\sqrt{n^4 + r^4}} - \frac{2nr^2}{(n^2 + r^2)\sqrt{n^4 + r^4}}$$ which simplifies to $$\sum_{r=1}^{\infty} \frac{1}{n} - \frac{2 \left( \frac{1}{n} \right)^2}{\sqrt{1 + \left( \frac{r}{n} \right)^4}} - \left( 1 + \left( \frac{r}{n} \right)^2 \right) \frac{1}{\sqrt{1 + \left( \frac{r}{n} \right)^4}}.$$ This leads to the integral $$\int_0^1 \frac{dx}{\sqrt{1 + x^4}} - \frac{2x^2 dx}{(1 + x^2)\sqrt{1 + x^4}}.$$ Simplifying further, we have $$\int_0^1 \frac{1 - x^2}{(1 + x^2)\sqrt{1 + x^4}} dx.$$ This can be rewritten as $$\int_0^1 \frac{1}{x^2} - 1 \left( x + \frac{1}{x} \right) \sqrt{x^2 + \frac{1}{x^2}}.$$ Thus, $$\int_0^1 \frac{1 - \frac{1}{x^2}}{\left( x + \frac{1}{x} \right) \sqrt{\left( x + \frac{1}{x} \right)^2 - 2}} dx.$$ Letting $x + \frac{1}{x} = t$, we have $$1 - \frac{1}{x^2} dx = dt.$$ Therefore, $$\int_\infty^{\sqrt{2}} \frac{dt}{t\sqrt{t^2 - 2}}.$$ This becomes $$\int_\infty^{\sqrt{2}} \frac{td\alpha}{(\alpha^2 + 2)\alpha}.$$ Simplifying, we have $$\int_\infty^{\sqrt{2}} \frac{d\alpha}{\alpha^2 + 2}.$$ This evaluates to $$\left[ -\frac{1}{\sqrt{2}} \tan^{-1} \frac{\alpha}{\sqrt{2}} \right]_\infty^{\sqrt{2}}.$$ Thus, $$-\frac{1}{\sqrt{2}} \{ \tan^{-1} 1 \} + \frac{1}{\sqrt{2}} \tan^{-1} \infty.$$ This simplifies to $$\frac{1}{\sqrt{2}} \left\{ \frac{\pi}{2} - \frac{\pi}{4} \right\}.$$ Therefore, $$\frac{\pi}{4\sqrt{2}} = \frac{\pi}{K}.$$ So, $K = 4\sqrt{2}$. Therefore, $K^2 = 32$.

Question 25

Maths · Binomial Theorem · Numerical

The remainder when $428^{2024}$ is divided by 21 is ________

Answer: 1

Solution

$(428)^{2024} = (420 + 8)^{2024}$ $= (21 \times 20 + 8)^{2024}$ $= 21m + 8^{2024}$ Now $8^{2024} = (8^2)^{1012}$ $= (64)^{1012}$ $= (63 + 1)^{1012}$ $= (21 \times 3 + 1)^{1012}$ $= 21n + 1$ $\Rightarrow$ Remainder is $1$.

Question 26

Maths · Continuity and Differentiability · Numerical

Let $f : (0, \pi) \to \mathbb{R}$ be a function given by $$ f(x) = \begin{cases} \left( \frac{8}{7} \right)^{\frac{\tan 8x}{\tan 7x}}, & 0 < x < \frac{\pi}{2} \\ a - 8, & x = \frac{\pi}{2} \\ \left(1 + |\cot x| \right)^{b} |\tan x|, & \frac{\pi}{2} < x < \pi \end{cases} $$ where $a, b \in \mathbb{Z}$. If $f$ is continuous at $x = \frac{\pi}{2}$, then $a^2 + b^2$ is equal to

Answer: 81

Solution

LHL at $x = \frac{\pi}{2}$ $$\lim_{x \to \frac{\pi}{2}} \left( \frac{8}{7} \right)^{\frac{\tan 8x}{\tan 7x}} = \left( \frac{8}{7} \right)^0 = 1$$ RHL at $x = \frac{\pi}{2}$ $$\lim_{x \to \frac{\pi}{2}} \left( 1 + |\cot x| \right)^{\frac{b}{\tan x}}$$ $$= e^{\lim_{x \to \frac{\pi}{2}} |\cot x| \frac{b}{a} |\tan x|} = e^{\frac{b}{a}}$$ $$\Rightarrow 1 = a - 8 = e^{\frac{b}{a}}$$ $$\Rightarrow a = 9, b = 0$$ $$\Rightarrow a^2 + b^2 = 81$$

Question 27

Maths · Matrices · Numerical

Let A be a non-singular matrix of order 3. If $\det(3 \; adj(2 \; adj((\det A)A))) = 3^{-13} \cdot 2^{-10}$ and $\det(3 \; adj(2A)) = 2^m \cdot 3^n$, then $|3m + 2n|$ is equal to

Answer: 14

Solution

Given $|3 \operatorname{adj}(2 \operatorname{adj}(|A|A))| = |3 \operatorname{adj}(2|A|^2 \operatorname{adj}(A))|$. This equals $|3 \cdot 2^2|A|^4 \operatorname{adj}(\operatorname{adj}(A))| = 2^6 3^3 |A|^{12} |A|^4$. Thus, $= 2^6 3^3 |A|^{16} = 2^{-10} 3^{-13}$. Therefore, $|A|^{16} = 2^{-16} 3^{-16}$ which implies $|A| = 2^{-1} 3^{-1}$. Now $|3 \operatorname{adj}(2 A)| = |3 \cdot 2^2 \operatorname{adj}(A)|$. This equals $= 2^6 3^3 |A|^2 = 2^{-m-3-n}$. Thus, $2^6 3^3 2^{-2} 3^{-2} = 2^{-m-3-n}$ which simplifies to $2^{-m-3-n} = 2^4 3^1$. This implies $m = -4, n = -1$. Finally, $|3m + 2n| = |-12 - 2| = 14$.

Question 28

Maths · Conic Sections · Numerical

Let the centre of a circle, passing through the points $(0, 0)$, $(1, 0)$ and touching the circle $x^2 + y^2 = 9$, be $(h, k)$. Then for all possible values of the coordinates of the centre $(h, k)$, $4 \left(h^2 + k^2\right)$ is equal to

Answer: 9

Solution

The equation of the circle is $(x-h)^2 + (y-k)^2 = h^2 + k^2$. Expanding, we get $x^2 + y^2 - 2hx - 2ky = 0$. Since it passes through $(1, 0)$, we have $1 + 0 - 2h = 0$, which implies $h = 1/2$. Therefore, $OC = \frac{\mathrm{OP}}{2}$. Solving $\sqrt{\left(\frac{1}{2}\right)^2 + k^2} = \frac{3}{2}$ gives $\frac{1}{4} + k^2 = \frac{9}{4}$, leading to $k^2 = 2$ and $k = \pm \sqrt{2}$. Possible coordinates of $c(h, k)$ are $\left(\frac{1}{2}, \sqrt{2}\right)$ and $\left(\frac{1}{2}, -\sqrt{2}\right)$. Calculating $4\left(h^2 + k^2\right) = 4\left(\frac{1}{4} + 2\right) = 4\left(\frac{9}{4}\right) = 9$.

Question 29

Maths · Relations and Functions · Numerical

If a function $f$ satisfies $f(m+n) = f(m) + f(n)$ for all $m, n \in \mathbb{N}$ and $f(1) = 1$, then the largest natural number $\lambda$ such that $\sum_{k=1}^{2022} f(\lambda + k) \leq (2022)^2$ is equal to

Answer: 1010

Solution

Given $f(m + n) = f(m) + f(n)$, we have $f(x) = kx$. Since $f(1) = 1$, it follows that $k = 1$. Therefore, $f(x) = x$. Now, $$\sum_{k=1}^{2022} f(\lambda + k) \leq (2022)^2$$ implies $$\sum_{k=1}^{2022} (\lambda + k) \leq (2022)^2.$$ This leads to $$2022\lambda + \frac{2022 \times 2023}{2} \leq (2022)^2.$$ Simplifying gives $$\lambda \leq 2022 - \frac{2023}{2}$$ which results in $$\lambda \leq 1010.5.$$ Therefore, the largest natural number $\lambda$ is 1010.

Question 30

Maths · Relations and Functions · Numerical

Let $A = \{2, 3, 6, 7\}$ and $B = \{4, 5, 6, 8\}$. Let $R$ be a relation defined on $A \times B$ by $(a_1, b_1) R (a_2, b_2)$ if and only if $a_1 + a_2 = b_1 + b_2$. Then the number of elements in $R$ is

Answer: 25

Solution

Given sets $A = \{2, 3, 6, 7\}$ and $B = \{2, 5, 6, 8\}$. The relation $(a_1, b_1) \, R \, (a_2, b_2)$ is defined by $a_1 + a_2 = b_1 + b_2$. The pairs are: 1. $(2, 4) \, R \, (6, 4)$ 2. $(2, 4) \, R \, (7, 5)$ 3. $(2, 5) \, R \, (7, 4)$ 4. $(3, 4) \, R \, (6, 5)$ 5. $(3, 5) \, R \, (6, 4)$ 6. $(3, 5) \, R \, (7, 5)$ 7. $(3, 6) \, R \, (7, 4)$ 8. $(3, 4) \, R \, (7, 6)$ 9. $(6, 5) \, R \, (7, 8)$ 10. $(6, 8) \, R \, (7, 5)$ 11. $(7, 8) \, R \, (7, 6)$ 12. $(6, 8) \, R \, (6, 4)$ 13. $(6, 6) \, R \, (6, 6)$ Total $24 + 1 = 25$

Physics

Question 31

Physics · Dual Nature of Radiation and Matter · Single correct

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:

  1. $\lambda_\alpha < \lambda_p < \lambda_e$
  2. $\lambda_e > \lambda_\alpha > \lambda_p$
  3. $\lambda_p > \lambda_e > \lambda_\alpha$
  4. $\lambda_p < \lambda_e < \lambda_\alpha$

Answer: (a)

Solution

Given $\lambda_{DB} = \frac{h}{p} = \frac{h}{\sqrt{2mk}}$. Therefore, $\lambda_{DB} \propto \frac{1}{\sqrt{m}}$. Thus, $\lambda_a < \lambda_p < \lambda_e$.

Question 32

Physics · Motion in a Straight Line · Single correct

A particle moving in a straight line covers half the distance with speed $6 \, \mathrm{m/s}$. The other half is covered in two equal time intervals with speeds $9 \, \mathrm{m/s}$ and $15 \, \mathrm{m/s}$ respectively. The average speed of the particle during the motion is:

  1. 8.8 m/s
  2. 10 m/s
  3. 9.2 m/s
  4. 8 m/s

Answer: (d)

Solution

Question 33

Physics · Electromagnetic Waves · Single correct

A plane EM wave is propagating along $x$ direction. It has a wavelength of $4 \, \mathrm{mm}$. If electric field is in $y$ direction with the maximum magnitude of $60 \, \mathrm{Vm}^{-1}$, the equation for magnetic field is :

  1. $B_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{2} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \mathrm{T}$
  2. $B_z = 60 \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \mathrm{T}$
  3. $B_x = 60 \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{i} \, \mathrm{T}$
  4. $B_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \mathrm{T}$

Answer: (a)

Solution

Given $E = BC$, $60 = B \times 3 \times 10^8$. Therefore, $B = 2 \times 10^{-7}$. Also, $C = f \lambda$. Thus, $3 \times 10^8 = f \times 4 \times 10^{-3}$. Therefore, $f = \frac{3}{4} \times 10^{11}$. Hence, $\omega = 2 \pi f = \frac{3}{4} \times 2 \pi \times 10^{11}$. Therefore, $\omega = \frac{\pi}{2} \times 10^3 C$. Thus, the electric field is in the $y$ direction. Propagation is in the $x$ direction. Magnetic field is in the $z$ direction.

Question 34

Physics · Ray Optics and Optical Instruments · Single correct

Given below are two statements : Statement (I) : When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side. Statement (II) : Concave lens always forms a virtual and erect image. 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: (c)

Solution

Given $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. $\frac{1}{v}-\frac{1}{-2f}=\frac{1}{-f}$ $\Rightarrow \frac{1}{v}=-\frac{1}{2f}\Rightarrow v=-2f$ $\frac{1}{v}=\frac{1}{u}+\frac{1}{f}\Rightarrow$ Virtual image of Real object. In statement II, it is not mentioned that object is real or virtual, hence Statement II is false.

Question 35

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

A light emitting diode (LED) is fabricated using GaAs semiconducting material whose band gap is $1.42 \, \mathrm{eV}$. The wavelength of light emitted from the LED is :

  1. 1400 nm
  2. 650 nm
  3. 875 nm
  4. 1243 nm

Answer: (c)

Solution

Given $\($ $\lambda$ = $\frac{1240}{1.42}$ = 875 $\mathrm{nm}$ $\)$ (Approx)

Question 36

Physics · Mechanical Properties of Fluids · Single correct

A sphere of relative density $\sigma$ and diameter $D$ has concentric cavity of diameter $d$. The ratio of $\frac{D}{d}$, if it just floats on water in a tank is:

  1. $\left(\frac{\sigma - 2}{\sigma + 2}\right)^{1/3}$
  2. $\left(\frac{\sigma}{\sigma - 1}\right)^{1/3}$
  3. $\left(\frac{\sigma - 1}{\sigma}\right)^{1/3}$
  4. $\left(\frac{\sigma + 1}{\sigma - 1}\right)^{1/3}$

Answer: (b)

Solution

Weight $(w) = \frac{4}{3} \pi \left( \frac{D^3 - d^3}{8} \right) \sigma g$. Buoyant force $(F_b) = 1 \times \frac{4}{3} \pi \left( \frac{D^3}{8} \right) \cdot g$. For just float, $w = F_b$. Therefore, $\left( D^3 - d^3 \right) \sigma = D^3$. This implies $1 - \frac{d^3}{D^3} = \frac{1}{\sigma}$. Hence, $1 - \frac{1}{\sigma} = \left( \frac{d}{D} \right)^3$. Therefore, $\left( \frac{\sigma}{\sigma - 1} \right)^{\frac{1}{3}} = \left( \frac{D}{d} \right)$.

Question 37

Physics · Electrostatic Potential and Capacitance · Single correct

A capacitor is made of a flat plate of area $A$ and a second plate having a stair-like structure as shown in figure. If the area of each stair is $\frac{A}{3}$ and the height is $d$, the capacitance of the arrangement is :

  1. $\frac{13 \varepsilon_0 A}{17 d}$
  2. $\frac{11 \varepsilon_0 A}{18 d}$
  3. $\frac{18 \varepsilon_0 A}{11 d}$
  4. $\frac{11 \varepsilon_0 A}{20 d}$

Answer: (b)

Solution

All capacitors are in parallel combination. Also effective area is common area only. $$\Rightarrow C_{eq} = C_1 + C_2 + C_3$$ $$\Rightarrow C_{eq} = \frac{A \varepsilon_0}{3d} + \frac{A \varepsilon_0}{3(2d)} + \frac{A \varepsilon_0}{3(3d)}$$ $$\Rightarrow C_{eq} = \frac{A \varepsilon_0}{3} \left( \frac{11}{6d} \right)$$ $$\Rightarrow C_{eq} = \frac{11 A \varepsilon_0}{18d}$$

Question 38

Physics · Laws of Motion · Single correct

A light unstretchable string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$. If the acceleration of the system is $\frac{g}{8}$, then the ratio of the masses $\frac{m_2}{m_1}$ is :

  1. 8 : 1
  2. 5 : 3
  3. 4 : 3
  4. 9 : 7

Answer: (d)

Solution

Given $a_{sys} = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g = \frac{g}{8}$. Therefore, $$\frac{m_2}{m_1} = \frac{9}{7}.$$

Question 39

Physics · Physical World, Units and Measurements · Single correct

The dimensional formula of latent heat is :

  1. $\mathrm{ML^2 \, T^{-2}}$
  2. $\mathrm{M^0 \, L^2 \, T^{-2}}$
  3. $\mathrm{MLT^{-2}}$
  4. $\mathrm{M^0LT^{-2}}$

Answer: (b)

Solution

Latent heat is specific heat. Therefore, $$\frac{ML^2 \, T^{-2}}{M} = M^0 \, L^2 \, T^{-2}$$

Question 40

Physics · Thermodynamics · Single correct

The volume of an ideal gas ($\gamma = 1.5$) is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is:

  1. $\frac{16}{25}$
  2. $\frac{4}{5}$
  3. $\frac{8}{5\sqrt{5}}$
  4. $\frac{2}{\sqrt{5}}$

Answer: (c)

Solution

For Adiabatic process $$P_i V_i = P_f V_f^\gamma$$ $$P_i (5)^{1.5} = P_f (4)^{1.5}$$ $$\frac{P_i}{P_f} = \left( \frac{4}{5} \right)^{\frac{3}{2}} = \frac{4}{5} \cdot \left( \frac{4}{5} \right)^{\frac{1}{2}} \Rightarrow \frac{8}{5\sqrt{5}}$$

Question 41

Physics · Nuclei · Single correct

The energy equivalent of 1 g of substance is :

  1. $5.6 \times 10^{12} \mathrm{MeV}$
  2. $5.6 \times 10^{26} \mathrm{MeV}$
  3. $11.2 \times 10^{24} \mathrm{MeV}$
  4. $5.6 \mathrm{eV}$

Answer: (b)

Solution

Given $E = mC^2$. Therefore, $E = (1 \times 10^{-3}) \times (3 \times 10^8)^2 \, \mathrm{J}$. Thus, $E = (10^{-3}) (9 \times 10^{16}) (6.241 \times 10^{18}) \, \mathrm{eV}$. $E = 56.169 \times 10^{31} \, \mathrm{eV}$. $E \approx 5.6 \times 10^{26} \, \mathrm{MeV}$.

Question 42

Physics · Gravitation · Single correct

An astronaut takes a ball of mass $m$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of $318.5 \, \mathrm{km}$. From earth's surface to the orbit, the change in total mechanical energy of the ball is $x \frac{GM_em}{2R_e}$. The value of $x$ is (take $R_e = 6370 \, \mathrm{km}$):

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

Answer: (d)

Solution

Given $h = 318.5 \approx \left( \frac{R_e}{20} \right)$. The initial total energy $T \cdot E_i = \frac{-GM_em}{R_e}$. The final total energy $T \cdot E_f = \frac{-GM_em}{2 \left( R_e + h \right)} = \frac{-GM_em}{2 \left( R_e + \frac{R_e}{20} \right)}$. Therefore, $T \cdot E_f = \frac{-10GM_em}{21R_e}$. The change in total mechanical energy is given by $$= TE_f - TE_i$$ $$= \frac{GM_em}{R_e} \left[ 1 - \frac{10}{21} \right] = \frac{11GM_em}{21R_e}.$$

Question 43

Physics · Moving Charges and Magnetism · Single correct

Given below are two statements : Statement (I) : When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account. Statement (II) : Ampere's circuital law does not depend on Biot-Savart's law. 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. Statement I is true but Statement II is false
  3. Both Statement I and Statement II are false
  4. Statement I is false but Statement II is true

Answer: (b)

Solution

Conceptual.

Question 44

Physics · Work, Energy and Power · Single correct

A particle of mass $m$ moves on a straight line with its velocity increasing with distance according to the equation $v = \alpha \sqrt{x}$, where $\alpha$ is a constant. The total work done by all the forces applied on the particle during its displacement from $x = 0$ to $x = d$, will be:

  1. $\frac{m}{2\alpha^2} d$
  2. $\frac{md}{2\alpha^2}$
  3. $2 \, m \alpha^2 \, d$
  4. $\frac{m \alpha^2 \, d}{2}$

Answer: (d)

Solution

Given $v = \alpha \sqrt{x}$. At $x = 0$: $v = 0$ and at $x = d$; $v = \alpha \sqrt{d}$. Work done $\mathrm{W.D} = K_f - K_i$. $$\mathrm{W.D} = \frac{1}{2} m (\alpha \sqrt{d})^2 - \frac{1}{2} m (0)^2$$ Therefore, $$\Rightarrow \mathrm{W.D} = \frac{m \alpha^2 d}{2}$$

Question 45

Physics · Current Electricity · Single correct

A galvanmeter has a coil of resistance $200\,\Omega$ with a full scale deflection at $20\,\mu\mathrm{A}$. The value of resistance to be added to use it as an ammeter of range $(0 - 20)\,\mathrm{mA}$ is ;

  1. 0.40$\Omega$
  2. 0.20$\Omega$
  3. 0.50$\Omega$
  4. 0.10$\Omega$

Answer: (b)

Solution

Given $G = 200 \, \Omega$ and $i_g = 20 \, \mu \mathrm{A}$. The current $i$ is given by: $$i = i_g \left( \frac{G}{S} + 1 \right)$$ Substituting the values: $$\Rightarrow 20 \times 10^{-3} = 20 \times 10^{-6} \left( \frac{200}{S} + 1 \right)$$ Solving for $S$: $$\Rightarrow \frac{200}{S} = 999$$ $$\Rightarrow S \approx 0.2 \, \Omega$$

Question 46

Physics · System of Particles and Rotational Motion · Single correct

A heavy iron bar, of weight $W$ is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle $\theta$ with the horizontal. The weight experienced by the person is :

  1. $W \cos \theta$
  2. $\frac{W}{2}$
  3. $W$
  4. $W \sin \theta$

Answer: (b)

Solution

Given $R = net reaction force by shoulder$. Balancing torque about point of contact on ground: $$W \left( \frac{L}{2} \cos \theta \right) = R (L \cos \theta)$$ $$\Rightarrow R = \frac{W}{2}$$

Question 47

Physics · Experimental Physics · Single correct

One main scale division of a vernier caliper is equal to $m$ units. If $n^{th}$ division of main scale coincides with $(n+1)^{th}$ division of vernier scale, the least count of the vernier caliper is:

  1. $\frac{n}{(n+1)}$
  2. $\frac{1}{(n+1)}$
  3. $\frac{m}{(n+1)}$
  4. $\frac{m}{n(n+1)}$

Answer: (c)

Solution

Given $nMSD = (n+1) \mathrm{VSD}$. Therefore, $$1 \mathrm{VSD} = \frac{n}{n+1} \mathrm{MSD}$$ L $\cdot$ C = 1 $\mathrm{MSD}$ - 1 $\mathrm{VSD}$ $$L \cdot C = m - m \left( \frac{n}{n+1} \right)$$ $$L \cdot C = m \left( \frac{n+1-n}{n+1} \right)$$ $$\Rightarrow L \cdot C = \left( \frac{m}{n+1} \right)$$

Question 48

Physics · Electrostatic Potential and Capacitance · Single correct

A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb:

  1. increases
  2. decreases
  3. remains same
  4. becomes zero

Answer: (a)

Solution

The impedance is given by $$Z = \sqrt{R^2 + X_C^2}$$ and $$X_C = \frac{1}{\omega C}$$ Due to the dielectric, $C \uparrow \Rightarrow X_C \downarrow \Rightarrow Z \downarrow$ So, current increases and thus the bulb will glow more brightly.

Question 49

Physics · Current Electricity · Single correct

The equivalent resistance between A and B is :

  1. 18$\Omega$
  2. 19$\Omega$
  3. 25$\Omega$
  4. 27$\Omega$

Answer: (b)

Solution

The equivalent resistance is calculated as follows: $$R_{eq} = 6\,\Omega + 5\,\Omega + 8\,\Omega = 19\,\Omega$$

Question 50

Physics · Thermodynamics · Single correct

A sample of 1 mole gas at temperature $T$ is adiabatically expanded to double its volume. If adiabatic constant for the gas is $\gamma = \frac{3}{2}$, then the work done by the gas in the process is:

  1. $\frac{R}{T} \left[ 2 - \sqrt{2} \right]$
  2. $\frac{T}{R} \left[ 2 + \sqrt{2} \right]$
  3. $RT \left[ 2 - \sqrt{2} \right]$
  4. $RT \left[ 2 + \sqrt{2} \right]$

Answer: (c)

Solution

Given $TV^{\gamma - 1} = constant$ $$\Rightarrow T(V)^{\frac{3}{2} - 1} = T_f(2\, V)^{\frac{3}{2} - 1}$$ $$\Rightarrow TV^{\frac{1}{2}} = T_f(2)^{\frac{1}{2}} (V)^{\frac{1}{2}}$$ $$\Rightarrow T_f = \left( \frac{T}{\sqrt{2}} \right)$$ Now, W.D. = $$\frac{nR\Delta T}{1 - \gamma} = \frac{1 \cdot R \left[ \frac{T}{\sqrt{2}} - T \right]}{1 - \frac{3}{2}}$$ $$\Rightarrow W.D. = 2RT \left[ 1 - \frac{1}{\sqrt{2}} \right]$$ $$\Rightarrow W.D. = RT[2 - \sqrt{2}]$$

Question 51

Physics · Mathematics in Physics · Numerical

If $\vec{a}$ and $\vec{b}$ makes an angle $\cos^{-1}\left(\frac{5}{9}\right)$ with each other, then $|\vec{a} + \vec{b}| = \sqrt{2}|\vec{a} - \vec{b}|$ for $|\vec{a}| = n|\vec{b}|$ The integer value of $n$ is ____

Answer: 3

Solution

Given $\cos \theta = \frac{5}{9}$. $$\frac{\vec{a} \cdot \vec{b}}{ab} = \frac{5}{9}$$ $$a^2 + b^2 + 2 \vec{a} \cdot \vec{b} = 2a^2 + 2b^2 - 4 \vec{a} \cdot \vec{b}$$ $$6 \vec{a} \cdot \vec{b} = a^2 + b^2$$ $$6 \times \frac{5}{9} ab = a^2 + b^2$$ $$\frac{10}{3} ab = a^2 + b^2 \& a = nb$$ $$\frac{10}{3} nb^2 = n^2 b^2 + b^2$$ $$3n^2 - 10n + 3 = 0$$ $$n = \frac{1}{3} and n = 3$$ The integer value $n = 3$.

Question 52

Physics · Electric Charges and Fields · Numerical

At the centre of a half ring of radius $R = 10 \, \mathrm{cm}$ and linear charge density $4n\mathrm{Cm}^{-1}$, the potential is $x\pi V$. The value of $x$ is

Answer: 36

Solution

Potential at centre of half ring $$V = \frac{KQ}{R}$$ $$V = \frac{K \lambda \pi R}{R}$$ $$V = K \lambda \pi \Rightarrow V = 9 \times 10^9 \times 4 \times 10^{-9} \pi$$ $$V = 36 \pi$$

Question 53

Physics · Nuclei · Numerical

A star has 100% helium composition. It starts to convert three $^4\mathrm{He}$ into one $^{12}\mathrm{C}$ via triple alpha process as $^4\mathrm{He} + ^4\mathrm{He} + ^4\mathrm{He} \rightarrow ^{12}\mathrm{C} + \mathrm{Q}$. The mass of the star is $2.0 \times 10^{32} \, \mathrm{kg}$ and it generates energy at the rate of $5.808 \times 10^{30} \, \mathrm{W}$. The rate of converting these $^4\mathrm{He}$ to $^{12}\mathrm{C}$ is $n \times 10^{42} \, \mathrm{s}^{-1}$, where $n$ is ________ [ Take, mass of $^4\mathrm{He} = 4.0026\, \mathrm{u}$, mass of $^{12}\mathrm{C} = 12\, \mathrm{u}$ ]

Answer: 15

Solution

The reaction is $^4\mathrm{He} + ^4\mathrm{He} + ^4\mathrm{He} \rightarrow ^{12}\mathrm{C} + Q$. The power generated is given by $\frac{N}{t} Q$, where $N$ is the number of reactions per second. The energy $Q$ is calculated as $Q = (3 \, m_{\mathrm{He}} - m_{\mathrm{C}}) C^2$. Substituting the values, $Q = (3 \times 4.0026 - 12) (3 \times 10^8)^2$. This gives $Q = 7.266 \, \mathrm{MeV}$. The rate of reactions is given by $$\frac{N}{t} = \frac{power}{Q} = \frac{5.808 \times 10^{30}}{7.266 \times 10^6 \times 1.6 \times 10^{-19}}$$ which simplifies to $$\frac{N}{t} = 5 \times 10^{42}$$. The rate of conversion of $^4\mathrm{He}$ into $^{12}\mathrm{C}$ is $15 \times 10^{42}$. Hence, $n = 15$.

Question 54

Physics · Wave Optics · Numerical

In a Young's double slit experiment, the intensity at a point is $\left( \frac{1}{4} \right)^{th}$ of the maximum intensity, the minimum distance of the point from the central maximum is _______ $\mu \mathrm{m}$. (Given : $\lambda = 600 \, \mathrm{nm}$, $d = 1.0 \, \mathrm{mm}$, $D = 1.0 \, \mathrm{m}$ )

Answer: 200

Solution

Given $I = I_0 \cos^2 \left( \frac{\Delta \phi}{2} \right)$. We have $\frac{I_0}{4} = \cos^2 \left( \frac{\Delta \phi}{2} \right)$. Therefore, $\Delta \phi = \frac{2\pi}{3}$. Using $\frac{2\pi}{\lambda} \left( \frac{yd}{D} \right) = \frac{2\pi}{3}$, we find $y = \frac{\lambda D}{3 \, d} = \frac{600 \times 10^{-9} \times 1}{3 \times 10^{-3}} = 2 \times 10^{-4} \, \mathrm{m}$.

Question 55

Physics · System of Particles and Rotational Motion · Numerical

A string is wrapped around the rim of a wheel of moment of inertia $0.40 \, \mathrm{kgm}^2$ and radius $10 \, \mathrm{cm}$. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of $40 \, \mathrm{N}$. The angular velocity of the wheel after $10 \, \mathrm{s}$ is $x \, \mathrm{rad/s}$, where $x$ is

Answer: 100

Solution

Given $\tau = \mathrm{FR} = I \alpha \Rightarrow 40 \times 0.1 = 0.4 \alpha$. $\alpha = 10 \, \mathrm{rad/s^2}$. $W_f = 10 \times 10 = 100 \, \mathrm{rad/s}$.

Question 56

Physics · Moving Charges and Magnetism · Numerical

A square loop of edge length 2 m carrying current of 2 A is placed with its edges parallel to the $x y$ axis. A magnetic field is passing through the $x - y$ plane and expressed as $\vec{B} = B_0(1 + 4x)\hat{k}$, where $B_0 = 5 \, \mathrm{T}$. The net magnetic force experienced by the loop is N.

Answer: 160

Solution

Given $B(x = 0) = B_0$, $B(x = 2) = 9 \, B_0$. Also, $F = i \ell B$. Therefore, $F_1 = i \ell B_0$ and $F_2 = 9 i \ell B_0$. The net force $F = F_2 - F_1 = 8 i \ell B_0 = 8 \times 2 \times 2 \times 5$. Thus, $F = 160 \, \mathrm{N}$.

Question 57

Physics · Mechanical Properties of Solids · Numerical

Two persons pull a wire towards themselves. Each person exerts a force of 200 N on the wire. Young's modulus of the material of wire is $1 \times 10^{11} \, \mathrm{N} \, \mathrm{m}^{-2}$. Original length of the wire is $2 \, \mathrm{m}$ and the area of cross section is $2 \, \mathrm{cm}^2$. The wire will extend in length by _____ $\mu \mathrm{m}$.

Answer: 20

Solution

Given $200 \, \mathrm{N} \leftrightarrow O$ $$\frac{F}{A} = Y \frac{\Delta \ell}{\ell} \implies \Delta \ell = \frac{F \ell}{A Y}$$ $$\Delta \ell = \frac{200 \times 2}{2 \times 10^{-4} \times 10^{11}} = 2 \times 10^{-5} = 20 \, \mu \mathrm{m}$$

Question 58

Physics · Alternating Current · Numerical

When a coil is connected across a 20 $\mathrm{V}$ $\mathrm{dc}$ supply, it draws a current of 5 $\mathrm{A}$. When it is connected across 20 $\mathrm{V}$, 50 $\mathrm{Hz}$ $\mathrm{ac}$ supply, it draws a current of 4 $\mathrm{A}$. The self inductance of the coil is _____ $\mathrm{mH}$. ( Take $\pi$ = 3 )

Answer: 10

Solution

Case-I: $i = \frac{20}{R} \implies R = 4 \Omega$ Case-II: $i = \frac{20}{Z}$ $4 = \frac{20}{\sqrt{R^2 + X_L^2}} \implies \sqrt{R^2 + X_L^2} = 5$ $R^2 + X_L^2 = 25 \implies X_L = 3 \Omega$ $L = \frac{3}{2 \pi f} = \frac{1}{2 \times 50} = \frac{1000}{100} \, \mathrm{mH}$ $L = 10 \, \mathrm{mH}$

Question 59

Physics · Oscillations · Numerical

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \, \mathrm{m}$, $2 \, \mathrm{ms^{-1}}$ and $16 \, \mathrm{ms^{-2}}$ at a certain instant. The amplitude of the motion is $\sqrt{x}$, m where $x$ is

Answer: 17

Solution

Given $x = 4 \, \mathrm{m}$, $V = 2 \, \mathrm{m/s}$, $a = 16 \, \mathrm{m/s^2}$. $|a| = \omega^2 x$ $$\Rightarrow 16 = \omega^2 (4)$$ $$\omega = 2 \, \mathrm{rad/s}$$ $$v = \omega \sqrt{A^2 - x^2}$$ $$A = \sqrt{\frac{v^2}{\omega^2} + x^2} \Rightarrow A = \sqrt{\frac{4}{4} + 16}$$ $$A = \sqrt{17} \, \mathrm{m}$$

Question 60

Physics · Current Electricity · Numerical

The current flowing through the 1$\Omega$ resistor is $\frac{n}{10}$ A. The value of n is

Answer: 25

Solution

The equations are given as follows: $$\frac{y - 5}{2} + \frac{y - 0}{2} + \frac{y - x + 10}{1} = 0$$ Simplifying, we have: $$y - 5 + y + 2y - 2x + 20 = 0$$ This simplifies to: $$4y - 2x + 15 = 0 \ldots (i)$$ Another equation is: $$\frac{x - 5}{4} + \frac{x - 0}{4} + \frac{x - 10 - y}{1} = 0$$ Simplifying, we have: $$x - 5 + x + 4x - 40 - 4y = 0$$ This simplifies to: $$6x - 4y - 45 = 0 \ldots (i)$$ Rearranging gives: $$-2x + 4y + 15 = 0 \ldots (ii)$$ Solving for $x$: $$4x - 30 = 0$$ Thus: $$x = \frac{15}{2} \& 4y - 15 + 15 = 0$$ Solving for $y$ gives: $$y = 0$$ Now, calculating $i$: $$i = \frac{y - x + 10}{1}$$ Substituting the values: $$i = \frac{0 - 7.5 + 10}{1}$$ This gives: $$i = 2.5 \mathrm{A} = \frac{n}{10} \mathrm{A}$$ Thus, $n = 25$

Chemistry

Question 61

Chemistry · Electrochemistry · Single correct

The molar conductivity for electrolytes $A$ and $B$ are plotted against $C^{1/2}$ as shown below. Electrolytes $A$ and $B$ respectively are:

  1. $A$: strong electrolyte ; $B$: weak electrolyte
  2. $A$: weak electrolyte ; $B$: weak electrolyte
  3. $A$: weak electrolyte ; $B$: strong electrolyte
  4. $A$: strong electrolyte ; $B$: strong electrolyte

Answer: (c)

Solution

Q13 A $\rightarrow$ Weak electrolyte B $\rightarrow$ Strong electrolyte

Question 62

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

Methods used for purification of organic compounds are based on :

  1. nature of compound and presence of impurity.
  2. neither on nature of compound nor on the impurity present.
  3. nature of compound only.
  4. presence of impurity only.

Answer: (a)

Solution

Organic compounds are purified based on their nature and impurity present in it.

Question 63

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

In the following sequence of reaction, the major products $B$ and $C$ respectively are:

Answer: (d)

Solution

The reaction starts with a compound containing $\mathrm{Cl}$ and $\mathrm{Br}$. It undergoes a Wurtz reaction with $\mathrm{Na/Et_2O}$ to form compound (A) with $\mathrm{Cl}$ groups. Compound (A) then undergoes a Swart reaction with $\mathrm{CoF}$ to form compound (C) with $\mathrm{F}$ groups. Alternatively, compound (A) can react with $\mathrm{Mg/Et_2O}$ followed by $\mathrm{D_2O}$ to form compound (B) with $\mathrm{D}$ groups.

Question 64

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

Correct order of basic strength of Pyrrole , Pyridine , and Piperidine is:

  1. Pyrrole > Piperidine > Pyridine
  2. Pyrrole > Pyridine > Piperidine
  3. Pyridine > Piperidine > Pyrrole
  4. Piperidine > Pyridine > Pyrrole

Answer: (d)

Solution

Order of basic strength is $$N (sp^3, localized lone pair) > N (sp^2, localized lone pair) > N (sp^2, delocalized lone pair, aromatic)$$ Therefore, Piperidine $>$ Pyridine $>$ Pyrrole

Question 65

Chemistry · Chemical Bonding and Molecular Structure · Single correct

In which one of the following pairs the central atoms exhibit $sp^2$ hybridization?

  1. $\mathrm{H_2O}$ and $\mathrm{NO_2}$
  2. $\mathrm{BF_3}$ and $\mathrm{NO_2^-}$
  3. $\mathrm{NH_2^-}$ and $\mathrm{H_2O}$
  4. $\mathrm{NH_2^-}$ and $\mathrm{BF_3}$

Answer: (b)

Solution

BF_3 $\rightarrow$ sp^2 NO_2^- $\rightarrow$ sp^2 H_2O $\rightarrow$ sp^3 NO_2 $\rightarrow$ sp^2 NH_2^- $\rightarrow$ sp^3

Question 66

Chemistry · Co-ordination Compounds · Single correct

The F^- ions make the enamel on teeth much harder by converting hydroxyapatite (the enamel on the surface of teeth) into much harder fluoroapatite having the formula.

  1. $\left[3\left(\mathrm{Ca_3(PO_4)_2}\right)\cdot\mathrm{Ca(OH)_2}\right]$
  2. $\left[3\left(\mathrm{Ca_3(PO_4)_2}\right)\cdot\mathrm{CaF_2}\right]$
  3. $\left[3\left(\mathrm{Ca_3(PO_4)_2}\right)\cdot\mathrm{Ca(OH)_2}\right]$
  4. $\left[3\left(\mathrm{Ca_3(PO_4)_2}\right)\cdot\mathrm{CaF_2}\right]$

Answer: (d)

Solution

Fluoroapatite implies $[3\mathrm{Ca}_3(\mathrm{PO}_4)_2 \cdot \mathrm{CaF}_2]$

Question 67

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

Relative stability of the contributing structures is :

  1. (I) > (II) > (III)
  2. (I) > (III) > (II)
  3. (II) > (I) > (III)
  4. (III) > (II) > (I)

Answer: (a)

Solution

(1) Neutral structures are more stable than charged ones. Therefore I is more stable than II and III. (2) +ve charge on less electronegative atom is more stable i.e., $\mathrm{C^\oplus}$ is more stable than $\mathrm{O^\oplus}$. Order is I $>$ III $>$ II.

Question 68

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

Given below are two statements : Statement (I) : The oxidation state of an element in a particular compound is the charge acquired by its atom on the basis of electron gain enthalpy consideration from other atoms in the molecule. Statement (II) : $p\pi - p\pi$ bond formation is more prevalent in second period elements over other periods. In the light of the above statements, choose the most appropriate answer from the options given below :

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

Answer: (c)

Solution

Oxidation state of an element in a particular compound is defined by the charge acquired by its atom on the basis of electronegativity consideration from other atoms in molecule.

Question 69

Chemistry · Haloalkanes and Haloarenes · Single correct

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason $(R)$. Assertion (A): $S_N2$ reaction of $C_6H_5CH_2Br$ occurs more readily than the $S_N2$ reaction of $CH_3CH_2Br$. Reason $(R)$: The partially bonded unhybridized p-orbital that develops in the trigonal bipyramidal transition state is stabilized by conjugation with the phenyl ring. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. (A) is correct but $(R)$ is not correct
  2. (A) is not correct but $(R)$ is correct
  3. Both (A) and $(R)$ are correct but $(R)$ is not the correct explanation of (A)
  4. Both (A) and $(R)$ are correct and $(R)$ is the correct explanation of (A)

Answer: (d)

Solution

The benzyl group acts in much the same way using the $\pi$-system of the benzene ring for conjugation with the p-orbital in the transition state.

Question 70

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

For the given compounds, the correct order of increasing $pK_a$ value:

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

Answer: (a)

Solution

Acidic strength order: B > D > C > A > E Correct pKa Order: B < D < C < A < E

Question 71

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

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Both rhombic and monoclinic sulphur exist as $\mathrm{S}_8$ while oxygen exists as $\mathrm{O}_2$. Reason (R) : Oxygen forms $p\pi - p\pi$ multiple bonds with itself and other elements having small size and high electronegativity like $\mathrm{C}, \mathrm{N}$, which is not possible for sulphur. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. (A) is correct but (R) is not correct
  2. (A) is not correct but (R) is correct
  3. Both (A) and (R) are correct and (R) is the correct explanation of (A)
  4. Both (A) and (R) are correct but (R) is not the correct explanation of (A)

Answer: (a)

Question 72

Chemistry · Co-ordination Compounds · Single correct

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason ( R ). Assertion (A) : The total number of geometrical isomers shown by $[\mathrm{Co(en)}_2\mathrm{Cl}_2]^+$ complex ion is three. Reason (R): $[\mathrm{Co(en)}_2\mathrm{Cl}_2]^+$ complex ion has an octahedral geometry. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Both (A) and (R) are correct but ( R ) is not the correct explanation of (A)
  2. is not correct but ( R ) is correct
  3. Both (A) and ( R ) are correct and ( R ) is the correct explanation of (A)
  4. is correct but ( R ) is not correct

Answer: (b)

Solution

[$\mathrm{Co(en)_2Cl_2}$]^+ has octahedral geometry with two geometrical isomers.

Question 73

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

The electronic configuration of Cu(II) is $3 \, d^9$ whereas that of Cu(I) is $3 \, d^{10}$. Which of the following is correct?

  1. Stability of Cu(I) and Cu(II) depends on nature of copper salts
  2. Cu(II) is more stable
  3. Cu(I) and Cu(II) are equally stable
  4. Cu(II) is less stable

Answer: (b)

Solution

Cu(II) is more stable than Cu(I) because hydration energy of $\mathrm{Cu^{+2}}$ ion compensates $\mathrm{IE_2}$ of Cu.

Question 74

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

What is the structure of C

Answer: (a)

Solution

Question 75

Chemistry · Structure of Atom · Single correct

Compare the energies of following sets of quantum numbers for multielectron system. (A) $n = 4, \ l = 1$ (B) $n = 4, \ l = 2$ ($C$) $n = 3, \ l = 1$ (D) $n = 3, \ l = 2$ (E) $n = 4, \ l = 0$ Choose the correct answer from the options given below :

  1. (B) > (A) > ($C$) > (E) > (D)
  2. (E) < ($C$) < (D) < (A) < (B)
  3. (E) > ($C$) > (A) > (D) > (B)
  4. ($C$) < (E) < (D) < (A) < (B)

Answer: (d)

Solution

Energy level can be determined by comparing $(n + \ell)$ values. (A) $n = 4$, $\ell = 1 \Rightarrow (n + \ell) = 5$ (B) $n = 4$, $\ell = 2 \Rightarrow (n + \ell) = 6$ (C) $n = 3$, $\ell = 1 \Rightarrow (n + \ell) = 4$ (D) $n = 3$, $\ell = 2 \Rightarrow (n + \ell) = 5$ (E) $n = 4$, $\ell = 0 \Rightarrow (n + \ell) = 4$ For same value of $(n + \ell)$, orbital having higher value of $n$, will have more energy. (B) > (A) > (D) > (E) > ($C$)

Question 76

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Identify major product "X" formed in the following reaction:

Answer: (c)

Solution

This is Gattermann-Koch reaction. $$Benzene + CO + HCl \xrightarrow{AlCl_3, CuCl} Benzaldehyde$$

Question 77

Chemistry · Hydrocarbons · Single correct

Identify the product A and product B in the following set of reactions.

Answer: (d)

Solution

(1) Hydration Reaction: $$\mathrm{CH_3 - CH = CH_2 + H^+ \longrightarrow CH_3 - \overset{+}{CH} - CH_3}$$ (More stable) $$\mathrm{\overset{+}{CH_3 - CH - CH_3} + H_2O \longrightarrow (CH_3 - \overset{\underset{OH}{|}}{CH} - CH_3) + H^+}$$ (A) (2) Hydroboration Oxidation Reaction: $$\mathrm{3CH_3 - CH = CH_2 + B_2H_6 \xrightarrow{THF} 2(CH_3CH_2CH_2)_3B}$$ $$\mathrm{(CH_3CH_2CH_2)_3B + 3H_2O_2 \xrightarrow{OH^-} 3CH_3CH_2CH_2OH + H_3BO_3}$$ (B)

Question 78

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

On reaction of Lead Sulphide with dilute nitric acid which of the following is not formed?

  1. Nitric oxide
  2. Nitrous oxide
  3. Lead nitrate
  4. Sulphur

Answer: (b)

Solution

The reaction is given by: $$\mathrm{PbS + HNO_3 \rightarrow Pb(NO_3)_2 + NO + S + H_2O}$$ Nitrous oxide ($\mathrm{N_2O}$) is not formed during the reaction.

Question 79

Chemistry · Analytical Chemistry · Single correct

Identify the incorrect statements regarding primary standard of titrimetric analysis. \\ (A) It should be purely available in dry form. \\ (B) It should not undergo chemical change in air. \\ (C) It should be hygroscopic and should react with another chemical instantaneously and stoichiometrically. \\ (D) It should be readily soluble in water. \\ (E) $KMnO_4$ \& $NaOH$ can be used as primary standard. \\ Choose the correct answer from the options given below :

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

Answer: (b)

Solution

$KMnO_4$ and NaOH are secondary standards. Primary standards should not be hygroscopic.

Question 80

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

0.05M $\mathrm{CuSO}_4$ when treated with 0.01M $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ gives green colour solution of $\mathrm{Cu}_2\mathrm{Cr}_2\mathrm{O}_7$. The two solutions are separated as shown below: Due to osmosis:

  1. Molarity of $\mathrm{CuSO}_4$ solution is lowered.
  2. Molarity of $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ solution is lowered.
  3. Green colour formation observed on side Y.
  4. Green colour formation observed on side X.

Answer: (a)

Solution

Only solvent molecules are allowed to pass through the SPM. The diagram shows $\mathrm{K_2Cr_2O_7}$ and $\mathrm{CuSO_4}$ separated by a semipermeable membrane (SPM). The volume $V$ decreases on the left side and increases on the right side. The molarity $M$ increases on the left side and decreases on the right side.

Question 81

Chemistry · Thermodynamics · Numerical

The heat of solution of anhydrous $\mathrm{CuSO}_4$ and $\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}$ are $-70 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$ and $+12 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$ respectively. The heat of hydration of $\mathrm{CuSO}_4$ to $\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}$ is $-x \, \mathrm{kJ}$. The value of $x$ is _____ (nearest integer).

Answer: 82

Solution

From (1) & (2) $$-70 = x + 12$$ $$x = -82$$

Question 82

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

Given below are two statements : Statement I: The rate law for the reaction $A + B \rightarrow C$ is rate $(r) = k[A]^2[B]$. When the concentration of both $A$ and $B$ is doubled, the reaction rate is increased " $x$ " times. Statement II : The figure is showing "the variation in concentration against time plot" for a " $y$ " order reaction. The Value of $x + y$ is ______

Answer: 8

Solution

Given $r = K[A]^2[B]$. If concentrations are doubled, then $r' = K[2A]^2[2B]^1$. This gives $r' = 8r \Rightarrow x = 8$. Therefore, zero order, $y = 0$. Thus, $x + y = 8$.

Question 83

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

How many compounds among the following compounds show inductive, mesomeric as well as hyperconjugation effects?

Answer: 4

Solution

Question 84

Chemistry · Electrochemistry · Numerical

The standard reduction potentials at $298\ K$ for the following half cells are given below : $Cr_2O_7^{2-}+14H^{+}+6e^{-}\rightarrow2Cr^{3+}+7H_2O,\ E^{\circ}=1.33\ V$ $Fe^{3+}_{(aq)}+3e^{-}\rightarrow Fe \qquad E^{\circ}=-0.04\ V$ $Ni^{2+}_{(aq)}+2e^{-}\rightarrow Ni \qquad E^{\circ}=-0.25\ V$ $Ag^{+}_{(aq)}+e^{-}\rightarrow Ag \qquad E^{\circ}=0.80\ V$ $Au^{3+}_{(aq)}+3e^{-}\rightarrow Au \qquad E^{\circ}=1.40\ V$ Consider the given electrochemical reactions, The number of metal(s) which will be oxidized by $Cr_2O_7^{2-}$, in aqueous solution is ______

Answer: 3

Solution

Fe, Ni, Ag will be oxidized due to lower S.R.P.

Question 85

Chemistry · Thermodynamics · Numerical

When equal volume of 1M $HCl$ and 1M $H_2SO_4$ are separately neutralised by excess volume of 1M $NaOH$ solution. $x$ and $y$ kJ of heat is liberated respectively. The value of $y/x$ is

Answer: 2

Solution

Given the reactions: $$\mathrm{H^+ + OH^- \rightarrow H_2O \Rightarrow x}$$ $$\mathrm{2H^+ + 2OH^- \rightarrow 2H_2O \Rightarrow 2x = y}$$ We have $y/x = 2$.

Question 86

Chemistry · Some Basic Concepts of Chemistry · Numerical

Molarity (M) of an aqueous solution containing $x$ g of anhyd. $\mathrm{CuSO_4}$ in 500 mL solution at 32$^\circ$C is $2 \times 10^{-1}$ M. Its molality will be _____ $\times 10^{-3}$ m. (nearest integer). [Given density of the solution = 1.25 g/mL]

Answer: 164

Solution

Given $M_{sol}^l = v_{sol}^n \times d_{sol}^n$. $$= 500 \times 1.25 = 625 \, g$$ Mass of solute $(x) = 0.2 \times 0.5 \times 159.5$ $$= 15.95$$ $n_{solute} = 0.1,$ Mass of solvent $=$ Mass of solution $-$ Mass of solute $$= 625 - 15.95$$ $$= 609.05$$ $$m = \frac{0.1}{\frac{609.05}{1000}}$$ $$m = 0.164 = 164 \times 10^{-3}$$

Question 87

Chemistry · Chemical Bonding and Molecular Structure · Numerical

The total number of species from the following in which one unpaired electron is present, is _______ $\mathrm{N_2}$, $\mathrm{O_2}$, $\mathrm{C_2^-}$, $\mathrm{O_2^-}$, $\mathrm{O_2^{2-}}$, $\mathrm{H_2^+}$, $\mathrm{CN^-}$, $\mathrm{He_2^+}$

Answer: 4

Solution

One unpaired $e^-$ is present in: $\mathrm{C_2^-}$; $\mathrm{O_2^-}$; $\mathrm{H_2^+}$; $\mathrm{He_2^+}$.

Question 88

Chemistry · Co-ordination Compounds · Numerical

Number of ambidentate ligands among the following is $\mathrm{NO_2^-}$, $\mathrm{SCN^-}$, $\mathrm{C_2O_4^{2-}}$, $\mathrm{NH_3}$, $\mathrm{CN^-}$, $\mathrm{SO_3^{2-}}$, $\mathrm{H_2O}$.

Answer: 3

Solution

Ligands which have two different donor sites but at a time connects with only one donor site to central metal are ambidentate ligands. Ambidentate ligands are $\mathrm{NO_2^-}$; $\mathrm{SCN^-}$; $\mathrm{CN^-}$.

Question 89

Chemistry · Biomolecules · Numerical

Total number of essential amino acid among the given list of amino acids is Arginine, Phenylalanine, Aspartic acid, Cysteine, Histidine, Valine, Proline

Answer: 4

Solution

Essential Amino acids are: Arginine, Phenylalanine, Histidine, Valine

Question 90

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

Number of colourless lanthanoid ions among the following is ________ $\mathrm{Eu}^{3+}, \mathrm{Lu}^{3+}, \mathrm{Nd}^{3+}, \mathrm{La}^{3+}, \mathrm{Sm}^{3+}$

Answer: 2

Solution

$\mathrm{La}^{3+} - [\mathrm{Xe}]\,4f^0$ $\mathrm{Nd}^{3+} - [\mathrm{Xe}]\,4f^3$ $\mathrm{Sm}^{3+} - [\mathrm{Xe}]\,4f^5$ $\mathrm{Eu}^{3+} - [\mathrm{Xe}]\,4f^6$ $\mathrm{Lu}^{3+} - [\mathrm{Xe}]\,4f^{14}$ $\mathrm{La}^{3+}$ and $\mathrm{Lu}^{3+}$ do not show any colour because no unpaired electron is present.