JEE Main 5 April 2024 Shift 1 question paper with solutions

JEE Main 5 April 2024 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 · Three Dimensional Geometry · Single correct

Let d be the distance of the point of intersection of the lines $\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1}$ and $\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2}$ from the point $(7, 8, 9)$. Then $d^2 + 6$ is equal to:

  1. 69
  2. 78
  3. 72
  4. 75

Answer: (d)

Solution

Given $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda$. $(1)$ $x=3\lambda-6,\ y=2\lambda,\ z=\lambda-1$ $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu$. $(2)$ $x=4\mu+7,\ y=3\mu+9,\ z=2\mu+4$ $3\lambda-6=4\mu+7 \Rightarrow 3\lambda-4\mu=13 \ldots (3)\times 2$ $2\lambda=3\mu+9 \Rightarrow 2\lambda-3\mu=9 \ldots (4)\times 3$ $\begin{aligned} 6\lambda-8\mu &= 26\\ 6\lambda-9\mu &= 27 \end{aligned}$ Subtracting, $\mu=-1$ Putting $\mu=-1$ in $(4)$, $2\lambda-3(-1)=9$ $2\lambda+3=9$ $2\lambda=6$ $\lambda=3$

Question 2

Maths · Applications of Derivatives · Single correct

Let a rectangle $ABCD$ of sides $2$ and $4$ be inscribed in another rectangle $PQRS$ such that the vertices of the rectangle $ABCD$ lie on the sides of the rectangle $PQRS$. Let $a$ and $b$ be the sides of the rectangle $PQRS$ when its area is maximum. Then $(a + b)^2$ is equal to:

  1. 72
  2. 60
  3. 64
  4. 80

Answer: (a)

Solution

Area $= (4 \cos \theta + 2 \sin \theta)(2 \cos \theta + 4 \sin \theta)$ $$= 8 \cos^2 \theta + 16 \sin \theta \cos \theta + 4 \sin \theta \cos \theta + 8 \sin^2 \theta$$ $$= 8 + 20 \sin \theta \cos \theta$$ $$= 8 + 10 \sin 2\theta$$ Max Area $= 8 + 10 = 18(\sin 2\theta = 1) \theta = 45^\circ$ $$(a + b)^2 = (4 \cos \theta + 2 \sin \theta + 2 \cos \theta + 4 \sin \theta)^2$$ $$= (6 \cos \theta + 6 \sin \theta)^2$$ $$= 36(\sin \theta + \cos \theta)^2$$ $$= 36(\sqrt{2})^2$$ $$= 72$$

Question 3

Maths · Straight Lines and Pair of Straight Lines · Single correct

Let two straight lines drawn from the origin O intersect the line $3x + 4y = 12$ at the points P and Q such that $\triangle OPQ$ is an isosceles triangle and $\angle POQ = 90^\circ$. If $l = OP^2 + PQ^2 + QO^2$, then the greatest integer less than or equal to $l$ is:

  1. 42
  2. 46
  3. 44
  4. 48

Answer: (b)

Solution

Given the equations: $$3x + 4y = 12$$ Substituting $x = r \cos \theta$ and $y = r \sin \theta$: $$3(r \cos \theta) + 4(r \sin \theta) = 12$$ This simplifies to: $$r(3 \cos \theta + 4 \sin \theta) = 12 \ldots (1)$$ Similarly, for the second equation: $$3(-r \sin \theta) + 4(r \cos \theta) = 12$$ This simplifies to: $$r(-3 \sin \theta + 4 \cos \theta) = 12 \ldots (2)$$ Now, consider: $$\left( \frac{12}{r} \right)^2 + \left( \frac{12}{r} \right)^2 = (3 \cos \theta + 4 \sin \theta)^2 + (-3 \sin \theta + 4 \cos \theta)^2$$ Simplifying: $$2 \left( \frac{12}{r} \right)^2 = 9 + 16$$ $$2 \times \frac{144}{r^2} = 25 \Rightarrow 288 = 25r^2$$ $$\Rightarrow \frac{288}{25} = r^2$$ $$\Rightarrow \sqrt{2 \left( \frac{12}{5} \right)} = r$$ Now, calculate $\ell$: $$\ell = OP^2 + PQ^2 + QO^2$$ $$\ell = r^2 + r^2 + r^2(\cos \theta + \sin \theta)^2 + r^2(\sin \theta + \cos \theta)^2$$ $$= 2r^2 + r^2(1 + \sin 2\theta + 1 - 2 \sin 2\theta)$$ $$= 2r^2 + 2r^2$$ $$= 4r^2$$ Substituting the value of $r^2$: $$= 4 \left( \frac{288}{25} \right) = \frac{1152}{25} = 46.08$$ Thus, $[\ell] = 46$

Question 4

Maths · Differential Equations · Single correct

If $y = y(x)$ is the solution of the differential equation $\frac{dy}{dx} + 2y = \sin(2x)$, $y(0) = \frac{3}{4}$, then $y\left(\frac{\pi}{8}\right)$ is equal to:

  1. $e^{\pi/8}$
  2. $e^{\pi/4}$
  3. $e^{-\pi/4}$
  4. $e^{-\pi/8}$

Answer: (c)

Solution

Given $\dfrac{dy}{dx}+2y=\sin 2x,\quad y(0)=\dfrac{3}{4}$. The integrating factor is $\mathrm{I.F.}=e^{\int 2\,dx}=e^{2x}$. Multiply through by the integrating factor: $y\cdot e^{2x}=\int e^{2x}\sin 2x\,dx$. Using integration by parts, $y\cdot e^{2x}=\dfrac{e^{2x}(2\sin 2x-2\cos 2x)}{4+4}+C$. Given $x=0,\ y=\dfrac{3}{4}$, $\dfrac{3}{4}=\dfrac{-1}{4}+C$. Solving for $C$, $C=1$. Thus, $y=\dfrac{2\sin 2x-2\cos 2x}{8}+e^{-2x}$. For $x=\dfrac{\pi}{8}$, $y=\dfrac{1}{8}\left(2\sin\frac{\pi}{4}-2\cos\frac{\pi}{4}\right)+e^{-2\left(\frac{\pi}{8}\right)}$. Therefore, $y=0+e^{-\frac{\pi}{4}}$. $\therefore\ y=e^{-\frac{\pi}{4}}$.

Question 5

Maths · Applications of Derivatives · Single correct

For the function $$f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x), where x \in \left[0, \frac{\pi}{2}\right],$$ consider the following two statements: (I) $f$ is increasing in $\left(0, \frac{\pi}{2}\right)$. (II) $f'$ is decreasing in $\left(0, \frac{\pi}{2}\right)$. Between the above two statements,

  1. only (II) is true.
  2. only (I) is true.
  3. neither (I) nor (II) is true.
  4. both (I) and (II) are true

Answer: (d)

Solution

Given $f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x)$ for $x \in \left[0, \frac{\pi}{2}\right]$. The derivative is $f'(x) = \cos x + 3 - \frac{2}{\pi}(2x + 1) > 0$, indicating $f(x)$ is increasing. At $x = 0$, $f'(x) = -\sin x + 0 - \frac{\pi}{2}(2)$. This simplifies to $-\sin x - \frac{4}{\pi} -\frac{2}{\pi} \left(2x + 1\right) > -\frac{2}{\pi} \left(\pi + 1\right)$$ $$3_{+ve} - \frac{2}{\pi} > 3 - \frac{2}{\pi} (2x + 1) > 3 - \frac{2}{\pi} (\pi_{+ve} + 1)$$

Question 6

Maths · Determinants · Single correct

If the system of equations $$11x + y + \lambda z = -5$$ $$2x + 3y + 5z = 3$$ $$8x - 19y - 39z = \mu$$ has infinitely many solutions, then $\lambda^4 - \mu$ is equal to:

  1. 51
  2. 45
  3. 47
  4. 49

Answer: (c)

Solution

Given the equations: $$11x + y + \lambda z = -5$$ $$2x + 3y + 5z = 3$$ $$8x - 19y - 39z = \mu$$ For infinite solutions, the determinant $D$ is given by: $$D = \begin{vmatrix} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{vmatrix} = 0$$ Calculating the determinant: $$\Rightarrow 11(-117 + 95) - 1(-78 - 40) + \lambda(-38 - 24)$$ $$\Rightarrow 11(-22) + 118 - \lambda(62) = 0$$ $$\Rightarrow 62\lambda = 118 - 242$$ $$\Rightarrow \lambda = \frac{-124}{62} = -2$$ Now, consider the determinant $D_1$: $$D_1 = \begin{vmatrix} -5 & 1 & -2 \\ 3 & 3 & 5 \\ \mu & -19 & -39 \end{vmatrix} = 0$$ Calculating $D_1$: $$\Rightarrow -5(-117 + 95) - 1(-117 - 5\mu) - 2(-57 - 3\mu) = 0$$ $$\Rightarrow -5(-22) + 117 + 5\mu + 114 + 6\mu = 0$$ $$\Rightarrow 11\mu = -110 - 231 = -341$$ $$\Rightarrow \mu = -31$$ Finally, calculate $\lambda^4 - \mu$: $$\lambda^4 - \mu = (-2)^4 - (-31) = 16 + 31 = 47$$

Question 7

Maths · Permutations and Combinations · Single correct

Let $A = \{1, 3, 7, 9, 11\}$ and $B = \{2, 4, 5, 7, 8, 10, 12\}$. Then the total number of one-one maps $f : A \rightarrow B$, such that $f(1) + f(3) = 14$, is :

  1. 480
  2. 240
  3. 120
  4. 180

Answer: (b)

Solution

Given sets: $A = \{1, 3, 7, 9, 11\}$ $B = \{2, 4, 5, 7, 8, 10, 12\}$ We have $f(1) + f(3) = 14$. (i) $2 + 12$ (ii) $4 + 10$ $2 \times (2 \times 5 \times 4 \times 3) = 240$

Question 8

Maths · Continuity and Differentiability · Single correct

If the function $f(x) = \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}$, $x \in \mathbb{R}$, is continuous at $x = 0$, then $f(0)$ is equal to:

  1. 2
  2. -2
  3. 4
  4. -4

Answer: (d)

Solution

Given $$f(x) = \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}$$ is continuous at $x = 0$. $$\lim_{x \to 0} \frac{3x - \frac{(3x)^3}{3} + \ldots + \alpha \left(x - \frac{x^3}{3} \ldots \right) - \beta \left(1 - \frac{(3x)^2}{2} \ldots \right)}{x^3} = f(0)$$ $$\lim_{x \to 0} \frac{-\beta + x(3 + \alpha) + \frac{9 \beta x^2}{2} + \left(\frac{-27}{3} - \frac{\alpha}{3} \right) x^3 \ldots}{x^3} = f(0)$$ For exist $$\beta = 0, \ 3 + \alpha = 0, \ \frac{-27}{3} - \frac{\alpha}{3} = f(0)$$ $$\alpha = -3, \ \frac{-27}{6} - \frac{(-3)}{6} = f(0)$$ $$f(0) = \frac{-27 + 3}{6} = -4$$

Question 9

Maths · Integrals · Single correct

The integral $\int_0^{\pi/4} \frac{136 \sin x}{3 \sin x + 5 \cos x} \, dx$ is equal to:

  1. $3\pi - 50 \log_e 2 + 20 \log_e 5$
  2. $3\pi - 25 \log_e 2 + 10 \log_e 5$
  3. $3\pi - 10 \log_e (2\sqrt{2}) + 10 \log_e 5$
  4. $3\pi - 30 \log_e 2 + 20 \log_e 5$

Answer: (a)

Solution

Given $$I = \int_0^{\pi/4} \frac{136 \sin x}{3 \sin x + 5 \cos x} \, dx$$ Assume $$136 \sin x = A(3 \sin x + 5 \cos x) + B(3 \cos x - 5 \sin x)$$ This gives the equations: $$136 = 3A - 5B \ldots (i)$$ $$0 = 5A + 3B \ldots (ii)$$ From equation (ii): $$3B = -5A \Rightarrow B = -\frac{5}{3}A$$ Substitute in equation (i): $$136 = 3A - 5 \left(-\frac{5}{3}A\right)$$ $$136 = 3A + \frac{25}{3}A$$ Combine terms: $$136 = \frac{34}{3}A$$ Solve for $A$: $$A = \frac{136 \times 3}{34} = 12$$ Then $B$ is: $$B = -\frac{5}{3}(12) = -20$$ Now, substitute back into the integral: $$I = \int_0^{\pi/4} \frac{A(3 \sin x + 5 \cos x)}{3 \sin x + 5 \cos x} + \int_0^{\pi/4} \frac{B(3 \cos x - 5 \sin x)}{3 \sin x + 5 \cos x}$$ This simplifies to: $$= A(x)\bigg|_0^{\pi/4} + B[\ln(3 \sin x + 5 \cos x)]\bigg|_0^{\pi/4} - \ln(0 + 5)$$ Substitute the values of $A$ and $B$: $$= 12 \left(\frac{\pi}{4}\right) - 20 \ln \left(\frac{3}{\sqrt{2}} + \frac{5}{\sqrt{2}}\right) - \ln(5)$$ Simplify: $$= 3\pi - 20 \ln 4\sqrt{2} + 20 \ln 5$$ $$= 3\pi - 20 \times \frac{5}{2} \ln 2 + 20 \ln 5$$ $$= 3\pi - 50 \ln 2 + 20 \ln 5$$

Question 10

Maths · Probability · Single correct

The coefficients $a, b, c$ in the quadratic equation $ax^2 + bx + c = 0$ are chosen from the set $\{$1, 2, 3, 4, 5, 6, 7, 8$\}$. The probability of this equation having repeated roots is:

  1. $\frac{1}{128}$
  2. $\frac{1}{64}$
  3. $\frac{3}{256}$
  4. $\frac{3}{128}$

Answer: (b)

Solution

Given the quadratic equation $ax^2 + bx + c = 0$ where $a, b, c \in \{1, 2, 3, 4, 5, 6, 7, 8\}$. For repeated roots, the discriminant $D = 0$. This implies $b^2 - 4ac = 0$ which leads to $b^2 = 4ac$. The probability is calculated as $$Prob = \frac{8}{8 \times 8 \times 8} = \frac{1}{64}$$ which implies the solutions for $(a, b, c)$ are: $(1, 2, 1); (2, 4, 2); (1, 4, 4); (4, 4, 1); (3, 6, 3); (2, 8, 8); (8, 8, 2); (4, 8, 4)$. There are 8 cases.

Question 11

Maths · Matrices · Single correct

Let A and B be two square matrices of order 3 such that $|A| = 3$ and $|B| = 2$. Then $$\left| A^T A (adj(2 \, A))^{-1} (adj(4 \, B))(adj(AB))^{-1} AA^T \right|$$ is equal to:

  1. 108
  2. 32
  3. 81
  4. 64

Answer: (d)

Solution

$|A|=3,\ |B|=2$ $|A^TA(adj(2A))^{-1}(adj(4B))(adj(AB))^{-1}AA^T|$ $=3\times3\times(adj(2A))^{-1}\times adj(4B)\times(adj(AB))^{-1}\times3\times3$ $=\frac{1}{|adj(2A)|}\times adj(4B)\times\frac{1}{|adj(AB)|}$ $=\frac{1}{2^6|adj\,A|}\times adj\,B\times adj\,A\times\frac{1}{2^2\cdot3^2}$ $=\frac{1}{2^6\cdot3^2}\times\frac{1}{2^2\cdot3^2}$ $\times2^{12}\cdot2^2$ $=3^4\cdot\frac{1}{2^6\cdot3^2}$ $\times\frac{2^{12}\cdot2^2}{2^2\cdot3^2}$ $=64$

Question 12

Maths · Conic Sections · Single correct

Let a circle $C$ of radius $1$ and closer to the origin be such that the lines passing through the point $(3, 2)$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle $C$ from the point $(5, 5)$ is :

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

Answer: (c)

Solution

Coordinates of the centre will be $(2, 1)$. Equation of circle will be $$ (x - 2)^2 + (y - 1)^2 = 1 $$ $$ QC = \sqrt{(5 - 2)^2 + (5 - 1)^2} $$ $$ QC = 5 $$ shortest distance $$ = RQ = CQ - CR $$ $$ = 5 - 1 $$ $$ = 4 $$

Question 13

Maths · Conic Sections · Single correct

Let the line $2x + 3y - k = 0$, $k > 0$, intersect the $x$-axis and $y$-axis at the points $A$ and $B$, respectively. If the equation of the circle having the line segment $AB$ as a diameter is $x^2 + y^2 - 3x - 2y = 0$ and the length of the latus rectum of the ellipse $x^2 + 9y^2 = k^2$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $2m + n$ is equal to

  1. 11
  2. 10
  3. 12
  4. 13

Answer: (a)

Solution

Centre of the circle = $\left$( $\frac{3}{2}$, 1 $\right$) Equation of diameter = 2x + 3y - k = 0 2 $\left$( $\frac{3}{2}$ $\right$) + 3(1) - k = 0 $\Rightarrow$ k = 6 Now, Equation of ellipse becomes $$x^2 + 9y^2 = 36$$ $$\frac{x^2}{6^2} + \frac{y^2}{2^2} = 1$$ length of $LR = \frac{2b^2}{a} = \frac{2 \cdot 2^2}{6} = \frac{8}{6} = \frac{m}{n}$ $\therefore 2m + n = 2(4) + 3 = 11$

Question 14

Maths · Complex Numbers and Quadratic Equations · Single correct

Consider the following two statements : Statement I : For any two non-zero complex numbers $z_1, z_2$, $$\left( |z_1| + |z_2| \right) \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq 2 \left( |z_1| + |z_2| \right)$$, and Statement II : If $x, y, z$ are three distinct complex numbers and $a, b, c$ are three positive real numbers such that $$\frac{a}{|y-z|} = \frac{b}{|z-x|} = \frac{c}{|x-y|}$$, then $$\frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y} = 1$$. Between the above two statements,

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

Answer: (a)

Solution

Statement I: $$\left( |z_1| + |z_2| \right) \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right|$$ Since $$\left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq \left| \frac{z_1}{|z_1|} \right| + \left| \frac{z_2}{|z_2|} \right|$$ $$\left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq 2$$ $$\left( |z_1| + |z_2| \right) \left( \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \right) \leq 2 \left( |z_1| + |z_2| \right)$$ Therefore, statement I is correct. For Statement II: $$\frac{a}{|y-z|} = \frac{b}{|z-x|} = \frac{c}{|x-y|}$$ $$\frac{a^2}{|y-z|^2} = \frac{b^2}{|z-x|^2} = \frac{c^2}{|x-y|^2} = \lambda$$ $$a^2 = \lambda \left( |y-z|^2 \right) = \lambda (y-z)(\bar{y} - \bar{z})$$ $$b^2 = \lambda (z-x)(\bar{z} - \bar{x}) and c^2 = \lambda (x-y)(\bar{x} - \bar{y})$$ $$\frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y} = \lambda (\bar{y} - \bar{z} + \bar{z} - \bar{x} + \bar{x} - \bar{y}) = 0$$ Statement II is false.

Question 15

Maths · Trigonometric Functions · Single correct

Suppose $\theta \in \left[ 0, \frac{\pi}{4} \right]$ is a solution of $4 \cos \theta - 3 \sin \theta = 1$. Then $\cos \theta$ is equal to:

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

Answer: (c)

Solution

Given $$4 \left( \frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} \right) - 3 \left( \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} \right) = 1$$ Let $\tan \frac{\theta}{2} = t$. $$\frac{4 - 4t^2 - 6t}{1 + t^2} = 1$$ $$4 - 4t^2 - 6t = 1 + t^2$$ $$\Rightarrow 5t^2 + 6t - 3 = 0$$ $$\Rightarrow t = \frac{-6 \pm \sqrt{36 - 4(5)(-3)}}{2(5)}$$ $$= \frac{-6 \pm \sqrt{96}}{10}$$ $$= \frac{-6 \pm 4\sqrt{6}}{10}$$ $$t = \frac{-3 + 2\sqrt{6}}{5}$$ $$\cos \theta = \frac{1 - t^2}{1 + t^2} = \frac{1 - \left( \frac{2\sqrt{6} - 3}{5} \right)^2}{1 + \left( \frac{2\sqrt{6} - 3}{5} \right)^2} = \frac{1 - \left( \frac{24 + 9 - 12\sqrt{6}}{25} \right)}{1 + \left( \frac{24 + 9 - 12\sqrt{6}}{25} \right)}$$ $$= \frac{25 - 33 + 12\sqrt{6}}{25 + 33 - 12\sqrt{6}} = \frac{12\sqrt{6} - 8}{58 - 12\sqrt{6}} = \frac{6\sqrt{6} - 4}{29 - 6\sqrt{6}} \times \frac{29 + 6\sqrt{6}}{29 + 6\sqrt{6}}$$ $$= \frac{100 + 150\sqrt{6}}{625} = \frac{4 + 6\sqrt{6}}{25} \times \frac{4 - 6\sqrt{6}}{4 - 6\sqrt{6}}$$ $$= \frac{-200}{25(4 - 6\sqrt{6})} = \frac{-8}{4 - 6\sqrt{6}} = \frac{4}{3\sqrt{6} - 2}$$

Question 16

Maths · Sequences and Series · Single correct

If $\frac{1}{\sqrt{1+\sqrt{2}}} + \frac{1}{\sqrt{2+\sqrt{3}}} + \cdots + \frac{1}{\sqrt{99+\sqrt{100}}} = m$ and $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{99 \cdot 100} = n$, then the point $(m, n)$ lies on the line

  1. $11(x - 1) - 100(y - 2) = 0$
  2. $11x - 100y = 0$
  3. $11(x - 2) - 100(y - 1) = 0$
  4. $11(x - 1) - 100y = 0$

Answer: (b)

Solution

The expression is given by $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \cdots + \frac{1}{\sqrt{99} + \sqrt{100}} = m$$ which simplifies to $$\frac{1}{\sqrt{1} - \sqrt{2}} + \frac{1}{\sqrt{2} - \sqrt{3}} + \cdots + \frac{1}{\sqrt{99} - \sqrt{100}} = m$$ Simplifying further, we have $$\sqrt{100} - 1 = m \Rightarrow m = 9$$ The next expression is $$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{99 \cdot 100} = n$$ which simplifies to $$1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} \cdots \frac{1}{99} - \frac{1}{100} = n$$ Thus, $$1 - \frac{1}{100} = n$$ which gives $$\frac{99}{100} = n$$ Therefore, $$(m, n) = \left(9, \frac{99}{100}\right)$$ This implies $$\Rightarrow 11(9) - 100\left(\frac{99}{100}\right)$$ which simplifies to $$= 99 - 99 = 0$$ The answer is $11x - 100y = 0$.

Question 17

Maths · Relations and Functions · Single correct

Let $f(x) = x^5 + 2x^3 + 3x + 1, x \in \mathbb{R}$, and $g(x)$ be a function such that $g(f(x)) = x$ for all $x \in \mathbb{R}$. Then $\frac{g(7)}{g'(7)}$ is equal to:

  1. 14
  2. 42
  3. 7
  4. 1

Answer: (a)

Solution

Given $f(x) = x^5 + 2x^3 + 3x + 1$. The derivative is $f'(x) = 5x^4 + 6x^2 + 3$. Evaluating at $x = 1$, we have $f'(1) = 5 + 6 + 3 = 14$. Given $g(f(x)) = x$, we have $g'(f(x)) f'(x) = 1$. For $f(x) = 7$, we solve $x^5 + 2x^3 + 3x + 1 = 7$, which gives $x = 1$. Thus, $g'(7) f'(1) = 1 \Rightarrow g'(7) = \frac{1}{f'(1)} = \frac{1}{14}$. With $x = 1$, $f(x) = 7 \Rightarrow g(7) = 1$. Therefore, $\frac{g(7)}{g'(7)} = \frac{1}{1/14} = 14$.

Question 18

Maths · Three Dimensional Geometry · Single correct

If $A(1, -1, 2)$, $B(5, 7, -6)$, $C(3, 4, -10)$ and $D(-1, -4, -2)$ are the vertices of a quadrilateral $ABCD$, then its area is:

  1. $48\sqrt{7}$
  2. $12\sqrt{29}$
  3. $24\sqrt{7}$
  4. $24\sqrt{29}$

Answer: (b)

Solution

Given points A(1, -1, 2), B(5, 7, -6), C(3, 4, -10), D(-1, -4, -2). The area is given by the formula: $$Area = \frac{1}{2} \left| \overrightarrow{AC} \times \overrightarrow{BD} \right| = \frac{1}{2} \left| (2\hat{i} + 5\hat{j} - 12\hat{k}) \times (6\hat{i} + 11\hat{j} - 4\hat{k}) \right|$$ $$= \frac{1}{2} \left| 112\hat{i} - 64\hat{j} - 8\hat{k} \right|$$ $$= 4 \left| 14\hat{i} - 8\hat{j} - \hat{k} \right|$$ $$= 4 \sqrt{196 + 64 + 1}$$ $$= 4 \sqrt{261}$$ $$= 12 \sqrt{29}$$

Question 19

Maths · Integrals · Single correct

The value of $\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^2 y} \, dy$ is :

  1. $2\pi^2$
  2. $\frac{\pi^2}{2}$
  3. $\frac{\pi}{2}$
  4. $\pi^2$

Answer: (d)

Solution

Given $$\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^2 y} \, dy$$ we can split it as $$\int_{-\pi}^{\pi} \frac{2y}{1+\cos^2 y} \, dy + \int_{-\pi}^{\pi} \frac{2y \sin y}{1+\cos^2 y} \, dy$$ where the first integral is odd and the second is even. This simplifies to $$0 + 2 \cdot 2 \int_0^{\pi} y \left( \frac{\sin y}{1+\cos^2 y} \right) \, dy$$. Let $$I = 4 \int_0^{\pi} \frac{y \sin y}{1+\cos^2 y} \, dy$$. Then $$I = 4 \int_0^{\pi} \frac{(\pi - y) \sin y}{1+\cos^2 y} \, dy$$. Adding these, $$2I = 4 \int_0^{\pi} \frac{\pi \sin y}{1+\cos^2 y} \, dy$$. Therefore, $$I = 2\pi \int_0^{\pi} \frac{\sin y}{1+\cos^2 y} \, dy$$. This evaluates to $$2\pi (- \tan^{-1}(\cos y)) \bigg|_0^{\pi}$$ which simplifies to $$-2\pi \left[ \left( -\frac{\pi}{4} \right) - \left( \frac{\pi}{4} \right) \right]$$. Thus, $$= -2\pi \left[ -\frac{2\pi}{4} \right] = \pi^2$$.

Question 20

Maths · Three Dimensional Geometry · Single correct

If the line $\frac{2-x}{3} = \frac{3y-2}{4\lambda+1} = 4 - z$ makes a right angle with the line $\frac{x+3}{3\mu} = \frac{1-2y}{6} = \frac{5-z}{7}$, then $4\lambda + 9\mu$ is equal to:

  1. 4
  2. 13
  3. 5
  4. 6

Answer: (d)

Solution

Given the equations: $$\frac{2-x}{3} = \frac{3y-2}{4\lambda+1} = 4-z \ldots (1)$$ $$\frac{x-2}{(-3)} = \frac{y-\frac{2}{3}}{\left(4\lambda+\frac{1}{3}\right)} = \frac{z-4}{(-1)}$$ From equation (2): $$\frac{x+3}{3\mu} = \frac{1-2y}{6} = \frac{5-z}{7} \ldots (2)$$ We have: $$\frac{x+3}{3\mu} = \frac{y-\frac{1}{2}}{(-3)} = \frac{z-5}{(-7)}$$ For a right angle, $$(-3)(3\mu) + \left(\frac{4\lambda+1}{3}\right)(-3) + (-1)(-7) = 0$$ Simplifying, $$-9\mu - 4\lambda - 1 + 7 = 0$$ Thus, $$4\lambda + 9\mu = 6$$

Question 21

Maths · Probability · Numerical

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $\sigma^2$, then $96\sigma^2$ is equal to .

Answer: 56

Solution

$X =$ denotes number of defective \begin{tabular}{|c|c|c|c|c|} \hline $x$ & $0$ & $1$ & $2$ & $3$ \\ \hline $P(x)$ & $\dfrac{7}{15}$ & $\dfrac{5}{12}$ & $\dfrac{5}{12}$ & $\dfrac{1}{12}$ \\[6pt] \hline $x_i^2$ & $0$ & $1$ & $4$ & $9$ \\ \hline $P_i x_i^2$ & $0$ & $\dfrac{5}{12}$ & $\dfrac{20}{12}$ & $\dfrac{9}{12}$ \\[6pt] \hline $p_i x_i$ & $0$ & $\dfrac{5}{12}$ & $\dfrac{10}{12}$ & $\dfrac{3}{12}$ \\[6pt] \hline \end{tabular} $\mu = \Sigma p_i x_i = \dfrac{18}{12}$ $\Sigma p_i x_i^2 = \dfrac{34}{12}$ $\sigma^2 = \Sigma p_i x_i^2 - (\mu)^2$ $= \dfrac{34}{12} - \left(\dfrac{18}{12}\right)^2 = \dfrac{17}{6} - \dfrac{9}{4}$ $\dfrac{34 - 27}{12} = \dfrac{7}{12}$ $96\sigma^2 = 96 \times \dfrac{7}{12} = 56$

Question 22

Maths · Binomial Theorem · Numerical

If the constant term in the expansion of $\left(1 + 2x - 3x^3\right)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$ is $p$, then $108p$ is equal to

Answer: 54

Solution

$(1 + 2x - 3x^3)\left(\dfrac{3}{2}x^2 - \dfrac{1}{3x}\right)^9$ General term $m\left(\dfrac{3}{2}x^2 - \dfrac{1}{3x}\right)^9$ $$= {}^9C_r \cdot \frac{3^{9-2r}}{2^{9-r}} \cdot (-1)^r \cdot x^{18-3r}$$ Put $r = 6$ to get coeff. of $x^0 = {}^9C_6 \cdot \dfrac{1}{6^3} \cdot x^0 = \dfrac{7}{18}x^0$ Put $r = 7$ to get coeff. of $x^{-3} = {}^9C_7 \cdot \dfrac{3^{-5}}{2^2} \cdot (-1)^7 \cdot x^{-3}$ $$= -{}^9C_7 \cdot \frac{1}{3^5 \cdot 2^2} \cdot x^{-3} = -\frac{1}{27}x^{-3}$$ $(1 + 2x - 3x^3)\left(\dfrac{7}{18}x^0 - \dfrac{1}{27}x^{-3}\right)$ $$\frac{7}{18} + \frac{27}{27} = \frac{7}{18} + \frac{1}{9} = \frac{7+2}{18} = \frac{9}{18} = \frac{1}{2}$$ $$\therefore\ 108 \cdot \frac{1}{2} = 54$$

Question 23

Maths · Applications of Integrals · Numerical

The area of the region enclosed by the parabolas $y = x^2 - 5x$ and $y = 7x - x^2$ is

Answer: 72

Solution

Given $y = x^2 - 5x$ and $y = 7x - x^2$. The integral is $$\int_0^6 (g(x) - f(x)) \, dx$$ This becomes $$\int_0^6 ((7x - x^2) - (x^2 - 5x)) \, dx$$ Simplifying, we have $$\int_0^6 (12x - 2x^2) \, dx = \left[ 12 \frac{x^2}{2} - 2 \frac{x^3}{3} \right]_0^6$$ Calculating, $$6(6)^2 - \frac{2}{3}(6)^3$$ This results in $$216 - 144 = 72 \, unit^2$$

Question 24

Maths · Permutations and Combinations · Numerical

The number of ways of getting a sum 16 on throwing a dice four times is _____.

Answer: 125

Solution

Given $\left(x^1 + x^2 + \cdots + x^6\right)^4$. We have: $$x^4 \cdot \left(\frac{1 - x^6}{1 - x}\right)^4$$ This simplifies to: $$x^4 \cdot (1 - x^6)^4 \cdot (1 - x)^{-4}$$ Expanding, we get: $$x^4 \left[1 - 4x^6 + 6x^{12} \cdots \right] \left[(1 - x)^{-4}\right]$$ Further simplification gives: $$(x^4 - 4x^{10} + 6x^{16} \cdots) (1 - x)^{-4}$$ This becomes: $$(x^4 - 4x^{10} + 6x^{16}) \left(1 + \binom{15}{12}x^{12} + \binom{9}{6}x^6 \cdots \right)$$ Calculating the coefficients: $$(\binom{15}{12} - 4 \cdot \binom{9}{6} + 6) x^{16}$$ Simplifying further: $$(\binom{15}{3} - 4 \cdot \binom{9}{6} + 6)$$ Finally, we calculate: $$= 35 \times 13 - 6 \times 8 \times 7 + 6$$ $$= 455 - 336 + 6$$ $$= 125$$

Question 25

Maths · Sets · Numerical

If $S = \{ a \in \mathbb{R} : |2a - 1| = 3[a] + 2\{a\} \}$, where $[t]$ denotes the greatest integer less than or equal to $t$ and $\{t\}$ represents the fractional part of $t$, then $72 \sum_{a \in S} a$ is equal to

Answer: 18

Solution

Given $|2a - 1| = 3[a] + 2\{a\}$. $|2a - 1| = [a] + 2a$. Case-1 : $a > \frac{1}{2}$. $$2a - 1 = [a] + 2a$$ $[a] = -1$ $\therefore$ $a \in [-1, 0)$ Reject. Case-2 : $a < \frac{1}{2}$. $$-2a + 1 = [a] + 2a$$ $a = I + f$. $$-2(I + f) + 1 = I + 2I + 2f$$ $I = 0, f = \frac{1}{4}$ $\therefore$ $a = \frac{1}{4}$. Hence $a = \frac{1}{4}$. $$72 \sum_{a \in S} a = 72 \times \frac{1}{4} = 18$$

Question 26

Maths · Continuity and Differentiability · Numerical

Let $f$ be a differentiable function in the interval $(0, \infty)$ such that $f(1) = 1$ and $\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t-x} = 1$ for each $x > 0$. Then $2f(2) + 3f(3)$ is equal to

Answer: 24

Solution

$\lim_{t\to x} \frac{t^2f(x)-x^2f(t)} {t-x} =1$ $\lim_{t\to x} \frac{2t\,f(x)-x^2f'(x)} {1} =1$ $2x\,f(x)-x^2f'(x)=1$ $\frac{dy}{dx}-\frac{2}{x}y$ $=-\frac1{x^2}$ I.F. $=e^{\int-\frac2x\,dx}$ $=\frac1{x^2}$ $\therefore\ \frac{y}{x^2}$ $=\int-\frac1{x^4}\,dx+C$ $\frac{y}{x^2}$ $=\frac1{3x^3}+C$ Put $\($ f(1) = 1 $\)$ $\($ C = $\frac{2}{3}$ $\)$ $\($ y = $\frac{1}{3x}$ + $\frac{2x^2}{3}$ $\)$ $\($ y = $\frac{2x^3 + 1}{3x}$ $\)$ $\($ f(2) = $\frac{17}{6}$ $\)$ $\($ f(3) = $\frac{55}{9}$ $\)$ $\($ 2f(2) + 3f(3) = $\frac{17}{3}$ + $\frac{55}{3}$ = $\frac{72}{3}$ = 24 $\)$

Question 27

Maths · Sequences and Series · Numerical

Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms. Let $A_k = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \ldots + a_{2k-1}^2 - a_{2k}^2$. If $A_3 = -153$, $A_5 = -435$ and $a_1^2 + a_2^2 + a_3^2 = 66$, then $a_{17} - A_7$ is equal to

Answer: 910

Solution

Let $d$ be the common difference. $A_k = -kd[2a + (2k - 1)d]$ $A_3 = -153$ This implies $153 = 13 \, d[2a + 5 \, d]$ Thus, $51 = d[2a + 5 \, d] \ldots (1)$ $A_5 = -435$ This implies $435 = 5 \, d[2a + 9 \, d]$ Thus, $87 = d[2a + 9 \, d]$ Subtracting (1) from (2): $$36 = 4 \, d^2$$ Therefore, $d = 3$, $a = 1$ Finally, $a_{17} - A_7 = 49 - [-7.3[2 + 39]] = 910$

Question 28

Maths · Vector Algebra · Numerical

Let $\vec{a}$ = $\hat{i}$ - 3 $\hat{j}$ + 7 $\hat{k}$ , $\vec{b}$ = 2 $\hat{i}$ - $\hat{j}$ + $\hat{k}$ and $\vec{c}$ be a vector such that ($\vec{a}$ - 2 $\vec{b}$) $\times$ $\vec{c}$ = 3($\vec{c}$ $\times$ $\vec{a}$) . If $\vec{a}$ $\cdot$ $\vec{c}$ = 130 , then $\vec{b}$ $\cdot$ $\vec{c}$ is equal to

Answer: 30

Solution

Given $\left( \vec{a} + 2 \vec{b} \right) \times \vec{c} = 3 \left( \vec{c} \times \vec{a} \right)$. Then $\left( 2 \vec{b} + 4 \vec{a} \right) \times \vec{c} = 0$. Thus, $\vec{c} = \lambda (4 \vec{a} + 2 \vec{b}) = \lambda (8 \hat{i} - 14 \hat{j} + 30 \hat{k})$. Now, $\vec{a} \cdot \vec{c} = 130$. So, $8 \lambda + 42 \lambda + 210 \lambda = 130$. Therefore, $\lambda = \frac{1}{2}$. Thus, $\vec{c} = 4 \hat{i} - 7 \hat{j} + 15 \hat{k}$. Finally, $\vec{b} \cdot \vec{c} = 8 + 7 + 15 = 30$.

Question 29

Maths · Complex Numbers and Quadratic Equations · Fill in the blank

The number of distinct real roots of the equation $|x||x + 2| - 5|x + 1| - 1 = 0$ is

Answer: 3

Solution

Case-1 $x \geq 0$ $$x^2 + 2x - 5x - 5 - 1 = 0$$ $$x^2 - 3x - 6 = 0$$ $$x = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2}$$ One positive root Case-2 $-1 \leq x < 0$ $$-x^2 - 2x - 5x - 5 - 1 = 0$$ $$x^2 + 7x + 6 = 0$$ $$(x + 6)(x + 1) = 0$$ $x = -1$ one root in range Case-3 $-2 \leq x < -1$ $$x^2 - 2x + 5x + 5 - 1 = 0$$ $$x^2 - 3x - 4 = 0$$ $$(x - 4)(x + 1) = 0$$ No root in range Case-4 $x < -2$ $$x^2 + 7x + 4 = 0$$ $$x = \frac{-7 \pm \sqrt{49 - 16}}{2} = \frac{7 \pm \sqrt{33}}{2}$$ one root in range Total number of distinct roots are 3

Question 30

Maths · Conic Sections · Numerical

Suppose $AB$ is a focal chord of the parabola $y^2 = 12x$ of length $l$ and slope $m < \sqrt{3}$. If the distance of the chord $AB$ from the origin is $d$, then $l \, d^2$ is equal to

Answer: 108

Solution

Given $\ell = 4a \cosec^2 \theta$. Also, $\ell = 12 \times \frac{9}{d^2}$. Therefore, $\ell d^2 = 108$.

Physics

Question 31

Physics · Wave Optics · Single correct

Light emerges out of a convex lens when a source of light kept at its focus. The shape of wavefront of the light is :

  1. both spherical and cylindrical
  2. plane
  3. spherical
  4. cylindrical

Answer: (b)

Solution

Light emerges parallel. Therefore, planar wavefront.

Question 32

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

Following gates section is connected in a complete suitable circuit. For which of the following combination, bulb will glow (ON):

  1. A = 0, B = 0, C = 0, D = 1
  2. A = 0, B = 1, C = 1, D = 1
  3. A = 1, B = 0, C = 0, D = 0
  4. A = 1, B = 1, C = 1, D = 0

Answer: (c)

Solution

Bulb will glow if bulb have potential drop on it. One end of bulb must be at high (1) and other must be at low (0). Option (3) satisfy this condition.

Question 33

Physics · Physical World, Units and Measurements · Single correct

If $G$ be the gravitational constant and $u$ be the energy density then which of the following quantity have the dimensions as that of the $\sqrt{uG}$:

  1. pressure gradient per unit mass
  2. Gravitational potential
  3. Energy per unit mass
  4. Force per unit mass

Answer: (d)

Solution

Given $$[\mathrm{uG}] = \left[ \left( \mathrm{M^1 \, L^{-1} \, T^{-2}} \right) \left( \mathrm{M^{-1} \, L^3 \, T^{-2}} \right) \right]$$ Simplifying, we have $$[\mathrm{uG}] = \left[ \mathrm{M^0 \, L^2 \, T^{-4}} \right]$$ Taking the square root, $$[\sqrt{\mathrm{uG}}] = \left[ \mathrm{L^1 \, T^{-2}} \right]$$

Question 34

Physics · Mechanical Properties of Fluids · Single correct

Given below are two statements: Statement I: When a capillary tube is dipped into a liquid, the liquid neither rises nor falls in the capillary. The contact angle may be $0^\circ$. Statement II: The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well. In the light of the above statements, choose the correct answer from the options given below.

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

Answer: (c)

Solution

Capillary rise $$h = \frac{2 \, T \, \cos \theta}{\rho g r};$$ If $\theta = 0^\circ$ then rise is non-zero. Therefore, Statement-1 is incorrect.

Question 35

Physics · Dual Nature of Radiation and Matter · Single correct

Given below are two statements: Statement I: Figure shows the variation of stopping potential with frequency $(\nu)$ for the two photosensitive materials $M_1$ and $M_2$. The slope gives value of $\frac{h}{e}$, where $h$ is Planck's constant, $e$ is the charge of electron. Statement II: $M_2$ will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency. 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 correct
  2. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are incorrect
  4. Statement I is correct and Statement II is incorrect

Answer: (d)

Solution

Given $eV_0 = h\nu - \phi$. $$V_0 = \frac{h}{e} \nu - \frac{\phi}{e}$$ $M_2$ material has higher work function, so statement-(II) is incorrect.

Question 36

Physics · Mathematics in Physics · Single correct

The angle between vector $\vec{Q}$ and the resultant of $(2\vec{Q} + 2\vec{P})$ and $(2\vec{Q} - 2\vec{P})$ is:

  1. $\tan^{-1}\left(\frac{(2\vec{Q} - 2\vec{P})}{2\vec{Q} + 2\vec{P}}\right)$
  2. $0^\circ$
  3. $\tan^{-1}(\mathrm{P}/\mathrm{Q})$
  4. $\tan^{-1}(2\mathrm{Q}/\mathrm{P})$

Answer: (b)

Solution

Given $$\vec{R} = (2\vec{Q} + 2\vec{P}) + (2\vec{Q} - 2\vec{P})$$ $$\vec{R} = 4\vec{Q}$$ Angle between $$\vec{Q}$$ and $$\vec{R}$$ is zero.

Question 37

Physics · Electric Charges and Fields · Single correct

In hydrogen like system the ratio of coulombian force and gravitational force between an electron and a proton is in the order of :

  1. $10^{39}$
  2. $10^{29}$
  3. $10^{19}$
  4. $10^{36}$

Answer: (a)

Solution

Given $$F_e = \frac{k Q_1 Q_2}{r^2} = \frac{9 \times 10^9 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{r^2}$$ $$F_g = \frac{G m_1 m_2}{r^2} = \frac{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-27}}{r^2}$$ $$\frac{F_e}{F_g} \cong 0.23 \times 10^{40} \cong 2.3 \times 10^{39}$$ Option (1)

Question 38

Physics · Moving Charges and Magnetism · Single correct

In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero:

  1. outside the cable
  2. inside the outer conductor
  3. inside the inner conductor
  4. in between the two conductors

Answer: (a)

Solution

The integral of the magnetic field $\vec{B}$ over a closed loop $\oint \vec{B} \cdot d\vec{\ell}$ is equal to $\mu_0 I_{enc} = 0$. Therefore, $B = 0$ outside the cable.

Question 39

Physics · Atoms · Single correct

An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is :

  1. E = U
  2. 2E = U
  3. 2E = 3U
  4. E = 2U

Answer: (b)

Solution

Given the force equation, $$F = \frac{k(Ze)(e)}{r^2} = \frac{mv^2}{r}$$ we can find the kinetic energy as $$KE = \frac{1}{2} mv^2 = \frac{1}{2} \frac{K(Ze)(e)}{r}$$ The potential energy is $$PE = -\frac{K(Ze)(e)}{r}$$ The total energy is $$TE = \frac{K(Ze)(e)}{2r} - \frac{K(Ze)(e)}{r} = -\frac{K(Ze)(e)}{2r}$$ Therefore, $$TE = \frac{PE}{2}$$ and $$2TE = PE$$

Question 40

Physics · Kinetic Theory · Single correct

If the collision frequency of hydrogen molecules in a closed chamber at 27°C is $Z$, then the collision frequency of the same system at 127°C is:

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

Answer: (b)

Solution

Assuming mean free path constant. $f \propto V \propto \sqrt{T}$ $$\frac{f_1}{f_2} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{400}}$$ $$f_2 = \sqrt{\frac{4}{3}} = f_1 = \frac{2}{\sqrt{3}} Z$$

Question 41

Physics · System of Particles and Rotational Motion · Single correct

Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis AB as shown in figure is $\sqrt{\frac{8}{x}}$. The value of $x$ is :

  1. 51
  2. 34
  3. 17
  4. 67

Answer: (d)

Solution

Given $I_{sphere} = \frac{2}{3} MR^2 = M k_1^2$. $I_{cylinder} = \frac{1}{12} M \left(4R^2\right) + \frac{1}{4} MR^2 + M(2R)^2$ $$= \frac{67}{12} MR^2 = M k_2^2$$ $$\frac{k_1}{k_2} = \sqrt{\frac{\frac{2}{3} \cdot 12}{67}} = \sqrt{\frac{8}{67}}$$

Question 42

Physics · Electromagnetic Induction · Single correct

Two conducting circular loops A and B are placed in the same plane with their centers coinciding as shown in figure. The mutual inductance between them is :

  1. $\frac{\mu_0 \pi b^2}{2a}$
  2. $\frac{\mu_0}{2\pi} \cdot \frac{b^2}{a}$
  3. $\frac{\mu_0}{2\pi} \cdot \frac{a^2}{b}$
  4. $\frac{\mu_0 \pi a^2}{2b}$

Answer: (d)

Solution

Given $\phi = Mi = BA$. Therefore, Mi = $\frac{\mu_0 i}{2 \, b}$ $\pi$ a^2$. Thus, $M = $\frac{\mu_0 \pi a^2}{2 \, b}$$.

Question 43

Physics · Gravitation · Single correct

Match List-I with List-II. \begin{tabular}{|c|l|c|c|} \hline & \textbf{List-I} & & \textbf{List-II} \\ \hline (A) & Kinetic energy of planet & (I) & $\displaystyle -\frac{GMm}{a}$ \\ \hline (B) & Gravitational potential energy of Sun-planet system & (II) & $\displaystyle \frac{GMm}{2a}$ \\ \hline (C) & Total mechanical energy of planet & (III) & $\displaystyle \frac{Gm}{r}$ \\ \hline (D) & Escape energy at surface of planet for unit mass object & (IV) & $\displaystyle \frac{GMm}{2a}$ \\ \hline \end{tabular} where $a$ = radius of planet orbit, $r$ = radius of planet, $M$ = mass of Sun, $m$ = mass of planet. Choose the correct answer from the options given below:

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

Answer: (a)

Solution

Question 44

Physics · Laws of Motion · Single correct

A wooden block of mass 5 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on the top of the block, the floor yields and the block and the cylinder together go down with an acceleration of 0.1 $\,$ $\mathrm{ms^{-2}}$. The action force of the system on the floor is equal to:

  1. 196 $\mathrm{N}$
  2. 291 $\mathrm{N}$
  3. 294 $\mathrm{N}$
  4. 297 $\mathrm{N}$

Answer: (b)

Solution

Taking $g = 9.8 \, \mathrm{m/s^2}$. The total weight is $30 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 294 \, \mathrm{N}$. Using the equation: $$294 - N = 30 \times 0.1$$ Solving for $N$: $$N = 291$$

Question 45

Physics · Gravitation · Single correct

A simple pendulum doing small oscillations at a place $R$ height above earth surface has time period of $T_1 = 4 \, \mathrm{s}$. $T_2$ would be it's time period if it is brought to a point which is at a height $2R$ from earth surface. Choose the correct relation $[R = radius of earth]$:

  1. $2 \, T_1 = T_2$
  2. $2 \, T_1 = 3 \, T_2$
  3. $T_1 = T_2$
  4. $3 \, T_1 = 2 \, T_2$

Answer: (d)

Solution

Given $$T_1 = 2\pi \sqrt{\frac{\ell}{GM} (2R)^2}$$ $$T_2 = 2\pi \sqrt{\frac{\ell}{GM} (3R)^2}$$ Therefore, $$\frac{T_1}{T_2} = \frac{2}{3}$$

Question 46

Physics · Work, Energy and Power · Single correct

A body of mass 50 kg is lifted to a height of 20 m from the ground in the two different ways as shown in the figures. The ratio of work done against the gravity in both the respective cases, will be:

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

Answer: (d)

Solution

Work done by gravity is independent of path. It depends only on vertical displacement so work done in both cases will be same.

Question 47

Physics · Physical World, Units and Measurements · Single correct

Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as 4.62 $\mathrm{s}$, 4.632 $\mathrm{s}$, 4.6 $\mathrm{s}$ and 4.64 $\mathrm{s}$. The arithmetic mean of these readings in correct significant figure is :

  1. 5 $\mathrm{s}$
  2. 4.623 $\mathrm{s}$
  3. 4.6 $\mathrm{s}$
  4. 4.62 $\mathrm{s}$

Answer: (c)

Solution

Sum of number by considering significant digits. $$sum = 4.6 + 4.6 + 4.6 + 4.6 = 18.4$$ Arithmetic Mean = $$\frac{sum}{4} = \frac{18.4}{4} = 4.6$$

Question 48

Physics · Thermodynamics · Single correct

The heat absorbed by a system in going through the given cyclic process is :

  1. 19.6 J
  2. 61.6 J
  3. 616 J
  4. 431.2 J

Answer: (b)

Solution

Given $\Delta U = 0$ (Cyclic process). $\Delta Q = W =$ area of $P - V$ curve. $$= \pi \times (140 \times 10^3 \, \mathrm{Pa}) \times (140 \times 10^{-6} \, \mathrm{m^3})$$ $\Delta Q = 61.6 \, \mathrm{J}$

Question 49

Physics · Current Electricity · Single correct

In the given figure $R_1 = 10\,\Omega$, $R_2 = 8\,\Omega$, $R_3 = 4\,\Omega$ and $R_4 = 8\,\Omega$. Battery is ideal with emf $12\,\mathrm{V}$. Equivalent resistant of the circuit and current supplied by battery are respectively:

  1. $10.5\,\Omega$ and $1.14\,\mathrm{A}$
  2. $12\,\Omega$ and $1\,\mathrm{A}$
  3. $10.5\,\Omega$ and $1\,\mathrm{A}$
  4. $12\,\Omega$ and $11.4\,\mathrm{A}$

Answer: (b)

Solution

Here $R_2$, $R_3$, $R_4$ are in parallel $$\frac{1}{R_{234}} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$$ $$R_{234} = 2\, \Omega$$ $R_{234}$ is in series with $R_1$ so $$R_{eq} = R_{234} + R_1 = 2 + 10 = 12\, \Omega$$ $$i = \frac{12}{12} = 1\, Amp$$

Question 50

Physics · Electromagnetic Waves · Single correct

An alternating voltage of amplitude 40 $\mathrm{V}$ and frequency 4 $\mathrm{kHz}$ is applied directly across the capacitor of 12 $\mu$$\mathrm{F}$. The maximum displacement current between the plates of the capacitor is nearly:

  1. 10 $\mathrm{A}$
  2. 12 $\mathrm{A}$
  3. 8 $\mathrm{A}$
  4. 13 $\mathrm{A}$

Answer: (b)

Solution

Displacement current is same as conduction current in capacitor. $$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$$ $$= \frac{1}{2\pi \times 4 \times 10^3 \times 12 \times 10^{-6}} = 3.317 \, \Omega$$ $$I = \frac{V}{X_C} = \frac{40}{3.317} = 12 \, \mathrm{A}$$

Question 51

Physics · Wave Optics · Numerical

In Young's double slit experiment, carried out with light of wavelength $5000 \, \mathrm{\AA}$, the distance between the slits is $0.3 \, \mathrm{mm}$ and the screen is at $200 \, \mathrm{cm}$ from the slits. The central maximum is at $x = 0 \, \mathrm{cm}$. The value of $x$ for third maxima is _____ mm.

Answer: 10

Solution

Given $\($ $\beta$ = $\frac{\lambda D}{d}$ = $\frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}}$ = $\frac{10 \times 10^{-3}}{3}$ $\)$ m. For 3rd maxima $\($ y_3 = 3 $\beta$ = 10 $\times$ $10^{-3} )$ m = 10 mm.

Question 52

Physics · Moving Charges and Magnetism · Numerical

A 2 A current carrying straight metal wire of resistance 1 $\Omega$, resistivity $2 \times 10^{-6} \, \Omega \mathrm{m}$, area of cross-section $10 \, \mathrm{mm}^2$ and mass $500 \, \mathrm{g}$ is suspended horizontally in mid air by applying a uniform magnetic field $\vec{B}$. The magnitude of $B$ is _____ $\times 10^{-1} \, \mathrm{T}$ (given, $g = 10 \, \mathrm{m/s}^2$).

Answer: 5

Solution

Given $R = \frac{\rho \ell}{A}$, we have $$\frac{2 \times 10^{-6} \times \ell}{10^{-5}} = 1 \Rightarrow \ell = 5$$ $$mg = Bi \ell$$ $$B = \frac{mg}{i \ell} = \frac{5}{2 \times 5} = 0.5 = 5 \times 10^{-1} Tesla$$

Question 53

Physics · Electrostatic Potential and Capacitance · Numerical

The electric field between the two parallel plates of a capacitor of $1.5 \, \mu \mathrm{F}$ capacitance drops to one third of its initial value in $6.6 \, \mu \mathrm{s}$ when the plates are connected by a thin wire. The resistance of this wire is _____ $\Omega$. (Given, $\log 3 = 1.1$)

Answer: 4

Solution

Given $E = \frac{E_0}{3}$, it follows that $V = \frac{V_0}{3}$. We have $\frac{V_0}{3} = V_0 e^{-\frac{t}{\tau}}$. Solving for $t$, we get $t = \tau \ln 3$. Given $6.6 \times 10^{-6} = R (1.5 \times 10^{-6}) (1.1)$, we find $R = \frac{6}{1.5} = 4 \, \Omega$.

Question 54

Physics · Laws of Motion · Numerical

Three blocks $M_1, M_2, M_3$ having masses $4 \, \mathrm{kg}, 6 \, \mathrm{kg}$ and $10 \, \mathrm{kg}$ respectively are hanging from a smooth pully using rope $1, 2$ and $3$ as shown in figure. The tension in the rope $1, T_1$ when they are moving upward with acceleration of $2 \, \mathrm{ms^{-2}}$ is $\mathrm{N}$ ( if $g = 10 \, \mathrm{m/s^2}$ ).

Answer: 240

Solution

FBD of $M_1$: $$T_1 - 200 = (4 + 6 + 10) \times 2$$ Therefore, $T_1 = 240$

Question 55

Physics · Mechanical Properties of Solids · Numerical

The density and breaking stress of a wire are $6 \times 10^4 \, \mathrm{kg/m^3}$ and $1.2 \times 10^8 \, \mathrm{N/m^2}$ respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity $g$ is $\frac{1}{3}$ of the value on the surface of earth. The maximum length of the wire with breaking is _____ m (take, $g = 10 \, \mathrm{m/s^2}$).

Answer: 600

Solution

Given $T = mg$. The stress $\sigma = \frac{T}{A} = \frac{mg}{A}$. The expression becomes $\frac{(\sigma A \ell) g}{A}$. Thus, $\ell = \frac{\sigma}{\rho g} = \frac{1.2 \times 10^8 \times 3}{6 \times 10^4 \times 10} = 600$.

Question 56

Physics · Motion in a Straight Line · Numerical

A body moves on a frictionless plane starting from rest. If $S_n$ is distance moved between $t = n - 1$ and $t = n$ and $S_{n-1}$ is distance moved between $t = n - 2$ and $t = n - 1$, then the ratio $\frac{S_{n-1}}{S_n}$ is $\left(1 - \frac{2}{x}\right)$ for $n = 10$. The value of $x$ is $\ldots$

Answer: 19

Solution

Question 57

Physics · Nuclei · Numerical

If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is _____ $\times 10^{-2} \, \mathrm{MeV}$. (Given $1 \, \mathrm{u} = 931 \, \mathrm{MeV}/c^2$, atomic mass of helium $= 4.002603 \, \mathrm{u}$ )

Answer: 727

Solution

Reaction: $$^4_2\mathrm{He} \longrightarrow ^{12}_6\mathrm{C} + \gamma rays$$ Mass defect $= \Delta m = (3 \, m_{\mathrm{He}} - m_{\mathrm{C}})$ $$= (3 \times 4.002603 - 12) = 0.007890 \, u$$ Energy released $$= 931 \Delta m \, MeV$$ $$= 7.27 \, MeV = 727 \times 10^{-2} \, MeV$$

Question 58

Physics · Alternating Current · Numerical

An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of $20\mu F$ is _____ V.

Answer: 50

Solution

Given $X_L = \omega L = 100 \times 1 = 100 \Omega$. $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 20 \times 10^{-6}} = 500 \Omega$. $$Z = \sqrt{(X_L - X_C)^2 + R^2}$$ $$\sqrt{(100 - 500)^2 + 300^2}$$ $Z = 500 \Omega$. $i_{rms} = \frac{V_{rms}}{Z} = \frac{50}{500} = 0.1 \, A$. RMS voltage across capacitor $V_{rms} = X_C i_{rms} = 500 \times 0.1 = 50 \, V$.

Question 59

Physics · Current Electricity · Numerical

In the experiment to determine the galvanometer resistance by half-deflection method, the plot of $1/\theta$ vs the resistance $(R)$ of the resistance box is shown in the figure. The figure of merit of the galvanometer is ____ $\times 10^{-1}$ $\mathrm{A/ division}$. [The source has emf 2V]

Answer: 5

Solution

Given $i = K \theta$. $$\frac{2}{G + R} = K \theta$$ Therefore, $$\frac{1}{\theta} = \frac{(G + R)K}{2} = R \left( \frac{K}{2} \right) + \frac{KG}{2}$$ The slope is $$\frac{K}{2} = \frac{1}{4} \Rightarrow K = 0.5 = 5 \times 10^{-1} \, \mathrm{A}$$

Question 60

Physics · Electrostatic Potential and Capacitance · Numerical

Three capacitors of capacitances $25\mu F$, $30\mu F$ and $45\mu F$ are connected in parallel to a supply of $100 \, V$. Energy stored in the above combination is $E$. When these capacitors are connected in series to the same supply, the stored energy is $\frac{9}{x}E$. The value of $x$ is _____.

Answer: 86

Solution

In parallel combination: Potential difference is same across all Energy $= \frac{1}{2} (C_1 + C_2 + C_3) V^2$ $$= \frac{1}{2} (25 + 30 + 45) \times (100)^2 \times 10^{-6} = 0.5 = E$$ In series combination: Charge is same on all. $$\frac{1}{C_{equ}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45}$$ $$\frac{1}{C_{equ}} = \frac{(18 + 15 + 10)}{450} = \frac{43}{450} \Rightarrow C_{equ} = \frac{450}{43}$$ Energy $= \frac{Q^2}{2C_1} + \frac{Q^2}{2C_2} + \frac{Q^2}{2C_3}$ $$= \frac{Q^2}{2} \left[ \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \right]$$ $$(V \times C_{equ})^2 \times \frac{1}{C_{equ}} = \frac{V^2 C_{equ}}{2}$$ $$\frac{(100)^2}{2} \times \frac{450}{43} \times 9 \times 10^{-6}$$ $$\Rightarrow \frac{4.5}{86} = \frac{9}{x} = E = \frac{9}{x} \times 0.5 \Rightarrow x = 86$$

Chemistry

Question 61

Chemistry · Some Basic Concepts of Chemistry · Single correct

The incorrect postulates of the Dalton's atomic theory are : (A) Atoms of different elements differ in mass. (B) Matter consists of divisible atoms. ($C$) Compounds are formed when atoms of different element combine in a fixed ratio. (D) All the atoms of given element have different properties including mass. (E) Chemical reactions involve reorganisation of atoms. Choose the correct answer from the options given below :

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

Answer: (b)

Solution

b,d

Question 62

Chemistry · Equilibrium · Single correct

The following reaction occurs in the Blast furnace where iron ore is reduced to iron metal $Fe_2O_3(s)+3CO(g)⇌Fe(l)+3CO_2(g)$ Using the Le-chatelier's principle, predict which one of the following will not disturb the equilibrium.

  1. Addition of $\mathrm{CO_2}$
  2. Removal of $\mathrm{CO_2}$
  3. Addition of $\mathrm{Fe_2O_3}$
  4. Removal of $\mathrm{CO}$

Answer: (c)

Solution

When solid added no effect on equilibrium.

Question 63

Chemistry · Alcohols, Phenols and Ethers · Single correct

Identify compound (Z) in the following reaction sequence.

Answer: (a)

Solution

The reaction starts with chlorobenzene reacting with $\mathrm{NaOH}$ at $623 \, \mathrm{K}$ and $300 \, \mathrm{atm}$ to form sodium phenoxide $(\mathrm{X})$. This is then treated with $\mathrm{HCl}$ to form phenol $(\mathrm{Y})$. Finally, phenol undergoes nitration with concentrated $\mathrm{HNO_3}$ to form 2,4,6-trinitrophenol $(\mathrm{Z})$.

Question 64

Chemistry · Thermodynamics · Single correct

Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason ($R$) Assertion (A) : Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always $-57 \, \mathrm{kJ \, mol^{-1}}$. Reason $(R)$ : Enthalpy of neutralisation is the amount of heat liberated when one mole of $\mathrm{H^+}$ ions furnished by acid combine with one mole of $\mathrm{OH^-}$ ions furnished by base to form one mole of water. In the light of the above statements, choose the correct answer from the options given below.

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

Answer: (c)

Solution

Enthalpy of neutralization of SA & SB is always $-57 \, \mathrm{kJ/mol}$ because strong monoacid gives one mole of $\mathrm{H}^+$ and strong mono base gives one mole of $\mathrm{OH}^-$ which form one mole of water.

Question 65

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

The statement(s) that are correct about the species $\mathrm{O^{2-}}$, $\mathrm{F^-}$, $\mathrm{Na^+}$ and $\mathrm{Mg^{2+}}$. (A) All are isoelectronic (B) All have the same nuclear charge ($C$) $\mathrm{O^{2-}}$ has the largest ionic radii (D) $\mathrm{Mg^{2+}}$ has the smallest ionic radii Choose the most appropriate answer from the options given below:

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

Answer: (c)

Solution

For $\mathrm{O}^{-2}$, $\mathrm{F}^{-}$, $\mathrm{Na}^{+}$, $\mathrm{Mg}^{+2}$, the number of electrons is 10 for each. The ionic radius order is $\mathrm{O}^{-2} > \mathrm{F}^{-} > \mathrm{Na}^{+} > \mathrm{Mg}^{+2}$. The effective nuclear charge (Zeff) order is $\mathrm{O}^{-2} < \mathrm{F}^{-} < \mathrm{Na}^{+} < \mathrm{Mg}^{+2}$.

Question 66

Chemistry · Alcohols, Phenols and Ethers · Single correct

For the compounds: The increasing order of boiling point is: Choose the correct answer from the options given below:

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

Answer: (b)

Solution

Question 67

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

Given below are two statements: Statement I: In group 13, the stability of +1 oxidation state increases down the group. Statement II: The atomic size of gallium is greater than that of aluminium. 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 correct
  2. Statement I is incorrect but Statement II is correct
  3. Both Statement I and Statement II are incorrect
  4. Statement I is correct but Statement II is incorrect

Answer: (d)

Solution

Statement I: Number of $d$ and $f$ electrons increases down the group and due to poor shielding of $d$ and $f$ electrons, stability of lower oxidation states increases down the group. Statement II: The atomic size of aluminium is greater than that of gallium.

Question 68

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

Number of $\sigma$ and $\pi$ bonds present in ethylene molecule is respectively:

  1. 4 and 1
  2. 5 and 2
  3. 3 and 1
  4. 5 and 1

Answer: (d)

Solution

Ethyene is it has 5 $\sigma$ bonds and 1 $\pi$ bond.

Question 69

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

Identify 'A' in the following reaction:

Answer: (d)

Solution

The reaction shown is a Wolff-Kishner reduction. The carbonyl group is reduced to a methylene group using hydrazine $\mathrm{(N_2H_4)}$ and ethylene glycol with potassium hydroxide $\mathrm{(KOH)}$. The product is an alkane.

Question 70

Chemistry · Electrochemistry · Single correct

The reaction at cathode in the cells commonly used in clocks involves.

  1. reduction of Mn from +7 to +2
  2. reduction of Mn from +4 to +3
  3. oxidation of Mn from +3 to +4
  4. oxidation of Mn from +2 to +7

Answer: (b)

Solution

In the cathode reaction manganese (Mn) is reduced from the $+4$ oxidation state to the $+3$ state.

Question 71

Chemistry · Co-ordination Compounds · Single correct

Which one of the following complexes will exhibit the least paramagnetic behaviour? [Atomic number, Cr = 24, Mn = 25, Fe = 26, Co = 27]

  1. $[\mathrm{Cr(H_2O)_6}]^{2+}$
  2. $[\mathrm{Fe(H_2O)_6}]^{2+}$
  3. $[\mathrm{Co(H_2O)_6}]^{2+}$
  4. $[\mathrm{Mn(H_2O)_6}]^{2+}$

Answer: (c)

Solution

The table shows the number of unpaired electrons and the magnetic moment $\mu = \sqrt{n(n+2)} \, \mathrm{B.M.}$ for different complexes. For $[\mathrm{Co(H_2O)_6}]^{2+}$, the number of unpaired electrons is 3, and $\mu = 3.87$. For $[\mathrm{Fe(H_2O)_6}]^{2+}$, the number of unpaired electrons is 4, and $\mu = 4.89$. For $[\mathrm{Mn(H_2O)_6}]^{2+}$, the number of unpaired electrons is 5, and $\mu = 5.92$. For $[\mathrm{Cr(H_2O)_6}]^{2+}$, the number of unpaired electrons is 4, and $\mu = 4.89$. The complex with the least paramagnetic behavior is $[\mathrm{Co(H_2O)_6}]^{2+}$.

Question 72

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

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason $(R)$. Assertion (A) : Cis form of alkene is found to be more polar than the trans form. Reason $(R)$: Dipole moment of trans isomer of 2-butene is zero. In the light of the above statements, choose the correct answer from the options given below :

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

Answer: (b)

Solution

Dipole moment is a vector quantity and for compound net dipole moment is the vector sum of all dipoles hence dipole moment of cis form is greater than trans form.

Question 73

Chemistry · Hydrocarbons · Single correct

Given below are two statements: 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. Both Statement I and Statement II are correct
  4. Both Statement I and Statement II are incorrect

Answer: (a)

Solution

In nitration of benzene concentrated $\mathrm{H_2SO_4}$ and $\mathrm{HNO_3}$ is used as reagent which generates electrophile $\mathrm{N_2}$ in following steps: $$\mathrm{H_2SO_4 + HNO_3 \rightleftharpoons HSO_4^- + H-O-NO_2^+}$$ $$\mathrm{HSO_4^- + H_2O + NO_2^+}$$ Lewis acids can promote the formation of electrophiles not Lewis base.

Question 74

Chemistry · Co-ordination Compounds · Single correct

The correct order of ligands arranged in increasing field strength.

  1. $\mathrm{F}^- < \mathrm{Br}^- < \mathrm{I}^- < \mathrm{NH}_3$
  2. $\mathrm{Br}^- < \mathrm{F}^- < \mathrm{H}_2\mathrm{O} < \mathrm{NH}_3$
  3. $\mathrm{H}_2\mathrm{O} < -\mathrm{OH} < \mathrm{CN}^- < \mathrm{NH}_3$
  4. $\mathrm{Cl}^- < -\mathrm{OH} < \mathrm{Br}^- < \mathrm{CN}^-$

Answer: (b)

Solution

Experimental order $\mathrm{Br}^- < \mathrm{F}^- < \mathrm{H_2O} < \mathrm{NH_3}$

Question 75

Chemistry · Biomolecules · Single correct

Which of the following gives a positive test with ninhydrin?

  1. Starch
  2. Egg albumin
  3. Polyvinyl chloride
  4. Cellulose

Answer: (b)

Solution

Ninhydrin test is a test of amino acids. Egg albumin contains protein which is a natural polymer of amino acids which will show positive ninhydrin test.

Question 76

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

The metal that shows highest and maximum number of oxidation state is :

  1. Fe
  2. Mn
  3. Co
  4. Ti

Answer: (b)

Solution

Mn shows highest oxidation state $\left( \mathrm{Mn}^{+7} \right)$ in 3 d series metals.

Question 77

Chemistry · Some Basic Concepts of Chemistry · Single correct

An organic compound has 42.1$\%$ carbon, 6.4$\%$ hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is :

  1. C_{11}H_{18}O_{12}
  2. C_{12}H_{20}O_{12}
  3. C_{12}H_{22}O_{11}
  4. C_{14}H_{20}O_{10}

Answer: (c)

Solution

Only $\mathrm{C_{12}H_{22}O_{11}}$ has $42.1\%$ carbon, $6.4\%$ hydrogen, and $51.5\%$ oxygen.

Question 78

Chemistry · Alcohols, Phenols and Ethers · Single correct

Given below are two statement: Statements I : Bromination of phenol in solvent with low polarity such as CHCl$_3$ or CS$_2$ requires Lewis acid catalyst. Statements II : The Lewis acid catalyst polarises the bromine to generate Br$^+$. 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. Statement I is true but Statement II is false
  3. Statement I is false but Statement II is true
  4. Both Statement I and Statement II are false

Answer: (c)

Solution

Phenol is a highly activated compound which can undergo bromination directly with Bromine without any Lewis acid.

Question 79

Chemistry · Electrochemistry · Single correct

Molar ionic conductivities of divalent cation and anion are $57 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$ and $73 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$ respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be :

  1. $187 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$
  2. $260 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$
  3. $130 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$
  4. $65 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$

Answer: (c)

Solution

Given $\Lambda_C^{+2} = 57 \, \mathrm{Scm^2 \, mol^{-1}}$ and $\Lambda_A^{+2} = 73 \, \mathrm{Scm^2 \, mol^{-1}}$. The solution is calculated as follows: $$\Lambda_{Solution} = \Lambda_C^{+2} + \Lambda_A^{-2}$$ $$= 57 + 73 = 130$$

Question 80

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

The number of neutrons present in the more abundant isotope of boron is '$x$'. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is '$y$'. The value of $x + y$ is _______

  1. 3
  2. 9
  3. 4
  4. 6

Answer: (b)

Solution

More abundant isotope = $\mathrm{B^{11}}$ [Number of neutrons = 6] $x = 6$ $\mathrm{B + O_2 \rightarrow B_2O_3}$ Oxidation state of B in $\mathrm{B_2O_3} = +3$ So, $y = 3$ Hence $x + y = 9$

Question 81

Chemistry · Structure of Atom · Numerical

The value of Rydberg constant ($R_H$) is $2.18 \times 10^{-18} \, \mathrm{J}$. The velocity of electron having mass $9.1 \times 10^{-31} \, \mathrm{kg}$ in Bohr's first orbit of hydrogen atom = \text{_____} $\times 10^5 \, \mathrm{ms}^{-1}$ (nearest integer).

Answer: 22

Solution

Given $$V = 2.18 \times 10^6 \times \frac{Z}{n}$$ Substituting the values, we have $$= 21.8 \times 10^5 \times \frac{1}{1} \approx 22 \times 10^5 (nearest)$$

Question 82

Chemistry · Analytical Chemistry · Numerical

In a borax bead test under hot condition, a metal salt (one from the given) is heated at point B of the flame, resulted in green colour salt bead. The spin-only magnetic moment value of the salt is _______ BM (Nearest integer) [Given atomic number of Cu = 29, Ni = 28, Mn = 25, Fe = 26]

Answer: 6

Solution

Fe^{+3} will give green coloured bead when heated at point B. Number of unpaired e^{-} in Fe^{+3} = 5. $$\mu = 5.92$$ Nearest integer = 6

Question 83

Chemistry · Thermodynamics · Fill in the blank

The heat of combustion of solid benzoic acid at constant volume is $-321.30 \, \mathrm{kJ}$ at $27^\circ \mathrm{C}$. The heat of combustion at constant pressure is $(-321.30 - xR) \, \mathrm{kJ}$, the value of $x$ is \text{_____}.

Answer: 150

Solution

For the reaction: $C_6H_5COOH$(S) + $\frac{15}{2}$ $O_2$(g) $\rightarrow$ 7 $CO_2$(g) + 3$H_2O$($\ell$) The change in enthalpy is given by: $$\Delta H = \Delta U + \Delta n_g RT$$ Substituting the values: $$= -321.30 - \frac{1}{2} \frac{R}{100} \times 300$$ $$= (-321.30 - 150R) kJ$$

Question 84

Chemistry · Alcohols, Phenols and Ethers · Numerical

Consider the given chemical reaction sequence: Total sum of oxygen atoms in Product A and Product B are _______

Answer: 14

Solution

Picric acid is prepared by treating phenol first with concentrated sulphuric acid which converts it to phenol-2,4-disulphonic acid and then with concentrated nitric acid to get 2, 4, 6 trinitrophenol.

Question 85

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

The spin-only magnetic moment value of the ion among $\mathrm{Ti}^{2+}$, $\mathrm{V}^{2+}$, $\mathrm{Co}^{3+}$ and $\mathrm{Cr}^{2+}$, that acts as strong oxidising agent in aqueous solution is ______ BM (Near integer). (Given atomic numbers : Ti : 22, V : 23, Cr : 24, Co : 27)

Answer: 5

Solution

Strong oxidising agent is $\mathrm{Co^{+3}}$. The number of unpaired electrons in $\mathrm{Co^{+3}} \left[ 3d^6 \right]$ is $4$. Hence $\mu = \sqrt{n(n+2)} = \sqrt{24} \mathrm{BM}$. Nearest integer is $5$.

Question 86

Chemistry · Chemical Kinetics and Nuclear Chemistry · Fill in the blank

During Kinetic study of reaction $2A + B \rightarrow C + D$, the following results were obtained: \begin{tabular}{|c|c|c|c|} \hline & A $[\mathrm{M}]$ & B $[\mathrm{M}]$ & Initial rate of formation of D \\ \hline I & 0.1 & 0.1 & $6.0 \times 10^{-3}$ \\ \hline II & 0.3 & 0.2 & $7.20 \times 10^{-2}$ \\ \hline III & 0.3 & 0.4 & $2.88 \times 10^{-1}$ \\ \hline IV & 0.4 & 0.1 & $2.40 \times 10^{-2}$ \\ \hline \end{tabular} Based on above data, overall order of the reaction is \_\_\_\_\_

Answer: 3

Solution

Given $r = K[A]^x[B]^y$. (I) $6 \times 10^{-3} = K[0.1]^x[0.1]^y$ (IV) $2.4 \times 10^{-2} = K[0.4]^x[0.1]^y$ Dividing (IV) by (I): $$4 = (4)^x$$ Thus, $x = 1$. Now, $r = K[A]^x[B]^y$. (III) $2.88 \times 10^{-1} = K[0.3]^x[0.4]^y$ (II) $7.2 \times 10^{-2} = K[0.3]^x[0.2]^y$ Dividing (III) by (II): $$4 = 2^y$$ Thus, $y = 2$. Overall order $= x + y = 1 + 2 = 3$

Question 87

Chemistry · Solutions · Numerical

An artificial cell is made by encapsulating $0.2\mathrm{M}$ glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a $0.05\mathrm{M}$ solution of $\mathrm{NaCl}$ at $300\mathrm{K}$ is ______ $\times 10^{-1}$ bar. (nearest integer). [Given : $R = 0.083\ \mathrm{Lbar}\ mol^{-1}\ K^{-1}$ Assume complete dissociation of $\mathrm{NaCl}$.

Answer: 25

Solution

NaCl dissociates into $\mathrm{Na^+}$ and $\mathrm{Cl^-}$. The concentrations are $0.05 \, \mathrm{M}$ for both ions. The total concentration $C_1$ is $0.05 + 0.05 = 0.1 \, \mathrm{M}$ (NaCl). The concentration $C_2$ is $0.2 \, \mathrm{M}$ (glucose). The osmotic pressure $\pi$ is given by $\pi = (C_2 - C_1) RT$. Substituting the values, we have $$\pi = (0.2 - 0.1) \times 0.083 \times 300$$ which equals $2.49 \, \mathrm{bar}$ or $24.9 \times 10^{-1} \, \mathrm{bar}$.

Question 88

Chemistry · Haloalkanes and Haloarenes · Numerical

The number of halobenzenes from the following that can be prepared by Sandmeyer's reaction is _______

Answer: 2

Solution

In Sandmeyer reaction only bromobenzene and chlorobenzene are prepared.

Question 89

Chemistry · Chemical Bonding and Molecular Structure · Numerical

In the lewis dot structure for $\mathrm{NO}_2^-$, total number of valence electrons around nitrogen is

Answer: 8

Solution

Number of valence $e^-$ around N-atom $= 8$

Question 90

Chemistry · Amines · Numerical

9.3 $\mathrm{g}$ of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product 'P'. The mass of product 'P' obtained is 26.4 $\mathrm{g}$. The percentage yield is \text{_____}\%.

Answer: 80

Solution

93 g of aniline produces 330 g of 2, 4, 6-tribromoaniline. Hence 9.3 g of aniline should produce 33 g of 2, 4, 6-tribromoaniline. Hence percentage yield $$\frac{26.4 \times 100}{33} = 80\%$$