JEE Main 25 February 2021 Shift 1 question paper with solutions
JEE Main 25 February 2021 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Probability · Single correct
When a missile is fired from a ship, the probability that it is intercepted is $\frac{1}{3}$ and the probability that the missile hits the target, given that it is not intercepted, is $\frac{3}{4}$. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:
$\frac{1}{8}$
$\frac{1}{27}$
$\frac{3}{4}$
$\frac{3}{8}$
Answer: (a)
Solution
Probability of not getting intercepted = $\frac{2}{3}$ Probability of missile hitting target = $\frac{3}{4}$ Therefore, probability that all 3 hit the target = $\left$( $\frac{2}{3}$ $\times$ $\frac{3}{4}$ $\right$)^3 = $\frac{1}{8}$
Question 2
Maths · Sequences and Series · Single correct
If $0 < \theta, \phi < \frac{\pi}{2}$, $x = \sum_{n=0}^{\infty} \cos^{2n} \theta$, $y = \sum_{n=0}^{\infty} \sin^{2n} \phi$ and $$z = \sum_{n=0}^{\infty} \cos^{2n} \theta \cdot \sin^{2n} \phi$$ then
$xyz = 4$
$xy - z = (x + y)z$
$xy + yz + zx = z$
$xy + z = (x + y)z$
Answer: (d)
Solution
Given $$x = 1 + \cos^2 \theta + \ldots \infty$$ We have $$x = \frac{1}{1 - \cos^2 \theta} = \frac{1}{\sin^2 \theta}$$ Similarly, $$y = 1 + \sin^2 \phi + \ldots$$ Thus, $$y = \frac{1}{1 - \sin^2 \phi} = \frac{1}{\cos^2 \phi}$$ Now, $$z = \frac{1}{1 - \cos^2 \theta \cdot \sin^2 \phi} = \frac{1}{1 - \left(1 - \frac{1}{x}\right) \left(1 - \frac{1}{y}\right)} = \frac{xy}{xy - (x-1)(y-1)}$$ We find $$xz + yz - z = xy$$ And $$xy + z = (x + y)z$$
Question 3
Maths · Relations and Functions · Single correct
Let f, g : $\mathbb{N} \to \mathbb{N}$ such that $f(n + 1) = f(n) + f(a)$ $\forall n \in \mathbb{N}$ and g be any arbitrary function. Which of the following statements is NOT true?
f is one-one
If fog is one-one, then g is one-one
If g is onto, then fog is one-one
If f is onto, then $f(n) = n \forall n \in \mathbb{N}$
Maths · Three Dimensional Geometry · Single correct
The equation of the line through the point (0,1,2) and perpendicular to the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}$ is:
$\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}$
$\frac{x}{3} = \frac{y-1}{4} = \frac{z-2}{3}$
$\frac{x}{3} = \frac{y-1}{-4} = \frac{z-2}{3}$
$\frac{x}{3} = \frac{y-1}{4} = \frac{z-2}{-3}$
Answer: (a)
Solution
Given $\($ $\frac{x-1}{2}$ = $\frac{y+1}{3}$ = $\frac{z-1}{-2}$ = $\lambda$ $\)$. Any point on this line is $\($(2$\lambda$ + 1, 3$\lambda$ - 1, -2$\lambda$ + 1)$\)$. Direction ratio of given line is $\($(2, 3, -2)$\)$. Direction ratio of line to be found is $\($(2$\lambda$ + 1, 3$\lambda$ - 2, -2$\lambda$ - 1)$\)$. Therefore, $\($ $\vec{d_1}$ $\cdot$ $\vec{d_2}$ = 0 $\)$. $\($ $\lambda$ = $\frac{2}{17}$ $\)$. Direction ratio of line $\($(21, -28, -21) $\equiv$ (3, -4, -3) $\equiv$ (-3, 4, 3)$\)$.
Question 5
Maths · Vector Algebra · Single correct
Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $1 + m - n = 0$ and $l^2 + m^2 - n^2 = 0$. Then the value of $\sin^4 \alpha + \cos^4 \alpha$ is:
$\frac{3}{4}$
$\frac{1}{2}$
$\frac{5}{8}$
$\frac{3}{8}$
Answer: (c)
Solution
Given $l^2 + m^2 + n^2 = 1$ Therefore, $2n^2 = 1$ $\Rightarrow n = \pm \frac{1}{\sqrt{2}}$ Thus, $l^2 + m^2 = \frac{1}{2}$ and $|l + m| = \frac{1}{\sqrt{2}}$ Squaring, $(l + m)^2 = \frac{1}{2}$ $\Rightarrow l^2 + m^2 + 2lm = \frac{1}{2}$ $\Rightarrow \frac{1}{2} + 2lm = \frac{1}{2}$ $\Rightarrow lm = 0$ Therefore, either $l = 0$ or $m = 0$. Hence, $l = 0,\; m = \frac{1}{\sqrt{2}}$ or $l = \frac{1}{\sqrt{2}},\; m = 0$ The vectors are $ $ or $ $ Therefore, $\cos\alpha = 0 + 0 + \frac{1}{2} = \frac{1}{2}$ Thus, $\sin^4\alpha + \cos^4\alpha = 1 - \frac{1}{2}\sin^2(2\alpha) = 1 - \frac{1}{2}\times\frac{3}{4} = \frac{5}{8}$
Question 6
Maths · Integrals · Single correct
The value of the integral $$\int \frac{\sin \theta \cdot \sin 2\theta (\sin^6 \theta + \sin^4 \theta + \sin^2 \theta) \sqrt{2 \sin^4 \theta + 3 \sin^2 \theta + 6}}{1 - \cos 2\theta} \, d\theta$$ is (where $c$ is a constant of integration)
A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is $30^\circ$ (Ignore man's height). After sailing for 20 seconds towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is $45^\circ$. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :
$10(\sqrt{3} - 1)$
$10\sqrt{3}$
$10$
$10(\sqrt{3} + 1)$
Answer: (d)
Solution
Given $\($ $\frac{h}{x+y}$ = $\tan$ 30^$\circ$ $\)$ $\($ x + y = $\sqrt{3}$h $\)$ Also $\($ $\frac{h}{y}$ = $\tan$ 45^$\circ$ $\)$ $\($ h = y $\)$ Put in (1) $\($ x + y = $\sqrt{3}$y $\)$ $\($ x = ($\sqrt{3}$ - 1)y $\)$ $\($ $\frac{x}{20}$ = v' $\)$ speed Therefore, time taken to reach Foot from B $\($ $\Rightarrow$ $\frac{y}{v'}$ $\)$ $\($ $\Rightarrow$ $\frac{\sqrt{\frac{x}{(\sqrt{3} - 1)x}} \times 20}{x}$ $\)$ $\($ $\Rightarrow$ 10($\sqrt{3}$ + 1) $\)$
Question 9
Maths · Conic Sections · Single correct
A tangent is drawn to the parabola $y^2 = 6x$ which is perpendicular to the line $2x + y = 1$. Which of the following points does NOT lie on it?
(0,3)
(-6,0)
(4,5)
(5,4)
Answer: (d)
Solution
Equation of tangent: $y = mx + \frac{3}{2m}$, $m_T = \frac{1}{2}$ (since perpendicular to line $2x + y = 1$). Therefore, tangent is: $y = \frac{x}{2} + 3 \Rightarrow x - 2y + 6 = 0$.
Question 10
Maths · Trigonometric Functions · Single correct
All possible values of $\theta \in [0, 2\pi]$ for which $\sin 2\theta + \tan 2\theta > 0$ lie in:
Maths · Complex Numbers and Quadratic Equations · Single correct
Let the lines $(2 - i)z = (2 + i)\bar{z}$ and $(i - 2)\bar{z} - 4i = 0$, (here $i^2 = -1$) be normal to a circle $C$. If the line $iz + \bar{z} + 1 + i = 0$ is tangent to this circle $C$, then its radius is:
$\frac{3}{\sqrt{2}}$
$3\sqrt{2}$
$\frac{3}{2\sqrt{2}}$
$\frac{1}{2\sqrt{2}}$
Answer: (c)
Solution
(2 - i)z = (2 + i)$\bar{z}$ $\Rightarrow$ (2 - i)(x + iy) = (2 + i)(x - iy) $\Rightarrow$ 2x - ix + 2iy + y = 2x + ix - 2 - iy + y $\Rightarrow$ 2ix - 4iy = 0 $\newline$ L_1 : x - 2y = 0 $\newline$ $\Rightarrow$ (2 + i)z + (i - 2)$\bar{z}$ - 4i = 0 $\Rightarrow$ (2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0 $\Rightarrow$ 2x + ix + 2iy - y + ix - 2x + y + 2iy - 4i = 0 $\Rightarrow$ 2ix + 4iy - 4i = 0 $\newline$ L_2 : x + 2y - 2 = 0 $\newline$ Solve L_1 and L_2 4y = 2, y = $\frac{1}{2}$ $\newline$ $\therefore$ x = 1 $\newline$ Centre $\left$( 1, $\frac{1}{2}$ $\right$) $\newline$ L_3 : iz + $\bar{z}$ + 1 + i = 0 $\newline$ $\Rightarrow$ i(x + iy) + x - iy + 1 + i = 0 $\Rightarrow$ ix - y + x - iy + 1 + i = 0 $\Rightarrow$ (x - y + 1) + i(x - y + 1) = 0 $\newline$ Radius = distance from $\left$( 1, $\frac{1}{2}$ $\right$) to x - y + 1 = 0 $\newline$ r = $\frac{1 - \frac{1}{2} + 1}{\sqrt{2}}$ $\newline$ r = $\frac{3}{2\sqrt{2}}$
Question 12
Maths · Straight Lines and Pair of Straight Lines · Single correct
The image of the point (3,5) in the line $x - y + 1 = 0$, lies on:
$(x - 2)^2 + (y - 4)^2 = 4$
$(x - 4)^2 + (y + 2)^2 = 16$
$(x - 4)^2 + (y - 4)^2 = 8$
$(x - 2)^2 + (y - 2)^2 = 12$
Answer: (a)
Solution
Image of P(3, 5) on the line $x - y + 1 = 0$ is $$\frac{x - 3}{1} = \frac{y - 5}{-1} = \frac{-2(3 - 5 + 1)}{2} = 1$$ $x = 4, y = 4$ Therefore, image is (4, 4) which lies on $$(x - 2)^2 + (y - 4)^2 = 4$$
Question 13
Maths · Applications of Derivatives · Single correct
If the curves, $\frac{x^2}{a} + \frac{y^2}{b} = 1$ and $\frac{x^2}{c} + \frac{y^2}{d} = 1$ intersect each other at an angle of $90^\circ$, then which of the following relations is true?
The coefficients $a$, $b$ and $c$ of the quadratic equation, $ax^2 + bx + c = 0$ are obtained by throwing a dice three times. The probability that this equation has equal roots is :
$\frac{1}{54}$
$\frac{1}{72}$
$\frac{1}{36}$
$\frac{5}{216}$
Answer: (d)
Solution
$ax^2 + bx + c = 0$ $a, b, c \in \{1,2,3,4,5,6\}$ $n(s) = 6 \times 6 \times 6 = 216$ $D = 0 \Rightarrow b^2 = 4ac$ $ac = \dfrac{b^2}{4}$ If $b = 2$, $ac = 1 \Rightarrow a = 1,\ c = 1$ If $b = 4$, $ac = 4 \Rightarrow a = 1,\ c = 4$ $\qquad\qquad\qquad\qquad a = 4,\ c = 1$ If $b = 6$, $ac = 9 \Rightarrow a = 2,\ c = 2$ $\therefore$ probability $= \dfrac{5}{216}$
Question 16
Maths · Permutations and Combinations · Single correct
The total number of positive integral solutions $(x, y, z)$ such that $xyz = 24$ is
36
45
24
30
Answer: (d)
Solution
Given $x \cdot y \cdot z = 24$ and $x \cdot y \cdot z = 2^3 \cdot 3^1$. Now using the beggars method. 3 things to be distributed among 3 persons. Each may receive none, one or more. Therefore, $\binom{5}{2}$ ways. Similarly for '1', $\binom{3}{2}$ ways. Total ways $= \binom{5}{2} \cdot \binom{3}{2} = 30$ ways.
Question 17
Maths · Complex Numbers and Quadratic Equations · Single correct
The integer 'k', for which the inequality $x^2 - 2(3k - 1)x + 8k^2 - 7 > 0$ is valid for every $x$ in $\mathbb{R}$ is:
If a curve passes through the origin and the slope of the tangent to it at any point $(x, y)$ is $\frac{x^2 - 4x + y + 8}{x - 2}$, then this curve also passes through the point:
(4,5)
(5,4)
(4,4)
(5,5)
Answer: (d)
Solution
Given $\($ $\frac{dy}{dx}$ = $\frac{(x-2)^2 + y + 4}{(x-2)}$ = (x-2) + $\frac{y+4}{(x-2)}$ $\)$. Let $\($ x - 2 = t $\Rightarrow$ dx = dt $\)$ and $\($ y + 4 = u $\Rightarrow$ dy = du $\)$. Then $\($ $\frac{dy}{dx}$ = $\frac{du}{dt}$ $\)$. $\($ $\frac{du}{dt}$ = t + $\frac{u}{t}$ $\Rightarrow$ $\frac{du}{dt}$ - $\frac{u}{t}$ = t $\)$. The integrating factor is $\($ $\mathrm{I.F}$ = e^{$\int$ $\frac{-1}{t}$ $\,$ dt} = e^{-$\ln$ t} = $\frac{1}{t}$ $\)$. Thus, $\($ $\frac{u}{t}$ = $\int$ t $\cdot$ $\frac{1}{t}$ $\,$ dt $\Rightarrow$ $\frac{u}{t}$ = t + c $\)$. Therefore, $\($ $\frac{y+4}{x-2}$ = (x-2) + c $\)$. Passing through $\($(0,0)$\)$ gives $\($ c = 0 $\)$. Hence, $\($ (y+4) = (x-2)^2 $\)$.
Question 19
Maths · Mathematical Reasoning · Single correct
The statement $A \rightarrow (B \rightarrow A)$ is equivalent to:
$A \rightarrow (A \land B)$
$A \rightarrow (A \lor B)$
$A \rightarrow (A \rightarrow B)$
$A \rightarrow (A \leftrightarrow B)$
Answer: (b)
Solution
Given $A \to (B \to A)$ $$\Rightarrow A \to (\sim B \lor A)$$ $$\Rightarrow \sim A \lor (\sim B \lor A)$$ $$\Rightarrow \sim B \lor (\sim A \lor A)$$ $$\Rightarrow \sim B \lor t$$ $= t$ (tautology) From options: (2) $A \to (A \lor B)$ $$\Rightarrow \sim A \lor (A \lor B)$$ $$\Rightarrow (\sim A \lor A) \lor B$$ $$\Rightarrow t \lor B$$ $$\Rightarrow t$$
Question 20
Maths · Applications of Derivatives · Single correct
If Rolle's theorem holds for the function $f(x) = x^3 - ax^2 + bx - 4, x \in [1, 2]$ with $f'\left(\frac{4}{3}\right) = 0$, then ordered pair $(a, b)$ is equal to :
(-5,8)
(5,8)
(5,-8)
(-5,-8)
Answer: (b)
Solution
Given $f(1) = f(2)$ $$\Rightarrow 1 - a + b - 4 = 8 - 4a + 2b - 4$$ $$3a - b = 7$$ The derivative is $f'(x) = 3x^2 - 2ax + b$ $$\Rightarrow f'\left(\frac{4}{3}\right) = 0 \Rightarrow 3 \times \frac{16}{9} - \frac{8}{3}a + b = 0$$ $$\Rightarrow -8a + 3b = -16$$ Solving the equations, $a = 5, b = 8$
Question 21
Maths · Applications of Derivatives · Numerical
Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x^6 is unity and it has extrema at x = -1 and x = 1. If $\lim$_{x $\to$ 0} $\frac{f(x)}{x^3}$ = 1, then 5 $\cdot$ f(2) is equal to
Maths · Continuity and Differentiability · Numerical
The number of points, at which the function $f(x) = |2x + 1| - 3|x + 2| + |x^2 + x - 2|$, $x \in \mathbb{R}$ is not differentiable, is
Answer: 2
Solution
Given $f(x) = |2x + 1| - 3|x + 2| + |x^2 + x - 2|$. $$f(x) = \begin{cases} x^2 - 7, & x > 1 \\ -x^2 - 2x - 3, & -\frac{1}{2} 1 \\ -2x - 3, & -\frac{1}{2} < x < 1 \\ -2x - 6, & -2 < x < -\frac{1}{2} \\ 2x + 2, & x < -2 \end{cases}$$ Check at $1$, $-2$ and $-\frac{1}{2}$. Non. Differentiable at $x = 1$ and $-\frac{1}{2}$.
Question 23
Maths · Applications of Integrals · Numerical
The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area $A$. Then $A^4$ is equal to
Answer: 64
Solution
The area A is given by the integral $$A = \int_{\pi/4}^{5\pi/4} (\sin x - \cos x) \, dx = \left[ -\cos x - \sin x \right]_{\pi/4}^{5\pi/4}$$ which simplifies to $$= - \left[ (\cos \frac{5\pi}{4} + \sin \frac{5\pi}{4}) - (\cos \frac{\pi}{4} + \sin \frac{\pi}{4}) \right]$$ $$= - \left[ \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) - \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) \right]$$ $$= \frac{4}{\sqrt{2}} = 2\sqrt{2}$$ Therefore, $$A^4 = (2\sqrt{2})^4 = 64$$
Question 24
Maths · Sequences and Series · Numerical
Let $A_1, A_2, A_3, \ldots$ be squares such that for each $n \geq 1$, the length of the side of $A_n$ equals the length of diagonal of $A_{n+1}$. If the length of $A_1$ is $12 \, \mathrm{cm}$, then the smallest value of $n$ for which area of $A_n$ is less than one, is
Answer: 9
Solution
Given $x = \frac{12}{\sqrt{2}}$ and $y = \frac{12}{(\sqrt{2})^2}$. Therefore, side lengths are in G.P. The general term is given by $$T_n = \frac{12}{(\sqrt{2})^{n-1}}.$$ Therefore, $$Area = \frac{144}{2^{n-1}} 144.$$ The smallest $n = 9$.
Question 25
Maths · Matrices · Numerical
Let $A = \begin{pmatrix} x & y & z \\ y & z & x \\ z & x & y \end{pmatrix}$, where $x, y$ and $z$ are real numbers such that $x + y + z > 0$ and $xyz = 2$. If $A^2 = I_3$, then the value of $x^3 + y^3 + z^3$ is
Answer: 7
Solution
Given $$A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}$$ Therefore, $$|A| = (x^3 + y^3 + z^3 - 3xyz)$$ Since $$A^2 = I_3$$ we have $$|A^2| = 1$$ Thus, $$(x^3 + y^3 + z^3 - 3xyz)^2 = 1$$ This implies $$x^3 + y^3 + z^3 - 3xyz = 1$$ Therefore, $$x^3 + y^3 + z^3 = 6 + 1 = 7$$
Question 26
Maths · Matrices · Numerical
If $A = \begin{bmatrix} 0 & -\tan\left(\frac{\theta}{2}\right) \\ \tan\left(\frac{\theta}{2}\right) & 0 \end{bmatrix}$ and $(I_2 + A)(I_2 - A)^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$, then $13 \left(a^2 + b^2\right)$ is equal to
The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is
Let $\vec{a}=\hat{i}+2\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r}\times\vec{a}=\vec{c}\times\vec{a}$ and $\vec{r}\cdot\vec{b}=0$, then $\vec{r}\cdot\vec{a}$ is equal to ________.
The locus of the point of intersection of the lines $\left(\sqrt{3}\right)kx + ky - 4\sqrt{3} = 0$ and $\sqrt{3}x - y - 4\left(\sqrt{3}\right)k = 0$ is a conic, whose eccentricity is
Answer: 2
Solution
Given the equations: $$\sqrt{3}kx + ky = 4\sqrt{3}$$ $$\sqrt{3}kx - ky = 4\sqrt{3}k^2$$ Adding equation (1) and (2): $$2\sqrt{3}kx = 4\sqrt{3}\left(k^2 + 1\right)$$ $$x = 2\left(k + \frac{1}{k}\right)$$ Subtracting equation (1) and (2): $$y = 2\sqrt{3}\left(\frac{1}{k} - k\right)$$ Therefore, $$\frac{x^2}{4} - \frac{y^2}{12} = 4$$ $$\frac{x^2}{16} - \frac{y^2}{48} = 1$$ This is a hyperbola. Therefore, $$e^2 = 1 + \frac{48}{16}$$ $$e = 2$$
Physics
Question 31
Physics · Moving Charges and Magnetism · Single correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : When a rod lying freely is heated, no thermal stress is developed in it. Reason R : On heating, the length of the rod increases. In the light of the above statements, choose the correct answer from the options given below :
A is true but R is false
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 false but R is true
Answer: (c)
Solution
When a rod is free and it is heated then there is no thermal stress produced in it. The rod will expand due to increase in temperature. so both a & R are true.
Question 32
Physics · Waves · Single correct
A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m/s. The zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is :
Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively. If $T_A$ and $T_B$ are the time periods of A and B respectively then the value of $T_B - T_A$ : [Given : radius of earth = 6400 km, mass of earth = $6 \times 10^{24}$ kg]
The angular frequency of alternating current in a L-C-R circuit is $100 \, \mathrm{rad/s}$. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.
$0.8 \, \mathrm{H}$ and $250 \, \mu \mathrm{F}$
$0.8 \, \mathrm{H}$ and $150 \, \mu \mathrm{F}$
$1.33 \, \mathrm{H}$ and $250 \, \mu \mathrm{F}$
$1.33 \, \mathrm{H}$ and $150 \, \mu \mathrm{F}$
Answer: (a)
Solution
Since key is open, circuit is series $$15 = i_{rms}(60)$$ Therefore, $i_{rms} = \frac{1}{4} A$ Now, $$20 = \frac{1}{4} X_L = \frac{1}{4} (\omega L)$$ Therefore, $L = \frac{4}{5} = 0.8 H$ And $$10 = \frac{1}{4} \frac{1}{(100C)}$$ $$C = \frac{1}{4000} F = 250 \mu F$$
Question 35
Physics · Thermal Properties of Matter · Single correct
A proton, a deuteron and an $\alpha$ particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces action on them is ____ and their speed is ____ in the ratio.
Given below are two statements: Statement I: A speech signal of $2 \, \mathrm{kHz}$ is used to modulate a carrier signal of $1 \, \mathrm{MHz}$. The bandwidth requirement for the signal is $4 \, \mathrm{kHz}$. Statement II: The side band frequencies are $1002 \, \mathrm{kHz}$ and $998 \, \mathrm{kHz}$. In the light of the above statements, choose the correct answer from the options given below:
Both statement I and statement II are false
Statement I is false but statement II is true
Statement I is true but statement II is false
Both statement I and statement II are true
Answer: (d)
Solution
Side band = (f_c - f_m) to (f_c + f_m) = (1000 - 2) $\mathrm{KHz}$ to (1000 + 2) $\mathrm{KHz}$ = 998 $\mathrm{KHz}$ to 1002 $\mathrm{kHz}$ Band width = 2f_m = 2 $\times$ 2 $\mathrm{KHz}$ = 4 $\mathrm{KHz}$ Both statements are true
Question 37
Physics · Oscillations · Single correct
If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :
$2\pi^2 \, \mathrm{ms}^{-2}$
$16 \, \mathrm{m/s}^2$
$9.8 \, \mathrm{ms}^{-2}$
$\pi^2 \, \mathrm{ms}^{-2}$
Answer: (a)
Solution
Given $T = 2\pi \sqrt{\frac{l}{g}}$. Squaring both sides, we have $$T^2 = \frac{4\pi^2 l}{g}.$$ Solving for $g$, we get $$g = \frac{4\pi^2 l}{T^2}.$$ Substituting the given values, $$g = \frac{4\pi^2 \times 2}{(2)^2} = 2\pi^2 \, \mathrm{ms^{-2}}.$$
Question 38
Physics · Mathematics in Physics · Single correct
The pitch of the screw gauge is $1 \, \mathrm{mm}$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lines $8$ divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while $72^{nd}$ division on circular scale coincides with the reference line. The radius of the wire is:
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
A 5 V battery is connected across the points X and Y. Assume $D_1$ and $D_2$ to be normal silicon diodes. Find the current supplied by the battery if the $+$ve terminal of the battery is connected to point X.
$\sim$ 0.86 $\mathrm{A}$
$\sim$ 0.5 $\mathrm{A}$
$\sim$ 0.43 $\mathrm{A}$
$\sim$ 1.5 $\mathrm{A}$
Answer: (c)
Solution
Since silicon diode is used so 0.7 Volt is drop across it, only $D_1$ will conduct so current through cell $$I = \frac{5 - 0.7}{10} = 0.43 \, \mathrm{A}$$
Question 40
Physics · Dual Nature of Radiation and Matter · Single correct
An $\alpha$ particle and a proton are accelerated from rest by a potential difference of $200 \, \mathrm{V}$. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\frac{\lambda_p}{\lambda_\alpha}$ is :
A diatomic gas, having $C_p = \frac{7}{2} R$ and $C_v = \frac{5}{2} R$, is heated at constant pressure. The ratio dU : dQ : dW
03:07:02
05:07:02
05:07:03
03:05:02
Answer: (b)
Solution
Given $C_p = \frac{7}{2} R$ and $C_v = \frac{5}{2} R$. $dU = nC_v dT$ $dQ = nC_p dT$ $dW = nR dT$ The ratio $dU : dQ : dW$ is equivalent to $C_v : C_p : R$. Substituting the values, we have: $$\frac{5}{2} R : \frac{7}{2} R : R$$ Simplifying gives the ratio $5 : 7 : 2$.
Question 42
Physics · Motion in a Straight Line · Single correct
An engine of a train, moving with uniform acceleration, passes the signal post with velocity $u$ and the last compartment with velocity $v$. The velocity with which middle point of the train passes the signal post is :
$\frac{\sqrt{v^2-u^2}}{2}$
$\frac{v-u}{2}$
$\frac{\sqrt{v^2+u^2}}{2}$
$\frac{u+v}{2}$
Answer: (c)
Solution
Given $a = uniform acceleration$ and $u = velocity of first compartment$, $v = velocity of last compartment$, $l = length of train$. Using the third equation of motion, $v^2 = u^2 + 2al$ $\ldots$ (1). For the middle compartment, $v_{middle}^2 = u^2 + 2a \cdot \frac{1}{2}$. Therefore, $v_{middle}^2 = u^2 + al$ $\ldots$ (2). From equations (1) and (2), $v_{middle}^2 = u^2 + \left( \frac{v^2 - u^2}{2} \right)$. This simplifies to $v_{middle}^2 = \frac{v^2 + u^2}{2}$. Thus, $v_{middle} = \sqrt{\frac{v^2 + u^2}{2}}$.
Question 43
Physics · Physical World, Units and Measurements · Single correct
Match List - I with List- II: $$ \begin{array}{ll} \text{List-I} & \text{List-II} \\ \hline (a) \, h \text{ (Planck's constant)} & (i) \, [M L T^{-1}] \\ (b) \, E \text{ (kinetic energy)} & (ii) \, [M L^2 T^{-1}] \\ (c) \, V \text{ (electric potential)} & (iii) \, [M L^2 T^{-2}] \\ (d) \, P \text{ (linear momentum)} & (iv) \, [M L^2 T^{-3} I^{-1}] \end{array} $$ Choose the correct answer from the options given below:
Physics · Moving Charges and Magnetism · Single correct
Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8:1. The radius of coil is
0.15 m
0.2 m
0.1 m
1.0 m
Answer: (c)
Solution
The magnetic field $B$ is given by the formula: $$B = \frac{\mu_0 N i R^2}{2(R^2 + x^2)^{3/2}}$$ At $x_1 = 0.05 \, \mathrm{m}$, $B_1$ is: $$B_1 = \frac{\mu_0 N i R^2}{2(R^2 + (0.05)^2)^{3/2}}$$ At $x_2 = 0.2 \, \mathrm{m}$, $B_2$ is: $$B_2 = \frac{\mu_0 N i R^2}{2(R^2 + (0.2)^2)^{3/2}}$$ The ratio $\frac{B_1}{B_2}$ is: $$\frac{B_1}{B_2} = \frac{(R^2 + 0.04)^{3/2}}{(R^2 + 0.0025)^{3/2}}$$ Simplifying, we have: $$\left(\frac{8}{1}\right)^{2/3} = \frac{R^2 + 0.04}{R^2 + 0.0025}$$ Solving for $R^2$: $$4(R^2 + 0.0025) = R^2 + 0.04$$ $$3R^2 = 0.04 - 0.0100$$ $$R^2 = \frac{0.03}{3} = 0.01$$ Therefore, $R = \sqrt{0.01} = 0.1 \, \mathrm{m}$.
Question 45
Physics · Gravitation · Single correct
A solid sphere of radius $R$ gravitationally attracts a particle placed at $3R$ from its centre with a force $F_1$. Now a spherical cavity of radius $\left( \frac{R}{2} \right)$ is made in the sphere (as shown in figure) and the force becomes $F_2$. The value of $F_1 : F_2$ is:
Two radioactive substances $X$ and $Y$ originally have $N_1$ and $N_2$ nuclei respectively. Half life of $X$ is half of the half life of $Y$. After there half lives of $Y$, number of nuclei of both are equal. The ratio $\frac{N_1}{N_2}$ will be equal to:
$\frac{8}{1}$
$\frac{1}{8}$
$\frac{3}{1}$
$\frac{1}{3}$
Answer: (a)
Solution
After n half life, the number of nuclei undecayed is $\frac{N_0}{2^n}$. Given $T_{\frac{1}{2}x} = \frac{T_{\frac{1}{2}y}}{2}$. So 3 half lives of $y = 6$ half lives of $x$. Given, $N_x = N_y$ (after 3 $T_{\frac{1}{2}y}$). $$\frac{N_1}{2^6} = \frac{N_2}{2^3}$$ $$\frac{N_1}{N_2} = \frac{2^6}{2^3} = 2^3 = \frac{8}{1}$$
Question 47
Physics · Mathematics in Physics · Single correct
In an octagon ABCDEFGH of equal side, what is the sum of $\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF} + \overrightarrow{AG} + \overrightarrow{AH}$ If, $\overrightarrow{AO} = 2\hat{i} + 3\hat{j} - 4\hat{k}$
Given below are two statements: one is labelled as Assertion A and the other is labelled as reason R. Assertion A: The escape velocities of planet A and B are same. But A and B are of unequal mass. Reason R: The product of their mass and radius must be same. $M_1R_1 = M_2R_2$ 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
A is correct but R is not correct
Both A and R are correct and R is the correct explanation of A
A is not correct but R is correct
Answer: (b)
Solution
Given $V_e = escape velocity$ $$v_e = \sqrt{\frac{2GM}{R}}$$ so for same $v_e$, $\frac{M_1}{R_1} = \frac{M_2}{R_2}$ $A$ is true but $R$ is false
Question 49
Physics · Current Electricity · Single correct
The current (i) at time $t = 0$ and $t = \infty$ respectively for the given circuit is :
$\frac{18E}{55}, \frac{5E}{18}$
$\frac{5E}{18}, \frac{18E}{55}$
$\frac{5E}{18}, \frac{10E}{33}$
$\frac{10E}{33}, \frac{5E}{18}$
Answer: (c)
Solution
At $t = 0$, inductor is removed, so circuit will look like this at $t = 0$: $$R_{eq} = \frac{5 \times 5}{5 + 5} + 1 + 4 = \frac{25}{10} + 5 = 2.5 + 5 = 7.5 \, \Omega$$ $I(t = 0) = \frac{E}{R_{eq}} = \frac{10}{7.5} = \frac{20}{15} = \frac{4}{3} \, \mathrm{A}$ At $t = \infty$, inductor is replaced by plane wire, so circuit will look like this at $t = \infty$: $$I(t = \infty) = \frac{E}{R_{eq}} = \frac{10}{3} \, \mathrm{A}$$ Now $$R_{eq} = \frac{1}{\frac{1}{5} + \frac{1}{5}} + \frac{1}{\frac{1}{1} + \frac{1}{4}} = \frac{5}{2} + \frac{4}{5} = \frac{25}{10} + \frac{8}{10} = \frac{33}{10} = 3.3 \, \Omega$$ $I = \frac{E}{R_{eq}} = \frac{10}{3.3} = 3 \, \mathrm{A}$
Question 50
Physics · Wave Optics · Single correct
Two coherent light sources having intensity in the ratio $2 \times$ produce an interference pattern. The ratio $\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$ will be:
A transmitting station releases waves of wavelength 960 m. A capacitor of 256 $\mu \mathrm{F}$ is used in the resonant circuit. The self inductance of coil necessary for resonance is ____ $\times 10^{-8} \mathrm{H}$.
The electric field in a region is given by $\vec{E} = \left(\dfrac{3}{5}E_0\,\hat{i} + \dfrac{4}{5}E_0\,\hat{j}\right)\dfrac{\mathrm{N}}{\mathrm{C}}$. The ratio of flux of reported field through the rectangular surface of area $0.2\,\mathrm{m}^2$ (parallel to $y$-$z$ plane) to that of the surface of area $0.3\,\mathrm{m}^2$ (parallel to $x$-$z$ plane) is $a:b$, where $a =$ (round off to nearest integer) [Here $\hat{i}$, $\hat{j}$ and $\hat{k}$ are unit vectors along $x$, $y$ and $z$-axes respectively]
In a certain thermodynamical process, the pressure of a gas depends on its volume as $kV^3$. The work done when the temperature changes from $100^\circ \mathrm{C}$ to $300^\circ \mathrm{C}$ will be ____ $nR$, where $n$ denotes number of moles of a gas.
Answer: 50
Solution
Given $P = kv^3$ and $pv^{-3} = k$. We have $x = -3$. The work done $w$ is given by: $$w = \frac{nR(T_1 - T_2)}{x - 1}$$ Substituting the values, we get: $$w = \frac{nR(100 - 300)}{-3 - 1}$$ $$= \frac{nR(-200)}{-4}$$ $$= 50nR$$
Question 54
Physics · Laws of Motion · Numerical
A small bob tied at one end of a thin string of length $1 \, \mathrm{m}$ is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio $5:1$. The velocity of the bob at the highest position is $\mathrm{m/s}$. (take $g = 10 \, \mathrm{m/s^2}$)
Physics · Experimental Physics · Fill in the blank
In the given circuit of potentiometer, the potential difference $E$ across $AB$ (10 m length) is larger than $E_1$ and $E_2$ as well. For key $K_1$ (closed), the jockey is adjusted to touch the wire at point $J_1$ so that there is no deflection in the galvanometer. Now the first battery ($E_1$) is replaced by second battery ($E_2$) for working by making $K_1$ open and $E_2$ closed. The galvanometer gives then null deflection at $J_2$. The value of $\frac{E_1}{E_2}$ is $\frac{a}{b}$, where $a =$
Physics · Ray Optics and Optical Instruments · Fill in the blank
The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is ____ cm.
Answer: 15
Solution
Given \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \tag{1} \] and \[ m = \frac{v}{u}. \tag{2} \] From equations (1) and (2), we get \[ m = \frac{f}{f+u}. \] Given that \[ m_1 = -m_2, \] we have \[ \frac{f}{f-10} = -\frac{f}{f-20}. \] Therefore, \[ f - 20 = -f + 10, \] which gives \[ 2f = 30. \] Hence, \[ f = 15\,\mathrm{cm}. \]
Question 57
Physics · Electric Charges and Fields · Numerical
512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ____ V.
Answer: 128
Solution
Let charge on each drop = q. radius = r $$v = \frac{kq}{r}$$ $$2 = \frac{kq}{r}$$ radius of bigger $$\frac{4}{3} \pi R^3 = 512 \times \frac{4}{3} \pi r^3$$ $$R = 8r$$ $$v = \frac{k(512)q}{R} = \frac{512}{8} \frac{kq}{r} = \frac{512}{8} \times 2$$ = 128 V
Question 58
Physics · Electromagnetic Induction · Numerical
A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by $V = 3t$ volt. (where $t$ is in second). If the voltage is applied when $t = 0$, then the energy stored in the coil after 4 s is ____ J.
Answer: 144
Solution
Given $L \frac{di}{dt} = \varepsilon$. This equals $3t$. Integrating, $L \int di = 3 \int t \, dt$. Thus, $Li = \frac{3t^2}{2}$. Therefore, $i = \frac{3t^2}{2L}$. The energy $E = \frac{1}{2} Li^2$. Substituting, $$E = \frac{1}{2} L \left( \frac{3t^2}{2L} \right)^2$$ This simplifies to $$= \frac{1}{2} \times \frac{9t^4}{4L}$$ Further simplifying, $$= \frac{9}{8} \times \frac{(4)^4}{4 \times 2} = 144 \, \mathrm{J}$$
Question 59
Physics · Kinetic Theory · Numerical
A monoatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with velocity 30 m/s. If container is suddenly stopped then change in temperature of the gas ( R = gas constant) is $\frac{x}{3R}$. Value of x is
The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as $$U = \frac{\alpha}{r^{10}} - \frac{\beta}{r^5} - 3$$ Where, a and b are positive constants. The equilibrium distance between two atoms will $$\left(\frac{2\alpha}{\beta}\right)^{\frac{a}{b}}$$. Where a =
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Given below are two statements: Statement-I : $CeO_2$ can be used for oxidation of aldehydes and ketones. Statement-II : Aqueous solution of $EuSO_4$ is a strong reducing agent.
Statement I is true, statement II is false
Statement I is false, statement II is true
Both Statement I and Statement II are false
Both Statement I and Statement II are true
Answer: (d)
Solution
$CeO_2$ can be used as oxidising agent like $SeO_2$. Similarly $EuSO_4$ used as a reducing agent.
Question 62
Chemistry · Chemical Bonding and Molecular Structure · Single correct
According to molecular orbital theory, the species among the following that does not exist is:
$\mathrm{He}_2^-$
$\mathrm{He}_2^+$
$\mathrm{O}_2^{2-}$
$\mathrm{Be}_2$
Answer: (d)
Solution
B. O. of $\mathrm{Be_2}$ is zero, so it does not exist.
Question 63
Chemistry · Amines · Single correct
\text{Which of the following reaction/s will not give } p\text{-aminoazobenzene?}
B only
A and B
C only
A only
Answer: (a)
Solution
Question 64
Chemistry · Hydrogen · Single correct
Which of the following equation depicts the oxidizing nature of $\mathrm{H_2O_2}$?
Given the reaction: $$2\mathrm{I}^- + \mathrm{H_2O_2} + 2\mathrm{H}^+ \rightarrow \mathrm{I_2} + 2\mathrm{H_2O}$$ Oxygen reduces from -1 to -2, so its reduction will take place. Hence it will behave as an oxidising agent or it shows oxidising nature. While in other option it changes from (-1) to 0.
Question 65
Chemistry · Hydrocarbons · Single correct
Identify A in the given chemical reaction.
Answer: (d)
Solution
Aromatization reaction or hydroforming reaction.
Question 66
Chemistry · Some Basic Concepts of Chemistry · Single correct
Complete combustion of $1.80\,\mathrm{g}$ of an oxygen containing compound ($\mathrm{C}_x\mathrm{H}_y\mathrm{O}_z$) gave $2.64\,\mathrm{g}$ of $\mathrm{CO}_2$ and $1.08\,\mathrm{g}$ of $\mathrm{H}_2\mathrm{O}$. The percentage of oxygen in the organic compound is:
63.53
53.33
51.63
50.33
Answer: (b)
Solution
Given, \[ n_{\mathrm{CO}_2}=\frac{2.64}{44}=0.06 \] \[ n_{\mathrm{C}}=0.06 \] \[ \text{Weight of carbon}=0.06\times12=0.72\,\mathrm{g} \] \[ n_{\mathrm{H}_2\mathrm{O}}=\frac{1.08}{18}=0.06 \] \[ n_{\mathrm{H}}=0.06\times2=0.12 \] \[ \text{Weight of hydrogen}=0.12\,\mathrm{g} \] Therefore, weight of oxygen in $\mathrm{C}_x\mathrm{H}_y\mathrm{O}_z$ \[ =1.80-(0.72+0.12) \] \[ =0.96\,\mathrm{g} \] Percentage by weight of oxygen \[ =\frac{0.96}{1.80}\times100 \] \[ =53.3\% \]
Question 67
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Which one of the following reactions will not form acetaldehyde?
The reaction involves the oxidation of ethanol to acetic acid using $\mathrm{CrO_3}$ and $\mathrm{H_2SO_4}$ as oxidizing agents. The chemical equation is: $$\mathrm{CH_3CH_2OH} \xrightarrow{\mathrm{CrO_3-H_2SO_4}} \mathrm{CH_3COOH}$$
Question 68
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
The correct statement about $\mathrm{B_2H_6}$ is:
All B-H-B angles are of $120^\circ$.
Its fragment, $\mathrm{BH_3}$, behaves as a Lewis base.
Terminal B-H bonds have less p-character when compared to bridging bonds.
The two B – H – B bonds are not of same length.
Answer: (c)
Solution
Terminal bond angle is greater than that of bridge bond angle. Bond angle is proportional to S-character. $$\propto \frac{1}{p-character}$$
Question 69
Chemistry · Structure of Atom · Single correct
The plots of radial distribution functions for various orbitals of hydrogen atom against 'r' are given below: The correct plot for 3 s orbital is:
D
B
A
C
Answer: (a)
Solution
3s orbital Number of radial nodes = $n - \ell - 1$ For 3 s orbital $n = 3$, $\ell = 0$ Number of radial nodes = $3 - 0 - 1 = 2$. It is correctly represented in graph of option D.
Question 70
Chemistry · Environmental Chemistry · Single correct
Given below are two statements: Statement-I : An allotrope of oxygen is an important intermediate in the formation of reducing smog. Statement-II : Gases such as oxides of nitrogen and sulphur present in troposphere contribute to the formation of photochemical smog. In the light of the above statements, choose the correct answer from the options given below:
Statement I and Statement II are true
Statement I is true about Statement II is false
Both Statement I and Statement II are false
Statement I is false but Statement II is true
Answer: (c)
Solution
Reducing smog acts as a reducing agent. The reducing character is due to the presence of sulphur dioxide and carbon particles.
Question 71
Chemistry · The d-and f-Block Elements · Single correct
In which of the following pairs, the outer most electronic configuration will be the same?
Which of the glycosidic linkage galactose and glucose is present in lactose?
C - 1 of glucose and C - 6 of galactose
C - 1 of galactose and C - 4 of glucose
C - 1 of glucose and C - 4 of galactose
C - 1 of galactose and C - 6 of glucose
Answer: (b)
Solution
The structure shown is a disaccharide composed of $\beta$-D-Galactose and $\beta$-D-Glucose linked together.
Question 73
Chemistry · Hydrocarbons · Single correct
Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are: A = B = C =
B and C only
B only
A and B only
C only
Answer: (a)
Solution
Compounds which are more acidic than $\mathrm{H_2CO_3}$ give $\mathrm{CO_2}$ gas on reaction with $\mathrm{NaHCO_3}$. Compound B i.e. Benzoic acid and compound C i.e. picric acid both are more acidic than $\mathrm{H_2CO_3}$.
Question 74
Chemistry · Co-ordination Compounds · Single correct
The hybridization and magnetic nature of $[\mathrm{Mn(CN)}_6]^{4-}$ and $[\mathrm{Fe(CN)}_6]^{3-}$, respectively are:
$d^2sp^3$ and paramagnetic
$sp^3d^2$ and paramagnetic
$d^2sp^3$ and diamagnetic
$sp^3d^2$ and diamagnetic
Answer: (a)
Solution
Question 75
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Ellingham diagram is a graphical representation of:
$\Delta G$ vs $T$
$(\Delta G - T \Delta S) vs T$
$\Delta H$ vs $T$
$\Delta G$ vs $P$
Answer: (a)
Solution
Ellingham diagram tells us about the spontaneity of a reaction with temperature.
Question 76
Chemistry · Equilibrium · Single correct
The solubility of AgCN in a buffer solution of pH = 3 is x. The value of X is: [Assume: No cyano complex is formed; $K_{sp}(\mathrm{AgCN}) = 2.2 \times 10^{-16}$ and $K_{a}(\mathrm{HCN}) = 6.2 \times 10^{-10}$]
In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption $\left( \frac{x}{m} \right)$ is directly proportional to $P^X$. The value of X is:
$\infty$
1
zero
$\frac{1}{n}$
Answer: (d)
Solution
Given $\frac{x}{m} = p^x$, the formula is $\frac{x}{m} = p^{1/n}$. Hence $x = \frac{1}{n}$. The value of $n$ is any natural number.
Question 78
Chemistry · Alcohols, Phenols and Ethers · Single correct
Identify A and B in the chemical reaction.
Answer: (d)
Solution
The reaction sequence involves the conversion of the iodine substituent to a chlorine substituent using $\mathrm{H^+Cl^-}$, resulting in compound (A). Then, compound (A) undergoes a substitution reaction with $\mathrm{NaI}$ in dry acetone to replace the chlorine with iodine, forming compound (B).
Question 79
Chemistry · Polymers · Single correct
Which statement is correct?
Buna-S is a synthetic and linear thermosetting polymer
Neoprene is addition copolymer used in plastic bucket manufacturing
Synthesis of Buna-S needs nascent oxygen
Buna-N is a natural polymer
Answer: (c)
Solution
Synthesis of Buna-S needs nascent oxygen.
Question 80
Chemistry · Amines · Single correct
The major product of the following chemical reaction is:
(CH_3CH_2CO)_2O
CH_3CH_2CHO
CH_3CH_2CH_3
CH_3CH_2CH_2OH
Answer: (b)
Solution
The reaction sequence starts with $\mathrm{CH_3CH_2CN}$, which is hydrolyzed with $\mathrm{H_3O^+}$ to form $\mathrm{CH_3CH_2COOH}$. This carboxylic acid is then treated with $\mathrm{SOCl_2}$ to form the corresponding acid chloride $\mathrm{CH_3CH_2COCl}$. Finally, the acid chloride is reduced using $\mathrm{Pd/BaSO_4}$ to yield the aldehyde $\mathrm{CH_3CH_2CHO}$.
Question 81
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
Among the following, the number of halide(s) which is/are inert to hydrolysis is \begin{enumerate} \item[(A)] $\mathrm{BF_3}$ \item[(B)] $\mathrm{SiCl_4}$ \item[(C)] $\mathrm{PCl_5}$ \item[(D)] $\mathrm{SF_6}$ \end{enumerate}
BF_3
SiCl_4
PCl_5
SF_6
Answer: (a)
Solution
Due to crowding $\mathrm{SF_6}$ is not hydrolysed.
Question 82
Chemistry · Solutions · Numerical
1 molal aqueous solution of an electrolyte $\mathrm{A}_2 \mathrm{B}_3$ is 60$\%$ ionised. The boiling point of the solution at 1 atm is _____ K. (Rounded-off to the nearest integer) [ Given $K_b$ for $(\mathrm{H}_2\mathrm{O})$ = 0.52 \, $\mathrm{K}$ \, $\mathrm{kg}$ \, $\mathrm{mol^{-1}}$ ]
Answer: 375
Solution
The reaction is given by $\mathrm{A_2 \, B_3 \rightarrow 2 \, A^{+3} + 3 \, B^{-2}}$. The number of ions is $2 + 3 = 5$. The van't Hoff factor $i$ is calculated as follows: $$i = 1 + (n - 1) \times \alpha$$ $$= 1 + (5 - 1) \times 0.6$$ $$= 1 + 4 \times 0.6 = 1 + 2.4 = 3.4$$ The boiling point elevation $\Delta T_b$ is given by $$\Delta T_b = K_b \times m \times i$$ $$= 0.52 \times 1 \times 3.4 = 1.768^\circ \mathrm{C}$$ The change in boiling point is $$\Delta T_b = (T_b)_{solution} - \left[(T_b)_{\mathrm{H_2O}}\right]_{Solution}$$ $$1.768 = (T_b)_{solution} - 100$$ Thus, $$(T_b)_{solution} = 101.768^\circ \mathrm{C}$$ Converting to Kelvin, $$= 375 \, \mathrm{K}$$
Question 83
Chemistry · Some Basic Concepts of Chemistry · Numerical
In basic medium $\mathrm{CrO_4^{2-}}$ oxidizes $\mathrm{S_2O_3^{2-}}$ to form $\mathrm{SO_4^{2-}}$ and itself changes into $\mathrm{Cr(OH)_4^-}$. The volume of $0.154\,\mathrm{MCrO_4^{2-}}$ required to react with $40\,\mathrm{mL}$ of $0.25\,\mathrm{MS_2O_3^{2-}}$ is ____ (Rounded-off to the nearest integer)
A car tyre is filled with nitrogen gas at $35 \, \mathrm{psi}$ at $27^\circ \mathrm{C}$. It will burst if pressure exceeds $40 \, \mathrm{psi}$. The temperature in $^\circ \mathrm{C}$ at which the car tyre will burst is ____ (Rounded-off to the nearest integer)
The combustion of cyanamide, $\mathrm{NH_2CN(s)}$, with oxygen was run in a bomb calorimeter and $\Delta U$ was found to be $-742.24\ \mathrm{kJ\,mol^{-1}}$. The magnitude of $\Delta H_{298}$ for the reaction $\mathrm{NH_2CN(s)+\frac{3}{2}O_2(g)\rightarrow N_2(g)+CO_2(g)+H_2O(l)}$ is \underline{\hspace{1cm}} kJ. (Rounded off to the nearest integer) [Assume ideal gases and $R=8.314\ J\,mol^{-1}\,K^{-1}$]
Using the provided information in the following paper chromatogram: The calculated $R_f$ value of A ____ $\times 10^{-1}$
Answer: 4
Solution
The retention factor $R_f$ is given by the formula: $$R_f = \frac{\text{Distance travelled by compound}}{\text{Distance travelled by solvent}}$$ On the chromatogram, the distance travelled by the compound is $2\,\mathrm{cm}$. The distance travelled by the solvent is $5\,\mathrm{cm}$. So, $R_f = \dfrac{2}{5} = 4 \times 10^{-1} = 0.4$.
Question 87
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For the reaction, $aA+bB\rightarrow cC+dD$, the plot of $\log k$ vs $\frac{1}{T}$ is given below: The temperature at which the rate constant of the reaction is $10^{-4}\,\mathrm{s}^{-1}$ is _____ K. [Rounded off to the nearest integer] [Given: The rate constant of the reaction is $10^{-5}\,\mathrm{s}^{-1}$ at $500$ K]
Answer: 526
Solution
Given $$\log_{10} K = \log_{10} A - \frac{E_a}{2.303RT}$$ The slope is $$\frac{E_a}{2.303R} = -10000$$ Thus, $$\log_{10} \frac{K_2}{K_1} = \frac{E_a}{2.303R} \times \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$$ Substituting the values, $$\log_{10} \frac{10^{-4}}{10^{-5}} = 10000 \times \left[ \frac{1}{500} - \frac{1}{T} \right]$$ This simplifies to $$1 = 10000 \times \left[ \frac{1}{500} - \frac{1}{T} \right]$$ Rearranging gives $$\frac{1}{10000} = \frac{1}{500} - \frac{1}{T}$$ Solving for $\($ $\frac{1}{T}$ $\)$, we have $$\frac{1}{T} = \frac{1}{500} - \frac{1}{10000}$$ Simplifying further, $$\frac{1}{T} = \frac{20 - 1}{10000} = \frac{19}{10000}$$ Finally, $$T = \frac{10000}{19} \Rightarrow 526 \, \mathrm{K}$$
Question 88
Chemistry · Redox Reactions · Numerical
0.4 g mixture of NaOH, Na$_2$CO$_3$ and some inert impurities was first titrated with $\frac{N}{10}$ HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage of Na$_2$CO$_3$ in the mixture is ____ (Rounded-off to the nearest integer)
Consider the following chemical reaction. $$\mathrm{HC} \equiv \mathrm{CH} \xrightarrow{\begin{array}{c}(1) Red hot Fe tube, 873 K \\ (b) CO+HCl/AlCl_3 \end{array}} Product$$ The number of $sp^2$ hybridized carbon atom(s) present in the product is
Answer: 7
Solution
All carbon atoms in benzaldehyde are $\mathrm{sp^2}$ hybridised.
Question 90
Chemistry · Equilibrium · Numerical
The ionization enthalpy of $\mathrm{Na^+}$ formation from $\mathrm{Na_{(g)}}$ is $495.8 \, \mathrm{kJ \, mol^{-1}}$, while the electron gain enthalpy of $\mathrm{Br}$ is $-325.0 \, \mathrm{kJ \, mol^{-1}}$. Given the lattice enthalpy of $\mathrm{NaBr}$ is $-728.4 \, \mathrm{kJ \, mol^{-1}}$. The energy for the formation of $\mathrm{NaBr}$ ionic solid is $(-)\, \_\_\_\_ \times 10^{-1} \, \mathrm{kJ \, mol^{-1}}$