JEE Main 27 July 2021 Shift 2 question paper with solutions
JEE Main 27 July 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 · Straight Lines and Pair of Straight Lines · Single correct
The point P(a, b) undergoes the following three transformations successively: (a) reflection about the line $y = x$. (b) translation through 2 units along the positive direction of $x$-axis. ($c$) rotation through angle $\frac{\pi}{4}$ about the origin in the anti-clockwise direction. If the co-ordinates of the final position of the point P are $\left( -\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right)$, then the value of $2a + b$ is equal to:
13
9
5
7
Answer: (b)
Solution
Image of $A(a, b)$ along $y = x$ is $B(b, a)$. Translating it 2 units it becomes $C(b + 2, a)$. Now, applying rotation theorem $$-\frac{1}{2} + \frac{7}{\sqrt{2}} i = ((b + 2) + ai) \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$$ $$-\frac{1}{\sqrt{2}} + \frac{7}{\sqrt{2}} i = \left( \frac{b+2}{\sqrt{2}} - \frac{a}{\sqrt{2}} \right) + i \left( \frac{b+2}{\sqrt{2}} + \frac{a}{\sqrt{2}} \right)$$ $\Rightarrow b - a + 2 = -1$ and $b + 2 + a = 7$ $\Rightarrow a = 4; b = 1$ $\Rightarrow 2a + b = 9$
Question 2
Maths · Binomial Theorem · Single correct
A possible value of '$x$', for which the ninth term in the expansion of $$\left\{ 3^{\log_3 \sqrt{25^{n-1} + 7}} + 3\left(-\frac{1}{8}\right)^{\log_3 (5^{x-1} + 1)} \right\}^{10}$$ in the increasing powers of $3\left(-\frac{1}{8}\right)^{\log_3 (5^{x-1} + 1)}$ is equal to 180, is:
0
-1
2
1
Answer: (d)
Solution
$\frac{{}^{10}C_{8}}{25^{x-1}+7}\times\left(5^{x-1}+1\right)^{-1}=180$ $\Rightarrow \frac{25^{x-1}+7}{5^{x-1}+1}=4$ Let $t=5^{x-1}$ $\Rightarrow \frac{t^{2}+7}{t+1}=4$ $\Rightarrow t^{2}-4t+3=0$ $\Rightarrow t=1,3$ $\Rightarrow 5^{x-1}=1$ (one of the possible values) $\Rightarrow x-1=0$ $\Rightarrow x=1$
Question 3
Maths · Three Dimensional Geometry · Single correct
For real numbers $\alpha$ and $\beta \neq 0$, if the point of intersection of the straight lines \[ \frac{x-\alpha}{1} = \frac{y-1}{2} = \frac{z-1}{3} and \frac{x-4}{\beta} = \frac{y-6}{3} = \frac{z-7}{3}, \] lies on the plane $x + 2y - z = 8$, then $\alpha - \beta$ is equal to:
5
9
3
7
Answer: (d)
Solution
First line is $(\phi+\alpha,\;2\phi+1,\;3\phi+1)$ and second line is $(q\beta+4,\;3q+6,\;3q+7)$ For intersection $\phi+\alpha=q\beta+4 \qquad \ldots (i)$ $2\phi+1=3q+6 \qquad \ldots (ii)$ $3\phi+1=3q+7 \qquad \ldots (iii)$ For (ii) and (iii), $\phi=1,\;q=-1$ So, from (i) $\alpha+\beta=3$ Now, point of intersection is $(\alpha+1,\;3,\;4)$ It lies on the plane. Hence, $\alpha=5 \ \&\ \beta=-2$
Question 4
Maths · Trigonometric Functions · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be defined as $$f(x+y) + f(x-y) = 2f(x)f(y), f\left(\frac{1}{2}\right) = -1.$$ Then the value of $$\sum_{k=1}^{20} \frac{1}{\sin(k) \sin(k+f(k))}$$ is equal to:
$\text{cosec}^2(21) \cos(20) \cos(2)$
$\sec^2(1) \sec(21) \cos(20)$
$\text{cosec}^2(1) \text{cosec}(21) \sin(20)$
$\sec^2(21) \sin(20) \sin(2)$
Answer: (c)
Solution
Given $f(x) = \cos \lambda x$. Therefore, $f\left(\frac{1}{2}\right) = -1$. So, $-1 = \cos \frac{\lambda}{2}$. This implies $\lambda = 2\pi$. Thus $f(x) = \cos 2\pi x$. Now $k$ is a natural number. Thus $f(k) = 1$. $$\sum_{k=1}^{20} \frac{1}{\sin k \sin(k+1)} = \frac{1}{\sin 1} \sum_{k=1}^{20} \left[ \frac{\sin((k+1)-k)}{\sin k \cdot \sin(k+1)} \right]$$ $$= \frac{1}{\sin 1} \sum_{k=1}^{20} (\cot k - \cot(k+1))$$ $$= \frac{\cot 1 - \cot 21}{\sin 1} = \cosec^2 1 \cdot \cosec(21) \cdot \sin 20$$
Question 5
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $\mathbb{C}$ be the set of all complex numbers. Let $$S_1 = \{ z \in \mathbb{C} : |z - 2| \leq 1 \}$$ and $$S_2 = \{ z \in \mathbb{C} : z(1+i) + \overline{z}(1-i) \geq 4 \}.$$ Then, the maximum value of $$\left| z - \frac{5}{2} \right|^2$$ for $$z \in S_1 \cap S_2$$ is equal to:
\frac{3+2\sqrt{2}}{4}
\frac{5+2\sqrt{2}}{2}
\frac{3+2\sqrt{2}}{2}
\frac{5+2\sqrt{2}}{4}
Answer: (d)
Solution
Given $|t - 2| \leq 1$. Put $t = x + iy$. $$(x - 2)^2 + y^2 \leq 1$$ Also, $t(1 + i) + \bar{t}(1 - i) \geq 4$ gives $x - y \geq 2$. Let point on circle be $A(2 + \cos \theta, \sin \theta)$. $$\theta \in \left[ -\frac{3\pi}{4}, \frac{\pi}{4} \right]$$ $$(AP)^2 = \left(2 + \cos \theta - \frac{5}{2}\right)^2 + \sin^2 \theta$$ $$= \cos^2 \theta - \cos \theta + \frac{1}{4} + \sin^2 \theta$$ $$= \frac{5}{4} - \cos \theta$$ For $(AP)^2$ maximum $\theta = -\frac{3\pi}{4}$. $$(AP)^2 = \frac{5}{4} + \frac{1}{\sqrt{2}} = \frac{5\sqrt{2} + 4}{4\sqrt{2}}$$
Question 6
Maths · Probability · Single correct
A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability. The smallest value of n, so that the probability of guessing at least 'n' correct answers is less than $\frac{1}{2}$, is
If $\tan\left(\frac{\pi}{9}\right)$, $x$, $\tan\left(\frac{7\pi}{18}\right)$ are in arithmetic progression and $\tan\left(\frac{\pi}{9}\right)$, $y$, $\tan\left(\frac{5\pi}{18}\right)$ are also in arithmetic progression, then $|x - 2y|$ is equal to :
Let the mean and variance of the frequency distribution \begin{tabular}{|l|l|l|l|l|} \hline x : & $x_1$ = 2 & $x_2$ = 6 & $x_3$ = 8 & $x_4$ = 9 \\ \hline f : & 4 & 4 & $\alpha$ & $\beta$ \\ \hline \end{tabular} be 6 and 6.8 respectively. If $x_3$ is changed from 8 to 7, then the mean for the new data will be:
4
5
$\frac{17}{3}$
$\frac{16}{3}$
Answer: (c)
Solution
Given $32 + 8\alpha + 9\beta = (8 + \alpha + \beta) \times 6$ which implies $2\alpha + 3\beta = 16$. Also, $4 \times 16 + 4 \times \alpha + 9\beta = (8 + \alpha + \beta) \times 6.8$ which implies $640 + 40\alpha + 90\beta = 544 + 68\alpha + 68\beta$. This simplifies to $28\alpha - 22\beta = 96$ and further to $14\alpha - 11\beta = 48$. From (i) and (ii), $\alpha = 5$ and $\beta = 2$. So, the new mean is $\frac{32 + 35 + 18}{15} = \frac{85}{15} = \frac{17}{3}$.
Question 9
Maths · Applications of Integrals · Single correct
The area of the region bounded by $y - x = 2$ and $x^2 = y$ is equal to :-
$\frac{16}{3}$
$\frac{2}{3}$
$\frac{9}{2}$
$\frac{4}{3}$
Answer: (c)
Solution
Given $y - x = 2$, $x^2 = y$. Now, $x^2 = 2 + x$. This implies $x^2 - x - 2 = 0$. Therefore, $(x + 1)(x - 2) = 0$. The area is given by $$\int_{-1}^{2} (2 + x - x^2) \, dx$$ which equals $$\left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^{2}$$. This simplifies to $$\left( 4 + 2 - \frac{8}{3} \right) - \left( -2 + \frac{1}{2} + \frac{1}{3} \right)$$ which equals $$6 - 3 + 2 - \frac{1}{2} = \frac{9}{2}$$.
Question 10
Maths · Differential Equations · Single correct
Let $y = y(x)$ be the solution of the differential equation $(x - x^3) \, dy = (y + yx^2 - 3x^4) \, dx, \, x > 2$ If $y(3) = 3$, then $y(4)$ is equal to:
4
12
8
16
Answer: (b)
Solution
(x - x^3) $\,$ dy = (y + yx^2 - 3x^4) $\,$ dx $\Rightarrow$ x $\,$ dy - y $\,$ dx = (yx^2 - 3x^4) $\,$ dx + x^3 $\,$ dy $\Rightarrow$ $\frac{x \, dy - y \, dx}{x^2}$ = (y $\,$ dx + x $\,$ dy) - 3x^2 $\,$ dx $\Rightarrow$ $\mathrm{d}$ $\left$( $\frac{y}{x}$ $\right$) = $\mathrm{d}$(xy) - $\mathrm{d}$(x^3) Integrate $\Rightarrow$ $\frac{y}{x}$ = xy - x^3 + c given f(3) = 3 $\Rightarrow$ $\frac{3}{3}$ = 3 $\times$ 3 - 3^3 + c $\Rightarrow$ c = 19 $\therefore$ $\frac{y}{x}$ = xy - x^3 + 19 at x = 4, $\frac{y}{4}$ = 4y - 64 + 19 15y = 4 $\times$ 45 $\Rightarrow$ y = 12
Question 11
Maths · Limits and Derivatives · Single correct
The value of $\lim_{x \to 0} \left( \frac{x}{\sqrt[3]{1-\sin x} - \sqrt[3]{1+\sin x}} \right)$ is equal to:
Maths · Straight Lines and Pair of Straight Lines · Single correct
Two sides of a parallelogram are along the lines $4x + 5y = 0$ and $7x + 2y = 0$. If the equation of one of the diagonals of the parallelogram is $11x + 7y = 9$, then other diagonal passes through the point:
(1, 2)
(2, 2)
(2, 1)
(1, 3)
Answer: (b)
Solution
Both the lines pass through origin. Point D is equal to the intersection of $4x + 5y = 0$ and $11x + 7y = 9$. So, coordinates of point D are $\left( \frac{5}{3}, -\frac{4}{3} \right)$. Also, point B is the point of intersection of $7x + 2y = 0$ and $11x + 7y = 9$. So, coordinates of point B are $\left( -\frac{2}{3}, \frac{7}{3} \right)$. Diagonals of the parallelogram intersect at the middle. Let the middle point of B, D be: $$\left( \frac{\frac{5}{3} - \frac{2}{3}}{2}, \frac{-\frac{4}{3} + \frac{7}{3}}{2} \right) = \left( \frac{1}{2}, \frac{1}{2} \right)$$ Equation of diagonal AC: $$\Rightarrow (y - 0) = \frac{\frac{1}{a} - 0}{\frac{1}{a} - 0}(\pi - 0)$$ $$y = x$$ Diagonal AC passes through $(2, 2)$.
Question 13
Maths · Relations and Functions · Single correct
Let $\alpha = \max_{x \in \mathbb{R}} \left\{ 8^{2 \sin 3x} \cdot 4^{4 \cos 3x} \right\}$ and $\beta = \min_{x \in \mathbb{R}} \left\{ 8^{2 \sin 3x} \cdot 4^{4 \cos 3x} \right\}$. If $8x^2 + bx + c = 0$ is a quadratic equation whose roots are $\alpha^{1/5}$ and $\beta^{1/5}$, then the value of $c - b$ is equal to:
42
47
43
50
Answer: (a)
Solution
Given ($\alpha=\max\left\{8^{2\sin3x}\cdot4^{4\cos3x}\right\}$\) \[ =\max\left\{2^{6\sin3x}\cdot2^{8\cos3x}\right\} \] \[ =\max\left\{2^{6\sin3x+8\cos3x}\right\} \] and \[ \beta=\min\left\{8^{2\sin3x}\cdot4^{4\cos3x}\right\} =\min\left\{2^{6\sin3x+8\cos3x}\right\} \] Now range of $\(6\sin3x+8\cos3x\)$ \[ \left[-\sqrt{6^2+8^2},\sqrt{6^2+8^2}\right] =[-10,10] \] \[ \alpha=2^{10},\qquad \beta=2^{-10} \] So, \[ \alpha^{1/5}=2^2=4 \] \[ \Rightarrow \beta^{1/5}=2^{-2}=\frac14 \] Quadratic $\(8x^2+bx+c=0\)$, \[ c-b=8\Big[(\text{product of roots})+(\text{sum of roots})\Big] \] \[ =8\left[4\cdot\frac14+4+\frac14\right] =8\left[\frac{21}{4}\right] =42 \]
Question 14
Maths · Continuity and Differentiability · Single correct
Let $f : [0, \infty) \to [0, 3]$ be a function defined by $f(x) = \begin{cases} \max\{\sin t : 0 \leq t \leq x\}, & 0 \leq x \leq \pi \\ 2 + \cos x, & x > \pi \end{cases}$ Then which of the following is true?
$f$ is continuous everywhere but not differentiable exactly at one point in $(0, \infty)$
$f$ is differentiable everywhere in $(0, \infty)$
$f$ is not continuous exactly at two points in $(0, \infty)$
$f$ is continuous everywhere but not differentiable exactly at two points in $(0, \infty)$
Answer: (b)
Solution
Graph of $\max \{ \sin t : 0 \leq t \leq x \}$ in $x \in [0, \pi]$. & graph of $\cos x$ for $x \in [\pi, \infty)$. So graph of $$f(x) = \begin{cases} \max \{ \sin t : 0 \leq t \leq x \}, & 0 \leq x \leq \pi \\ 2 + \cos x, & x > h \end{cases}$$ $f(x)$ is differentiable everywhere in $(0, \infty)$.
Question 15
Maths · Sets · Single correct
Let $\mathbb{N}$ be the set of natural numbers and a relation R on $\mathbb{N}$ be defined by $$R = \{(x,y) \in \mathbb{N} \times \mathbb{N} : x^3 - 3x^2y - xy^2 + 3y^3 = 0\}$$ Then the relation R is :
symmetric but neither reflexive nor transitive
reflexive but neither symmetric nor transitive
reflexive and symmetric, but not transitive
an equivalence relation
Answer: (b)
Solution
Given the equation $$x^3 - 3x^2y - xy^2 + 3y^3 = 0$$ we have: $$\Rightarrow x \left(x^2 - y^2\right) - 3y \left(x^2 - y^2\right) = 0$$ $$\Rightarrow (x - 3y)(x - y)(x + y) = 0$$ Now, $x = y$ for all $(x, y) \in \mathbb{N} \times \mathbb{N}$ so reflexive. But not symmetric and transitive. See, $(3, 1)$ satisfies but $(1, 3)$ does not. Also $(3, 1)$ and $(1, -1)$ satisfies but $(3, -1)$ does not.
Question 16
Maths · Mathematical Reasoning · Single correct
Which of the following is the negation of the statement "for all $M > 0$, there exists $x \in S$ such that $x \geq M$"?
there exists $M > 0$, such that $x < M$ for all $x \in S$
there exists $M > 0$, there exists $x \in S$ such that $x \geq M$
there exists $M > 0$, there exists $x \in S$ such that $x < M$
there exists $M > 0$, such that $x \geq M$ for all $x \in S$
Answer: (a)
Solution
P: for all $M > 0$, there exists $x \in S$ such that $x \geq M$. Not P: there exists $M > 0$, for all $x \in S$ such that $x < m$. Negation of 'there exists' is 'for all'.
Question 17
Maths · Conic Sections · Single correct
Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept $6\sqrt{5}$ on the x-axis. Then the radius of the circle C is equal to:
$\sqrt{53}$
9
8
$\sqrt{82}$
Answer: (b)
Solution
The radius $r$ is calculated using the Pythagorean theorem. $$r = \sqrt{6^2 + (3\sqrt{5})^2}$$ $$= \sqrt{36 + 45} = 9$$
Question 18
Maths · Vector Algebra · Single correct
Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a} = \vec{b} \times (\vec{b} \times \vec{c})$. If magnitudes of the vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ are $\sqrt{2}$, $1$ and $2$ respectively and the angle between $\vec{b}$ and $\vec{c}$ is $\theta \left(0 < \theta < \frac{\pi}{2}\right)$, then the value of $1 + \tan \theta$ is equal to:
Let A and B be two $3 \times 3$ real matrices such that $$\left( A^2 - B^2 \right)$$ is invertible matrix. If $A^5 = B^5$ and $A^3 B^2 = A^2 B^3$, then the value of the determinant of the matrix $A^3 + B^3$ is equal to:
Maths · Applications of Derivatives · Single correct
Let $f : (a, b) \to \mathbb{R}$ be twice differentiable function such that $f(x) = \int_a^x g(t)\,dt$ for a differentiable function $g(x)$. If $f(x) = 0$ has exactly five distinct roots in $(a, b)$, then $g(x)g'(x) = 0$ has at least :
twelve roots in (a, b)
five roots in (a, b)
seven roots in (a, b)
three roots in (a, b)
Answer: (c)
Solution
Given $f(x) = \int_a^x g(t) \, dt$. As $f(x) \to 5$, $f'(x) \to 4$, $g(x) \to 4$, and $g'(x) \to 3$.
Question 21
Maths · Vector Algebra · Numerical
Let $\vec{a}=\hat{i}-\alpha\hat{j}+\beta\hat{k}$, $\vec{b}=3\hat{i}+\beta\hat{j}-\alpha\hat{k}$, and $\vec{c}=-\alpha\hat{i}-2\hat{j}+\hat{k}$, where $\alpha$ and $\beta$ are integers. If $\vec{a}\cdot\vec{b}=-1$ and $\vec{b}\cdot\vec{c}=10$, then $(\vec{a}\times\vec{b})\cdot\vec{c}$ is equal to $\underline{\hspace{2cm}}$.
The distance of the point $P(3, 4, 4)$ from the point of intersection of the line joining the points $Q(3, -4, -5)$ and $R(2, -3, 1)$ and the plane $2x + y + z = 7$, is equal to
Answer: 7
Solution
$\overrightarrow{QR}:\ \frac{x-3}{1}=\frac{y+4}{-1}=\frac{z+5}{-6}=r$ $\Rightarrow (x,y,z)=(r+3,\,-r-4,\,-6r-5)$ Now, satisfying it in the given plane. We get $r=-2$. So, required point of intersection is $T(1,\,-2,\,7)$. Hence, $PT=7$.
Question 23
Maths · Complex Numbers and Quadratic Equations · Numerical
If the real part of the complex number $z = \frac{3 + 2i \cos \theta}{1 - 3i \cos \theta}, \theta \in \left(0, \frac{\pi}{2}\right)$ is zero, then the value of $\sin^2 3\theta + \cos^2 \theta$ is equal to
Answer: 1
Solution
Given $\mathrm{Re}(z)=\frac{3-6\cos^2\theta}{1+9\cos^2\theta}=0$ $\Rightarrow 3-6\cos^2\theta=0$ $\Rightarrow \cos^2\theta=\frac{1}{2}$ $\Rightarrow \theta=\frac{\pi}{4}$ Hence, $\sin^2(3\theta)+\cos^2\theta=1$
Question 24
Maths · Conic Sections · Numerical
Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3, -4)$, one focus at $(4, -4)$ and one vertex at $(5, -4)$. If $mx - y = 4$, $m > 0$ is a tangent to the ellipse E, then the value of $5 \, m^2$ is equal to
Answer: 3
Solution
Given $C(3, -4)$, $S(4, -4)$ and $A(5, -4)$. Hence, $a = 2$ and $ae = 1$. Therefore, $e = \frac{1}{2}$ which implies $b^2 = 3$. So, $E : \frac{(x-3)^2}{4} + \frac{(y+4)^2}{3} = 1$. Intersecting with given tangent. $$\frac{x^2 - 6x + 9}{4} + \frac{m^2 x^2}{3} = 1$$ Now, $D = 0$ (as it is tangent). So, $5m^2 = 3$.
Question 25
Maths · Integrals · Numerical
If $\int_{0}^{\pi} (\sin^3 x) e^{-\sin^2 x} dx = \alpha - \frac{\beta}{e} \int_{0}^{1} \sqrt{t e^t} dt$, then $\alpha + \beta$ is equal to
Maths · Complex Numbers and Quadratic Equations · Numerical
The number of real roots of the equation $$e^{4x} - e^{3x} - 4e^{2x} - e^{x} + 1 = 0$$ is equal to
Answer: 2
Solution
Given $t^4 - t^3 - 4t^2 - t + 1 = 0$, $e^x = t > 0$. $$\Rightarrow t^2 - t - 4 - \frac{1}{t} + \frac{1}{t^2} = 0$$ $$\Rightarrow \alpha^2 - \alpha - 6 = 0, \alpha = t + \frac{1}{t} \geq 2$$ $$\Rightarrow \alpha = 3, -2 (reject)$$ $$\Rightarrow t + \frac{1}{t} = 3$$ Therefore, the number of real roots $= 2$.
Question 27
Maths · Differential Equations · Numerical
Let y = y(x) be the solution of the differential equation dy = e^{(x+y)} dx; $\alpha\in\mathbb{N}$. If y($\log_e$ 2) = $\log_e$ 2 and y(0) = $\log_e$ $( \frac{1}{2})$, then the value of $\alpha$ is equal to
Answer: 2
Solution
Given $$\int e^{-y} \, dy = \int e^{ax} \, dx$$ $$\Rightarrow e^{-y} = \frac{e^{ax}}{\alpha} + c$$ Put $$(x, y) = (\ln 2, \ln 2)$$ $$-\frac{1}{2} = \frac{2^\alpha}{\alpha} + C$$ Put $$(x, y) \equiv (0, -\ln 2)$$ in (i) $$-2 = \frac{1}{\alpha} + C$$ (ii) - (iii) $$\frac{2^\alpha - 1}{\alpha} = \frac{3}{2}$$ $$\Rightarrow \alpha = 2 (as \alpha \in \mathbb{N})$$
Question 28
Maths · Permutations and Combinations · Numerical
Let n be a non-negative integer. Then the number of divisors of the form "4n + 1" of the number $$(10)^{10} \cdot (11)^{11} \cdot (13)^{13}$$ is equal to
Answer: 924
Solution
Given $N = 2^{10} \times 5^{10} \times 11^{11} \times 13^{13}$. Now, power of 2 must be zero, power of 5 can be anything, power of 13 can be anything. But, power of 11 should be even. So, required number of divisors is $$1 \times 11 \times 14 \times 6 = 924$$
Question 29
Maths · Sets · Numerical
Let $A = \{ n \in \mathbb{N} \mid n^2 \leq n + 10,000 \}$, $B = \{ 3k + 1 \mid k \in \mathbb{N} \}$ and $C = \{ 2k \mid k \in \mathbb{N} \}$, then the sum of all the elements of the set $A \cap (B - C)$ is equal to
Answer: 832
Solution
B - C $\equiv$ $\{$7, 13, 19, $\ldots$, 97, $\ldots$$\}$ Now, $n^2 - n \leq 100 \times 100$ $\($$\Rightarrow$ n(n - 1) $\leq$ 100 $\times$ 100$\)$ $\($$\Rightarrow$ A = $\{$1, 2, $\ldots$, 100$\}$$\)$ So, $\($A $\cap$ (B - C) = $\{$7, 13, 19, $\ldots$, 97$\}$$\)$ Hence, sum = $\($$\frac{16}{2}$(7 + 97) = 832$\)$
Question 30
Maths · Matrices · Numerical
If $A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ and $M = A + A^2 + A^3 + \ldots + A^{20}$ then the sum of all the elements of the matrix $M$ is equal to
Physics · Dual Nature of Radiation and Matter · Single correct
An electron and proton are separated by a large distance. The electron starts approaching the proton with energy $3 \, \mathrm{eV}$. The proton captures the electrons and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength $4000 \, \mathrm{\mathring{A}}$. What is the maximum kinetic energy of the emitted photoelectron?
$7.61 \, \mathrm{eV}$
$1.41 \, \mathrm{eV}$
$3.3 \, \mathrm{eV}$
No photoelectron would be emitted
Answer: (b)
Solution
Initially, energy of electron $= +3 \, \mathrm{eV}$ finally, in $2^{nd}$ excited state, energy of electron $= -\frac{(13.6 \, \mathrm{eV})}{3^2}$ $= -1.51 \, \mathrm{eV}$ Loss in energy is emitted as photon. So, photon energy $\frac{hc}{\lambda} = 4.51 \, \mathrm{eV}$ $\mathrm{KE_{max}} = \frac{hc}{\lambda} - \phi = 4.51 - \left( \frac{hc}{\lambda_{th}} \right)$ $= 4.51 \, \mathrm{eV} - \frac{12400 \, \mathrm{eV} \cdot \AA}{4000 \AA}$ $= 1.41 \, \mathrm{eV}$
Question 32
Physics · Ray Optics and Optical Instruments · Single correct
The expected graphical representation of the variation of angle of deviation ' $\delta$ ' with angle of incidence ' $i$ ' in a prism is :
Answer: (a)
Solution
Standard graph between angle of deviation and incident angle.
Question 33
Physics · Mechanical Properties of Fluids · Single correct
A raindrop with radius $R = 0.2 \, \mathrm{mm}$ falls from a cloud at a height $h = 2000 \, \mathrm{m}$ above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is: [Density of water $f_w = 1000 \, \mathrm{kg} \, \mathrm{m}^{-3}$ and Density of air $f_a = 1.2 \, \mathrm{kg} \, \mathrm{m}^{-3}$, $g = 10 \, \mathrm{m/s}^2$ Coefficient of viscosity of air $= 1.8 \times 10^{-5} \, \mathrm{Nsm}^{-2}$]
One mole of an ideal gas is taken through an adiabatic process where the temperature rises from $27^\circ \mathrm{C}$ to $37^\circ \mathrm{C}$. If the ideal gas is composed of polyatomic molecule that has $4$ vibrational modes, which of the following is true? \[ \mathrm{R} = 8.314 \, \mathrm{J} \, \mathrm{mol}^{-1} \mathrm{K}^{-1} \]
The work done by the gas is close to \(332\,\mathrm{J}\).
The work done on the gas is close to \(582\,\mathrm{J}\).
The work done by the gas is close to \(582\,\mathrm{J}\).
The work done on the gas is close to \(332\,\mathrm{J}\).
Answer: (b)
Solution
Since, each vibrational mode, corresponds to two degrees of freedom, hence, $f = 3 (trans.) + 3 (rot.) + 8 (vib.) = 14$. $\gamma = 1 + \frac{2}{f}$ $$\gamma = 1 + \frac{2}{14} = \frac{8}{7}$$ $$W = \frac{nR\Delta T}{\gamma - 1} = -582$$ As $W < 0$, work is done on the gas.
Question 35
Physics · Oscillations · Single correct
An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $t = \frac{T}{4}$ s starting from mean position. Assume that the initial phase of the oscillation is zero.
0.62 J
6.2 $\times$ 10^{-3} J
1.2 $\times$ 10^{3} J
6.2 $\times$ 10^{3} J
Answer: (a)
Solution
Given $T = 2\pi \sqrt{\frac{m}{k}}$. $0.2 = 2\pi \sqrt{\frac{0.5}{k}}$. $k = 50\pi^2$. $\approx 500$. $x = A \sin(\omega t + \phi)$. $= 5 \, cm \sin\left(\frac{\omega T}{4} + 0\right)$. $= 5 \, cm \sin\left(\frac{\pi}{2}\right)$. $= 5 \, cm$. $PE = \frac{1}{2} k x^2$. $= \frac{1}{2} (500) \left(\frac{5}{100}\right)^2$. $= 0.6255$.
Question 36
Physics · Physical World, Units and Measurements · Single correct
Match List I with List II. $$ \begin{array}{ll} \text{List-I} & \text{List-II} \\ \hline (a) \text{ Capacitance, } C & (i) \, M^1 L^1 T^{-3} A^{-1} \\ (b) \text{ Permittivity of free space, } \varepsilon_0 & (ii) \, M^{-1} L^{-3} T^4 A^2 \\ (c) \text{ Permeability of free space, } \mu_0 & (iii) \, M^{-1} L^{-2} T^4 A^2 \\ (d) \text{ Electric field, } E & (iv) \, M^1 L^1 T^{-2} A^{-2} \end{array} $$ Choose the correct answer from the options given below
Given below is the plot of a potential energy function $U(x)$ for a system, in which a particle is in one dimensional motion, while a conservative force $F(x)$ acts on it. Suppose that $E_{mech} = 8 \, \mathrm{J}$, the incorrect statement for this system is :
at $x > x_4$, K. E. is constant throughout the region.
at $x < x_1$, K. E. is smallest and the particle is moving at the slowest speed.
at $x = x_2$, KE. is greatest and the particle is moving at the fastest speed.
at $x = x_3$, K. E. = 4 J.
Answer: (b)
Solution
Given $E_{mech.} = 8 \, J$. (A) At $x > x_4$, $U = constant = 6 \, J$. $K = E_{mech.} - U = 2 \, J = constant$. (B) At $x < x_1$, $U = constant = 8 \, J$. $K = E_{mech.} - U = 8 - 8 = 0 \, J$. Particle is at rest. (C) At $x = x_2$, $U = 0 \implies E_{mech.} = K = 8 \, J$. KE is greatest, and particle is moving at fastest speed. (D) At $x = x_3$, $U = 4 \, J$. $U + K = 8 \, J$. $K = 4 \, J$.
Question 38
Physics · Alternating Current · Single correct
A 100 $\Omega$ resistance, a 0.1 $\mu\mathrm{F}$ capacitor and an inductor are connected in series across a 250 $\mathrm{V}$ supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 $\mathrm{Hz}$.
0.70 $\mathrm{H}$
70.3 $\mathrm{mH}$
7.03 $\times 10^{-5} \mathrm{H}$
70.3 $\mathrm{H}$
Answer: (d)
Solution
Given $C = 0.1 \, \mu \mathrm{F} = 10^{-7} \, \mathrm{F}$. Resonant frequency $= 60 \, \mathrm{Hz}$, $\omega_0 = \frac{1}{\sqrt{LC}}$. $$2 \pi f_0 = \frac{1}{\sqrt{LC}} \Rightarrow L = \frac{1}{4 \pi^2 f_0^2 C}$$ by putting values $L = 70.3 \, \mathrm{Hz}$.
Question 39
Physics · Electric Charges and Fields · Single correct
A simple pendulum of mass ' m ', length ' l ' and charge '+q' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be
Let $E$ be the electric field in air. $T \sin \theta = qE$. $T \cos \theta = mg$. $$\tan \theta = \frac{qE}{mg}$$ $$Q = \left[ \frac{C_1 C_2}{C_1 + C_2} \right] [V_1 + V_2]$$ $$E = \frac{Q}{A \varepsilon_0} = \left[ \frac{C_1 C_2}{C_1 + C_2} \right] \frac{[V_1 + V_2]}{A \varepsilon_0}$$ $$C_1 = \frac{\varepsilon_0 A}{d-t} \implies E = \frac{C_2 [V_1 + V_2]}{(C_1 + C_2)(d-t)}$$ Now $$\theta = \tan^{-1} \left[ \frac{qE}{mg} \right]$$ $$\theta = \tan^{-1} \left[ \frac{q}{mg} \times \frac{C_2 (V_1 + V_2)}{(C_1 + C_2)(d-t)} \right]$$
Question 40
Physics · Thermodynamics · Single correct
Two Carnot engines A and B operate in series such that engine A absorbs heat at $T_1$ and rejects heat to a sink at temperature $T$. Engine B absorbs half of the heat rejected by Engine A and rejects heat to the sink at $T_3$. When workdone in both the cases is equal, to value of $T$ is :
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Find the truth table for the function Y of A and B represented in the following figure.
Answer: (b)
Solution
Question 42
Physics · Moving Charges and Magnetism · Single correct
Figure A and B shown two long straight wires of circular cross-section (a and b with a < b), carrying current I which is uniformly distributed across the cross-section. The magnitude of magnetic field B varies with radius r and can be represented as:
Answer: (c)
Question 43
Physics · Gravitation · Single correct
Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :
$\sqrt{\frac{G}{2R^3}}$
$\frac{1}{2} \sqrt{\frac{G}{R^3}}$
$\frac{1}{2R} \sqrt{\frac{1}{G}}$
$\sqrt{\frac{2G}{R^3}}$
Answer: (b)
Solution
The force $F$ is given by $$F = \frac{Gm^2}{(2R)^2} = mR\omega^2.$$ Solving for $\omega$, we have $$\omega = \frac{1}{2} \sqrt{\frac{G}{R^3}}.$$
Question 44
Physics · Nuclei · Single correct
Consider the following statements: A. Atoms of each element emit characteristics spectrum. B. According to Bohr's Postulate, an electron in a hydrogen atom, revolves in a certain stationary orbit. C. The density of nuclear matter depends on the size of the nucleus. D. A free neutron is stable but a free proton decay is possible. E. Radioactivity is an indication of the instability of nuclei. Choose the correct answer from the options given below :
A, B, C, D and E
A, B and E only
B and D only
A, C and E only
Answer: (b)
Solution
(A) True, atom of each element emits characteristic spectrum. (B) True, according to Bohr's postulates $$mvr = \frac{nh}{2\pi}$$ and hence electron resides into orbits of specific radius called stationary orbits. ($C$) False, density of nucleus is constant (D) False, A free neutron is unstable decays into proton and electron and antineutrino. (E) True unstable nucleus show radioactivity.
Question 45
Physics · Electric Charges and Fields · Single correct
What will be the magnitude of electric field at point O as shown in figure? Each side of the figure is $l$ and perpendicular to each other?
A physical quantity 'y' is represented by the formula $y = m^2 r^{-4} g^{-x} l^{-\frac{3}{2}}$ If the percentage errors found in $y, m, l$ and $g$ are $18, 1, 0.5, 4$ and $p$ respectively, then find the value of $x$ and $p$.
5 and $\pm 2$
4 and $\pm 3$
$\frac{16}{3}$ and $\pm \frac{3}{2}$
8 and $\pm 2$
Answer: (c)
Solution
Given $$\frac{\Delta y}{y} = \frac{2 \Delta m}{m} + \frac{4 \Delta r}{r} + \frac{x \Delta g}{g} + \frac{3}{2} \frac{\Delta \ell}{\ell}$$ $$18 = 2(1) + 4(0.5) + xp + \frac{3}{2}(4)$$ $$8 = xp$$ By checking from options, $$x = \frac{16}{3}, p = \pm \frac{3}{2}$$
Question 47
Physics · Work, Energy and Power · Single correct
An automobile of mass ' m ' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by:
The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of $9.0 \times 10^3$ km. Find the mass of Mars. $\{$ Given $\frac{4\pi^2}{G}$ = 6 $\times$ $10^{11} N^{-1} m^{-2} kg^2$ $\}$
$5.96 \times 10^{19}$ kg
$3.25 \times 10^{21}$ kg
$7.02 \times 10^{25}$ kg
$6.00 \times 10^{23}$ kg
Answer: (d)
Solution
Option D is correct $$T^2 = \frac{4\pi^2}{GM} \cdot r^3$$ $$M = \frac{4\pi^2}{G} \cdot \frac{r^3}{T^2}$$ by putting values $$M = 6 \times 10^{23}$$
Question 49
Physics · Motion in a Straight Line · Single correct
A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation $$ F = F_0 \left[ 1 - \left( \frac{t - T}{T} \right)^2 \right] $$ Where $F_0$ and $T$ are constants. The force acts only for the time interval $2T$. The velocity $v$ of the particle after time $2T$ is :
$2 F_0 T/M$
$F_0 T/2M$
$4 F_0 T/3M$
$F_0 T/3M$
Answer: (c)
Solution
Given $t = 0$, $u = 0$. $$a = \frac{F_0}{M} - \frac{F_0}{MT^2}(t - T)^2 = \frac{dv}{dt}$$ $$\int_0^v dv = \int_{t=0}^{2T} \left( \frac{F_0}{M} - \frac{F_0}{MT^2}(t - T)^2 \right) dt$$ $$V = \left[ \frac{F_0}{M} t \right]_0^{2T} - \frac{F_0}{MT^2} \left[ \frac{t^3}{3} - t^2 T + T^2 t \right]_0^{2T}$$ $$V = \frac{4 F_0 T}{3M}$$
Question 50
Physics · Current Electricity · Single correct
The resistance of a conductor at $15^{\circ}C$ is $16\,\Omega$ and at $100^{\circ}C$ is $20\,\Omega$. What will be the temperature coefficient of resistance of the conductor?
Physics · System of Particles and Rotational Motion · Numerical
In the given figure, two wheels $P$ and $Q$ are connected by a belt $B$. The radius of $P$ is three times as that of $Q$. In case of same rotational kinetic energy, the ratio of rotational inertias $\left( \frac{I_1}{I_2} \right)$ will be $x : 1$. The value of $x$ will be
Answer: 9
Solution
The equation for rotational kinetic energy is given by $$\frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} I_2 \omega_2^2.$$ By substituting the expressions for angular velocity, we have $$I_1 \left( \frac{v}{3R} \right)^2 = I_2 \left( \frac{v}{R} \right)^2.$$ Solving for the ratio of moments of inertia, we find $$\frac{I_1}{I_2} = \left( \frac{3R}{R} \right)^2 = \frac{9}{1}.$$
Question 52
Physics · Wave Optics · Numerical
The difference in the number of waves when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is mm. [Refractive index of air = 1.0003, wavelength of yellow light in vacuum = 6000$\AA$...]
The maximum amplitude for an amplitude modulated wave is found to be $12 \, \mathrm{V}$ while the minimum amplitude is found to be $3 \, \mathrm{V}$. The modulation index is $0.6x$ where $x$ is
Answer: 1
Solution
Given $A_{max} = A_c + A_m = 12$ and $A_{min} = A_c - A_m = 3$. Therefore, $A_c = \frac{15}{2}$ and $A_m = \frac{9}{2}$. The modulation index is given by $\frac{A_m}{A_c} = \frac{9/2}{15/2} = 0.6$. Thus, $x = 1$.
Question 54
Physics · Electromagnetic Induction · Numerical
In the given figure the magnetic flux through the loop increases according to the relation $\phi_B(t) = 10t^2 + 20t$, where $\phi_B$ is in milliwebers and $t$ is in seconds. The magnitude of current through $R = 2\Omega$ resistor at $t = 5 \, \mathrm{s}$ is_____ mA.
A particle executes simple harmonic motion represented by displacement function as $$x(t) = A \sin(\omega t + \phi)$$ If the position and velocity of the particle at $t = 0 \, \mathrm{s}$ are $2 \, \mathrm{cm}$ and $2\omega \, \mathrm{cm/s}^{-1}$ respectively, then its amplitude is $x\sqrt{2} \, \mathrm{cm}$ where the value of $x$ is
Answer: 2
Solution
Given $$x(t) = A \sin(\omega t + \phi)$$ $$v(t) = A \omega \cos(\omega t + \phi)$$ From equation (1): $$2 = A \sin \phi$$ From equation (2): $$2\omega = A \omega \cos \phi$$ From (1) and (2), we have $\($ $\tan$ $\phi$ = 1 $\)$. Thus, $\($ $\phi$ = 45^$\circ$ $\)$. Putting the value of $\($ $\phi$ $\)$ in equation (1): $$2 = A \left\{ \frac{1}{\sqrt{2}} \right\}$$ Therefore, $$A = 2\sqrt{2}$$ Finally, $$x = 2$$
Question 56
Physics · Motion in a Plane · Numerical
A swimmer wants to cross a river from point $A$ to point $B$. Line $AB$ makes an angle of $30^\circ$ with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle $\theta$ with the line $AB$ should be $\underline{\hspace{1cm}}$, so that the swimmer reaches point $B$.
Answer: 30
Solution
Both velocity vectors are of same magnitude, therefore the resultant would pass exactly midway through them. \[ \theta = 30^\circ \]
Question 57
Physics · Current Electricity · Numerical
For the circuit shown, the value of current at time $t=3.2\ \mathrm{s}$ will be $\underline{\qquad}\ \mathrm{A}.$ [Voltage distribution $V(t)$ is shown by Fig.\,(1) and the circuit is shown in Fig.\,(2)]
Answer: A
Solution
From graph voltage at t $= 3.2sec$ is $6 \, volt$. $$i = \frac{6 - 5}{1}$$ $i = 1 \, A$
Question 58
Physics · Motion in a Plane · Numerical
A small block slides down from the top of hemisphere of radius $R = 3 \, \mathrm{m}$ as shown in the figure. The height 'h' at which the block will lose contact with the surface of the sphere is m. (Assume there is no friction between the block and the hemisphere)
Answer: 2
Solution
Given $mg \cos \theta = \frac{mv^2}{R}$ and $\cos \theta = \frac{h}{R}$. Using energy conservation, $mg\{R - h\} = \frac{1}{2} mv^2$. From (1) and (2), $mg \left\{ \frac{h}{R} \right\} = \frac{2mg\{R-h\}}{R}$. Solving for $h$, we get $h = \frac{2R}{3} = 2 \, \mathrm{m}$.
Question 59
Physics · Atoms · Numerical
The $K_\alpha$ X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms with a $K$ electron knocked out is 27.5 keV, the energy of this atom when an $L$ electron is knocked out will be keV. (Round off to the nearest integer) $$[h = 4.14 \times 10^{-15} \, \mathrm{eVs}, c = 3 \times 10^8 \, \mathrm{ms^{-1}}]$$
The water is filled upto height of 12 m in a tank having vertical sidewalls. A hole is made in one of the walls at a depth 'h' below the water level. The value of 'h' for which the emerging stream of water strikes the ground at the maximum range is m.
Answer: 6
Solution
Given the problem, we have: $$R = \sqrt{2gh} \times \sqrt{\frac{(12-h) \times 2}{g}}$$ Simplifying, we get: $$\sqrt{4h(12-h)} = R$$ For maximum $R$, we set the derivative to zero: $$\frac{dR}{dh} = 0$$ This implies $h = 6 \, \mathrm{m}$.
Chemistry
Question 61
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Which one of the following set of elements can be detected using sodium fusion extract?
Sulfur, Nitrogen, Phosphorous, Halogens
Phosphorous, Oxygen, Nitrogen, Halogens
Nitrogen, Phosphorous, Carbon, Sulfur
Halogens, Nitrogen, Oxygen, Sulfur
Answer: (a)
Solution
By sodium fusion extract we can detect sulphur, nitrogen, Phosphorous and halogens, because they are converted into their ionic form with sodium metal.
Question 62
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Consider the above reaction, the major product $"P"$ formed is
Answer: (b)
Solution
Question 63
Chemistry · Hydrogen · Single correct
The number of neutrons and electrons, respectively, present in the radioactive isotope of hydrogen is :-
1 and 1
3 and 1
2 and 1
2 and 2
Answer: (c)
Solution
Radioactive isotope of hydrogen is Tritium $\left( ^3_1 \mathrm{T} \right)$. No. of neutrons $(A - Z) = 3 - 1 = 2$ No. of electrons $= 1$
Question 64
Chemistry · Co-ordination Compounds · Single correct
Match List - I with List II :
A. (a)- (v), (b) - (i), (c ) - (ii), (d) - (iv)
B. (a)- (v), (b) - (ii), (c ) - (iv), (d) - (i)
C. (a)- (iv), (b) - (iii), (c ) - (i), (d) - (ii)
D. (a)- (v), (b) - (iii), (c ) - (ii), (d) - (i)
Answer: (d)
Solution
Li makes alloy with Lead to make white metal bearings for motor engines. Liquid Na metal is used as coolant in fast breeder nuclear reactor. K is a very absorbent of $\mathrm{CO_2}$. Cs is used in making photoelectric cell.
Question 65
Chemistry · Surface Chemistry · Single correct
Given below are two statement : one is labelled as Assertion $\textbf{A}$ and the other is labelled as Reason $\textbf{R}$. Assertion $\textbf{A}$ : $\mathrm{SO_2 (g)}$ is adsorbed to a large extent than $\mathrm{H_2 (g)}$ on activated charcoal. Reason $\textbf{R}$ : $\mathrm{SO_2 (g)}$ has a higher critical temperature than $\mathrm{H_2 (g)}$. In the light of the above statements, choose the most appropriate answer from the options given below.
Both A and R are correct but R is not the correct explanation fo A
Both A and R are correct and R is the correct explanation of A.
A is not correct but R is correct.
A is correct but R is not correct.
Answer: (b)
Question 66
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The CORRECT order of first ionisation enthalpy is :
Mg < S < Al < P
Mg < Al < S < P
Al < Mg < S < P
Mg < Al < P < S
Answer: (a)
Solution
Given the elements Mg, Al, P, and S, the ionization energy order is Al < Mg < S < P. The valence configuration is $[\mathrm{Ne}] : 3s^2 3p^1 3s^2 3p^3 3s^2 3p^4$. Properties include full and filled stable half-filled shells.
Question 67
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Given below are two statements: Statement I : Hyperconjugation is a permanent effect. Statement II : Hyperconjugation in ethyl cation ($\text{CH}_3 - \text{CH}_2^+$) involves the overlapping of $\text{C}_{sp^2} - \text{H}_{1s}$ bond with empty 2p orbital of other carbon. Choose the correct option: (1) Both statement I and statement II are false (2) Statement I is incorrect but statement II is true (3) Statement I is correct but statement II is false (4) Both Statement I and statement II are true.
Both statement I and statement II are false
Statement I is incorrect but statement II is true
Statement I is correct but statement II is false
Both Statement I and statement II are true.
Answer: (c)
Solution
Statement I: It is correct statement. Statement II: $\mathrm{CH_3 - CH_2}$ involve $\mathrm{C_{sp^3} - H_{1s}}$ bond with empty $2p$ orbital hence given statement is false.
Question 68
Chemistry · Co-ordination Compounds · Single correct
Given below are two statements: Statement I: $[\mathrm{Mn(CN)}_6]^{3-}$, $[\mathrm{Fe(CN)}_6]^{3-}$ and $[\mathrm{Co(C_2O_4)}_3]^{3-}$ are $d^2sp^3$ hybridised. Statement II: $[\mathrm{MnCl}_6]^{3-}$ and $[\mathrm{FeF}_6]^{3-}$ are paramagnetic and have 4 and 5 unpaired electrons, respectively. In the light of the above statements, choose the correct answer from the options given below:
Statement I is correct but statement II is false
Both statement I and statement II are false
Statement I is incorrect but statement II is true
Both statement I and statement II are true
Answer: (d)
Solution
$[\mathrm{Mn(CN)_6}]^{3-}$ $[\mathrm{Fe(CN)_6}]^{3-}$ $[\mathrm{Co(C_2O_4)_3}]^{3-}$ $\downarrow$ $\mathrm{Mn}^{3+},\ \mathrm{CN^-}$ $\mathrm{Fe}^{3+},\ \mathrm{CN^-}$ $\mathrm{Co}^{3+},\ \mathrm{C_2O_4^{2-}}$ $d^4$ configuration, strong field ligand (SFL) $d^5$ configuration, strong field ligand (SFL) $d^6$ configuration, chelating ligand $\Rightarrow$ All have large crystal field splitting; hence all undergo $d^2sp^3$ hybridisation. $[\mathrm{MnCl_6}]^{3-}$ and $[\mathrm{FeF_6}]^{3-}$ $d^4$ configuration, $\mathrm{Cl^-}$ (WFL) $d^5$ configuration, $\mathrm{F^-}$ (WFL) $\Rightarrow\ [\mathrm{MnCl_6}]^{3-}$ has $4$ unpaired electrons. $\Rightarrow\ [\mathrm{FeF_6}]^{3-}$ has $5$ unpaired electrons.
Question 69
Chemistry · Analytical Chemistry · Single correct
To an aqueous solution containing ions such as $\mathrm{Al}^{3+}$, $\mathrm{Zn}^{2+}$, $\mathrm{Ca}^{2+}$, $\mathrm{Fe}^{3+}$, $\mathrm{Ni}^{2+}$, $\mathrm{Ba}^{2+}$ and $\mathrm{Cu}^{2+}$ was added conc. HCl, followed by $\mathrm{H}_2 \mathrm{S}$. The total number of cations precipitated during this reaction is/are:
1
3
4
2
Answer: (a)
Solution
$Al^{3+}$ and $Fe^{3+}$ sulphides hydrolyse in water. $Ni^{2+}$ and $Zn^{2+}$ require basic medium with $H_2 S$ to form ppt. $Ca^{2+}$ and $Ba^{2+}$ sulphides are soluble hence we will receive only $CuS$ ppt.
Question 70
Chemistry · Chemistry in Everyday Life · Single correct
Given below are two statements: Statement I: Penicillin is a bacteriostatic type antibiotic. Statement II: The general structure of Penicillin is:
Both statement I and statement II are false
Statement I is incorrect but statement II is true
Both statement I and statement II are true
Statement I is correct but statement II is false
Answer: (b)
Solution
Statement I: Penicillin is bactericidal not bacteriostatic hence given statement is false. Statement II: Structure of penicillin given is correct.
Question 71
Chemistry · Biomolecules · Single correct
Compound A gives D-Galactose and D-Glucose on hydrolysis. The compound A is:
Amylose
Sucrose
Maltose
Lactose
Answer: (d)
Solution
Lactose: It is a disaccharide of $\beta - \mathrm{D}$-Galactose and $\beta - \mathrm{D}$-Glucose with $C_1$ of galactose and $C_4$ of glucose link. Lactose: $\beta - \mathrm{D}$-Galactose $+ \beta - \mathrm{D}$-Glucose
Question 72
Chemistry · Amines · Single correct
R - CN $\xrightarrow{\begin{array}{c} (i) \ DIBAL-H \\ (ii) \ H_2O \end{array}}$ R - Y Consider the above reaction and identify "Y"
-CH_2NH_2
-CONH_2
-CH
-COO
Answer: (c)
Solution
The reaction given is: $$\mathrm{R{-}C \equiv N \xrightarrow{(1)DiBAL-H}{(2)H_2O} R{-}C{-}H}$$ Here Y is an aldehyde.
Question 73
Chemistry · Hydrocarbons · Single correct
consider the above reaction, and choose the correct statement:
The reaction is not possible in acidic medium
Both compounds A and B are formed equally
Compound A will be the major product
Compound B will be the major product
Answer: (c)
Solution
The reaction begins with the protonation of the alcohol group by $\mathrm{H^+}$ from $\mathrm{H_2SO_4}$, forming a water molecule attached to the benzene ring. This intermediate loses water ($\mathrm{-H_2O}$) to form a carbocation. The carbocation undergoes rearrangement to form a more stable structure. The elimination of $\mathrm{H_2SO_4}$ leads to the formation of an alkene. The reaction shows geometric isomerism (GI) resulting in two products: (A) Trans (more stable product, Saytzeff's alkene, major) and (B) Cis.
Question 74
Chemistry · Co-ordination Compounds · Single correct
Match List - I with List - II : Choose the correct answer from the options given below:
(a)- (iii), (b) - (iv), (c) - (i), (d) - (ii)
(a)- (iv), (b) - (i), (c) - (iii), (d) - (ii)
(a)- (i), (b) - (ii), (c) - (iii), (d) - (iv)
(a)- (iii), (b) - (iv), (c) - (ii), (d) - (i)
Answer: (a)
Question 75
Chemistry · Structure of Atom · Single correct
If the Thompson model of the atom was correct, then the result of Rutherford's gold foil experiment would have been:
All of the $\alpha$-particles pass through the gold foil without decrease in speed.
$\alpha$-Particles are deflected over a wide range of angles.
All $\alpha$-particles get bounced back by $180^\circ$
$\alpha$-Particles pass through the gold foil deflected by small angles and with reduced speed.
Answer: (d)
Solution
As in Thomson model, protons are diffused (charge is not centred) $\alpha$-particles deviate by small angles and due to repulsion from protons, their speed decreases.
Question 76
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Number of Cl = O bonds in chlorous acid, chloric acid and perchloric acid respectively are:
3,1 and 1
4,1 and 0
1,1 and 3
1,2 and 3
Answer: (c)
Solution
Question 77
Chemistry · The Solid State · Multiple correct
"Select the correct statements. (A) Crystalline solids have long range order. (B) Crystalline solids are isotropic. $(C)$ Amorphous solid are sometimes called pseudo solids. (D) Amorphous solids soften over a range of temperatures. (E) Amorphous solids have a definite heat of fusion. Choose the most appropriate answer from the options given below."
$(A), (B), (E)$ only
$(B), (D)$ only
$(C), (D)$ only
$(A), (C), (D)$ only
Answer: (c)
Solution
Crystalline solids have definite arrangement of constituent particles and have long range order. Different constituent particles of an amorphous solid have different bond strengths and soften over a range of temperatures.
Question 78
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
What is A in the following reaction?
Answer: (d)
Solution
Question 79
Chemistry · Hydrocarbons · Single correct
The correct sequence of correct reagents for the following transformation is :-
The reaction sequence starts with a nitrobenzene compound where $\mathrm{NO_2}$ is meta directing. The first step involves chlorination using $\mathrm{Cl_2/FeCl_3}$, resulting in the addition of a chlorine atom to the benzene ring. Next, reduction is carried out using $\mathrm{Fe/HCl}$, converting the $\mathrm{NO_2}$ group to an $\mathrm{NH_2}$ group. Following this, diazotisation is performed using $\mathrm{NaNO_2 + HCl}$ at $0^\circ \mathrm{C}$, forming a diazonium salt. Finally, the diazonium group is replaced by an $\mathrm{OH}$ group using $\mathrm{H_2O/H^+}$, resulting in the final product.
Question 80
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The addition of silica during the extraction of copper from its sulphide ore :-
converts copper sulphide into copper silicate
converts iron oxide into iron silicate
reduces copper sulphide into metallic copper
reduces the melting point of the reaction mixture
Answer: (b)
Solution
Silica is used to remove FeO impurity from the ore of copper. $$\mathrm{FeO + SiO_2 \rightarrow FeSiO_3}$$ iron silicate (Slag)
Question 81
Chemistry · Equilibrium · Numerical
The equilibrium constant for the reaction $$\mathrm{A(s) \rightleftharpoons M(s) + \frac{1}{2}O_2(g)}$$ is $K_p = 4$. At equilibrium, the partial pressure of $O_2$ is $\_$$\_$$\_$ atm. (Round off to the nearest integer)
Answer: 16
Solution
Given $k_p = P_{\mathrm{O}_2}^{1/2} = 4$. Therefore, $P_{\mathrm{O}_2} = 16 bar = 16 atm$.
Question 82
Chemistry · Thermodynamics · Numerical
When $400 \, \mathrm{mL}$ of $0.2 \, \mathrm{M}$ $\mathrm{H_2SO_4}$ solution is mixed with $600 \, \mathrm{mL}$ of $0.1 \, \mathrm{M}$ $\mathrm{NaOH}$ solution, the increase in temperature of the final solution is _______ $\times 10^{-2} \, \mathrm{K}$. (Round off to the nearest integer). $\left[ \text{Use: } \mathrm{H^+} (\mathrm{aq}) + \mathrm{OH^-} (\mathrm{aq}) \rightarrow \mathrm{H_2O} : \Delta_r H = -57.1 \, \mathrm{kJ \, mol^{-1}} \right]$ Specific heat of $\mathrm{H_2O} = 4.18 \, \mathrm{J \, K^{-1} \, g^{-1}}$ density of $\mathrm{H_2O} = 1.0 \, \mathrm{g \, cm^{-3}}$ Assume no change in volume of solution on mixing.
Chemistry · Some Basic Concepts of Chemistry · Numerical
$2\mathrm{SO_2}(g)+\mathrm{O_2}(g)\rightarrow2\mathrm{SO_3}(g)$ The above reaction is carried out in a vessel starting with partial pressure $P_{\mathrm{SO_2}}=250\ \mathrm{mbar}$, $P_{\mathrm{O_2}}=750\ \mathrm{mbar}$ and $P_{\mathrm{SO_3}}=0\ \mathrm{bar}$. When the reaction is complete, the total pressure in the reaction vessel is $\mathrm{mbar}$. (Round off to the nearest integer).
Answer: 375
Solution
The reaction is given by: $$2\mathrm{SO_2} (\mathrm{g}) + \mathrm{O_2} (\mathrm{g}) \rightarrow 2\mathrm{SO_3} (\mathrm{g})$$ Initial pressures are 250 $\mathrm{\, mbar}$ for $\mathrm{SO_2}$ and 750 $\mathrm{\, mbar}$ for $\mathrm{O_2}$. Since $\mathrm{SO_2}$ is the limiting reactant (L. R.), the final pressures are: Final pressure of $\mathrm{SO_2}$ = -250 $\mathrm{\, mbar}$ Final pressure of $\mathrm{O_2}$ = -125 $\mathrm{\, mbar}$ Final pressure of $\mathrm{SO_3}$ = 250 $\mathrm{\, mbar}$ Therefore, the final total pressure is: $$625 + 250 = 875 \mathrm{\, mbar}$$
Question 84
Chemistry · Redox Reactions · Numerical
$10.0\,\mathrm{mL}$ of $0.05\,\mathrm{M}$ $\mathrm{KMnO_4}$ solution was consumed in a titration with $10.0\,\mathrm{mL}$ of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is \_\_\_\_ $\times 10^{-2}\,\mathrm{g/L}$. (Round off to the nearest integer)
Chemistry · Chemical Bonding and Molecular Structure · Numerical
The total number of electrons in all bonding molecular orbitals of $\mathrm{O}_2^{2-}$ is ........ (Round off to the nearest integer)
Answer: 10
Solution
M. O. Configuration of $\mathrm{O_2^{2-}}$ (18$\bar{e}$) $$\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 = \pi 2p_y^2$$ $$\pi 2p_x^2 = \pi 2p_y^2$$ Total B.M.O electrons = 10
Question 86
Chemistry · Co-ordination Compounds · Numerical
Three moles of a metal complex with formula $\mathrm{Co(en)_2Cl_3}$ give $3$ moles of Silver Chloride on treatment with excess Silver Nitrate. The secondary valency of Co in the complex is $\underline{\qquad}$. (Round off to the nearest integer)
Answer: 6
Solution
The reaction is given by: $$3 \ [\mathrm{Co(en)_2Cl_2}] \ Cl + \mathrm{AgNO_3} \xrightarrow{excess} 3 \ \mathrm{AgCl} \ (white ppt)$$ The secondary valency of Co is 6. (C. N.)
Question 87
Chemistry · Solutions · Numerical
In a solvent 50$\%$ of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is $\times 10^{-2}$ (Round off to the nearest integer)
Answer: 125
Solution
The reaction is given by: $$2\mathrm{HA} \rightleftharpoons \mathrm{H_2} \mathrm{A_2HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A}$$ Initial moles are $a \times \frac{50}{100}$ for $\mathrm{HA}$, and $0$ for $\mathrm{H^+}$ and $\mathrm{A}$. Final moles are $0$ for $\mathrm{HA}$, $0.25a$ for $\mathrm{H^+}$, and $0.5a$ for $\mathrm{A}$. Now, $$i = \frac{final moles}{initial moles} = \frac{0.25a + 0.5a + 0.5a}{0.5a + 0.5a}$$ $$= 1.25 = 125 \times 10^{-2}$$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Fill in the blank
For the first order reaction $\mathrm{A\rightarrow 2B}$, $1$ mole of reactant $\mathrm{A}$ gives $0.2$ moles of $\mathrm{B}$ after $100$ minutes. The half-life of the reaction is $\underline{\hspace{1cm}}\,\mathrm{min}$. (Round off to the nearest integer). [Use: $\ln 2=0.69,\ \ln 10=2.3$. Properties of logarithms: $\ln x^y=y\ln x$, $\ln\left(\frac{x}{y}\right)=\ln x-\ln y$]
Answer: 300
Solution
A $\rightarrow 2 \, \mathrm{B}$ At $t = 0$, $1 \, mole$ of A and $0$ of B. At $t = 100 \, min$, $1 - x$ moles of A and $2x$ moles of B. $= 0.9 \, mol$ of A and $= 0.2 \, mol$ of B. Now, $t = \frac{t_{1/2}}{\ln 2} \times \frac{[A_0]}{[A_t]}$ $$100 = \frac{t_{1/2}}{\ln 2} \times \ln \frac{1}{0.9} \implies t_{1/2} = 690 \, min.$$ (taking $\ln 3 = 1.11$) Ans. 600 to 700
Question 90
Chemistry · Electrochemistry · Numerical
For the cell $\text{Cu(s)} \mid \text{Cu}^{2+}\text{(aq)(0.1M)} \| \text{Ag}^+\text{(aq) (0.01M)} \mid \text{Ag(s)}$ the cell potential $E_1 = 0.3095$ V For the cell $\text{Cu(s)} \mid \text{Cu}^{2+}\text{(aq) (0.01M)} \| \text{Ag}^+\text{(aq) (0.001M)} \mid \text{Ag(s)}$ the cell potential $= x \times 10^{-2}$ V. (Round off the Nearest Integer). $\left[\text{Use : } \dfrac{2.303RT}{F} = 0.059\right]$