JEE Main 30 January 2024 Shift 1 question paper with solutions

JEE Main 30 January 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 · Straight Lines and Pair of Straight Lines · Single correct

A line passing through the point $A(9, 0)$ makes an angle of $30°$ with the positive direction of $x$-axis. If this line is rotated about $A$ through an angle of $15°$ in the clockwise direction, then its equation in the new position is

  1. $\frac{y}{\sqrt{3}-2} + x = 9$
  2. $\frac{x}{\sqrt{3}-2} + y = 9$
  3. $\frac{x}{\sqrt{3}+2} + y = 9$
  4. $\frac{y}{\sqrt{3}+2} + x = 9$

Answer: (a)

Solution

Equation: $y - 0 = \tan 15^\circ (x - 9)$ implies $y = (2 - \sqrt{3})(x - 9)$

Question 2

Maths · Sequences and Series · Single correct

Let $S_a$ denote the sum of first $n$ terms an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145$, then $S_{15} - S_5$ is

  1. 395
  2. 390
  3. 405
  4. 410

Answer: (a)

Solution

Given $S_{20} = \frac{20}{2} [2a + 19d] = 790$. $2a + 19d = 79 \ldots (i)$ $S_{10} = \frac{10}{2} [2a + 9d] = 145$. $2a + 9d = 29 \ldots (2)$ From (1) and (2) $a = -8$, $d = 5$. $S_{15} - S_{5} = \frac{15}{2} [2a + 14d] - \frac{5}{2} [2a + 4d]$ $$= \frac{15}{2} [-16 + 70] - \frac{5}{2} [-16 + 20]$$ $$= 405 - 10$$ $$= 395$$

Question 3

Maths · Complex Numbers and Quadratic Equations · Single correct

If $z = x + iy$, $xy \neq 0$, satisfies the equation $z^2 + i\bar{z} = 0$, then $|z^2|$ is equal to:

  1. 9
  2. 1
  3. 4
  4. $\frac{1}{4}$

Answer: (b)

Solution

Given $z^2 = -\overline{iz}$. $|z^2| = |iz|$ $|z^2| = |z|$ $|z|^2 - |z| = 0$ $|z|(|z| - 1) = 0$ $|z| = 0$ (not acceptable) Therefore, $|z| = 1$ Thus, $|z|^2 = 1$

Question 4

Maths · Vector Algebra · Single correct

Let $\vec{a}$ = $a_1$ $\hat{i}$ + $a_2$ $\hat{j}$ + $a_3$ $\hat{k}$ and $\vec{b}$ = $b_1$ $\hat{i}$ + $b_2$ $\hat{j}$ + $b_3$ $\hat{k}$ be two vectors such that |$\vec{a}$| = 1; $\vec{a}$ $\cdot$ $\vec{b}$ = 2 and |$\vec{b}$| = 4. If $\vec{c}$ = 2($\vec{a}$ $\times$ $\vec{b}$) - 3$\vec{b}$, then the angle between $\vec{b}$ and $\vec{c}$ is equal to :

  1. $\cos^{-1}$ ( $\frac{2}{\sqrt{3}}$ )
  2. $\cos^{-1}$ ( -$\frac{1}{\sqrt{3}}$ )
  3. $\cos^{-1}$ ( -$\frac{\sqrt{3}}{2}$ )
  4. $\cos^{-1}$ ( $\frac{2}{3}$ )

Answer: (c)

Solution

Given $|\vec{a}| = 1$, $|\vec{b}| = 4$, $\vec{a} \cdot \vec{b} = 2$. $$\vec{c} = 2(\vec{a} \times \vec{b}) - 3 \vec{b}$$ Dot product with $\vec{a}$ on both sides $$\vec{c} \cdot \vec{a} = -6 \ldots (1)$$ Dot product with $\vec{b}$ on both sides $$\vec{b} \cdot \vec{c} = -48 \ldots (2)$$ $$\vec{c} \cdot \vec{c} = 4 \left[ |\vec{a} \times \vec{b}|^2 + 9 |\vec{b}|^2 \right]$$ $$|\vec{c}|^2 = 4 \left[ |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 + 9 \right] |\vec{b}|^2$$ $$|\vec{c}|^2 = 4 \left[ (1)(4)^2 - (4) \right] + 9(16)$$ $$|\vec{c}|^2 = 4[12] + 144$$ $$|\vec{c}|^2 = 48 + 144$$ $$|\vec{c}|^2 = 192$$ $$\therefore \cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|}$$ $$\therefore \cos \theta = \frac{-48}{\sqrt{192} \cdot 4}$$ $$\therefore \cos \theta = \frac{-48}{8\sqrt{3} \cdot 4}$$ $$\therefore \cos \theta = \frac{-3}{2\sqrt{3}}$$ $$\therefore \cos \theta = \frac{-\sqrt{3}}{2} \Rightarrow \theta = \cos^{-1} \left( \frac{-\sqrt{3}}{2} \right)$$

Question 5

Maths · Applications of Derivatives · Single correct

The maximum area of a triangle whose one vertex is at $(0, 0)$ and the other two vertices lie on the curve $y = -2x^2 + 54$ at points $(x, y)$ and $(-x, y)$ where $y > 0$ is:

  1. 88
  2. 122
  3. 92
  4. 108

Answer: (d)

Solution

Area of $\Delta$ is given by the determinant: $$\frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ x & y & 1 \\ -x & y & 1 \end{vmatrix}$$ This simplifies to: $$\Rightarrow \left| \frac{1}{2} (xy + xy) \right| = |xy|$$ The area of $\Delta$ is $|xy|$, which is $|x(-2x^2 + 54)|$. Differentiating with respect to $x$: $$\frac{d(\Delta)}{dx} = \left| (-6x^2 + 54) \right| \Rightarrow \frac{d\Delta}{dx} = 0 at x = 3$$ Substituting $x = 3$: $$Area = 3(-2 \times 9 + 54) = 108$$

Question 6

Maths · Limits and Derivatives · Single correct

The value of $\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n^3}{(n^2+k^2)(n^2+3k^2)}$ is:

  1. $\frac{(2\sqrt{3}+3)\pi}{24}$
  2. $\frac{13\pi}{8(4\sqrt{3}+3)}$
  3. $\frac{13(2\sqrt{3}-3)\pi}{8}$
  4. $\frac{\pi}{8(2\sqrt{3}+3)}$

Answer: (b)

Solution

The limit is given by $$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n^3}{n^4 \left(1 + \frac{k^2}{n^2}\right) \left(1 + \frac{3k^2}{n^2}\right)} $$ which simplifies to $$ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{n^3}{\left(1 + \frac{k^2}{n^2}\right) \left(1 + \frac{3k^2}{n^2}\right)} $$ This is equivalent to the integral $$ \int_{0}^{1} \frac{dx}{3 \left(1 + x^2\right) \left(\frac{1}{3} + x^2\right)} $$ which can be rewritten as $$ \int_{0}^{1} \frac{1}{\frac{1}{3}} \times \frac{3}{2} \left(\frac{x^2 + 1}{\left(1 + x^2\right) \left(x^2 + \frac{1}{3}\right)}\right) dx $$ This simplifies to $$ \frac{1}{2} \int_{0}^{1} \left[ \frac{1}{x^2 + \left(\frac{1}{\sqrt{3}}\right)^2} - \frac{1}{1 + x^2} \right] dx $$ Evaluating the integral gives $$ \frac{1}{2} \left[ \sqrt{3} \tan^{-1}(\sqrt{3}x) \right]_{0}^{1} - \frac{1}{2} \left(\tan^{-1} x \right)_{0}^{1} $$ which results in $$ \frac{\sqrt{3}}{2} \left(\frac{\pi}{3}\right) - \frac{1}{2} \left(\frac{\pi}{4}\right) = \frac{\pi}{2\sqrt{3}} - \frac{\pi}{8} $$ Finally, this simplifies to $$ \frac{13\pi}{8 \cdot (4\sqrt{3} + 3)} $$

Question 7

Maths · Continuity and Differentiability · Single correct

Let $g:\mathbb{R}\to\mathbb{R}$ be a non constant twice differentiable such that $g'\left(\frac{1}{2}\right)=g'\left(\frac{3}{2}\right)$. If a real valued function $f$ is defined as $f(x)=\frac{1}{2}\left[g(x)+g(2-x)\right]$, then

  1. $f''(x) = 0$ for atleast two $x$ in $(0, 2)$
  2. $f''(x) = 0$ for exactly one $x$ in $(0, 1)$
  3. $f''(x) = 0$ for no $x$ in $(0, 1)$
  4. $f' \left( \frac{3}{2} \right) + f' \left( \frac{1}{2} \right) = 1$

Answer: (a)

Solution

Given $f'(x) = \frac{g'(x) - g'(2-x)}{2}$, $f'\left(\frac{3}{2}\right) = \frac{g'\left(\frac{3}{2}\right) - g'\left(\frac{1}{2}\right)}{2} = 0$. Also $f'\left(\frac{1}{2}\right) = \frac{g'\left(\frac{1}{2}\right) - g'\left(\frac{3}{2}\right)}{2} = 0$, $f'\left(\frac{1}{2}\right) = 0$. Therefore, $f'\left(\frac{3}{2}\right) = f'\left(\frac{1}{2}\right) = 0$. This implies roots in $\left(\frac{1}{2}, 1\right)$ and $\left(1, \frac{3}{2}\right)$. Therefore, $f''(x)$ is zero at least twice in $\left(\frac{1}{2}, \frac{3}{2}\right)$.

Question 8

Maths · Applications of Integrals · Single correct

The area (in square units) of the region bounded by the parabola $y^2 = 4(x - 2)$ and the line $y = 2x - 8$

  1. 8
  2. 9
  3. 6
  4. 7

Answer: (b)

Solution

Let $X = x - 2$. Given $y^2 = 4x$, $y = 2(x + 2) - 8$. Simplifying, we have $y^2 = 4x$, $y = 2x - 4$. The area $A$ is given by the integral $$A = \int_{-2}^{4} \frac{y^2}{4} - \frac{y + 4}{2} \, dy.$$ The result of the integration is $9$.

Question 9

Maths · Differential Equations · Single correct

Let $y = y(x)$ be the solution of the differential equation $\sec x \, dy + \{2(1-x) \tan x + x(2-x)\} \, dx = 0$ such that $y(0) = 2$. Then $y(2)$ is equal to:

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

Answer: (a)

Solution

Given $\($ $\frac{dy}{dx}$ = 2(x-1) $\sin$ x + (x^2 - 2x) $\cos$ x $\)$. Now both side integrate $$y(x) = \int 2(x-1) \sin x \, dx + \left[ (x^2 - 2x)(\sin x) - \int (2x-2) \sin x \, dx \right]$$ $\($ y(x) = (x^2 - 2x) $\sin$ x + $\lambda$ $\)$ $\($ y(0) = 0 + $\lambda$ $\Rightarrow$ 2 = $\lambda$ $\)$ $\($ y(x) = (x^2 - 2x) $\sin$ x + 2 $\)$ $\($ y(2) = 2 $\)$

Question 10

Maths · Three Dimensional Geometry · Single correct

Let $(\alpha, \beta, \gamma)$ be the foot of perpendicular from the point $(1, 2, 3)$ on the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. Then $19(\alpha + \beta + \gamma)$ is equal to:

  1. 102
  2. 101
  3. 99
  4. 100

Answer: (b)

Solution

Let foot $P(5k - 3, 2k + 1, 3k - 4)$. DR's $\rightarrow$ AP: $5k - 4, 2k - 1, 3k - 7$. DR's $\rightarrow$ Line: $5, 2, 3$. Condition of perpendicular lines $(25k - 20) + (4k - 2) + (9k - 21) = 0$. Then $k = \frac{43}{38}$. Then $19(\alpha + \beta + \gamma) = 101$.

Question 11

Maths · Probability · Single correct

Two integers x and y are chosen with replacement from the set $\{0,1,2,3,\ldots,10\}$. Then the probability that $$|x - y| > 5$$ is:

  1. $\frac{30}{121}$
  2. $\frac{62}{121}$
  3. $\frac{60}{121}$
  4. $\frac{31}{121}$

Answer: (a)

Solution

If $x=0,\ y=6,7,8,9,10$ If $x=1,\ y=7,8,9,10$ If $x=2,\ y=8,9,10$ If $x=3,\ y=9,10$ If $x=4,\ y=10$ If $x=5,\ y=$ no possible value Total possible ways $=(5+4+3+2+1)\times2$ $=30$ Required probability $=\frac{30}{11\times11}$ $=\frac{30}{121}$

Question 12

Maths · Inverse Trigonometric Functions · Single correct

If the domain of the function $f(x) = \cos^{-1}\left(\frac{2-|x|}{4}\right) + \left(\log_e(3-x)\right)^{-1}$ is $[-\alpha, \beta) - \{y\}$, then $\alpha + \beta + \gamma$ is equal to:

  1. 12
  2. 9
  3. 11
  4. 8

Answer: (c)

Solution

Given $$-1 \leq \left| \frac{2 - |x|}{4} \right| \leq 1$$ This implies $$\left| \frac{2 - |x|}{4} \right| \leq 1$$ $$-4 \leq 2 - |x| \leq 4$$ $$-6 \leq -|x| \leq 2$$ $$-2 \leq |x| \leq 6$$ $$|x| \leq 6$$ Thus, $$x \in [-6, 6] \ldots (1)$$ Now, $3 - x \neq 1$ And $x \neq 2 \ldots (2)$ and $3 - x > 0$ $$x < 3 \ldots (3)$$ From (1), (2) and (3) $$\Rightarrow x \in [-6, 3) - \{2\}$$ $$\alpha = 6$$ $$\beta = 3$$ $$\gamma = 2$$ $$\alpha + \beta + \gamma = 11$$

Question 13

Maths · Determinants · Single correct

Consider the system of linear equation $x + y + z = 4\mu, x + 2y + 2\lambda z = 10\mu, x + 3y + 4\lambda^2 z = \mu^2 + 15$, where $\lambda, \mu \in \mathbb{R}$. Which one of the following statements is NOT correct?

  1. The system has unique solution if $\lambda \neq \frac{1}{2}$ and $\mu \neq 1, 15$
  2. The system is inconsistent if $\lambda = \frac{1}{2}$ and $\mu \neq 1$
  3. The system has infinite number of solutions if $\lambda = \frac{1}{2}$ and $\mu = 15$
  4. The system is consistent if $\lambda \neq \frac{1}{2}$

Answer: (b)

Solution

Given the equations $x + y + z = 4\mu$, $x + 2y + 2\lambda z = 10\mu$, $x + 3y + 4\lambda^2 z = \mu^2 + 15$. The determinant $\Delta$ is given by $$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix} = (2\lambda - 1)^2$$ For a unique solution, $\Delta \neq 0$, $2\lambda - 1 \neq 0$, $\left( \lambda \neq \frac{1}{2} \right)$. Let $\Delta = 0$, $\lambda = \frac{1}{2}$. $$\Delta_y = 0, \Delta_x = \Delta_z = \begin{vmatrix} 4\mu & 1 \\ 10\mu & 2 \\ \mu^2 + 15 & 3 \end{vmatrix} = (\mu - 15)(\mu - 1)$$ For an infinite solution $\lambda = \frac{1}{2}$, $\mu = 1$ or $15$.

Question 14

Maths · Conic Sections · Single correct

If the circles $(x+1)^2 + (y+2)^2 = r^2$ and $x^2 + y^2 - 4x - 4y + 4 = 0$ intersect at exactly two distinct points, then

  1. $5 < r < 9$
  2. $0 < r < 7$
  3. $3 < r < 7$
  4. $\frac{1}{2} < r < 7$

Answer: (c)

Solution

If two circles intersect at two distinct points, then $$|r_1 - r_2| 5$$ $$-5 3 \ldots (2)$$ $$-3 < r < 7 \ldots (1)$$ From (1) and (2) $$3 < r < 7$$

Question 15

Maths · Conic Sections · Single correct

If the length of the minor axis of ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :

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

Answer: (d)

Solution

Given $2b = ae$. $$\frac{b}{a} = \frac{e}{2}$$ $$e = \sqrt{1 - \frac{e^2}{4}}$$ $$e = \frac{2}{\sqrt{5}}$$

Question 16

Maths · Statistics · Single correct

Let $M$ denote the median of the following frequency distribution. \begin{tabular}{|c|c|c|c|c|c|} \hline Class & $0-4$ & $4-8$ & $8-12$ & $12-16$ & $16-20$ \\ \hline Frequency & $3$ & $9$ & $10$ & $8$ & $6$ \\ \hline \end{tabular} Then $20M$ is equal to:

  1. 416
  2. 104
  3. 52
  4. 208

Answer: (d)

Solution

\begin{tabular}{|c|c|c|} \hline Class & Frequency & Cumulative frequency \\ \hline $0-4$ & $3$ & $\mathbf{3}$ \\ \hline $4-8$ & $9$ & $\mathbf{12}$ \\ \hline $8-12$ & $10$ & $\mathbf{22}$ \\ \hline $12-16$ & $8$ & $\mathbf{30}$ \\ \hline $16-20$ & $6$ & $\mathbf{36}$ \\ \hline \end{tabular} $$M = 1 + \left(\frac{\frac{N}{2} - C}{f}\right)h$$ $$M = 8 + \frac{18 - 12}{10} \times 4$$ $$M = 10.4$$ $$20M = 208$$

Question 17

Maths · Determinants · Numerical

If $f(x)=\begin{vmatrix}2\cos^4 x & 2\sin^4 x & 3+\sin^2 2x \\ 3+2\cos^4 x & 2\sin^4 x & \sin^2 2x \\ 2\cos^4 x & 3+2\sin^4 x & \sin^2 2x\end{vmatrix}$ then $\frac{1}{5}f'(0)$ is equal to _____

  1. 0
  2. 1
  3. 2
  4. 6

Answer: (a)

Solution

Given the matrix: $$\begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^2 4x & \sin^2 2x \end{vmatrix}$$ Perform the row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$ to get: $$\begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{vmatrix}$$ Given $f(x) = 45$ and $f'(x) = 0$.

Question 18

Maths · Vector Algebra · Single correct

Let $A(2, 3, 5)$ and $C(-3, 4, -2)$ be opposite vertices of a parallelogram $ABCD$ if the diagonal $\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$ then the area of the parallelogram is equal to

  1. $\frac{1}{2} \sqrt{410}$
  2. $\frac{1}{2} \sqrt{474}$
  3. $\frac{1}{2} \sqrt{586}$
  4. $\frac{1}{2} \sqrt{306}$

Answer: (b)

Solution

Area = $\left$| $\overrightarrow{AC}$ $\times$ $\overrightarrow{BD}$ $\right$| $$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -1 & 7 \\ 1 & 2 & 3 \end{vmatrix}$$ $$= \frac{1}{2} \left| -17\hat{i} - 8\hat{j} + 11\hat{k} \right| = \frac{1}{2} \sqrt{474}$$

Question 19

Maths · Trigonometric Functions · Single correct

If $2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0$ has exactly 3 solutions in the interval $\left[ 0, \frac{n\pi}{2} \right]$, $n \in \mathbb{N}$, then the roots of the equation $x^2 + nx + (n-3) = 0$ belong to:

  1. (0, $\infty$)
  2. (-$\infty$, 0)
  3. $(-\\frac{\\sqrt{17}}{2}, \\frac{\\sqrt{17}}{2})$
  4. $Z$

Answer: (b)

Solution

Given the equation $2 \sin^3 x + 2 \sin x \cdot \cos^2 x + 4 \sin x - 4 = 0$. Simplifying, we have $2 \sin^3 x + 2 \sin x \cdot (1 - \sin^2 x) + 4 \sin x - 4 = 0$. This reduces to $6 \sin x - 4 = 0$. Solving for $\sin x$, we get $\sin x = \frac{2}{3}$. The number of solutions $n = 5$ (in the given interval). For the quadratic equation $x^2 + 5x + 2 = 0$, the solutions are $x = \frac{-5 \pm \sqrt{17}}{2}$. The required interval is $(-\infty, 0)$.

Question 20

Maths · Limits and Derivatives · Single correct

Let $f : \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \rightarrow \mathbb{R}$ be a differentiable function such that $f(0) = \frac{1}{2}$, If the $\lim_{x \to 0} \frac{x \int_{0}^{x} f(t) \, dt}{e^{x^2} - 1} = \alpha$, then $8\alpha^2$ is equal to:

  1. 16
  2. 2
  3. 1
  4. 4

Answer: (b)

Solution

Given $$\lim_{x \to 0} \frac{x \int_0^x f(t) \, dt}{\left(\frac{e^{x^2} - 1}{x^2}\right) \times x^2}$$ We have $$\lim_{x \to 0} \frac{\int_0^x f(t) \, dt}{x} \left( \lim_{x \to 0} \frac{e^{x^2} - 1}{x^2} = 1 \right)$$ This simplifies to $$= \lim_{x \to 0} \frac{f(x)}{1} (using L'Hospital)$$ Given $$f(0) = \frac{1}{2}$$ Therefore, $$\alpha = \frac{1}{2}$$ And $$8 \alpha^2 = 2$$

Question 21

Maths · Sets · Numerical

A group of 40 students appeared in an examination of 3 subjects – Mathematics, Physics $\&$ Chemistry. It was found that all students passed in at least one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, at most 11 students passed in both Mathematics and Physics, at most 15 students passed in both Physics and Chemistry, at most 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is $\ldots$

Answer: 10

Solution

Question 22

Maths · Three Dimensional Geometry · Numerical

If $d_1$ is the shortest distance between the lines $x + 1 = 2y = -12z, x = y + 2 = 6z - 6$ and $d_2$ is the shortest distance between the lines $\frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}, \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}$, then the value of $\frac{32\sqrt{3} \, d_1}{d_2}$ is :

Answer: 16

Solution

Given $$L_1 : \frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12}, L_2 : \frac{x+2}{1} = \frac{y+2}{1} = \frac{z-1}{1/6}$$ $d_1$ is the shortest distance between $L_1$ and $L_2$. $$d_1 = \frac{\left| (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \right|}{\left| \vec{b_1} \times \vec{b_2} \right|}$$ $$d_1 = 2$$ Given $$L_3 : \frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}, L_4 : \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}$$ $d_2$ is the shortest distance between $L_3$ and $L_4$. $$d_2 = \frac{12}{\sqrt{3}}$$ Hence, $$\frac{32\sqrt{3} \, d_1}{d_2} = \frac{32\sqrt{3} \times 2}{\frac{12}{\sqrt{3}}} = 16$$

Question 23

Maths · Conic Sections · Numerical

Let the latus rectum of the hyperbola $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$ subtend an angle of $\frac{\pi}{3}$ at the centre of the hyperbola. If $b^2$ is equal to $\frac{l}{m}(1 + \sqrt{n})$, where $l$ and $m$ are co-prime numbers, then $l^2 + m^2 + n^2$ is equal to

Answer: 182

Solution

Given that LR subtends $60^\circ$ at the center. $$\Rightarrow \tan 30^\circ = \frac{b^2/a}{ae} = \frac{b^2}{a^2 e} = \frac{1}{\sqrt{3}}$$ $$\Rightarrow e = \frac{\sqrt{3} \, b^2}{9}$$ Also, $e^2 = 1 + \frac{b^2}{9} \Rightarrow 1 + \frac{b^2}{9} = \frac{3 \, b^4}{81}$ $$\Rightarrow b^4 = 3 \, b^2 + 27$$ $$\Rightarrow b^4 - 3 \, b^2 - 27 = 0$$ $$\Rightarrow b^2 = \frac{3}{2} (1 + \sqrt{13})$$ $$\Rightarrow \ell = 3, \; m = 2, \; n = 13$$ $$\Rightarrow \ell^2 + m^2 + n^2 = 182$$

Question 24

Maths · Sets · Numerical

Let $A = \{1, 2, 3, \ldots 7\}$ and let $P(1)$ denote the power set of $A$. If the number of functions $f : A \rightarrow P(A)$ such that $a \in f(a)$, $\forall a \in A$ is $m^n$, $m$ and $n \in \mathbb{N}$ and $m$ is least, then $m + n$ is equal to

Answer: 44

Solution

Given $f: A \to \mathcal{P}(A)$ and $a \in f(a)$. That means 'a' will connect with subset which contain element 'a'. Total options for 1 will be $2^6$. (Because $2^6$ subsets contains 1) Similarly, for every other element. Hence, total is $2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 = 2^{42}$. Ans. $2 + 42 = 44$

Question 25

Maths · Integrals · Numerical

The value $9 \int_{0}^{9} \left[ \sqrt{\frac{10x}{x+1}} \right] \, dx$, where $[t]$ denotes the greatest integer less than or equal to $t$, is_____

Answer: 155

Solution

Given $\($ $\frac{10x}{x+1}$ = 1 $\)$, it implies $\($ x = $\frac{1}{9}$ $\)$. Given $\($ $\frac{10x}{x+1}$ = 4 $\)$, it implies $\($ x = $\frac{2}{3}$ $\)$. Given $\($ $\frac{10x}{x+1}$ = 9 $\)$, it implies $\($ x = 9 $\)$. $\[$ I = 9 $\left$( $\int$_{0}^{1/9} 0 $\,$ dx + $\int$_{1/9}^{2/3} 1 $\,$ dx + $\int$_{2/3}^{9} 2 $\,$ dx $\right$) $\]$ $\[$ = 155 $\]$

Question 26

Maths · Binomial Theorem · Numerical

Number of integral terms in the expansion of $$\left\{ 7^{\left( \frac{1}{2} \right)} + 11^{\left( \frac{1}{6} \right)} \right\}^{824}$$ is equal to

Answer: 138

Solution

General term in expansion of $\left((7)^{1/2} + (11)^{1/6}\right)^{824}$ is $$t_{r+1} = {}^{824}C_r\, (7)^{\frac{824-r}{2}}\, (11)^{r/6}$$ For integral term, $r$ must be multiple of $6$. Hence $r = 0, 6, 12, \ldots, 822$

Question 27

Maths · Differential Equations · Numerical

Let $y=y(x)$ be the solution of the differential equation $(1-x^2)\,dy=\left[xy+(x^3+2)\sqrt{3(1-x^2)}\right]dx$, $-1<x<1$, $y(0)=0$. If $y\left(\frac{1}{2}\right)=\frac{m}{n}$, $m$ and $n$ are coprime numbers, then $m+n$ is equal to _____.

Answer: 97

Solution

Given $\dfrac{dy}{dx}-\dfrac{xy}{1-x^2}=\dfrac{(x^3+2)\sqrt{3(1-x^2)}}{1-x^2}$ If $\mathrm{I.F.}=e^{\int -\frac{x}{1-x^2}\,dx}$ $=e^{\frac{1}{2}\ln(1-x^2)}$ $=\sqrt{1-x^2}$ then $y\sqrt{1-x^2}=\sqrt{3}\int (x^3+2)\,dx$ $y\sqrt{1-x^2}=\sqrt{3}\left(\dfrac{x^4}{4}+2x\right)+c$ $\Rightarrow y(0)=0$ $\therefore c=0$ At $x=\dfrac{1}{2}$, $y\left(\dfrac{1}{2}\right)=\dfrac{\sqrt{3}\left(\dfrac{1}{64}+1\right)}{\sqrt{1-\dfrac{1}{4}}}$ $=\dfrac{\sqrt{3}\left(\dfrac{65}{64}\right)}{\dfrac{\sqrt{3}}{2}}$ $=\dfrac{65}{32}$ $=\dfrac{m}{n}$ $\therefore m+n=97$

Question 28

Maths · Complex Numbers and Quadratic Equations · Numerical

Let $\alpha, \beta \in \mathbb{N}$ be roots of equation $x^2 - 70x + \lambda = 0$, where $\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbb{N}$. If $\lambda$ assumes the minimum possible value, then $\frac{(\sqrt{\alpha-1} + \sqrt{\beta-1})(\lambda + 35)}{|\alpha - \beta|}$ is equal to:

Answer: 60

Solution

Given the equation $x^2 - 70x + \lambda = 0$. We have $\alpha + \beta = 70$ and $\alpha \beta = \lambda$. Therefore, $\alpha(70 - \alpha) = \lambda$. Since 2 and 3 do not divide $\lambda$, we have $\alpha = 5$, $\beta = 65$, $\lambda = 325$. By putting the values of $\alpha$, $\beta$, and $\lambda$, we get the required value 60.

Question 29

Maths · Continuity and Differentiability · Fill in the blank

If the function $f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 2 \\ ax^2 + 2b, & |x| < 2 \end{cases}$ is differentiable on $\mathbb{R}$, then $48(a+b)$ is equal to ____

Answer: 15

Solution

Given $$f(x) = \begin{cases} \frac{1}{x}; & x \geq 2 \\ ax^2 + 2b; & -2 < x < 2 \\ -\frac{1}{x}; & x \leq -2 \end{cases}$$ Continuous at $x = 2$ implies $$\frac{1}{2} = \frac{a}{4} + 2b$$ Continuous at $x = -2$ implies $$\frac{1}{2} = \frac{a}{4} + 2b$$ Since it is differentiable at $x = 2$ $$-\frac{1}{x^2} = 2ax$$ Differentiable at $x = 2$ implies $$-\frac{1}{4} = 4a \Rightarrow a = -\frac{1}{16}, b = \frac{3}{8}$$

Question 30

Maths · Sequences and Series · Numerical

Let $\alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \ldots$ up to 10 terms and $\beta = \sum_{n=1}^{10} n^4$. If $4\alpha - \beta = 55k + 40$, then $k$ is equal to

Answer: 353

Solution

Given $\alpha = 1^2 + 4^2 + 8^2 \ldots$ and $t_n = a^2 + bn + c$. We have the equations: $$1 = a + b + c$$ $$4 = 4a + 2b + c$$ $$8 = 9a + 3b + c$$ On solving we get, $a = \frac{1}{2}$, $b = \frac{3}{2}$, $c = -1$. Now, $$\alpha = \sum_{n=1}^{10} \left( \frac{n^2}{2} + \frac{3n}{2} - 1 \right)^2$$ $$4\alpha = \sum_{n=1}^{10} \left( n^2 + 3n - 2 \right)^2, \beta = \sum_{n=1}^{10} n^4$$ $$4\alpha - \beta = \sum_{n=1}^{10} \left( 6n^3 + 5n^2 - 12n + 4 \right) = 55(353) + 40$$

Physics

Question 31

Physics · Physical World, Units and Measurements · Single correct

Match List-I with List-II. \begin{tabular}{|c|l|c|c|} \hline & \textbf{List-I} & & \textbf{List-II} \\ \hline A. & Coefficient of viscosity & I. & $[ML^2T^{-2}]$ \\ \hline B. & Surface Tension & II. & $[ML^2T^{-1}]$ \\ \hline C. & Angular momentum & III. & $[ML^{-1}T^{-1}]$ \\ \hline D. & Rotational kinetic energy & IV. & $[ML^0T^{-2}]$ \\ \hline \end{tabular}

  1. A–II, B–I, C–IV, D–III
  2. A–I, B–III, C–III, D–IV
  3. A–III, B–IV, C–II, D–I
  4. A–IV, B–III, C–II, D–I

Answer: (c)

Solution

Question 32

Physics · Laws of Motion · Single correct

All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass 2 kg is :

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

Answer: (b)

Solution

Given the equations: $$40 - 2T = 4a$$ $$T - 10 = 4a \Rightarrow 20 = 12a$$ Solving for $a$ gives: $$a = \frac{5}{3} \Rightarrow 2a = \frac{g}{3}$$

Question 33

Physics · Current Electricity · Single correct

A potential divider circuit is shown in figure. The output voltage $V_0$ is

  1. 4 \, $\mathrm{V}$
  2. 2 \, $\mathrm{mV}$
  3. 0.5 \, $\mathrm{V}$
  4. 12 \, $\mathrm{mV}$

Answer: (c)

Solution

Given $R_{eq} = 4000 \, \Omega$. $i = \frac{4}{4000} = \frac{1}{1000} \, \mathrm{A}$. $V_0 = i \cdot R = \frac{1}{1000} \times 500 = 0.5 \, \mathrm{V}$.

Question 34

Physics · Experimental Physics · Single correct

Young's modules of material of a wire of length ' L ' and cross-sectional area A is Y. If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be :

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

Answer: (c)

Solution

Young's modulus depends on the material not length and cross sectional area. So young's modulus remains same.

Question 35

Physics · Dual Nature of Radiation and Matter · Single correct

The work function of a substance is $3.0 \, \mathrm{eV}$. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:

  1. 215 nm
  2. 414 nm
  3. 400 nm
  4. 200 nm

Answer: (b)

Solution

For P.E.E.: $\lambda \leq \frac{hc}{W_e}$ $$\lambda \leq \frac{1240 \, \mathrm{nm} - \mathrm{eV}}{3 \, \mathrm{eV}}$$ $$\lambda \leq 413.33 \, \mathrm{nm}$$ $\lambda_{\max} \approx 414 \, \mathrm{nm}$ for P.E.E.

Question 36

Physics · Atoms · Single correct

The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the $5^{th}$ excited state of a hydrogen atom is:

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

Answer: (c)

Solution

Given $\frac{1}{2} |PE| = KE$ for each value of $n$ (orbit). Therefore, $$\frac{KE}{|PE|} = \frac{1}{2}.$$

Question 37

Physics · Work, Energy and Power · Single correct

A particle is placed at the point A of a frictionless track ABC as shown in figure. It is gently pushed toward right. The speed of the particle when it reaches the point B is: (Take $g = 10 \, \mathrm{m/s^2}$).

  1. 20 m/s
  2. $\sqrt{10}$ $\,$ $\mathrm{m/s}$
  3. 2$\sqrt{10}$ $\,$ $\mathrm{m/s}$
  4. 10 $\,$ $\mathrm{m/s}$

Answer: (b)

Solution

By conservation of mechanical energy (COME), we have: $$KE_A + U_A = KE_B + U_B$$ Substituting the values, we get: $$0 + mg(1) = \frac{1}{2}mv^2 + mg \times 0.5$$ Solving for $v$, we find: $$v = \sqrt{g} = \sqrt{10} \, \mathrm{m/s}$$

Question 38

Physics · Electromagnetic Waves · Single correct

The electric field of an electromagnetic wave in free space is represented as $\vec{E} = E_0 \cos(\omega t - kz) \hat{i}$. The corresponding magnetic induction vector will be:

  1. $\vec{B} = E_0 C \cos(\omega t - kz) \hat{j}$
  2. $\vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j}$
  3. $\vec{B} = E_0 \cos(\omega t + kz) \hat{j}$
  4. $\vec{B} = \frac{E_0}{C} \cos(\omega t + kz) \hat{j}$

Answer: (b)

Solution

Given $\vec{E} = E_0 \cos(\omega t - kz) \hat{i}$. $$\vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j}$$ $$\hat{C} = \hat{E} \times \hat{B}$$

Question 39

Physics · Electromagnetic Induction · Single correct

Two insulated circular loop $A$ and $B$ radius '$a$' carrying a current of '$I$' in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be :

  1. $\frac{\sqrt{2} \mu_0 I}{a}$
  2. $\frac{\mu_0 I}{2a}$
  3. $\frac{\mu_0 I}{\sqrt{2} a}$
  4. $\frac{2 \mu_0 I}{a}$

Answer: (c)

Solution

Given $B_A = \frac{\mu_0 I}{2a}$ and $B_B = \frac{\mu_0 I}{2a}$. Therefore, $B_{net} = \frac{\sqrt{2} \mu_0 I}{2a}$.

Question 40

Physics · Wave Optics · Single correct

The diffraction pattern of a light of wavelength $400 \, nm$ diffracting from a slit of width $0.2 \, mm$ is focused on the focal plane of a convex lens of focal length $100 \, cm$. The width of the $1^{st}$ secondary maxima will be :

  1. 2 mm
  2. 2 cm
  3. 0.02 mm
  4. 0.2 mm

Answer: (a)

Solution

Width of 1st secondary maxima = $\frac{\lambda}{a}$ $\cdot$ D Here $a = 0.2 \times 10^{-3} \, \mathrm{m}$ $\lambda = 400 \times 10^{-9} \, \mathrm{m}$ $D = 100 \times 10^{-2}$ Width of 1st secondary maxima $$= \frac{400 \times 10^{-9}}{0.2 \times 10^{-3}} \times 100 \times 10^{-2}$$ $$= 2 \, \mathrm{mm}$$

Question 41

Physics · Alternating Current · Single correct

Primary coil of a transformer is connected to 220 V ac. Primary and secondary turns of the transforms are 100 and 10 respectively. Secondary coil of transformer is connected to two series resistance shown in shown in figure. The output voltage ($V_0$) is :

  1. 7 V
  2. 15 V
  3. 44 V
  4. 22 V

Answer: (a)

Solution

Given $$\frac{\varepsilon_1}{\varepsilon_2} = \frac{N_1}{N_2} = \frac{100}{10}$$ which implies $$\varepsilon_2 = 22 \, \mathrm{V}$$. The current is $$I = \frac{22}{22 \times 10^3} = 1 \, \mathrm{mA}$$ and $$V_0 = 7 \, \mathrm{V}$$.

Question 42

Physics · Gravitation · Single correct

The gravitational potential at a point above the surface of earth is $-5.12 \times 10^7 \, \mathrm{J/kg}$ and the acceleration due to gravity at that point is $6.4 \, \mathrm{m/s^2}$. Assume that the mean radius of earth to be $6400 \, \mathrm{km}$. The height of this point above the earth's surface is :

  1. 1600 km
  2. 540 km
  3. 1200 km
  4. 1000 km

Answer: (a)

Solution

Given $$-\frac{GM_E}{R_E + h} = -5.12 \times 10^{-7} (i)$$ $$\frac{GM_E}{(R_E + h)^2} = 6.4 (ii)$$ By (i) and (ii) $$\Rightarrow h = 16 \times 10^5 \, \mathrm{m} = 1600 \, \mathrm{km}$$

Question 43

Physics · Current Electricity · Single correct

An electric toaster has resistance of $60\, \Omega$ at room temperature $(27^\circ \mathrm{C})$. The toaster is connected to a $220\, \mathrm{V}$ supply. If the current flowing through it reaches $2.75\, \mathrm{A}$, the temperature attained by toaster is around : (if $\alpha = 2 \times 10^{-4}/^\circ \mathrm{C}$)

  1. $694^\circ \mathrm{C}$
  2. $1235^\circ \mathrm{C}$
  3. $1694^\circ \mathrm{C}$
  4. $1667^\circ \mathrm{C}$

Answer: (c)

Solution

Given $R_{T=27} = 60 \, \Omega$, $R_T = \frac{220}{2.75} = 80 \, \Omega$. $R = R_0 (1 + \alpha \Delta T)$ $$80 = 60 \left[ 1 + 2 \times 10^{-4} (T - 27) \right]$$ $T \approx 1694^\circ \mathrm{C}$

Question 44

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

A Zener diode of breakdown voltage 10 V is used as a voltage regulator as shown in the figure. The current through the Zener diode is

  1. 50 $\mathrm{mA}$
  2. 0
  3. 30 $\mathrm{mA}$
  4. 20 $\mathrm{mA}$

Answer: (c)

Solution

Zener is in breakdown region. $I_3 = \frac{10}{500} = \frac{1}{50}$ $I_1 = \frac{10}{200} = \frac{1}{20}$ $I_2 = I_1 - I_3$ $I_2 = \left( \frac{1}{20} - \frac{1}{50} \right) = \left( \frac{3}{100} \right) = 30 \, \mathrm{mA}$

Question 45

Physics · Kinetic Theory · Single correct

Two thermodynamical process are shown in the figure. The molar heat capacity for process $A$ and $B$ are $C_A$ and $C_B$. The molar heat capacity at constant pressure and constant volume are represented by $C_P$ and $C_V$, respectively. Choose the correct statement.

  1. $C_B = \infty, C_A = 0$
  2. $C_A = 0$ and $C_B = \infty$
  3. $C_P > C_B > C_A > C_V$
  4. $C_A > C_P > C_V$

Answer: (c)

Solution

For process A $$\log P = \gamma \log V \Rightarrow P = V^\gamma, (\gamma > 1)$$ $$PV^{-\gamma} = Constant$$ $$C_A = C_V + \frac{R}{1+\gamma} \ldots (i)$$ Likewise for process B $\rightarrow PV^{-1} = Constant$ $$C_B = C_v + \frac{R}{1+1}$$ $$C_B = C_v + \frac{R}{2} \ldots (ii)$$ $$C_P = C_v + R \ldots (iii)$$ By (i), (ii) $\&$ (iii) $$C_P > C_B > C_A > C_v [No answer matching]$$

Question 46

Physics · Electric Charges and Fields · Single correct

The electrostatic potential due to an electric dipole at a distance ' r ' varies as:

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

Answer: (b)

Solution

Given $$V = \frac{kP \cos \theta}{r^2}$$ It can also be checked dimensionally.

Question 47

Physics · System of Particles and Rotational Motion · Single correct

A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by: (given $g = 9.8 \, \mathrm{m/s^2}$)

  1. 4.32 kg ms$^{-1}$
  2. 43.2 kg ms$^{-1}$
  3. 23.9 kg ms$^{-1}$
  4. 2.39 kg ms$^{-1}$

Answer: (d)

Solution

Given $\vec{I} = \Delta \vec{P} = \vec{P}_f - \vec{P}_i$. $M = 0.1 \, \mathrm{kg}$. $I = \Delta P = 0.1 \left( \sqrt{2} \times 9.8 \times 5 - \left( \sqrt{2} \times 9.8 \times 10 \right) \right)$ $= 0.1 (14 + 7 \sqrt{2}) \approx 2.39 \, \mathrm{kg \, ms^{-1}}$

Question 48

Physics · System of Particles and Rotational Motion · Single correct

A particle of mass m projected with a velocity ' u ' making an angle of $30^\circ$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is :

  1. $\frac{\sqrt{3}}{16} \frac{mu^3}{g}$
  2. $\frac{\sqrt{3}}{2} \frac{mu^2}{g}$
  3. $\frac{mu^3}{\sqrt{2g}}$
  4. zero

Answer: (a)

Solution

Given $$L = mu \cos \theta H$$ $$= mu \cos \theta \times \frac{u^2 \sin^2 \theta}{2g}$$ $$= \frac{mu^3}{2g} \times \frac{\sqrt{3}}{2} \times \left(\frac{1}{2}\right)^2 = \frac{\sqrt{3} mu^3}{16g}$$

Question 49

Physics · Kinetic Theory · Single correct

At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\circ} \mathrm{C}$ ?

  1. 80\,\mathrm{K}
  2. -73\,\mathrm{K}
  3. 4\,\mathrm{K}
  4. 20\,\mathrm{K}

Answer: (d)

Solution

Given $$\sqrt{\frac{3RT}{2}} = \sqrt{\frac{3R(320)}{32}}$$. Therefore, $$T = \frac{320}{16} = 20 \, \mathrm{K}$$.

Question 50

Physics · Alternating Current · Single correct

A series L,R circuit connected with an ac source $E = (25 \sin 1000t) V$ has a power factor of $\frac{1}{\sqrt{2}}$. If the source of emf is changed to $E = (20 \sin 2000 \, t) \, V$, the new power factor of the circuit will be:

  1. $\frac{1}{\sqrt{2}}$
  2. $\frac{1}{\sqrt{3}}$
  3. $\frac{1}{\sqrt{5}}$
  4. $\frac{1}{\sqrt{7}}$

Answer: (c)

Solution

Given $E = 25 \sin(1000t)$ and $\cos \theta = \frac{1}{\sqrt{2}}$. For an LR circuit, initially $$\frac{R}{\omega_1 L} = \frac{1}{\tan \theta} = \frac{1}{\tan 45^\circ} = 1.$$ Therefore, $X_L = \omega_1 L$. Given $\omega_2 = 2\omega_1$, we have $$\tan \theta' = \frac{\omega_2 L}{R} = \frac{2\omega_1 L}{R}.$$ Thus, $\tan \theta' = 2$ and $\cos \theta' = \frac{1}{\sqrt{5}}$.

Question 51

Physics · Magnetism and Matter · Numerical

The horizontal component of earth's magnetic field at a place is $3.5 \times 10^{-5} \, T$. A very long straight conductor carrying current of $\sqrt{2} \, A$ in the direction from South east to North West is placed. The force per unit length experienced by the conductor is....... $\times 10^{-6} \, \mathrm{N/m}$.

Answer: 35

Solution

Given $B_H = 3.5 \times 10^{-5} \, \mathrm{T}$. $F = i \ell B \sin \theta$, where $i = \sqrt{2} \, \mathrm{A}$. The force per unit length is given by $$\frac{F}{\ell} = i B \sin \theta = \sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}}$$ This simplifies to $$= 35 \times 10^{-6} \, \mathrm{N/m}$$

Question 52

Physics · Current Electricity · Numerical

Two cells are connected in opposition as shown. Cell $E_1$ is of $8 \, \mathrm{V}$ emf and $2\Omega$ internal resistance; the cell $E_2$ is of $2 \, \mathrm{V}$ emf and $4\Omega$ internal resistance. The terminal potential difference of cell $E_2$ is:

Answer: 6

Solution

The current $I$ is calculated as follows: $$I = \frac{8 - 2}{2 + 4} = \frac{6}{6} = 1 \, \mathrm{A}$$ Applying Kirchhoff's law from C to B: $$V_C - 2 - 4 \times 1 = V_B$$ $$V_C - V_B = 6 \, \mathrm{V}$$ Therefore, $V_C - V_B = 6 \, \mathrm{V}$.

Question 53

Physics · Atoms · Numerical

A electron of hydrogen atom on an excited state is having energy $E_n = -0.85 \, \mathrm{eV}$. The maximum number of allowed transitions to lower energy level is ...... .

Answer: 6

Solution

Given $\($ E_n = -$\frac{13.6}{n^2}$ = -0.85 $\)$. Therefore, $\($ n = 4 $\)$. The number of transitions is given by \[ \frac{n(n-1)}{2} = \frac{4(4-1)}{2} = 6 \]

Question 54

Physics · Mechanical Properties of Solids · Numerical

Each of three blocks P, Q and R shown in figure has a mass of $3 \, \mathrm{kg}$. Each of the wire A and B has cross-sectional area $0.005 \, \mathrm{cm^2}$ and Young's modulus $2 \times 10^{11} \, \mathrm{N \, m^{-2}}$. Neglecting friction, the longitudinal strain on wire B is _____ $\times 10^{-4}$. (Take $g = 10 \, \mathrm{m/s^2}$)

Answer: 2

Solution

Given the acceleration, we have $$a = \frac{10}{3} \, \mathrm{m/s^2}$$ Using the equation of motion, $$30 - T_1 = 3 \times a$$ Solving for $T_1$, we get $$T_1 = 20 \, \mathrm{N}$$ The strain is given by $$strain = \frac{stress}{Y}$$ Substituting the values, $$= 2 \times 10^{-4}$$

Question 55

Physics · Ray Optics and Optical Instruments · Fill in the blank

The distance between object and its two times magnified real image as produced by a convex lens is 45 cm. The focal length of the lens used is ____ cm.

Answer: 10

Solution

Given $\frac{v}{u}=-2$. $v=-2u$ ...(i) $v-u=45$ ...(ii) $\Rightarrow u=-15\,\mathrm{cm}$ $v=30\,\mathrm{cm}$ $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ $f=+10\,\mathrm{cm}$

Question 56

Physics · Motion in a Straight Line · Numerical

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t + 1)\, \mathrm{s}$ are $125\, \mathrm{m}$ and $50\, \mathrm{m/s}$, respectively. The distance travelled by the particle in $(t + 2)^{th}\, \mathrm{s}$ is _____ $\mathrm{m}$.

Answer: 175

Solution

Considering acceleration is constant $$v = u + at$$ $$u + 50 = u + a \Rightarrow a = 50 \, \mathrm{m/s^2}$$ $$125 = ut + \frac{1}{2}at^2$$ $$125 = u + \frac{a}{2}$$ $$\Rightarrow u = 100 \, \mathrm{m/s}$$ $$\therefore S_{nin} = u + \frac{a}{2}[2n - 1]$$ $$= 175 \, \mathrm{m}$$

Question 57

Physics · Electrostatic Potential and Capacitance · Numerical

A capacitor of capacitance $C$ and potential $V$ has energy $E$. It is connected to another capacitor of capacitance $2C$ and potential $2 \, \mathrm{V}$. Then the loss of energy is $\frac{x}{3}E$, where $x$ is_____

Answer: 2

Solution

Energy loss = $\frac{1}{2}$ $\frac{C_1 C_2}{C_1 + C_2}$ (V_1 - V_2)^2 = $\frac{2}{3}$ $\cdot$ E $\therefore$ x = 2

Question 58

Physics · System of Particles and Rotational Motion · Numerical

Consider a Disc of mass 5 kg, radius 2 m, rotating with angular velocity of 10 $\mathrm{rad/s}$ about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is $\underline{\hspace{1cm}}\,\mathrm{J}$.

Answer: 250

Solution

Given $\vec{L}_i = I \omega_i = \frac{MR^2}{2} \cdot \omega = 100 \mathrm{kgm^2/s}$. $E_i = \frac{1}{2} \cdot \frac{MR^2}{2} \cdot \omega^2 = 500 \, \mathrm{J}$. $\vec{L}_i = \vec{L}_f \Rightarrow 100 = 2I \omega_f$. $\omega_f = 5 \, \mathrm{rad/sec}$. $E_f = 2 \times \frac{1}{2} \cdot \frac{5(2)^2}{2} \cdot (5)^2 = 250 \, \mathrm{J}$. $\Delta E = 250 \, \mathrm{J}$.

Question 59

Physics · Waves · Numerical

In a closed organ pipe, the frequency of fundamental note is $30 \, \mathrm{Hz}$. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to $110 \, \mathrm{Hz}$. If the organ pipe has a cross-sectional area of $2 \, \mathrm{cm}^2$, the amount of water poured in the organ tube is_____ g. (Take speed of sound in air is $330 \, \mathrm{m/s}$)

Answer: 400

Solution

Given $\dfrac{V}{4\ell_1} = 30 \Rightarrow \ell_1 = \dfrac{11}{4}\,\mathrm{m}$ and $\dfrac{V}{4\ell_2} = 110 \Rightarrow \ell_2 = \dfrac{3}{4}\,\mathrm{m}$. The change in length $\Delta\ell = 2\,\mathrm{m}$. Change in volume $= A\,\Delta\ell = 400\,\mathrm{cm^3}$. $M = 400\,\mathrm{g}$; $(\therefore \rho = 1\,\mathrm{g/cm^3})$

Question 60

Physics · Magnetism and Matter · Numerical

A ceiling fan having 3 blades of length 80 $\mathrm{cm}$ each is rotating with an angular velocity of 1200 $\mathrm{rpm}$. The magnetic field of earth in that region is 0.5 $\mathrm{G}$ and angle of dip is $30^\circ$. The emf induced across the blades is $N \pi \times 10^{-5} \mathrm{V}$. The value of $N$ is_____

Answer: 32

Solution

Given $$B_v = B \sin 30 = \frac{1}{4} \times 10^{-4}$$ $$\omega = 2\pi \times f = \frac{2\pi}{60} \times 1200 \, \mathrm{rad/s}$$ $$\varepsilon = \frac{1}{2} B_v \omega \ell^2$$ $$= 32\pi \times 10^{-5} \, \mathrm{V}$$

Chemistry

Question 61

Chemistry · Analytical Chemistry · Single correct

Given below are two statements:\ Statement-I: The gas liberated on warming a salt with dil $\mathrm{H_2SO_4}$, turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion.\ Statement-II: In statement-I the colour of paper turns black because of formation of lead sulphite.\ 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 false
  2. Statement-I is false but Statement-II is true
  3. Statement-I is true but Statement-II is false
  4. Both Statement-I and Statement-II are true.

Answer: (c)

Solution

The reaction of $\mathrm{Na_2S}$ with $\mathrm{H_2SO_4}$ produces $\mathrm{Na_2SO_4}$ and $\mathrm{H_2S}$. The reaction of $\mathrm{(CH_3COO)_2Pb}$ with $\mathrm{H_2S}$ produces $\mathrm{PbS}$ and $2\mathrm{CH_3COOH}$. The $\mathrm{PbS}$ formed is black lead sulphide. The given reaction involves the reduction of the carbonyl group to an aldehyde using $\mathrm{H_2}$ and $\mathrm{Pd-BaSO_4}$ catalyst.

Question 62

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct

This reduction reaction is known as:

  1. Rosenmund reduction
  2. Wolff-Kishner reduction
  3. Stephen reduction
  4. Etard reduction

Answer: (a)

Solution

It is known as Rosenmund reduction that is the partial reduction of acid chloride to aldehyde.

Question 63

Chemistry · Biomolecules · Single correct

Sugar which does not give reddish brown precipitate with Fehling's reagent is:

  1. Sucrose
  2. Lactose
  3. Glucose
  4. Maltose

Answer: (a)

Solution

Sucrose does not contain a hemiacetal group. Hence it does not give a test with Fehling solution. While all others give a positive test with Fehling solution.

Question 64

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

Given below are the two statements: one is labeled as Assertion (A) and the other is labeled as Reason ($R$).\ Assertion (A): There is a considerable increase in covalent radius from N to P. However from As to Bi only a small increase in covalent radius is observed.\ Reason ($R$): Covalent and ionic radii in a particular oxidation state increases down the group.\ In the light of the above statement, choose the most appropriate answer from the options given below:

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

Answer: (b)

Solution

According to NCERT, Statement-I: Factual data, Statement-II is true. But correct explanation is presence of completely filled d and f-orbitals of heavier members.

Question 65

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

Which of the following molecule/species is most stable?

Answer: (a)

Solution

The given species is a cyclopropenyl cation. It is aromatic because it satisfies Huckel's rule of $4n + 2$ pi electrons, where $n = 0$. Therefore, it has $2$ pi electrons, which makes it aromatic.

Question 66

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

Diamagnetic Lanthanoid ions are:

  1. $\mathrm{Nd}^{3+}$ and $\mathrm{Eu}^{3+}$
  2. $\mathrm{La}^{3+}$ and $\mathrm{Ce}^{4+}$
  3. $\mathrm{Nd}^{3+}$ and $\mathrm{Ce}^{4+}$
  4. $\mathrm{Lu}^{3+}$ and $\mathrm{Eu}^{3+}$

Answer: (b)

Solution

Ce: $[\mathrm{Xe}]4f^15 \, 5d^1 \, 6s^2$; $\mathrm{Ce}^{4+}$ diamagnetic La: $[\mathrm{Xe}]4f^0 \, 5d^1 \, 6s^2$; $\mathrm{La}^{3+}$ diamagnetic

Question 67

Chemistry · Co-ordination Compounds · Single correct

Aluminium chloride in acidified aqueous solution forms an ion having geometry

  1. Octahedral
  2. Square Planar
  3. Tetrahedral
  4. Trigonal bipyramidal

Answer: (a)

Solution

AlCl$_3$ in acidified aqueous solution forms octahedral geometry $[\mathrm{Al(H_2O)_6}]^{3+}$.

Question 68

Chemistry · Structure of Atom · Single correct

Given below are two statements: Statement-I: The orbitals having same energy are called as degenerate orbitals. Statement-II: In hydrogen atom, 3p and 3 d orbitals are not degenerate orbitals. In the light of the above statements, choose the most appropriate answer from the options given

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

Answer: (a)

Solution

For single electron species the energy depends upon principal quantum number $'n'$ only. So, statement II is false. Statement I is correct definition of degenerate orbitals.

Question 69

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

Example of vinylic halide is

Answer: (a)

Solution

Vinyl carbon is $\mathrm{sp^2}$ hybridized aliphatic carbon is vinyl halide while is aryl halide and are allyl halide.

Question 70

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

Structure of 4-Methylpent-2-enal is

Answer: (d)

Solution

The structure shown is 4-Methylpent-2-enal.

Question 71

Chemistry · Chemical Bonding and Molecular Structure · Single correct

Match List-I with List-II List-I \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \multicolumn{2}{|c|}{Molecule} & \multicolumn{2}{c|}{Shape} \\ \hline (A) & BrF$_5$ & (I) & T-shape \\ \hline (B) & H$_2$O & (II) & See saw \\ \hline (C) & ClF$_3$ & (III) & Bent \\ \hline (D) & SF$_4$ & (IV) & Square pyramidal \\ \hline \end{tabular}

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

Answer: (d)

Solution

The molecular geometry of the given compounds are as follows: For $\mathrm{BrF_5}$, the geometry is square pyramidal. For $\mathrm{H_2O}$, the geometry is bent. For $\mathrm{ClF_3}$, the geometry is T-shape. For $\mathrm{SF_4}$, the geometry is see-saw.

Question 72

Chemistry · Haloalkanes and Haloarenes · Single correct

The final product $A$, formed in the following multistep reaction sequence is:

Answer: (b)

Solution

The reaction sequence begins with the conversion of bromobenzene to phenylmagnesium bromide using $Mg, ether$. This Grignard reagent then reacts with carbon dioxide $\left( O = C = O \right)$ to form benzoic acid after hydrolysis with $H^+$. The benzoic acid is then converted to benzamide using $Br_2 / NaOH$ in the Hoffmann bromamide reaction, which results in the formation of aniline.

Question 73

Chemistry · Hydrocarbons · Single correct

In the given reactions identify the reagent $A$ and reagent $B$

  1. $A - CrO_3$ $B - CrO_3$
  2. $A - CrO_3$ $B - CrO_2Cl_2$
  3. $A - CrO_2Cl_2$ $B - CrO_2Cl_2$
  4. $A - CrO_2Cl_2$ $B - CrO_3$

Answer: (b)

Solution

The given reaction is an Etard reaction. The starting compound is toluene, which is oxidized using $\mathrm{CrO_2Cl_2}$ in $\mathrm{CS_2}$ to form benzaldehyde. The intermediate formed is $\mathrm{CH[OCrCl_2(OH)]_2}$, which upon hydrolysis with $\mathrm{H_3O^+}$ gives benzaldehyde ($\mathrm{CHO}$).

Question 74

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

Given below are two statement one is labeled as Assertion (A) and the other is labeled as Reason $(R)$.\ Assertion (A): $CH_2 = CH - CH_2 - Cl$ is an example of allyl halide.\ Reason $(R)$: Allyl halides are the compounds in which the halogen atom is attached to $sp^2$ hybridised carbon atom.\ In the light of the two above statements, choose the most appropriate answer from the options given below:

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

Answer: (a)

Solution

CH$_2$ = CH - CH$_2$ - Cl It is allyl carbon and sp$^3$ hybridized

Question 75

Chemistry · Solutions · Single correct

What happens to freezing point of benzene when small quantity of napthalene is added to benzene?

  1. Increases
  2. Remains unchanged
  3. First decreases and then increases
  4. Decreases

Answer: (d)

Solution

On addition of naphthalene to benzene there is depression in freezing point of benzene.

Question 76

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

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

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

Answer: (b)

Solution

For Chromium ($\mathrm{Cr}$), the electron configuration is $[\mathrm{Ar}] 3d^5 4s^1$. For $\mathrm{Cr}^{2+}$, it becomes $[\mathrm{Ar}] 3d^4$. For Manganese ($\mathrm{Mn}$), the electron configuration is $[\mathrm{Ar}] 3d^5 4s^2$. For $\mathrm{Mn}^{+}$, it becomes $[\mathrm{Ar}] 3d^5 4s^1$. For Nickel ($\mathrm{Ni}$), the electron configuration is $[\mathrm{Ar}] 3d^8 4s^2$. For $\mathrm{Ni}^{2+}$, it becomes $[\mathrm{Ar}] 3d^8$. For Vanadium ($\mathrm{V}$), the electron configuration is $[\mathrm{Ar}] 3d^3 4s^2$. For $\mathrm{V}^{+}$, it becomes $[\mathrm{Ar}] 3d^3 4s^1$.

Question 77

Chemistry · Hydrocarbons · Single correct

Compound $A$ formed in the following reaction reacts with $B$ gives the product $C$. Find out $A$ and $B$.

  1. $A=CH_3-C\equiv CNa,\quad B=CH_3-CH_2-CH_2-Br$
  2. $A = \mathrm{CH_3-CH=CH_2}$, $B = \mathrm{CH_3-CH_2-CH_2-Br}$
  3. $A = \mathrm{CH_3-CH_2-CH_3}$, $B = \mathrm{CH_3-C \equiv CH}$
  4. $A = \mathrm{CH_3-C \equiv C^+a}$, $B = \mathrm{CH_3-CH_2-CH_3}$

Answer: (a)

Solution

The reaction starts with $\mathrm{CH_3-C \equiv CH}$ reacting with sodium ($\mathrm{Na}$) to form $\mathrm{CH_3-C \equiv C^-Na^+}$. This intermediate then reacts with $\mathrm{CH_3CH_2CH_2-Br}$ to form $\mathrm{NaBr}$ and $\mathrm{CH_3-C \equiv C-CH_2CH_2CH_3}$.

Question 78

Chemistry · Amines · Single correct

Following is a confirmatory test for aromatic primary amines. Identify reagent (A) and (B)

Answer: (d)

Solution

The reaction begins with aniline, which is treated with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ at $0 - 5^\circ \mathrm{C}$ to form a diazonium salt. This diazonium salt is then reacted with $\beta$-naphthol in the presence of $\mathrm{NaOH}$ to produce a scarlet red dye.

Question 79

Chemistry · Analytical Chemistry · Single correct

The Lassaigne's extract is boiled with dil $\mathrm{HNO_3}$ before testing for halogens because,

  1. $\mathrm{AgCN}$ is soluble in $\mathrm{HNO_3}$
  2. Silver halides are soluble in $\mathrm{HNO_3}$
  3. $\mathrm{Ag_2S}$ is soluble in $\mathrm{HNO_3}$
  4. $\mathrm{Na_2S}$ and $\mathrm{NaCN}$ are decomposed by $\mathrm{HNO_3}$

Answer: (d)

Solution

If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium during Lassaigne's test.

Question 80

Chemistry · Co-ordination Compounds · Single correct

Choose the correct Statements from the following: (A) Ethane-1 2-diamine is a chelating ligand. (B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite. (C) Cyanide ion is used as ligand for leaching of silver. (D) Phosphine act as a ligand in Wilkinson catalyst. (E) The stability constants of $\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$ are similar with EDTA complexes. Choose the correct answer from the options given below:

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

Answer: (c)

Solution

Bidentate, chelating. Based on Hall-Heroults process. $[\mathrm{Rh(PPh_3)_3Cl}]$ Wilkinson's catalyst. $$\mathrm{Ag_2S + NaCN \xrightleftharpoons[Air]{} Na[Ag(CN)_2] + Na_2S}$$ $\mathrm{Ca^{++}}$ ion forms more stable complex with EDTA.

Question 81

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

The rate of first order reaction is $0.04 \, \mathrm{mol \, L^{-1} \, s^{-1}}$ at 10 minutes and $0.03 \, \mathrm{mol \, L^{-1} \, s^{-1}}$ at 20 minutes after initiation. Half life of the reaction is $\underline{\phantom{0000}}$ minutes. (Given $\log 2 = 0.3010, \log 3 = 0.4771$)

Answer: 24

Solution

Given $$0.04 = k[A]_0 e^{-k \times 10 \times 60} \ldots (1)$$ $$0.03 = k[A]_0 e^{-k \times 20 \times 60} \ldots (2)$$ Dividing (1) by (2), we have $$\frac{4}{3} = e^{600k(2-1)}$$ $$\frac{4}{3} = e^{600k}$$ Taking the natural logarithm on both sides, $$\ln \frac{4}{3} = 600k$$ Substituting for $k$, $$\ln \frac{4}{3} = 600 \times \frac{\ln 2}{t_{1/2}}$$ Solving for $t_{1/2}$, $$t_{1/2} = 600 \frac{\ln 2}{\ln \frac{4}{3}} sec$$ Converting to minutes, $$t_{1/2} = 600 \times \frac{\log 2}{\log 4 - \log 3} sec = 10 \times \frac{0.3010}{0.6020 - 0.477} min$$ Finally, $$t_{1/2} = 24.08 min$$ Ans. 24

Question 82

Chemistry · Equilibrium · Numerical

The pH at which $\mathrm{Mg(OH)_2}$ $\left[ K_{sp} = 1 \times 10^{-11} \right]$ begins to precipitate from a solution containing $0.10 \mathrm{M} \ Mg^{2+}$ ions is

Answer: 9

Solution

Precipitation when $Q_{sp} = K_{sp}$ $$[\mathrm{Mg^{2+}}][\mathrm{OH^{-}}]^2 = 10^{-11}$$ $$0.1 \times [\mathrm{OH^{-}}]^2 = 10^{-11} \Rightarrow [\mathrm{OH^{-}}] = 10^{-5}$$ $$\Rightarrow pOH = 5 \Rightarrow pH = 9$$

Question 83

Chemistry · Thermodynamics · Fill in the blank

An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A $\rightarrow$ B $\rightarrow$ C $\rightarrow$ A as shown in the diagram. The total work done in the process is $\underline{\phantom{0000}}\,\mathrm{J}$.

Answer: 200

Solution

Work done is given by area enclosed in the P vs V cyclic graph or V vs P cyclic graph. Sign of work is positive for clockwise cyclic process for V vs P graph. $$W = \frac{1}{2} \times (30 - 10) \times (30 - 10) = 200 \, \mathrm{kPa} - \mathrm{dm^3}$$ $$= 200 \times 1000 \, \mathrm{Pa} - \mathrm{L} = 2 \, \mathrm{L} - \mathrm{bar} = 200 \, \mathrm{J}$$

Question 84

Chemistry · Classification of Elements and Periodicity in Properties · Numerical

If IUPAC name of an element is "Ununnunium" then the element belongs to nth group of periodic table. The value of $n$ is

Answer: 11

Solution

11 belongs to 11th group

Question 85

Chemistry · Chemical Bonding and Molecular Structure · Numerical

The total number of molecular orbitals formed from 2 s and 2p atomic orbitals of a diatomic molecule

Answer: 8

Solution

Two molecular orbitals $\sigma 2s$ and $\sigma^* 2s$. Six molecular orbitals $\sigma 2p_z$ and $\sigma^* 2p_z$, $\pi 2p_x$, $\pi 2p_y$ and $\pi^* 2p_x$, $\pi^* 2p_y$.

Question 86

Chemistry · Analytical Chemistry · Fill in the blank

On a thin layer chromatographic plate, an organic compound moved by $3.5 \, \mathrm{cm}$, while the solvent moved by $5 \, \mathrm{cm}$. The retardation factor of the organic compound is ______ $\times 10^{-1}$

Answer: 7

Solution

Retardation factor is given by the formula: $$Retardation factor = \frac{Distance travelled by sample/organic compound}{Distance travelled by solvent}$$ Substituting the given values: $$= \frac{3.5}{5} = 7 \times 10^{-1}$$

Question 87

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Fill in the blank

The compound formed by the reaction of ethanal with semicarbazide contains _______ number of nitrogen atoms.

Answer: 3

Solution

The given compound is an amide with the structure $\mathrm{CH_3{-}CH{=}N{-}NH{-}C{-}NH_2}$.

Question 88

Chemistry · Some Basic Concepts of Chemistry · Fill in the blank

0.05 cm thick coating of silver is deposited on a plate of $0.05 \, \mathrm{m}^2$ area. The number of silver atoms deposited on plate are _______ $\times 10^{23}$. (At mass Ag = 108, d = 7.9 g cm$^{-3}$)

Answer: 11

Solution

Volume of silver coating = $0.05 \times 0.06 \times 10000 = 25 \, \mathrm{cm^3}$ Mass of silver deposited = $25 \times 7.9 \, \mathrm{g}$ Moles of silver atoms = $\frac{25 \times 7.9}{108}$ Number of silver atoms = $\frac{25 \times 7.9}{108} \times 6.023 \times 10^{23} = 11.01 \times 10^{23}$ Ans. 11

Question 89

Chemistry · Redox Reactions · Fill in the blank

$2\mathrm{MnO}_4^- + b\mathrm{I}^- + c\mathrm{H}_2\mathrm{O} \rightarrow x\mathrm{I}_2 + y\mathrm{MnO}_2 + z\mathrm{OH}^-$ If the above equation is balanced with integer coefficients, the value of $z$ is $\underline{\hspace{1cm}}$.

Answer: 8

Solution

Reduction Half $$2\mathrm{MnO_4^-} \rightarrow 2\mathrm{MnO_2}$$ $$2\mathrm{MnO_4^-} + 4\mathrm{H_2O} + 6e^- \rightarrow 2\mathrm{MnO_2} + 8\mathrm{OH^-}$$ Oxidation Half $$2\mathrm{I^-} \rightarrow \mathrm{I_2} + 2e^-$$ $$6\mathrm{I^-} \rightarrow 3\mathrm{I_2} + 6e^-$$ Adding oxidation half and reduction half, net reaction is $$2\mathrm{MnO_4^-} + 6\mathrm{I^-} + 4\mathrm{H_2O} \rightarrow 3\mathrm{I_2} + 2\mathrm{MnO_2} + 8\mathrm{OH^-}$$ $$\Rightarrow z = 8$$ $$\Rightarrow Ans 8$$

Question 90

Chemistry · Solutions · Numerical

The mass of sodium acetate ($\mathrm{CH_3COONa}$) required to prepare $250 \, \mathrm{mL}$ of $0.35 \, \mathrm{M}$ aqueous solution is $\underline{\phantom{0000}}$ ${g}$. (Molar mass of $\mathrm{CH_3COONa}$ is $82.02 \, \mathrm{g \, mol}^{-1}$)

Answer: 7

Solution

Moles = Molarity $\times$ Volume in litres $$= 0.35 \times 0.25$$ Mass = moles $\times$ molar mass $$= 0.35 \times 0.25 \times 82.02 = 7.18 \, \mathrm{g}$$ Ans. 7