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.
Register free to attempt this shift as a timed 180-minute test with instant scoring and chapter-wise analysis.
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?
$(-\\frac{1}{3}, 1, -1)$
$(-\\frac{1}{3}, -1, 1)$
$(-\\frac{1}{3}, 1, 1)$
$(-\\frac{1}{3}, -1, -1)$
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:
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:
$2x + 3y = 9$
$2x + 3y = -9$
$2x + 3y = -6$
$2x + 3y = 6$
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}$$\right$) Thus, 2x + 3y = 9
Question 4
Maths · Straight Lines and Pair of Straight Lines · Single correct
A ray of light coming from the point $P(1, 2)$ gets reflected from the point $Q$ on the $x$-axis and then passes through the point $R(4, 3)$. If the point $S(h, k)$ is such that $PQRS$ is a parallelogram, then $hk^2$ is equal to:
70
80
60
90
Answer: (a)
Solution
Image of $P$ with respect to the $x$-axis will be $P'(1, -2)$. The equation of the line joining $P'R$ will be $$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:
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:
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:
7
4
1
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 :
30
25
40
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:
15$\pi$
18$\pi$
6$\pi$
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:
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 :
32
40
24
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:
10
5
15
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:
73
62
51
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:
$25x^2 + 25y^2 - 2x + 2y - 60 = 0$
$5x^2 + 5y^2 - 4x + 2y - 12 = 0$
$5x^2 + 5y^2 - 4x - 2y - 12 = 0$
$25x^2 + 25y^2 - 20x + 2y - 60 = 0$
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:
$\frac{178}{\sqrt{563}}$
$\frac{187}{\sqrt{563}}$
$\frac{185}{\sqrt{563}}$
$\frac{179}{\sqrt{563}}$
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:
The solution of the differential equation $(x^2 + y^2) \, \mathrm{dx} - 5xy \, \mathrm{dy} = 0, y(1) = 0$, is :
$|x^2 - 2y^2|^6 = x$
$|x^2 - 4y^2|^6 = x$
$|x^2 - 4y^2|^5 = x^2$
$|x^2 - 2y^2|^5 = x^2$
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:
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:
$x^2 - 190x + 9466 = 0$
$x^2 - 180x + 9506 = 0$
$x^2 - 195x + 9506 = 0$
$x^2 - 195x + 9466 = 0$
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.
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
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 ________
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
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:
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:
8.8 m/s
10 m/s
9.2 m/s
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 :
$B_z = 60 \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \mathrm{T}$
$B_x = 60 \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{i} \, \mathrm{T}$
$B_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \mathrm{T}$
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 :
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
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 :
1400 nm
650 nm
875 nm
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:
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 :
$\frac{13 \varepsilon_0 A}{17 d}$
$\frac{11 \varepsilon_0 A}{18 d}$
$\frac{18 \varepsilon_0 A}{11 d}$
$\frac{11 \varepsilon_0 A}{20 d}$
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 :
8 : 1
5 : 3
4 : 3
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 :
$\mathrm{ML^2 \, T^{-2}}$
$\mathrm{M^0 \, L^2 \, T^{-2}}$
$\mathrm{MLT^{-2}}$
$\mathrm{M^0LT^{-2}}$
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:
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 $R_e = 6370 \, \mathrm{km}$):
10
12
9
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 :
Both Statement I and Statement II are true
Statement I is true but Statement II is false
Both Statement I and Statement II are false
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:
$\frac{m}{2\alpha^2} d$
$\frac{md}{2\alpha^2}$
$2 \, m \alpha^2 \, d$
$\frac{m \alpha^2 \, d}{2}$
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 ;
0.40$\Omega$
0.20$\Omega$
0.50$\Omega$
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 :
$W \cos \theta$
$\frac{W}{2}$
$W$
$W \sin \theta$
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:
$\frac{n}{(n+1)}$
$\frac{1}{(n+1)}$
$\frac{m}{(n+1)}$
$\frac{m}{n(n+1)}$
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:
increases
decreases
remains same
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 :
18$\Omega$
19$\Omega$
25$\Omega$
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:
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 : $\lambda = 600 \, \mathrm{nm}$, $d = 1.0 \, \mathrm{mm}$, $D = 1.0 \, \mathrm{m}$ )
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
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 _____ $\mu \mathrm{m}$.
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 )
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
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:
$A$: strong electrolyte ; $B$: weak electrolyte
$A$: weak electrolyte ; $B$: weak electrolyte
$A$: weak electrolyte ; $B$: strong electrolyte
$A$: 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 :
nature of compound and presence of impurity.
neither on nature of compound nor on the impurity present.
nature of compound only.
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:
Pyrrole > Piperidine > Pyridine
Pyrrole > Pyridine > Piperidine
Pyridine > Piperidine > Pyrrole
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?
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.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Relative stability of the contributing structures is :
(I) > (II) > (III)
(I) > (III) > (II)
(II) > (I) > (III)
(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 :
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Statement I is incorrect but Statement II is correct
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:
(A) is correct but $(R)$ is not correct
(A) is not correct but $(R)$ is correct
Both (A) and $(R)$ are correct but $(R)$ is not the correct explanation of (A)
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:
(B) < (D) < $(C)$ < (A) < (E)
(D) < (E) < $(C)$ < (B) < (A)
(E) < (D) < $(C)$ < (B) < (A)
(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 :
(A) is correct but (R) is not correct
(A) is not correct but (R) is correct
Both (A) and (R) are correct and (R) is the correct explanation of (A)
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 :
Both (A) and (R) are correct but ( R ) is not the correct explanation of (A)
is not correct but ( R ) is correct
Both (A) and ( R ) are correct and ( R ) is the correct explanation of (A)
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?
Stability of Cu(I) and Cu(II) depends on nature of copper salts
Cu(II) is more stable
Cu(I) and Cu(II) are equally stable
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 :
(B) > (A) > ($C$) > (E) > (D)
(E) < ($C$) < (D) < (A) < (B)
(E) > ($C$) > (A) > (D) > (B)
($C$) < (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.
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?
Nitric oxide
Nitrous oxide
Lead nitrate
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 :
(A) and (B) only
($C$) and (E) only
(B) and (E) only
($C$) 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:
Molarity of $\mathrm{CuSO}_4$ solution is lowered.
Molarity of $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ solution is lowered.
Green colour formation observed on side Y.
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, $x + y = 8$.
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
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.