JEE Main 25 June 2022 Shift 2 question paper with solutions
JEE Main 25 June 2022 Shift 2: 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 · Sets · Single correct
Let $A = \{ x \in \mathbb{R} : |x + 1| < 2 \}$ and $B = \{ x \in \mathbb{R} : |x - 1| \geq 2 \}$. Then which one of the following statements is NOT true?
$A - B = (-1, 1)$
$B - A = \mathbb{R} - (-3, 1)$
$A \cap B = (-3, -1]$
$A \cup B = \mathbb{R} - [1, 3]$
Answer: (b)
Solution
Given sets A and B: A: $x \in (-3, 1)$ B: $x \in (-\infty, -1] \cup [3, \infty)$ The set difference $B - A$ is calculated as: $$B - A = (-\infty, -3] \cup [3, \infty) = \mathbb{R} - (-3, 3)$$
Question 2
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $a,b \in \mathbb{R}$ be such that the equation $ax^2 - 2bx + 15 = 0$ has a repeated root $\alpha$. If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2bx + 21 = 0$, then $\alpha^2 + \beta^2$ is equal to:
The coefficient of $x^{101}$ in the expression $$(5+x)^{500} + x(5+x)^{499} + x^2(5+x)^{498} + \ldots x^{500},$$ $x > 0$, is
$^{501}C_{101}(5)^{399}$
$^{501}C_{101}(5)^{400}$
$^{501}C_{100}(5)^{400}$
$^{500}C_{101}(5)^{399}$
Answer: (a)
Solution
The expression is $(5+x)^{500} + x(5+x)^{499} + x^2(5+x)^{498} + \ldots + x^{500}$. This can be rewritten as: $$\frac{(5+x)^{501} - x^{501}}{(5+x) - x} = \frac{(5+x)^{501} - x^{501}}{5}.$$ The coefficient of $x^{101}$ in the given expression is: $$\frac{{501 \choose 101} 5^{400}}{5} = {501 \choose 101} 5^{399}.$$
Question 8
Maths · Sequences and Series · Single correct
The sum $1 + 2 \cdot 3 + 3 \cdot 3^2 + \ldots + 10 \cdot 3^9$ is equal to
Maths · Three Dimensional Geometry · Single correct
Let P be the plane passing through the intersection of the planes $$ \vec{r} \cdot \left( \hat{i} + 3\hat{j} - \hat{k} \right) = 5 $$ and $$ \vec{r} \cdot \left( 2\hat{i} - \hat{j} + \hat{k} \right) = 3 $$, and the point $ (2,1,-2) $. Let the position vectors of the points X and Y be $\hat{i}$ - 2$\hat{j}$ + 4$\hat{k}$ and $5\hat{i}$ - $\hat{j}$ + 2$\hat{k}$ respectively. Then the points
X and X + Y are on the same side of P
Y and Y - X are on the opposite sides of P
X and Y are on the opposite sides of P
X + Y and X - Y are on the same side of P
Answer: (c)
Solution
Given $\mathbf{P}_1 + \lambda \mathbf{P}_2 = 0$. This implies $$(x + 3y - z - 5) + \lambda (2x - y + z - 3) = 0.$$ The point $(2, 1, -2)$ lies on this plane. Therefore, $\lambda = 1$ implies the plane is $3x + 2y - 8 = 0$.
Question 10
Maths · Conic Sections · Single correct
A circle touches both the y-axis and the line $x + y = 0$. Then the locus of its center is
$y = \sqrt{2}x$
$x = \sqrt{2}y$
$y^2 - x^2 = 2xy$
$x^2 - y^2 = 2xy$
Answer: (d)
Solution
Let $(h, k)$ be the centre of the circle. $$\left| \frac{h-k}{\sqrt{2}} \right| = |h|$$ $$k^2 - h^2 + 2hk = 0$$ Therefore, the equation of the locus is $y^2 - x^2 + 2xy = 0$.
Question 11
Maths · Applications of Derivatives · Single correct
Water is being filled at the rate of $1 \, \mathrm{cm}^3/\mathrm{sec}$ in a right circular conical vessel (vertex downwards) of height $35 \, \mathrm{cm}$ and diameter $14 \, \mathrm{cm}$. When the height of the water level is $10 \, \mathrm{cm}$, the rate (in $\mathrm{cm}^2/\mathrm{sec}$) at which the wet conical surface area of the vessel increases is
5
$\frac{\sqrt{21}}{5}$
$\frac{\sqrt{26}}{5}$
$\frac{\sqrt{26}}{10}$
Answer: (c)
Solution
From figure $\frac{r}{h} = \frac{7}{35} \Rightarrow h = 5r$. Given $\frac{dV}{dt} = 1 \Rightarrow \frac{d}{dt} \left( \frac{\pi r^2 h}{3} \right) = 1$. $$\Rightarrow \frac{d}{dt} \left( \frac{5\pi}{3} r^3 \right) = 1 \Rightarrow r^2 \frac{dr}{dt} = \frac{1}{5\pi}$$ Let wet conical surface area $= S$. $$= \pi r \ell = \pi r \sqrt{h^2 + r^2}$$ $$= \sqrt{26} \pi r^2 \Rightarrow \frac{dS}{dt} = 2 \sqrt{26} \pi r \frac{dr}{dt}$$ When $h = 10$ then $r = 2$ $$\Rightarrow \frac{dS}{dt} = \frac{2 \sqrt{26}}{10}$$
Question 12
Maths · Integrals · Single correct
If $b_n = \int_0^{\frac{\pi}{2}} \frac{\cos^2 nx}{\sin x} \, dx$, $n \in \mathbb{N}$, then
$b_3 - b_2, \ b_4 - b_3, \ b_5 - b_4$ are in an A.P. with common difference $-2$
$\frac{1}{b_3 - b_2}, \ \frac{1}{b_4 - b_3}, \ \frac{1}{b_5 - b_4}$ are in an A.P. with common difference $2$
$b_3 - b_2, \ b_4 - b_3, \ b_5 - b_4$ are in a G.P.
$\frac{1}{b_3 - b_2}, \ \frac{1}{b_4 - b_3}, \ \frac{1}{b_5 - b_4}$ are in an A.P. with common difference $-2$
Answer: (d)
Solution
Given $$b_n = \int_0^{\pi/2} \frac{1 + \cos 2nx}{\sin x} \, dx$$ We have $$b_{n+1} - b_n = \int_0^{\pi/2} \frac{\cos^2(n+1)x - \cos^2 nx}{\sin x} \, dx$$ This simplifies to $$= \int_0^{\pi/2} \frac{-\sin(2n+1)x \sin x}{\sin x} \, dx$$ Which further simplifies to $$= \left( \frac{\cos(2n+1)x}{2n+1} \right)_0^{\pi/2} = \frac{-1}{2n+1}$$ Thus, $$\frac{1}{b_3 - b_2}, \frac{1}{b_4 - b_3}, \frac{1}{b_5 - b_4}$$ are in A.P. with c.d. = -2
Question 13
Maths · Differential Equations · Single correct
If $y = y(x)$ is the solution of the differential equation $2x^2 \frac{dy}{dx} - 2xy + 3y^2 = 0$ such that $y(e) = \frac{e}{3}$, then $y(1)$ is equal to
$\frac{1}{3}$
$\frac{2}{3}$
$\frac{3}{2}$
3
Answer: (b)
Solution
Given \[ \frac{dy}{dx}-\frac{y}{x}=-\frac{3}{2}\left(\frac{y}{x}\right)^2 \] Let \[ y=vx \] Then \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute into the equation: \[ v+x\frac{dv}{dx}-v=-\frac{3}{2}v^2 \] \[ x\frac{dv}{dx}=-\frac{3}{2}v^2 \] \[ \frac{dv}{v^2}=-\frac{3\,dx}{2x} \] Integrating: \[ -\frac{1}{v}=-\frac{3}{2}\ln|x|+C \] Since \[ v=\frac{y}{x}, \] we get \[ -\frac{x}{y}=-\frac{3}{2}\ln|x|+C \] When \[ x=e,\qquad y=\frac{e}{3}, \] \[ -\frac{e}{e/3}=-\frac{3}{2}\ln e+C \] \[ -3=-\frac{3}{2}+C \] \[ C=-\frac{3}{2} \] Therefore, \[ -\frac{x}{y}=-\frac{3}{2}\ln|x|-\frac{3}{2} \] When \[ x=1, \] \[ -\frac{1}{y}=-\frac{3}{2}\ln 1-\frac{3}{2} \] \[ -\frac{1}{y}=-\frac{3}{2} \] \[ y=\frac{2}{3} \]
Question 14
Maths · Applications of Derivatives · Single correct
If the angle made by the tangent at the point $(x_0,y_0)$ on the curve $x=12(t+\sin t\cos t)$, $y=12(1+\sin t)^2$, $0<t<\frac{\pi}{2}$, with the positive x-axis is $\frac{\pi}{3}$, then $y_0$ is equal to
6(3 + 2$\\sqrt{2}$)
3(7 + 4$\\sqrt{3}$)
27
48
Answer: (c)
Solution
Question 15
Maths · Trigonometric Functions · Single correct
The value of $2\sin(12^\circ) - \sin(72^\circ)$ is:
$\frac{\sqrt{5}(1-\sqrt{3})}{4}$
$\frac{1-\sqrt{5}}{8}$
$\frac{\sqrt{3}(1-\sqrt{5})}{2}$
$\frac{\sqrt{3}(1-\sqrt{5})}{4}$
Answer: (d)
Solution
Given $\sin 12^\circ + \sin 12^\circ - \sin 72^\circ$. This is equal to $\sin 12^\circ - 2 \cos 42^\circ \sin 30^\circ$. Simplifying further, we have $\sin 12^\circ - \sin 48^\circ$. This can be rewritten as $-2 \cos 30^\circ \sin 18^\circ$. Substituting the values, we get $$-2 \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{5} - 1}{4}$$ Finally, this simplifies to $$\frac{\sqrt{3}}{4} (1 - \sqrt{5})$$
Question 16
Maths · Probability · Single correct
A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark $n$ is $\frac{1}{n}$. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is
$\frac{7}{2^{11}}$
$\frac{7}{2^{12}}$
$\frac{3}{2^{10}}$
$\frac{13}{2^{12}}$
Answer: (d)
Solution
Given $P(n) = \frac{1}{n}$. $P(2) = \frac{1}{2}$, $P(8) = \frac{1}{8}$, $P(4) = \frac{1}{4}$, $P(16) = \frac{1}{16}$, $P(32) = \frac{2}{32}$. Possible cases: 16, 16, 16 and 32, 8, 8 Probability = $$\frac{1}{16^3} + \frac{2}{32} \times \frac{1}{8} \times 3 = \frac{13}{16^3}$$
Question 17
Maths · Mathematical Reasoning · Single correct
The negation of the Boolean expression $((\sim q) \land p) \Rightarrow ((\sim p) \lor q)$ is logically equivalent to
$p \Rightarrow q$
$q \Rightarrow p$
$\sim (p \Rightarrow q)$
$\sim (q \Rightarrow p)$
Answer: (c)
Solution
Given $\sim p \lor q \equiv p \rightarrow q$. $\sim q \land p \equiv \sim (p \rightarrow q)$. Negation of $\sim (p \rightarrow q) \rightarrow (p \rightarrow q)$ is $\sim (p \rightarrow q) \land (\sim (p \rightarrow q))$ i.e. $\sim (p \rightarrow q)$.
Question 18
Maths · Conic Sections · Single correct
If the line $y = 4 + kx$, $k > 0$, is the tangent to the parabola $y = x - x^2$ at the point $P$ and $V$ is the vertex of the parabola, then the slope of the line through $P$ and $V$ is:
\frac{3}{2}
\frac{26}{9}
\frac{5}{2}
\frac{23}{6}
Answer: (c)
Solution
Slope of tangent at P = Slope of line AP $$y'\big|_P = 1 - 2\alpha = \frac{\alpha - \alpha^2 - 4}{\alpha}$$ Solving $\alpha = -2 \Rightarrow \mathrm{P}(-2, -6)$ Slope of PV = $\frac{5}{2}$
Question 19
Maths · Inverse Trigonometric Functions · Single correct
The value of $\tan^{-1} \left( \frac{\cos \left( \frac{15\pi}{4} \right) - 1}{\sin \left( \frac{\pi}{4} \right)} \right)$ is equal to
$-\frac{\pi}{4}$
$-\frac{\pi}{8}$
$-\frac{5\pi}{12}$
$-\frac{4\pi}{9}$
Answer: (b)
Solution
Given $$\tan^{-1} \left[ \frac{\cos \left( 4\pi - \frac{\pi}{4} \right) - 1}{\sin \frac{\pi}{4}} \right]$$ This simplifies to $$\tan^{-1} \left( \frac{\cos \frac{\pi}{4} - 1}{\sin \frac{\pi}{4}} \right)$$ Further simplifying, we have $$\tan^{-1} \left( \frac{1 - \sqrt{2}}{1} \right) = -\frac{\pi}{8}$$
Question 20
Maths · Conic Sections · Single correct
The line $y = x + 1$ meets the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ at two points P and Q. If $r$ is the radius of the circle with $PQ$ as diameter then $(3r)^2$ is equal to
20
12
11
8
Answer: (a)
Solution
Ellipse $x^2 + 2y^2 = 4$. Line $y = x + 1$. Point of intersection $x^2 + 2(x+1)^2 = 4$. $$3x^2 + 4x - 2 = 0$$ $$|x_1 - x_2| = \frac{\sqrt{40}}{3}$$ $$AB = 2r = |x_1 - x_2| \sqrt{1 + m^2},$$ where $m$ is the slope of the given line. $$AB = \frac{\sqrt{40}}{3} \sqrt{1 + 1}$$ $$2r = \frac{\sqrt{80}}{3} \implies r = \frac{\sqrt{80}}{6}$$ $$(3r)^2 = \left(3 \times \frac{\sqrt{80}}{6}\right)^2 = \frac{80}{4} = 20$$
Question 21
Maths · Matrices · Numerical
Let $A = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix}$. Then the number of elements in the set $\{(n, m) : n, m \in \{1, 2, \ldots, 10\}$ and $nA^n + mB^m = I\}$ is ____
Answer: 1
Solution
Given $A^2 = A$ and $B^2 = B$. Therefore equation $nA^n + mB^m = I$ becomes $nA + mB = I$, which gives $m = n = 1$. Only one set possible.
Question 22
Maths · Continuity and Differentiability · Numerical
Let $f(x) = [2x^2 + 1]$ and $g(x) = \begin{cases} 2x - 3, & x < 0 \\ 2x + 3, & x \geq 0 \end{cases}$, where $[t]$ is the greatest integer $\leq t$. Then, in the open interval $(-1, 1)$, the number of points where $fog$ is discontinuous is equal to
Answer: 62
Solution
Given $f(g(x)) = \left[ 2g^2(x) \right] + 1$. $$= \begin{cases} \left[ 2(2x - 3)^2 \right] + 1; & x < 0 \\ \left[ 2(2x + 3)^2 \right] + 1; & x \geq 0 \end{cases}$$ Therefore, $f \circ g$ is discontinuous whenever $2(2x - 3)^2$ or $2(2x + 3)^2$ belongs to integer except $x = 0$. Thus, 62 points of discontinuity.
Question 23
Maths · Integrals · Numerical
The value of $b > 3$ for which $$12 \int_{3}^{b} \frac{1}{(x^2 - 1)(x^2 - 4)} \, dx = \log_e \left( \frac{49}{40} \right),$$ is equal to
Answer: 6
Solution
Given $$\frac{12}{3} \left[ \int_{3}^{b} \left( \frac{1}{x^2 - 4} - \frac{1}{x^2 - 1} \right) \, \mathrm{dx} \right] = \log \frac{49}{40}$$ We have $$\frac{12}{3} \left[ \frac{1}{4} \ln \left| \frac{x-2}{x+2} \right| - \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| \right]_{3}^{b} = \log \frac{49}{40}$$ This simplifies to $$\ln \frac{(b-2)(b+1)^2}{(b+2)(b-1)^2} = \ln \frac{49}{50}$$ Thus, $$b = 6$$
Question 24
Maths · Binomial Theorem · Numerical
If the sum of the coefficients of all the positive even powers of $x$ in the binomial expansion of $$\left(2x^3 + \frac{3}{x}\right)^{10}$$ is $5^{10} - \beta \cdot 3^9$, then $\beta$ is equal to
Answer: 83
Solution
Given $$T_{r+1} = \binom{10}{r} (2x^3)^{10-r} \left( \frac{3}{x} \right)^r$$ This simplifies to: $$= \binom{10}{r} 2^{10-r} 3^r x^{30-4r}$$ Put $r = 0, 1, 2, \ldots, 7$ and we get $\beta = 83$.
Question 25
Maths · Statistics · Fill in the blank
If the mean deviation about the mean of the numbers 1, 2, 3, ....., n, where n is odd, is $\frac{5(n+1)}{n}$, then n is equal to _____
Answer: 21
Solution
Mean deviation about mean of first n natural numbers is $$\frac{n^2 - 1}{4n}$$. Therefore, $$n = 21$$.
Question 26
Maths · Vector Algebra · Numerical
Let $\vec{b} = \hat{i} + \hat{j} + \lambda \hat{k}, \lambda \in \mathbb{R}$. If $\vec{a}$ is a vector such that $$\vec{a} \times \vec{b} = 13 \hat{i} - \hat{j} - 4 \hat{k}$$ and $$\vec{a} \cdot \vec{b} + 21 = 0,$$ then $$(\vec{b} - \vec{a}) \cdot (\hat{k} - \hat{j}) + (\vec{b} + \vec{a}) \cdot (\hat{i} - \hat{k})$$ is equal to
The total number of three-digit numbers, with one digit repeated exactly two times, is
Answer: 243
Solution
If 0 taken twice then ways = 9 If 0 taken once then $^9C_1 \times 2 = 18$ If 0 not taken then $^9C_1 \cdot ^8C_1 \cdot 3 = 216$ Total = 243
Question 28
Maths · Applications of Derivatives · Numerical
Let $f(x) = \left| (x-1)(x^2 - 2x - 3) \right| + x - 3, \ x \in \mathbb{R}$. If $m$ and $M$ are respectively the number of points of local minimum and local maximum of $f$ in the interval $(0, 4)$, then $m + M$ is equal to ____
Answer: 3
Solution
Given $$f(x) = \begin{cases} \left( x^2 - 1 \right)(x-3) + (x-3), & x \in (0,1) \cup (3,4) \\ -\left( x^2 - 1 \right)(x-3) + (x-3), & x \in [1,3] \end{cases}$$ Differentiating, we have $$f'(x) = \begin{cases} 3x^2 - 6x, & x \in (0,1) \cup (3,4) \\ -3x^2 + 6x + 2, & x \in (1,3) \end{cases}$$ The function $f(x)$ is non-derivable at $x = 1$ and $x = 3$. Also, $f'(x) = 0$ at $x = 1 + \frac{\sqrt{5}}{3}$, which implies $m + M = 3$.
Question 29
Maths · Conic Sections · Numerical
Let the eccentricity of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $\frac{5}{4}$. If the equation of the normal at the point $\left( \frac{8}{\sqrt{5}}, \frac{12}{5} \right)$ on the hyperbola is $8\sqrt{5}x + \beta y = \lambda$, then $\lambda - \beta$ is equal to
Let $l_1$ be the line in $xy$-plane with $x$ and $y$ intercepts $\frac{1}{8}$ and $\frac{1}{4\sqrt{2}}$ respectively, and $l_2$ be the line in $zx$-plane with $x$ and $z$ intercepts $-\frac{1}{8}$ and $-\frac{1}{6\sqrt{3}}$ respectively. If $d$ is the shortest distance between the line $l_1$ and $l_2$, then $d^{-2}$ is equal to
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. If A and B reached the maximum height $h_1$ and $h_2$ respectively, then $R = 4\sqrt{h_1 h_2}$ Reason R: Product of said heights. $$h_1 h_2 = \left( \frac{u^2 \sin^2 \theta}{2g} \right) \cdot \left( \frac{u^2 \cos^2 \theta}{2g} \right)$$ Choose the CORRECT answer :
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is NOT the correct explanation of A.
Physics · Motion in a Straight Line · Single correct
Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by $X_P(t) = \alpha t + \beta t^2$ and $X_Q(t) = ft - t^2$. At what time, both the buses have same velocity?
$\frac{\alpha - f}{1 + \beta}$
$\frac{\alpha + f}{2(\beta - 1)}$
$\frac{\alpha + f}{2(1 + \beta)}$
$\frac{f - \alpha}{2(1 + \beta)}$
Answer: (d)
Solution
Given $$X_P(t) = \alpha t + \beta t^2$$ $$X_Q = ft - t^2$$ The velocities are $$V_P(t) = \alpha + 2\beta t$$ $$V_Q = f - 2t$$ Setting the velocities equal, $$V_P = V_Q$$ This gives $$\alpha + 2\beta t = f - 2t$$ Solving for $t$, $$t = \frac{f - \alpha}{2\beta + 2}$$
Question 33
Physics · Laws of Motion · Single correct
A disc with a flat small bottom beaker placed on it at a distance $R$ from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $\omega$. The coefficient of static friction between the bottom of the beaker and the surface of the disc is $\mu$. The beaker will revolve with the disc if:
$R \leq \frac{\mu g}{2\omega^2}$
$R \leq \frac{\mu g}{\omega^2}$
$R \geq \frac{\mu g}{2\omega^2}$
$R \geq \frac{\mu g}{\omega^2}$
Answer: (b)
Solution
For the beaker to move with the disc, the static friction force is given by $$f_s = m \omega^2 R$$ We know that $$f_s \leq f_{s_{max}}$$ Therefore, $$m \omega^2 R \leq \mu m g$$ Simplifying, we get $$R \leq \frac{\mu g}{\omega^2}$$
Question 34
Physics · Thermal Properties of Matter · Single correct
A solid metallic cube having total surface area 24 $\mathrm{m}^2$ is uniformly heated. If its temperature is increased by 10$^\circ\mathrm{C}$, calculate the increase in volume of the cube (Given: $\alpha = 5.0 \times 10^{-4} \, ^\circ\mathrm{C}^{-1}$)
$2.4 \times 10^6\,\mathrm{cm^3}$
$1.2 \times 10^5\,\mathrm{cm^3}$
$6.0 \times 10^4\,\mathrm{cm^3}$
$4.8 \times 10^5\,\mathrm{cm^3}$
Answer: (b)
Solution
Increase in volume $\Delta V = \gamma V_0 \Delta T$. $\gamma = 3 \alpha$ So $\Delta V = (3 \alpha) V_0 \Delta T$. Total surface area $= 6a^2$, where $a$ is side length. $24 = 6a^2$ $a = 2 \, \mathrm{m}$ Volume $V_0 = (2)^3 = 8 \, \mathrm{m^3}$ $\Delta V = (3 \times 5 \times 10^{-4})(8) \times 10$ $= 1.2 \times 10^5 \, \mathrm{cm^3}$
Question 35
Physics · Thermal Properties of Matter · Single correct
A copper block of mass 5.0 kg is heated to a temperature of 500°C and is placed on a large ice block. What is the maximum amount of ice that can melt? [Specific heat of copper: 0.39 J g⁻¹ °C⁻¹ and latent heat of fusion of water : 335 J g⁻¹]
1.5 kg
5.8 kg
2.9 kg
3.8 kg
Answer: (c)
Solution
Heat given by block to get $0^\circ \mathrm{C}$ temperature $$\Delta Q_1 = 5 \times \left(0.39 \times 10^3\right) \times (500 - 0)$$ $$= 975 \times 10^3 \, \mathrm{J}$$ Heat absorbed by ice to melt mass $m$ $$\Delta Q_2 = m \times \left(335 \times 10^3\right) \, \mathrm{J}$$ $$\Delta Q_1 = \Delta Q_2$$ $$m \times \left(335 \times 10^3\right) = 975 \times 10^3$$ $$m = \frac{975}{335} = 2.910 \, \mathrm{kg}$$
Question 36
Physics · Kinetic Theory · Single correct
The ratio of specific heats ( $\frac{C_P}{C_V}$ ) in terms of degree of freedom (f) is given by:
( 1 + $\frac{f}{3}$ )
( 1 + $\frac{2}{f}$ )
( 1 + $\frac{f}{2}$ )
( 1 + $\frac{1}{f}$ )
Answer: (b)
Solution
Molar heat capacity at constant volume $C_V = \frac{fR}{2}$ where $f$ is degree of freedom. Molar heat capacity at constant pressure can be written as $C_P = R + C_V = R + \frac{fR}{2} = \left(1 + \frac{f}{2}\right) R$. So $$\frac{C_P}{C_V} = 1 + \frac{2}{f}$$
Question 37
Physics · Motion in a Plane · Single correct
For a particle in uniform circular motion, the acceleration $\vec{a}$ at any point $P(R, \theta)$ on the circular path of radius $R$ is (when $\theta$ is measured from the positive $x$-axis and $v$ is uniform speed):
Physics · Electrostatic Potential and Capacitance · Single correct
Two metallic plates form a parallel plate capacitor. The distance between the plates is $d$. A metal sheet of thickness $\frac{d}{2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?
Two cells of same emf but different internal resistances $r_1$ and $r_2$ are connected in series with a resistance $R$. The value of resistance $R$, for which the potential difference across second cell is zero, is
$r_2 - r_1$
$r_1 - r_2$
$r_1$
$r_2$
Answer: (a)
Solution
Given $$I = \frac{2E}{R + r_1 + r_2} .....(i)$$ But $V_A - V_B = E - Ir_2 = 0$ $$\Rightarrow I = \frac{E}{r_2} .....(ii)$$ Comparing values of $I$ from (i) and (ii) $$\frac{E}{r_2} = \frac{2E}{R + r_1 + r_2}$$ $$\Rightarrow R = r_2 - r_1$$
Question 40
Physics · Magnetism and Matter · Single correct
Given below are two statements: Statement – I : Susceptibilities of paramagnetic and ferromagnetic substances increase with decrease in temperature. Statement – II: Diamagnetism is a result of orbital motions of electrons developing magnetic moments opposite to the applied magnetic field. 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: (a)
Solution
According to Curie's law, magnetic susceptibility is inversely proportional to temperature for a fixed value of external magnetic field i.e. $\chi = \frac{C}{T}$. The same is applicable for ferromagnet and the relation is given as $\chi = \frac{C}{T - T_C}$ ($T_C$ is Curie temperature). Diamagnetism is due to non-cooperative behaviour of orbiting electrons when exposed to external magnetic field. Hence option (A).
Question 41
Physics · Moving Charges and Magnetism · Single correct
A long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to
$B$
$2B$
$4B$
$\frac{B}{2}$
Answer: (a)
Solution
Given $\mathbf{B}_1 = \mu_0 n I$. $\mathbf{B}_2 = \mu_0 \left( \frac{n}{2} \right) (2I)$. Therefore, $\mathbf{B}_1 = \mathbf{B}_2$.
Question 42
Physics · Alternating Current · Single correct
A sinusoidal voltage $V(t) = 210 \sin 3000t$ volt is applied to a series LCR circuit in which $L = 10 \, \mathrm{mH}$, $C = 25 \, \mu\mathrm{F}$ and $R = 100\Omega$. The phase difference $(\Phi)$ between the applied voltage and resultant current will be:
The electromagnetic waves travel in a medium at a speed of $2.0 \times 10^8 \, \mathrm{m/s}$. The relative permeability of the medium is $1.0$. The relative permittivity of the medium will be:
The interference pattern is obtained with two coherent light sources of intensity ratio $4 : 1$. And the ratio $\frac{I_{\text{max}} + I_{\text{min}}}{I_{\text{max}} - I_{\text{min}}}$ is $\frac{5}{x}$. Then, the value of $x$ will be equal to:
Physics · Ray Optics and Optical Instruments · Single correct
A light whose electric field vectors are completely removed by using a good Polaroid, allowed to incident on the surface of the prism at Brewster’s angle. Choose the most suitable option for the phenomenon related to the prism.
Reflected and refracted rays will be perpendicular to each other
Wave will propagate along the surface of prism
No refraction, and there will be total reflection of light.
No reflection and there will be total transmission of light.
Answer: (d)
Solution
But as the incident light electric field vectors are completely removed so there will be no reflection and there will be total transmission of light, explained by an experiment in NCERT. [Reference NCERT Part-2 Pg-380, (A special case of total transmission)] Note: Since direction of polarization is not mentioned hence most suitable option (D) corresponding to case in which electric field is absent perpendicular to plane consisting incident and normal.
Question 46
Physics · Dual Nature of Radiation and Matter · Single correct
A proton, a neutron, an electron and an $\alpha$-particle have same energy. If $\lambda_p, \lambda_n, \lambda_e$ and $\lambda_\alpha$ are the de Broglie's wavelengths of proton, neutron, electron and $\alpha$ particle respectively, then choose the correct relation from the following:
The wavelength $\lambda$ is given by $$\lambda = \frac{h}{\sqrt{2Em}}$$ where $h$ is Planck's constant, $E$ is energy, and $m$ is mass. Therefore, $$\lambda \propto \frac{1}{\sqrt{m}}$$ This implies that the wavelength is inversely proportional to the square root of the mass. Thus, $$\lambda_e > \lambda_p > \lambda_n > \lambda_\alpha$$
Question 47
Physics · Atoms · Single correct
Which of the following figure represents the variation of $\ln \left( \frac{R}{R_0} \right)$ with $\ln A$ (If $R =$ radius of a nucleus and $A =$ its mass number)
Answer: (b)
Solution
Given $R = R_0 A^{\frac{1}{3}}$. Taking the natural logarithm on both sides, we have $$\ln \frac{R}{R_0} = \frac{1}{3} \ln A.$$ The graph of $\ln \frac{R}{R_0}$ versus $\ln A$ is a straight line.
Question 48
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Identify the logic operation performed by the given circuit:
AND gate
OR gate
NOR gate
NAND gate
Answer: (a)
Solution
The expression is simplified as follows: $$= \overline{\left[ \overline{A + A} + \overline{B + B} \right]}$$ Using De Morgan's Law: $$Y = \overline{\overline{A} + \overline{B}}$$ Therefore, the result is: $$Y = AB$$
Question 49
Physics · Communication Systems · Single correct
Match List I with List II Choose the correct answer from the following options :
A –IV, B-III, C-II, D-I
A-I, B-IV, C-II, D-III
A –IV, B-II, C-III, D-I
A-I, B-II, C-III, D-IV
Answer: (b)
Solution
Question based on the theory given in NCERT.
Question 50
Physics · Current Electricity · Single correct
If n represents the actual number of deflections in a converted galvanometer of resistance G and shunt resistance S. Then the total current I when its figure of merit is K will be:
For $z = a^2 x^3 y^{\frac{1}{2}}$, where $'a'$ is a constant. If percentage error in measurement of $'x'$ and $'y'$ are $4\%$ and $12\%$, respectively, then the percentage error for $'z'$ will be $\%$.
A curved in a level road has a radius 75m. The maximum speed of a car turning this curved road can be 30 m/s without skidding. If radius of curved road is changed to 48 m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be __ m/s.
Answer: 24
Solution
The maximum static friction force is given by $$f_{s max} = \frac{mv^2}{R}$$. The frictional force can also be expressed as $$\mu mg = \frac{mv^2}{R}$$. Solving for velocity, we have $$v = \sqrt{\mu R g}$$. The ratio of velocities is given by $$\frac{v_2}{v_1} = \sqrt{\frac{R_2}{R_1}}$$. Substituting the given values, $$\frac{v_2}{30} = \sqrt{\frac{48}{75}}$$. Solving for $v_2$, we find $$v_2 = 24 \, \mathrm{m/s}$$.
Question 53
Physics · Laws of Motion · Numerical
A block of mass 200 $\,$ $\mathrm{g}$ is kept stationary on a smooth inclined plane by applying a minimum horizontal force $F = \sqrt{x}N$ as shown in figure. The value of $x =$ .
Answer: 12
Solution
Given $mg = 2N$. The forces acting on the block are resolved into components along the incline. The equation is given by: $$\sqrt{x} \cdot \frac{1}{2} = \frac{2\sqrt{3}}{2}$$ Solving for $x$, we find: $$x = 12$$
Question 54
Physics · System of Particles and Rotational Motion · Numerical
Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows: $I_1$ = M.I. of solid sphere about its diameter $I_2$ = M.I. of solid cylinder about its axis $I_3$ = M.I. of solid circular disc about its diameter $I_4$ = M.I. of thin circular ring about its diameter If $2(I_2 + I_3) + I_4 = x \cdot I_1$ then the value of $x$ will be
Answer: 5
Solution
Given $$I_1 = \frac{2}{5} M (2R)^2 = \frac{8}{5} MR^2$$ $$I_2 = \frac{1}{2} M (2R)^2 = 2MR^2$$ $$I_3 = \frac{M (2R)^2}{4} = MR^2$$ $$I_4 = \frac{M (2R)^2}{2} = 2MR^2$$ The equation is $$2(I_2 + I_3) + I_4 = x \, I_1$$ Substituting the values, we get $$8MR^2 = x \frac{8}{5} MR^2$$ Solving for $x$, we find $$x = 5$$
Question 55
Physics · Gravitation · Numerical
Two satellites $S_1$ and $S_2$ are revolving in circular orbits around a planet with radius $R_1 = 3200 \, \mathrm{km}$ and $R_2 = 800 \, \mathrm{km}$ respectively. The ratio of speed of satellite $S_1$ to the speed of satellite $S_2$ in their respective orbits would be $\frac{1}{x}$ where $x =$
Answer: 2
Solution
Given $V = \frac{GM}{r}$, it follows that $\frac{V_1}{V_2} = \sqrt{\frac{800}{3200}} = \frac{1}{2}$.
Question 56
Physics · Kinetic Theory · Numerical
When a gas filled in a closed vessel is heated by raising the temperature by 1°C, its pressure increase by 0.4$\%$. The initial temperature of the gas is_____ K.
Answer: 250
Solution
Given the equation $pV = nRT$. The change in pressure and temperature is given by $\Delta P \cdot V = nR \Delta T$. Therefore, $$\frac{\Delta P}{P} = \frac{\Delta T}{T} = \frac{0.4}{100}$$ which implies $$T = \frac{100 \times 1}{0.4} = 250 \, \mathrm{K}$$
Question 57
Physics · Electric Charges and Fields · Numerical
27 identical drops are charged at 22V each. They combine to form a bigger drop. The potential of the bigger drop will be ____ V.
The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ____%.
In a series LCR circuit, the inductance, capacitance and resistance are $L = 100 \, \mathrm{mH}$, $C = 100 \, \mu \mathrm{F}$ and $R = 10 \, \Omega$ respectively. They are connected to an AC source of voltage $220 \, \mathrm{V}$ and frequency of $50 \, \mathrm{Hz}$. The approximate value of current in the circuit will be ____ A.
In an experiment of CE configuration of n-p-n transistor, the transfer characteristics are observed as given in figure. If the input resistance is $200 \, \Omega$ and output resistance is $60 \, \Omega$ the voltage gain in this experiment will be
Answer: 15
Solution
Voltage Gain = $\frac{I_C}{I_B}$ $\times$ $\frac{R_0}{R_I}$ = $\frac{10 \times 10^{-3}}{200 \times 10^{-6}}$ $\times$ $\frac{60}{200}$ = 15
Chemistry
Question 61
Chemistry · Structure of Atom · Single correct
The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is: [Given: The threshold frequency of platinum is $1.3 \times 10^{15} \, \mathrm{s}^{-1}$ and $h = 6.6 \times 10^{-34} \, \mathrm{J} \, \mathrm{s}$.]
$3.21 \times 10^{-14} \, \mathrm{J}$
$6.24 \times 10^{-16} \, \mathrm{J}$
$8.58 \times 10^{-19} \, \mathrm{J}$
$9.76 \times 10^{-20} \, \mathrm{J}$
Answer: (c)
Solution
Given $W = h \nu$ $$= 6.6 \times 10^{-34} \times 1.3 \times 10^{15}$$ $$= 8.58 \times 10^{-19} \, \mathrm{J}$$
Question 62
Chemistry · Thermodynamics · Single correct
At 25°C and 1 atm pressure, the enthalpy of combustion of benzene (l) and acetylene (g) are $-3268 \, \mathrm{kJ \, mol^{-1}}$ and $-1300 \, \mathrm{kJ \, mol^{-1}}$, respectively. The change in enthalpy for the reaction $$3 \, \mathrm{C_2H_2(g)} \rightarrow \mathrm{C_6H_6(l)}$$ is
$+324 \, \mathrm{kJ \, mol^{-1}}$
$+632 \, \mathrm{kJ \, mol^{-1}}$
$-632 \, \mathrm{kJ \, mol^{-1}}$
$-732 \, \mathrm{kJ \, mol^{-1}}$
Answer: (c)
Solution
Given $\Delta H=\sum\Delta H_{\mathrm{Combustion}}(\mathrm{Reactant})-\sum\Delta H_{\mathrm{Combustion}}(\mathrm{Product})$. $=3\times(-1300)-[3268]$ $=-632\ \mathrm{kJ\ mol^{-1}}$
Question 63
Chemistry · Solutions · Single correct
Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by $0.2^\circ\mathrm{C}$. The percentage association of solute A in water is: Given: Molar mass of A = $93\ \mathrm{g\,mol^{-1}}$ and molal depression constant of water = $1.86\ \mathrm{K\,kg\,mol^{-1}}$.
The $K_{sp}$ for bismuth sulphide $(\mathrm{Bi}_2\mathrm{S}_3)$ is $1.08 \times 10^{-73}$. The solubility of $\mathrm{Bi}_2\mathrm{S}_3$ in $\mathrm{mol} \, \mathrm{L}^{-1}$ at $298 \, \mathrm{K}$ is
$1.0 \times 10^{-15}$
$2.7 \times 10^{-12}$
$3.2 \times 10^{-10}$
$4.2 \times 10^{-8}$
Answer: (a)
Solution
Question 65
Chemistry · Chemistry in Everyday Life · Single correct
Match List I with List II. Choose the correct answer from the options given below:
A-II, B-III, C-I, D-IV
A-II, B-III, C-IV, D-I
A-III, B-II, C-IV, D-I
A-III, B-II, C-I, D-IV
Answer: (b)
Solution
Zymase naturally occurs in yeast. Diastase is found in malt. Urease is found in soyabean. Pepsin is found in stomach.
Question 66
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The correct order of electron gain enthalpies of Cl, F, Te and Po is
F < Cl < Te < Po
Po < Te < F < Cl
Te < Po < Cl < F
Cl < F < Te < Po
Answer: (b)
Solution
As Cl has maximum electron affinity among all elements. \begin{tabular}{|c|c|} \hline Element & $\Delta_{eg}H$ ($\mathrm{kJ\,mol}^{-1}$) \\ \hline F & $-328$ \\ \hline Cl & $-349$ \\ \hline Te & $-190$ \\ \hline Po & $-174$ \\ \hline \end{tabular}
Question 67
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Given below are two statements. Statement I: During electrolytic refining, blister copper deposits precious metals Statement II: In the process of obtaining pure copper by electrolysis method, copper blister is used to make the anode. 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: (a)
Solution
In the electro-refining, impure metal (here blister copper) is used as an anode while precious metal like Au, Pt get deposited as anode mud.
Question 68
Chemistry · Equilibrium · Single correct
Given below are two statements one is labelled as Assertion A and the other is labelled as Reason R: Assertion A : The amphoteric nature of water is explained by using Lewis acid/base concept. Reason R : Water acts as an acid with $NH_3$ and as a base with $H_2S$. In the light of the above statements choose the correct answer from the options given below :
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is NOT the correct explanation of A.
A is true but R is false.
A is false but R is true.
Answer: (d)
Solution
The first reaction is $\mathrm{H_2S} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{HS^-}$. Here, $\mathrm{H_2S}$ acts as an acid and $\mathrm{H_2O}$ acts as a base. The second reaction is $\mathrm{H_2O} + \mathrm{NH_3} \rightleftharpoons \mathrm{NH_4OH}$. Here, $\mathrm{H_2O}$ acts as an acid and $\mathrm{NH_3}$ acts as a base.
Question 69
Chemistry · Electrochemistry · Single correct
The correct order of reduction potentials of the following pairs is: A. $\mathrm{Cl}_2/\mathrm{Cl}^{-}$ B. $\mathrm{I}_2/\mathrm{I}^{-}$ C. $\mathrm{Ag}^{+}/\mathrm{Ag}$ D. $\mathrm{Na}^{+}/\mathrm{Na}$ E. $\mathrm{Li}^{+}/\mathrm{Li}$ Choose the correct answer from the options given below.
A > C > B > D > E
A > B > C > D > E
A > C > B > E > D
A > B > C > E > D
Answer: (a)
Solution
The standard electrode potentials are given as follows: $$E^\circ_{\mathrm{Cl_2/Cl^-}} = +1.36 \, \mathrm{V}$$ $$E^\circ_{\mathrm{I_2/I^-}} = +0.54 \, \mathrm{V}$$ $$E^\circ_{\mathrm{Ag^+/Ag}} = +0.80 \, \mathrm{V}$$ $$E^\circ_{\mathrm{Na^+/Na}} = -2.71 \, \mathrm{V}$$ $$E^\circ_{\mathrm{Li^+/Li}} = -3.05 \, \mathrm{V}$$
Question 70
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
The number of bridged oxygen atoms present in compound B formed from the following reactions is $$\mathrm{Pb(NO_3)_2} \xrightarrow{673 \, \mathrm{K}} \mathrm{A} + \mathrm{PbO} + \mathrm{O_2}$$ $$\mathrm{A} \xrightarrow{Dimerise} \mathrm{B}$$
0
1
2
3
Answer: (a)
Solution
Question 71
Chemistry · The d-and f-Block Elements · Single correct
The metal ion (in gaseous state) with lowest spin-only magnetic moment value is
Chemistry · Environmental Chemistry · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: Polluted water may have a value of BOD of the order of 17 ppm. Reason R: BOD is a measure of oxygen required to oxidise both the biodegradable and non-biodegradable organic material in water. In the light of the above statements, choose the most appropriate answer from the options given below.
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.
A is correct but R is not correct.
A is not correct but R is correct.
Answer: (c)
Solution
Clean water have BOD less than 5 ppm while highly polluted water has BOD greater or equal to 17 ppm. So, assertion is correct. BOD is measure of oxygen required to oxidise only bio-degradable organic matter. So, reason is false.
Question 73
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: A mixture contains benzoic acid and napthalene. The pure benzoic acid can be separated out by the use of benzene. Reason R: Benzoic acid is soluble in hot water. In the light of the above statements, choose the most appropriate answer from the options given below.
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is NOT the correct explanation of A.
A is true but R is false.
A is false but R is true.
Answer: (d)
Solution
Benzoic acid and Napthalene can be effectively separated by crystallization. Benzoic acid is soluble in hot water whereas Napthalene is insoluble. Hence assertion is incorrect but reason is correct.
Question 74
Chemistry · Amines · Single correct
During halogen test, sodium fusion extract is boiled with concentrated $\mathrm{HNO_3}$ to
remove unreacted sodium
decompose cyanide or sulphide of sodium
extract halogen from organic compound
maintain the pH of extract
Answer: (b)
Solution
Sodium fusion extract is boiled with concentrated $\mathrm{HNO_3}$ to remove sodium cyanide and sodium sulphide.
Question 75
Chemistry · Alcohols, Phenols and Ethers · Single correct
Amongst the following, the major product of the given chemical reaction is
Answer: (a)
Solution
The reaction involves the addition of $\mathrm{Br_2}$ to the alkene, forming a bromonium ion intermediate. The bromonium ion is then attacked by $\mathrm{CH_3OH}$, leading to the formation of the final product with a bromine and methoxy group added across the double bond.
Question 76
Chemistry · Haloalkanes and Haloarenes · Single correct
In the given reaction ‘A’ can be
benzyl bromide
bromobenzene
cyclohexyl bromide
methyl bromide
Answer: (b)
Solution
The reaction begins with the formation of phenylmagnesium bromide ($PhMgBr$) from bromobenzene using magnesium in THF. This Grignard reagent then reacts with the ester $Ph-C(O)-O-CH_3$. The carbonyl carbon is attacked by the nucleophilic carbon of the Grignard reagent, leading to the formation of an intermediate alkoxide. This intermediate then rearranges to form a ketone $Ph-C(O)-Ph$. Another equivalent of the Grignard reagent attacks the carbonyl carbon of the ketone, forming a tertiary alcohol after hydrolysis with water ($H_2O/H^+$). The final product is a triphenylmethanol derivative.
Question 77
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Which of the following conditions or reaction sequence will NOT give acetophenone as the major product?
Answer: (c)
Solution
Question 78
Chemistry · Alcohols, Phenols and Ethers · Single correct
The major product formed in the following reaction, is
Answer: (d)
Solution
The reaction begins with the protonation of the epoxide, forming a more reactive oxonium ion. The hydroxide ion then attacks the less hindered carbon, opening the epoxide ring. This results in the formation of a secondary alcohol. The final product is a cyclic ether after the elimination of a proton.
Question 79
Chemistry · Amines · Single correct
Which of the following ketone will NOT give enamine on treatment with secondary amines? [where $t-Bu$ is $-C(CH_3)_3$]
Answer: (c)
Solution
Question 80
Chemistry · Chemistry in Everyday Life · Single correct
An antiseptic dettol is a mixture of two compounds 'A' and 'B' where A has $6\pi$ electrons and B has $2\pi$ electrons. What is 'B'?
Bithionol
Terpineol
Chloroxylenol
Chloramphenicol
Answer: (b)
Solution
Dettol is a mixture of chloroxylenol (Compound A) and terpineol (Compound B). Chloroxylenol has $6\pi e^{-}$ and terpineol has $2\pi e^{-}$. Hence compound 'B' is terpineol.
Question 81
Chemistry · Some Basic Concepts of Chemistry · Numerical
A protein ‘A’ contains 0.30$\%$ of glycine (molecular weight 75). The minimum molar mass of the protein ‘A’ is ______ $\times$ $10^3$ $\mathrm{g \, mol^{-1}}$ [nearest integer]
Answer: 25
Solution
0.30$\%$ glycine is equal to 75. 1$\%$ $\rightarrow$ $\frac{75}{0.30}$ 100$\%$ $\rightarrow$ $\frac{75}{0.30}$ $\times$ 100 = 25000 $\mathrm{g}$
Question 82
Chemistry · States of Matter · Numerical
A rigid nitrogen tank stored inside a laboratory has a pressure of 30 atm at 06:00 am when the temperature is 27 $\degree$ C. At 03:00 pm, when the temperature is 45$\degree$ C, the pressure in the tank will be _________ atm. [nearest integer]
Answer: 32
Solution
Given $\frac{P_1}{T_1}=\frac{P_2}{T_2}$. $\frac{30}{300}=\frac{P_2}{318}$ $P_2=\frac{30}{300}\times318$ $=\frac{1}{10}\times318$ $=32$
Question 83
Chemistry · Chemical Bonding and Molecular Structure · Numerical
Amongst $\mathrm{BeF_2}$, $\mathrm{BF_3}$, $\mathrm{H_2O}$, $\mathrm{NH_3}$, $\mathrm{CCl_4}$ and $\mathrm{HCl}$, the number of molecules with non-zero net dipole moment is
Answer: 3
Solution
For $\mathrm{BeF_2}$, $\mathrm{BF_3}$, and $\mathrm{CCl_4}$, the net dipole moment $\mu_{net} = 0$. For $\mathrm{H_2O}$, $\mathrm{NH_3}$, and $\mathrm{HCl}$, the net dipole moment $\mu_{net} \neq 0$.
Question 84
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 $\mathrm{kPa}$ was 340 $\mathrm{s}$. When the pressure was 27.8 $\mathrm{kPa}$, the half life was fund to be 170 $\mathrm{s}$. The order of the reaction is__________. [integer answer]
Answer: 0
Solution
The half-life $t_{1/2}$ is given by the equation: $$t_{1/2} \times \frac{1}{[P_0]^{n-1}}$$ The ratio of times $\frac{t_1}{t_2}$ is: $$\frac{t_1}{t_2} = \frac{(P_2)^{n-1}}{(P_1)^{n-1}}$$ Substituting the given values: $$\frac{340}{170} = \left(\frac{27.8}{55.5}\right)^{n-1}$$ This implies: $$2 = \frac{1}{(2)^{n-1}}$$ Solving for $n$ gives: $$n = 0$$
Question 85
Chemistry · Electrochemistry · Numerical
A solution of $\mathrm{Fe_2(SO_4)_3}$ is electrolyzed for 'x' min with a current of $1.5 \, \mathrm{A}$ to deposit $0.3482 \, \mathrm{g}$ of $\mathrm{Fe}$. The value of $x$ is ______. [nearest integer] Given : $1 \, \mathrm{F} = 96500 \, \mathrm{C \, mol^{-1}}$ Atomic mass of $\mathrm{Fe} = 56 \, \mathrm{g \, mol^{-1}}$
Answer: 20
Solution
The reaction is $\mathrm{Fe^{3+} + 3e^- \rightarrow Fe}$. 3F corresponds to 1 mole Fe being deposited. For 56 g, the required charge is $3 \times 96500$. For 1 g, the required charge is $\frac{3 \times 96500}{56}$. For 0.3482 g, the required charge is $\frac{3 \times 96500}{56} \times 0.3482 = 1800.06$. Using $Q = it$, we have $$1800.06 = 1.5 \, t$$ Solving for $t$, we find $$t = 20 \, min$$
Question 86
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
Consider the following reactions: $$\mathrm{PCl_3 + H_2O \rightarrow A + HCl}$$ $$\mathrm{A + H_2O \rightarrow B + HCl}$$ number of ionisable protons present in the product B ________.
Answer: 2
Solution
$PCl_3+H_2O \xrightarrow{\substack{\text{Partial}\\\text{hydrolysis}}} PCl_2(OH)\ \text{(or)}\ PCl(OH)_2 +HCl$ no. of ionisable protons in B = 2
Question 87
Chemistry · Co-ordination Compounds · Numerical
Amongst $\mathrm{FeCl_3.3H_2O}$, $\mathrm{K_3[Fe(CN)_6]}$ and $\mathrm{[Co(NH_3)_6]Cl_3}$, the spin-only magnetic moment value of the inner-orbital complex that absorbs light at shortest wavelength is _______ B.M. [nearest integer]
Answer: 2
Solution
[$\mathrm{Fe(H_2O)_3Cl_3}$], $\underline{\mathrm{K_3[Fe(CN)_6]}}$, $\underline{\mathrm{[Co(NH_3)_6]Cl_3}}$ inner orbital complexes $\mathrm{K_3[Fe(CN)_6]}$ has more value of $\Delta$_0 than that of $\mathrm{[Co(NH_3)_6]Cl_3}$; as $\overline{\mathrm{CN}}$ is stronger ligand. More $\Delta$_0 $\Rightarrow$ smaller value of absorbed $\lambda$ $\mathrm{K_3[Fe(CN)_6]}$ : $\mathrm{Fe^{3+}}$ : 3d^5 4s^0 4p^0 $\begin{array}{cccccc}$ $\uparrow$$\downarrow$ & $\uparrow$$\downarrow$ & $\uparrow$$\downarrow$ & $\uparrow$ & & $\end{array}$ 1 unpaired e^- d^2sp^3 Spin only magnetic moment ($\mu$) = $\sqrt{3}$ BM = 1.732 BM Rounding off $\Rightarrow$ 2
Question 88
Chemistry · Polymers · Numerical
The Novolac polymer has mass of $963 \, \mathrm{g}$. The number of monomer units present in it are
Answer: 9
Solution
Monomer unit of Novolac is its molecular mass is 124 amu. Upon considering molecular weight of polymer as 963 amu (In question it is given as 963 gram). Now if during formation of Novolac, $(n-1)$ unit of water are removed then $$n \times 124 = 963 + \left[ 18 \times (n-1) \right]$$ $n = 9$
Question 89
Chemistry · Biomolecules · Numerical
How many of the given compounds will give a positive Biuret test _____ ? Glycine, Glycylalanine, Tripeptide, Biuret
Answer: 2
Solution
Biuret test is given by all proteins and peptides having at least two peptide linkages. Hence positive test must be given by tripeptide and Biuret.
Question 90
Chemistry · Redox Reactions · Numerical
The neutralization occurs when 10 mL of 0.1 M acid 'A' is allowed to react with 30 mL of 0.05 M base M(OH)$_2$. The basicity of the acid 'A' is______. [M is a metal]
Answer: 3
Solution
$\text{Acid} + \text{Base} \rightarrow \text{Salt} + \mathrm{H_2O}$ $0.1\,\mathrm{M}\ \mathrm{M(OH)_2}\ 10\,\mathrm{mL} \hspace{1cm} 0.05\,\mathrm{M}\ 30\,\mathrm{mL}$ At equivalence point: equivalents of acid $=$ equivalents of base $0.1 \times 10 \times n = 30 \times 0.05 \times 2$ $n = 3$