JEE Main 17 March 2021 Shift 2 question paper with solutions
JEE Main 17 March 2021 Shift 2: 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 · Integrals · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = e^{-x} \sin x$. If $F : [0, 1] \to \mathbb{R}$ is a differentiable function such that $F(x) = \int_0^x f(t) \, dt$, then the value of $\int_0^1 \left( F'(x) + f(x) \right) e^x \, dx$ lies in the interval
If the integral $$\int_0^{10} \frac{\lfloor \sin 2 \pi x \rfloor}{e^{x - \lfloor x \rfloor}} \, dx = \alpha e^{-1} + \beta e^{-\frac{1}{2}} + \gamma,$$ where $\alpha, \beta, \gamma$ are integers and $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$, then the value of $\alpha + \beta + \gamma$ is equal to:
Let y = y(x) be the solution of the differential equation $$\cos x (3 \sin x + \cos x + 3) dy = (1 + y \sin x (3 \sin x + \cos x + 3)) dx$$ $$0 \leq x \leq \frac{\pi}{2}, y(0) = 0.$$ Then, $y\left(\frac{\pi}{3}\right)$ is equal to:
The value of $\lim_{n \to \infty} \frac{[r] + [2r] + \ldots + [nr]}{n^2}$, where $r$ is non-zero real number and $[r]$ denotes the greatest integer less than or equal to $r$, is equal to:
$\frac{r}{2}$
$r$
$2r$
$0$
Answer: (a)
Solution
We know that $$r \leq [r] < r + 1$$ and $$2r \leq [2r] < 2r + 1$$ $$3r \leq [3r] < 3r + 1$$ Now, $$\lim_{n \to \infty} \frac{n(n+1) \cdot r}{2 \cdot n^2} = \frac{r}{2}$$ and $$\lim_{n \to \infty} \frac{n(n+1)r}{2n^2} + \frac{n}{n^2} = \frac{r}{2}$$ So, by Sandwich Theorem, we can conclude that $$\lim_{n \to \infty} \frac{[r] + [2r] + \ldots + [nr]}{n^2} = \frac{r}{2}$$
Question 6
Maths · Inverse Trigonometric Functions · Single correct
The number of solutions of the equation $\sin^{-1}\left[x^2 + \frac{1}{3}\right] + \cos^{-1}\left[x^2 - \frac{2}{3}\right] = x^2$ for $x \in [-1, 1]$, and $[x]$ denotes the greatest integer less than or equal to $x$, is:
2
0
4
Infinite
Answer: (b)
Solution
Given equation $$\sin^{-1}\left[x^2 + \frac{1}{3}\right] + \cos^{-1}\left[x^2 - \frac{2}{3}\right] = x^2$$ Now, $\sin^{-1}\left[x^2 + \frac{1}{3}\right]$ is defined if $$-1 \leq x^2 + \frac{1}{3} < 2 \Rightarrow -\frac{4}{3} \leq x^2 < \frac{5}{3}$$ $$\Rightarrow 0 \leq x^2 < \frac{5}{3}$$ and $\cos^{-1}\left[x^2 - \frac{2}{3}\right]$ is defined if $$-1 \leq x^2 - \frac{2}{3} \leq 2 \Rightarrow -\frac{1}{3} \leq x^2 < \frac{8}{3}$$ $$\Rightarrow 0 \leq x^2 < \frac{8}{3}$$ So, from (1) and (2) we can conclude $$0 \leq x^2 < \frac{5}{3}$$ Case - I if $0 \leq x^2 < \frac{2}{3}$ $$\sin^{-1}(0) + \cos^{-1}(-1) = x^2$$ $$\Rightarrow x + \pi = x^2$$ $$\Rightarrow x^2 = \pi$$ but $\pi \notin \left[0, \frac{2}{3}\right]$ $$\Rightarrow$$ No value of 'x' Case - II if $\frac{2}{3} \leq x^2 < \frac{5}{3}$ $$\sin^{-1}(1) + \cos^{-1}(0) = x^2$$ $$\Rightarrow \frac{\pi}{2} + \frac{\pi}{2} = x^2$$ $$\Rightarrow x^2 = \pi$$ but $\pi \notin \left[\frac{2}{3}, \frac{5}{3}\right]$ $$\Rightarrow$$ No value of 'x' So, number of solutions of the equation is zero.
Question 7
Maths · Probability · Single correct
Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be $\frac{1}{2}$ and probability of occurrence of 0 at the odd place be $\frac{1}{3}$. Then the probability that '10' is followed by '01' is equal to:
The number of solutions of the equation $$x + 2 \tan x = \frac{\pi}{2}$$ in the interval $[0, 2\pi]$ is:
3
4
2
5
Answer: (a)
Solution
Given the equation $x + 2 \tan x = \frac{\pi}{2}$. This implies $2 \tan x = \frac{\pi}{2} - x$. Therefore, $\tan x = -\frac{1}{2}x + \frac{\pi}{4}$. The number of solutions of the given equation is '3'.
Question 9
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $S_1, S_2$ and $S_3$ be three sets defined as $$S_1 = \{ z \in \mathbb{C} : |z - 1| \leq \sqrt{2} \}$$ $$S_2 = \{ z \in \mathbb{C} : \mathrm{Re}((1-i)z) \geq 1 \}$$ $$S_3 = \{ z \in \mathbb{C} : \mathrm{Im}(z) \leq 1 \}$$ Then the set $S_1 \cap S_2 \cap S_3$
is a singleton
has exactly two elements
has infinitely many elements
has exactly three elements
Answer: (c)
Solution
Question 10
Maths · Differential Equations · Single correct
If the curve $y = y(x)$ is the solution of the differential equation $$2 \left( x^2 + x^{5/4} \right) dy - y \left( x + x^{1/4} \right) dx = 2x^{9/4} dx, \; x > 0$$ which passes through the point $\left( 1, 1 - \frac{4}{3} \log_e 2 \right)$, then the value of $y(16)$ is equal to:
Maths · Permutations and Combinations · Single correct
If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to:
364
240
333
360
Answer: (c)
Solution
Total Number of triangles formed $={}^{14}C_{3}-3\times{}^{8}C_{3}-{}^{5}C_{3}-{}^{6}C_{3}$ $=333$
Question 12
Maths · Determinants · Single correct
If $x, y, z$ are in arithmetic progression with common difference $d$, $x \neq 3d$, and the determinant of the matrix $$\begin{vmatrix} 3 & 4\sqrt{2} & x \\ 4 & 5\sqrt{2} & y \\ 5 & k & z \end{vmatrix}$$ is zero, then the value of $k^2$ is
72
12
36
6
Answer: (a)
Solution
Given the matrix equation: $$\begin{vmatrix} 3 & 4\sqrt{2} & x \\ 4 & 5\sqrt{2} & y \\ 5 & k & z \end{vmatrix}$$ Perform the row operation $R_2 \rightarrow R_1 + R_3 - 2R_2$: $$\begin{vmatrix} 3 & 4\sqrt{2} & x \\ 0 & k - 6\sqrt{2} & 0 \\ 5 & k & z \end{vmatrix} = 0$$ This implies: $$(k - 6\sqrt{2})(3z - 5x) = 0$$ If $3z - 5x = 0$ then $3(x + 2d) - 5x = 0$. Thus, $x = 3d$ (Not possible) implies $k = 6\sqrt{2}$, leading to $k^2 = 72$.
Question 13
Maths · Vector Algebra · Single correct
Let $\mathbf{O}$ be the origin. Let $\overrightarrow{OP}$ = x$\hat{i}$ + y$\hat{j}$ - $\hat{k}$ and $\overrightarrow{OQ}$ = -$\hat{i}$ + 2$\hat{j}$ + 3x$\hat{k}$, x, y $\in$ $\mathbb{R}$, x > 0, be such that | $\overrightarrow{PQ}$ | = $\sqrt{20}$ and the vector $\overrightarrow{OP}$ is perpendicular to $\overrightarrow{OQ}$. If $\overrightarrow{OR}$ = 3$\hat{i}$ + z$\hat{j}$ - 7$\hat{k}$, z $\in$ $\mathbb{R}$, is coplanar with $\overrightarrow{OP}$ and $\overrightarrow{OQ}$, then the value of $x^2$ + $y^2$ + $z^2$ is equal to
Two tangents are drawn from a point P to the circle $x^2 + y^2 - 2x - 4y + 4 = 0$, such that the angle between these tangents is $\tan^{-1}\left(\frac{12}{5}\right) \in (0, \pi)$. If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of $\triangle PAB$ and $\triangle CAB$ is :
11: 4
9: 4
3: 1
2: 1
Answer: (b)
Solution
Given $\tan \theta = \frac{12}{5}$. $PA = \cot \frac{\theta}{2}$. Therefore, the area of $\Delta PAB = \frac{1}{2} (PA)^2 \sin \theta = \frac{1}{2} \cot^2 \frac{\theta}{2} \sin \theta$. $$= \frac{1}{2} \left( \frac{1 + \cos \theta}{1 - \cos \theta} \right) \sin \theta$$ The area of $\Delta CAB = \frac{1}{2} \sin \theta = \frac{1}{2} \left( \frac{12}{13} \right) = \frac{6}{13}$. Therefore, $\frac{area of \Delta PAB}{area of \Delta CAB} = \frac{9}{4}$.
Question 15
Maths · Applications of Derivatives · Single correct
Consider the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \begin{cases} \left(2 - \sin\left(\frac{1}{x}\right)\right)|x|, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f$ is
monotonic on $(-\infty, 0) \cup (0, \infty)$
not monotonic on $(-\infty, 0)$ and $(0, \infty)$
monotonic on $(0, \infty)$ only
monotonic on $(-\infty, 0)$ only
Answer: (b)
Solution
Given $$f(x) = \begin{cases} -x \left( 2 - \sin \left( \frac{1}{x} \right) \right) & x 0 \end{cases}$$ The derivative is $$f'(x) = \begin{cases} - \left( 2 - \sin \frac{1}{x} \right) - x \left( - \cos \frac{1}{x} \cdot \left( - \frac{1}{x^2} \right) \right) & x 0 \end{cases}$$ Simplifying, we have $$f'(x) = \begin{cases} -2 + \sin \frac{1}{x} - \frac{1}{x} \cos \frac{1}{x} x & x 0 \end{cases}$$ $f'(x)$ is an oscillating function which is non-monotonic in $(-\infty, 0) \cup (0, \infty)$.
Question 16
Maths · Conic Sections · Single correct
Let L be a tangent line to the parabola $y^2 = 4x - 20$ at $(6, 2)$. If L is also a tangent to the ellipse $$\frac{x^2}{2} + \frac{y^2}{b} = 1$$, then the value of $b$ is equal to:
11
14
16
20
Answer: (b)
Solution
Tangent to parabola $2y = 2(x + 6) - 20$ $$\Rightarrow y = x - 4$$ Condition of tangency for ellipse. $16 = 2(1)^2 + b$ $$\Rightarrow b = 14$$
Question 17
Maths · Limits and Derivatives · Single correct
The value of the limit $\lim_{\theta \to 0} \frac{\tan(\pi \cos^2 \theta)}{\sin(2\pi \sin^2 \theta)}$ is equal to :
$-\frac{1}{2}$
$-\frac{1}{4}$
$0$
$\frac{1}{4}$
Answer: (a)
Solution
Given the limit expression: $$\lim_{\theta \to 0} \frac{\tan(\pi(1 - \sin^2 \theta))}{\sin(2\pi \sin^2 \theta)}$$ We simplify it as follows: $$= \lim_{\theta \to 0} \frac{-\tan(\pi \sin^2 \theta)}{\sin(2\pi \sin^2 \theta)}$$ $$= \lim_{\theta \to 0} - \left( \frac{\tan(\pi \sin^2 \theta)}{\pi \sin^2 \theta} \right) \left( \frac{2\pi \sin^2 \theta}{\sin(2\pi \sin^2 \theta)} \right) \times \frac{1}{2}$$ $$= \frac{-1}{2}$$
Question 18
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let the tangent to the circle $x^2 + y^2 = 25$ at the point $R(3, 4)$ meet $x$-axis and $y$-axis at point $P$ and $Q$, respectively. If $r$ is the radius of the circle passing through the origin $O$ and having centre at the incentre of the triangle $OPQ$, then $r^2$ is equal to
$\($ $\frac{529}{64}$ $\)$
$\($ $\frac{125}{72}$ $\)$
$\($ $\frac{625}{72}$ $\)$
$\($ $\frac{585}{66}$ $\)$
Answer: (c)
Solution
Tangent to circle $3x + 4y = 25$. OP + OQ + OR = 25 Incentre = $\left$( $\frac{25}{4}$ $\times$ $\frac{25}{3}$ $\frac{25}{25}$, $\frac{25}{4}$ $\times$ $\frac{25}{3}$ $\frac{1}{25}$ $\right$) = $\left$( $\frac{25}{12}$, $\frac{25}{12}$ $\right$) $\therefore$ r^2 = 2 $\left$( $\frac{25}{12}$ $\right$)^2 = 2 $\times$ $\frac{625}{144}$ = $\frac{625}{72}$
Question 19
Maths · Mathematical Reasoning · Single correct
If the Boolean expression $(p \land q) \odot (p \otimes q)$ is a tautology, then $\odot$ and $\otimes$ are respectively given by
$\rightarrow, \rightarrow$
$\land, \lor$
$\lor, \rightarrow$
$\land, \rightarrow$
Answer: (a)
Solution
Option (1) $$(p \land q) \longrightarrow (p \rightarrow q)$$ $$= \sim (p \land q) \lor (\sim p \lor q)$$ $$= (\sim p \lor \sim q) \lor (\sim p \lor q)$$ $$= \sim p \lor (\sim q \lor q)$$ $$= \sim p \lor t$$ $$= t$$ Option (2) $$(p \land q) \land (p \lor q) = (p \land q) \quad \text{(Not a tautology)}$$ Option (3) $$(p \land q) \lor (p \rightarrow q)$$ $$= (p \land q) \lor (\sim p \lor q)$$ $$= \sim p \lor q$$ (Not a tautology) Option (4) $$(p \land q) \land (p \rightarrow q)$$ $$= (p \land q) \land (\sim p \lor q)$$ $$= p \land q$$ (Not a tautology)
Question 20
Maths · Three Dimensional Geometry · Single correct
If the equation of plane passing through the mirror image of a point $(2,3,1)$ with respect to line $\frac{x-2}{3} = \frac{1-y}{2} = \frac{z+1}{1}$ and containing the line $\frac{x+1}{2} = \frac{y-3}{1} = \frac{z+2}{-1}$ is $\alpha x + \beta y + \gamma z = 24$ then $\alpha + \beta + \gamma$ is equal to:
If 1, $\log_{10}(4^x - 2)$ and $\log_{10}\left(4^x + \frac{18}{5}\right)$ are in arithmetic progression for a real number $x$, then the value of the determinant $$\begin{vmatrix} 2\left(x - \frac{1}{2}\right) & x - 1 & x^2 \\ 1 & 0 & x \\ x & 1 & 0 \end{vmatrix}$$ is equal to :
Let $f : [-1, 1] \to \mathbb{R}$ be defined as $f(x) = ax^2 + bx + c$ for all $x \in [-1, 1]$, where $a, b, c \in \mathbb{R}$ such that $f(-1) = 2$, $f'(-1) = 1$ and for $x \in (-1, 1)$ the maximum value of $f''(x)$ is $\frac{1}{2}$. If $f(x) \leq \alpha$ $x \in [-1, 1]$, then the least value of $\alpha$ is equal to .
Answer: 5
Solution
Given $f : [-1, 1] \to \mathbb{R}$ $f(x) = ax^2 + bx + c$ $f(-1) = a - b + c = 2$ $f'(-1) = -2a + b = 1$ $f''(x) = 2a$ Therefore, the maximum value of $f''(x) = 2a = \frac{1}{2}$ Thus, $a = \frac{1}{4}$, $b = \frac{3}{2}$, $c = \frac{13}{4}$ Therefore, $f(x) = \frac{x^2}{4} + \frac{3}{2}x + \frac{13}{4}$ For $x \in [-1, 1]$, $2 \leq f(x) \leq 5$ Therefore, the least value of $\alpha$ is $5$
Question 23
Maths · Applications of Integrals · Numerical
Let f : [-3, 1] $\to$ $\mathbb{R}$ be given as $$f(x) = \begin{cases} \min \{(x+6), x^2\}, & -3 \leq x \leq 0 \\ \max \{\sqrt{x}, x^2\}, & 0 \leq x \leq 1 \end{cases}$$ If the area bounded by $y = f(x)$ and $x$-axis is $A$, then the value of $6A$ is equal to ____.
Answer: 41
Solution
Given the function $f : [-3, 1] \to \mathbb{R}$ defined by $$f(x) = \begin{cases} \min \{ (x+6), x^2 \}, & -3 \leq x \leq 0 \\ \max \{ \sqrt{x}, x^2 \}, & 0 \leq x \leq 1 \end{cases}$$ The area bounded by $y = f(x)$ and the $x$-axis is given by $$\int_{-3}^{-2} (x+6) \, dx + \int_{-2}^{0} x^2 \, dx + \int_{0}^{1} \sqrt{x} \, dx$$ Calculating the area, we find $$A = \frac{41}{6}$$ Thus, $6A = 41$.
Question 24
Maths · Straight Lines and Pair of Straight Lines · Numerical
Let $\tan \alpha$, $\tan \beta$ and $\tan \gamma$; $\alpha$, $\beta$, $\gamma \neq \frac{(2n-1)\pi}{2}$, $n \in \mathbb{N}$ be the slopes of three line segments $OA$ $OB$ and $OC$, respectively, where $O$ is origin. If circumcentre of $\triangle ABC$ coincides with origin and its orthocentre lies on $y$-axis, then the value of $\left( \frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos \alpha \cos \beta \cos \gamma} \right)^2$ is equal to :
Consider a set of $3n$ numbers having variance $4$. In this set, the mean of first $2n$ numbers is $6$ and the mean of the remaining $n$ numbers is $3$. A new set is constructed by adding $1$ into each of first $2n$ numbers, and subtracting $1$ from each of the remaining $n$ numbers. If the variance of the new set is $k$, then $9k$ is equal to ____.
Let the coefficients of third, fourth and fifth terms in the expansion of $\left( x + \frac{a}{x^2} \right)^n$, $x \neq 0$, be in the ratio 12: 8: 3. Then the term independent of $x$ in the expansion, is equal to ____.
Answer: 4
Solution
Given $$T_{r+1} = nC_r(x)^{n-r} \left( \frac{a}{x^2} \right)^r$$ $$= nC_ra^rx^{n-3r}$$ $$nC_2a^2 : nC_3a^3 : nC_4a^4 = 12 : 8 : 3$$ After solving $$n = 6, a = \frac{1}{2}$$ For term independent of 'x' implies $$n = 3r$$ $$r = 2$$ Therefore, Coefficient is $$6C_2 \left( \frac{1}{2} \right)^2 = \frac{15}{4}$$ Nearest integer is 4
Question 27
Maths · Matrices · Fill in the blank
Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and and $B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}\ne\begin{pmatrix}0\\0\end{pmatrix}$ such that $AB=B$ and $a+d=2021$, then the value of $ad-bc$ is equal to $\underline{\hspace{1cm}}$.
Answer: 2020
Solution
Given matrices $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$, with $AB = B$. This implies $$(A - I)B = O$$ which further implies $$|A - I| = O, since B \neq O$$ Therefore, $$\begin{vmatrix} (a - 1) & b \\ c & (d - 1) \end{vmatrix} = 0$$ which gives $$ad - bc = 2020$$
Question 28
Maths · Vector Algebra · Fill in the blank
Let $\vec{x}$ be a vector in the plane containing vectors $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$. If the vector $\vec{x}$ is perpendicular to $(3\hat{i}+2\hat{j}-\hat{k})$, and its projection on $\vec{a}$ is $\frac{17\sqrt{6}}{2}$, then the value of $|\vec{x}|^2$ is equal to $\underline{\hspace{2cm}}$.
Answer: 486
Solution
Let $\vec{x} = \lambda \vec{a} + \mu \vec{b}$ ($\lambda$ and $\mu$ are scalars). $$\vec{x} = \hat{i}(2\lambda + \mu) + \hat{j}(2\mu - \lambda) + \hat{k}(\lambda - \mu)$$ Since $\vec{x} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 0$ $$3\lambda + 8\mu = 0 \ldots$$ Also, the projection of $\vec{x}$ on $\vec{a}$ is $\frac{17\sqrt{6}}{2}$ $$\frac{\vec{x} \cdot \vec{a}}{|\vec{a}|} = \frac{17\sqrt{6}}{2}$$ $$6\lambda - \mu = 51$$ From (1) and (2) $\lambda = 8$, $\mu = -3$ $$\vec{x} = 13\hat{i} - 14\hat{j} + 11\hat{k}$$ $$|\vec{x}|^2 = 486$$
Question 29
Maths · Integrals · Numerical
Let $I_n = \int_{1}^{e} x^{19} (\log |x|)^n \, dx$, where $n \in \mathbb{N}$. If $(20) I_{10} = \alpha I_9 + \beta I_8$, for natural numbers $\alpha$ and $\beta$, then $\alpha - \beta$ equal to ____.
Answer: 1
Solution
Let $\vec{x} = \lambda \vec{a} + \mu \vec{b}$ ($\lambda$ and $\mu$ are scalars). $\vec{x} = \hat{i}(2\lambda + \mu) + \hat{j}(2\mu - \lambda) + \hat{k}(\lambda - \mu)$. Since $\vec{x} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 0$, $$3\lambda + 8\mu = 0$$ Also, the projection of $\vec{x}$ on $\vec{a}$ is $\frac{17\sqrt{6}}{2}$. $$\frac{\vec{x} \cdot \vec{a}}{|\vec{a}|} = \frac{17\sqrt{6}}{2}$$ $$6\lambda - \mu = 51$$ From (1) and (2), $\lambda = 8, \mu = -3$. $\vec{x} = 13\hat{i} - 14\hat{j} + 11\hat{k}$. $$|\vec{x}|^2 = 486$$
Question 30
Maths · Three Dimensional Geometry · Numerical
Let $P$ be an arbitrary point having sum of the squares of the distance from the planes $$x + y + z = 0, \, lx - nz = 0 and x - 2y + z = 0$$ equal to 9. If the locus of the point $P$ is $x^2 + y^2 + z^2 = 9$, then the value of $l - n$ is equal to ___.
Answer: 0
Solution
Let point P is $(\alpha, \beta, \gamma)$. $$\left( \frac{\alpha + \beta + \gamma}{\sqrt{3}} \right)^2 + \left( \frac{\ell \alpha - n \gamma}{\sqrt{\ell^2 + n^2}} \right)^2 + \left( \frac{\alpha - 2\beta + \gamma}{\sqrt{6}} \right)^2 = 9$$ Locus is $$\frac{(x+y+z)^2}{3} + \frac{(\ell x - nz)^2}{\ell^2 + n^2} + \frac{(x - 2y + z)^2}{6} = 9$$ $$x^2 \left( \frac{1}{2} + \frac{\ell^2}{\ell^2 + n^2} \right) + y^2 + z^2 \left( \frac{1}{2} + \frac{n^2}{\ell^2 + n^2} \right) + 2zx \left( \frac{1}{2} - \frac{\ell n}{\ell^2 + n^2} \right) - 9 = 0$$ Since it's given that $x^2 + y^2 + z^2 = 9$ After solving $\ell = n$
Physics
Question 31
Physics · Motion in a Straight Line · Single correct
A rubber ball is released from a height of $5\,\mathrm{m}$ above the floor. It bounces back repeatedly, always rising to $\frac{81}{100}$ of the height through which it falls. Find the average speed of the ball. (Take $g = 10\,\mathrm{m\,s^{-2}}$)
$3.0\,\mathrm{m\,s^{-1}}$
$3.50\,\mathrm{m\,s^{-1}}$
$2.0\,\mathrm{m\,s^{-1}}$
$2.50\,\mathrm{m\,s^{-1}}$
Answer: (d)
Solution
Given $v_0 = \sqrt{2gh}$. $v = e \sqrt{2gh} = \sqrt{2gh}$. This implies $e = 0.9$. $s = h + 2e^2 h + 2e^4 h + \ldots$ $t = \sqrt{\frac{2h}{g}} + 2e \sqrt{\frac{2h}{g}} + 2e^2 \sqrt{\frac{2h}{g}} + \ldots$ $v_{av} = \frac{s}{t} = 2.5 \, \mathrm{m/s}$
Question 32
Physics · Kinetic Theory · Single correct
If one mole of the polyatomic gas is having two vibrational modes and $\beta$ is the ratio of molar specific heats for polyatomic gas $\left( \beta = \frac{C_P}{C_V} \right)$ then the value of $\beta$ is :
A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 $\mathrm{cm}$. After 2 minutes the amplitude decreases to 6 $\mathrm{cm}$. Determine the value of the damping constant for this motion. (take $\ln$ 2 = 0.693)
Given $A = A_0 e^{-\gamma t}$. We have $\ln 2 = \frac{b}{2m} \times 120$. Solving for $b$, we get: $$\frac{0.693 \times 2 \times 1}{120} = b$$ The result is $1.16 \times 10^{-2} \, \mathrm{kg/sec}$.
Question 34
Physics · Communication Systems · Single correct
Which one of the following will be the output of the given circuit?
NOR Gate
AND Gate
AND Gate
XOR Gate
Answer: (d)
Solution
Conceptual
Question 35
Physics · Mechanical Properties of Fluids · Single correct
An object is located at 2 km beneath the surface of the water. If the fractional compression $\frac{\Delta V}{V}$ is 1.36$\%$, the ratio of hydraulic stress to the corresponding hydraulic strain will be [Given : density of water is $1000 \, \mathrm{kg} \, \mathrm{m}^{-3}$ and $g = 9.8 \, \mathrm{ms}^{-2}$.]
A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of $11R$ above the surface of 'P', $R$ being the radius of 'P'. The time period of another satellite in hours at a height of $2R$ from the surface of 'P' is $\_$$\_$$\_$$\_$$\_$ has the time period of 24 hours.
$6\sqrt{2}$
$\frac{6}{\sqrt{2}}$
3
5
Answer: (c)
Solution
Given $T \propto R^{3/2}$. $$\frac{24}{T} = \left(\frac{12R}{3R}\right)^{3/2} \Rightarrow T = 3 \, \mathrm{hr}$$
Question 37
Physics · Waves · Single correct
A sound wave of frequency $245~\mathrm{Hz}$ travels with a speed of $300~\mathrm{m~s^{-1}}$ along the positive $x$-axis. Each point of the wave moves to and fro through a total distance of $6~\mathrm{cm}$. What will be the mathematical expression of this travelling wave?
Which one is the correct option for the two different thermodynamic processes?
$(C)$ and $(A)$
$(C)$ and $(D)$
$(A)$ only
$(B)$ and $(C)$
Answer: (b)
Solution
Question 39
Physics · Motion in a Straight Line · Single correct
The velocity of a particle is $v = v_0 + gt + Ft^2$ Its position is $x = 0$ at $t = 0$; then its displacement after time $(t = 1)$ is :
$v_0 + g + F$
$v_0 + \frac{g}{2} + \frac{F}{3}$
$v_0 + \frac{g}{2} + F$
$v_0 + 2g + 3F$
Answer: (b)
Solution
Given $v = v_0 + gt + Ft^2$. The derivative $\frac{ds}{dt} = v_0 + gt + Ft^2$. Integrating both sides: $$\int ds = \int_0^1 (v_0 + gt + Ft^2) \, dt$$ This gives: $$s = \left[ v_0 t + \frac{gt^2}{2} + \frac{Ft^3}{3} \right]_0^1$$ Evaluating the integral: $$s = v_0 + \frac{g}{2} + \frac{F}{3}$$
Question 40
Physics · Communication Systems · Single correct
A carrier signal $C(t) = 25 \sin(2.512 \times 10^{10} t)$ is amplitude modulated by a message signal $m(t) = 5 \sin(1.57 \times 10^8 t)$ and transmitted through an antenna. What will be the bandwidth of the modulated signal?
Two cells of emf $2E$ and $E$ with internal resistance $r_1$ and $r_2$ respectively are connected in series to an external resistor $R$ (see figure). The value of $R$, at which the potential difference across the terminals of the first cell becomes zero is
$r_1 + r_2$
$\frac{r_1}{2} - r_2$
$\frac{r_1}{2} + r_2$
$r_1 - r_2$
Answer: (b)
Solution
The current $i$ is given by $$i = \frac{3E}{R + r_1 + r_2}.$$ The terminal potential difference (TPD) is $$TPD = 2E - ir_1 = 0.$$ Therefore, $$2E = ir_1.$$ Substituting for $i$, we have $$2E = \frac{3E \times r_1}{R + r_1 + r_2}.$$ Simplifying, $$2R + 2r_1 + 2r_2 = 3r_1.$$ Solving for $R$, we get $$R = \frac{r_1}{2} - r_2.$$
Question 42
Physics · Moving Charges and Magnetism · Single correct
A hairpin like shape as shown in figure is made by bending a long current carrying wire. What is the magnitude of a magnetic field at point P which lies on the centre of the semicircle?
The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15$\Omega$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 $\mathrm{V}$ is maintained across AC.
2.44$\mu$$\mathrm{A}$
2.44 $\mathrm{mA}$
4.87 $\mathrm{mA}$
4.87$\mu$$\mathrm{A}$
Answer: (c)
Solution
Given the circuit, we have the following equations: $$\frac{x - 10}{100} + \frac{x - y}{15} + \frac{x - 0}{10} = 0$$ $$53x - 20y = 30 \ldots (1)$$ $$\frac{y - 10}{60} + \frac{y - x}{15} + \frac{y - 0}{5} = 0$$ $$17y - 4x = 10 \ldots (2)$$ On solving equations (1) and (2), we find: $$x = 0.865$$ $$y = 0.792$$ The voltage difference is given by: $$\Delta V = 0.073R = 15\Omega$$ The current is: $$i = 4.87 \, \mathrm{mA}$$
Question 44
Physics · Oscillations · Single correct
Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_1$ and $K_2$ respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of $A$ and $B$ is
Match List-I with List-II List-I \begin{tabular}{|c|p{7cm}|c|p{5cm}|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \hline (a) & Phase difference between current and voltage in a purely resistive AC circuit & (i) & $\dfrac{\pi}{2}$ ; current leads voltage \\ \hline (b) & Phase difference between current and voltage in a pure inductive AC circuit & (ii) & zero \\ \hline (c) & Phase difference between current and voltage in a pure capacitive AC circuit & (iii) & $\dfrac{\pi}{2}$ ; current lags voltage \\ \hline (d) & Phase difference between current and voltage in an LCR series AC circuit & (iv) & $\tan^{-1}\left(\dfrac{X_C-X_L}{R}\right)$ \\ \hline \end{tabular}
In option (a), the voltage $V$ is equal to the resistive voltage $V_R$ with current $I$ in the same direction. In option (b), the inductive voltage $V_L$ is perpendicular to the current $I$. In option (c), the capacitive voltage $V_C$ is opposite to the current $I$. In option (d), the tangent of the phase angle $\phi$ is given by $$\tan \phi = \frac{V_L - V_C}{V_R} = \frac{X_L - X_C}{R}.$$
Question 46
Physics · System of Particles and Rotational Motion · Single correct
Two identical blocks $A$ and $B$ each of mass $m$ resting on the smooth horizontal floor are connected by a light spring of natural length $L$ and spring constant $K$. A third block $C$ of mass $m$ moving with a speed $v$ along the line joining $A$ and $B$ collides with $A$. The maximum compression in the spring is
$\sqrt{\frac{M}{2K}}$
$\sqrt{\frac{mv}{2K}}$
$\sqrt{\frac{mv}{K}}$
$\sqrt{\frac{m}{2K}}$
Answer: (a)
Solution
C comes to rest $V_{cm}$ of $A \& B = \frac{v}{2}$ $$\Rightarrow \frac{1}{2} is v_{ret}^2 = \frac{1}{2} k x^2$$ $$x = \sqrt{\frac{\mu \times v^2}{k}} = \sqrt{\frac{m}{2k}} v$$
Question 47
Physics · Atoms · Single correct
The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region?
Brackett series
Paschen series
Lyman series
Balmer series
Answer: (d)
Solution
Conceptual
Question 48
Physics · Dual Nature of Radiation and Matter · Single correct
Two identical photocathodes receive the light of frequencies $f_1$ and $f_2$ respectively. If the velocities of the photo-electrons coming out are $v_1$ and $v_2$ respectively, then
Given $\dfrac{1}{2}mv_1^2 = hf_1 - \phi$ and $\dfrac{1}{2}mv_2^2 = hf_2 - \phi$. Subtracting the second equation from the first, we get: $$v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$$
Question 49
Physics · Alternating Current · Single correct
What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved?
Both, inductive reactance and current will be halved.
Inductive reactance will be halved and current will be doubled.
Inductive reactance will be doubled and current will be halved.
Both, inductive reactance and current will be doubled.
Answer: (b)
Solution
Given $X_L = \omega L$. Therefore, $$i = \frac{v_0}{\omega L}$$
Question 50
Physics · System of Particles and Rotational Motion · Single correct
A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 $\mathrm{ms}^{-1}$ goes up an inclined plane which makes an angle of $30^\circ$ with the horizontal plane, without slipping. How low will the sphere take to return to the starting point A ?
The electric field intensity produced by the radiation coming from a $100\,\mathrm{W}$ bulb at a distance of $3\,\mathrm{m}$ is $E$. The electric field intensity produced by the radiation coming from $60\,\mathrm{W}$ at the same distance is $\sqrt{\dfrac{x}{5}}\,E$. Where the value of $x =$
A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction $\frac{1}{\sqrt{3}}$. It is desired to make the body move by applying the minimum possible force $F_N$. The value of $F$ will be ______ (Round off to the Nearest Integer) [Take $g = 10 \, \mathrm{ms^{-2}}$]
Answer: 5
Solution
Given the equations: $$F \cos \theta = \mu N$$ $$F \sin \theta + N = mg$$ We can solve for $F$: $$\Rightarrow F = \frac{\mu mg}{\cos \theta + \mu \sin \theta}$$ The minimum force $F_{\min}$ is given by: $$F_{\min} = \frac{\mu mg}{\sqrt{1 + \mu^2}} = \frac{1}{\sqrt{3}} \times 10 = \frac{2}{\sqrt{3}} = 5$$
Question 53
Physics · Laws of Motion · Numerical
A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is N. (Round off to the Nearest Integer) [Take $g = 10 \, \mathrm{ms^{-2}}$]
Answer: 30
Solution
Given the equations: $$N + T = 90$$ $$T = \mu N = 0.5(90 - T)$$ Solving for $T$: $$1.5 \, T = 45$$ Therefore, $$T = 30$$
Question 54
Physics · Mechanical Properties of Fluids · Numerical
Suppose you have taken a dilute solution of oleic acid such that its concentration becomes $0.01\,\mathrm{cm^3}$ of oleic acid per $\mathrm{cm^3}$ of the solution. Then you make a thin film of this solution (of monomolecular thickness) having an area of $4\,\mathrm{cm^2}$ by using 100 spherical drops of radius $\left(\frac{3}{40\pi}\right)^{\frac{1}{3}} \times 10^{-3}\,\mathrm{cm}$. Then the thickness of the oleic acid layer will be $x \times 10^{-14}\,\mathrm{m}$, where $x$ is:
A particle of mass $m$ moves in a circular orbit in a central potential field $U(r) = U_0 r^4$. If Bohr's quantization conditions are applied, radii of possible orbitals $r_n$ vary with $n^{1/\alpha}$, where $\alpha$ is ______
Answer: 3
Solution
Given the force equation $$F = -\frac{dU}{dr} = -4U_0 r^3 = \frac{mv^2}{r}$$ we have $$mv^2 = 4U_0 r^4$$ which implies $$v \propto r^2$$ Using the angular momentum equation $$mvr = \frac{nh}{2\pi}$$ we find $$r^3 \propto n$$ Therefore, $$r \propto n^{1/3}$$ Thus, the answer is $$= 3$$
Question 56
Physics · Electric Charges and Fields · Numerical
The electric field in a region is given by $$\vec{E} = \frac{2}{5} E_0 \hat{i} + \frac{3}{5} E_0 \hat{j}$$ with $E_0 = 4.0 \times 10^3 \, \mathrm{N/C}$. The flux of this field through a rectangular surface area $0.4 \, \mathrm{m}^2$ parallel to the $Y - Z$ plane is ______ $\mathrm{Nm}^2\mathrm{C}^{-1}$
Physics · System of Particles and Rotational Motion · Numerical
The disc of mass $M$ with uniform surface mass density $\sigma$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $\frac{x}{3} \frac{a}{\pi}, \frac{x}{3} \frac{a}{\pi}$ where $x$ is ________ (Round off to the Nearest Integer) [a \text{ is an area as shown in the figure}]
Answer: 4
Solution
C.O.M of quarter disc is at $\frac{4a}{3\pi}$, $\frac{4a}{3\pi}$ $$= 4$$
Question 58
Physics · Ray Optics and Optical Instruments · Numerical
The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at $\frac{2rd}{3}$ of the distance of the object from the surface. The wavelength of light inside the surface is $\frac{2}{3}$ times the wavelength in air. The radius of the curved surface is $\frac{x}{13}$ m. the value of 'x' is ______
Answer: 30
Solution
Given $\lambda_m = \frac{\lambda_a}{\mu}$, it implies $\mu = \frac{3}{2}$. The equation $\frac{\mu}{v} - \frac{1}{u} = \frac{\mu - 1}{R}$ can be rewritten as: $$\frac{3}{2 \times 10} + \frac{1}{15} = \frac{\frac{3}{2} - 1}{R}$$ Solving for $R$, we find: $$R = \frac{30}{13}$$ Thus, the final result is $= 30$.
Question 59
Physics · Electrostatic Potential and Capacitance · Numerical
A $2\,\mu\mathrm{F}$ capacitor $C_1$ is first charged to a potential difference of $10\,\mathrm{V}$ using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor $C_2$ of $8\,\mu\mathrm{F}$. The charge in $C_2$ on equilibrium condition is $\_\_\_\,\mu\mathrm{C}$. (Round off to the Nearest Integer)
Answer: 16
Solution
Given $20 = (C_1 + C_2) V \Rightarrow V = 2 \, volt$. $Q_2 = C_2 V = 16 \, \mu C$. Therefore, the answer is $16$.
Question 60
Physics · Current Electricity · Numerical
Seawater at a frequency $f = 9 \times 10^2 \, \mathrm{Hz}$, has permittivity $\varepsilon = 80\varepsilon_0$ and resistivity $\rho = 0.25 \, \Omega \mathrm{m}$. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source $V(t) = V_0 \sin(2\pi ft)$. Then the conduction current density becomes $10^x$ times the displacement current density after time t = $\frac{1}{800}$ $\mathrm{s}$. The value of x is ______ $(Given : \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \mathrm{Nm^2C^{-2}})$
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The set of elements that differ in mutual relationship from those of the other sets is:
$\mathrm{Li} - \mathrm{Mg}$
$\mathrm{B} - \mathrm{Si}$
$\mathrm{Be} - \mathrm{Al}$
$\mathrm{Li} - \mathrm{Na}$
Solution
Li - $\mathrm{Mg}$, $\mathrm{B}$ - $\mathrm{Si}$, $\mathrm{Be-Al}$ show diagonal relationship but Li and Na do not show diagonal relationship as both belongs to same group and not placed diagonally.
Question 63
Chemistry · The s-Block Elements · Single correct
The functional groups that are responsible for the ion-exchange property of cation and anion exchange resins, respectively, are:
$-\mathrm{SO}_3\mathrm{H}$ and $-\mathrm{NH}_2$
$-\mathrm{SO}_3\mathrm{H}$ and $-\mathrm{COOH}$
$-\mathrm{NH}_2$ and $-\mathrm{COOH}$
$-\mathrm{NH}_2$ and $-\mathrm{SO}_3\mathrm{H}$
Answer: (a)
Solution
Cation exchanger contains $-\mathrm{SO_3H}$ or $-\mathrm{COOH}$ groups while anion exchanger contains basic groups like $-\mathrm{NH_2}$.
Question 64
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Match List-I and List-II : \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List-I} & \multicolumn{2}{c|}{List-II} \\ \hline (a) & Haematite & (i) & Al$_2$O$_3\cdot x$H$_2$O \\ \hline (b) & Bauxite & (ii) & Fe$_2$O$_3$ \\ \hline (c) & Magnalite & (iii) & CuCO$_3\cdot$Cu(OH)$_2$ \\ \hline (d) & Malachite & (iv) & Fe$_3$O$_4$ \\ \hline \end{tabular} Choose the correct answer from the options given below :
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The correct pair(s) of the ambident nucleophiles is (are): (A) $\mathrm{AgCN/KCN}$ (B) $\mathrm{RCOOAg/RCOOK}$ (C) $\mathrm{AgNO_2/KNO_2}$ (D) $\mathrm{AgI/KI}$
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
The set that represents the pair of neutral oxides of nitrogen is:
$NO$ and $N_2O$
$N_2O$ and $N_2O_3$
$N_2O$ and $NO_2$
$NO$ and $NO_2$
Answer: (a)
Question 67
Chemistry · Co-ordination Compounds · Single correct
Match List-I with List-II : Choose the correct answer from the options given below :
(a) - (iii), (b) - (i), (c) - (ii), (d) - (iv)
(a) -(iv), (b)-(ii), (c)-(iii), (d)-(i)
(a) -(ii), (b)-(i), (c)-(iii), (d)-(iv)
(a) -(i), (b)-(ii), (c)-(iii), (d)-(iv)
Answer: (a)
Solution
The complexes and their corresponding types of isomerism are as follows: (a) $[\mathrm{Co(NH_3)_6}] [\mathrm{Cr(CN)_6}]$ exhibits co-ordination isomerism. (b) $[\mathrm{Co(NH_3)_3} (\mathrm{NO_2})_3]$ exhibits linkage isomerism. (c) $[\mathrm{Cr(H_2O)_6}]\mathrm{Cl_3}$ exhibits solvate isomerism. (d) $\textit{cis}$-$[\mathrm{CrCl_2(ox)}]_3^-$ exhibits optical isomerism.
Question 68
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Primary, secondary and tertiary amines can be separated using :-
Para-Toluene sulphonyl chloride
Chloroform and KOH
Benzene sulphonic acid
Acetyl amide
Answer: (a)
Solution
Primary amines react with Para Toluene sulfonyl chloride to form a precipitate that is soluble in $\mathrm{NaOH}$. Secondary amines react with para toluene sulfonyl chloride to give a precipitate that is insoluble in $\mathrm{NaOH}$. Tertiary amines do not react with para toluen.
Question 69
Chemistry · The d-and f-Block Elements · Single correct
The common positive oxidation states for an element with atomic number 24, are:
+2 to +6
+1 and +3 to +6
+1 and +3
+1 to +6
Answer: (a)
Solution
Cr $(Z = 24)$ $[\mathrm{Ar}] \, 4s^1 \, 3d^5$ Cr shows common oxidation states starting from $+2$ to $+6$.
Question 70
Chemistry · Chemistry in Everyday Life · Single correct
Match List-I with List-II : Choose the correct match :
Given below are two statements: Statement-I: 2-methylbutane on oxidation with $\mathrm{KMnO_4}$ gives 2-methylbutan-2-ol. Statement-II: n-alkanes can be easily oxidised to corresponding alcohol with $\mathrm{KMnO_4}$. Choose the correct option:
Both statement I and statement II are correct
Both statement I and statement II are incorrect
Statement I is correct but Statement II is incorrect
Statement I is incorrect but Statement II is correct
Answer: (c)
Solution
Alkanes are very less reactive, tertiary hydrogen can oxidize to alcohol with $\mathrm{KMnO_4}$.
Question 72
Chemistry · Analytical Chemistry · Single correct
Nitrogen can be estimated by Kjeldahl's method for which of the following compound?
Answer: (b)
Solution
Kjeldahl method is not applicable to compounds containing nitrogen in nitrogroup, Azo groups and nitrogen present in the ring (e.g Pyridine) as nitrogen of these compounds does not change to Ammonium sulphate under these conditions.
Question 73
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Amongst the following, the linear species is:
$\mathrm{NO}_2$
$\mathrm{Cl}_2\mathrm{O}$
$\mathrm{O}_3$
$\mathrm{N}_3^-$
Answer: (d)
Solution
The first molecule is $\mathrm{NO_2}$, which has a bent shape. The second molecule is $\mathrm{Cl_2O}$, which also has a bent shape. The third molecule is $\mathrm{O_3}$, which has a bent shape as well. The fourth molecule is $\mathrm{N_3^-}$, which is linear.
Question 74
Chemistry · Biomolecules · Single correct
$$ \underset{\mathrm{Sucrose}}{\mathrm{C_{12}H_{22}O_{11}+H_2O}} \xrightarrow{\mathrm{Enzyme\ A}} \underset{\mathrm{Glucose}}{\mathrm{C_6H_{12}O_6}} + \underset{\mathrm{Fructose}}{\mathrm{C_6H_{12}O_6}} $$ $$ \underset{\mathrm{Glucose}}{\mathrm{C_6H_{12}O_6}} \xrightarrow{\mathrm{Enzyme\ B}} \mathrm{2C_2H_5OH+2CO_2} $$ In the above reactions, the enzyme A and enzyme B respectively are:
Amylase and Invertase
Invertase and Amylase
Invertase and Zymase
Zymase and Invertase
Answer: (c)
Solution
The reaction catalyzed by invertase is: $$\mathrm{C_{12}H_{22}O_{11} + H_2O \xrightarrow{Invertase} C_6H_{12}O_6 + C_6H_{12}O_6}$$ This produces glucose and fructose. The reaction catalyzed by zymase is: $$\mathrm{C_6H_{12}O_6 \xrightarrow{Zymase} 2C_2H_5OH + 2CO_2}$$
Question 75
Chemistry · The s-Block Elements · Single correct
One of the by-products formed during the recovery of $\mathrm{NH}_3$ from Solvay process is :
$\mathrm{Ca(OH)_2}$
$\mathrm{NaHCO_3}$
$\mathrm{CaCl_2}$
$\mathrm{NH_4Cl}$
Answer: (c)
Solution
{ "rawText": "", "answer": "3" }
Question 76
Chemistry · Amines · Single correct
$C_7H_7N_2OCl$ + $C_2H_3OH$ $\rightarrow$ ${ + N_2 + "X" + "Y"}$ In the above reaction, the structural formula of (A), "X" and "Y" respectively are:
Answer: (a)
Solution
The reaction involves the decomposition of diazonium salt in the presence of ethanol. The products formed are anisole, acetaldehyde, hydrochloric acid, and nitrogen gas. The reaction can be represented as: $$C_6H_5N_2^+Cl^- + CH_3CH_2OH \rightarrow C_6H_5OCH_3 + CH_3CHO + HCl + N_2$$
Question 77
Chemistry · Surface Chemistry · Single correct
For the coagulation of a negative sol, the species below, that has the highest flocculating power is:
$\mathrm{SO}_4^{2-}$
$\mathrm{Ba}^{2+}$
$\mathrm{Na}^{+}$
$\mathrm{PO}_4^{3-}$
Answer: (b)
Solution
To coagulate negative sol, cation with higher charge has higher coagulation value.
Question 78
Chemistry · Environmental Chemistry · Single correct
Which of the following statement(s) is (are) incorrect reason for eutrophication?
(A) only
(C) only
(B) and (D) only
(D) only
Answer: (d)
Solution
The process in which nutrient enriched water bodies support a dense plant population which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity is known as eutrophication.
Question 79
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Choose the correct statement regarding the formation of carbocations A and B given :-
Carbocation B is more stable and formed relatively at faster rate
Carbocation A is more stable and formed relatively at slow rate
Carbocation B is more stable and formed relatively at slow rate
Carbocation A is more stable and formed relatively at faster rate
Answer: (a)
Solution
This is more stable due to secondary cation formation and formed with faster rate due to low activation energy.
Question 80
Chemistry · Thermodynamics · Multiple correct
During which of the following processes, does entropy decrease? $(A)$ Freezing of water to ice at $0^\circ C$ $(B)$ Freezing of water to ice at $-10^\circ C$ $(C)$ $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightarrow 2\mathrm{NH_3(g)}$ $(D)$ Adsorption of $\mathrm{CO(g)}$ and lead surface $(E)$ Dissolution of $\mathrm{NaCl}$ in water
$(A)$, $(B)$, $(C)$ and $(D)$ only
$(B)$ and $(C)$ only
$(A)$ and $(E)$ only
$(A)$, $(C)$ and $(E)$ only
Answer: (a)
Solution
(A) Water $0^{\circ} \mathrm{C} \rightarrow$ ice; $\Delta S = -\mathrm{ve}$ (B) Water $-10^{\circ} \mathrm{C} \rightarrow$ ice; $\Delta S = -\mathrm{ve}$ (C) $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$; $\Delta S = -\mathrm{ve}$ (D) Adsorption; $\Delta S = -\mathrm{ve}$ (E) $\mathrm{NaCl(s) \rightarrow Na^+(aq) + Cl^-(aq)}$; $\Delta S = +\mathrm{ve}$
Question 81
Chemistry · Electrochemistry · Numerical
A KCL solution of conductivity $0.14 \, \mathrm{S} \, \mathrm{m}^{-1}$ shows a resistance of $4.19 \, \Omega$ in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to $1.03 \, \Omega$. The conductivity of the HCl solution is ____ $\times 10^{-2} \, \mathrm{S} \, \mathrm{m}^{-1}$. (Round off to the Nearest Integer).
Answer: 57
Solution
Given $$\kappa = \frac{1}{R} \cdot G^*$$. For same conductivity cell, $G^*$ is constant and hence $$\kappa \cdot R = constant$$. Therefore, $$0.14 \times 4.19 = \kappa \times 1.03$$ or, $$\kappa = \frac{0.14 \times 4.19}{1.03}$$ $$= 0.5695 \, \mathrm{S} \mathrm{m}^{-1}$$ $$= 56.95 \times 10^{-2} \, \mathrm{S} \mathrm{m}^{-1} \approx 57 \times 10^{-2} \, \mathrm{S} \mathrm{m}^{-1}$$
Question 82
Chemistry · The d-and f-Block Elements · Numerical
On complete reaction of $\mathrm{FeCl_3}$ with oxalic acid in aqueous solution containing $\mathrm{KOH}$, resulted in the formation of product A. The secondary valency of Fe in the product A is ____
Answer: 6
Solution
The reaction is: $$\mathrm{Fe^{3+} }+ 3\ K^+ + 3\ C_2O_4^{2-} \rightarrow K_3[Fe(C_2O_4)_3]$$ (A) Secondary valency of Fe in 'A' is 6.
Question 83
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The reaction $2 \, \mathrm{A} + \mathrm{B}_2 \rightarrow 2 \mathrm{AB}$ is an elementary reaction. For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ____. (Round off to the Nearest Integer).
Answer: 27
Solution
Reaction: $2 \mathrm{A} + \mathrm{B}_2 \rightarrow 2 \mathrm{AB}$ As the reaction is elementary, the rate of reaction is $r = K [\mathrm{A}]^2 [\mathrm{B}_2]$ On reducing the volume by a factor of $3$, the concentrations of $\mathrm{A}$ and $\mathrm{B}_2$ will become $3$ times and hence, the rate becomes $3^2 \times 3 = 27$ times of initial rate.
Question 84
Chemistry · Hydrocarbons · Numerical
The total number of C - C sigma bond/s in mesityl oxide ($C_6H_{10}O$) is ____ (Round off to the Nearest Integer).
A 1 molal $\mathrm{K_4Fe(CN)_6}$ solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute $A$. The molar mass of $A$ is _____ u. (Round off to the Nearest Integer). [Density of water = $1.0 \, \mathrm{g \, cm^{-3}}$]
Answer: 85
Solution
The reaction is given by: $$\mathrm{K_4Fe(CN)_6 \rightleftharpoons 4K^+ + Fe(CN)_6^{4-}}$$ Initial concentration: $1 \, \mathrm{m}$ for $\mathrm{K_4Fe(CN)_6}$ and $0$ for products. Final concentration: $(1 - 0.4) \, \mathrm{m} = 0.6 \, \mathrm{m}$ for $\mathrm{K_4Fe(CN)_6}$, $4 \times 0.4 = 1.6 \, \mathrm{m}$ for $\mathrm{K^+}$, and $0.4 \, \mathrm{m}$ for $\mathrm{Fe(CN)_6^{4-}}$. Effective molality $= 0.6 + 1.6 + 0.4 = 2.6 \, \mathrm{m}$. For the same boiling point, the molality of another solution should also be $2.6 \, \mathrm{m}$. Now, $18.1$ weight percent solution means $18.1 \, \mathrm{gm}$ solute is present in $100 \, \mathrm{gm}$ solution and hence, $(100 - 18.1 =) 81.9 \, \mathrm{gm}$ water. Now, $$2.6 = \frac{18.1/M}{81.9/1000}$$ Therefore, the molar mass of solute, $M = 85$.
Question 86
Chemistry · Structure of Atom · Numerical
In the ground state of atomic Fe(Z = 26), the spin-only magnetic moment is ____ $\times$ 10^{-1} BM. (Round off to the Nearest Integer). [Given: $\sqrt{3} = 1.73$, $\sqrt{2} = 1.41$]
Answer: 49
Solution
Fe $\rightarrow [\mathrm{Ar}] \, 4s^2 3d^6$ Number of unpaired $e^- = 4$ $$\mu = \sqrt{4(4+2)} \, \mathrm{B \cdot M}$$ $$\mu = \sqrt{24} \, \mathrm{B \cdot M}$$ $$\mu = 4.89 \, \mathrm{B \cdot M}$$ $$\mu = 48.9 \times 10^{-1} \, \mathrm{B \cdot M}$$ Nearest integer value will be 49.
Question 87
Chemistry · States of Matter · Numerical
The number of chlorine atoms in 20 mL of chlorine gas at $STP$ is ____ $10^{21}$. (Round off to the Nearest Integer). [Assume chlorine is an ideal gas at $STP$ $R = 0.083 \, \mathrm{L \, bar \, mol^{-1} \, K^{-1}}$, $N_A = 6.023 \times 10^{23}$]
Answer: 1
Solution
Given the equation $PV = nRT$, we have: $$1.0 \times \frac{20}{1000} = \frac{N}{6.023 \times 10^{23}} \times 0.083 \times 273$$ Therefore, the number of $\mathrm{Cl_2}$ molecules, $N = 5.3 \times 10^{20}$. Hence, the number of Cl-atoms $= 1.06 \times 10^{21}$. $$\approx 1 \times 10^{21}$$
Question 88
Chemistry · Some Basic Concepts of Chemistry · Numerical
KBr is doped with $10^{-5}$ mole percent of $\mathrm{SrBr_2}$. The number of cationic vacancies in $1 \, \mathrm{g}$ of $\mathrm{KBr}$ crystal is _____ $10^{14}$. (Round off to the Nearest Integer). [Atomic Mass : $\mathrm{K} : 39.1\, \mathrm{u}$, $\mathrm{Br} : 79.9\, \mathrm{u}$ $N_A = 6.023 \times 10^{23}$]
Answer: 5
Solution
1 mole KBr ( = 119 $\mathrm{gm}$ ) have $\frac{10^{-5}}{100}$ moles SrBr_2 and hence, 10^{-7} moles cation vacancy (as 1Sr^{2+} will result 1 cation vacancy). Therefore, required number of cation vacancies $$= \frac{10^{-7} \times 6.023 \times 10^{23}}{119} = 5.06 \times 10^{14} \simeq 5 \times 10^{14}$$
Question 89
Chemistry · Equilibrium · Numerical
Consider the reaction $\mathrm{N_2O_4 (g) \rightleftharpoons 2NO_2 (g)}$. The temperature at which $K_C = 20.4$ and $K_p = 600.1$, is ____ K. (Round off to the Nearest Integer). [Assume all gases are ideal and $R = 0.0831 \, \mathrm{L \, bar \, K^{-1} \, mol^{-1}}$]
Answer: 354
Solution
Given the reaction $\mathrm{N_2O_4\,(g) \rightleftharpoons 2NO_2\,(g)}$, we have $\Delta n_g = 2 - 1 = 1$. Now, $K_p = K_c \cdot (RT)^{\Delta_{gg}}$. Therefore, $$600.1 = 20.4 \times (0.0831 \times T)^1$$ Thus, $T = 353.99 \, \mathrm{K} \approx 354 \, \mathrm{K}$.
Question 90
Chemistry · Some Basic Concepts of Chemistry · Numerical
Consider the above reaction. The percentage yield of amide product is (Round off to the Nearest Integer). (Given : Atomic mass: C : 12.0 u, H : 1.0 u N : 14.0 u, O : 16.0 u, Cl : 35.5 u)
Answer: 77
Solution
Given 0.140 gm of reactant, calculate the limiting reagent. $$\frac{169}{140.5} \times 0.140 = 0.168 \, gm < 0.388 \, gm (excess)$$ Therefore, the theoretical amount of given product formed is $$\frac{273}{140.5} \times 0.140 = 0.272 \, gm$$ But its actual amount formed is 0.210 gm. Hence, the percentage yield of product is $$\frac{0.210}{0.272} \times 100 = 77.20 \approx 77$$ OR Mole of $Ph - CoCl = \frac{0.140}{140} = 10^{-3} \, mol$ Mole of $Ph - C - N(Ph)_2$, that should be obtained by mol-mol analysis $= 10^{-3} \, mol$ Theoretical mass of product $= 10^{-3} \times 273 = 273 \times 10^{-3} \, g$ Observed mass of product $= 210 \times 10^{-3} \, g$ $\%$ yield of product $= \frac{210 \times 10^{-3}}{273 \times 10^{-3}} \times 100 = 76.9\% \approx 77$