JEE Main 25 July 2021 Shift 1 question paper with solutions

JEE Main 25 July 2021 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Maths

Question 1

Maths · Heights and Distances · Single correct

A spherical gas balloon of radius 16 meter subtends an angle $60^\circ$ at the eye of the observer $A$ while the angle of elevation of its center from the eye of $A$ is $75^\circ$. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is:

  1. 8(2 + 2$\sqrt{3}$ + $\sqrt{2}$)
  2. 8($\sqrt{6}$ + $\sqrt{2}$ + 2)
  3. 8($\sqrt{2}$ + 2 + $\sqrt{3}$)
  4. 8($\sqrt{6}$ - $\sqrt{2}$ + 2)

Answer: (b)

Solution

O is the centre of the sphere. P, Q are the points of contact of tangents from A. Let T be the topmost point of the balloon and R be the foot of the perpendicular from O to the ground. From triangle OAP, $OA = 16 \csc 30^\circ = 32$. From triangle ABO, $OR = OA \sin 75^\circ = 32 \frac{(\sqrt{3}+1)}{2\sqrt{2}}$. So the level of the topmost point is $OR + OT = 8(\sqrt{6} + \sqrt{2} + 2)$.

Question 2

Maths · Applications of Derivatives · Single correct

Let \[ f(x)=3\sin^4x+10\sin^3x+6\sin^2x-3,\qquad x\in\left[-\frac{\pi}{6},\frac{\pi}{2}\right]. \] Then, $f$ is:

  1. increasing in $\left(-\frac{\pi}{6}, \frac{\pi}{2}\right)$
  2. decreasing in $\left(0, \frac{\pi}{2}\right)$
  3. increasing in $\left(-\frac{\pi}{6}, 0\right)$
  4. decreasing in $\left(-\frac{\pi}{6}, 0\right)$

Answer: (d)

Solution

Given $f(x) = 3 \sin^4 x + 10 \sin^3 x + 6 \sin^2 x - 3$, $x \in \left[ -\frac{\pi}{6}, \frac{\pi}{2} \right]$. Differentiating, we have $$f'(x) = 12 \sin^3 x \cos x + 30 \sin^2 x \cos x + 12 \sin x \cos x$$ This simplifies to $$= 6 \sin x \cos x \left( 2 \sin^2 x + 5 \sin x + 2 \right)$$ Further simplifying, $$= 6 \sin x \cos x (2 \sin x + 1)(\sin x + 2)$$ The function is decreasing in $\left( -\frac{\pi}{6}, 0 \right)$.

Question 3

Maths · Sequences and Series · Single correct

Let $S_n$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3n} = 3 \ S_{2n}$, then the value of $\frac{S_{4n}}{S_{2n}}$ is :

  1. 6
  2. 4
  3. 2
  4. 8

Answer: (a)

Solution

Let $a$ be the first term and $d$ be the common difference of this A.P. Given $S_{3a} = 3S_{2n}$. Therefore, $$\frac{3n}{2} \left[ 2a + (3n - 1)d \right] = 3 \frac{2n}{2} \left[ 2a + (2n - 1)d \right]$$ which implies $$2a + (3n - 1)d = 4a + (4n - 2)d$$ leading to $$2a + (n - 1)d = 0.$$ Now, $$\frac{S_{4n}}{S_{2n}} = \frac{\frac{4n}{2} \left[ 2a + (4n - 1)d \right]}{\frac{2n}{2} \left[ 2a + (2n - 1)d \right]} = \frac{2 \left[ 2a + (n - 1)d + 3nd \right]}{\left[ 2a + (n - 1)d + nd \right]}$$ which simplifies to $$\frac{6nd}{nd} = 6.$$

Question 4

Maths · Conic Sections · Single correct

The locus of the centroid of the triangle formed by any point P on the hyperbola $16x^2 - 9y^2 + 32x + 36y - 164 = 0$, and its foci is:

  1. $16x^2 - 9y^2 + 32x + 36y - 36 = 0$
  2. $9x^2 - 16y^2 + 36x + 32y - 144 = 0$
  3. $16x^2 - 9y^2 + 32x + 36y - 144 = 0$
  4. $9x^2 - 16y^2 + 36x + 32y - 36 = 0$

Answer: (a)

Solution

Given hyperbola is $16(x+1)^2 - 9(y-2)^2 = 164 + 16 - 36 = 144$. $$\frac{(x+1)^2}{9} - \frac{(y-2)^2}{16} = 1$$ Eccentricity, $e = \sqrt{1 + \frac{16}{9}} = \frac{5}{3}$. Foci are $(4, 2)$ and $(-6, 2)$. Let the centroid be $(h, k)$. Let $A(\alpha, \beta)$ be a point on the hyperbola. So $h = \frac{\alpha - 6 + 4}{3}$, $k = \frac{\beta + 2 + 2}{3}$. $$\Rightarrow \alpha = 3h + 2, \beta = 3k - 4$$ $(\alpha, \beta)$ lies on the hyperbola so $16(3h + 2 + 1)^2 - 9(3k - 4 - 2)^2 = 144$. $$\Rightarrow 144(h + 1)^2 - 81(k - 2)^2 = 144$$ $$\Rightarrow 16(h^2 + 2h + 1) - 9(k^2 - 4k + 4) = 16$$ $$\Rightarrow 16x^2 - 9y^2 + 32x + 36y - 36 = 0$$

Question 5

Maths · Vector Algebra · Single correct

Let the vectors $$(2 + a + b)\hat{i} + (a + 2b + c)\hat{j} - (b + c)\hat{k}, (1 + b)\hat{i} + 2\hat{b} - b\hat{k}$$ and $$(2 + b)\hat{i} + 2\hat{b}\hat{j} + (1 - b)\hat{k} a, b, c, \in \mathbb{R}$$ be co-planar. Then which of the following is true?

  1. $2b = a + c$
  2. $3c = a + b$
  3. $a = b + 2c$
  4. $2a = b + c$

Answer: (a)

Solution

If the vectors are co-planar, $$\begin{vmatrix} a + b + 2 & a + 2b + c & -b - c \\ b + 1 & 2b & -b \\ b + 2 & 2b & 1 - b \end{vmatrix} = 0$$ Now $R_3 \rightarrow R_3 - R_2$, $R_1 \rightarrow R_1 - R_2$ So $$\begin{vmatrix} a + 1 & a + c & -c \\ b + 1 & 2b & -b \\ 1 & 0 & 1 \end{vmatrix} = 0$$ $$= (a + 1)2b - (a + c)(2b + 1) - c(-2b)$$ $$= 2ab + 2b - 2ab - a - 2bc - c + 2bc$$ $$= 2b - a - c = 0$$

Question 6

Maths · Continuity and Differentiability · Single correct

Let $f : \mathbb{R} \to \mathbb{R}$ be defined as $$f(x) = \begin{cases} \frac{\lambda |x^2 - 5x + 6|}{\mu (5x - x^2 - 6)}, & x 2 \\ \mu, & x = 2 \end{cases}$$ where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x = 2$, then $\lambda + \mu$ is equal to

  1. $e(-e + 1)$
  2. $e(e - 2)$
  3. $2e - 1$

Answer: (a)

Solution

Given $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} e^{\frac{\sin(x-2)}{x-2}} = e^1$. $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{-\lambda(x-2)(x-3)}{\mu(x-2)(x-3)} = -\frac{\lambda}{\mu}$. For continuity $\mu = e = -\frac{\lambda}{\mu} \implies \mu = e, \lambda = -e^2$. $\lambda + \mu = e(-e + 1)$

Question 7

Maths · Integrals · Single correct

The value of the definite integral $$\int_{x/24}^{5x/24} \frac{dx}{1 + \sqrt[3]{\tan 2x}}$$ is

  1. $\frac{\pi}{3}$
  2. $\frac{\pi}{6}$
  3. $\frac{\pi}{12}$
  4. $\frac{\pi}{18}$

Answer: (c)

Solution

Let $I = \int_{\pi/24}^{5\pi/24} \frac{(\cos 2x)^{1/3}}{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}} \, dx$ ....(i) Therefore, $I = \int_{\pi/2 - 4}^{\pi/24} \frac{\left( \cos \left\{ 2 \left( \frac{\pi}{4} - x \right) \right\} \right)^{1/3}}{\left( \cos \left\{ 2 \left( \frac{\pi}{4} - x \right) \right\} \right)^{1/3} + \left( \sin \left\{ 2 \left( \frac{\pi}{4} - x \right) \right\} \right)^{1/3}} \, dx$ $$\left\{ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \right\}$$ So $I = \int_{\pi/24}^{5\pi/24} \frac{(\sin 2x)^{1/3}}{(\sin 2x)^{1/3} + (\cos 2x)^{1/3}} \, dx$ Hence $2I = \int_{\pi/24}^{5\pi/24} \, dx [(i) + (ii)]$ Therefore, $2I = \frac{4\pi}{24} \Rightarrow I = \frac{\pi}{12}$

Question 8

Maths · Binomial Theorem · Single correct

If b is very small as compared to the value of a, so that the cube and other higher powers of $\frac{b}{a}$ can be neglected in the identity $\frac{1}{a-b} + \frac{1}{a-2b} + \frac{1}{a-3b} + \ldots + \frac{1}{a-nb} = \alpha n + \beta n^2 + \gamma n^3$, then the value of $\gamma$ is:

  1. $\frac{a^2+b}{3a^3}$
  2. $\frac{a+b}{3a^2}$
  3. $\frac{b^2}{3a^3}$
  4. $\frac{a+b^2}{3a^3}$

Answer: (c)

Solution

Given $$(a-b)^{-1} + (a-2b)^{-1} + \ldots + (a-nb)^{-1}$$ This is equal to $$\frac{1}{a} \sum_{t=1}^{n} \left(1 - \frac{rb}{a}\right)^{-1}$$ This simplifies to $$\frac{1}{a} \sum_{i=1}^{n} \left\{ \left(1 + \frac{rb}{a} + \frac{r^2b^2}{a^2}\right) + (terms to be neglected) \right\}$$ This is equal to $$\frac{1}{a} \left[ n + \frac{n(n+1)}{2} \cdot \frac{b}{a} + \frac{n(n+1)(2n+1)}{6} \cdot \frac{b^2}{a^2} \right]$$ This simplifies to $$\frac{1}{a} \left[ n^3 \left( \frac{b^2}{3a^2} \right) + \ldots \right]$$ So $\gamma = \frac{b^2}{3a^3}$

Question 9

Maths · Differential Equations · Single correct

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = 1 + xe^{y-x}$, $-\sqrt{2} < x < \sqrt{2}, y(0) = 0$ then, the minimum value of $y(x), x \in (-\sqrt{2}, \sqrt{2})$ is equal to:

  1. $(2 - \sqrt{3}) - \log_e 2$
  2. $(2 + \sqrt{3}) + \log_e 2$
  3. $(1 + \sqrt{3}) - \log_e(\sqrt{3} - 1)$
  4. $(1 - \sqrt{3}) - \log_e(\sqrt{3} - 1)$

Answer: (d)

Solution

Given $\($ $\frac{dy-dx}{e^{y-x}}$ = x dx $\)$ $\($ $\Rightarrow$ $\frac{dy-dx}{e^{y-x}}$ = x dx $\)$ $\($ $\Rightarrow$ -e^{x-y} = $\frac{x^2}{2}$ + c $\)$ At $\($ x = 0, y = 0 $\Rightarrow$ c = -1 $\)$ $\($ $\Rightarrow$ e^{x-y} = $\frac{2-x^2}{2}$ $\)$ $\($ $\Rightarrow$ y = x - $\ln$ $\left$( $\frac{2-x^2}{2}$ $\right$) $\)$ $\($ $\Rightarrow$ $\frac{dy}{dx}$ = 1 + $\frac{2x}{2-x^2}$ = $\frac{2+2x-x^2}{2-x^2}$ $\)$ $\($ $\sqrt{1 - \sqrt{1 + \sqrt{2} + \sqrt{2}}}$ $\)$ So minimum value occurs at $\($ x = 1 - $\sqrt{3}$ $\)$ $\($ y(1 - $\sqrt{3}$) = (1 - $\sqrt{3}$) - $\ln$ $\left$( $\frac{2-(4-2\sqrt{3})}{2}$ $\right$) $\)$

Question 10

Maths · Mathematical Reasoning · Single correct

The Boolean expression $(p \Rightarrow q) \land (q \Rightarrow \sim p)$ is equivalent to :

  1. $\sim$ q
  2. q
  3. p
  4. p

Answer: (d)

Solution

(p $\to$ q) $\land$ (q $\to$ $\sim$ p) = ($\sim$ p $\lor$ q) $\land$ ($\sim$ q $\lor$ $\sim$ p) $\{$p $\to$ q = $\sim$ p $\lor$ q$\}$ = ($\sim$ p $\lor$ q) $\land$ ($\sim$ p $\lor$ $\sim$ q) $\{$commutative property$\}$ = $\sim$ p $\lor$ (q $\land$ $\sim$ q) $\{$distributive property$\}$ = $\sim$ p

Question 11

Maths · Applications of Integrals · Single correct

The area (in sq. units) of the region, given by the set $\{(x, y) \in \mathbb{R} \times \mathbb{R} \mid x \geq 0, 2x^2 \leq y \leq 4 - 2x\}$ is :

  1. $\frac{8}{3}$
  2. $\frac{17}{3}$
  3. $\frac{13}{3}$
  4. $\frac{7}{3}$

Answer: (d)

Solution

Required area = $\int$_0^1 (4 - 2x - 2x^2) $\,$ dx = 4x - x^2 - $\frac{2x^3}{3}$ $\bigg$|_0^1 = 4 - 1 - $\frac{2}{3}$ = $\frac{7}{3}$

Question 12

Maths · Trigonometric Functions · Single correct

The sum of all values of $x$ in $[0, 2\pi]$, for which $$\sin x + \sin 2x + \sin 3x + \sin 4x = 0,$$ is equal to:

  1. $8\pi$
  2. $11\pi$
  3. $12\pi$
  4. $9\pi$

Answer: (d)

Solution

Given $$ (\sin x+\sin 4x)+(\sin 2x+\sin 3x)=0 $$ This implies $$ 2\sin\frac{5x}{2}\left(\cos\frac{3x}{2}+\cos\frac{x}{2}\right)=0 $$ which simplifies to $$ 2\sin\frac{5x}{2}\left(2\cos x\cos\frac{x}{2}\right)=0 $$ Thus, $$ \sin\frac{5x}{2}=0 \implies \frac{5x}{2}=0,\pi,2\pi,3\pi,4\pi,5\pi $$ This gives $$ x=0,\frac{2\pi}{5},\frac{4\pi}{5},\frac{6\pi}{5},\frac{8\pi}{5},2\pi $$ For $$ \cos\frac{x}{2}=0, $$ we have $$ \frac{x}{2}=\frac{\pi}{2} \implies x=\pi $$ For $$ \cos x=0, $$ we have $$ x=\frac{\pi}{2},\frac{3\pi}{2} $$ So the sum is $$ \left(0+\frac{2\pi}{5}+\frac{4\pi}{5}+\frac{6\pi}{5}+\frac{8\pi}{5}+2\pi\right) +\pi +\left(\frac{\pi}{2}+\frac{3\pi}{2}\right) =6\pi+\pi+2\pi =9\pi $$

Question 13

Maths · Relations and Functions (Advanced) · Single correct

Let $g : \mathbb{N} \to \mathbb{N}$ be defined as $$g(3n + 1) = 3n + 2$$ $$g(3n + 2) = 3n + 3$$ $g(3n + 3) = 3n + 1$, for all n $\geq 0$ Then which of the following statements is true ?

  1. There exists an onto function $f : \mathbb{N} \to \mathbb{N}$ such that fog = f
  2. There exists a one-one function $f : \mathbb{N} \to \mathbb{N}$ such that fog = f
  3. gogog = g
  4. There exists a function $f : \mathbb{N} \to \mathbb{N}$ such that gof = f

Answer: (a)

Solution

Given $g : \mathbb{N} \to \mathbb{N}$ with $g(3n + 1) = 3n + 2$. Then $g(3n + 2) = 3n + 3$ and $g(3n + 3) = 3n + 1$. The function $g(x)$ is defined as: $$g(x) = \begin{cases} x + 1 & x = 3k + 1 \\ x + 1 & x = 3k + 2 \\ x - 2 & x = 3k + 3 \end{cases}$$ For $g(g(x))$, we have: $$g(g(x)) = \begin{cases} x + 2 & x = 3k + 1 \\ x - 1 & x = 3k + 2 \\ x - 1 & x = 3k + 3 \end{cases}$$ And for $g(g(g(x)))$, we have: $$g(g(g(x))) = \begin{cases} x & x = 3k + 1 \\ x & x = 3k + 2 \\ x & x = 3k + 3 \end{cases}$$ If $f : \mathbb{N} \to \mathbb{N}$ is a one-one function such that $f(g(x)) = f(x) \Rightarrow g(x) = x$, which is not the case. If $f : \mathbb{N} \to \mathbb{N}$ is an onto function such that $f(g(x)) = f(x)$, one possibility is $f(x) = \begin{cases} n & x = 3n + 1 \\ n & x = 3n + 2 \\ n & x = 3n + 3 \end{cases}$ where $n \in \mathbb{N}_0$. Here $f(x)$ is onto, also $f(g(x)) = f(x) \forall x \in \mathbb{N}$.

Question 14

Maths · Integrals · Single correct

Let $f : [0, \infty) \to [0, \infty)$ be defined as $$f(x) = \int_0^x [y] \, dy$$ where $[x]$ is the greatest integer less than or equal to $x$. Which of the following is true?

  1. $f$ is continuous at every point in $[0, \infty)$ and differentiable except at the integer points.
  2. $f$ is both continuous and differentiable except at the integer points in $[0, \infty)$.
  3. $f$ is continuous everywhere except at the integer points in $[0, \infty)$.
  4. $f$ is differentiable at every point in $[0, \infty)$. Official

Answer: (a)

Solution

Given $f : [0, \infty) \to [0, \infty)$, $f(x) = \int_0^x [y] \, dy$. Let $x = n + f$, $f \in (0, 1)$. So $f(x) = 0 + 1 + 2 + \ldots + (n - 1) + \int_n^{n+f} n \, dy$. $$f(x) = \frac{n(n-1)}{2} + n f$$ $$= \frac{[x]([x]-1)}{2} + [x]\{x\}$$ Note $\lim_{x \to a^+} f(x) = \frac{n(n-1)}{2}$, $\lim_{x \to \infty} f(x) = \frac{(n-1)(n-2)}{2} + (n-1)$ $$= \frac{n(n-1)}{2}$$ $$f(x) = \frac{n(n-1)}{2} (n \in \mathbb{N}_0)$$ So $f(x)$ is continuous for all $x \geq 0$ and differentiable except at integer points.

Question 15

Maths · Determinants · Single correct

The values of $a$ and $b$, for which the system of equations $2x + 3y + 6z = 8$ $x + 2y + az = 5$ $3x + 5y + 9z = b$ has no solution, are:

  1. $a = 3, b \neq 13$
  2. $a \neq 3, b \neq 13$
  3. $a \neq 3, b = 3$
  4. $a = 3, b = 13$

Answer: (a)

Solution

Given $$D = \begin{vmatrix} 2 & 3 & 6 \\ 1 & 2 & a \\ 3 & 5 & 9 \end{vmatrix} = 3 - a$$ $$D = \begin{vmatrix} 2 & 3 & 8 \\ 1 & 2 & 5 \\ 3 & 5 & b \end{vmatrix} = b - 13$$ If $a = 3$, $b \neq 13$, no solution.

Question 16

Maths · Probability · Single correct

Let 9 distinct balls be distributed among 4 boxes, $B_1$, $B_2$, $B_3$ and $B_4$. If the probability than $B_3$ contains exactly 3 balls is $k \left( \frac{3}{4} \right)^9$ then $k$ lies in the set:

  1. $\{ x \in \mathbb{R} : |x - 3| < 1 \}$
  2. $\{ x \in \mathbb{R} : |x - 2| \leq 1 \}$
  3. $\{ x \in \mathbb{R} : |x - 1| < 1 \}$
  4. $\{ x \in \mathbb{R} : |x - 5| \leq 1 \}$

Answer: (a)

Solution

required probability = $\frac{{^9C_3 \cdots 6}}{{4^9}}$ = $\frac{{^9C_3}}{{27}}$ $\cdot$ $\left$($\frac{3}{4}$$\right$)^9 = $\frac{28}{9}$ $\cdot$ $\left$($\frac{3}{4}$$\right$)^9 $\Rightarrow$ k = $\frac{28}{9}$ Which satisfies |x - 3| < 1

Question 17

Maths · Conic Sections · Single correct

Let a parabola $P$ be such that its vertex and focus lie on the positive $x$-axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from $O(0, 0)$ to the parabola $P$ which meet $P$ at $S$ and $R$, then the area (in sq. units) of $\triangle SOR$ is equal to

  1. $16\sqrt{2}$
  2. 16
  3. 32
  4. $8\sqrt{2}$

Answer: (b)

Solution

Clearly RS is latus-rectum. Therefore, $VF = 2 = a$. Thus, $RS = 4a = 8$. Now $OF = 2a = 4$. Therefore, the area of triangle ORS is 16.

Question 18

Maths · Applications of Derivatives · Single correct

The number of real roots of the equation $e^{6x} - e^{4x} - 2e^{3x} - 12e^{2x} + e^x + 1 = 0$ is:

  1. 2
  2. 4
  3. 6
  4. 1

Answer: (a)

Solution

Given the equation: $$e^{6x} - e^{4x} - 2e^{3x} - 12e^{2x} + e^x + 1 = 0$$ This implies: $$\left(e^{3x} - 1\right)^2 - e^x \left(e^{3x} - 1\right) = 12e^{2x}$$ Simplifying further: $$\left(e^{3x} - 1\right)^2 \left(e^x - e^{-x} - e^{-2x}\right) = 12$$

Question 19

Maths · Conic Sections · Single correct

Let an ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a^2 > b^2$, passes through $\left( \frac{\sqrt{3}}{2}, 1 \right)$ and has eccentricity $\frac{1}{\sqrt{3}}$. If a circle, centered at focus $F(\alpha, 0), \alpha > 0$, of $E$ and radius $\frac{2}{\sqrt{3}}$, intersects $E$ at two points $P$ and $Q$, then $PQ^2$ is equal to:

  1. $\frac{8}{3}$
  2. $\frac{4}{3}$
  3. $\frac{16}{3}$
  4. 3

Answer: (c)

Solution

Given $\($ $\frac{3}{2a^2}$ + $\frac{1}{b^2}$ = 1 $\)$ and $\($ 1 - $\frac{b^2}{a^2}$ = $\frac{1}{3}$ $\)$. $\($ $\Rightarrow$ a^2 = 3b^2 = 3 $\)$ $\($ $\Rightarrow$ $\frac{x^2}{3}$ + $\frac{y^2}{2}$ = 1 $\ldots$ (i) $\)$ Its focus is $\($(1, 0)$\)$. Now, the equation of the circle is $\($ (x - 1)^2 + y^2 = $\frac{4}{3}$ $\ldots$ (ii) $\)$ Solving (i) and (ii) we get $\($ y = $\pm$ $\frac{2}{\sqrt{3}}$, x = 1 $\)$ $\($ $\Rightarrow$ PQ^2 = $\left$( $\frac{4}{\sqrt{3}}$ $\right$)^2 = $\frac{16}{3}$ $\)$

Question 20

Maths · Three Dimensional Geometry · Single correct

Let the foot of perpendicular from a point P(1, 2, -1) to the straight line L : $\frac{x}{1} = \frac{y}{0} = \frac{z}{-1}$ be N. Let a line be drawn from P parallel to the plane $x + y + 2z = 0$ which meets L at point Q. If $\alpha$ is the acute angle between the lines PN and PQ, then $\cos \alpha$ is equal to

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

Answer: (c)

Solution

Given $\overrightarrow{\mathrm{PN}} \cdot (\hat{i} - \hat{k}) = 0$. This implies $\mathrm{N}(1, 0, -1)$. Now, $\overrightarrow{\mathrm{PQ}} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 0$. This implies $\mu = -1$. Therefore, $\mathrm{Q}(-1, 0, 1)$. $\overrightarrow{\mathrm{PN}} = 2\hat{j}$ and $\overrightarrow{\mathrm{PQ}} = 2\hat{i} + 2\hat{j} - 2\hat{k}$. Thus, $\cos \alpha = \frac{1}{\sqrt{3}}$.

Question 21

Maths · Differential Equations · Numerical

Let $y = y(x)$ be solution of the following differential equation $$e^y \frac{dy}{dx} - 2e^y \sin x + \sin x \cos^2 x = 0, \ y\left(\frac{\pi}{2}\right) = 0$$ If $y(0) = \log_c \left(\alpha + \beta e^{-2}\right)$, then $4(\alpha + \beta)$ is equal to

Answer: 4

Solution

Let $e^y = t$. $$\frac{dt}{dx} - (2 \sin x) t = - \sin x \cos^2 x$$ I.F. $= e^{2 wax}$ $$t \cdot e^{2 \cos x} = \int e^{2 \cos x} \cdot \left( - \sin x \cos^2 x \right) \, dx$$ At $x = \frac{\pi}{2}$, $y = 0 \Rightarrow C = \frac{3}{4}$. $$e^y = \frac{1}{2} \cos^2 x - \frac{1}{2} \cos x + \frac{1}{4} + \frac{3}{4} e^{-2 \omega x}$$ $$y = \log \left[ \frac{\cos^2 x}{2} - \frac{\cos x}{2} + \frac{1}{4} + \frac{3}{4} e^{-2 \cos x} \right]$$ Put $x = 0$. $$y = \log \left[ \frac{1}{4} + \frac{3}{4} e^{-2} \right] \Rightarrow \alpha = \frac{1}{4}, \beta = \frac{3}{4}$$

Question 22

Maths · Sequences and Series · Numerical

If the value of $$\left(1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \ldots upto \infty \right) \log_{0.25} \left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots upto \infty \right)$$ is $l$, then $l^2$ is equal to

Answer: 3

Solution

Given $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \cdots$ Then, $\frac{S}{3} = \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \cdots$ Subtracting, $S - \frac{S}{3} = 1 + \frac{1}{3} + \frac{4}{3^2} + \frac{4}{3^3} + \cdots$ $\frac{2S}{3} = \frac{4}{3} + \frac{4}{3^2} + \frac{4}{3^3} + \cdots$ Solving, $S = \frac{3}{2}\left(\frac{\frac{4}{3}}{1-\frac{1}{3}}\right) = 3$ Now, $\ell = 3^{\log_{9}\left(10+1\right)\log_{11}\left(\frac{1}{2}\right)}$ $\ell = 3^{1/2} = \sqrt{3}$ Therefore, $\ell^2 = 3$

Question 23

Maths · Statistics · Numerical

Consider the following frequency distribution \begin{tabular}{|l|l|l|l|l|l|} \hline class : & 10 - 20 & 20 - 30 & 30 - 40 & 40 - 50 & 50 - 60 \\ \hline Frequency : & $\alpha$ & 110 & 54 & 30 & $\beta$ \\ \hline \end{tabular} If the sum of all frequencies is 584 and median is 45, then $|\alpha - \beta|$ is equal to

Answer: 164

Solution

Sum of frequencies = 584 $\($ $\Rightarrow$ $\alpha$ + $\beta$ = 390 $\)$ Now, Median is at $\($ $\frac{584}{2}$ = 292^{th} $\)$ Median = 45 (lies in class 40 – 50) $\($ $\Rightarrow$ $\alpha$ + 110 + 54 + 15 = 292 $\)$ $\($ $\Rightarrow$ $\alpha$ = 113, $\beta$ = 277 $\)$ $\($ $\Rightarrow$ |$\alpha$ - $\beta$| = 164 $\)$

Question 24

Maths · Vector Algebra · Fill in the blank

Let $\vec{p}=2\hat{i}+3\hat{j}+\hat{k}$ and $\vec{q}=\hat{i}+2\hat{j}+\hat{k}$ be two vectors. If a vector $\vec{r}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ is perpendicular to each of the vectors $(\vec{p}+\vec{q})$ and $(\vec{p}-\vec{q})$, and $|\vec{r}|=\sqrt{3}$, then $|\alpha|+|\beta|+|\gamma|$ is equal to ________.

Answer: 3

Solution

Given $\vec{p} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{q} = \hat{i} + 2\hat{j} + \hat{k}$. Now $\left(\vec{p} + \vec{q}\right) \times \left(\vec{p} - \vec{q}\right)$ is given by the determinant: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ 1 & 1 & 0 \end{vmatrix}$$ which simplifies to $-2\hat{i} - 2\hat{j} - 2\hat{k}$. Therefore, $$\vec{r} = \pm \sqrt{3} \frac{\left(\vec{p} + \vec{q}\right) \times \left(\vec{p} - \vec{q}\right)}{\left|\left(\vec{p} + \vec{q}\right) \times \left(\vec{p} - \vec{q}\right)\right|} = \pm \frac{\sqrt{3}(-2\hat{i} - 2\hat{j} - 2\hat{k})}{\sqrt{2^2 + 2^2 + 2^2}}$$ which results in $$\vec{r} = \pm (-\hat{i} - \hat{j} - \hat{k}).$$ According to the question, $\vec{r} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$. So $|\alpha| = 1$, $|\beta| = 1$, $|\gamma| = 1$. Therefore, $|\alpha| + |\beta| + |\gamma| = 3$.

Question 25

Maths · Binomial Theorem · Numerical

The ratio of the coefficient of the middle term in the expansion of $(1 + x)^{20}$ and the sum of the coefficients of two middle terms in expansion of $(1 - x)^{19}$ is

Answer: 1

Solution

Coefficient of middle term in $(1 + x)^{20} = \binom{20}{10}$. Sum of coefficients of two middle terms in $(1 + x)^{19} = \binom{19}{9} + \binom{19}{10}$. So required ratio $$= \frac{\binom{20}{10}}{\binom{9}{9} + \binom{19}{10}} = \frac{\binom{2}{10}}{\binom{2}{10}} = 1.$$

Question 26

Maths · Determinants · Numerical

Let $M = \left\{ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in \{ \pm 3, \pm 2, \pm 1, 0 \} \right\}$ Define $f : M \rightarrow \mathbb{Z}$, as $f(A) = \det(A)$, for all $A \in M$ where $\mathbb{Z}$ is set of all integers. Then the number of $A \in M$ such that $f(A) = 15$ is equal to

Answer: 16

Solution

Given $|A| = ad - bc = 15$ where $a, b, c, d \in \{ \pm 3, \pm 2, \pm 1, 0 \}$. Case I: $ad = 9$ and $bc = -6$. For $ad$, possible pairs are $(3, 3), (-3, -3)$. For $bc$, possible pairs are $(3, -2), (-3, 2), (-2, 3), (2, -3)$. So total matrix $= 2 \times 4 = 8$. Case II: $ad = 6$ and $bc = -9$. Similarly, total matrix $= 2 \times 4 = 8$. Therefore, total such matrices are $= 16$.

Question 27

Maths · Permutations and Combinations · Numerical

There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100k, then k is equal to

Answer: 238

Solution

Class $10^{th} \ 11^{th} \ 12^{th}$ Total student $5 \ 6 \ 8 \ 2 \ 3 \ 5 \Rightarrow {^5C_2} \times {^6C_3} \times {^8C_5}$ Number of selection $2 \ 2 \ 6 \Rightarrow {^5C_2} \times {^6C_2} \times {^3C_6}$ $3 \ 2 \ 5 \Rightarrow {^5C_3} \times {^6C_2} \times {^8C_5}$ $\Rightarrow$ Total number of ways $= 23800$ According to question $100 \ K = 23800$ $\Rightarrow K = 238$

Question 28

Maths · Complex Numbers and Quadratic Equations · Numerical

If $\alpha, \beta$ are roots of the equation $x^2 - 5(\sqrt{2})x + 10 = 0$, $\alpha > \beta$ and $P_n = \alpha^n - \beta^n$ for each positive integer $n$, then the value of $$\left( \frac{P_{17}P_{20} + 5\sqrt{2}P_{17}P_{19}}{P_{18}P_{19} + 5\sqrt{2}P_{18}^2} \right)$$ is equal to

Answer: 1

Solution

Given $x^2 + 5\sqrt{2}x + 10 = 0$ and $x_p = \alpha^n - \beta^n$ (Given). Now, $$\frac{P_{17}P_{20} + 5\sqrt{2}P_{17}P_{19}}{P_{18}P_{19} + 5\sqrt{2}P_{18}^2} = \frac{P_{17}\left(P_{20} + 5\sqrt{2}P_{19}\right)}{P_{18}\left(P_{19} + 5\sqrt{2}P_{18}\right)}$$ $$\frac{P_{17}\left(\alpha^{20} - \beta^{20} + 5\sqrt{2}(\alpha^{19} - \beta^{19})\right)}{P_{18}\left(\alpha^{19} - \beta^{19} + 5\sqrt{2}(\alpha^{18} - \beta^{18})\right)}$$ $$\frac{P_{17}\left(\alpha^{19}(\alpha + 5\sqrt{2}) - \beta^{19}(\beta + 5\sqrt{2})\right)}{P_{18}\left(\alpha^{18}(\alpha + 5\sqrt{2}) - \beta^{18}(\beta + 5\sqrt{2})\right)}$$ Since $\alpha + 5\sqrt{2} = -10/\alpha \ldots (1)$ and $\beta + 5\sqrt{2} = -10/\beta$. Now put these values in the above expression: $$= -\frac{10P_{17}P_{18}}{-10P_{18}P_{17}} = 1$$

Question 29

Maths · Binomial Theorem · Numerical

The term independent of $x$ in the expansion of $$\left( \frac{x+1}{x^{2/3} - x^{3} + 1} - \frac{x-1}{x-x^{1/2}} \right)^{10}$$, where $x \neq 0, 1$ is equal to

Answer: 210

Solution

$\left(\left(x^{1/3}+1\right)-\frac{x^{1/2}+1}{x^{1/2}}\right)^{10}$ $=\left(x^{1/3}-\frac{1}{x^{1/2}}\right)^{10}$ Now General Term $T_{r+1}={}^{10}C_{r}\left(x^{1/3}\right)^{10-r} \left(-\frac{1}{x^{1/2}}\right)^{r}$ For independent term $\frac{10-r}{3}-\frac{r}{2}=0$ $\Rightarrow r=4$ $\Rightarrow T_{5}={}^{10}C_{4}=210$

Question 30

Maths · Matrices · Numerical

Let $$ S = \left\{ n \in \mathbb{N} \left( \begin{array}{cc} 0 & i \\ 1 & 0 \end{array} \right)^n \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \forall a, b, c, d \in \mathbb{R} \right\} $$ where $i = \sqrt{-1}$. Then the number of 2-digit numbers in the set $S$ is

Answer: 11

Solution

Let $X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $A = \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix}$. Therefore, $AX = IX$. This implies $A = I$. Thus, $\left( \begin{pmatrix} 0 & i \\ 1 & 0 \end{pmatrix} \right)^n = I$. Therefore, $A^8 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Hence, $n$ is a multiple of 8. So the number of 2 digit numbers in the set $S = 11(16, 24, 32, \ldots, 96)$.

Physics

Question 31

Physics · Kinetic Theory · Single correct

For a gas $C_P - C_V = R$ in a state P and $C_P - C_V = 1.10R$ in a state Q, $T_P$ and $T_Q$ are the temperatures in two different states P and Q respectively. Then

  1. $T_P = T_Q$
  2. $T_P < T_Q$
  3. $T_P = 0.9 \, T_Q$
  4. $T_P > T_Q$

Answer: (d)

Solution

Given $C_P - C_V = R$ for an ideal gas and the gas behaves as an ideal gas at high temperature, so $T_P > T_Q$.

Question 32

Physics · System of Particles and Rotational Motion · Single correct

Given below are two statements: one is labelled as Assertion $\textbf{A}$ and the other is labelled as Reason $\textbf{R}$. Assertion $\textbf{A}$ : Moment of inertia of a circular disc of mass 'M' and radius 'R' about X, Y axes (passing through its plane) and Z-axis which is perpendicular to its plane were found to be $I_x$, $I_y$ and $I_z$ respectively. The respective radii of gyration about all the three axes will be the same. Reason $\textbf{R}$ : A rigid body making rotational motion has fixed mass and shape. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both A and R are correct but R is NOT the correct explanation of A.
  2. A is not correct but R is correct.
  3. A is correct but R is not correct.
  4. Both A and R are correct and R is the correct explanation of A

Answer: (b)

Solution

Given $I_z = I_x + I_y$ (using perpendicular axis theorem) and $I = mK^2$ ($K$: radius of gyration) so $mK_z^2 = mK_x^2 + mK_y^2$. Therefore, $K_z^2 = K_x^2 + K_y^2$. So radius of gyration about axes $x$, $y$ and $z$ won't be the same. Hence assertion A is not correct. Reason R is a correct statement (property of a rigid body).

Question 33

Physics · Dual Nature of Radiation and Matter · Single correct

What should be the order of arrangement of de-Broglie wavelength of electron ($\lambda_e$), an $\alpha$-particle ($\lambda_\alpha$) and proton ($\lambda_p$) given that all have the same kinetic energy?

  1. $\lambda_e = \lambda_p = \lambda_\alpha$
  2. $\lambda_e < \lambda_p < \lambda_\alpha$
  3. $\lambda_e > \lambda_p > \lambda_\alpha$
  4. $\lambda_e = \lambda_p > \lambda_\alpha$

Answer: (c)

Solution

Given $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \propto \frac{1}{\sqrt{m}}$. $m_\alpha > m_p > m_e$ so $\lambda_e > \lambda_p > \lambda_\alpha$

Question 34

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

Identify the logic operation carried out.

  1. OR
  2. AND
  3. NOR
  4. NAND

Answer: (b)

Solution

The circuit uses two NOT gates and an OR gate. The expression for the output is $\overline{\overline{A} + \overline{B}}$. By De Morgan's laws, this simplifies to $A \cdot B$. So, the circuit functions as an AND gate.

Question 35

Physics · Dual Nature of Radiation and Matter · Single correct

A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be :

  1. 1 : 3
  2. 3 : 1
  3. 1 : $\sqrt{3}$
  4. 1 : 1

Answer: (d)

Solution

Given $$\lambda = \frac{h}{p}$$ both the particles will move with momentum same in magnitude and opposite in direction. So De-Broglie wavelength of both will be same i.e. ratio 1 : 1

Question 36

Physics · Nuclei · Single correct

Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as: (where $\lambda$ is the decay constant)

  1. $\frac{1}{2} \frac{\ln 2}{\lambda}$
  2. $\frac{\ln 2}{\lambda}$
  3. $\frac{2 \ln 2}{\lambda}$
  4. $\frac{\ln d}{2}$

Answer: (d)

Solution

Given $$\frac{3N_0}{4} = N_0 e^{-\lambda t_1}$$ $$\frac{N_0}{2} = N_0 e^{-\lambda t_2}$$ $$\ln(3/4) = -\lambda t_1$$ $$\ln(1/2) = -\lambda t_2 \cdots (i)$$ $$\ln(3/4) - \ln(1/2) = \lambda (t_2 - t_1)$$ $$\Delta t = \frac{\ln(3/2)}{\lambda}$$

Question 37

Physics · Mathematics in Physics · Single correct

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

  1. $(a) \to (iv), (b) \to (i), (c) \to (iii), (d) \to (ii)$
  2. $(a) \to (iv), (b) \to (iii), (c) \to (i), (d) \to (ii)$
  3. $(a) \to (iii), (b) \to (ii), (c) \to (iv), (d) \to (i)$
  4. $(a) \to (i), (b) \to (iv), (c) \to (ii), (d) \to (iii)$

Answer: (b)

Solution

Option (iv) is correct. (a) $\vec{C} = \vec{A} + \vec{B}$ (b) $\vec{A} = \vec{B} + \vec{C} = \vec{C} + \vec{B}$ (c) $\vec{B} = \vec{A} + \vec{C}$ (d) $\vec{A} + \vec{B} + \vec{C} = 0$

Question 38

Physics · Electrostatic Potential and Capacitance · Single correct

A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as: $$\varepsilon(x) = \varepsilon_0 + kx, for \left(0 < x \leq \frac{d}{2}\right)$$ $$\varepsilon(x) = \varepsilon_0 + k(d-x), for \left(\frac{d}{2} \leq x \leq d\right)$$

  1. $\left( \varepsilon_0 + \frac{kd}{2} \right)^{2/kA}$
  2. $\frac{kA}{2 \ln \left( \frac{2\varepsilon_0 + kd}{2\varepsilon_0} \right)}$
  3. 0
  4. $\frac{kA}{2} \ln \left( \frac{2\varepsilon_0}{2\varepsilon_0 - kd} \right)$

Answer: (b)

Solution

Taking an element of width $dx$ at a distance $x (x < d/2)$ from left plate $$dc = \frac{(\varepsilon_0 + kx)A}{dx}$$ Capacitance of half of the capacitor $$\frac{1}{C} = \int_0^{\Delta/2} \frac{1}{dc} = \frac{1}{A} \int_0^{dn} \frac{dx}{\varepsilon_0 + kx}$$ $$\frac{1}{C} = \frac{1}{kA} \ln \left( \frac{\varepsilon_0 + kd/2}{\varepsilon_0} \right)$$ Capacitance of second half will be same $$C_{eq} = \frac{C}{2} = \frac{kA}{2 \ln \left( \frac{2\varepsilon_0 + kd}{2\varepsilon_0} \right)}$$

Question 39

Physics · Thermodynamics · Single correct

A monoatomic ideal gas, initially at temperature $T_1$ is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature $T_2$ by releasing the piston suddenly. If $l_1$ and $l_2$ are the lengths of the gas column, before and after the expansion respectively, then the value of $\frac{T_1}{T_2}$ will be

  1. $\left( \frac{l_1}{l_2} \right)^{\frac{2}{3}}$
  2. $\left( \frac{l_2}{l_1} \right)^{\frac{2}{3}}$
  3. $\frac{l_2}{l_1}$
  4. $\frac{l_1}{l_2}$

Answer: (b)

Solution

Question 40

Physics · Ray Optics and Optical Instruments · Single correct

A ray of laser of a wavelength 630 $\mathrm{nm}$ is incident at an angle of 30^$\circ$ at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is 2.42 and that of air is 1. Choose the correct option.

  1. angle of refraction is 24.41^$\circ$
  2. angle of refraction is 30^$\circ$
  3. refraction is not possible
  4. angle of refraction is 53.4^$\circ$

Answer: (c)

Solution

Given $\sin \theta_C = \frac{1}{\mu} = \frac{1}{2\mu_2} \sin \theta_C$. $\theta > \theta_c$. Total internal reflection will happen.

Question 41

Physics · Mechanical Properties of Solids · Single correct

Two wires of same length and radius are joined end to end and loaded. The Young's moduli of the materials of the two wires are $Y_1$ and $Y_2$. The combination behaves as a single wire then its Young's modulus is :

  1. $Y = \frac{2Y_1 Y_2}{3(Y_1 + Y_2)}$
  2. $Y = \frac{2Y_1 Y_2}{Y_1 + Y_2}$
  3. $Y = \frac{Y_1 Y_2}{2(Y_1 + Y_2)}$
  4. $Y = \frac{Y_1 Y_2}{Y_1 + Y_2}$

Answer: (b)

Solution

In series combination $\Delta l = \ell_1 + \ell_2$ $$Y = \frac{F/A}{\Delta \ell / \ell} \implies \Delta \ell = \frac{F \ell}{AY}$$ $$\implies \Delta \ell \propto \frac{\ell}{Y}$$ Equivalent length of rod after joining is $2\ell$. As lengths are same and force is also same in series $$\Delta \ell = \Delta \ell_1 + \Delta \ell_2$$ $$\frac{\ell_{eq}}{Y_{eq}} = \frac{\ell}{Y_1} + \frac{\ell}{Y_2} \implies \frac{2\ell}{Y} = \frac{\ell}{Y_1} + \frac{\ell}{Y_2}$$ $$Y = \frac{2Y_1Y_2}{Y_1 + Y_2}$$

Question 42

Physics · Nuclei · Single correct

The half-life of $^{198}\mathrm{Au}$ is 3 days. If atomic weight of $^{198}\mathrm{Au}$ is 198 g/mol then the activity of 2mg of $^{198}\mathrm{Au}$ is [in disintegration/second] :

  1. $2.67 \times 10^{12}$
  2. $6.06 \times 10^{18}$
  3. $32.36 \times 10^{12}$
  4. $16.18 \times 10^{12}$

Answer: (d)

Solution

Given $A = \lambda N$. $$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{3 \times 24 \times 60 \times 60} \, \mathrm{sec^{-1}} = 2.67 \times 10^{-6} \, \mathrm{sec^{-1}}$$ $N =$ Number of atoms in $2 \, \mathrm{mg}$ Au $$= \frac{2 \times 10^{-3}}{198} \times 6 \times 10^{23} = 6.06 \times 10^{15}$$ $$A = \lambda N = 1.618 \times 10^{13} = 16.18 \times 10^{12} \, \mathrm{dps}$$

Question 43

Physics · Laws of Motion · Single correct

Two billiard balls of equal mass 30 g strike a rigid wall with same speed of 108 $\mathrm{kmph}$ (as shown) but at different angles. If the balls get reflected with the same speed then the ratio of the magnitude of impulses imparted to ball 'a' and ball 'b' by the wall along 'X' direction is :

  1. 1 : 1
  2. $\sqrt{2}$ : 1
  3. 2 : 1
  4. 1 : $\sqrt{2}$

Answer: (b)

Solution

Impulse equals change in momentum. Ball (a) $|\Delta \vec{p}| = 2mu = J_1$ Ball (b) $|\Delta \vec{p}| = 2mu \cos 45^\circ = J_2$ $$\frac{J_1}{J_2} = \frac{1}{\cos 45^\circ} = \sqrt{2}$$

Question 44

Physics · Wave Optics · Single correct

In the Young's double slit experiment, the distance between the slits varies in time as $d(t) = d_0 + a_0 \sin \omega t$ ; where $d_0$, $\omega$ and $a_0$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

  1. $\frac{2 \lambda D (d_0)}{(d_0^2 - a_0^2)}$
  2. $\frac{2 \lambda D a_0}{(d_0^2 - a_0^2)}$
  3. $\frac{\lambda D}{d_0^2} a_0$
  4. $\frac{\lambda D}{d_0 + a_0}$

Answer: (b)

Solution

Fringe Width, $\beta = \frac{\lambda D}{d}$ $\beta_{aax} \Rightarrow d_{min}$ and $\beta_{min} \Rightarrow d_{max}$ $d = d_0 + a_0 \sin \omega t$ $d_{max} = d_0 + a_0$ and $d_{min} = d_0 - a_0$ $\beta_{min} = \frac{\lambda D}{d_0 + a_0}$ and $\therefore \beta_{max} = \frac{\lambda D}{d_0 - a_0}$

Question 45

Physics · Thermal Properties of Matter · Single correct

Two different metal bodies A and B of equal mass are heated at a uniform rate under similar conditions. The variation of temperature of the bodies is graphically represented as shown in the figure. The ratio of specific heat capacities is :

  1. $\frac{8}{3}$
  2. $\frac{3}{8}$
  3. $\frac{3}{4}$
  4. $\frac{4}{3}$

Answer: (b)

Solution

The rate of heat transfer for A is equal to the rate of heat transfer for B: $$\left( \frac{\Delta Q}{\Delta t} \right)_A = \left( \frac{\Delta Q}{\Delta t} \right)_B$$ The mass and specific heat capacity relation for A and B is given by: $$\mathrm{mS}_A \left( \frac{\Delta T}{\Delta t} \right)_A = \mathrm{mS}_B \left( \frac{\Delta T}{\Delta t} \right)_B$$

Question 46

Physics · Electromagnetic Waves · Single correct

A linearly polarized electromagnetic wave in vacuum is $$\mathbf{E} = 3.1 \cos \left[(1.8)z - \left(5.4 \times 10^6\right)t \right] \hat{i} \mathrm{N/C}$$ is incident normally on a perfectly reflecting wall at $z = a$. Choose the correct option

  1. The wavelength is 5.4 m
  2. The frequency of electromagnetic wave is $54 \times 10^4$ Hz
  3. The transmitted wave will be $$3.1 \cos \left[(1.8)z - \left(5.4 \times 10^6\right)t \right] \hat{i} \mathrm{N/C}$$
  4. The reflected wave will be $1 \cos \left[(1.8)z + \left(5.4 \times 10^6\right)t \right] \hat{i} \mathrm{N/C}$

Answer: (d)

Solution

Reflected wave will have direction opposite to incident wave.

Question 47

Physics · Current Electricity · Single correct

In the given figure, there is a circuit of potentiometer of length AB = 10 $\mathrm{m}$. The resistance per unit length is 0.1 $\Omega$ per cm. Across AB, battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is :

  1. 5 $\mathrm{V}$
  2. 2.25 $\mathrm{V}$
  3. 6 $\mathrm{V}$
  4. 2.75 $\mathrm{V}$

Answer: (a)

Solution

Max, voltage that can be measured by this potentiometer will be equal to potential drop across AB. $$R_{AB} = 10 \times 0.1 \times 100 = 100 ohm.$$ $$V_{AB} = \frac{6}{20+100} \times 100 = 6 \times \frac{100}{120} = 5 \, V$$

Question 48

Physics · Communication Systems · Single correct

In amplitude modulation, the message signal $V_m(t) = 10 \sin(2\pi \times 10^5 t)$ volts and Carrier signal $V_c(t) = 20 \sin(2\pi \times 10^7 t)$ volts The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is $\alpha \mathrm{kHz}$. The value of $\alpha$ is :

  1. 200$\mathrm{kHz}$
  2. 50$\mathrm{kHz}$
  3. 100$\mathrm{kHz}$
  4. 0

Answer: (a)

Solution

Bandwidth $= 2 \times f_m$ $$= 2 \times 10^5 \, \mathrm{Hz} = 200 \, \mathrm{kHz}$$

Question 49

Physics · Motion in a Straight Line · Single correct

Water droplets are falling from an open tap at a constant rate. The spacing between a droplet observed at the 4$^{\text{th}}$ second after its fall and the next droplet is $34.3\,\mathrm{m}$. At what rate are the droplets falling from the tap? (Take $g = 9.8\,\mathrm{m\,s^{-2}}$)

  1. 3 drops /2 seconds
  2. 2 drops / second
  3. 1 drop / second
  4. 1 drop /7 seconds

Answer: (c)

Solution

In 4 sec, 1st drop will travel $\Rightarrow \frac{1}{2} \times (9.8) \times (4)^2 = 78.4 \, \mathrm{m}$. Therefore, 2nd drop would have travelled $\Rightarrow 78.4 - 34.3 = 44.1 \, \mathrm{m}$. Time for 2nd drop $\frac{1}{2} (9.8) t^2 = 44.1$. $t = 3 \, \mathrm{sec}$. Each drop has a time gap of 1 sec. Therefore, 1 drop per sec.

Question 50

Physics · Gravitation · Single correct

The minimum and maximum distances of a planet revolving around the Sun are $x_1$ and $x_2$. If the minimum speed of the planet on its trajectory is $v_0$ then its maximum speed will be:

  1. $\frac{v_0 x_1^2}{x_2^2}$
  2. $\frac{v_0 x_2^2}{x_1^2}$
  3. $\frac{v_0 x_1}{x_2}$
  4. $\frac{v_0 x_2}{x_1}$

Answer: (d)

Solution

Angular momentum conservation equation $\mathbf{v}_0 x_2 = \mathbf{v}_1 x_1$ $$\mathbf{v}_1 = \frac{\mathbf{v}_0 x_2}{x_1}$$

Question 51

Physics · System of Particles and Rotational Motion · Numerical

A body of mass 2 kg moving with a speed of 4 m/s. makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial speed. The speed of the two body centre of mass is $\frac{x}{10} \, \mathrm{m/s}$. Then the value of x is

Answer: 25

Solution

Given $p_i = p_f$. $$2 \times 4 = 2 \times 1 + m_2 \times v_2$$ $m_2 V_2 = 6$ By coefficient of restitution $1 = \frac{v_2 - 1}{4} \implies v_2 = 5 \, \mathrm{m/s}$ By (i) $m_2 \times 5 = 6$ $m_2 = 1.2 \, \mathrm{kg}$ $v_{\mathrm{em}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$ $$v_{\mathrm{cm}} = \frac{2 \times 1 + 1.2 \times 5}{2 + 1.2} = \frac{8}{3.2} = \frac{25}{10}$$ $x = 25$

Question 52

Physics · Experimental Physics · Fill in the blank

Student A and Student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is Given pitch = 0.1 cm.

Answer: 13

Solution

For (A) Reading = MSR + CSR + Error $$0.322 = 0.300 + CSR + 5 \times LC$$ $$0.322 = 0.300 + CSR + 0.005$$ $$CSR = 0.017$$ For B Reading = MSR + CSR + Error $$0.322 = 0.200 + CSR + 0.092$$ $$CSR = 0.030$$ Difference = $$0.030 - 0.017 = 0.013 cm$$ Division on circular scale = $$\frac{0.013}{0.001} = 13$$

Question 53

Physics · Alternating Current · Numerical

An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 kΩ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 μs is $\frac{x}{100}$ mA. Then x is equal to (Take $e^{-1} = 0.37$)

Answer: 74

Solution

$I_{\max}=\dfrac{V}{R} =\dfrac{20\,V}{10\,K\Omega} =2\,mA$ For LR-decay circuit $I=I_{\max}e^{-Rr h}$ $\frac{-10\times10^{3}\times(x)|0^{+}}{4}$ $I=2\,mA\,e$ $I=2\,mA\,e^{-1}$ $\therefore\ I=2\times0.37\,mA$ $I=\frac{74}{100}\,\mathrm{mA}$ $x=74$

Question 54

Physics · Electromagnetic Induction · Numerical

A circular conducting coil of radius 1 m is being heated by the change of magnetic field $\vec{B}$ passing perpendicular to the plane in which the coil is laid. The resistance of the coil is $2 \, \mu \Omega$. The magnetic field is slowly switched off such that its magnitude changes in time as $B = \frac{4}{\pi} \times 10^{-3} \, \mathrm{T} \left( 1 - \frac{t}{100} \right)$ The energy dissipated by the coil before the magnetic field is switched off completely is $E = mJ$.

Answer: 80

Solution

Given $\phi = \vec{B} \vec{S}$. $$\phi = \frac{4}{\pi} \times 10^{-3} \left(1 - \frac{t}{100}\right) \cdot \pi R^2$$ $$\phi = 4 \times 10^{-3} \times (1)^2 \left(1 - \frac{t}{100}\right)$$ The electromotive force $\varepsilon$ is given by $$\varepsilon = -\frac{d\phi}{dt}$$ $$\varepsilon = -\frac{d}{dt} \left(4 \times 10^{-3} \left(1 - \frac{t}{100}\right)\right)$$ $$\varepsilon = 4 \times 10^{-3} \left(\frac{1}{100}\right) = 4 \times 10^{-5} \, \mathrm{V}$$ When $B = 0$, $1 - \frac{t}{100} = 0$ Therefore, $t = 100 \, \mathrm{sec}$ The heat generated is given by $$Heat = \frac{\varepsilon^2}{R} t$$ $$Heat = \frac{(4 \times 10^{-5})^2}{2 \times 10^{-6}} \times 100 \, \mathrm{J}$$ $$Heat = \frac{16 \times 10^{-10} \times 100}{2 \times 10^{-6}} \, \mathrm{J}$$ $$Heat = 0.08 \, \mathrm{J}$$ Therefore, the heat is $80 \, \mathrm{mJ}$.

Question 55

Physics · Oscillations · Numerical

In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be rad/s when $k = 20 \, \mathrm{N/m}$.

Answer: 10

Solution

Given $\omega = \sqrt{\frac{k_{eq}}{\mu}}$. $\mu$ is the reduced mass. Springs are in series connection. $$k_{eq} = \frac{k_1 k_2}{k_1 + k_2}$$ $$k_{eq} = \frac{k \times 4k}{5k} = \frac{4k}{5}$$ $$k_{eq} = \frac{4 \times 20}{5} \, \mathrm{N/m} = 16 \, \mathrm{N/m}$$ $$\mu = \frac{m_1 \, m_2}{m_1 + m_2} = \frac{0.2 \times 0.8}{0.2 + 0.8} = 0.16 \, \mathrm{kg}$$ $$\omega = \sqrt{\frac{16}{0.16}} = \sqrt{100} = 10$$

Question 56

Physics · Magnetism and Matter · Numerical

The value of aluminium susceptibility is $2.2 \times 10^{-5}$. The percentage increase in the magnetic field if space within a current carrying toroid is filled with aluminium is $\frac{x}{10^4}$. Then the value of $x$ is

Answer: 22

Solution

Given $B = \mu \cdot (H + I)$. $B = \mu \cdot H \left(1 + \frac{I}{H}\right)$. $B = B_0(1 + x)$. $B - B_0 = B_0 x$. $$\frac{B - B_0}{B_0} = x$$ $$\frac{B - B_b}{B_0} \times 100 = 100x$$ $$= 2.2 \times 10^{-3} = \frac{22}{10^4}$$

Question 57

Physics · Electric Charges and Fields · Numerical

A particle of mass 1 mg and charge $q$ is lying at the mid-point of two stationary particles kept at a distance '2 m' when each is carrying same charge '$q$'. If the free charged particle is displaced from its equilibrium position through distance '$x$' $(x \ll 1 \, \mathrm{m})$. The particle executes SHM. Its angular frequency of oscillation will be $\times 10^5 \, \mathrm{rad/s}$ if $q^2 = 10 \, \mathrm{C}^2$

Answer: 6

Solution

Net force on free charged particle $F = \frac{kq^2}{(d+x)^2} - \frac{kq^2}{(d-x)^2}$. $$F = -kq^2 \left[ \frac{4 \, dx}{(d^2 - x^2)^2} \right]$$ $$a = -\frac{4kq^2 d}{m} \left( \frac{x}{d^4} \right)$$ $$a = -\left( \frac{4kq^2}{md^3} \right) x$$ So, angular frequency $\omega = \sqrt{\frac{4kq^2}{md^3}}$. $$\omega = \sqrt{\frac{4 \times 9 \times 10^9 \times 10}{1 \times 10^{-6} \times 1^3}}$$ $$\omega = 6 \times 10^8 \, \mathrm{rad/sec}$$

Question 58

Physics · Current Electricity · Numerical

An electric bulb rated as 200 $\mathrm{W}$ at 100 $\mathrm{V}$ is used in a circuit having 200 $\mathrm{V}$ supply. The resistance 'R' that must be put in series with the bulb so that the bulb delivers the same power is $\Omega$.

Answer: 50

Solution

Power $P = \frac{V^2}{R_B}$ $R_B = \frac{V^2}{P} = \frac{100 \times 100}{200}$ $R_B = 50 \Omega$ To produce the same power, the same voltage (i.e. $100 \, \mathrm{V}$) should be across the bulb. Hence, $R = R_B$ $R = 50 \Omega$

Question 59

Physics · Oscillations · Numerical

A pendulum bob has a speed of $3 \, \mathrm{m/s}$ at its lowest position. The pendulum is $50 \, \mathrm{cm}$ long. The speed of bob, when the length makes an angle of $60^\circ$ to the vertical will be $(g = 10 \, \mathrm{m/s^2})$ m/s

Answer: 2

Solution

Applying work energy theorem: $$w_g + w_T = \Delta K$$ $$-mg l (1 - \cos 60^\circ) = \frac{1}{2} mv^2 - \frac{1}{2} mu^2$$ $$v^2 = u^2 - 2gl(1 - \cos 60^\circ)$$ $$v^2 = 9 - 2 \times 10 \times 0.5 \left( \frac{1}{2} \right)$$ $$v^2 = 4$$ $$v = 2 \, \mathrm{m/s}$$

Question 60

Physics · System of Particles and Rotational Motion · Numerical

A particle of mass ' m ' is moving in time ' t ' on a trajectory given by $$\vec{r} = 10 \alpha t^2 \hat{i} + 5 \beta (t - 5) \hat{j}$$ Where $\alpha$ and $\beta$ are dimensional constants. The angular momentum of the particle becomes the same as it was for $t = 0$ at time $t =$ seconds.

Answer: 10

Solution

Given $$\vec{r} = 10\alpha t^2 \hat{i} + 5\beta (t-5)\hat{j}$$ and $$\vec{v} = 20\alpha t \hat{i} + 5\beta \hat{j}.$$ The angular momentum is $$\vec{L} = m(\vec{r}\times\vec{v}).$$ Therefore, $$\vec{L} = m\left[10\alpha t^2 \hat{i} + 5\beta (t-5)\hat{j}\right] \times \left[20\alpha t \hat{i} + 5\beta \hat{j}\right].$$ $$\vec{L} = m\left[50\alpha\beta t^2 \hat{k} - 100\alpha\beta t(t-5)\hat{k}\right].$$ At $$t=0,$$ $$\vec{L}=\vec{0}.$$ For $$50\alpha\beta t^2 - 100\alpha\beta t(t-5)=0,$$ we get $$t-2(t-5)=0.$$ Hence, $$t=10\,\mathrm{s}.$$

Chemistry

Question 61

Chemistry · Polymers · Single correct

is a repeating unit for:

  1. Novolac
  2. Buna-N
  3. Acrilan
  4. Neoprene

Answer: (a)

Solution

Question 62

Chemistry · Co-ordination Compounds · Single correct

Which one of the following species responds to a external magnetic field?

  1. $[\mathrm{Fe(H_2O)_6}]^{3+}$
  2. $[\mathrm{Ni(CN)_4}]^{2-}$
  3. $[\mathrm{Co(CN)_6}]^{3-}$
  4. $\mathrm{Ni(CO)_4}$

Answer: (a)

Solution

1. $[\mathrm{Fe(H_2O)_6}]^{3+}$ $\mathrm{Fe^{3+}} : [\mathrm{Ar}]3d^5$ Hybridisation: $sp^3d^2$ Magnetic nature: Paramagnetic (so this complex responds to external magnetic field) 2. $[\mathrm{Ni(CN)_4}]^{2-}$ $\mathrm{Ni^{2+}} : [\mathrm{Ar}]3d^8$ Hybridisation: $dsp^2$ Magnetic nature: diamagnetic 3. $[\mathrm{Co(CN)_6}]^{3-}$ $\mathrm{Co^{3+}} : [\mathrm{Ar}]3d^6$ Hybridisation: $d^2sp^3$ Magnetic nature: diamagnetic 4. $[\mathrm{Ni(CO)_4}]$ $\mathrm{Ni} : [\mathrm{Ar}]3d^84s^2$ Hybridisation: $sp^3$ Magnetic nature: diamagnetic

Question 63

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Consider the above reaction, the major product 'P' is:

Answer: (c)

Solution

The reaction begins with the 1,2-addition of $\mathrm{CH_3CH_2MgBr}$ to the carbonyl group. This forms an alkoxide intermediate. Upon protonation with $\mathrm{H^+}$, the alkoxide is converted to an alcohol. The final step involves the addition of $\mathrm{HCl}$, which results in the formation of the chlorinated alcohol product.

Question 64

Chemistry · Chemistry in Everyday Life · Single correct

Sodium stearate $CH_3(CH_2)_{16}COO^-Na^+$ is an anionic surfactant which forms micelles in oil. Choose the correct statement for it from the following:

  1. It forms spherical micelles with $\mathrm{CH_3(CH_2)_{16}}$ group pointing towards the centre of sphere.
  2. It forms non-spherical micelles with $\mathrm{-COO^-}$ group pointing outwards on the surface.
  3. It forms spherical micelles with $\mathrm{CH_3(CH_2)_{16}}$ group pointing outwards on the surface of sphere.
  4. It forms non-spherical micelles with $\mathrm{CH_3(CH_2)_{16}}$ group pointing towards the centre.

Answer: (a)

Solution

Forms spherical micelles with $\mathrm{CH_3(CH_2)_{16}}$ group pointing towards the centre of sphere

Question 65

Chemistry · Biomolecules · Single correct

The water soluble protein is

  1. Fibrin
  2. Albumin
  3. Myosin
  4. Collagen

Answer: (b)

Solution

Albumin is water soluble.

Question 66

Chemistry · Hydrogen · Single correct

At $298.2 \, \mathrm{K}$ the relationship between enthalpy of bond dissociation (in $\mathrm{kJ \, mol^{-1}}$) for hydrogen $(E_H)$ and its isotope, deuterium $(E_D)$, is best described by:

  1. $E_H = \frac{1}{2} E_D$
  2. $E_H = E_D$
  3. $E_H = E_D - 7.5$
  4. $E_H = 2E_D$

Answer: (c)

Solution

Enthalpy of bond dissociation (kJ/mole) at 298.2 $\mathrm{K}$. For hydrogen, it is 435.88. For deuterium, it is 443.35. $$E_\mathrm{H} \simeq E_\mathrm{D} - 7.5$$

Question 67

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Consider the given reaction, the product 'X' is:

Answer: (d)

Solution

The reaction begins with the formation of an enolate ion from the ketone in the presence of a base, $OH^-$. This enolate ion then undergoes a rate-determining step (RDS) to form a new intermediate. The intermediate reacts with water to form the final product. The filtrate is treated with $CH_3I$ and $I_2$, $NaOH$ to perform the iodoform test, resulting in a yellow precipitate. Finally, the filtrate is acidified with HCl to yield the carboxylic acid.

Question 68

Chemistry · Alcohols, Phenols and Ethers · Multiple correct

The given reaction can occur in the presence of: (a) Bromine water (b) Br$_2$ in CS$_2$, 273 K (c) Br$_2$/FeBr$_3$ (d) Br$_2$ in CHCl$_3$, 273 K

  1. (b) and (d) only
  2. (a) and (c) only
  3. (b), (c) and (d) only
  4. (a), (b) and (d) only

Answer: (c)

Solution

Bromine water gives tribromo products, other gives monobromo products in which para is major product.

Question 69

Chemistry · Amines · Single correct

Given below are two statements, one is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (R) : Gabriel phthal imide synthesis cannot be used to prepare aromatic primary amines. Reason : Aryl halides do not undergo nucleophilic substitution reaction. In the light of the above statements, choose the correct answer from the options given below:

  1. Both (A) and (R) true but (R) is not the correct explanation of (A).
  2. (A) is false but (R) is true.
  3. Both (A) and (R) true and (R) is correct explanation of (A).
  4. (A) is true but (R) is false.

Answer: (c)

Solution

The Gabriel phthalimide synthesis involves the following steps. First, the phthalimide reacts with $\mathrm{KOH}$ to form a potassium salt. This salt then undergoes an $\mathrm{S_N^2}$ reaction with an alkyl halide $\mathrm{R-X}$ to form an N-alkyl phthalimide. This step is not given by aryl halide. Finally, hydrolysis with $\mathrm{HOH/H^+}$ yields a primary aliphatic amine $\mathrm{R-NH_2}$ and phthalic acid.

Question 70

Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct

For the following graphs, Choose from the options given below, the correct one regarding order of reaction is:

  1. (b )zero order ($c$) and (e) First order (b) Zero order
  2. (a) and (e) First order
  3. (b) and (d) Zero order (e) First order
  4. (a) and (b) Zero order $(c)$ and (e) First order

Answer: (a)

Solution

For zero order reaction's rate $= K [Reactant]^0$ $r = k$. Reactant concentration after time $t \rightarrow C_t = C_0 \, e^{-kt}$

Question 71

Chemistry · Amines · Single correct

Which one of the products of the following reactions does not react with Hinsberg reagent to form sulphonamide?

Answer: (b)

Solution

The reaction involves the conversion of a cyano group $\mathrm{CN}$ to an imino group $\mathrm{CH{=}NH}$ using $\mathrm{SnCl_2 + HCl}$. The imino group does not react with Hinsberg reagent.

Question 72

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

The ionic radii of $\mathrm{K}^+$, $\mathrm{Na}^+$, $\mathrm{Al}^{3+}$ and $\mathrm{Mg}^{2+}$ are in the order:

  1. $\mathrm{Na}^+ < \mathrm{K}^+ < \mathrm{Mg}^{2+} < \mathrm{Al}^{3+}$
  2. $\mathrm{Al}^{3+} < \mathrm{Mg}^{2+} < \mathrm{K}^+ < \mathrm{Na}^+$
  3. $\mathrm{Al}^{3+} < \mathrm{Mg}^{2+} < \mathrm{Na}^+ < \mathrm{K}^+$
  4. $\mathrm{K}^+ < \mathrm{Al}^{3+} < \mathrm{Mg}^{2+} < \mathrm{Na}$

Answer: (c)

Solution

$Al^{3+}$, $\mathrm{Mg}^{2+}$ and $\mathrm{Na}^{+}$ are isoelectronic ionic species. For monoatomic ionic isoelectronic species as positive charge increases ionic size decreases. The order of size of $\mathrm{Na}^{+}$ and $\mathrm{K}^{+}$ is $\mathrm{Na}^{+}$ < $\mathrm{K}^{+}$, therefore, order of ionic radii is: $\mathrm{Al}^{3+}$ < $\mathrm{Mg}^{2+}$ < $\mathrm{Na}^{+}$ < $\mathrm{K}^{+}$

Question 73

Chemistry · The p-Block Elements (Group-13 and 14) · Single correct

Which one of the following compounds of Group- 14 elements is not known?

  1. $[\mathrm{GeCl}_6]^{2-}$
  2. $[\mathrm{Sn(OH)}_6]^{2-}$
  3. $[\mathrm{SiCl}_6]^{2-}$
  4. $[\mathrm{SiF}_6]^{2-}$

Answer: (c)

Solution

[$\mathrm{SiCl_6}$]^{2-} does not exist due to steric crowding of surrounding atoms.

Question 74

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

Which one among the following resonating structures is not correct?

Answer: (a)

Solution

It is unstable RS (due to similar charge on adjacent atom)

Question 75

Chemistry · The s-Block Elements · Single correct

Given below are two statements Statement I : None of the alkaline earth metal hydroxides dissolve in alkali. Statement II : Solubility of alkaline earth metal hydroxides in water increases down the group. In the light of the above statements, choose the most appropriate answer from the options given below

  1. Statement I is correct but Statement II is incorrect.
  2. Statement I is incorrect but Statement II is correct.
  3. Statement I and Statement II both are incorrect.
  4. Statement I and Statement II both are correct.

Answer: (b)

Solution

Statement-I is incorrect. $\mathrm{Be(OH)_2}$ dissolves in alkali due to its amphoteric nature. Statement-II is correct. Solubility of alkaline earth metal hydroxide in water increases down the group due to rapid decreases in lattice energy as compared to hydration energy.

Question 76

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

The correct order of following 3 d metal oxides, according to their oxidation numbers is: (a) $\mathrm{CrO}_3$ (b) $\mathrm{Fe}_2\mathrm{O}_3$ (c ) $\mathrm{MnO}_2$ (d) $\mathrm{V}_2\mathrm{O}_5$ (e) $\mathrm{Cu}_2\mathrm{O}$

  1. (d) > (a) > (b) > (c ) > (e)
  2. (a) > (c ) > (d) > (b) > (e)
  3. (a) > (d) > (c ) > (b) > (e)
  4. (c ) > (a) > (d) > (e) > (b)

Answer: (c)

Solution

So order of oxidation state $a > d > c > b > e$

Question 77

Chemistry · Environmental Chemistry · Single correct

Which one of the following chemical agent is not being used for dry-cleaning of clothes?

  1. H_2O_2
  2. CCl_4
  3. Liquid CO_2
  4. Cl_2C = CCl_2

Answer: (b)

Solution

$\text{CO}_2$, $\text{CCl}_4$ and $\text{Cl}_2\text{C} = \text{CCl}_2$ are used as dry cleaning agents for clothes. $\text{H}_2\text{O}_2$ is used as bleaching agent in laundry.

Question 78

Chemistry · Amines · Single correct

Which one of the following compounds will liberate $CO_2$, when treated with $NaHCO_3$?

  1. $(CH_3)_3NHCl
  2. $(CH_3)_4N^+OH^-
  3. $CH_3NH_2

Answer: (a)

Solution

The reaction is as follows: $\mathrm{(CH_3)_3NHCl} + \mathrm{NaHCO_3} \rightarrow \mathrm{H_2CO_3} + \mathrm{(CH_3)_3N} + \mathrm{NaCl}$. The $\mathrm{H_2CO_3}$ further decomposes to $\mathrm{CO_2} + \mathrm{H_2O}$.

Question 79

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

In the leaching of alumina from bauxite, the ore expected to leach out in the process by reacting with NaOH is :

  1. $\mathrm{TiO}_{2}$
  2. $\mathrm{Fe}_{2}\mathrm{O}_{3}$
  3. $\mathrm{ZnO}$
  4. $\mathrm{SiO}_{2}$

Answer: (d)

Solution

In bauxite impurities of $\mathrm{Fe_2O_3}$, $\mathrm{TiO_2}$ and $\mathrm{SiO_2}$ are present, $\mathrm{Fe_2O_3}$ and $\mathrm{TiO_2}$ are basic oxides therefore does not react with or dissolve in NaOH whereas $\mathrm{SiO_2}$ is acidic oxide it gets dissolve in NaOH, hence leach out $$\mathrm{SiO_2} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2SiO_3(aq.)} + \mathrm{H_2O}$$

Question 80

Chemistry · Hydrocarbons · Single correct

An organic compound ' $A'C_4H_8$ on treatment with $KMnO_4/H^+$ yields compound ' $B'C_3H_6O$. Compound 'A' also yields compound 'B' an ozonolysis. Compound 'A' is :

  1. 2-Methylpropene
  2. 1-Methylcyclopropane
  3. But-2-ene
  4. Cyclobutane

Answer: (a)

Solution

The compound 2-methylpropene undergoes two reactions. First, it reacts with $\mathrm{KMnO_4/H^+}$ to form a carboxylic acid. Second, it undergoes ozonolysis to form an aldehyde.

Question 81

Chemistry · Hydrocarbons · Numerical

The number of sigma bonds in is

Answer: 10

Solution

The number of $\sigma$ bonds is 10.

Question 82

Chemistry · Co-ordination Compounds · Numerical

Three moles of $\mathrm{AgCl}$ are precipitated when one mole of an octahedral coordination compound with empirical formula $\mathrm{CrCl_3\cdot3NH_3\cdot3H_2O}$ reacts with excess $\mathrm{AgNO_3}$. The number of chloride ions satisfying the secondary valency of the metal ion is $\underline{\qquad}$.

Answer: 0

Solution

Mole of AgCl precipitated is equal to the mole of Cl^- present in ionization sphere. $$\mathrm{[Cr(H_2O)_4(NH_3)_2]Cl_3 \rightarrow [Cr(H_2O)_4(NH_3)_2]^{3+} + 3Cl^-}$$ 1 mole $\hspace{2cm}$ 1 mole $\hspace{2cm}$ 3 mole AgNO_3 (Excess) $$\mathrm{AgCl (1 \, mole)}$$ Since none of Cl^- is present in the co-ordination sphere. Therefore answer is zero.

Question 83

Chemistry · Structure of Atom · Numerical

A source of monochromatic radiation of wavelength 400 $\mathrm{nm}$ provides 1000 $\mathrm{J}$ of energy in 10 seconds. When this radiation falls on the surface of sodium, x $\times 10^{20}$ electrons are ejected per second. Assume that wavelength 400 $\mathrm{nm}$ is sufficient for ejection of electron from the surface of sodium metal. The value of x is (Nearest integer) $( h = 6.626$ $\times 10^{-34}\mathrm{Js}$)"

Answer: 2

Solution

Total energy provided by Source per second is $\frac{1000}{10} = 100 \, \mathrm{J}$. Energy required to eject electron is $\frac{hc}{\lambda}$. $$= \frac{6.626 \times 10^{-34}}{400 \times 10^{-9}} \times 3 \times 10^8$$ Number of electrons ejected is $$= \frac{400 \times 10^{-7} \times 10^{26}}{6.626 \times 3}$$ $$= \frac{40 \times 10^{-20}}{6.626 \times 3}$$ $$= 2.01 \times 10^{20}$$

Question 84

Chemistry · Solutions · Numerical

$CO_2$ gas is bubbled through water during a soft drink manufacturing process at 298 $\mathrm{K}$. If $CO_2$ exerts a partial pressure of 0.835 $\mathrm{bar}$ then x mmol of $CO_2$ would dissolve in 0.9 $\mathrm{L}$ of water. The value of x is . (Nearest integer) (Henry's law constant for $CO_2$ at 298 $\mathrm{K}$ is 1.67 $\times 10^3$ $\mathrm{bar}$)

Answer: 25

Solution

From Henry's law $$P_{gas} = K_H \cdot X_{gas}$$ $$n(\mathrm{CO_2}) = 0.025$$ Millimoles of $\mathrm{CO_2} = 0.025 \times 1000 = 25$$

Question 85

Chemistry · Equilibrium · Numerical

For the reaction $$\mathrm{A + B \rightleftharpoons 2C}$$ the value of equilibrium constant is 100 at 298 $\mathrm{K}$. If the initial concentration of all the three species is 1 $\mathrm{M}$ each, then the equilibrium concentration of $\mathrm{C}$ is $x \times 10^{-1}$ $\mathrm{M}$. The value of $x$ is (Nearest integer)

Answer: 25

Solution

The reaction is given by: $$\mathrm{A} + \mathrm{B} \rightleftharpoons 2\mathrm{C}$$ Initial concentrations are 1 for A and B, and 1 for C. The changes in concentration are $-x$ for A and B, and $2x$ for C. At equilibrium, the concentrations are $1-x$ for A and B, and $1+2x$ for C. The equilibrium constant $K$ is given by: $$K = \frac{[\mathrm{C}]_{eq}^2}{[\mathrm{A}]_{eq}[\mathrm{B}]_{eq}} = \frac{(1+2x)^2}{(1-x)(1-x)}$$ Given $K = 100$, we have: $$100 = \left(\frac{1+2x}{1-x}\right)^2$$ Taking the square root: $$\frac{1+2x}{1-x} = 10$$ Solving for $x$: $$x = \frac{3}{4}$$ The equilibrium concentration of C is: $$[\mathrm{C}]_{eq} = 1 + 2x$$ Substituting $x = \frac{3}{4}$: $$= 1 + 2 \left(\frac{3}{4}\right)$$ $$= 2.5 \, \mathrm{M}$$ This is equivalent to $25 \times 10^{-1} \, \mathrm{M}$.

Question 86

Chemistry · Electrochemistry · Numerical

Consider the cell at $25^\circ$C $\text{Zn}\left|\text{Zn}^{2+}\text{(aq), (1M)}\middle|\middle|\text{Fe}^{3+}\text{(aq), Fe}^{2+}\text{(aq)}\right|\text{Pt(s)}$ The fraction of total iron present as $\text{Fe}^{3+}$ ion at the cell potential of 1.500 V is $x \times 10^{-2}$. The value of $x$ (Nearest integer) $\left(\text{Given: } E^0_{\text{Fe}^{2+}/\text{Fe}^{2+}} = 0.77\,\text{V},\ E^0_{\text{Za}^2/\text{Zan}} = -0.76\,\text{V}\right)$

Answer: 24

Solution

The reactions are as follows: $$\mathrm{Zn} \rightarrow \mathrm{Zn^{2+}} + 2e^-$$ $$2\mathrm{Fe^{3+}} \rightarrow 2e^- + 2\mathrm{Fe^{2+}}$$ Overall reaction: $$\mathrm{Zn} + 2\mathrm{Fe^{3+}} \rightarrow \mathrm{Zn^{2+}} + 2\mathrm{Fe^{2+}}$$ The standard cell potential is given by: $$E^0_{cell} = 0.77 - (0.76)$$ $$= 1.53 \, \mathrm{V}$$ Now, we have: $$1.50 = 1.53 - \frac{0.06}{2} \log \left( \frac{\mathrm{Fe^{2+}}}{\mathrm{Fe^{3+}}} \right)^2$$ Taking the logarithm: $$\log \left( \frac{\mathrm{Fe^{2+}}}{\mathrm{Fe^{3+}}} \right) = \frac{0.03}{0.06} = \frac{1}{2}$$ Thus: $$\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]} = 10^{1/2} = \sqrt{10}$$ Therefore: $$\frac{[\mathrm{Fe^{3+}}]}{[\mathrm{Fe^{2+}}]} = \frac{1}{\sqrt{10}}$$ And: $$\frac{[\mathrm{Fe^{3+}}]}{[\mathrm{Fe^{2+}}] + [\mathrm{Fe^{3+}}]} = \frac{1}{1 + \sqrt{10}} = \frac{1}{4.16}$$ This gives: $$= 0.2402$$ Finally: $$= 24 \times 10^{-2}$$

Question 87

Chemistry · Thermodynamics · Numerical

At 298 K, the enthalpy of fusion of a solid (X) is $2.8 \, \mathrm{kJ \, mol^{-1}}$ and the enthalpy of vaporisation of the liquid (X) is $98.2 \, \mathrm{kJ \, mol^{-1}}$. The enthalpy of sublimation of the substance (X) in $\mathrm{kJ \, mol^{-1}}$ is (in nearest integer)

Answer: 101

Solution

Q5 $$\Delta H_{ab} = \Delta H_{fis} + \Delta H_{vqp}$$ $$= 2.8 + 98.2$$ $$= 101 \, \mathrm{kJ/mol}$$

Question 88

Chemistry · States of Matter · Numerical

A home owner uses $4.00 \times 10^3\,\mathrm{m}^3$ of methane ($\mathrm{CH_4}$) gas (assume $\mathrm{CH_4}$ is an ideal gas) in a year to heat his home. Under the pressure of $1.0\,\mathrm{atm}$ and $300\,\mathrm{K}$, mass of gas used is $x \times 10^5\,\mathrm{g}$. The value of $x$ is _____. (Nearest integer) (Given: $R = 0.083\,\mathrm{L\,atm\,K}^{-1}\,\mathrm{mol}^{-1}$)

Answer: 26

Solution

The number of moles of $\mathrm{CH_4}$ is given by $$n(\mathrm{CH_4}) = \frac{PV}{RT}$$ Substituting the values, we have $$= \frac{1 \times 4 \times 10^3 \times 1000}{0.083 \times 300}$$ The weight of $\mathrm{CH_4}$ is $$Weight of \mathrm{CH_4} = \frac{40 \times 16 \times 10^5}{0.083 \times 300} \, \mathrm{gm}$$ Simplifying, we get $$= 25.7 \times 10^5 \, \mathrm{gm}$$

Question 89

Chemistry · Redox Reactions · Numerical

When $10\,\mathrm{mL}$ of an aqueous solution of $\mathrm{Fe^{2+}}$ ions was titrated in the presence of dil $\mathrm{H_2SO_4}$ using diphenylamine indicator, $15\,\mathrm{mL}$ of $0.02\,\mathrm{M}$ solution of $\mathrm{K_2Cr_2O_7}$ was required to get the end point. The molarity of the solution containing $\mathrm{Fe^{2+}}$ ions is $x \times 10^{-2}\,\mathrm{M}$. The value of $x$ is (Nearest integer)

Answer: 18

Solution

Milli-equivalents of $\mathrm{Fe^{2+}}$ = milli-equivalents of $\mathrm{K_2Cr_2O_7}$. $M \times 10 \times 1 = 0.02 \times 15 \times 6$ $M = 0.18 = 18 \times 10^{-2} M$

Question 90

Chemistry · Some Basic Concepts of Chemistry · Numerical

Consider the complete combustion of butane, the amount of butane utilized to produce 72.0 g of water is x $10^{-1}$ g. (in nearest integer)

Answer: 464

Solution

The reaction is given by: $$\mathrm{C_4H_{10}} + \frac{13}{2} \mathrm{O_2} \rightarrow 4 \mathrm{CO_2} + 5 \mathrm{H_2O}$$ Moles of $\mathrm{H_2O} = \frac{72}{18} = 4$ Moles of $\mathrm{C_4H_{10}}$ used $= \frac{1}{5} \times 4$ Weight of $\mathrm{C_4H_{10}}$ used $= \frac{4}{5} \times 58$ $= 46.4 \mathrm{gm}$