JEE Main 28 July 2022 Shift 1 question paper with solutions

JEE Main 28 July 2022 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 · Differential Equations · Single correct

Let the solution curve of the differential equation $$x \, dy = \left( \sqrt{x^2 + y^2} + y \right) \, dx, \; x > 0,$$ intersect the line $$x = 1$$ at $$y = 0$$ and the line $$x = 2$$ at $$y = \alpha$$. Then the value of $$\alpha$$ is:

  1. $\frac{1}{2}$
  2. $\frac{3}{2}$
  3. $-\frac{3}{2}$
  4. $\frac{5}{2}$

Answer: (b)

Solution

Given the equation $x \, dy = \left( \sqrt{x^2 + y^2 + y} \right) dx$. Rewriting, we have $$\frac{x \, dy}{x} = y \, dx - \frac{\sqrt{x^2 + y^2}}{x} dx.$$ This simplifies to $$\frac{d \left( \frac{y}{x} \right)}{\sqrt{1 + \left( \frac{y}{x} \right)^2}} = \frac{dx}{x}.$$ Integrating both sides, we get $$\ln \left( \frac{y}{x} + \sqrt{\left( \frac{y}{x} \right)^2 + 1} \right) = \ln x + R.$$ Thus, $$\frac{y + \sqrt{y^2 + x^2}}{x} = cx.$$ Therefore, $$y + \sqrt{y^2 + x^2} = cx^2.$$ Setting $x = 1, y = 0$, we find $0 + 1 = C \Rightarrow C = 1$. The curve is $y + \sqrt{x^2 + y^2} = x^2$. For $x = 2, y = \alpha$, we have $$2 + \sqrt{4 + \alpha^2} = 4.$$ Solving, $$4 + \alpha^2 = 16 + \alpha^2 = 8 \alpha.$$ Therefore, $\alpha = \frac{3}{2}$.

Question 2

Maths · Inverse Trigonometric Functions · Single correct

Considering only the principal values of the inverse trigonometric functions, the domain of the function $f(x) = \cos^{-1}\left(\frac{x^2 - 4x + 2}{x^2 + 3}\right)$ is:

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

Answer: (b)

Solution

Given $\($ $\left$| $\frac{x^2 + 4x + 2}{x^2 + 3}$ $\right$| $\leq$ 1 $\)$ This implies $\($ (x^2 - 4x + 2)^2 $\leq$ (x^2 + 3)^2 $\)$ This implies $\($ (x^2 - 4x + 2)^2 - (x^2 + 3)^2 $\leq$ 0 $\)$ This implies $\($ (2x^2 - 4x + 5)(-4x - 1) $\leq$ 0 $\)$ This implies $\($ -4x - 1 $\leq$ 0 $\rightarrow$ x $\geq$ -$\frac{1}{4}$ $\)$

Question 3

Maths · Vector Algebra · Single correct

Let the vectors $\vec{a} = (1+t)\hat{i} + (1-t)\hat{j} + \hat{k}$, $\vec{b} = (1-t)\hat{i} + (1+t)\hat{j} + 2\hat{k}$ and $\vec{c} = t\hat{i} - t\hat{j} + \hat{k}$, $t \in \mathbb{R}$ be such that for $\alpha, \beta, \gamma \in \mathbb{R}$, $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = \vec{0}$ $\Rightarrow \alpha = \beta = \gamma = 0$. Then, the set of all values of $t$ is:

  1. a non-empty finite set
  2. equal to $\mathbb{N}$
  3. equal to $\mathbb{R} - \{0\}$
  4. equal to $\mathbb{R}$

Answer: (c)

Solution

By its given condition, $\vec{a}, \vec{b}, \vec{c}$ are linearly independent vectors. Therefore, $$\left[ \vec{a} \; \vec{b} \; \vec{c} \right] \neq 0 (i)$$ Now, $$\left[ \begin{array}{ccc} 1+t & 1-t & 1 \\ 1-t & 1+t & 2 \\ t & -t & 1 \end{array} \right]$$ Applying $C_2 \rightarrow C_1 + C_2$, we get $$\left[ \begin{array}{ccc} 1+t & 2 & 1 \\ 1-t & 2 & 2 \\ t & 0 & 1 \end{array} \right]$$ $$= 2 \left| \begin{array}{ccc} 1+t & 1 & 1 \\ 1-t & 1 & 2 \\ t & 0 & 1 \end{array} \right|$$ $$= 2 \left[ (1+t) - (1-t) + t \right]$$ $$= 2[3t] = 6t$$ Therefore, $$\left[ \vec{a} \; \vec{b} \; \vec{c} \right] \neq 0 \Rightarrow t \neq 0$$

Question 4

Maths · Inverse Trigonometric Functions · Single correct

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $\cos^{-1}(x) - 2\sin^{-1}(x) = \cos^{-1}(2x)$ is equal to:

  1. 0
  2. 1
  3. $\frac{1}{2}$
  4. $-\frac{1}{2}$
Solution

Given $\cos^{-1} x = 2 \sin^{-1} x = \cos^{-1} 2x$. $$\cos^{-1} x - 2 \left( \frac{\pi}{2} - \cos^{-1} x \right) = \cos^{-1} 2x$$ $$\cos^{-1} x - \pi + 2 \cos^{-1} x = \cos^{-1} 2x$$ $$3 \cos^2 x = \pi + \cos^{-1} 2x$$ $$\cos \left( 3 \cos^{-1} x \right) = \cos \left( \pi + \cos^{-1} 2x \right)$$ $$4x^3 - 3x = -2x$$ $$4x^3 = x \implies x = 0, \pm \frac{1}{2}$$ All satisfy the original equation. Sum $= -\frac{1}{2}$ to $+\frac{1}{2} = 0$

Question 5

Maths · Mathematical Reasoning · Single correct

Let the operations $*$, $\odot \in \{\land, \lor\}$. If $(p * q) \odot (p \odot \sim q)$ is a tautology, then the ordered pair $(*, \odot)$ is:

  1. $(\lor, \land)$
  2. $(\lor, \lor)$
  3. $(\land, \land)$
  4. $(\land, \lor)$
Solution

Well check each option. For A $\pi = \lor$ of $0 = \Lambda$. $$(p \lor q) \land (p \lor \sim q)$$ $$\equiv p \lor (q \land \sim q)$$ $$\equiv p \lor (c) \equiv p$$ For B: $* = \lor$, $O = \lor$. $$(p \lor q) \lor (p \lor \sim q) \equiv t$$ using Venn Diagrams.

Question 6

Maths · Vector Algebra · Single correct

Let a vector $\vec{a}$ has a magnitude 9. Let a vector $\vec{b}$ be such that for every $(x,y) \in \mathbb{R} \times \mathbb{R} - \{(0,0)\}$, the vector $(x\vec{a} + y\vec{b})$ is perpendicular to the vector $(6y\vec{a} - 18x\vec{b})$. Then the value of $|\vec{a} \times \vec{b}|$ is equal to:

  1. $9\sqrt{3}$
  2. $27\sqrt{3}$
  3. 9
  4. 81

Answer: (b)

Solution

Given $|\bar{a}| = 9$ and $(\bar{x} \bar{a} + \bar{y} \bar{b}) \cdot (6y \bar{a} - 18x \bar{b}) = 0$. $$\Rightarrow \ 6xy |\bar{a}|^2 - 18x^2 (\bar{a} \cdot \bar{b}) + 6y^2 (\bar{a} \cdot \bar{b}) - 18xy |\bar{b}|^2 = 0$$ $$\Rightarrow \ 6xy \left( |\bar{a}|^2 - 3 |\bar{b}|^2 \right) + (\bar{a} \cdot \bar{b}) (y^2 - 3x^2) = 0$$ This should hold $\forall \ x, y \in \mathbb{R} \times \mathbb{R}$. Therefore, $|\bar{a}|^2 = 3 |\bar{b}|^2$ and $(\bar{a} \cdot \bar{b}) = 0$. Now $|\bar{a} \times \bar{b}|^2 = |\bar{a}|^2 |\bar{b}|^2 - (\bar{a} \cdot \bar{b})^2$. $$= |\bar{a}|^2 \cdot \frac{|\bar{a}|^2}{3}$$ Therefore, $|\bar{a} \times \bar{b}| = \frac{|\bar{a}|^2}{\sqrt{3}} = \frac{81}{\sqrt{3}} = 27 \sqrt{3}$.

Question 7

Maths · Properties of Triangles · Single correct

For $t \in (0, 2\pi)$, if $ABC$ is an equilateral triangle with vertices $A(\sin t, -\cos t)$, $B(\cos t, \sin t)$ and $C(a, b)$ such that its orthocentre lies on a circle with centre $\left(1, \frac{1}{3}\right)$, then $(a^2 - b^2)$ is equal to:

  1. $\frac{8}{3}$
  2. 8
  3. $\frac{77}{9}$
  4. $\frac{80}{9}$

Answer: (b)

Solution

Given $s \equiv \sin t$, $c \equiv \cos t$. Let orthocentre be $(h, k)$. Since it is an equilateral triangle, hence orthocentre coincides with centroid. Therefore, $a + s + c = 3h$, $b + s - c = 3k$. Thus, $$(3h - a)^2 + (3k - b)^2 = (s + c)^2 + (s - c)^2 = 2(s^2 + c^2) = 2$$ Therefore, $$\left( h - \frac{a}{3} \right)^2 + \left( k - \frac{b}{3} \right)^2 = \frac{2}{9},$$ circle centre at $\left( \frac{a}{3}, \frac{b}{3} \right)$. Gives, $\frac{a}{3} = 1$, $\frac{b}{3} = \frac{1}{3} \implies a = 3, b = 1$ Thus, $a^2 - b^2 = 8$

Question 8

Maths · Relations and Functions · Single correct

For $\alpha \in \mathbb{N}$, consider a relation $R$ on $\mathbb{N}$ given by $R = \{(x,y) : 3x + \alpha y$ is a multiple of $7\}$. The relation $R$ is an equivalence relation if and only if:

  1. $\alpha = 14$
  2. $\alpha$ is a multiple of $4$
  3. $4$ is the remainder when $\alpha$ is divided by $10$
  4. $4$ is the remainder when $\alpha$ is divided by $7$

Answer: (d)

Solution

For R to be reflexive $\Rightarrow \ x \, R \, x$ $$\Rightarrow \ 3x + \alpha \, x = 7x \Rightarrow (3 + \alpha) \, x = 7K$$ $$\Rightarrow \ 3 + \alpha = 7\lambda \Rightarrow \alpha = 7\lambda - 3 = 7N + 4, \, K, \, \lambda, \, N \in \mathbb{I}$$ Therefore, when $\alpha$ divided by 7, remainder is 4. R to be symmetric $xRy \Rightarrow yRx$ $$3x + \alpha y = 7N_1, \, 3y + \alpha x = 7N_2$$ $$\Rightarrow (3 + \alpha)(x + y) = 7(N_1 + N_2) = 7N_3$$ Which holds when $3 + \alpha$ is multiple of 7 Therefore, $\alpha = 7N + 4$ (as did earlier) R to be transitive $xRy \& \, yRz \Rightarrow xRz.$ $$3x + \alpha y = 7N_1 \& 3y + \alpha z = 7N_2 and$$ $$3x + \alpha z = 7N_3$$ Therefore, $3x + 7N_2 - 3y = 7N_3$ Therefore, $7N_1 - \alpha y + 7N_2 - 3y = 7N_3$ Therefore, $7(N_1 + N_2) - (3 + \alpha) y = 7N_3$ Therefore, $(3 + \alpha) y = 7N$ Which is true again when $3 + \alpha$ divisible by 7, i.e. when $\alpha$ divided by 7, remainder is 4.

Question 9

Maths · Probability · Single correct

Out of 60$\%$ female and 40$\%$ male candidates appearing in an exam, 60$\%$ candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :

  1. $\frac{2}{3}$
  2. $\frac{11}{16}$
  3. $\frac{23}{32}$
  4. $\frac{13}{16}$

Answer: (a)

Solution

Probability that chosen candidate is female = $\frac{40}{60}$ = $\frac{2}{3}$

Question 10

Maths · Differential Equations · Single correct

If $$ y = y(x), \; x \in \left(0, \frac{\pi}{2}\right) \,$$ be the solution curve of the differential equation $$\left(\sin^2 2x\right) \frac{dy}{dx} + \left(8 \sin^2 2x + 2 \sin 4x\right) y = 2e^{-4x} \left(2 \sin 2x + \cos 2x\right), \; with \; y\left(\frac{\pi}{4}\right) = e^{-\pi},$$ then $$\, y\left(\frac{\pi}{6}\right) \,$$ is equal to:

  1. $\frac{2}{\sqrt{3}} e^{-2\pi/3}$
  2. $\frac{2}{\sqrt{3}} e^{2\pi/3}$
  3. $\frac{1}{\sqrt{3}} e^{-2\pi/3}$
  4. $\frac{1}{\sqrt{3}} e^{2\pi/3}$

Answer: (a)

Solution

Given differential equation can be re-written as $$\frac{dy}{dx} + \left(8 + 4 \cot 2x \right) y = \frac{2e^{-4x}}{\sin^2 2x} \left(2 \sin x + \cos 2x \right)$$ which is a linear diff. equation. I.f. = $$e^{\int (8 + 4 \cot 2x) \, dx} = e^{8x + 2 \ln(\sin 2x)}$$ = $$e^{8x} \cdot \sin^2 2x$$ Therefore, solution is $$y \left(e^{8x} \cdot \sin^2 2x \right) = \int 2e^{4x} \left(2 \sin 2x + \cos 2x \right) dx + C$$ = $$e^{4x} \cdot \sin 2x + C$$ Given $$y \left(\frac{\pi}{4} \right) = e^{-\pi} \implies C = 0$$ Therefore, $$y = \frac{e^{-4x}}{\sin 2x}$$ Thus, $$y \left(\frac{\pi}{6} \right) = \frac{e^{-\frac{4 \cdot \pi}{6}}}{\sin \left(2 \cdot \frac{\pi}{6} \right)} = \frac{2}{\sqrt{3}} e^{\frac{2\pi}{3}}$$

Question 11

Maths · Conic Sections · Single correct

If the tangents drawn at the points P and Q on the parabola $y^2 = 2x - 3$ intersect at the point R(0, 1), then the orthocentre of the triangle PQR is:

  1. (0, 1)
  2. (2, -1)
  3. (6, 3)
  4. (2, 1)

Answer: (b)

Solution

Given $y^2 = 2x - 3$ (i). Equation of chord of contact is $y = x - 3$ (2). From (1) and (2), $$(x \cdot 3)^2 = 2x - 3$$ $$x^2 - 8x + 12 = 0$$ $$(x - 2)(x - 6) = 0$$ $$x = 2 or 6$$ $$y = -1 or 3$$ MQR is $$\frac{2}{6} = \frac{1}{3}$$ MPR is $$\frac{2}{6} = \frac{1}{3}$$ MPR is $$\frac{2}{-2} = -1$$ Therefore, $MPQ \times MPR = -1 \implies PQ \perp PR$. Orthocentre is $P(2, -1)$.

Question 12

Maths · Conic Sections · Single correct

Let C be the centre of the circle $x^2 + y^2 - x + 2y = \frac{11}{4}$ and P be a point on the circle. A line passes through the point C, makes an angle of $\frac{\pi}{4}$ with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit$^2$) is :

  1. 2
  2. 2$\sqrt{2}$
  3. 8\sin\left(\frac{\pi}{8}\right)
  4. 8\cos\left(\frac{\pi}{8}\right)

Answer: (b)

Solution

Given the equation $x^2 + y^2 - x + 2y = \frac{11}{4}$. Rewriting it, we have: $$\left(x - \frac{1}{2}\right)^2 + (y + 1)^2 = (2)^2$$ Or consider $\triangle PQR$. From the diagram, we have: $$= 4 \cdot 6 \sin \frac{\pi}{8}$$ $$PQ = QR \cos 22 \frac{1}{2}$$ $$= 4 \cos \frac{\pi}{8}$$ As $\triangle PQR = \frac{1}{2} PR \times PQ$, we have: $$= \frac{1}{2} \left(4^2 \sin \frac{\pi}{6}\right) \left(4 \cos \frac{\pi}{8}\right)$$ $$= 4 \sin \frac{\pi}{4} = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$$

Question 13

Maths · Binomial Theorem · Single correct

The remainder when $7^{2022} + 3^{2022}$ is divided by 5 is:

  1. 0
  2. 2
  3. 3
  4. 4

Answer: (c)

Solution

Given $7^{2022} + 3^{2022}$. $$= (49)^{1011} + (9)^{1011}$$ $$= (50 - 1)^{1011} + (10 - 1)^{1011}$$ $$= 5\lambda - 1 + 5K - 1$$ $$= 5m - 2$$ Remainder $= 5 - 2 = 3$

Question 14

Maths · Matrices · Single correct

Let the matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ and the matrix $B_0 = A^{49} + 2A^{98}$. If $B_n = Adj(B_{n-1})$ for all $n \geq 1$, then $\det(B_4)$ is equal to:

  1. $3^{28}$
  2. $3^{30}$
  3. $3^{32}$
  4. $3^{36}$

Answer: (c)

Solution

Given $$A^2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$$ $$= \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$ Swap $a \leftrightarrow R_2$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$ Swap $R_2 \leftrightarrow R_3$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$$ $$B_0 = A^{49} + 2A^{98}$$ $$= A + 2I$$ $$= |B_0|^{(n-1)^4}$$ $$= |B_0|^{16}$$ $$B_0 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$ $$= \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{bmatrix}$$ $$= 2(4-0) - 1(0-1)$$ $$= 9$$ $$B_4(9)^{16} = (3)^{32}$$

Question 15

Maths · Complex Numbers and Quadratic Equations · Single correct

Let $S_1 = \left\{ z_1 \in \mathbb{C} : |z_1 - 3| = \frac{1}{2} \right\}$ and $S_2 = \left\{ z_2 \in \mathbb{C} : |z_2 - |z_2 + 1|| = |z_2 + |z_2 - 1|| \right\}$. Then, for $z_1 \in S_1$ and $z_2 \in S_2$, the least value of $|z_2 - z_1|$ is :

  1. 0
  2. $\frac{1}{2}$
  3. $\frac{3}{2}$
  4. $\frac{5}{2}$

Answer: (c)

Solution

Given $|z_2 + |z_2 - 1||^2 = |z_2 - |z_2 + 1||^2$. $$\Rightarrow |z_2 + |z_2 - 1||(\bar{z}_2 + |z_2 + 1|) = (z_2 - |z_2 + 1|)(\bar{z}_2 - (z_2 + 1))$$ $$\Rightarrow z_2 \bar{z}_2 + 1 z_2 - 1| - (z_2 - |z_2 + 1|) + \bar{z}_2 (|z_2 - 1| + |z_2 + 1|)$$ $$= |z_2 + 1|^2 = |z_2 - 1|^2$$ $$\Rightarrow [z_2 + \bar{z}_2)(|z_2 - 1|) + (z_2 + 1)] = 2(z_2 + \bar{z}_2)$$ $$\Rightarrow (z_2 + \bar{z}_2)(|z_2 - 1| + |z_2 + 1| - 2) = 0$$ Therefore, $z_2 + \bar{z}_2 = 0$ or $|z_2 - 1| + |z_2 + 1| - 2 = 0$. Thus, $z_2$ lie on imaginary axis. Or on real axis within $[-1, 1]$. Also $|z_1 - 3| = \frac{1}{2}$ lie on circle having centre 3 and radius $\frac{1}{2}$. Clearly $|z_1 - z_2|_{\min} = \frac{5}{2} - 1 = \frac{3}{2}$.

Question 16

Maths · Three Dimensional Geometry · Single correct

The foot of the perpendicular from a point on the circle $x^2 + y^2 = 1$, $z = 0$ to the plane $2x + 3y + z = 6$ lies on which one of the following curves?

  1. $(6x + 5y - 12)^2 + 4(3x + 7y - 8)^2 = 1$, $z = 6 - 2x - 3y$
  2. $(5x + 6y - 12)^2 + 4(3x + 5y - 9)^2 = 1$, $z = 6 - 2x - 3y$
  3. $(6x + 5y - 14)^2 + 9(3x + 5y - 7)^2 = 1$, $z = 6 - 2x - 3y$
  4. $(5x + 6y - 14)^2 + 9(3x + 7y - 8)^2 = 1$, $z = 6 - 2x - 3y$

Answer: (b)

Solution

The given equations are: $$\frac{h - \cos \theta}{2} = \frac{k - \sin \theta}{3} = \frac{w - 0}{1}$$ $$= \frac{-1(2 \cos \theta + 3 \sin \theta - 6)}{14}$$ $$h = \cos \frac{-2(2 \cos \theta + 3 \sin \theta - 6)}{14}$$ $$= \frac{10 \cos \theta - 6 \sin \theta + 12}{14}$$ $$k = \sin \theta - \frac{3}{14}(2 \cos \theta + 3 \sin \theta - 6)$$ $$k = \frac{5 \sin \theta - 6 \cos \theta + 18}{14}$$ Elementary $\sin \theta$ and $\cos \theta$ $$ (5h + 6k - 12)^2 + 4(3h + 5k - 9)^2 = 1 $$

Question 17

Maths · Applications of Derivatives · Single correct

If the minimum value of $f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5}$, $x > 0$, is 14, then the value of $\alpha$ is equal to:

  1. 32
  2. 64
  3. 128
  4. 256

Answer: (c)

Solution

Given the expression: $$\frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{\alpha}{2x^5} + \frac{\alpha}{2x^5}$$ is greater than or equal to $$7 \left( \frac{\alpha^2}{2^7} \right)^{\frac{1}{7}}$$. Solving, we have: $$\frac{7 \cdot (\alpha)^{2/7}}{2} = 14$$ $$\left(\alpha^2\right)^{1/7} = 2^2$$ $$\alpha = \left(2^2\right)^{7/2} = 2^7$$ Therefore, $$\alpha = 128$$

Question 18

Maths · Relations and Functions (Advanced) · Single correct

Let $\alpha$, $\beta$ and $\gamma$ be three positive real numbers. Let $f(x) = \alpha x^5 + \beta x^3 + \gamma x$, $x \in \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ be such that $g(f(x)) = x$ for all $x \in \mathbb{R}$. If $a_1, a_2, a_3, \ldots, a_n$ be in arithmetic progression with mean zero, then the value of $f\left(g\left(\dfrac{1}{n}\sum_{i=1}^{n} f(a_i)\right)\right)$ is equal to:

  1. 0
  2. 3
  3. 9
  4. 27

Answer: (a)

Solution

Consider a case when $\alpha = \beta = 0$ then $f(x) = yx$ $g(x) = \frac{x}{y}$ $$\frac{1}{n} \sum_{i=1}^{n} f(a_i) \Rightarrow \frac{y}{n} (a_1 + a_2 + \ldots + a_n)$$ $$= 0$$ $$\Rightarrow f(g(0)) \Rightarrow f(0)$$ $$\Rightarrow 0$$

Question 19

Maths · Sequences and Series · Single correct

Consider the sequence $a_1, a_2, a_3, \ldots$ such that $$a_1 = 1, \ a_2 = 2 \ and \ a_{n+2} = \frac{2}{a_{n+1}} + a_n \ for \ n = 1, 2, 3, \ldots$$ If $$\left( \frac{a_1 + \frac{1}{a_2}}{a_3} \right) \cdot \left( \frac{a_2 + \frac{1}{a_3}}{a_4} \right) \cdot \left( \frac{a_3 + \frac{1}{a_4}}{a_5} \right) \cdot \ldots \cdot \left( \frac{a_{30} + \frac{1}{a_{31}}}{a_{32}} \right) = 2^\alpha \left( ^{61}C_{31} \right),$$ then $\alpha$ is equal to:

  1. $-30$
  2. $-31$
  3. $-60$
  4. $-61$

Answer: (c)

Solution

Given $a_{n+2} \cdot a_{n+1} - a_{n+1} \cdot a_n = 2$. Series will satisfy $$\frac{a_n + \frac{1}{a_{n+1}}}{a_{n+2}} = \frac{a_{n+2} - \frac{1}{a_{n+1}}}{a_{n+2}}$$ $$= 1 - \frac{1}{a_{n+1} a_{n+2}}$$ $$= 1 - \frac{1}{2(r+1)}$$ $$= \frac{2r+1}{2(r+1)}$$ Now proof is given by $$\prod_{r=1}^{30} \frac{(2r+1)}{2(r+1)}$$ $$= \frac{(1 \cdot 3 \cdot 5 \cdot \ldots \cdot 61)}{2^{30} \cdot (2 \cdot 3 \cdot \ldots \cdot 31)}$$ $$\Rightarrow \frac{(1 \cdot 3 \cdot 5 \cdot \ldots \cdot 61)}{31 \cdot 2^{30}} \times \frac{2^{30} \times 30}{2^{30} \times 30}$$ $\alpha = -60$

Question 20

Maths · Integrals · Single correct

The minimum value of the twice differentiable function $f(x) = \int_{0}^{x} e^{x-t} f'(t) \, dt - (x^2 - x + 1) e^x$, $x \in \mathbb{R}$, is:

  1. -$\frac{2}{\sqrt{e}}$
  2. -2$\sqrt{e}$
  3. -$\sqrt{e}$
  4. $\frac{2}{\sqrt{e}}$

Answer: (a)

Solution

Given $$f(x) = e^x \cdot \int_0^x \frac{f'(t)}{e^t} \, dt$$ We have $$f'(x) = e^x \cdot \int_0^x \frac{f'(t)}{e^t} \, dt + e^x \cdot \frac{f'(x)}{e^x}$$ This simplifies to $$-\left[ (2x-1) \cdot e^x + (x^2 - x + 1) \cdot e^x \right]$$ The integral $$\int_0^x \frac{f'(t)}{e^t} \, dt = x^2 + x$$ Thus, $$\frac{f'(x)}{e^x} = 2x + 1$$ So, $$f'(x) = (2x + 1) \cdot e^x$$ Setting $$f'(x) = 0 \Rightarrow x = -\frac{1}{2}$$ Then, $$f(x) = (2x + 1) \cdot e^x - 2e^x + C$$ Given $$f(0) = -1$$ We have $$-1 = 1 - 2 + C$$ Thus, $$C = 0$$ Therefore, $$f(x) = e^x (2x - 1)$$ Finally, $$f\left(-\frac{1}{2}\right) = \frac{-2}{\sqrt{e}}$$

Question 21

Maths · Permutations and Combinations · Fill in the blank

Let S be the set of all passwords which are six to eight characters long, where each character is either an alphabet from $\{$A, B, C, D, E$\}$ or a number from {1, 2, 3, 4, 5} with the repetition of characters allowed. If the number of passwords in S whose at least one character is a number from {1, 2, 3, 4, 5} is $\alpha \times 5^6$, then $\alpha$ is equal to _____.

Answer: 7073

Solution

Required no. = Total - no character from $\{$1, 2, 3, 4, 5$\}$ $$= (10^6 - 5^6) + (10^7 - 5^7) + (10^8 - 5^8)$$ $$= 10^6 (1 + 10 + 100) - 5^6 (1 + 5 + 25)$$ $$= 10^6 \times 111 - 5^6 \times 31$$ $$= 2^6 \times 5^6 \times 111 - 5^6 \times 31$$ $$= 5^6 (2^6 \times 111 - 31)$$ $$= 5^6 \times \underbrace{7073}_{\alpha}$$ Therefore, $\($ $\alpha$ = 7073 $\)$

Question 22

Maths · Three Dimensional Geometry · Fill in the blank

Let P(-2, -1, 1) and Q $(\frac{56}{17}, \frac{43}{17}, \frac{111}{17})$ be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are $\alpha$, -1, $\beta$, where both $\alpha$ and $\beta$ are integers of minimum absolute values, then $\alpha^2$ + $\beta^2$ is equal to ______.

Answer: 450

Solution

DR of $RS \equiv (\alpha,\ -1,\ \beta)$ DR of $PQ \equiv \left(\dfrac{56}{17} + 2,\ \dfrac{43}{17} + 1,\ \dfrac{111}{17} - 1\right) \equiv \left(\dfrac{90}{17},\ \dfrac{60}{17},\ \dfrac{94}{17}\right)$ $$\frac{90}{17}\alpha + \frac{60}{17}(-1) + \frac{94}{17}\beta = 0$$ $$90\alpha + 94\beta = 60$$ $$\beta = \frac{60 - 90\alpha}{94}$$ $$\beta = \frac{30(2 - 3\alpha)}{94}$$ $$\beta = -30\cdot\frac{(3\alpha - 2)}{94}$$ $$\beta = \frac{-15}{47}(3\alpha - 2)$$ $$\Rightarrow \frac{\beta}{-15} = \frac{3\alpha - 2}{47}$$ $$\Rightarrow \beta = -15,\quad \alpha = -15$$ $$\alpha^2 + \beta^2 = 225 + 225 = 450$$

Question 23

Maths · Applications of Derivatives · Numerical

Let $f : [0, 1] \to \mathbb{R}$ be a twice differentiable function in $(0, 1)$ such that $f(0) = 3$ and $f(1) = 5$. If the line $y = 2x + 3$ intersects the graph of $f$ at only two distinct points in $(0, 1)$, then the least number of points $x \in (0, 1)$, at which $f''(x) = 0$, is ______.

Answer: 2

Solution

Given $f'(a) = f'(b) = f'(c) = 2$. This implies $f''(x)$ is zero for at least $x_1 \in (a, b)$ and $x_2 \in (b, c)$.

Question 24

Maths · Integrals · Numerical

If $$\int_{0}^{\sqrt{3}} \frac{15x^3}{\sqrt{1+x^2 + \sqrt{\left(1+x^2\right)^3}}} \, dx = \alpha \sqrt{2} + \beta \sqrt{3}$$, where $\alpha$, $\beta$ are integers, then $\alpha + \beta$ is equal to

Answer: 10

Solution

Put $1 + x^2 = t^2$. Then $2x \, dx = 2t \, dt$. The integral becomes $$15 \int_1^2 \frac{t(t^2 - 1)}{t \sqrt{1 + t}} \, dt$$ Put $1 + t = u^2$, then $dt = 2u \, du$. The integral becomes $$15 \int_{\sqrt{2}}^{\sqrt{3}} \frac{(u^2 - 1)^2 - 1}{u} \times 2u \, du$$ $$30 \int_{\sqrt{2}}^{\sqrt{3}} (u^4 - 2u^2) \, du$$ $$30 \left[ \frac{u^5}{5} - \frac{2u^3}{3} \right]_{\sqrt{2}}^{\sqrt{3}}$$ $$30 \left[ \frac{1}{5} (\sqrt{3}^5 - \sqrt{2}^5) - \frac{2}{3} (\sqrt{3}^3 - \sqrt{2}^3) \right]$$ $$30 \left[ \frac{1}{5} (9\sqrt{3} - 4\sqrt{2}) - \frac{2}{3} (3\sqrt{3} - 2\sqrt{2}) \right]$$ $$30 \left[ -\frac{1}{5} \times \sqrt{3} + \frac{8}{15} \sqrt{2} \right]$$ $$-6\sqrt{3} + 16\sqrt{2} = \alpha \sqrt{2} + \beta \sqrt{3}$$ $$\alpha = 16, \beta = -6$$ Therefore, $\alpha + \beta = 10$

Question 25

Maths · Matrices · Numerical

Let $A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$, $\alpha, \beta \in \mathbb{R}$. Let $\alpha_1$ be the value of $\alpha$ which satisfies $(A+B)^2 = A^2 + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ and $\alpha_2$ be the value of $\alpha$ which satisfies $(A+B)^2 = B^2$. Then $|\alpha_1 - \alpha_2|$ is equal to _______.

Answer: 2

Solution

Given $$A + B = \begin{bmatrix} \beta + 1 & 0 \\ 3 & \alpha \end{bmatrix}$$ $$(A + B)^2 = \begin{bmatrix} \beta + 1 & 0 \\ 3 & \alpha \end{bmatrix} \begin{bmatrix} \beta + 1 & 0 \\ 3 & \alpha \end{bmatrix}$$ $$= \begin{bmatrix} (\beta + 1)^2 & 0 \\ 3(\beta + 1) + 3\alpha & \alpha^2 \end{bmatrix}$$ $$A^2 = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$$ $$= \begin{bmatrix} -1 & -1 - \alpha \\ 2 + 2\alpha & \alpha^2 - 2 \end{bmatrix}$$ Therefore, $$\begin{bmatrix} 1 & -\alpha + 1 \\ 2\alpha + 4 & \alpha^2 \end{bmatrix} = \begin{bmatrix} (\beta + 1)^2 & 0 \\ 3(\alpha + \beta + 1) & \alpha^2 \end{bmatrix}$$ $$\boxed{\alpha = 1} = \alpha_1$$ $$B^2 = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$$ $$= \begin{bmatrix} \beta^2 + 1 & \beta \\ \beta & 1 \end{bmatrix} = \begin{bmatrix} (\beta + 1)^2 & 0 \\ 3(\beta + 1) + 3\alpha & \alpha^2 \end{bmatrix}$$ Therefore, $$\beta = 0, \alpha = -1 = \alpha_2$$ $$|\alpha_1 - \alpha_2| = |1 - (-1)| = 2$$

Question 26

Maths · Sequences and Series · Numerical

For $p, q \in \mathbb{R}$, consider the real valued function $f(x) = (x - p)^2 - q$, $x \in \mathbb{R}$ and $q > 0$. Let $a_1, a_2, a_3$ and $a_4$ be in an arithmetic progression with mean $p$ and positive common difference. If $\left| f(a_i) \right| = 500$ for all $i = 1, 2, 3, 4$, then the absolute difference between the roots of $f(x) = 0$ is

Answer: 50

Solution

Given $f(x) = 0 \Rightarrow (x - p)^2 - q = 0$. Roots are $p + \sqrt{q}$, $p - \sqrt{q}$, absolute difference between roots $2\sqrt{q}$. Now, $|f(a_i)| = 500$. Let $a_1, a_2, a_3, a_4$ are $a_1$, $a + d$, $a + 2d$, $a + 3d$. $|f(a_4)| = 500$. $|(a_1 - p)^2 - q| = 500$. $\Rightarrow (a_1 - p)^2 - q = 500$. $$\Rightarrow \frac{9}{4}d^2 - q = 500 (1)$$ and $|f(a_1)|^2 = |f(a_2)|^2$. $((a_1 - p)^2 - q)^2 = ((a_2 - p)^2 - q)^2$. $\Rightarrow ((a_1 - p)^2 - (a_2 - p)^2)((a_1 - p)^2 - q + (a_2 - p)^2 - q) = 0$. $$\Rightarrow \frac{9}{4}d^2 - q + \frac{d^2}{4} - q = 0$$ $$2q = \frac{10d^2}{4} \Rightarrow q = \frac{5d^2}{4}$$ $$\Rightarrow d^2 = \frac{4q}{5}$$ From equation (1) $\frac{9}{4} \cdot \frac{4q}{5} - q = 500$. $$\frac{4q}{5} = 500$$ $$\frac{4q}{5} = 500$$

Question 27

Maths · Conic Sections · Fill in the blank

For the hyperbola H : $x^2-y^2=1$ and the ellipse E : $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b>0$, let the (1) eccentricity of E be reciprocal of the eccentricity of H, and (2) the line $y=\sqrt{\frac{5}{2}}x+K$ be a common tangent of E and H. Then $4(a^2+b^2)$ is equal to ______.

Answer: 3

Solution

Given $e_E = \sqrt{1 - \frac{b^2}{a^2}}$, $e_H = \sqrt{2}$. If $e_E = \frac{1}{e_H}$, then $$\frac{a^2 - b^2}{a^2} = \frac{1}{2}$$ $$2a^2 - 2b^2 = a^2$$ $$a^2 = 2b^2$$ and $y = \sqrt{\frac{5}{2}} x + k$ is tangent to the ellipse, then $$K^2 = a^2 \times \frac{5}{2} + b^2 = \frac{3}{2}$$ $$6b^2 = \frac{3}{2} \Rightarrow b^2 = \frac{1}{4} and a^2 = \frac{1}{2}$$ Therefore, $4(a^2 + b^2) = 3$.

Question 28

Maths · Sequences and Series · Numerical

Let $x_1, x_2, x_3, \ldots, x_{20}$ be in geometric progression with $x_1 = 3$ and the common ratio $\frac{1}{2}$. A new data is constructed replacing each $x_i$ by $(x_i - i)^2$. If $\overline{x}$ is the mean of new data, then the greatest integer less than or equal to $\overline{x}$ is

Answer: 142

Solution

The sum from $x_0$ to $x_{20}$ is given by: $$\sum_{x_0}^{20} 3 \left( 1 - \left( \frac{1}{2} \right) \right)^{20} = \frac{1 - \frac{1}{2}}{1 - \frac{1}{2}} = 6 \left( 1 - \frac{1}{2^{20}} \right)$$ This equals: $$\sum_{i=1}^{20} (x_{i-1})^2$$ $$= \sum_{i=1}^{20} (x_i)^2 + (i)^2 - 2x_i i$$ Now, $$\sum_{i=1}^{20} (x_i)^2 = \frac{9 \left( 1 - \left( \frac{1}{4} \right) \right)^{20}}{1 - \frac{1}{4}} = 12 \left( 1 - \frac{1}{2^{40}} \right)$$ $$\sum_{i=1}^{20} i^2 = \frac{1}{6} \times 20 \times 21 \times 41 = 2870$$ $$\sum_{i=1}^{20} x_i \cdot i = s = 3 + 2.3 \cdot \frac{1}{2} + 3.3 \cdot \frac{1}{2^2} + 4.3 \cdot \frac{1}{2^3} + \ldots AGP$$ $$= 6 \left( 2 - \frac{22}{2^{20}} \right)$$ $$x = \frac{12 - \frac{12}{2^{40}} + 2870 - 12 \left( 2 - \frac{22}{2^{20}} \right)}{20}$$ $$-x = \frac{2858 + \left( -\frac{12}{2^{40}} + \frac{22}{2^{20}} \right) \times \frac{1}{20}}{20}$$ $$\left[ x \right] = 142$$

Question 29

Maths · Limits and Derivatives · Numerical

$\lim_{x\to0} \left( \frac{ (x+2\cos x)^3 +2(x+2\cos x)^2 +3\sin(x+2\cos x) }{ (x+2)^3 +2(x+2)^2 +3\sin(x+2) } \right)^{\frac{100}{x}}$ is equal to ______.

Answer: 1

Solution

Given the limit $\lim_{x\to0} \left( \frac{ (x+2\cos x)^3 +2(x+2\cos x)^2 +3\sin(x+2\cos x) }{ (x+2)^3 +2(x+2)^2 +3\sin(x+2) } \right)^{100/x}$ Form $1^\infty$. We have $\exp\!\left[ \lim_{x\to0} \frac{100}{x} \left( \frac{ (x+2\cos x)^3 +2(x+2\cos x)^2 +3\sin(x+2\cos x) }{ (x+2)^3 +2(x+2)^2 +3\sin(x+2) } -1 \right) \right]$ $=\exp\!\left[ \lim_{x\to0} \frac{100}{x} \cdot \frac{ (x+2\cos x)^3 +2(x+2\cos x)^2 +3\sin(x+2\cos x) }{ (x+2)^3 +2(x+2)^2 +3\sin(x+2) } \right]$ $=\exp\!\left[ \lim_{x\to0} \frac{ 100\Big( (x+2\cos x)^3 +2(x+2\cos x)^2 +3\sin(x+2\cos x) \Big) }{ x\Big( (x+2)^3 +2(x+2)^2 +3\sin(x+2) \Big) } \right]$

Question 30

Maths · Complex Numbers and Quadratic Equations · Numerical

The sum of all real values of $x$ for which $$\frac{3x^2 - 9x + 17}{x^2 + 3x + 10} = \frac{5x^2 - 7x + 19}{3x^2 + 5x + 12}$$ is equal to ___.

Answer: 6

Solution

Given $\($ $\frac{3x^2 - 9x + 17}{x^2 + 3x + 10}$ = $\frac{5x^2 - 7x + 19}{3x^2 + 5x + 12}$ $\)$. Multiply both sides by $\($ x^2 + 3x + 10 $\)$ and $\($ 3x^2 + 5x + 12 $\)$: $$ \frac{x^2 + 3x + 10 + 2x^2 - 12x + 7}{x^2 + 3x + 10} = \frac{3x^2 + 5x + 12 + 2x^2 - 12x + 7}{3x^2 + 5x + 12} $$ Simplify: $$ 1 + \frac{2x^2 - 12x + 7}{x^2 + 3x + 10} = 1 + \frac{2x^2 - 12x + 7}{3x^2 + 5x + 12} $$ This implies: $$ \left( 2x^2 - 12x + 7 \right) \left( \frac{1}{x^2 + 3x + 10} - \frac{1}{3x^2 + 5x + 12} \right) = 0 $$ Thus, either: $$ 2x^2 - 12x + 7 = 0 OR 3x^2 + 5x + 12 = x^2 + 3x + 10 $$ Solving the first equation: $$ x = \frac{12 \pm \sqrt{D}}{4} $$ Solving the second equation: $$ 2x^2 + 2x + 2 = 0 $$ Which simplifies to: $$ x^2 + x + 1 = 0 $$ Sum of Roots = 6. No solution.

Physics

Question 31

Physics · Physical World, Units and Measurements · Single correct

The dimensions of $\frac{B^2}{\mu_0}$ will be: (if $\mu_0$: permeability of free space and B: magnetic field)

  1. $[M\,L^{2}\,T^{-2}]$
  2. $[M\,L\,T^{-2}]$
  3. $[M\,L^{-1}\,T^{-2}]$
  4. $[M\,L^{2}\,T^{-2}\,A^{-1}]$

Answer: (c)

Solution

Given $$u = \frac{B^2}{2\mu_0}$$ $u$ represents energy per unit volume. The dimensional formula is $$\left[ \frac{B^2}{\mu_0} \right] = [u] = \left[ \frac{\mathrm{ML^2T^{-2}}}{\mathrm{L^3}} \right] = [\mathrm{ML^{-1}T^{-2}}]$$

Question 32

Physics · Motion in a Plane · Single correct

A NCC parade is going at a uniform speed of 9 km/h under a mango tree on which a monkey is sitting at a height of 19.6 m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is: (Given g = 9.8 m/s²)

  1. 5 m
  2. 10 m
  3. 19.8 m
  4. 24.5 m

Answer: (a)

Solution

Monkey Time taken by mango = $\sqrt{\frac{2n}{g}}$ = $\sqrt{\frac{2 \times 19.6}{9.8}}$ = 2 second Distance = vt = 9 $\times$ $\frac{5}{18}$ $\times$ 2 = 5 m

Question 33

Physics · Laws of Motion · Single correct

In two different experiments, an object of mass $5 \, \mathrm{kg}$ moving with a speed of $25 \, \mathrm{ms}^{-1}$ hits two different walls and comes to rest within (i) $3 \, \mathrm{seconds}$, (ii) $5 \, \mathrm{seconds}$, respectively. Choose the correct option out of the following:

  1. Impulse and average force acting on the object will be same for both the cases.
  2. Impulse will be same for both the cases but the average force will be different.
  3. Average force will be same for both the cases but the impulse will be different.
  4. Average force and impulse will be different for both the cases.

Answer: (b)

Solution

Impulse equals change in momentum. $I = \Delta P$ $F_{avg} = \frac{\Delta P}{\Delta t}$ $\Delta t_1 = 3$ $\Delta t_2 = 5$ $\Delta P_1 = \Delta P_2$ $I_1 = I_2$ $F_{avg}$ in case (i) is more than (ii).

Question 34

Physics · Laws of Motion · Single correct

A balloon has mass of 10 g in air. The air escapes from the balloon at a uniform rate with velocity $4.5 \, \mathrm{cm/s}$. If the balloon shrinks in $5 \, \mathrm{s}$ completely. Then, the average force acting on that balloon will be (in dyne).

  1. 3
  2. 9
  3. 12
  4. 18

Answer: (b)

Solution

The force is given by the equation $$ F = \frac{dm}{dt} v $$. Substituting the given values, we have $$ = \frac{10 \, \mathrm{g}}{5 \, \mathrm{s}} \left( \frac{4.5 \, \mathrm{cm}}{\mathrm{s}} \right) = 9 \, \frac{\mathrm{g} \cdot \mathrm{cm}}{\mathrm{s}^2} = 9 \, \mathrm{dyne} $$.

Question 35

Physics · Gravitation · Single correct

If the radius of earth shrinks by 2$\%$ while its mass remains same. The acceleration due to gravity on the earth’s surface will approximately :

  1. decrease by 2$\%$
  2. decrease by 4$\%$
  3. increase by 2$\%$
  4. increase by 4$\%$

Answer: (d)

Solution

Given $g = \frac{GM}{R^2}$. Since $M = constant$, $g \propto \frac{1}{R^2}$. Therefore, $$100 \frac{\Delta g}{g} = -2 \frac{\Delta R}{R} 100$$ The percentage change is $-2 (-2)$. Thus, the percentage change in $g$ is $4\%$. This results in an increase by $4\%$.

Question 36

Physics · Mechanical Properties of Solids · Single correct

The force required to stretch a wire of cross-section $1 \, \mathrm{cm}^2$ to double its length will be: (Given Yong's modulus of the wire $= 2 \times 10^{11} \, \mathrm{N/m}^2$)

  1. $1 \times 10^7 \, \mathrm{N}$
  2. $1.5 \times 10^7 \, \mathrm{N}$
  3. $2 \times 10^7 \, \mathrm{N}$
  4. $2.5 \times 10^7 \, \mathrm{N}$

Answer: (c)

Solution

The force is given by the formula: $$F = \gamma A \frac{\Delta \ell}{\ell}$$ Substituting the values: $$= 2 \times 10^{11} \times 10^{-4} \left( \frac{2\ell - \ell}{\ell} \right)$$ Simplifying gives: $$= 2 \times 10^7 \, \mathrm{N}$$

Question 37

Physics · Thermodynamics · Single correct

A Carnot engine has efficiency of 50$\%$. If the temperature of sink is reduced by $40^\circ$$\mathrm{C}$, its efficiency increases by 30$\%$. The temperature of the source will be:

  1. $166.7\,\mathrm{K}$
  2. $255.1\,\mathrm{K}$
  3. $266.7\,\mathrm{K}$
  4. $367.7\,\mathrm{K}$

Answer: (c)

Solution

Question 38

Physics · Kinetic Theory · Single correct

Given below are two statements : Statement I : The average momentum of a molecule in a sample of an ideal gas depends on temperature. Statement II : The rms speed of oxygen molecules in a gas is $v$. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become $2v$. In the light of the above statements, choose the correct answer from the options given below :

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is false but Statement II is true

Answer: (d)

Solution

Given $[P_{avg} = 0]$ (due to random motion) $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ $$T_{new} = 2T$$ $$M_{new} = \frac{M}{2}$$ $$\frac{v_{new}}{v} = \frac{\sqrt{\frac{2T}{M/2}}}{\sqrt{\frac{T}{M}}}$$ $$v_{new} = 2v$$

Question 39

Physics · Waves · Single correct

In the wave equation $$y = 0.5 \sin \frac{2\pi}{\lambda} (400\, t - x)\, \mathrm{m}$$ the velocity of the wave will be:

  1. 200 $\mathrm{m/s}$
  2. 200$\sqrt{2}$ $\mathrm{m/s}$
  3. 400 $\mathrm{m/s}$
  4. 400$\sqrt{2}$ $\mathrm{m/s}$

Answer: (c)

Solution

Given the equation of the wave: $$y = 0.5 \sin \left( \frac{2\pi}{\lambda} 400t - \frac{2\pi}{\lambda} x \right)$$ The angular frequency is given by: $$\omega = \frac{2\pi}{\lambda} 400$$ The wave number is: $$K = \frac{2\pi}{\lambda}$$ The velocity of the wave is: $$v = \frac{\omega}{k}$$ Thus, $$v = 400 \, \mathrm{m/s}$$

Question 40

Physics · Electrostatic Potential and Capacitance · Single correct

Two capacitors, each having capacitance $40 \, \mu\mathrm{F}$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24 \, \mu\mathrm{F}$. The value of $K$ will be:

  1. 1.5
  2. 2.5
  3. 1.2
  4. 3

Answer: (a)

Solution

The equivalent capacitance is given by $$C_{eq} = \frac{C(KC)}{C + KC} = \frac{KC}{K + 1}.$$ Solving for the given condition, $$24 = \frac{K \cdot 40}{K + 1}.$$ From this, we find $$K = 1.5.$$

Question 41

Physics · Current Electricity · Single correct

A wire of resistance $R_1$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

  1. 9 : 1
  2. 1 : 9
  3. 4 : 1
  4. 3 : 1

Answer: (a)

Solution

Given $\($ R_1 = $\rho$ $\frac{L_1}{A_1}$ $\)$. $\($ R_2 = $\rho$ $\left$( $\frac{3L_1}{A_1/3}$ $\right$) = 9 $\rho$ $\frac{L_1}{A_1}$ $\)$. Therefore, $\($ $\frac{R_2}{R_1}$ = 9 $\)$.

Question 42

Physics · Moving Charges and Magnetism · Multiple correct

The current sensitivity of a galvanometer can be increased by: (A) decreasing the number of turns (B) increasing the magnetic field (C) decreasing the area of the coil (D) decreasing the torsional constant of the spring Choose the most appropriate answer from the options given below:

  1. and (C) only
  2. and (D) only
  3. and (C) only
  4. and (D) only

Answer: (d)

Solution

Given $i = \left( \frac{K}{NAB} \right) \theta$. Therefore, $$\frac{d\theta}{di} = \frac{NAB}{K}.$$

Question 43

Physics · Moving Charges and Magnetism · Single correct

As shown in the figure, a metallic rod of linear density $0.45 \, \mathrm{kg \, m^{-1}}$ is lying horizontally on a smooth incline plane which makes an angle of $45^\circ$ with the horizontal. The minimum current flowing in the rod required to keep it stationary, when $0.15 \, \mathrm{T}$ magnetic field is acting on it in the vertical upward direction, will be: $\{$Use $g = 10 \, \mathrm{m/s^2}$$\}$

  1. 30 A
  2. 15 A
  3. 10 A
  4. 3 A

Answer: (a)

Solution

Given the equation $mg \sin 45^\circ = ILB \cos 45^\circ$. Therefore, $$I = \left( \frac{m}{L} \right) \frac{g}{B}$$ $$= \frac{(0.45)(10)}{0.15} = 30 \, \mathrm{A}$$

Question 44

Physics · Alternating Current · Single correct

The equation of current in a purely inductive circuit is $5\sin(49\pi t - 30^\circ)$. If the inductance is $30\,\mathrm{mH}$ then the equation for the voltage across the inductor, will be: $$\left\{\text{Let } \pi = \frac{22}{7}\right\}$$

  1. $1.47 \sin(49 \pi t - 30^\circ)$
  2. $1.47 \sin(49 \pi t + 60^\circ)$
  3. $23.1 \sin(49 \pi t - 30^\circ)$
  4. $23.1 \sin(49 \pi t + 60^\circ)$

Answer: (d)

Solution

Given $v_0 = i_0 x_L$. $$= i_0 (wL)$$ $$= (5)(49\pi)(30 \times 10^{-3})$$ $$= 23.1$$ Voltage will lead current by $90^\circ$. Therefore, $V = 23.1 \sin(49 \pi t + 60^\circ)$.

Question 45

Physics · Ray Optics and Optical Instruments · Single correct

As shown in the figure, after passing through the medium 1. The speed of light $v_2$ in medium 2 will be: (Given $c = 3 \times 10^8 \, \mathrm{ms}^{-1}$)

  1. $1.0 \times 10^8 \, \mathrm{ms}^{-1}$
  2. $0.5 \times 10^8 \, \mathrm{ms}^{-1}$
  3. $1.5 \times 10^8 \, \mathrm{ms}^{-1}$
  4. $3.0 \times 10^8 \, \mathrm{ms}^{-1}$

Answer: (a)

Solution

Given $\($ $\frac{\mu_2}{\mu_{air}}$ = $\frac{C}{v_2}$ $\)$. Therefore, $\($ $\frac{\sqrt{\mu_{r_2} \varepsilon_{r_2}}}{(1)}$ = $\frac{C}{v_2}$ $\)$. Thus, $\($ $\sqrt{(1)(9)}$ = $\frac{C}{v_2}$ $\)$. Therefore, $\($ v_2 = $\frac{C}{3}$ $\)$.

Question 46

Physics · Ray Optics and Optical Instruments · Single correct

In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30 $\mathrm{cm}$. The focal length of the objective, when the angular magnification of the telescope is 2, will be:

  1. 20 $\mathrm{cm}$
  2. 30 $\mathrm{cm}$
  3. 10 $\mathrm{cm}$
  4. 15 $\mathrm{cm}$

Answer: (a)

Solution

Given $f_0 + f_e = 30$. $m = \frac{f_0}{f_e}$. $2 = \frac{f_0}{f_e} \Rightarrow f_0 = 2f_e$. So $f_0 + \frac{f_0}{2} = 30$. $f_0 = 20 \, \mathrm{cm}$.

Question 47

Physics · Dual Nature of Radiation and Matter · Single correct

The equation $\lambda = \frac{1.227}{x} \, \mathrm{nm}$ can be used to find the de-Brogli wavelength of an electron. In this equation $x$ stands for: Where, m = mass of electron P = momentum of electron K = Kinetic energy of electron V = Accelerating potential in volts for electron

  1. $\sqrt{mK}$
  2. $\sqrt{P}$
  3. $\sqrt{K}$
  4. $\sqrt{V}$

Answer: (d)

Solution

Given the de-Broglie's wavelength, $$\lambda = \frac{h}{m \nu}$$. We have $$\lambda = \frac{h}{\sqrt{2m(K \cdot E)}}$$. This can be rewritten as $$h = \frac{h}{\sqrt{2mqV}}$$. Putting the values of $m$ and $q$, we get $$\lambda = \frac{1 \cdot 22}{\sqrt{V}} \, \mathrm{nm}$$.

Question 48

Physics · Nuclei · Single correct

The half life period of a radioactive substance is 60 days. The time taken for $\frac{7}{8}$ of its original mass to disintegrate will be:

  1. 120 days
  2. 130 days
  3. 180 days
  4. 20 days

Answer: (c)

Solution

7/8 disintegrates means 1/8 remains. Or $$\left( \frac{1}{2} \right)^3$$. Therefore, 3 half lives = 180 days.

Question 49

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

Identify the solar cell characteristics from the following options:

Answer: (b)

Solution

Conceptual / theory

Question 50

Physics · Communication Systems · Single correct

In the case of amplitude modulation to avoid distortion the modulation index ($\mu$) should be:

  1. $\mu \leq 1$
  2. $\mu \geq 1$
  3. $\mu = 2$
  4. $\mu = 0$

Answer: (a)

Solution

Given $$\mu = \frac{A_m}{A_c}$$ $$\mu \leq 1$$ to avoid distortion because $$\mu > 1$$ will result in interference between carrier frequency and message frequency.

Question 51

Physics · Mathematics in Physics · Numerical

If the projection of $2\hat{i} + 4\hat{j} - 2\hat{k}$ on $\hat{i} + 2\hat{j} + \alpha \hat{k}$ is zero. Then, the value of $\alpha$ will be

Answer: 5

Solution

Given $\vec{a} \cdot \vec{b} = 0$. Therefore, $\vec{a} \cdot \vec{b} = 0$. Thus, $2 \times 1 + 4 \times 2 - 2 \times a = 0$. Therefore, $a = 5$.

Question 52

Physics · Nuclei · Numerical

A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be ______ hours.

Answer: 15

Solution

Given $A = A_0 \times 2^{-t/T}$. $$\frac{A_0}{64} = A_0 \times 2^{-t/T}$$ Therefore, $t = 6T = 6 \times 2 \cdot 5 = \boxed{15}$ hours.

Question 53

Physics · Wave Optics · Numerical

In a Young's double slit experiment, a laser light of 560 $\mathrm{nm}$ produces an interference pattern with consecutive bright fringes' separation of 7.2 $\mathrm{mm}$. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 $\mathrm{mm}$. The wavelength of second light is $\mathrm{nm}$.

Answer: 630

Solution

Given $\beta \propto \lambda$. $$\lambda_2 = \frac{9}{8} \lambda_1$$ Therefore, $$\beta_2 = \frac{9}{8} \beta_1 = \frac{9}{8} \times 560 = 630 \, \mathrm{nm}.$$

Question 54

Physics · Alternating Current · Numerical

The frequencies at which the current amplitude in an LCR series circuit becomes $\frac{1}{\sqrt{2}}$ times its maximum value, are $212 \, \mathrm{rad} \, \mathrm{s}^{-1}$ and $232 \, \mathrm{rad} \, \mathrm{s}^{-1}$. The value of resistance in the circuit is $R = 5 \, \Omega$. The self inductance in the circuit is ______ mH.

Answer: 250

Solution

Band width $= 232 - 212 = \frac{R}{L}$ Therefore, $L = \frac{5}{20} = 250 \, \mathrm{mH}$

Question 55

Physics · Current Electricity · Numerical

As shown in the figure, a potentiometer wire of resistance $20\,\Omega$ and length $300\,\mathrm{cm}$ is connected with resistance box (R.B.) and a standard cell of emf $4\,\mathrm{V}$. For a resistance ‘R’ of resistance box introduced into the circuit, the null point for a cell of $20\,\mathrm{mV}$ is found to be $60\,\mathrm{cm}$. The value of ‘R’ is _________ $\Omega$.

Answer: 780

Solution

Given the equation for E: $$E = \frac{AC}{AB} (V_A - V_B)$$ Therefore, $$20 \times 10^{-3} = \frac{60}{300} \times \frac{4 \times 20}{R + 20}$$ Therefore, $$R = \boxed{780} \, \Omega$$

Question 56

Physics · Electric Charges and Fields · Numerical

Two electric dipoles of dipole moments $1.2 \times 10^{-30} \, \mathrm{cm}$ and $2.4 \times 10^{-30} \, \mathrm{cm}$ are placed in two different uniform electric fields of strengths $5 \times 10^{4} \, \mathrm{NC}^{-1}$ and $15 \times 10^{4} \, \mathrm{NC}^{-1}$ respectively. The ratio of maximum torque experienced by the electric dipoles will be $\frac{1}{x}$. The value of $x$ is _______.

Answer: 6

Solution

Given $|\tau|_{max} = PE$. $$\frac{\tau_1}{\tau_2} = \frac{P_1 E_1}{P_2 E_2} = \frac{1 \cdot 2 \times 10^{-30} \times 5 \times 10^4}{2 \cdot 4 \times 10^{-30} \times 15 \times 10^4} = \frac{1}{6}$$ Hence $x = 6$.

Question 57

Physics · Waves · Fill in the blank

The frequency of echo will be $\underline{\hspace{1cm}}$ Hz if the train blowing a whistle of frequency $320$ Hz is moving with a velocity of $36$ km/h towards a hill from which an echo is heard by the train driver. Velocity of sound in air is $330$ m/s.

Answer: 340

Solution

The hill will be a secondary source. $f_1$ = frequency of the car with respect to the hill $$f_1 = \left( \frac{v}{v - v_s} \right) f = \left( \frac{330}{320} \right) \times 320 = 330 \, \mathrm{Hz}$$ $f_2$ = Frequency of the sound reflected by the hill with respect to the car (echo) $$f_2 = \left( \frac{v + v_0}{v} \right) f_1 = \left( \frac{330 + 10}{330} \right) \times 330 = 340 \, \mathrm{Hz}$$

Question 58

Physics · Mechanical Properties of Fluids · Numerical

The diameter of an air bubble which was initially 2 $\mathrm{mm}$, rises steadily through a solution of density $1750 \, \mathrm{kg \, m^{-3}}$ at the rate of $0.35 \, \mathrm{cm \, s^{-1}}$. The coefficient of viscosity of the solution is ______ poise (in nearest integer). (the density of air is negligible).

Answer: 11

Solution

As the bubble is rising steadily the net force acting on it will be zero. (Because of density of air the value of mg can be neglected) So $B = F \implies \frac{4\pi}{3} R^3 \rho g = 6\pi \eta R v$ Putting $R = 1 \, \mathrm{mm} = 10^{-3} \, \mathrm{m}$ $\rho = 1.75 \times 10^3 \, \mathrm{kg/m^3}$ $g = 10 \, \mathrm{m/s^2}$ $v = 0.35 \times 10^{-2} \, \mathrm{m/s}$ $\eta = \frac{10}{9} \cong 1.11 \, \mathrm{SI \, unit} = 11 \, \mathrm{poise \, (CGS)}$

Question 59

Physics · Work, Energy and Power · Numerical

A block of mass 'm' (as shown in figure) moving with kinetic energy $E$ compresses a spring through a distance $25 \, \mathrm{cm}$ when, its speed is halved. The value of spring constant of used spring will be $nE \, \mathrm{Nm}^{-1}$ for n = _____________.

Answer: 24

Solution

Using work-energy theorem $$W_{net} = (K_f - K_i)$$ $$\Rightarrow -\frac{1}{2} K x^2 = \frac{1}{2} m \left( \frac{v}{2} \right)^2 - \frac{1}{2} m v^2 = \frac{E}{4} - E$$ $$\Rightarrow \frac{1}{2} K x^2 = \frac{3E}{4} \Rightarrow K = \frac{3E}{2x^2}$$ $$\Rightarrow K = \frac{3E}{2 \times \left( \frac{1}{4} \right)^2} = 24E$$ $n = 24$

Question 60

Physics · System of Particles and Rotational Motion · Numerical

Four identical discs each of mass 'M' and diameter 'a' are arranged in a small plane as shown in figure. If the moment of inertia of the system about OO' is $\frac{x}{4}Ma^2$. Then, the value of $x$ will be _____.

Answer: 3

Solution

Given $I_1 = I_3 = \frac{MR^2}{4}$. $I_2 = \frac{MR^2}{4} + MR^2 = \frac{5}{4}MR^2 = I_4$. So $I = I_1 + I_2 + I_3 + I_4$. $$= \frac{MR^2}{2} + \frac{5}{2}MR^2$$ $$= 3MR^2$$, Putting $R = \frac{a}{2}$ $I = \frac{3Ma^2}{4}$, So $x = 3$

Chemistry

Question 61

Chemistry · Structure of Atom · Single correct

Identify the incorrect statement from the following.

  1. A circular path around the nucleus in which an electron moves is proposed as Bohr's orbit.
  2. An orbital is the one electron wave function ($\Psi$) in an atom.
  3. The existence of Bohr's orbits is supported by hydrogen spectrum.
  4. Atomic orbital is characterised by the quantum numbers $n$ and $l$ only.

Answer: (d)

Solution

Atomic orbital is characterised by $n$, $l$, $m$.

Question 62

Chemistry · Thermodynamics · Single correct

Which of the following relation is not correct?

  1. $\Delta H = \Delta U - P \Delta V$
  2. $\Delta U = q + W$
  3. $\Delta S_{sys} + \Delta S_{surr} \geq 0$
  4. $\Delta G = \Delta H - T \Delta S$

Answer: (a)

Solution

If $U + Pv$ (By definition) $$\Delta I4 = \Delta U + \Delta (Pr)$$ at constant pressure $$\Delta H = \Delta U + P \Delta V$$

Question 63

Chemistry · Electrochemistry · Single correct

Match List-I with List-II. Choose the correct answer from the options given below :

  1. $(A)(a) - (I), (b) - (II), (c) - (III), (b) - (IV)$
  2. $(B)(a) - (II), (b) - (I), (c) - (IV), (d) - (III)$
  3. $(C)(a) - (II), (b) - (I), (c) - (IV), (d) - (III)$
  4. $(D)(a) - (IV), (b) - (II), (c) - (I), (d) - (IV)$

Answer: (c)

Solution

$(a)$ $\mathrm{Cd(s) + 2Ni(OH)_3(s) \rightarrow CdO(s) + 2Ni(OH)_2(s) + H_2O(l)}$ Discharge of secondary Battery $(b)$ $\mathrm{Zn(Hg) + HgO(s) \rightarrow ZnO(s) + Hg(l)}$ (Primary Battery Mercury cell) $(c)$ $\mathrm{2PbSO_4(s) + 2H_2O(l) \rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq)}$ Charging of secondary Battery $(d)$ $\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)}$ – Fuel cell

Question 64

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

Match List-I with List-II. \begin{tabular}{|c|p{7cm}|c|l|} \hline \multicolumn{2}{|c|}{\textbf{List-I}} & \multicolumn{2}{c|}{\textbf{List-II}} \\ \multicolumn{2}{|c|}{\textbf{Reaction}} & \multicolumn{2}{c|}{\textbf{Catalyst}} \\ \hline (A) & $4NH_3(g)+5O_2(g)\rightarrow4NO(g)+6H_2O(g)$ & (I) & NO(g) \\ \hline (B) & $N_2(g)+3H_2(g)\rightarrow2NH_3(g)$ & (II) & H$_2$SO$_4(l)$ \\ \hline (C) & $C_{12}H_{22}O_{11}(aq)+H_2O(l)\rightarrow C_6H_{12}O_6$ (Glucose) $+$ C$_6$H$_{12}$O$_6$ (Fructose) & (III) & Pt(s) \\ \hline (D) & $2SO_2(g)+O_2(g)\rightarrow2SO_3(g)$ & (IV) & Fe(s) \\ \hline \end{tabular} Choose the correct answer from the options given below:

  1. (A) - (II), (B) - (III), $(C)$ - (I), (D) - (IV)
  2. (A) - (III), (B) - (II), $(C)$ - (I), (D) - (IV)
  3. (A) - (III), (B) - (IV), $(C)$ - (II), (D) - (I)
  4. (A) - (III), (B) - (II), $(C)$ - (IV), (D) - (I)

Answer: (c)

Solution

(a) $4\mathrm{NH_3}(g) + 5\mathrm{O_2}(g) \xrightarrow{\mathrm{Pt(s)}} 4\mathrm{NO}(g) + 6\mathrm{H_2O}(g)$ Ostwald process 500 K (b) $\mathrm{N_2} + 3\mathrm{H_2} \xrightarrow{\mathrm{Fe(s)}} 2\mathrm{NH_3}(g)$ Haber’s process $(c)$ $\mathrm{C_{12}H_{22}O_{11}}(aq.) + \mathrm{H_2O}(\ell) \xrightarrow{\mathrm{H^+}} \mathrm{C_6H_{12}O_6} + \mathrm{C_6H_{12}O_6}$ Inversion of sugar cane (d) $2\mathrm{SO_2}(g) + \mathrm{O_2}(g) \xrightarrow{\mathrm{NO(g)}} 2\mathrm{SO_3}(g)$

Question 65

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

In which of the following pairs, electron gain enthalpies of constituent elements are nearly the same or identical? (A) Rb and Cs (B) Na and K $(C)$ Ar and Kr (D) I and At Choose the correct answer from the options given below:

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

Answer: (c)

Solution

Rb and Cs have nearly same electron gain enthalpy. Electron gain enthalpy is $-46 \, \mathrm{kJ/mol}$. Ar and Kr have same $\Delta H_{eq}$. Value is $+96 \, \mathrm{kJ/mol}$.

Question 66

Chemistry · General Principles and Processes of Isolation of Elements · Single correct

Which of the reaction is suitable for concentrating ore by leaching process?

  1. $2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$
  2. $Fe_3O_4 + CO \rightarrow 3FeO + CO_2$
  3. $Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]$
  4. $Al_2O_3 + 6Mg \rightarrow 6MgO + 4Al$

Answer: (c)

Solution

The reaction is given by $$\mathrm{Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]}$$. This process is known as leaching.

Question 67

Chemistry · Hydrogen · Single correct

The metal salts formed during softening of hardwater using Clark’s method are:

  1. $Ca(OH)_2$ and $Mg(OH)_2$
  2. $CaCO_3$ and $Mg(OH)_2$
  3. $Ca(OH)_2$ and $MgCO_3$
  4. $CaCO_3$ and $MgCO_3$

Answer: (b)

Solution

Clark's Method Reaction $$\mathrm{Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 + 2H_2O}$$ $$\mathrm{Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 + Mg(OH)_2 + 2H_2O}$$

Question 68

Chemistry · The s-Block Elements · Single correct

Which of the following statement is incorrect ?

  1. Low solubility of LiF in water is due to its small hydration enthalpy.
  2. $\mathrm{KO_2}$ is paramagnetic.
  3. Solution of sodium in liquid ammonia is conducting in nature.
  4. Sodium metal has higher density than potassium metal

Answer: (a)

Solution

Low solubility of LiF in water is due to high lattice enthalpy.

Question 69

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

Match List-I with List-II, match the gas evolved during each reaction. \begin{tabular}{|c|p{5.5cm}|c|c|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \hline (A) & $(NH_4)_2Cr_2O_7\xrightarrow{\Delta}$ & (I) & H$_2$ \\ \hline (B) & $KMnO_4+HCl\rightarrow$ & (II) & N$_2$ \\ \hline (C) & $Al+NaOH+H_2O\rightarrow$ & (III) & O$_2$ \\ \hline (D) & $NaNO_3\xrightarrow{\Delta}$ & (IV) & Cl$_2$ \\ \hline \end{tabular} Choose the correct answer from the options given below :

  1. (A) - (II), (B) - (III), $C)$ - (I), (D) - (IV)
  2. (A)- (III), (B) - (I), $C)$ - (IV), (D) - (II)
  3. (A)- (II), (B) - (IV), $C)$ - (I), (D) - (III)
  4. (A)- (III), (B) - (IV), $C)$ - (I), (D) - (II)

Answer: (c)

Solution

The given reactions are: $$(\mathrm{NH_4})_2 \mathrm{Cr_2O_7} \xrightarrow{\Delta} \mathrm{N_2} + \mathrm{Cr_2O_3} + 4\mathrm{H_2O}$$ $$\mathrm{KMnO_4} + \mathrm{HCl} \rightarrow \mathrm{MnCl_2} + \mathrm{KCl} + \mathrm{Cl_2} + \mathrm{H_2O}$$ $$\mathrm{Al} + \mathrm{NaOH} + \mathrm{H_2O} \rightarrow \mathrm{H_2} + \mathrm{Na} \left[ \mathrm{Al(OH)_4} \right]$$ $$\mathrm{NaNO_3} \longrightarrow \mathrm{NaNO_2} + \mathrm{O_2}$$

Question 70

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

Which of the following has least tendency to liberate $\mathrm{H}_2$ from mineral acids?

  1. Cu
  2. Mn
  3. Ni
  4. Zn

Answer: (a)

Solution

Copper is least electropositive among the given metals and it lies below H in reactivity series.

Question 71

Chemistry · Environmental Chemistry · Single correct

Given below are two statements : Statement I : In polluted water values of both dissolved oxygen and BOD are very low. Statement II : Eutrophication results in decrease in the amount of dissolved oxygen. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is false but Statement II is true

Answer: (d)

Solution

Since eutrophication is the result of excessive growth of weed in water bodies, which consume dissolved oxygen of water bodies. Therefore, eutrophication decreases the amount of dissolved oxygen in water bodies. Polluted water has a low value of dissolved oxygen, but a high value of BOD (Biological oxygen demand), since chemical and organic matter requires dissolved oxygen to decompose.

Question 72

Chemistry · Co-ordination Compounds · Single correct

Match List-I with List-II. Choose the correct answer from the options given below:

  1. (A) - (II), (B) - (I), ($C$) - (IV), (D) - (III)
  2. (A) - (IV), (B) - (III), ($C$) - (I), (D) - (II)
  3. (A) - (III), (B) - (IV), ($C$) - (I), (D) - (II)
  4. (A) - (IV), (B) - (III), ($C$) - (II), (D) - (I)

Answer: (c)

Solution

Question 73

Chemistry · Hydrocarbons · Single correct

Choose the correct option for the following reactions.

  1. 'A' and 'B' are both Markovnikov addition products.
  2. 'A' is Markovnikov product and 'B' is anti-Markovnikov product.
  3. 'A' and 'B' are both anti-Markovnikov products.
  4. 'B' is Markovnikov and 'A' is anti-Markovnikov product.

Answer: (b)

Solution

The reaction of the given alkene with $\mathrm{Hg(OAc)_2}$ and $\mathrm{H_2O}$ followed by $\mathrm{NaBH_4}$ results in the Markovnikov product (A). The reaction with $\mathrm{B_2H_6}$ and $\mathrm{H_2O_2/ OH^-}$ results in the anti-Markovnikov product (B).

Question 74

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

Among the following marked proton of which compound shows lowest $pK_a$ value?

Answer: (c)

Solution

Question 75

Chemistry · Alcohols, Phenols and Ethers · Single correct

Identify the major product A and B for the below given reaction sequence.

Answer: (b)

Solution

The reaction sequence involves the following steps: 1. Friedel-Crafts alkylation of benzene with isopropyl chloride in the presence of AlCl3 to form isopropylbenzene. 2. Oxidation of isopropylbenzene with oxygen in acidic medium to form phenol (P). 3. Bromination of phenol in carbon disulfide to form bromophenol (B). 4. Oxidation of phenol with Na2Cr2O7 and H2SO4 to form benzoquinone (A).

Question 76

Chemistry · Amines · Single correct

Identify the correct statement for the below given transformation.

  1. $A - CH_3CH_2CH = CH-CH_3$, $B - CH_3CH_2CH_2CH = CH_2$, Saytzeff products
  2. $A - CH_3CH_2CH = CH-CH_3$, $B - CH_3CH_2CH_2CH = CH_2$, Hafmann products
  3. $A - CH_3CH_2CH_2CH = CH_2$, $B - CH_3CH_2CH = CHCH_3$, Hofmann products
  4. $A - CH_3CH_2CH_2CH = CH_2$, $B - CH_3CH_2CH = CHCH_3$, Saytzeff products

Answer: (c)

Solution

Question 77

Chemistry · Polymers · Single correct

Terylene polymer is obtained by condensation of:

  1. Ethane-1, 2-diol and Benzene-1, 3 dicarboxylic acid
  2. Propane-1, 2-diol and Benzene-1, 4 dicarboxylic acid
  3. Ethane-1, 2-diol and Benzene-1, 4 dicarboxylic acid
  4. Ethane-1, 2-diol and Benzene-1, 2 dicarboxylic acid

Answer: (c)

Solution

Question 78

Chemistry · Biomolecules · Single correct

For the below given cyclic hemiacetal (X), the correct pyranose structure is:

Answer: (d)

Solution

The correct pyranose structure is shown. The corresponding aldopyranose is depicted in the cyclic form as a hemiacetal.

Question 79

Chemistry · Chemistry in Everyday Life · Single correct

Statements about Enzyme Inhibitor Drugs are given below : (A) There are Competitive and Non-competitive inhibitor drugs. (B) These can bind at the active sites and allosteric sites. (C) Competitive Drugs are allosteric site blocking drugs. (D) Non-competitive Drugs are active site blocking drugs. Choose the correct answer from the options given below :

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

Answer: (c)

Solution

Enzyme inhibitors can be competitive inhibitors (inhibit the attachment of substrate on active site of enzyme) and non-competitive inhibitor (changes the active site of enzyme after binding at allosteric site).

Question 80

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

For kinetic study of the reaction of iodide ion with $H_2O_2$ at room temperature: (A) Always use freshly prepared starch solution. (B) Always keep the concentration of sodium thiosulphate solution less than that of KI solution. (C) Record the time immediately after the appearance of blue colour. (D) Record the time immediately before the appearance of blue colour. (E) Always keep the concentration of sodium thiosulphate solution more than that of KI solution. Choose the correct answer from the options given below:

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

Answer: (a)

Solution

The is recorded immediately after the blue colour appears. $\mathrm{Na_2S_2O_3}$ is kept in limited amount.

Question 81

Chemistry · Some Basic Concepts of Chemistry · Numerical

In the given reaction, $$\mathrm{X + Y + 3Z \rightleftharpoons XYZ_3}$$ if one mole of each of X and Y with 0.05 mol of Z gives compound XYZ$_3$. (Given : Atomic masses of X, Y and Z are 10, 20 and 30 amu, respectively). The yield of XYZ$_3$ is _________ g. (Nearest integer)

Answer: 2

Solution

Given the reaction: $$\mathrm{X + Y + 3Z \rightleftharpoons XYZ_3}$$ with initial moles: $1 mol$ of X, $1 mol$ of Y, and $0.05 mol$ of Z. Z is the limiting reagent (L.R.). The calculation for moles of $\mathrm{XYZ_3}$ is: $$\frac{0.05}{3} = 1 mole of \mathrm{XYZ_3}$$ The mass of $\mathrm{XYZ_3}$ is calculated as: $$Mass of \mathrm{XYZ_3} = \frac{0.05}{3} \times (10 + 20 + 30 \times 3)$$ This simplifies to: $$= 2 g$$

Question 82

Chemistry · The Solid State · Numerical

An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is $6.0 \, \mathrm{g \, cm^{-3}}$. The number of atoms present in 180 g of the element is _____ $\times \, 10^{23}$. (Nearest integer)

Answer: 22

Solution

M is body centred cubic, therefore $Z = 2$. Let mass of 1 atom of M be $A$. Edge length $= 300 \, \mathrm{pm}$. Density $= 6 \, \mathrm{g/cm^3}$. Therefore, $$6 \, \mathrm{g/cm^3} = \frac{Z \times A}{(300 \times 10^{-10})^3} = \frac{2 \times A}{27 \times 10^{-24}}$$ $A = 81 \times 10^{-24} \, \mathrm{g}$. Therefore, Atomic mass $= 48.6 \, \mathrm{g}$. Therefore, Mole in $180 \, \mathrm{g} = \frac{180}{48.6} = 3.7$ moles. Atoms of M $= 3.7 \times 6 \times 10^{23} = 22.22 \times 10^{23}$ atoms.

Question 83

Chemistry · Chemical Bonding and Molecular Structure · Numerical

The number of paramagnetic species among the following is __________. $\mathrm{B_2}$, $\mathrm{Li_2}$, $\mathrm{C_2}$, $\mathrm{C_2^-}$, $\mathrm{O_2^{2-}}$, $\mathrm{O_2}$ and $\mathrm{He_2^+}$

Answer: 4

Solution

Paramagnetic $\mathrm{B_2}$, $\mathrm{C_2^-}$, $\mathrm{O_2^+}$, $\mathrm{He_2^+}$

Question 84

Chemistry · Solutions · Numerical

$150\,\mathrm{g}$ of acetic acid was contaminated with $10.2\,\mathrm{g}$ of ascorbic acid ($\mathrm{C_6H_8O_6}$) to lower down its freezing point by $(x \times 10^{-1})°\mathrm{C}$. The value of $x$ is \_\_\_\_. (Nearest integer) [Given $K_f = 3.9\,\mathrm{K\,kg\,mol^{-1}}$; Molar mass of ascorbic acid $= 176\,\mathrm{g\,mol^{-1}}$]

Answer: 15

Solution

$150\,\mathrm{g}\ \mathrm{CH_3COOH}$, $10.2\,\mathrm{g}$ ascorbic acid $\Rightarrow 0.058$ moles $\Delta T_f = \left(x \times 10^{-1}\right)°\mathrm{C}$ $\Delta T_f = K_f \cdot \text{molality} = 3.9 \times \dfrac{0.058}{150} \times 1000 = 1.5°\mathrm{C} = 15 \times 10^{-1}\,\mathrm{C}$

Question 85

Chemistry · Equilibrium · Numerical

$K_a$ for butyric acid $(\mathrm{C_3H_7COOH})$ is $2 \times 10^{-5}$. The pH of $0.2 \, \mathrm{M}$ solution of butyric acid is $\_\_\_ \times 10^{-1}$. (Nearest integer) [Given $\log 2 = 0.30$]

Answer: 27

Solution

Given $K_a$ of Butyric acid $\Rightarrow 2 \times 10^{-5}$, $\mathrm{p}K_a = 4.7$. Find the pH of a $0.2 \, \mathrm{M}$ solution. $$\mathrm{pH} = \frac{1}{2} \mathrm{p}K_a - \frac{1}{2} \log C$$ Substitute the values: $$= \frac{1}{2} (4.7) - \frac{1}{2} \log (0.2)$$ Calculate: $$= 2.35 + 0.35 = 2.7$$ Thus, $$\mathrm{pH} = 27 \times 10^{-1}$$

Question 86

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

For the given first order reaction A $\rightarrow$ B the half life of the reaction is 0.3010 min. The ratio of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to . (Nearest integer)

Answer: 100

Solution

Given the reaction $\mathrm{A} \rightarrow \mathrm{B}$ with $t_{1/2} = 0.3010 \, \mathrm{min}$. We need to find $A_0/A_t$ at time $2 \, \mathrm{min}$. The rate constant $K$ is given by $$K = \frac{2.303}{t} \log \left[ \frac{A_0}{A_t} \right].$$ Thus, $$\frac{0.693}{t_{1/2}} = \frac{2 \cdot 2.303}{2} \log \left( \frac{A_0}{A_t} \right).$$ Or, $$\frac{2.303 \times 0.3010}{0.3010} = \frac{2.303}{2} \log \frac{A_0}{A_t}.$$ This simplifies to $$\log \frac{A_0}{A_t} = 2.$$ Therefore, $$\frac{A_0}{A_t} = 10^2 = 100.$$

Question 87

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

The number of interhalogens from the following having square pyramidal structure is : $ClF_3, IF_7, BrF_5, BrF_3, I_2Cl_6, IF_5, ClF, ClF_5$

Answer: 3

Solution

Square pyramidal structures are $\mathrm{BrF_5}$, $\mathrm{IF_5}$ and $\mathrm{ClF_5}$.

Question 88

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

The disproportionation of $\mathrm{MnO}_4^{2-}$ in acidic medium resulted in the formation of two manganese compounds A and B. If the oxidation state of Mn in B is smaller than that of A, then the spin-only magnetic moment ($\mu$) value of B in BM is ___________. (Nearest integer)

Answer: 4

Solution

The reaction is given by: $$\mathrm{MnO_4^{2-} \xrightarrow{H^+} MnO_4^- + MnO_2}$$ The number of unpaired electrons is 3. Therefore, the magnetic moment $\mu$ is calculated as: $$\mu = \sqrt{15} = 3.877$$ The nearest integer is 4.

Question 89

Chemistry · Co-ordination Compounds · Numerical

Total number of relatively more stable isomer(s) possible for octahedral complex $[Cu(en)_2(SCN)_2]$ will be __________.

Answer: 3

Solution

The compound $[\mathrm{Cu(en)_2(SCN)_2}]$ can exhibit different geometrical isomers. The diagrams show possible configurations of the ligands around the copper center. The ethylenediamine (en) ligands are bidentate, and the thiocyanate (SCN) ligands can bind in different orientations, leading to isomerism.

Question 90

Chemistry · Some Basic Concepts of Chemistry · Numerical

On complete combustion of $0.492$ g of an organic compound containing C, H and O, $0.7938$ g of $\mathrm{CO_2}$ and $0.4428$ g of $\mathrm{H_2O}$ was produced. The % composition of oxygen in the compound is _____.

Answer: 46

Solution

$0.492\,\mathrm{g}$ of $\mathrm{C_xH_yO_z}$ gives $0.7938\,\mathrm{g}$ of $\mathrm{CO_2}=0.018$ moles. $0.4428\,\mathrm{g}$ of $\mathrm{H_2O}=0.0246$ moles. So moles of C $=0.018\Rightarrow0.216\,\mathrm{g}$. Moles of H $=0.049\Rightarrow0.049\,\mathrm{g}$. $\therefore$ wt. of Oxygen $=0.492-0.216-0.049=0.227\,\mathrm{g}$. $\%$ of Oxygen $=\frac{0.227}{0.492}\times100\approx46$.