JEE Main 18 March 2021 Shift 2 question paper with solutions
JEE Main 18 March 2021 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 · Differential Equations · Single correct
Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = (y + 1) \left( (y + 1)e^{x^2/2} - x \right)$, $0 < x < 2.1$, with $y(2) = 0$. Then the value of $\frac{dy}{dx}$ at $x = 1$ is equal to:
$\frac{-e^{3/2}}{(e^2+1)^2}$
$\frac{-2e^2}{(1+e^2)^2}$
$\frac{e^{5/2}}{(1+e^2)^2}$
$\frac{5e^{1/2}}{(e^2+1)^2}$
Answer: (a)
Solution
Let $y + 1 = Y$. Therefore, $\frac{dY}{dx} = Y^2 e^{\frac{x^2}{2}} - xY$. Put $\frac{1}{Y} = k$. Thus, $\frac{dk}{dx} + k(-x) = e^{\frac{x^2}{2}}$. The integrating factor is $I.F. = e^{-\frac{x^2}{2}}$. Therefore, $k = (x + c)e^{x^2/2}$. Put $k = -\frac{1}{y+1}$. Thus, $y + 1 = -\frac{1}{(x+c)e^{x^2/2}}$. When $x = 2$, $y = 0$, then $c = -2 - \frac{1}{e^2}$. Differentiate the equation (i) and put $x = 1$. We get $$\left( \frac{dy}{dx} \right)_{x=1} = -\frac{e^{3/2}}{(1+e^2)^2}$$
Question 2
Maths · Vector Algebra · Single correct
In a triangle ABC, if $|\overrightarrow{BC}| = 8$, $|\overrightarrow{CA}| = 7$, $|\overrightarrow{AB}| = 10$, then the projection of the vector $\overrightarrow{AB}$ on $\overrightarrow{AC}$ is equal to:
$\frac{25}{4}$
$\frac{85}{14}$
$\frac{127}{20}$
$\frac{115}{16}$
Answer: (b)
Solution
Given $|\vec{a}| = 8$, $|\vec{b}| = 7$, $|\vec{c}| = 10$. $\cos \theta = \frac{|\vec{b}|^2 + |\vec{c}|^2 - |\vec{a}|^2}{2|\vec{b}||\vec{c}|} = \frac{17}{28}$. Projection of $\vec{c}$ on $\vec{b}$ is $$= |\vec{c}| \cos \theta$$ $$= 10 \times \frac{17}{28}$$ $$= \frac{85}{14}$$
Question 3
Maths · Determinants · Single correct
Let the system of linear equations $$4x + \lambda y + 2z = 0$$ $$2x - y + z = 0$$ $$\mu x + 2y + 3z = 0, \lambda, \mu \in \mathbb{R}$$ has a non-trivial solution. Then which of the following is true?
$\mu = 6, \lambda \in \mathbb{R}$
$\lambda = 2, \mu \in \mathbb{R}$
$\lambda = 3, \mu \in \mathbb{R}$
$\mu = -6, \lambda \in \mathbb{R}$
Answer: (a)
Solution
For non-trivial solution $$\begin{vmatrix} 4 & \lambda & 2 \\ 2 & -1 & 1 \\ \mu & 2 & 3 \end{vmatrix} = 0$$ $$\Rightarrow 2\mu - 6\lambda + \lambda \mu = 12$$ when $\mu = 6$, $12 - 6\lambda + 6\lambda = 12$ which is satisfied by all $\lambda$
Question 4
Maths · Relations and Functions · Single correct
Let $f : \mathbb{R} - \{3\} \to \mathbb{R} - \{1\}$ be defined by $f(x) = \frac{x-2}{x-3}$. Let $g : \mathbb{R} \to \mathbb{R}$ be given as $g(x) = 2x - 3$. Then, the sum of all the values of $x$ for which $f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$ is equal to
7
2
5
3
Answer: (c)
Solution
Given $f(x) = y = \frac{x-2}{x-3}$. Therefore, $x = \frac{3y-2}{y-1}$. So, $f^{-1}(x) = \frac{3x-2}{x-1}$. And $g(x) = y = 2x - 3$. Therefore, $x = \frac{y+3}{2}$. So, $g^{-1}(x) = \frac{x+3}{2}$. Thus, $f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$. Therefore, $x_1, x_2$. Thus, $x^2 - 5x + 6 = 0$. Therefore, sum of roots $x_1 + x_2 = 5$.
Question 5
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line $x + y = 3$. If $R$ and $r$ be the radius of circumcircle and incircle respectively of $\triangle ABC$, then $(R + r)$ is equal to:
$\frac{9}{\sqrt{2}}$
$7\sqrt{2}$
$2\sqrt{2}$
$3\sqrt{2}$
Answer: (a)
Solution
Given $r = OM = \frac{3}{\sqrt{2}}$. Since $\sin 30^\circ = \frac{1}{2} = \frac{r}{R}$, it follows that $R = \frac{6}{\sqrt{2}}$. Therefore, $r + R = \frac{9}{\sqrt{2}}$.
Question 6
Maths · Conic Sections · Single correct
Consider a hyperbola $H : x^2 - 2y^2 = 4$. Let the tangent at a point $P(4, \sqrt{6})$ meet the $x$-axis at $Q$ and latus rectum at $R(x_1, y_1)$, $x_1 > 0$. If $F$ is a focus of $H$ which is nearer to the point $P$, then the area of $\triangle QFR$ is equal to
$4\sqrt{6}$
$\sqrt{6} - 1$
$\frac{7}{\sqrt{6}} - 2$
$4\sqrt{6} - 1$
Answer: (c)
Solution
Given the hyperbola equation $\($ $\frac{x^2}{4}$ - $\frac{y^2}{2}$ = 1 $\)$. The eccentricity $\($ e = $\sqrt{1 + \frac{b^2}{a^2}}$ = $\sqrt{\frac{3}{2}}$ $\)$. The focus $\($ F(ae, 0) $\Rightarrow$ F($\sqrt{6}$, 0) $\)$. The equation of the tangent at $\($ P $\)$ to the hyperbola is $\($ 2x - y$\sqrt{6}$ = 2 $\)$. The tangent meets the x-axis at $\($ Q(1, 0) $\)$. The latus rectum $\($ x = $\sqrt{6}$ $\)$ at $\($ R $\left$( $\sqrt{6}$, $\frac{2}{\sqrt{6}}$($\sqrt{6}$ - 1) $\right$) $\)$. Therefore, the area of $\($ $\triangle$ QFR = $\frac{1}{2}$($\sqrt{6}$ - 1) $\cdot$ $\frac{2}{\sqrt{6}}$($\sqrt{6}$ - 1) $\)$ which simplifies to $\($ $\frac{7}{\sqrt{6}}$ - 2 $\)$.
Question 7
Maths · Mathematical Reasoning · Single correct
If $P$ and $Q$ are two statements, then which of the following compound statement is a tautology?
Given $((P \rightarrow Q) \land \sim Q)$. $$\equiv (\sim P \lor Q) \land \sim Q$$ $$\equiv (\sim P \land \sim Q) \lor (Q \land \sim Q)$$ $$\equiv \sim P \land \sim Q$$ LHS of all the options are same i.e. $(A)$ $(P \land \sim Q) \rightarrow Q$ $$\equiv \sim (\sim P \land \sim Q) \lor Q$$ $$\equiv (P \lor Q) \lor Q \neq \text{tautology}$$ $(B)$ $(\sim P \land \sim Q) \rightarrow \sim P$ $$\equiv \sim (\sim P \land \sim Q) \lor \sim P$$ $$\equiv (P \lor Q) \lor \sim P$$ $$\Rightarrow \text{Tautology}$$ $(C)$ $(\sim P \land \sim Q) \rightarrow P$ $$\equiv (P \lor Q) \lor P \neq \text{Tautology}$$ $(D)$ $(\sim P \land \sim Q) \rightarrow (P \land Q)$ $$\equiv (P \lor Q) \lor (P \land Q) \neq \text{Tautology}$$
Question 8
Maths · Integrals · Single correct
Let $g(x) = \int_0^x f(t) \, dt$, where $f$ is continuous function in $[0,3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0,1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1,3]$ The largest possible interval in which $g(3)$ lies is:
$\left[ -1, -\frac{1}{2} \right]$
$\left[ -\frac{3}{2}, -1 \right]$
$\left[ \frac{1}{3}, 2 \right]$
$[1,3]$
Answer: (c)
Solution
Given $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$. Now, $g(3) = \int_0^3 f(t) dt = \int_0^1 f(t) dt + \int_1^3 f(t) dt$. Therefore, $\int_0^1 \frac{1}{3} dt \leq \int_0^1 f(t) dt \leq \int_0^1 1 \cdot dt \ldots (1)$ and $\int_1^3 0 dt \leq \int_1^3 f(1) dt \leq \int_1^3 \frac{1}{2} dt \ldots (2)$. Adding, we get $\frac{1}{3} + 0 \leq g(3) \leq 1 + \frac{1}{2} (3 - 1)$. Thus, $\frac{1}{3} \leq g(3) \leq 2$.
Question 9
Maths · Sequences and Series · Single correct
Let $S_1$ be the sum of first $2n$ terms of an arithmetic progression. Let $S_2$ be the sum of first $4n$ terms of the same arithmetic progression. If $(S_2 - S_1)$ is $1000$, then the sum of the first $6n$ terms of the arithmetic progression is equal to:
Maths · Complex Numbers and Quadratic Equations · Single correct
Let a complex number be $w = 1 - \sqrt{3}i$. Let another complex number $z$ be such that $|zw| = 1$ and $\arg(z) - \arg(w) = \frac{\pi}{2}$. Then the area of the triangle with vertices origin, $z$ and $w$ is equal to:
4
$\frac{1}{2}$
$\frac{1}{4}$
2
Answer: (b)
Solution
Given $w = 1 - \sqrt{3} \cdot i$, we have $|w| = 2$. Now, $|z| = \frac{1}{|w|} \Rightarrow |z| = \frac{1}{2}$ and $amp(z) = \frac{\pi}{2} + amp(w)$. Therefore, the area of the triangle is $\frac{1}{2} \cdot OP \cdot OQ$. This equals $\frac{1}{2} \cdot 2 \cdot \frac{1}{2} = \frac{1}{2}$.
Question 11
Maths · Statistics · Single correct
Let in a series of $2n$ observations, half of them are equal to $a$ and remaining half are equal to $-a$. Also by adding a constant $b$ in each of these observations, the mean and standard deviation of new set become $5$ and $20$, respectively. Then the value of $a^2 + b^2$ is equal to:
425
650
250
925
Answer: (a)
Solution
Let observations are denoted by $x_i$ for $1 \leq i < 2n$. $$\bar{x} = \frac{\sum x_i}{2n} = \frac{(a + a + \ldots + a) - (a + a + \ldots + a)}{2n}$$ $$\Rightarrow \bar{x} = 0$$ and $$\sigma_x^2 = \frac{\sum x_i^2}{2n} - (\bar{x})^2 = \frac{a^2 + a^2 + \ldots + a^2}{2n} - 0 = a^2$$ $$\Rightarrow \sigma_x = a$$ Now, adding a constant $b$ then $\bar{y} = \bar{x} + b = 5$ $$\Rightarrow b = 5$$ and $$\sigma_y = \sigma_x$$ (No change in S.D.) $$\Rightarrow a = 20 \Rightarrow a^2 + b^2 = 425$$
Question 12
Maths · Conic Sections · Single correct
Let $S_1 : x^2 + y^2 = 9$ and $S_2 : (x - 2)^2 + y^2 = 1$ Then the locus of center of a variable circle $S$ which touches $S_1$ internally and $S_2$ externally always passes through the points:
Given $S_1: x^2 + y^2 = 9$ with $r_1 = 3$ and $A(0, 0)$, and $S_2: (x - 2)^2 + y^2 = 1$ with $r_2 = 1$ and $B(2, 0)$. Since $C_1C_2 = r_1 - r_2$, the given circles are touching internally. Let a variable circle with center $P$ and radius $r$. Then $PA = r_1 - r$ and $PB = r_2 + r$. Therefore, $PA + PB = r_1 + r_2$. Hence, $PA + PB = 4 (> AB)$. The locus of $P$ is an ellipse with foci at $A(0, 0)$ and $B(2, 0)$ and the length of the major axis is $2a = 4$, so $e = \frac{1}{2}$. The center is at $(1, 0)$ and $b^2 = a^2 (1 - e^2) = 3$. If it is an $x$-ellipse, then $E: \frac{(x-1)^2}{4} + \frac{y^2}{3} = 1$, which is satisfied by $\left(2, \pm \frac{3}{2}\right)$.
Question 13
Maths · Vector Algebra · Single correct
Let $\vec{a}$ and $\vec{b}$ be two non-zero vectors perpendicular to each other and $|\vec{a}| = |\vec{b}|$. If $|\vec{a} \times \vec{b}| = |\vec{a}|$, then the angle between the vectors $(\vec{a} + \vec{b} + (\vec{a} \times \vec{b}))$ and $\vec{a}$ is equal to:
Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to:
Maths · Applications of Derivatives · Single correct
Let a tangent be drawn to the ellipse $\frac{x^2}{27} + y^2 = 1$ at $\left(3\sqrt{3} \cos \theta, \sin \theta\right)$ where $\theta \in \left(0, \frac{\pi}{2}\right)$. Then the value of $\theta$ such that the sum of intercepts on axes made by this tangent is minimum is equal to:
Define a relation R over a class of $n \times n$ real matrices A and B as "ARB iff there exists a non-singular matrix P such that $PAP^{-1} = B$". Then which of the following is true?
R is symmetric, transitive but not reflexive,
R is reflexive, symmetric but not transitive
R is an equivalence relation
R is reflexive, transitive but not symmetric
Answer: (c)
Solution
A and B are matrices of $n \times n$ order and $ARB$ iff there exists a non-singular matrix $P(\det(P) \neq 0)$ such that $PAP^{-1} = B$. For reflexive $ARA \Rightarrow PAP^{-1} = A$ (1) must be true. For $P = I$, Eq.(1) is true so 'R' is reflexive. For symmetric $ARB \Rightarrow PAP^{-1} = B$ (1) is true for $BRA$ iff $PBP^{-1} = A$ (2) must be true. Therefore, $PAP^{-1} = B$. $$P^{-1}PAP^{-1} = P^{-1}B$$ $$IAP^{-1}P = P^{-1}BP$$ $$A = P^{-1}BP$$ from (2) and (3) $PBP^{-1} = P^{-1}BP$ can be true some $P = P^{-1} \Rightarrow P^2 = I(\det(P) \neq 0)$ So 'R' is symmetric. For transitive $ARB \Rightarrow PAP^{-1} = B$ is true. $BRC \Rightarrow PBP^{-1} = C$ is true. Now $PPAP^{-1}P^{-1} = C$ $$P^2 A \left(P^2\right)^{-1} = C \Rightarrow ARC$$ So 'R' is transitive relation. Therefore, R is equivalence.
Question 17
Maths · Heights and Distances · Single correct
A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from each corner of the park be $\frac{\pi}{3}$. If the radius of the circumcircle of $\Delta ABC$ is $2$, then the height of the pole is equal to:
$\frac{2\sqrt{3}}{3}$
$2\sqrt{3}$
$\sqrt{3}$
$\frac{1}{\sqrt{3}}$
Answer: (b)
Solution
Let $PD = h$, $R = 2$. As angle of elevation of top of pole from $A$, $B$, $C$ are equal, so $D$ must be circumcentre of $\triangle ABC$. $$\tan\left(\frac{\pi}{3}\right) = \frac{PD}{R} = \frac{h}{R}$$ $$h = R \tan\left(\frac{\pi}{3}\right) = 2\sqrt{3}$$
Question 18
Maths · Trigonometric Functions · Single correct
If $$ 15\sin^4\alpha + 10\cos^4\alpha = 6, $$ for some $\alpha \in \mathbb{R}$, then the value of $$ 27\sec^6\alpha + 8\operatorname{cosec}^6\alpha $$ is equal to:
350
500
400
250
Answer: (d)
Question 19
Maths · Applications of Integrals · Single correct
The area bounded by the curve $4y^2 = x^2 (4-x)(x-2)$ is equal to:
Maths · Continuity and Differentiability · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined as $$f(x) = \begin{cases} \frac{\sin(a+1)x + \sin 2x}{2x}, & if x 0 \end{cases}$$ If $f$ is continuous at $x = 0$, then the value of $a + b$ is equal to:
$-\frac{5}{2}$
$-2$
$-3$
$-\frac{3}{2}$
Answer: (d)
Solution
Given that $f(x)$ is continuous at $x = 0$, we have: $$\lim_{x \to 0^+} f(x) = f(0) = \lim_{x \to 0^-} f(x) \ldots (1)$$ From this, $f(0) = b \ldots (2)$ Next, consider: $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left( \frac{(\sin(a+1)x)}{2x} + \frac{\sin 2x}{2x} \right)$$ This simplifies to: $$= \frac{a+1}{2} + 1 \ldots (3)$$ Now, consider: $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x + bx^3} - \sqrt{x}}{bx^{5/2}}$$ This becomes: $$= \lim_{x \to 0^+} \frac{(x + bx^3 - x)}{bx^{5/2} \left( \sqrt{x + bx^3 + \sqrt{x}} \right)}$$ Further simplifying gives: $$= \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{x \left( \sqrt{1 + bx^2} + 1 \right)}} = \frac{1}{2} \ldots (4)$$ Using equations (2), (3), and (4) in equation (1), we have: $$\frac{1}{2} = b = \frac{a+1}{2} + 1$$ This implies: $$\Rightarrow b = \frac{1}{2}, a = -2$$ Thus, $a + b = -\frac{3}{2}$.
Question 21
Maths · Complex Numbers and Quadratic Equations · Fill in the blank
If $f(x)$ and $g(x)$ are two polynomials such that the polynomial $P(x) = f(x^3) + xg(x^3)$ is divisible by $x^2 + x + 1$, then $P(1)$ is equal to ___
Answer: 0
Solution
Given $P(x) = f(x^3) + xg(x^3)$. $P(1) = f(1) + g(1)$ ...(1) Now $P(x)$ is divisible by $x^2 + x + 1$ $$\Rightarrow P(x) = Q(x)(x^2 + x + 1)$$ $P(w) = 0 = P(w^2)$ where $w, w^2$ are non-real cube roots of unity. $P(x) = f(x^3) + xg(x^3)$ $P(w) = f(w^3) + wg(w^3) = 0$ $f(1) + wg(1) = 2$ ...(2) $P(w^2) = f(w^6) + w^2g(w^6) = 0$ $f(1) + w^2g(1) = 0$ ...(3) (2) + (3) $$\Rightarrow 2f(1) + (w + w^2)g(1) = 0$$ $2f(1) = g(1)$ ...(4) (2) - (3) $$\Rightarrow (w - w^2)g(1) = 0$$ $g(1) = 0 = f(1)$ from (4) from (1) $P(1) = f(1) + g(1) = 0$
Question 22
Maths · Matrices · Numerical
Let I be an identity matrix of order 2 $\times$ 2 and $$P = \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix}$$. Then the value of $n \in \mathbb{N}$ for which $P^n = 5I - 8P$ is equal to ___
If $\sum_{r=1}^{10} r!\left(r^3 + 6r^2 + 2r + 5\right) = \alpha(11!)$, then the value of $\alpha$ is equal to
Answer: 160
Solution
Given $$\sum_{r=1}^{10} r! \{ (r+1)(r+2)(r+3) - 9(r+1) + 8 \}$$ This is equal to $$\sum_{r=1}^{10} \{ (r+3)! - (r+1)! \} - 8 \{ (r+1)! - r! \}$$ Simplifying, we have $$= (13! + 12! - 2! - 3!) - 8(11! - 1)$$ Further simplification gives $$= (12 \cdot 13 + 12 - 8) \cdot 11! - 8 + 8$$ Finally, $$= (160)(11!)$$ Hence $\alpha = 160$
Question 24
Maths · Binomial Theorem · Numerical
The term independent of $x$ in the expansion of $$\left[ \frac{x+1}{x^{2/3} - x^{1/3} + 1} - \frac{x-1}{x-x^{1/2}} \right]^{10}, \ x \neq 1,$$ is equal to
Answer: 210
Solution
The expression is given by $$\left( \left( x^{1/3} + 1 \right) - \left( \frac{\sqrt{x} + 1}{\sqrt{x}} \right) \right)^{10} \left( x^{1/3} - x^{-1/2} \right)^{10}$$. The term $T_{r+1}$ is $$T_{r+1} = 10C_r \left( x^{1/3} \right)^{10-r} \left( -x^{-1/2} \right)^r$$. Solving $$\frac{10-r}{3} - \frac{r}{2} = 0 \Rightarrow 20 - 2r - 3r = 0$$ gives $$\Rightarrow r = 4$$. Therefore, $$T_5 = 10C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$.
Question 25
Maths · Applications of Derivatives · Numerical
Let P(x) be a real polynomial of degree 3 which vanishes at $x = -3$. Let P(x) have local minima at $x = 1$, local maxima at $x = -1$ and $\int_{-1}^{1} P(x) \, dx = 18$, then the sum of all the coefficients of the polynomial P(x) is equal to ___
Answer: 8
Solution
Let $p'(x) = a(x-1)(x+1) = a\left(x^2 - 1\right)$. $p(x) = a \int \left(x^2 - 1\right) \, dx + c$ $$= a\left(\frac{x^3}{3} - x\right) + c$$ Now $p(-3) = 0$ $$\Rightarrow a\left(-\frac{27}{3} + 3\right) + c = 0$$ $$\Rightarrow -6a + c = 0$$ Now $\int_{-1}^{1} \left(a\left(\frac{x^3}{3} - x\right) + c\right) \, dx = 18$ $$= 2c = 18 \Rightarrow c = 9 \cdots (2)$$ From (1) $\&$ (2) $\Rightarrow -6a + 9 = 0 \Rightarrow a = \frac{3}{2}$ $$\Rightarrow p(x) = \frac{3}{2}\left(\frac{x^3}{3} - x\right) + 9$$ Sum of coefficient $$= \frac{1}{2} - \frac{3}{2} + 9$$ $$= 8$$
Question 26
Maths · Three Dimensional Geometry · Numerical
Let the mirror image of the point (1, 3, a) with respect to the plane $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) - b = 0$ be $(-3, 5, 2)$. Then the value of $|a + b|$ is equal to ___
Answer: 1
Solution
Plane $=2x-y+z=b$ $R\equiv\left(-1,\ 4,\ \dfrac{a+2}{2}\right)\ \rightarrow\ \text{on plane}$ $\therefore\ 2(-1)-4+\dfrac{a+2}{2}=b$ $\Rightarrow a+2=2b+12$ $\Rightarrow a=2b+10 \qquad \ldots (i)$ $\overrightarrow{PQ}=\langle -4,\ 2,\ 2-a\rangle$ Direction ratios of normal to the plane are $\langle 2,\ -1,\ 1\rangle$ Since $\overrightarrow{PQ}\perp$ plane, $\langle -4,\ 2,\ 2-a\rangle \parallel \langle 2,\ -1,\ 1\rangle$ $\therefore\ \dfrac{-4}{2}=\dfrac{2}{-1}=\dfrac{2-a}{1}$ $\Rightarrow -2=2-a$ $\Rightarrow a=4$ From (i), $4=2b+10$ $\Rightarrow b=-3$ $\therefore\ |a+b|=|4+(-3)|=1$
Question 27
Maths · Continuity and Differentiability · Numerical
Let $f : \mathbb{R} \to \mathbb{R}$ satisfy the equation $f(x + y) = f(x) \cdot f(y)$ for all $x, y \in \mathbb{R}$ and $f(x) \neq 0$ for any $x \in \mathbb{R}$. If the function $f$ is differentiable at $x = 0$ and $f'(0) = 3$, then $\lim_{h \to 0} \frac{1}{h} (f(h) - 1)$ is equal to ___
Let ${}^{n}C_{r}$ denote the binomial coefficient of $x^r$ in the expansion of $(1+x)^n$. If$$ \sum_{k=0}^{10}(2^2+3k)^n{}^{n}C_{k}=\alpha\cdot3^{10}+\beta\cdot2^{10},\qquad \alpha,\beta\in\mathbb{R}, $$ then $\alpha+\beta$ is equal to $\underline{\hspace{2cm}}$.
Let P be a plane containing the line $\frac{x-1}{3} = \frac{y+6}{4} = \frac{z+5}{2}$ and parallel to the line $\frac{x-3}{4} = \frac{y-2}{-3} = \frac{z+5}{7}$. If the point $(1, -1, \alpha)$ lies on the plane P, then the value of $|5\alpha|$ is equal to ___
Let $y = y(x)$ be the solution of the differential equation $x dy - y dx = \sqrt{x^2 - y^2} dx, x \geq 1$, with $y(1) = 0$. If the area bounded by the line $x = 1$, $x = e^\pi$, $y = 0$ and $y = y(x)$ is $\alpha e^{2\pi} + \beta$, then the value of $10(\alpha + \beta)$ is equal to ___
Answer: 4
Solution
Given $$xdy - ydx = \sqrt{x^2 - y^2} dx$$ This implies $$\frac{xdy - ydx}{x^2} = \frac{1}{x} \sqrt{1 - \frac{y^2}{x^2}} dx$$ Therefore, $$\int \frac{d\left(\frac{y}{x}\right)}{\sqrt{1 - \left(\frac{y}{x}\right)^2}} = \int \frac{dx}{x}$$ This leads to $$\sin^{-1}\left(\frac{y}{x}\right) = \ln|x| + c$$ At $x = 1$, $y = 0$ implies $c = 0$. Thus, $$y = x \sin(\ln x)$$ The area $A$ is given by $$A = \int_1^e x \sin(\ln x) dx$$ Let $a = e^t$, then $dx = e^t dt$ which implies $$\int_0^a e^{2t} \sin(t) dt = A$$ Therefore, $$a = \frac{1}{5}, \beta = \frac{1}{5}$$ So, $$10(\alpha + \beta) = 4$$
Physics
Question 31
Physics · Magnetism and Matter · Single correct
Which of the following statements are correct? $(A)$ Electric monopoles do not exist whereas magnetic monopoles exist. $(B)$ Magnetic field lines due to a solenoid at its ends and outside cannot be completely straight and confined. $(C)$ Magnetic field lines are completely confined within a toroid. $(D)$ Magnetic field lines inside a bar magnet are not parallel. $(E)$ $\chi = -1$ is the condition for a perfect diamagnetic material, where $\chi$ is its magnetic susceptibility. Choose the correct answer from the options given below:
$(C)$ and $(E)$ only
$(B)$ and $(D)$ only
$(A)$ and $(B)$ only
$(B)$ and $(C)$ only
Answer: (a)
Solution
Statement $(C)$ is correct because, the magnetic field outside the toroid is zero and they form closed loops inside the toroid itself. Statement $(E)$ is correct because we know that superconductors are materials inside which the net magnetic field is always zero and they are perfect diamagnetic. $\mu_r = 1 + \chi$ $$\chi = -1$$ $$\mu_r = 0$$ For superconductors.
Question 32
Physics · System of Particles and Rotational Motion · Single correct
An object of mass $m_1$ collides with another object of mass $m_2$, which is at rest. After the collision the objects move with equal speeds in opposite direction. The ratio of the masses $m_2 : m_1$ is :
3: 1
2: 1
1: 2
1: 1
Answer: (a)
Solution
Given the equation of motion: $$m_1 v_1 = -m_1 v + m_2 v$$ Solving for $v_1$: $$v_1 = -v + \frac{m_2}{m_1} v$$ Rearranging gives: $$\frac{(v_1 + v)}{v} = \frac{m_2}{m_1}$$ The coefficient of restitution $e$ is given by: $$e = \frac{2v}{v_1} = 1$$ Solving for $v$ gives: $$v = \frac{v_1}{2}$$ Substituting back: $$\frac{v_1 + v_1/2}{v_1/2} = \frac{m_2}{m_1}$$ This simplifies to: $$3 = \frac{m_2}{m_1}$$
Question 33
Physics · Thermodynamics · Single correct
For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to (where $\gamma$ is the ratio of specific heats):
$-\gamma \frac{dV}{V}$
$-\gamma \frac{V}{dV}$
$-\frac{1}{\gamma} \frac{dV}{V}$
$\frac{dV}{V}$
Answer: (a)
Solution
Question 34
Physics · Moving Charges and Magnetism · Single correct
A proton and an $\alpha$-particle, having kinetic energies $K_p$ and $K_\alpha$, respectively, enter into a magnetic field at right angles. The ratio of the radii of trajectory of proton to that of $\alpha$-particle is $2 : 1$. The ratio of $K_p : K_\alpha$ is :
1 : 8
8 : 1
1 : 4
4 : 1
Answer: (d)
Solution
Given $\;r=\frac{mv}{qB}=\frac{p}{qB}$ and $\frac{m_\alpha}{m_p}=4$. $\frac{r_p}{r_\alpha}=\frac{p_p}{q_p}\frac{q_\alpha}{p_\alpha}=\frac{2}{1}$ $\frac{p_p}{p_\alpha}=\frac{2q_p}{q_\alpha}=2\left(\frac{1}{2}\right)$ $\frac{p_p}{p_\alpha}=1$ $\frac{K_p}{K_\alpha}=\frac{p_p^2}{p_\alpha^2}\frac{m_\alpha}{m_p}=(1)(4)$
Question 35
Physics · Electromagnetic Waves · Single correct
A plane electromagnetic wave propagating along y-direction can have the following pair of electric field $\left( \vec{E} \right)$ and magnetic field $\left( \vec{B} \right)$ components.
E_y, B_y or E_z, B_z
E_y, B_x or E_x, B_y
E_x, B_z or E_z, B_x
E_x, B_y or E_y, B_x
Answer: (c)
Solution
The diagrams show two different orientations of electric field $\mathbf{E}$, magnetic field $\mathbf{B}$, and wave vector $\mathbf{K}$. In the left diagram, $\mathbf{E}$ is along the $z$-axis, $\mathbf{B}$ is in the $xy$-plane, and $\mathbf{K}$ is along the $y$-axis. In the right diagram, $\mathbf{B}$ is along the $z$-axis, $\mathbf{E}$ is in the $xy$-plane, and $\mathbf{K}$ is along the $y$-axis.
Question 36
Physics · System of Particles and Rotational Motion · Single correct
Consider a uniform wire of mass $M$ and length $L$. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is :
$\frac{1}{4} \frac{ML^2}{\pi^2}$
$\frac{2}{5} \frac{ML^2}{\pi^2}$
$\frac{ML^2}{\pi^2}$
$\frac{1}{2} \frac{ML^2}{\pi^2}$
Answer: (c)
Solution
Given $\pi r = L$, we have $r = \frac{L}{\pi}$. The moment of inertia is $I = Mr^2 = \frac{ML^2}{\pi^2}$.
Question 37
Physics · Motion in a Straight Line · Single correct
The velocity-displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by :
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The correct relation between $\alpha$ (ratio of collector current to emitter current) and $\beta$ (ratio of collector current to base current) of a transistor is:
Physics · Ray Optics and Optical Instruments · Single correct
Three rays of light, namely red (R), green (G) and blue (B) are incident on the face PQ of a right angled prism PQR as shown in figure. The refractive indices of the material of the prism for red, green and blue wavelength are 1.27, 1.42 and 1.49 respectively. The colour of the ray(s) emerging out of the face PR is :
green
red
blue and green
blue
Answer: (b)
Solution
Assuming that the right angled prism is an isosceles prism, so the other angles will be $45^\circ$ each. Each incident ray will make an angle of $45^\circ$ with the normal at face PR. The wavelength corresponding to which the incidence angle is less than the critical angle, will pass through PR. Therefore, $\theta_C = critical angle$. $\theta_C = \sin^{-1}\left(\frac{1}{\mu}\right)$. If $\theta_C \geq 45^\circ$ the light ray will pass. $$(\theta_C)_{Red} = \sin^{-1}\left(\frac{1}{1.27}\right) = 51.94^\circ$$ Red will pass. $$(\theta_C)_{Green} = \sin^{-1}\left(\frac{1}{1.42}\right) = 44.76^\circ$$ Green will not pass. $$(\theta_C)_{Blue} = \sin^{-1}\left(\frac{1}{1.49}\right) = 42.15^\circ$$ Blue will not pass. So only red will pass through PR.
Question 40
Physics · Gravitation · Single correct
If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately : (Take : $g = 10 \, \mathrm{ms^{-2}}$, the radius of earth, $R = 6400 \times 10^3 \, \mathrm{m}$, Take $\pi = 3.14$)
60 minutes
does not change
1200 minutes
84 minutes
Answer: (d)
Solution
For objects to float $mg=m\omega^{2}R$ $\omega=$ angular velocity of earth $R=$ radius of earth $\omega=\sqrt{\frac{g}{R}} \qquad \cdots (1)$ Duration of day $=T$ $T=\frac{2\pi}{\omega} \qquad \cdots (2)$ $\Rightarrow T=2\pi\sqrt{\frac{R}{g}}$ $=2\pi\sqrt{\frac{6400\times10^{3}}{10}}$ $\Rightarrow \frac{T}{60}=83.775\ \text{minutes}$ $\simeq84\ \text{minutes}$
Question 41
Physics · Nuclei · Single correct
The decay of a proton to neutron is :
not possible as proton mass is less than the neutron mass
possible only inside the nucleus
not possible but neutron to proton conversion is possible
always possible as it is associated only with $\beta^+$ decay
Answer: (b)
Solution
It is possible only inside the nucleus and not otherwise.
Question 42
Physics · Alternating Current · Single correct
In a series LCR circuit, the inductive reactance ($X_L$) is $10\,\Omega$ and the capacitive reactance ($X_C$) is $4\,\Omega$. The resistance ($R$) in the circuit is $6\,\Omega$. The power factor of the circuit is:
$\frac{1}{2}$
$\frac{1}{2\sqrt{2}}$
$\frac{1}{\sqrt{2}}$
$\frac{\sqrt{3}}{2}$
Answer: (c)
Solution
We know that power factor is $\cos \phi$, $$\cos \phi = \frac{R}{Z} \ldots (1)$$ $$Z = \sqrt{R^2 + (X_L - X_C)^2} \ldots (2)$$ $$(\omega L - 1/\omega C)$$ $$\Rightarrow Z = \sqrt{6^2 + (10 - 4)^2}$$ $$\Rightarrow Z = 6\sqrt{2} \mid \cos \phi = \frac{6}{6\sqrt{2}}$$ $$\cos \phi = \frac{1}{\sqrt{2}}$$
Question 43
Physics · Gravitation · Single correct
The angular momentum of a planet of mass $M$ moving around the sun in an elliptical orbit is $\vec{L}$. The magnitude of the areal velocity of the planet is:
$\frac{4L}{M}$
$\frac{L}{M}$
$\frac{2L}{M}$
$\frac{L}{2M}$
Answer: (d)
Solution
For small displacement $ds$ of the planet, its area can be written as $$dA = \frac{1}{2} r d\ell$$ $$= \frac{1}{2} r ds \sin \theta$$ A. vel $= \frac{dA}{dt} = \frac{1}{2} r \sin \theta \frac{ds}{dt} = \frac{Vr \sin \theta}{2}$ $$\frac{dA}{dt} = \frac{1}{2} \frac{mVr \sin \theta}{m} = \frac{L}{2m}$$
Question 44
Physics · Oscillations · Single correct
The function of time representing a simple harmonic motion with a period of $\frac{\pi}{\omega}$ is:
Time period $T = \frac{2\pi}{\omega'}$ $$\frac{\pi}{\omega'} = \frac{2\pi}{\omega}$$ $$\omega' = 2\omega \rightarrow Angular frequency of SHM$$ Option (c) $$\sin^2 \omega t = \frac{1}{2} (2 \sin^2 \omega t) = \frac{1}{2} (1 - \cos 2\omega t)$$ Angular frequency of $\left( \frac{1}{2} - \frac{1}{2} \cos 2\omega t \right)$ is $2\omega$ Option (d) Angular frequency of SHM $$3 \cos \left( \frac{\pi}{4} - 2\omega t \right)$$ is $2\omega$. So option (c) $\&$ (d) both have angular frequency $2\omega$ but option (d) is direct answer.
Question 45
Physics · Laws of Motion · Single correct
A solid cylinder of mass $m$ is wrapped with an inextensible light string and, is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is : [The coefficient of static friction, $\mu_s$, is 0.4]
$\frac{7}{2} mg$
$5mg$
$\frac{mg}{5}$
0
Answer: (c)
Solution
Let's take solid cylinder is in equilibrium. $T + f = mg \sin 60 \ldots (i)$ $TR - fR = 0 \ldots (ii)$ Solving we get $$T = f_{req} = \frac{mg \sin \theta}{2}$$ But limiting friction $<$ required friction $$\mu mg \cos 60^\circ < \frac{mg \sin 60^\circ}{2}$$ Therefore, the cylinder will not remain in equilibrium. Hence $f = kinetic$ $$= \mu_k N$$ $$= \mu_k mg \cos 60^\circ$$ $$= \frac{mg}{5}$$
Question 46
Physics · Moving Charges and Magnetism · Single correct
The time taken for the magnetic energy to reach 25$\%$ of its maximum value, when a solenoid of resistance $R$, inductance $L$ is connected to a battery, is :
$\frac{L}{R} \ln 5$
infinite
$\frac{L}{R} \ln 2$
$\frac{L}{R} \ln 10$
Answer: (c)
Solution
Magnetic energy = $\frac{1}{2} L i^2 = 25\%$ ME $\Rightarrow 25\% \Rightarrow i = \frac{i_0}{2}$ $i = i_0 \left(1 - e^{-Rt/L}\right)$ for charging $t = \frac{L}{R} \ln 2$
Question 47
Physics · Work, Energy and Power · Single correct
A particle of mass $m$ moves in a circular orbit under the central potential field, $U(r) = \frac{-C}{r}$, where $C$ is a positive constant. The correct radius - velocity graph of the particle's motion is :
Answer: (a)
Solution
Given $U = -\frac{C}{r}$. The force $F$ is given by $F = -\frac{dU}{dr} = -\frac{C}{r^2}$. The magnitude of the force is $|F| = \frac{mv^2}{r}$. Equating the expressions for force, we have $\frac{C}{r^2} = \frac{mv^2}{r}$. This implies $v^2 \propto \frac{1}{r}$.
Question 48
Physics · Thermodynamics · Single correct
An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is $S_1$ and that of the other part is $S_2$. Given that $S_1 > S_2$. If the piston is removed then the total entropy of the system will be:
$S_1 \times S_2$
$S_1 - S_2$
$\frac{S_1}{S_2}$
$S_1 + S_2$
Answer: (d)
Solution
Question 49
Physics · Kinetic Theory · Single correct
Consider a sample of oxygen behaving like an ideal gas. At 300 K, the ratio of root mean square (rms) velocity to the average velocity of gas molecule would be: (Molecular weight of oxygen is 32 g/mol; $R = 8.3 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$)
$\sqrt{\frac{3}{3}}$
$\sqrt{\frac{8}{3}}$
$\sqrt{\frac{3\pi}{8}}$
$\sqrt{\frac{8\pi}{3}}$
Answer: (c)
Solution
Given $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ $$v_{avg} = \sqrt{\frac{8}{\pi} \frac{RT}{M}}$$ The ratio is $$\frac{v_{rms}}{v_{avg}} = \sqrt{\frac{3\pi}{8}}$$
Question 50
Physics · Wave Optics · Single correct
The speed of electrons in a scanning electron microscope is $1 \times 10^7 \, \mathrm{ms}^{-1}$. If the protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of:
1837
$\frac{1}{1837}$
$\sqrt{1837}$
$\frac{1}{\sqrt{1837}}$
Answer: (a)
Solution
Resolving power (RP) is proportional to $\frac{1}{\lambda}$. $$\lambda = \frac{h}{p} = \frac{h}{mv}$$ So, (RP) is proportional to $\frac{mv}{h}$. RP is proportional to $P$. RP is proportional to $mv$. RP is proportional to $m$.
Question 51
Physics · System of Particles and Rotational Motion · Numerical
The projectile motion of a particle of mass $5 \, \mathrm{g}$ is shown in the figure. The initial velocity of the particle is $5 \sqrt{2} \, \mathrm{ms^{-1}}$ and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points $A$ and $B$ is $x \times 10^{-2} \, \mathrm{kgms^{-1}}$. The value of $x$, to the nearest integer, is ___.
Answer: 5
Solution
Given $|\vec{u}| = |\vec{v}|$...(1) $\vec{u} = u \cos 45^\circ \hat{i} + u \sin 45^\circ \hat{j}$...(2) $\vec{v} = v \cos 45^\circ \hat{i} - v \sin 45^\circ \hat{j}$...(3) $|\Delta \vec{P}| = |m(\vec{v} - \vec{u})|$...(4) $\Delta P = 2mu \sin 45^\circ$ $= 2 \times 5 \times 10^{-3} \times 5 \sqrt{2} \times \frac{1}{\sqrt{2}}$ $= 50 \times 10^{-3}$ $= 5 \times 10^{-2}$
Question 52
Physics · Work, Energy and Power · Numerical
A ball of mass $4\,\mathrm{kg}$, moving with a velocity of $10\,\mathrm{m\,s^{-1}}$, collides with a spring of length $8\,\mathrm{m}$ and force constant $100\,\mathrm{N\,m^{-1}}$. The length of the compressed spring is $x\,\mathrm{m}$. The value of $x$, to the nearest integer, is:
Answer: 6
Solution
Let's say the compression in the spring by: $y$. So, by work energy theorem we have $$\frac{1}{2}mv^2 = \frac{1}{2}ky^2$$ $$\Rightarrow y = \sqrt{\frac{m}{k}} \cdot v$$ $$\Rightarrow y = \sqrt{\frac{4}{100}} \times 10$$ $$\Rightarrow y = 2\, \mathrm{m}$$ $$\Rightarrow final length of spring = 8 - 2 = 6\, \mathrm{m}$$
The typical output characteristics curve for a transistor working in the common-emitter configuration is shown in the figure. The estimated current gain from the figure is
Physics · Mechanical Properties of Fluids · Numerical
Consider a water tank as shown in the figure. It's cross-sectional area is $0.4 \, \mathrm{m}^2$. The tank has an opening B near the bottom whose cross-section area is $1 \, \mathrm{cm}^2$. A load of $24 \, \mathrm{kg}$ is applied on the water at the top when the height of the water level is $40 \, \mathrm{cm}$ above the bottom, the velocity of water coming out the opening B is $v \, \mathrm{ms}^{-1}$. The value of $v$, to the nearest integer, is ___. [Take value of $g$ to be $10 \, \mathrm{ms}^{-2}$]
A TV transmission tower antenna is at a height of 20 $\mathrm{m}$. Suppose that the receiving antenna is at. (i) ground level (ii) a height of 5 $\mathrm{m}$. The increase in antenna range in case (ii) relative to case (i) is $n\%$ The value of $n$, to the nearest integer, is.
Answer: 50
Solution
Range $= \sqrt{2Rh}$ Range (i) $= \sqrt{2Rh}$ Range (ii) $= \sqrt{2Rh} + \sqrt{2Rh'}$ where $h = 20 \, \mathrm{m}$ and $h' = 5 \, \mathrm{m}$ Ans $= \frac{\sqrt{2Rh'}}{\sqrt{2Rh}} \times 100\% = \frac{\sqrt{5}}{\sqrt{20}} \times 100\% = 50\%$
Question 56
Physics · Mathematics in Physics · Numerical
The radius of a sphere is measured to be $(7.50 \pm 0.85)\, \mathrm{cm}$. Suppose the percentage error in its volume is $x$. The value of $x$, to the nearest $x$, is ___.
Answer: 34
Solution
Given $v = \frac{4}{3} \pi r^3$. Taking log and then differentiate, $$\frac{dV}{V} = 3 \frac{dr}{r}$$ $$= \frac{3 \times 0.85}{7.5} \times 100\% = 34\%$$
Question 57
Physics · Electrostatic Potential and Capacitance · Numerical
An infinite number of point charges, each carrying $1 \, \mu \mathrm{C}$ charge, are placed along the $y$-axis at $y = 1 \, \mathrm{m}, 2 \, \mathrm{m}, 4 \, \mathrm{m}, 8 \, \mathrm{m} \ldots$ The total force on a $1 \, \mathrm{C}$ point charge, placed at the origin, is $x \times 10^3 \, \mathrm{N}$. The value of $x$, to the nearest integer, is ___. [Take $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \mathrm{Nm^2/C^2}$]
Answer: 12
Solution
The force $F$ is given by the formula: $$F = k(1 \, \mathrm{C})(1 \, \mu \mathrm{C}) \left[ 1 + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \ldots \right]$$ Calculating the series: $$= 9 \times 10^3 \left[ \frac{1}{1 - \frac{1}{4}} \right] = 12 \times 10^3 \, \mathrm{N}$$
Question 58
Physics · Experimental Physics · Numerical
Consider a 72 cm long wire AB as shown in the figure. The galvanometer jockey is placed at P on AB at a distance $x cm$ from A. The galvanometer shows zero deflection. The value of $x$, to the nearest integer, is
Two wires of same length and thickness having specific resistances $6\, \Omega \mathrm{cm}$ and $3\, \Omega \mathrm{cm}$ respectively are connected in parallel. The effective resistivity is $\rho \Omega \mathrm{cm}$. The value of $\rho$ to the nearest integer, is
A galaxy is moving away from the earth at a speed of $286 \, \mathrm{km/s}$. The shift in the wavelength of a red line at $630 \, \mathrm{nm}$ is $x \times 10^{-10} \, \mathrm{m}$. The value of $x$, to the nearest integer, is ___. [Take the value of speed of light $c$, as $3 \times 10^8 \, \mathrm{m/s}$]
Answer: 6
Solution
Given $\frac{\Delta\lambda}{\lambda}=\frac{v}{c}$ $\Delta\lambda=\frac{v}{c}\times\lambda$ $=\frac{286}{3\times10^5}\times630\times10^{-9}$ $=6\times10^{-10}$
Chemistry
Question 61
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
The oxidation states of nitrogen in $NO, NO_2, N_2O$ and $NO_3^-$ are in the order of:
NO_3^- > NO_2 > NO > N_2O
NO_2 > NO_3^- > NO > N_2O
N_2O > NO_2 > NO > NO_3^-
NO > NO_2 > N_2O > NO_3^-
Answer: (a)
Question 62
Chemistry · Hydrogen · Single correct
In basic medium, $\mathrm{H_2O_2}$ exhibits which of the following reactions? $(A)$ $\mathrm{Mn^{2+} \to Mn^{4+}}$ $(B)$ $\mathrm{I_2 \to I^-}$ $(C)$ $\mathrm{PbS \to PbSO_4}$ Choose the most appropriate answer from the options given below:
$(A), (C)$ only
$(A)$ only
$(B)$ only
$(A), (B)$ only
Answer: (d)
Solution
In basic medium, oxidising action of $\mathrm{H_2O_2} \; \mathrm{Mn^{2+}} + \mathrm{H_2O_2} \rightarrow \mathrm{Mn^{4+}} + 2\mathrm{OH}$. In basic medium, reducing action of $\mathrm{H_2O_2} \; \mathrm{I_2} + \mathrm{H_2O_2} + 2\mathrm{OH}^- \rightarrow 2\mathrm{I}^- + 2\mathrm{H_2O} + \mathrm{O_2}$. In acidic medium, oxidising action of $\mathrm{H_2O_2} \; \mathrm{PbS(s)} + 4\mathrm{H_2O_2(aq)} \rightarrow \mathrm{PbSO_4(s)} + 4\mathrm{H_2O(\ell)}$. Hence correct option (d).
Question 63
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
In the reaction of hypobromite with amide, the carbonyl carbon is lost as :
$\mathrm{CO_3^{2-}}$
$\mathrm{HCO_3^{-}}$
$\mathrm{CO_2}$
$\mathrm{CO}$
Answer: (a)
Solution
The reaction is as follows: $$\mathrm{R{-}C{-}NH_2 + Br_2 + 4NaOH \rightarrow}$$ $$\mathrm{R{-}NH_2 + Na_2CO_3 + 2NaBr + 2H_2O}$$ Mechanism: 1. $$\mathrm{R{-}C{-}NH_2 + OH^- \rightarrow R{-}C{-}NH + Br^-}$$ 2. $$\mathrm{R{-}C{-}NH + Br^- \rightarrow R{-}C{-}N{-}Br}$$ 3. $$\mathrm{R{-}C{-}N{-}Br \rightarrow R{-}N{=}C{=}O}$$ 4. $$\mathrm{R{-}N{=}C{=}O + H_2O \rightarrow R{-}NH_2 + Na_2CO_3}$$ The reaction involves the formation of an intermediate isocyanate, which hydrolyzes to give the final products.
Question 64
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The oxide that shows magnetic property is :
$\mathrm{SiO_2}$
$\mathrm{Mn_3O_4}$
$\mathrm{Na_2O}$
$\mathrm{MgO}$
Answer: (b)
Solution
$\mathrm{Mn_3O_4}$ shows magnetic properties.
Question 65
Chemistry · Amines · Single correct
Main Products formed during a reaction of 1-methoxy naphthalene with hydroiodic acid are:
Answer: (b)
Solution
Question 66
Chemistry · Biomolecules · Single correct
Deficiency of vitamin K causes:
Increase in blood clotting time
Increase in fragility of RBC's
Cheilosis
Decrease in blood clotting time
Answer: (a)
Solution
Due to deficiency of Vitamin K, there is an increase in blood clotting time. Note: Vitamin K is related to blood factor.
Question 67
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
An organic compound "A" on treatment with benzene sulphonyl chloride gives compound B. B is soluble in dil. NaOH solution. Compound A is :
$C_6H_5 - N - (CH_3)_2$
$C_6H_5 - NHCH_2CH_3$
$C_6H_5 - CH_2NHCH_3$
Answer: (d)
Solution
Hinsberg reagent (Benzene sulphonyl chloride) gives reaction product with $1^\circ$ amine and it is soluble in dil. NaOH. $$R - \overset{\cdot \cdot}{N}H_2 + Cl - \overset{O}{\overset{\parallel}{S}} - O \rightarrow R - N - \overset{O}{\overset{\parallel}{S}} - O$$ (A) ($1^\circ$ amine) $$dil. NaOH \rightarrow R - \overset{O}{\overset{\parallel}{S}} - O$$ (B)
Question 68
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The first ionization energy of magnesium is smaller as compared to that of elements $X$ and $Y$, but higher than that of $Z$. the elements $X$, $Y$ and $Z$, respectively, are:
chlorine, lithium and sodium
argon, lithium and sodium
argon, chlorine and sodium
neon, sodium and chlorine
Answer: (c)
Solution
The 1st IE order of 3rd period is $\mathrm{Na} < \mathrm{Al} < \mathrm{Mg} < \mathrm{Si} < \mathrm{S} < \mathrm{P} < \mathrm{Cl} < \mathrm{Ar}$. $X \& Y$ are $\mathrm{Ar} \& \mathrm{Cl}$. $Z$ is sodium (Na).
Question 69
Chemistry · The s-Block Elements · Single correct
The secondary valency and the number of hydrogen bonded water molecule(s) in $\mathrm{CuSO_4 \cdot 5H_2O}$, respectively, are
6 and 4
4 and 1
6 and 5
5 and 1
Answer: (b)
Solution
Hydrogen bonded water molecule = 1 Secondary valency = 4
Question 70
Chemistry · Structure of Atom · Single correct
Given below are two statements: Statement I: Bohr's theory accounts for the stability and line spectrum of $\mathrm{Li}^{+}$ ion. Statement II: Bohr's theory was unable to explain the splitting of spectral lines in the presence of a magnetic field. In the light of the above statements, choose the most appropriate answer from the options given below:
Both statement I and statement II are true.
Statement I is false but statement II is true.
Both statement I and statement II are false.
Statement I is true but statement II is false.
Answer: (b)
Solution
Statement-I is false since Bohr's theory accounts for the stability and spectrum of single electronic species (e.g.: $\mathrm{He^+}$, $\mathrm{Li^{2+}}$ etc). Statement II is true.
Question 71
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
C > A > B
B > C > A
A > C > B
C > B > A
Answer: (d)
Solution
Aniline reacts with $\mathrm{HNO_3} + \mathrm{H_2SO_4}$ at $288 \, \mathrm{K}$ to form three products. The products are: (A) with a yield of $2\%$, (B) with a yield of $47\%$, and (C) with a yield of $51\%$. The percentage yield order is $C > B > A$.
Question 72
Chemistry · Surface Chemistry · Single correct
The charges on the colloidal CdS sol and $TiO_2$ sol are, respectively:
positive and positive
positive and negative
negative and negative
negative and positive
Answer: (d)
Solution
CdS sol leads to negative sol. $TiO_2$ sol leads to positive sol.
Question 73
Chemistry · Chemistry in Everyday Life · Single correct
Match the list -I with list - II \begin{tabular}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ (Class of Chemicals) & (Example) \\ \hline (a) Antifertility drug & (i) Meprobamate \\ \hline (b) Antibiotic & (ii) Alitame \\ \hline (c) Tranquilizer & (iii) Norethindrone \\ \hline (d) Artificial Sweetener & (iv) Salvarsan \\ \hline \end{tabular}
(a) - (ii), (b)-(iii), ($c$)-(i v), (d)-(i)
(a) - (iv), (b)-(iii), ($c$)-(ii), (d)-(i)
(a) - (iii), (b)-(iv), ($c$)-(i), (d)-(ii)
(a) - (ii), (b)-(iv), ($c$)-(i), (d)-(iii)
Answer: (c)
Solution
(A) Antifertility drug $\rightarrow$ (iii) Nor ethindrone (B) Antibiotic $\rightarrow$ (iv) Salvarsan (C) Tranquilizer $\rightarrow$ (i) Meprobamate (D) Artificial sweetener $\rightarrow$ (ii) Alitame A-iii, B-iv, C-i, D-ii
Question 74
Chemistry · Alcohols, Phenols and Ethers · Single correct
Consider the above reaction, the product 'X' and 'Y' respectively are:
Answer: (c)
Solution
The reaction begins with the deprotonation of the carbonyl compound by hydroxide ion, resulting in the formation of an enolate ion. This enolate ion then undergoes an intramolecular nucleophilic attack, leading to the formation of a cyclic intermediate. The intermediate loses water to form a new compound. Upon heating (denoted by $\Delta$), the compound (X) undergoes a rearrangement to form compound (Y).
Question 75
Chemistry · The s-Block Elements · Single correct
Match list-I with list-II: List-I List-II Choose the most appropriate answer the option given below:
a - iv, b - iii, c - i, d - ii
a - iv, b - iii, c - ii, d - i
a - iii, b - iv, c - v, d - ii
a - iii, b - iv, c - ii, d - v
Answer: (b)
Solution
(a) Be → it is used in the Windows of X-ray tubes (b) Mg → it is used in the Incendiary bombs and signals (c) Ca → it is used in the Extraction of metals (d) Ra → it is used in the Treatment of cancer
Question 76
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Given below are two statements: Statement I: $\mathrm{C_2H_5OH}$ and $\mathrm{AgCN}$ both can generate nucleophiles. Statement II: $\mathrm{KCN}$ and $\mathrm{AgCN}$ both generate nitrile nucleophiles under all reaction conditions.
Statement I is true but statement II is false
Both statement I and statement II are true
Statement I is false but statement II is true
Both statement I and statement II are false
Answer: (a)
Solution
The given equation is $\frac{1}{2} \times 2 \times 2 = 2$. Therefore, the answer is $2$.
Question 77
Chemistry · Environmental Chemistry · Single correct
Given below are two statements: Statement I: Non-biodegradable wastes are generated by the thermal power plants. Statement II: Bio-degradable detergents leads to eutrophication. In the light of the above statements, choose the most appropriate answer from the option given below
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
Both statement I and statement II are true.
Answer: (d)
Solution
Non-biodegradable wastes are generated by the thermal power plants which produces fly ash. Detergents which are biodegradable causes problem called eutrophication which kills animal life by depriving it of oxygen.
Question 78
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Match list-I with list-II : \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \hline (a) & Mercury & (i) & Vapour phase refining \\ \hline (b) & Copper & (ii) & Distillation refining \\ \hline (c) & Silicon & (iii) & Electrolytic refining \\ \hline (d) & Nickel & (iv) & Zone refining \\ \hline \end{tabular} Choose the most appropriate answer from the option given below :
Chemistry · Chemical Bonding and Molecular Structure · Single correct
In the following molecules, Hybridisation of carbon a, b and c respectively are :
sp^3, sp, sp
sp^3, sp^2, sp
sp^3, sp^2, sp^2
sp^3, sp, sp^2
Answer: (c)
Solution
The hybridization states of the carbon atoms in the given structure are as follows: Carbon labeled as $a$ is $sp^3$ hybridized. Carbon labeled as $b$ is $sp^2$ hybridized. The oxygen labeled as $c$ is $sp^2$ hybridized.
Question 80
Chemistry · Surface Chemistry · Single correct
A hard substance melts at high temperature and is an insulator in both solid and in molten state. This solid is most likely to be a / an :
Ionic solid
Molecular solid
Metallic solid
Covalent solid
Answer: (d)
Solution
Covalent or network solids have very high melting points and they are insulators in their solid and molten form.
Question 81
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
A reaction has a half life of 1 min. The time required for 99.9$\%$ completion of the reaction is_____ min. (Round off to the Nearest integer) [ Use : $\ln$ 2 = 0.69, $\ln$ 10 = 2.3]
The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 $\mathrm{Scm}^2 \mathrm{mol}^{-1}$ respectively. The molar conductivity at infinite dilution of barium sulphate is ______ $\mathrm{Scm}^2 \mathrm{mol}^{-1}$ (Round off to the Nearest Integer).
Chemistry · Chemical Bonding and Molecular Structure · Numerical
The number of species below that have two lone pairs of electrons in their central atom is _______ (Round off to the Nearest integer) $\mathrm{SF_4}$, $\mathrm{BF_4^-}$, $\mathrm{ClF_3}$, $\mathrm{AsF_3}$, $\mathrm{PCl_5}$, $\mathrm{BrF_5}$, $\mathrm{XeF_4}$, $\mathrm{SF_6}$
Answer: 2
Solution
Two lone pairs on the central atom are in $\mathrm{ClF_3}$ and $\mathrm{XeF_4}$.
Question 84
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
A xenon compound 'A' upon partial hydrolysis gives $\mathrm{XeO_2F_2}$. The number of lone pair of electrons present in compound A is _______ (Round off to the Nearest integer)
Answer: 19
Question 85
Chemistry · Equilibrium · Numerical
For the gas-phase reaction $\mathrm{2A(g) \rightleftharpoons A_2(g)}$ at $400\,\mathrm{K}$, $\Delta G^\circ = +25.2\,\mathrm{kJ\,mol^{-1}}$. The equilibrium constant $K_C$ for this reaction is \underline{\hspace{1cm}} $\times 10^{-2}$. (Round off to the nearest integer.) Use: $R = 8.3\,\mathrm{J\,mol^{-1}\,K^{-1}}$, $\ln 10 = 2.3$, $\log_{10} 2 = 0.30$, $1\,\mathrm{atm} = 1\,\mathrm{bar}$, $\mathrm{antilog}(-0.3) = 0.501$.
Chemistry · Alcohols, Phenols and Ethers · Numerical
In Tollen's test for aldehyde, the overall number of electron(s) transferred to the Tollen's reagent formula $[\mathrm{Ag(NH_3)_2}]^+$ per aldehyde group to form silver mirror is _______. (Round off to the Nearest integer)
Answer: 2
Solution
Question 87
Chemistry · Equilibrium · Numerical
The solubility of CdSO$_4$ in water is $8.0 \times 10^{-4} \, \mathrm{mol \, L^{-1}}$. Its solubility in $0.01 \, \mathrm{M} \, \mathrm{H_2SO_4}$ solution is _____ (Round off to the Nearest integer) (Assume that solubility is much less than $0.01 \, \mathrm{M}$ )
Answer: 64
Solution
In pure water, $K_{sp}=S^2=\left(8\times10^{-4}\right)^2$ $=64\times10^{-8}$ In $0.01\,\mathrm{M}\ \mathrm{H_2SO_4}$, $\mathrm{H_2SO_4(aq)\rightarrow 2H^+_{(aq)}+SO_4^{2-}_{(aq)}}$ $0.02 \hspace{2cm} 0.01$ $\mathrm{BaSO_4(s)\rightleftharpoons Ba^{2+}_{(aq)}+SO_4^{2-}_{(aq)}}$ $x \hspace{2.2cm} x+0.01$ $K_{sp}=x(x+0.01)$ $=64\times10^{-8}$ $x+0.01\approx0.01\,\mathrm{M}$ So, $x(0.01)=64\times10^{-8}$ $x=64\times10^{-6}\,\mathrm{M}$
Question 88
Chemistry · Solutions · Numerical
A solute a dimerizes in water. The boiling point of a 2 molar solution of A is $100.52^{\circ} \mathrm{C}$. The percentage association of A is ________ (Round off to the Nearest integer) [Use : $K_b$ for water $= 0.52 \, \mathrm{K \, kg \, mol^{-1}}$ Boiling point of water $= 100^{\circ} \mathrm{C}$]
$10.0\,\mathrm{mL}$ of $\mathrm{Na_2CO_3}$ solution is titrated against $0.2\,\mathrm{M}$ $\mathrm{HCl}$ solution. The following titre values were obtained in 5 readings: $4.8\,\mathrm{mL}$, $4.9\,\mathrm{mL}$, $5.0\,\mathrm{mL}$, $5.0\,\mathrm{mL}$ and $5.0\,\mathrm{mL}$. Based on these readings, and convention of titrimetric estimation, the concentration of $\mathrm{Na_2CO_3}$ solution is \_\_\_\_ $\mathrm{mM}$ (Round off to the Nearest integer)
Answer: 50
Solution
Most precise volume of $\mathrm{HCl} = 5\,\mathrm{mL}$ at equivalence point. $\text{meq. of } \mathrm{Na_2CO_3} = \text{meq. of } \mathrm{HCl}$ Let molarity of $\mathrm{Na_2CO_3}$ solution $= M$, then $$M \times 10 \times 2 = 0.2 \times 5 \times 1$$ $$M = 0.05\,\mathrm{mol/L}$$ $$= 0.05 \times 1000$$
Question 90
Chemistry · Some Basic Concepts of Chemistry · Numerical
Consider the above reaction where $6.1 \, \mathrm{g}$ of benzoic acid is used to get $7.8 \, \mathrm{g}$ of m-bromo benzoic acid. The percentage yield of the product is ____ (Round off to the Nearest integer) [Given : Atomic masses : $C = 12.0\mathrm{u}$, $H = 1.0\mathrm{u}$, $O : 16.0\mathrm{u}$, $Br = 80.0\mathrm{u}$]
Answer: 78
Solution
Moles of Benzoic acid $= \frac{6.1}{122}$ moles of m-bromobenzoic acid. So, weight of m-bromobenzoic acid $$= \frac{6.1}{122} \times 201 \, \mathrm{gm}$$ $$= 10.05 \, \mathrm{gm}$$ Percentage yield $= \frac{Actual weight}{Theoretical weight} \times 100$ $$= \frac{7.8}{10.05} \times 100$$ $$= 77.61\%$$