JEE Main 25 July 2022 Shift 1 question paper with solutions
JEE Main 25 July 2022 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 · Relations and Functions · Single correct
The total number of functions, $$f : \{1, 2, 3, 4\} \rightarrow \{1, 2, 3, 4, 5, 6\}$$ such that $f(1) + f(2) = f(3)$, is equal to :
Maths · Complex Numbers and Quadratic Equations · Single correct
If $\alpha$, $\beta$, $\gamma$, $\delta$ are the roots of the equation $x^4 + x^3 + x^2 + x + 1 = 0$, then $\alpha^{2021}$ + $\beta^{2021}$ + $\gamma^{2021}$ + $\delta^{2021}$ is equal to .
-4
-1
1
4
Answer: (b)
Solution
Given that $\alpha, \beta, \gamma, \delta$ are roots of the equation $$x^4 + x^3 + x^2 + x + 1 = 0$$ which are 5th roots of unity except 1. Then $$\alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021} =$$ Since $\alpha + \beta + \gamma + \delta = -1$, the expression simplifies to $-1$.
Question 3
Maths · Complex Numbers and Quadratic Equations · Single correct
For $n \in \mathbb{N}$, let $S_n = \left\{ z \in \mathbb{C} : |z - 3 + 2i| = \frac{n}{4} \right\}$ and $$T_n = \left\{ z \in \mathbb{C} : |z - 2 + 3i| = \frac{1}{n} \right\}.$$ Then the number of elements in the set $$\{ n \in \mathbb{N} : S_n \cap T_n = \emptyset \}$$ is :
0
2
3
Infinite
Answer: (d)
Solution
Given $S_n : |z - (3 - 2i)| = \frac{n}{4}$ is a circle center $C_1(3, -2)$ and radius $n/4$. $T_n : |z - (2 - 3i)| = \frac{1}{n}$ is a circle center $C_2(2, -3)$ and radius $1/n$. Here $S_n \cap T_n = \phi$. Both circles do not intersect each other. Case-1: $C_1C_2 > n/4 + 1/n$ $$\sqrt{2} > \frac{n}{4} + \frac{1}{n}$$ then $n = 1, 2, 3, 4$ Case-2: $C_1C_2 < \left| \frac{n}{4} - \frac{1}{n} \right|$ $$\Rightarrow \sqrt{2} < \frac{n^2 - 4}{4n}$$ $\Rightarrow n$ has infinite solutions for $n \in \mathbb{N}$
Question 4
Maths · Determinants · Single correct
The number of $\theta \in (0, 4\pi)$ for which the system of linear equations $$3 (\sin 3\theta) x - y + z = 2$$ $$3 (\cos 2\theta) x + 4y + 3z = 3$$ $$6x + 7y + 7z = 9$$ has no solution is :
6
7
8
9
Answer: (b)
Solution
The system of equation has no solution. $$\begin{vmatrix} 3 \sin 3\theta & -1 & 1 \\ 3 \cos 2\theta & 4 & 3 \\ 6 & 7 & 7 \end{vmatrix} = 0$$ $$21 \sin 3\theta + 42 \cos 2\theta - 42 = 0$$ $$\sin 3\theta + 2 \cos 2\theta - 2 = 0$$ Number of solution is 7 in $(0, 4\pi)$
Question 5
Maths · Limits and Derivatives · Single correct
If $\lim_{n \to \infty} \left( \sqrt{n^2 - n - 1 + n \alpha + \beta} \right) = 0$ then $8(\alpha + \beta)$ is equal to:
Maths · Applications of Derivatives · Single correct
If the absolute maximum value of the function $f(x)=(x^2-2x+7)\mathrm{e}^{4x^3-12x^2-180x+31}$ in the interval $[-3,0]$ is $f(\alpha)$, then:
$\alpha$ = 0
$\alpha$ = -3
$\alpha$ $\in$ (-1,0)
$\alpha$ $\in$ (-3,-1)
Answer: (b)
Solution
Given $f'(x) = e^{(4x^3 - 12x^2 - 180x + 31)} \left(12(x^2 - 2x + 7)(x + 3)(x - 5) + 2(x - 1)\right)$. For $x \in [-3, 0]$, $f'(x) < 0$. Therefore, $f(x)$ is a decreasing function on $[-3, 0]$. The absolute maximum value of the function $f(x)$ is at $x = -3$. Thus, $\alpha = -3$.
Question 7
Maths · Applications of Derivatives · Single correct
The curve $y(x) = ax^3 + bx^2 + cx + 5$ touches the x-axis at the point $P(-2, 0)$ and cuts the y-axis at the point $Q$, where $y'$ is equal to $3$. Then the local maximum value of $y(x)$ is:
$\frac{27}{4}$
$\frac{29}{4}$
$\frac{37}{4}$
$\frac{9}{2}$
Answer: (a)
Solution
Given $y(x) = ax^3 + bx^2 + cx + 5$ is passing through $(-2,0)$ then $8a - 4b + 2c = 5$ (1). $y'(x) = 3ax^2 + 2bx + c$ touches the x-axis at $(-2,0)$ $$12a - 4b + c = 0$$ (2). Again, for $x = 0$, $y'(x) = 3$ $$c = 3$$ (3). Solving eq. (1), (2) & (3) $a = -\frac{1}{2}$, $b = -\frac{3}{4}$. $$y'(x) = -\frac{3}{2}x^2 - \frac{3}{2}x + 3$$ $y(x)$ has local maxima at $x = 1$. $$y(1) = \frac{27}{4}$$
Question 8
Maths · Applications of Integrals · Single correct
The area of the region given by $$A=\{(x, y) : x^2 \leq y \leq \min \{x+2, 4-3x\}\}$$ is :
$\frac{31}{8}$
$\frac{17}{6}$
$\frac{19}{6}$
$\frac{27}{8}$
Answer: (b)
Solution
The area A is given by the integral: $$A = \int_{-1}^{\frac{1}{2}} (x + 2 - x^2) \, dx + \int_{\frac{1}{2}}^{1} (4 - 3x - x^2) \, dx = \frac{17}{6}$$
Question 9
Maths · Integrals · Single correct
For any real number $x$, let $[x]$ denote the largest integer less than equal to $x$. Let $f$ be a real valued function defined on the interval $[-10, 10]$ by $$f(x) = \begin{cases} x - [x], & if (x) is odd \\ 1 + [x] - x, & if (x) is even \end{cases}$$ Then the value of $$\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x \, dx$$ is:
4
2
1
0
Answer: (a)
Solution
Given that $f(x)$ is a periodic function whose period is $2$. $$\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x \, dx = \frac{\pi^2}{10} \times 10 \int_{0}^{2} f(x) \cos \pi x \, dx$$ $$= \pi^2 \left( \int_{0}^{1} (1-x) \cos \pi x \, dx + \int_{1}^{2} (x-1) \cos \pi x \, dx \right)$$ Using by parts, $$= \pi^2 \times \frac{4}{\pi^2} = 4$$
Question 10
Maths · Differential Equations · Single correct
The slope of the tangent to a curve C : y =y(x) at any point [x, y) on it is $\frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$. If C passes through the points $\left( 0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}} \right)$ and $\left( \alpha, \frac{1}{2} e^{2\alpha} \right)$ then $e^{\alpha}$ is equal to :
Maths · Straight Lines and Pair of Straight Lines · Single correct
A line, with the slope greater than one, passes through the point A(4, 3) and intersects the line $x - y - 2 = 0$ at the point B. If the length of the line segment AB is $\frac{\sqrt{29}}{3}$, then B also lies on the line:
2x + y = 9
3x - 2y = 7
x + 2y = 6
2x - 3y = 3
Answer: (c)
Solution
Let $B(x_1, x_1 - 2)$. $$\sqrt{(x_1 - 4)^2 + (x_1 - 2 - 3)^2} = \frac{\sqrt{29}}{3}$$ Squaring on both sides $$18x_1^2 - 162x_1 + 340 = 0$$ $$x_1 = \frac{51}{9} or x_1 = \frac{10}{3}$$ $$y_1 = \frac{33}{9} or y_1 = \frac{4}{3}$$ Option (C) will satisfy $\left( \frac{10}{3}, \frac{4}{3} \right)$
Question 13
Maths · Conic Sections · Single correct
Let the locus of the centre $(\alpha, \beta)$, $\beta > 0$, of the circle which touches the circle $x^2 + (y - 1)^2 = 1$ externally and also touches the $x$-axis be $L$. Then the area bounded by $L$ and the line $y = 4$ is:
$\frac{32\sqrt{2}}{3}$
$\frac{40\sqrt{2}}{3}$
$\frac{64}{3}$
$\frac{32}{3}$
Answer: (c)
Solution
Given $(\alpha - 0)^2$ + $(\beta - 1)^2$ = $(\beta + 1)^2$. We have $\alpha^2 = 4\beta$ and $x^2 = 4y$. The area $A$ is given by $$ A = 2 \int_0^4 \left( 4 - \frac{x^2}{4} \right) \, dx = \frac{64}{3}. $$
Question 14
Maths · Three Dimensional Geometry · Single correct
Let P be the plane containing the straight line $\frac{x-3}{9} = \frac{y+4}{-1} = \frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{x}{2} = \frac{y}{3} = \frac{z}{5}$ and $\frac{x}{3} = \frac{y}{7} = \frac{z}{8}$. If $d$ is the distance of P from the point (2, -5, 11), then $d^2$ is equal to:
$\frac{147}{2}$
96
$\frac{32}{3}$
54
Answer: (c)
Solution
Given the equation $a(x - 3) + b(y + 4) + c(z - 7) = 0$. The conditions are $9a - b - 5c = 0$ and $-11a - b + 5c = 0$. After solving, the direction ratios are proportional to $(1, -1, 2)$. The equation of the plane is $x - y + 2z = 21$. The distance $d$ is given by $$d = \frac{8}{\sqrt{6}}$$ and $$d^2 = \frac{32}{3}$$.
Question 15
Maths · Vector Algebra · Single correct
Let ABC be a triangle such that $\overline{BC} = \vec{a}$, $\overline{CA} = \vec{b}$, $\overline{AB} = \vec{c}$, $|\vec{a}| = 6\sqrt{2}$, $|\vec{b}| = 2\sqrt{3}$ and $\vec{b} \cdot \vec{c} = 12$ Consider the statements : (S1) : $| (\vec{a} \times \vec{b}) + (\vec{c} \times \vec{b}) | - |\vec{c}| = 6(2\sqrt{2} - 1)$ (S2) : $\angle ABC = \cos^{-1}\left(\frac{\sqrt{2}}{3}\right)$. Then
If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :
$\frac{33}{2^{32}}$
$\frac{33}{2^{29}}$
$\frac{33}{2^{28}}$
$\frac{33}{2^{27}}$
Answer: (c)
Solution
Given the equations: $$np + npq = 24 ...(1)$$ $$np \cdot npq = 128 ...(2)$$ Solving (1) and (2): We get $p = \frac{1}{2}$, $q = \frac{1}{2}$, $n = 32$. Now, $$P(X = 1) + P(X = 2)$$ $$= \binom{32}{1} pq^{31} + \binom{32}{2} p^2 q^{30}$$ $$= \frac{33}{2^{28}}$$
Question 17
Maths · Complex Numbers and Quadratic Equations · Single correct
If the numbers appeared on the two throws of a fair six faced die are $\alpha$ and $\beta$, then the probability that $x^2 + \alpha x + \beta > 0$, for all $x \in \mathbb{R}$, is :
$\frac{17}{36}$
$\frac{4}{9}$
$\frac{1}{2}$
$\frac{19}{36}$
Answer: (a)
Solution
Given $x^2 + \alpha x + \beta > 0$, for all $x \in \mathbb{R}$. The discriminant $D = \alpha^2 - 4\beta < 0$. Therefore, $\alpha^2 < 4\beta$. Total cases $= 6 \times 6 = 36$. Favorable cases are $\beta = 1$, $\alpha = 1$; $\beta = 2$, $\alpha = 1, 2$; $\beta = 3$, $\alpha = 1, 2, 3$; $\beta = 4$, $\alpha = 1, 2, 3$; $\beta = 5$, $\alpha = 1, 2, 3, 4$; $\beta = 6$, $\alpha = 1, 2, 3, 4$. Total favorable cases $= 17$. The probability $P(x) = \frac{17}{36}$.
Question 18
Maths · Trigonometric Functions · Single correct
The number of solutions of $|\cos x| = \sin x$, such that $-4\pi \leq x \leq 4\pi$ is :
4
6
8
12
Answer: (c)
Solution
2 solutions in $(0, 2\pi)$ So, 8 solutions in $[-4\pi, 4\pi]$
Question 19
Maths · Heights and Distances · Single correct
A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that $QR = 15 \, \mathrm{m}$. If from a point A on the ground the angle of elevation of R is $60^\circ$ and the part PR of the tower subtends an angle of $15^\circ$ at A, then the height of the tower is :
Let $$ A=\begin{bmatrix}2&-1&-1\\1&0&-1\\1&-1&0\end{bmatrix} $$ and $B=A-I$. If $$ \omega=\frac{\sqrt{3}i-1}{2}, $$ then the number of elements in the set $$ \left\{n\in\{1,2,\ldots,100\}:A+(\omega B)^n=A+B\right\} $$ is equal to ______.
Answer: 17
Solution
Given $$A = \begin{bmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{bmatrix}$$ we have $$A^2 = A \implies A^n = A.$$ This holds for all $$n \in \{1, 2, \ldots, 100\}.$$ Now, $$B = A - I = \begin{bmatrix} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{bmatrix}$$ We find $$B^2 = -B$$ which implies $$B^3 = -B^2 = B$$ and $$B^5 = B$$ thus $$B^{99} = B.$$ Also, $$\omega^{3k} = 1.$$ So, $n$ is the common element of $$\{1, 3, 5, \ldots, 99\}$$ and $$\{3, 6, 9, \ldots, 99\}$$ which is $$17.$$
Question 22
Maths · Permutations and Combinations · Numerical
The letters of the word 'MANKIND' are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word 'MANKIND' is ______.
Answer: 1492
Solution
\begin{tabular}{|c|c|c|c|c|c|c|} \hline M & A & N & K & I & N & D \\ \hline \end{tabular} $4\times\frac{6!}{2!}$ $+(5!\times0)$ $+\frac{4!\times3}{2!}$ $+(3!\times2)$ $+(2!\times1)$ $+(1!\times1)$ $+(0!\times0)$ $+1$ $\Downarrow$ $1492$
Question 23
Maths · Binomial Theorem · Numerical
If the maximum value of the term independent of $t$ in the expansion of $$\left( t^2 x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{15}, \ x \geq 0,$$ is $K$, then $8K$ is equal to .
Answer: 6006
Solution
Given $$\left( t^2 x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{15}$$ $$T_{r+1} = \binom{15}{r} \left( t^2 x^{\frac{1}{5}} \right)^{15-r} \cdot \frac{(1-x)^{\frac{r}{10}}}{t^r}$$ For independent of $t$, $$30 - 2r - r = 0$$ $$\Rightarrow r = 10$$ So, Maximum value of $$\binom{15}{10} x(1-x)$$ will be at $$x = \frac{1}{2}$$ i.e. 6006
Question 24
Maths · Complex Numbers and Quadratic Equations · Numerical
Let a, b be two non-zero real numbers. If p and r are the roots of the equation $x^2 - 8ax + 2a = 0$ and q and s are the roots of the equation $x^2 + 12bx + 6b = 0$, such that $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ are in A.P., then $a^{-1} - b^{-1}$ is equal to .
Answer: 38
Solution
Given $x^2 - 8ax + 2a = 0$ and $x^2 + 12bx + 6b = 0$. For the first equation, $p + r = 8a$ and $pr = 2a$. For the second equation, $q + s = -12b$ and $qs = 6b$. We have $\frac{1}{p} + \frac{1}{r} = 4$ and $\frac{1}{q} + \frac{1}{s} = -2$. Solving $\frac{2}{q} = 4$, we get $q = \frac{1}{2}$. Solving $\frac{2}{r} = -2$, we get $r = -1$. Thus, $p = \frac{1}{5}$ and $s = -\frac{1}{4}$. Now, $\frac{1}{a} - \frac{1}{b} = \frac{2}{pr} - \frac{6}{qs} = 38$.
Question 25
Maths · Sequences and Series · Fill in the blank
Let $a_1 = b_1 = 1$, $a_n = a_{n-1} + 2$ and $b_n = a_n + b_{n-1}$ for every natural number $n \geq 2$. Then $\sum_{n=1}^{15} a_n \cdot b_n$ is equal to _______.
Maths · Continuity and Differentiability · Numerical
Let $f(x) = \begin{cases} |4x^2 - 8x + 5|, & if \ 8x^2 - 6x + 1 \geq 0 \\ [4x^2 - 8x + 5], & if \ 8x^2 - 6x + 1 < 0 \end{cases},$ where $[\alpha]$ denotes the greatest integer less than or equal to $\alpha$. Then the number of points in $\mathbb{R}$ where $f$ is not differentiable is _______.
Answer: 3
Solution
The function is not defined (ND) at $\frac{1}{4}$, $x_1$, and $\frac{1}{2}$.
Question 27
Maths · Limits and Derivatives · Numerical
If $\lim_{n\to\infty} \frac{(n+1)^{\,k-1}} {n^{\,k+1}} \Big[(nk+1)+(nk+2)+\cdots+$ $(nk+n)\Big] =33,$ $\lim_{n\to\infty} \frac{1} {n^{\,k+1}} \Big[1^k+2^k+3^k+\cdots+n^k\Big],$ then the integral value of $k$ is equal to ______.
Let the equation of two diameters of a circle $x^2 + y^2 - 2x + 2fy + 1 = 0$ be $2px - y = 1$ and $2x + py = 4p$. Then the slope $m \in (0, \infty)$ of the tangent to the hyperbola $3x^2 - y^2 = 3$ passing through the centre of the circle is equal to _______.
Answer: 2
Solution
Given the equations: $$2p + f - 1 = 0 (1)$$ $$2 - pf - 4p = 0 (2)$$ From equation (2): $$2 = p(f + 4)$$ Solving for $p$: $$p = \frac{2}{f + 4}$$ Substitute $p$ in equation (1): $$2p = 1 - f$$ $$\frac{4}{f + 4} = 1 - f$$ $$f^2 + 3f = 0$$ $$f = 0 or -3$$ For the hyperbola: $$3x^2 - y^2 = 3, x^2 - \frac{y^2}{3} = 1$$ The equation of the line is: $$y = mx \pm \sqrt{m^2 - 3}$$ It passes through $(1, 0)$: $$0 = m \pm \sqrt{m^2 - 3}$$ As $m$ tends to $\infty$, it passes through $(1, 3)$: $$3 = m \pm \sqrt{m^2 - 3}$$ Solving: $$(3 - m)^2 = m^2 - 3$$ $$m = 2$$
Question 29
Maths · Conic Sections · Numerical
The sum of diameters of the circles that touch (i) the parabola $75x^2 = 64(5y - 3)$ at the point $\left(\frac{8}{5}, \frac{6}{5}\right)$ and (ii) the y-axis, is equal to .
Answer: 10
Solution
Given $$x^2 = \frac{64.5}{75} \left( y - \frac{3}{5} \right)$$ Equation of tangent at $$\left( \frac{8}{5}, \frac{6}{5} \right)$$ $$x \cdot \frac{8}{5} = \frac{64}{15} \left( y + \frac{6}{5} - \frac{3}{5} \right)$$ $$3x - 4y = 0$$ Equation of family of circle is $$\left( x - \frac{8}{5} \right)^2 + \left( y - \frac{6}{5} \right)^2 + \lambda (3x - 4y) = 0$$ It touches y-axis so $f^2 = c$ $$x^2 + y^2 + x \left( 3\lambda - \frac{16}{5} \right) + y \left( -4\lambda - \frac{12}{5} \right) + 4 = 0$$ $$\frac{(4\lambda + \frac{12}{5})^2}{4} = 4$$ $$\lambda = \frac{2}{5} or \lambda = -\frac{8}{5}$$ $$\lambda = \frac{2}{5}, r = 1$$ $$\lambda = -\frac{8}{5}, r = 4$$ $$d_1 + d_2 = 10$$
Question 30
Maths · Three Dimensional Geometry · Numerical
The line of shortest distance between the lines $\frac{x-2}{0} = \frac{y-1}{1} = \frac{z}{1}$ and $\frac{x-3}{2} = \frac{y-5}{2} = \frac{z-1}{1}$ makes an angle of $\cos^{-1} \left( \sqrt{\frac{2}{27}} \right)$ with the plane $P : ax - y - z = 0$, $(a > 0)$. If the image of the point $(1, 1, -5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta - \gamma$ is equal to ______. (Enter 0 if no such plane exists)
Answer: 0
Solution
DR's of line of shortest distance $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 2 & 2 & 1 \end{vmatrix} = -\hat{i} + 2\hat{j} - 2\hat{k}$$ Angle between line and plane is $\cos^{-1} \sqrt{\frac{2}{27}} = \alpha$ $$\cos \alpha = \sqrt{\frac{2}{27}}, \sin \alpha = \frac{5}{3\sqrt{3}}$$ DR's normal to plane $(1, -1, -1)$ $$\sin \alpha = \frac{\begin{vmatrix} -a - 2 + 2 \end{vmatrix}}{\sqrt{4 + 4 + 1} \sqrt{a^2 + 1 + 1}} = \frac{5}{3\sqrt{3}}$$ $$\sqrt{3} |a| = 5\sqrt{a^2 + 2}$$ $$3a^2 = 25a^2 + 50$$ No value of $(a)$ (No such plane exists)
Physics
Question 31
Physics · Physical World, Units and Measurements · Single correct
If momentum [P], area [A] and time [T] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is:
A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60 $\mathrm{m}$, then the magnitude of displacement would be : (Given $\cos$ $135^\circ$ = -0.7)
42 $\mathrm{m}$
47 $\mathrm{m}$
19 $\mathrm{m}$
40 $\mathrm{m}$
Answer: (b)
Solution
Given $d = R \theta$. $60 = R \left( \frac{3\pi}{4} \right)$ $$R = \frac{60 \times 4}{3\pi} = \frac{80}{\pi} \, \mathrm{m}$$ Displacement is given by $$\sqrt{R^2 + R^2 - 2R^2 \cos 135}$$ This simplifies to $$\Rightarrow \sqrt{2R^2 - 2R^2 (-0.7)}$$ $$\Rightarrow \sqrt{3.4R^2} = \sqrt{3.4 \left( \frac{80}{\pi} \right)^2}$$ This is approximately $47 \, \mathrm{m}$.
Question 34
Physics · Work, Energy and Power · Single correct
A body of mass 0.5 $\mathrm{kg}$ travels on straight line path with velocity $v = (3x^2 + 4) \, \mathrm{m/s}$. The net workdone by the force during its displacement from $x = 0$ to $x = 2 \, \mathrm{m}$ is:
64 $\mathrm{J}$
60 $\mathrm{J}$
120 $\mathrm{J}$
128 $\mathrm{J}$
Answer: (b)
Solution
Given $v_i = 3(0^2) + 4 = 4$ when $x = 0$. For $v_F = 3(2)^2 + 4$, we have $v_F = 16$ when $x = 2$. The work done $W = \Delta K = \frac{1}{2} m (16^2 - 4^2)$. This simplifies to $= \frac{1}{2} \times \frac{1}{2} (256 - 16)$. Finally, $= \frac{240}{4} = 60 \, \mathrm{J}$.
Question 35
Physics · System of Particles and Rotational Motion · Single correct
A solid cylinder and a solid sphere, having same mass $M$ and radius $R$, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be:
$\sqrt{\frac{5}{3}}$
$\sqrt{\frac{4}{5}}$
$\sqrt{\frac{3}{5}}$
$\sqrt{\frac{14}{15}}$
Answer: (d)
Solution
The velocity is given by the equation: $$V = \sqrt{\frac{2gH}{1 + k^2 / R^2}}$$ The ratio of the volumes is given by: $$\frac{V_{cylinder}}{V_{sphere}} = \sqrt{\frac{(1 + k^2 / R^2)_{sphere}}{(1 + k^2 / R^2)_{cylinder}}}$$ Simplifying further: $$= \sqrt{\frac{1 + 2/5}{1 + 1/2}} = \sqrt{\frac{7}{5} \times \frac{2}{3}} = \sqrt{\frac{14}{15}}$$
Question 36
Physics · Gravitation · Single correct
Three identical particle A, B and C of mass 100 kg each are placed in a straight line with AB = BC = 13 m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13 m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately:
21 G
100 G
59 G
42 G
Answer: (b)
Solution
The force is given by the equation: $$F = \frac{GMM}{r^2} + \sqrt{2} \frac{GMM}{(\sqrt{2}r)^2}$$ This simplifies to: $$= \frac{GMM}{r^2} \left( 1 + \frac{1}{\sqrt{2}} \right)$$ Substituting the given values: $$= \frac{G \times 10^4}{13^2} \left( 1 + \frac{1}{\sqrt{2}} \right)$$ Finally, the force is approximately: $$F \approx 100G$$
Question 37
Physics · Thermodynamics · Single correct
A certain amount of gas of volume $V$ at $27^\circ \mathrm{C}$ temperature and pressure $2 \times 10^7 \, \mathrm{Nm}^{-2}$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use $\gamma = 1.5$)
$3.536 \times 10^5 \, \mathrm{Pa}$
$3.536 \times 10^6 \, \mathrm{Pa}$
$1.25 \times 10^6 \, \mathrm{Pa}$
$1.25 \times 10^5 \, \mathrm{Pa}$
Answer: (b)
Solution
Question 38
Physics · Kinetic Theory · Single correct
Following statements are given : (1) The average kinetic energy of a gas molecule decreases when the temperature is reduced. (2) The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature. (3) The average kinetic energy of a gas molecule decreases with increases in volume. (4) Pressure of a gas increases with increase in temperature at constant pressure. (5) The volume of gas decreases with increase in temperature. Choose the correct answer from the options given below :
(1) and (4) only
(1), (2) and (4) only
(2) and (4) only
(1), (2) and (5) only
Answer: (a)
Solution
The average kinetic energy is given by $$\mathrm{KE_{avg} = \frac{3}{2} k T}$$. The pressure is given by $$P = \frac{1}{3} \rho V_{\mathrm{rms}}^2$$. Note: Statement (4) is correct only if we consider it at constant volume and not constant pressure. Ideally, this question must be bonus but most appropriate answer is option (A).
Question 39
Physics · Oscillations · Single correct
In figure (A), mass $2m$ is fixed on mass $m$, which is attached to two springs of spring constant $k$. In figure (B), mass $m$ is attached to two springs of spring constant $k$ and $2k$. If mass $m$ in (A) and (B) are displaced by distance $x$ horizontally and then released, then time periods $T_1$ and $T_2$ corresponding to (A) and (B), respectively, follow the relation:
$\frac{T_1}{T_2} = \frac{3}{\sqrt{2}}$
$\frac{T_1}{T_2} = \frac{\sqrt{3}}{2}$
$\frac{T_1}{T_2} = \sqrt{\frac{2}{3}}$
$\frac{T_1}{T_2} = \frac{\sqrt{2}}{3}$
Answer: (a)
Solution
Given $$T_1 = 2\pi \sqrt{\frac{3m}{2k}}$$ $$T_2 = 2\pi \sqrt{\frac{m}{3k}}$$ The ratio is $$\frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{3m}{2k}}}{2\pi \sqrt{\frac{m}{3k}}} = \frac{3}{\sqrt{2}}$$
Question 40
Physics · Electrostatic Potential and Capacitance · Single correct
A condenser of 2 \, $\mu$ $\mathrm{F}$ capacitance is charged steadily from 0 to 5C. Which of the following graph represents correctly the variation of potential difference (V) across it's plates with respect to the charge (Q) on the condenser?
Answer: (a)
Solution
Given $Q = CV$. Then $V = \frac{1}{C} Q$. This represents a straight line with slope $= \frac{1}{C}$. Slope $= \frac{1}{C} = \frac{1}{2 \times 10^{-6}} = 5 \times 10^{5}$.
Question 41
Physics · Moving Charges and Magnetism · Single correct
Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is 6 : 5 and their respective masses ratio is 9 : 4. Then, the ratio of their charges will be :
To increase the resonant frequency in series LCR circuit,
Source frequency should be increased
Another resistance should be added in series with the first resistance.
Another capacitor should be added in series with the first capacitor
The source frequency should be decreased
Answer: (c)
Solution
The frequency is given by the formula $$f = \frac{1}{2\pi \sqrt{LC}}.$$ To increase the resonating frequency, the product of $L$ and $C$ should decrease. By joining capacitors in series, the capacitance will decrease.
Question 43
Physics · Electromagnetic Induction · Single correct
A small square loop of wire of side $l$ is placed inside a large square loop of wire $L$ $(L \gg l)$. Both loops are coplanar and their centres coincide at point $O$ as shown in figure. The mutual inductance of the system is:
$\frac{2\sqrt{2}\mu_0 L^2}{\pi l}$
$\frac{\mu_0 l^2}{2\sqrt{2}\pi L}$
$\frac{2\sqrt{2}\mu_0 l^2}{\pi L}$
$\frac{\mu_0 L^2}{2\sqrt{2}\pi l}$
Answer: (c)
Solution
Assuming current $I$ in outer loop, magnetic field at centre is given by $$4 \times \frac{\mu_0 i}{4 \pi \times \frac{L}{2}} \times (2 \sin 45^\circ) = \frac{2 \sqrt{2} \mu_0 i}{\pi L}$$ The mutual inductance $M$ is given by $$M = \frac{Flux through inner loop}{i}$$ Therefore, $$M = \frac{2 \sqrt{2} \mu_0 \ell^2}{\pi L}$$
Question 44
Physics · Alternating Current · Single correct
The rms value of conduction current in a parallel plate capacitor is $6.9 \, \mu \mathrm{A}$. The capacity of this capacitor, if it is connected to $230 \, \mathrm{V}$ ac supply with an angular frequency of $600 \, \mathrm{rad/s}$, will be:
5 pF
50 pF
100 pF
200 pF
Answer: (b)
Question 45
Physics · Ray Optics and Optical Instruments · Single correct
Which of the following statement is correct ?
In primary rainbow, observer sees red colour on the top and violet on the bottom
In primary rainbow, observer sees violet colour on the top and red on the bottom
In primary rainbow, light wave suffers total internal reflection twice before coming out of water drops
Primary rainbow is less bright than secondary rainbow.
Answer: (a)
Solution
In primary rainbow, red colour is at top and violet is at bottom. Intensity of secondary rainbow is less in comparison to primary rainbow.
Question 46
Physics · Ray Optics and Optical Instruments · Single correct
Time taken by light to travel in two different materials A and B of refractive indices $\mu_A$ and $\mu_B$ of same thickness is $t_1$ and $t_2$ respectively. If $t_2 - t_1 = 5 \times 10^{-10} \, \mathrm{s}$ and the ratio of $\mu_A$ to $\mu_B$ is $1 : 2$. Then the thickness of material, in meter is: (Given $v_A$ and $v_B$ are velocities of light in A and B materials respectively).
$5 \times 10^{-10} \, v_A \, \mathrm{m}$
$5 \times 10^{-10} \, \mathrm{m}$
$1.5 \times 10^{-10} \, \mathrm{m}$
$5 \times 10^{-10} \, v_B \, \mathrm{m}$
Answer: (a)
Solution
Given $\frac{\mu_A}{\mu_B} = \frac{c/V_A}{c/V_B} = \frac{V_B}{V_A} = \frac{1}{2}$. Let the thickness is $d$. $$\frac{d}{v_B} - \frac{d}{v_A} = 5 \times 10^{-10}$$ $$d = \frac{5 \times 10^{-10} \times v_A v_B}{v_A - v_B}$$ As $v_A = 2v_B \Rightarrow d = 5 \times 10^{-10} \times 2v_B$. Or $d = 5 \times 10^{-10} \times v_A$.
Question 47
Physics · Dual Nature of Radiation and Matter · Single correct
A metal exposed to light of wavelength 800 nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The work function of the metal is (Take $hc = 1230 \, \mathrm{eV} \cdot \mathrm{nm}$).
1.537 eV
2.46 eV
0.615 eV
1.23 eV
Answer: (c)
Solution
Given $$k_1 = \frac{1230}{800} - \phi ...(1)$$ $$k_2 = 2k_1 = \frac{1230}{500} - \phi ...(2)$$ Eliminating $k_1$ from (1) and (2) we get $$0 = \frac{1230}{500} - \frac{1230}{400} + \phi$$ $$\phi = 0.615 \, eV$$
Question 48
Physics · Atoms · Single correct
The momentum of an electron revolving in $n^{th}$ orbit is given by: (Symbols have their usual meanings)
$\frac{nh}{2\pi r}$
$\frac{nh}{2r}$
$\frac{nh}{2\pi}$
$\frac{2\pi r}{nh}$
Answer: (a)
Solution
Angular momentum is integral multiple of $\frac{h}{2\pi}$. $$mvr = \frac{nh}{2\pi}$$ So momentum $mv = \frac{nh}{2\pi r}$
Question 49
Physics · Atoms · Single correct
The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by:
$\vec{\mu}_L = \frac{e \vec{L}}{2m}$
$\vec{\mu}_L = -\frac{e \vec{L}}{2m}$
$\vec{\mu}_1 = -\frac{e \vec{L}}{m}$
$\vec{\mu}_1 = \frac{2e \vec{L}}{m}$
Answer: (b)
Solution
Ratio of magnetic moment and angular momentum $$\frac{\vec{\mu}}{\vec{L}} = \frac{q}{2m}$$ For $e^-$ $$\vec{\mu} = -\frac{e}{2m} \vec{L}$$
Question 50
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
In the circuit, the logical value of $A = 1$ or $B = 1$ when potential at $A$ or $B$ is $5\,\mathrm{V}$ and the logical value of $A = 0$ or $B = 0$ when potential at $A$ or $B$ is $0\,\mathrm{V}$. The truth table of the given circuit will be:
1 0 0
1 0 1
1 0 0
1 0 1
Answer: (a)
Solution
When both A and B have logical value '1' both diode are reverse bias and current will flow in resistor hence output will be 5 volt i.e. logical value '1'. In all other case conduction will take place, hence output will be zero volt i.e. logical value '0'. So truth table is A B Y 0 0 0 0 1 0 (AND gate) 1 0 0 1 1 1
Question 51
Physics · Motion in a Straight Line · Numerical
A car is moving with speed of $150 \, \mathrm{km/h}$ and after applying the brake it will move $27 \, \mathrm{m}$ before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling ________ m distance.
Answer: 3
Solution
Stopping distance $= \frac{v^2}{2a} = d$ If speed is made $\frac{1}{3} rd$ $d' = \frac{1}{9} d$. $d' = \frac{27}{9} = 3$. Braking acceleration remains same
Question 52
Physics · Laws of Motion · Numerical
Four forces are acting at a point P in equilibrium as shown in the figure. The ratio of force $F_1$ to $F_2$ is $1:x$, where $x = \_\_\_\_$.
Physics · Mechanical Properties of Solids · Numerical
A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by 5 cm. Another wire of the same material of length 4L and radius 4r is pulled by a force 4F under same conditions. The increase in length of this wire is ______ cm.
Physics · Thermal Properties of Matter · Fill in the blank
A unit scale is to be prepared whose length does not change with temperature and remains 20 cm, using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that difference between their lengths remains constant. If length of brass is 40 cm and length of iron will be _____ cm. ($\alpha_{iron} = 1.2 \times 10^{-5} \, \mathrm{K}^{-1}$ and $\alpha_{brass} = 1.8 \times 10^{-5} \, \mathrm{K}^{-1}$).
Answer: 60
Solution
Given the equation $\ell_B (1 + \alpha_B \Delta T) - \ell_i (1 + \alpha_i \Delta T) = \ell_B - \ell_i$. We have $\alpha_B \ell_B = \ell_i \alpha_i$. Substituting the values, $1.8 \times 10^{-5} \times 40 = \ell_i \times 1.2 \times 10^{-5}$. Solving for $\ell_i$, we get $$\ell_i = \frac{1.8 \times 10^{-5} \times 40}{1.2 \times 10^{-5}} = \frac{3 \times 40}{2} = 60.$$ Therefore, $\ell_i = 60 \, \mathrm{cm}$.
Question 55
Physics · Waves · Numerical
An observer is riding on a bicycle and moving towards a hill at 18 $\mathrm{kmh}^{-1}$. He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by source is 640 $\mathrm{Hz}$ and velocity of the sound in air is 320 $\mathrm{m/s}$, the beat frequency between the two sounds heard by observer will be $\mathrm{Hz}$.
Answer: 20
Solution
Given $V_S = 0$, $V_{ob} = 5 \, \mathrm{m/s}$. The direct frequency is calculated as: $$f_{direct} = \left( \frac{320 - 5}{320} \right) 640 = 630 \, \mathrm{Hz}$$ The reflected frequency is calculated as: $$f_{reflected} = \left( \frac{320 + 5}{320} \right) 640 = 650 \, \mathrm{Hz}$$ The beat frequency is: $$f_{beat} = 650 - 630 = 20 \, \mathrm{Hz}$$
Question 56
Physics · Electric Charges and Fields · Numerical
The volume charge density of a sphere of radius $6 \, \mathrm{m}$ is $2 \, \mu \mathrm{C} \, \mathrm{cm}^{-3}$. The number of lines of force per unit surface area coming out from the surface of the sphere is _______ $\times 10^{10} \, \mathrm{NC}^{-1}$. [Given: Permittivity of vacuum $\varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C}^2 \, \mathrm{N}^{-1} \, \mathrm{m}^{-2}$]
Answer: 45
Solution
Number of electric field lines per unit area equals electric field. $$E = \frac{\rho r}{3 \varepsilon_0}, for r = R$$ $$E = \frac{\rho R}{3 \varepsilon_0} = \frac{2 \times 6}{3 \times 8.85 \times 10^{-12}} = 0.45 \times 10^{12} \, \mathrm{NC^{-1}}$$ $$= 45 \times 10^{10} \, \mathrm{N/C}$$
Question 57
Physics · Current Electricity · Numerical
In the given figure, the value of $V_0$ will be V.
Eight copper wire of length $l$ and diameter $d$ are joined in parallel to form a single composite conductor of resistance $R$. If a single copper wire of length $2l$ have the same resistance $(R)$ then its diameter will be $d$.
Answer: 4
Solution
Each wire has resistance $= \rho \frac{4\ell}{\pi d^2} = r$. Eight wires in parallel, then equivalent resistance is $$\frac{r}{8} = \frac{\rho \ell}{2 \pi d^2}$$ Single copper wire of length $2\ell$ has resistance $$R = \rho \frac{2\ell \times 4}{\pi d_1^2} = \frac{\rho \ell}{2 \pi d^2}$$ Therefore, $d_1 = 4d$.
Question 59
Physics · Dual Nature of Radiation and Matter · Numerical
The energy band gap of semiconducting material to produce violet (wavelength = 4000 $\,$ $\mathrm{\AA}$) LED is _____ eV. (Round off to the nearest integer).
Answer: 3
Solution
Given $E_g = \frac{hc}{\lambda} = \frac{1242}{\lambda (\mathrm{nm})} = \frac{1242}{400} = 3.105$. Answer rounded to $3 \, \mathrm{eV}$.
Question 60
Physics · Communication Systems · Numerical
The required height of a TV tower which can cover the population of 6.03 lakh is h. If the average population density is 100 per square km and the radius of earth is 6400 km, then the value of h will be _______ m.
Answer: 150
Solution
Given $$d = \sqrt{2Rh}$$ Substituting the values, $$d = \sqrt{2 \times 6400 \times h \times 10^{-3}} (h in m)$$ The area is given by $$Area = \pi d^2$$ Substituting for $d$, $$= (\pi \times 2 \times 6400 \times h \times 10^{-3}) \ km^2$$ Simplifying, $$6.03 \times 100000 = 100 \times \pi \times 2 \times 6400 \times 10^{-3} \times h$$ Solving for $h$, $$h = \frac{6.03 \times 10^5}{10 \times \pi \times 128}$$ Finally, $$h = 150 \, m$$
Chemistry
Question 61
Chemistry · Redox Reactions · Single correct
$\mathrm{SO_2Cl_2}$ on reaction with excess of water results into acidic mixture. $$\mathrm{SO_2Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl}$$ $16$ moles of $\mathrm{NaOH}$ is required for the complete neutralisation of the resultant acidic mixture. The number of moles of $\mathrm{SO_2Cl_2}$ used is:
Which of the following sets of quantum numbers is not allowed?
n = 3, l = 2, m_l = 0, s = +$\frac{1}{2}$
n = 3, l = 2, m_l = -2, s = +$\frac{1}{2}$
n = 3, l = 3, m_l = -3, s = -$\frac{1}{2}$
n = 3, l = 0, m_l = 0, s = -$\frac{1}{2}$
Answer: (c)
Solution
Given $l = 0, 1, 2, \ldots, (n-1)$. Therefore, for $n = 3$, $l = 0, 1, 2$. This implies $l = 3$, which is not possible for $n = 3$.
Question 63
Chemistry · Solutions · Single correct
The depression in freezing point observed for a formic acid solution of concentration $0.5 \, \mathrm{mL} \, \mathrm{L}^{-1}$ is $0.0405^\circ \mathrm{C}$. Density of formic acid is $1.05 \, \mathrm{g} \, \mathrm{mL}^{-1}$. The Van't Hoff factor of the formic acid solution is nearly: (Given for water $k_f = 1.86 \, \mathrm{K} \, \mathrm{kg} \, \mathrm{mol}^{-1}$)
20 $\mathrm{mL}$ of 0.1 $\mathrm{M}$ $\mathrm{NH_4OH}$ is mixed with 40 $\mathrm{mL}$ of 0.05 $\mathrm{M}$ $\mathrm{HCl}$. The pH of the mixture is nearest to: (Given: $K_b(\mathrm{NH_4OH}$) = 1 $\times 10^{-5}$, $\log 2$ = 0.30, $\log 3$ = 0.48, $\log 5$ = 0.69, $\log 7$ = 0.84, $\log 11$ =1.04)
3.2
4.2
5.2
6.2
Answer: (c)
Solution
The reaction is given by $\($ $\mathrm{NH_4OH + HCl \rightarrow NH_4Cl + H_2O}$ $\)$. The millimoles are 2 for both $\($ $\mathrm{NH_4OH}$ $\)$ and $\($ $\mathrm{HCl}$ $\)$. After the reaction, 2 millimoles of $\($ $\mathrm{NH_4Cl}$ $\)$ are formed. The concentration of $\($ [$\mathrm{NH_4^+}$] $\)$ is calculated as follows: $$ [\mathrm{NH_4^+}] = \frac{2 mmol}{60 ml} = \frac{1}{30} M $$ The pH is calculated using the formula: $$ pH = \frac{\mathrm{pK_w} - \mathrm{pK_b} - \log C}{2} = \frac{14 - 5 + 1.48}{2} = 5.24 $$
Question 65
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Match List - I with List - II \begin{tabular}{|c|l|c|l|} \hline \multicolumn{2}{|c|}{List - I} & \multicolumn{2}{c|}{List - II} \\ \hline (A) & $N_2(g)+3H_2(g)\rightarrow2NH_3(g)$ & (I) & Cu \\ \hline (B) & $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ & (II) & Cu/ZnO-Cr$_2$O$_3$ \\ \hline (C) & $CO(g)+H_2(g)\rightarrow HCHO(g)$ & (III) & Fe$_x$O$_y$ + K$_2$O + Al$_2$O$_3$ \\ \hline (D) & $CO(g)+2H_2(g)\rightarrow CH_3OH(g)$ & (IV) & Ni \\ \hline \end{tabular} Choose the correct answer from the options given below:
(A)- (II), (B) - (IV), $(C)$ - (I), (D) - (III)
(A)- (II), (B) - (I), $(C)$ - (IV), (D) - (III)
(A)- (III), (B) - (IV), $(C)$ - (I), (D) - (II)
(A)- (III), (B) - (I), $(C)$ - (IV), (D) - (II)
Answer: (c)
Solution
Factual
Question 66
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The IUPAC nomenclature of an element with electronic configuration $[\mathrm{Rn}]5f^{14}6d^17s^2$ is:
Unnilbium
Unnilunium
Unnilquadium
Unniltrium
Answer: (d)
Solution
Atomic Number 103
Question 67
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The compound(s) that is(are) removed as slag during the extraction of copper is : (1) CaO (2) FeO (3) $Al_2O_3$ (4) ZnO (5) NiO Choose the correct answer from the options given below :
(3) (4) Only
(1), (2), (5) Only
(1), (2) Only
(2) Only
Answer: (d)
Solution
The reaction is given by: $$\mathrm{FeO} + \mathrm{SiO_2} \rightarrow \mathrm{FeSiO_3}.$$
Question 68
Chemistry · The d-and f-Block Elements · Single correct
The reaction of $\mathrm{H_2O_2}$ with potassium permanganate in acidic medium leads to the formation of mainly:
$\mathrm{Mn^{2+}}$
$\mathrm{Mn^{4+}}$
$\mathrm{Mn^{3+}}$
$\mathrm{Mn^{6+}}$
Answer: (a)
Solution
The unbalanced chemical equation is given as: $$\mathrm{H_2O_2} + \mathrm{MnO_4^-} \rightarrow \mathrm{Mn^{2+}} + \mathrm{O_2}.$$
Question 69
Chemistry · The s-Block Elements · Single correct
Choose the correct order of density of the alkali metals:
Li < K < $\mathrm{Na}$ < $\mathrm{Rb}$ < $\mathrm{Cs}$
Li < $\mathrm{Na}$ < K < $\mathrm{Rb}$ < $\mathrm{Cs}$
$\mathrm{Cs}$ < $\mathrm{Rb}$ < K < $\mathrm{Na}$ < $\mathrm{Li}$
Li < $\mathrm{Na}$ < K < $\mathrm{Cs}$ < $\mathrm{Rb}$
Answer: (a)
Solution
Q1 (A) Factual
Question 70
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
The geometry around boron in the product 'B' formed from the following reaction is $BF_3+NaH \xrightarrow{450K} A+NaF$ $$\mathrm{A + NMe_3 \rightarrow B}$$
Chemistry · Environmental Chemistry · Single correct
The photochemical smog does not generally contain:
NO
$\mathrm{NO}_2$
$\mathrm{SO}_2$
$\mathrm{HCHO}$
Answer: (c)
Solution
Q1 (C) Factual
Question 73
Chemistry · Hydrocarbons · Single correct
A compound 'A' on reaction with 'X' and 'Y' produces the same major product but different by product 'a' and 'b'. Oxidation of 'a' gives a substance produced by ants. 'X' and 'Y' respectively are :
$\mathrm{KMnO_4/H^+}$ and dil. $\mathrm{KMnO_4}$, $273\ \mathrm{K}$
$\mathrm{KMnO_4}$(dilute), $273\ \mathrm{K}$ and $\mathrm{KMnO_4/H^+}$
$\mathrm{KMnO_4/H^+}$ and $\mathrm{O_3}$, $\mathrm{H_2O/Zn}$
$\mathrm{O_3}$, $\mathrm{H_2O/Zn}$ and $\mathrm{KMnO_4/H^+}$
Answer: (d)
Solution
Let initial moles of reactant taken $= n$ Total moles obtained for benzene sulphonic acid (with $\%$ yield $= 60\%) = 0.6n$
Question 74
Chemistry · Alcohols, Phenols and Ethers · Single correct
Most stable product of the following reaction is: $$(i) \mathrm{H_3C} - \mathrm{SO_2Cl, Pyridine}$$ $$(ii) \mathrm{NaCN, DMF}$$
Answer: (b)
Solution
The reaction involves the conversion of an alcohol to a tosylate using TsCl and pyridine, followed by an $\mathrm{S_N^2}$ reaction with $\mathrm{CN^-}$ to form a nitrile.
Question 75
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Which one of the following reactions does not represent correct combination of substrate and product under the given conditions?
Answer: (d)
Solution
OH is formed.
Question 76
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
An organic compound 'A' on reaction with $\mathrm{NH_3}$ followed by heating gives compound B. Which on further strong heating gives compound C ($\mathrm{C_8H_5NO_2}$). Compound C on sequential reaction with ethanolic KOH, alkyl chloride and hydrolysis with alkali gives a primary amine. The compound A is :
Answer: (c)
Solution
Question 77
Chemistry · Polymers · Single correct
Melamine polymer is formed by the condensation of:
Answer: (a)
Solution
Melamine: Formaldehyde HCHO. Melamine formaldehyde resin is melamine polymer.
Question 78
Chemistry · Biomolecules · Single correct
During the denaturation of proteins, which of these structures will remain intact?
Primary
Secondary
Tertiary
Quaternary
Answer: (a)
Solution
Primary structure remains intact during denaturation of proteins.
Question 79
Chemistry · Chemistry in Everyday Life · Single correct
Drugs used to bind to receptors, inhibiting its natural function and blocking a message are called:
Agonists
Antagonists
Allosterists
Anti histaminists
Answer: (b)
Solution
Q1 (B) Factual
Question 80
Chemistry · Alcohols, Phenols and Ethers · Single correct
Given below are two statements : Statement I : On heating with $KHSO_4$, glycerol is dehydrated and acrolein is formed. Statement II : Acrolein has fruity odour and can be used to test glycerol's presence. 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
Acrolein has a pungent, suffocating odour. Acrolein is used to detect presence of glycerol.
Question 81
Chemistry · Chemical Bonding and Molecular Structure · Numerical
Among the following species $\mathrm{N_2}$, $\mathrm{N_2^+}$, $\mathrm{N_2^-}$, $\mathrm{N_2^{2-}}$, $\mathrm{O_2}$, $\mathrm{O_2^-}$, $\mathrm{O_2^{2-}}$ the number of species showing diamagnetism is
Answer: 2
Solution
Diamagnetic species are: $\mathrm{N_2}$, $\mathrm{O_2^{2-}}$
Question 82
Chemistry · Thermodynamics · Numerical
The enthalpy of combustion of propane, graphite and dihydrogen at 298 $\mathrm{K}$ are: $-2220.0 \, \mathrm{kJ \, mol^{-1}}$, $-393.5 \, \mathrm{kJ \, mol^{-1}}$ and $-285.8 \, \mathrm{kJ \, mol^{-1}}$ respectively. The magnitude enthalpy of formation of propane ($\mathrm{C_3H_8}$) is $\ldots \ldots \ldots \ldots \, \mathrm{kJ \, mol^{-1}}$. (Nearest integer)
Answer: 104
Solution
The reaction is given by $$3\mathrm{C_{(gr)}} + 4\mathrm{H_{2(g)}} \rightarrow \mathrm{C_3H_{8(g)}}$$ The enthalpy change is $$= -103.7 \, \mathrm{kJ \, mol^{-1}}$$
Question 83
Chemistry · States of Matter · Numerical
The pressure of a moist gas at $27^{\circ}C$ is $4 \, \mathrm{atm}$. The volume of the container is doubled at the same temperature. The new pressure of the moist gas is $\ldots \times 10^{-1} \, \mathrm{atm}$. (Nearest integer) (Given : The vapour pressure of water at $27^{\circ}C$ is $0.4 \, \mathrm{atm}$)
Answer: 22
Solution
Given $\left[ P_{gas} \right]_0 + V.P. = 4$. $\left[ P_{gas} \right]_0 = 4 - 0.4 = 3.6$. As volume is doubled, $\left[ P_{gas} \right]_{new} = 1.8 \, atm$. New Total Pressure $= 1.8 + 0.4 = 2.2 \, atm$.
Question 84
Chemistry · Electrochemistry · Numerical
The cell potential for $\mathrm{Zn|Zn^{2+}(aq)||Sn^{x+}|Sn}$ is $0.801 \, \mathrm{V}$ at $298 \, \mathrm{K}$. The reaction quotient for the above reaction is $10^{-2}$. The number of electrons involved in the given electrochemical cell reaction is _______ (Given $E^0_{\mathrm{Zn^{2+}|Zn}} = -0.763 \, \mathrm{V}$, $E^0_{\mathrm{Sn^{x+}|Sn}} = +0.008 \, \mathrm{V}$ and $\frac{2.303RT}{F} = 0.06 \, \mathrm{V}$)
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The half life for the decomposition of gaseous compound A is 240 $\mathrm{s}$ when the gaseous pressure was 500 $\mathrm{Torr}$ initially. When the pressure was 250 $\mathrm{Torr}$, the half life was found to be 4.0 $\mathrm{min}$. The order of the reaction is....... (Nearest integer)
Answer: 1
Solution
Given $\left(t_{1/2}\right)_{500 \, \mathrm{torr}} = 240 \, \mathrm{sec} = 4 \, \mathrm{min}$. $\left(t_{1/2}\right)_{250 \, \mathrm{torr}} = 4 \, \mathrm{min}$. $t_{1/2} \propto a^{1-n}$ As $t_{1/2}$ is independent of initial pressure. Hence, order is 1st order.
Question 86
Chemistry · Co-ordination Compounds · Numerical
Consider the following metal complexes : $$[Co(NH_3)_6]^{3+}$$ $$[CoCl(NH_3)_5]^{2+}$$ $$[Co(CN)_6]^{3-}$$ $$[Co(NH_3)_5(H_2O)]^{3+}$$ The spin-only magnetic moment value of the complex that absorbs light with shortest wavelength is B.M. (Nearest integer)
Answer: 0
Solution
Given $\Delta_0 \propto \frac{1}{\lambda}$. Here, CN being SFL will have maximum CFSE. So, $[\mathrm{Co(CN)_6}]^{3-}$ will be $d^2sp^3$, $\mu = 0$.
Question 87
Chemistry · The d-and f-Block Elements · Numerical
Among $\mathrm{Co}^{3+}$, $\mathrm{Ti}^{2+}$, $\mathrm{V}^{2+}$ and $\mathrm{Cr}^{2+}$ ions, one if used as a reagent cannot liberate $\mathrm{H}_2$ from dilute mineral acid solution, its spin-only magnetic moment in gaseous state is ......B.M. (Nearest integer)
Answer: 5
Solution
$\mathrm{Co^{3+}}$ can't liberate $\mathrm{H_2}$. It has $d^6$ configuration, Number of unpaired electrons $= 4$ $$\mu = \sqrt{4 \times 6} = 4.92 \, \mathrm{B.M.}$$
Question 88
Chemistry · Analytical Chemistry · Numerical
While estimating the nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from 0.25 g of the compound neutralized 2.5 mL of 2 M $H_2SO_4$. The percentage of nitrogen present in organic compound is ......
Answer: 56
Solution
The percentage of nitrogen is calculated using the formula: %N = $\frac{1.4(N_1 V_1)}{\text{mass of organic compound}}$ Substituting the given values: $$\%N = \frac{1.4(2.5 \times 2 \times 2)}{0.25} = 56$$
The number of $sp^3$ hybridised carbons in an acyclic neutral compound with molecular formula $\mathrm{C_4H_5N}$ is :
Answer: 1
Solution
The degree of unsaturation (DU) is calculated as follows: $$DU = 4 + 1 - \left(\frac{5 - 1}{2}\right) = 3$$ The first compound is $\mathrm{H_3C{-}CH{=}CH{-}C{\equiv}N}$, which has one $\mathrm{sp^3}$ carbon. The second compound is $\mathrm{CH_2{=}C{=}CH{=}CH{=}NH}$, which has zero $\mathrm{sp^3}$ carbon.
Question 90
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical
In the given reaction (Where Et is $-C_2H_5$) The number of chiral carbon/s in product A is
Answer: 2
Solution
The reaction involves the formation of a bicyclic compound with two chiral centers. The mechanism shows the movement of electrons leading to the formation of the product. The final structure has two chiral carbons, indicated by the asterisks.