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
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 :
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:
88
122
92
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:
$\frac{(2\sqrt{3}+3)\pi}{24}$
$\frac{13\pi}{8(4\sqrt{3}+3)}$
$\frac{13(2\sqrt{3}-3)\pi}{8}$
$\frac{\pi}{8(2\sqrt{3}+3)}$
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
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 $y = 2x - 8$
8
9
6
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:
2
2$\{$1 - $\sin$(2)$\}$
2$\{$$\sin$(2) + 1$\}$
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:
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:
$\frac{30}{121}$
$\frac{62}{121}$
$\frac{60}{121}$
$\frac{31}{121}$
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:
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?
The system has unique solution if $\lambda \neq \frac{1}{2}$ and $\mu \neq 1, 15$
The system is inconsistent if $\lambda = \frac{1}{2}$ and $\mu \neq 1$
The system has infinite number of solutions if $\lambda = \frac{1}{2}$ and $\mu = 15$
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
$5 < r < 9$
$0 < r < 7$
$3 < r < 7$
$\frac{1}{2} < r < 7$
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 :
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:
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 _____
0
1
2
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
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:
(0, $\infty$)
(-$\infty$, 0)
$(-\\frac{\\sqrt{17}}{2}, \\frac{\\sqrt{17}}{2})$
$Z$
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:
16
2
1
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
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
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}
A–II, B–I, C–IV, D–III
A–I, B–III, C–III, D–IV
A–III, B–IV, C–II, D–I
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 :
$g$
$\frac{g}{3}$
$\frac{g}{2}$
$\frac{g}{4}$
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
4 \, $\mathrm{V}$
2 \, $\mathrm{mV}$
0.5 \, $\mathrm{V}$
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 :
$\frac{Y}{4}$
$4Y$
$Y$
$2Y$
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:
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:
4
$\frac{1}{4}$
$\frac{1}{2}$
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 $g = 10 \, \mathrm{m/s^2}$).
20 m/s
$\sqrt{10}$ $\,$ $\mathrm{m/s}$
2$\sqrt{10}$ $\,$ $\mathrm{m/s}$
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:
$\vec{B} = E_0 C \cos(\omega t - kz) \hat{j}$
$\vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j}$
$\vec{B} = E_0 \cos(\omega t + kz) \hat{j}$
$\vec{B} = \frac{E_0}{C} \cos(\omega t + kz) \hat{j}$
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 :
$\frac{\sqrt{2} \mu_0 I}{a}$
$\frac{\mu_0 I}{2a}$
$\frac{\mu_0 I}{\sqrt{2} a}$
$\frac{2 \mu_0 I}{a}$
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 :
2 mm
2 cm
0.02 mm
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 :
7 V
15 V
44 V
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 :
1600 km
540 km
1200 km
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 $\alpha = 2 \times 10^{-4}/^\circ \mathrm{C}$)
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.
$C_B = \infty, C_A = 0$
$C_A = 0$ and $C_B = \infty$
$C_P > C_B > C_A > C_V$
$C_A > C_P > C_V$
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:
r
$\frac{1}{r^2}$
$\frac{1}{r^3}$
$\frac{1}{r}$
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 $g = 9.8 \, \mathrm{m/s^2}$)
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 :
$\frac{\sqrt{3}}{16} \frac{mu^3}{g}$
$\frac{\sqrt{3}}{2} \frac{mu^2}{g}$
$\frac{mu^3}{\sqrt{2g}}$
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 $47^{\circ} \mathrm{C}$ ?
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:
$\frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{3}}$
$\frac{1}{\sqrt{5}}$
$\frac{1}{\sqrt{7}}$
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.
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 $\underline{\hspace{1cm}}\,\mathrm{J}$.
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_____
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:
Both Statement-I and Statement-II are false
Statement-I is false but Statement-II is true
Statement-I is true but Statement-II is false
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:
Rosenmund reduction
Wolff-Kishner reduction
Stephen reduction
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:
Sucrose
Lactose
Glucose
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:
is false but ($R$ ) is true
Both (A) and ($R$ ) are true but ($R$ ) is not the correct explanation of (A)
is true but ($R$ ) is false
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
Chemistry · Co-ordination Compounds · Single correct
Aluminium chloride in acidified aqueous solution forms an ion having geometry
Octahedral
Square Planar
Tetrahedral
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
Statement-I is true but Statement-II is false
Both Statement-I and Statement-II are true.
Both Statement-I and Statement-II are false
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
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 $B$
$A - CrO_3$ $B - CrO_3$
$A - CrO_3$ $B - CrO_2Cl_2$
$A - CrO_2Cl_2$ $B - CrO_2Cl_2$
$A - CrO_2Cl_2$ $B - CrO_3$
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:
is true but $(R)$ is false
Both (A) and $(R)$ are true but $(R)$ is not the correct explanation of (A)
is false but $(R)$ is true
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?
Increases
Remains unchanged
First decreases and then increases
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:
(A)-I, (B)-II, (C)-III, (D)-IV
(A)-II, (B)-IV, (C)-I, (D)-III
(A)-III, (B)-I, (C)-IV, (D)-II
(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 $B$.
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,
$\mathrm{AgCN}$ is soluble in $\mathrm{HNO_3}$
Silver halides are soluble in $\mathrm{HNO_3}$
$\mathrm{Ag_2S}$ is soluble in $\mathrm{HNO_3}$
$\mathrm{Na_2S}$ 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:
, (C), (E) only
, (D), (E) only
, (B), (C) only
, (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
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 $\underline{\hspace{1cm}}$.
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