JEE Main 24 June 2022 Shift 1 question paper with solutions
JEE Main 24 June 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 · Complex Numbers and Quadratic Equations · Single correct
If the sum of the squares of the reciprocals of the roots $\alpha$ and $\beta$ of the equation $3x^2 + \lambda x - 1 = 0$ is $15$, then $6(\alpha^3 + \beta^3)^2$ is equal to:
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $A = \{ z \in \mathbb{C} : 1 \leq |z - (1 + i)| \leq 2 \}$ and $B = \{ z \in A : |z - (1 - i)| = 1 \}$. Then, B :
is an empty set
contains exactly two elements
contains exactly three elements
is an infinite set
Answer: (d)
Solution
Set A is defined as $A = \{ z \in \mathbb{C} : 1 \leq |z - (1+i)| \leq 2 \}$. Set B is defined as $B = \{ z \in A : |z - (1-i)| = 1 \}$. The intersection $A \cap B$ has an infinite set.
Question 3
Maths · Sequences and Series · Single correct
If $\{ a_i \}_{i=1}^{n}$ where $n$ is an even integer, is an arithmetic progression with common difference 1, and $$\sum_{i=1}^{n} a_i = 192, \sum_{i=1}^{n/2} a_{2i} = 120,$$ then $n$ is equal to:
Let $x^2 + y^2 + Ax + By + C = 0$ be a circle passing through $(0, 6)$ and touching the parabola $y = x^2$ at $(2, 4)$. Then $A + C$ is equal to_____.
16
$\frac{88}{5}$
72
-8
Answer: (a)
Solution
The equation $x^2 + y^2 + Ax + By + C = 0$ is passing through $(0,6)$. Therefore, $6B + C = -36$. The tangent of the parabola $y = x^2$ at $(2,4)$ is given by $$4x - y - 4 = 0 ----(1)$$ The tangent of the circle $x^2 + y^2 + Ax + By + C = 0$ at $(2,4)$ is $$(4 + A)x + (8 + B)y + 2A + 4B + 2C = 0 ----(2)$$ From Equation (1) and (2), $$\frac{4 + A}{4} = \frac{8 + B}{-1} = \frac{2A + 4B + 2C}{-4}$$ Therefore, $A + 4B = -36$ ---(3) and $3A + 4B + 2C = -4$ ---(4). From equation (3) and (4), $A + C = 16$.
Question 7
Maths · Conic Sections · Single correct
Let $\lambda x - 2y = \mu$ be a tangent to the hyperbola $a^2 x^2 - y^2 = b^2$. Then $\left( \frac{\lambda}{a} \right)^2 - \left( \frac{\mu}{b} \right)^2$ is equal to:
-2
-4
2
4
Answer: (d)
Solution
Given $\lambda x - 2y = \mu$ is a tangent to the curve $a^2 x^2 - y^2 = b^2$, then $$a^2 x^2 - \left( \frac{\lambda x - \mu}{2} \right)^2 = b^2$$ $$\left(4a^2 - \lambda^2 \right)x^2 + 2\lambda \mu x - \mu^2 - 4b^2 = 0$$ Discriminant $= 0$ $$4\lambda^2 \mu^2 + 4\left(4a^2 - \lambda^2 \right)\left(\mu^2 + 4b^2 \right) = 0$$ $$4\lambda^2 b^2 - 4a^2 \mu^2 = 16a^2 b^2$$ $$\frac{\lambda^2}{a^2} - \frac{\mu^2}{b^2} = 4$$
Question 8
Maths · Mathematical Reasoning · Single correct
The number of choices of $\Delta \in \{\land, \lor, \Rightarrow, \Leftrightarrow\}$, such that $(p \Delta q) \Rightarrow ((p \Delta \sim q) \lor ((\sim p) \Delta q))$ is a tautology, is
1
2
3
4
Answer: (b)
Solution
For tautology $((p \Delta \sim q) \lor ((\sim p) \Delta q))$ must be true. This is possible only when $\Delta = \lor \& \Rightarrow$
Question 9
Maths · Matrices · Single correct
Given the set $S=\{\sqrt{n}:1\le n\le 50,\ \text{and n is odd}\}$. This is equal to $\{ \sqrt{1}, \sqrt{3}, \sqrt{5}, \ldots, \sqrt{49} \}$, which contains 25 terms. The determinant $|A| = 1 + a^2$. The sum of determinants is given by $$\sum_{a \in S} \det(adjA) = \sum_{a \in S} |A|^2 = \sum (1 + a^2)^2.$$ This results in $$= 22100 = 100 \lambda.$$ Solving for $\lambda$, we find $$\lambda = 221.$$
218
221
663
1717
Answer: (b)
Solution
Given the set $S = \{ \sqrt{n} : 1 \leq n \leq 50 and n is odd \}$. This is equal to $\{ \sqrt{1}, \sqrt{3}, \sqrt{5}, \ldots, \sqrt{49} \}$, which contains 25 terms. The determinant $|A| = 1 + a^2$. The sum of determinants is given by $$\sum_{a \in S} \det(adjA) = \sum_{a \in S} |A|^2 = \sum (1 + a^2)^2.$$ This results in $$= 22100 = 100 \lambda.$$ Solving for $\lambda$, we find $$\lambda = 221.$$
Question 10
Maths · Determinants · Single correct
The number of values of $\alpha$ for which the system of equations : $x + y + z = \alpha$ $\alpha x + 2\alpha y + 3z = -1$ $x + 3\alpha y + 5z = 4$ is inconsistent, is
0
1
2
3
Answer: (b)
Solution
Given the equations: $$x + y + z = \alpha$$ $$\alpha x + 2\alpha y + 3z = -1$$ $$x + 3\alpha y + 5z = 4$$ The system has an inconsistent solution if: $$D = \begin{vmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3 \\ 1 & 3\alpha & 5 \end{vmatrix} = 0$$ This implies: $$(\alpha - 1)^2 = 0$$ Thus, $\alpha = 1$. For $\alpha = 1$: $$D_1 = \begin{vmatrix} 1 & 1 & 1 \\ -1 & 2 & 3 \\ 4 & 3 & 5 \end{vmatrix}$$ Calculating the determinant: $$(10 - 9) - (-5 - 12) + (-3 - 8)$$ $$= 1 + 17 - 11 \neq 0$$ For $\alpha = 1$, the system of equations has an inconsistent solution.
Question 11
Maths · Inverse Trigonometric Functions · Single correct
The set of all values of $k$ for which $(\tan^{-1} x)^3 + (\cot^{-1} x)^3 = k \pi^3, x \in \mathbb{R}$, is the interval:
$[\\frac{1}{32}, \\frac{7}{8})$
$(\\frac{1}{24}, \\frac{13}{16})$
$[\\frac{1}{48}, \\frac{13}{16}]$
$[\\frac{1}{32}, \\frac{9}{8})$
Solution
Let $S = (\tan^{-1} x)^3 + (\cot^{-1} x)^3$. This equals $\left( \tan^{-1} x + \cot^{-1} x \right)^3$ $- 3 \tan^{-1} x \cdot \cot^{-1} x \left( \tan^{-1} x + \cot^{-1} x \right)$. This simplifies to $$= \frac{\pi^3}{8} - \frac{3\pi}{2} \tan^{-1} x \left( \frac{\pi}{2} - \tan^{-1} x \right)$$ Further simplifying, we get $$= \frac{3\pi}{2} \left( \tan^{-1} x - \frac{\pi}{4} \right)^2 + \frac{\pi^3}{32}$$ Thus, $$\Rightarrow \frac{\pi^3}{32} \leq S < \frac{7}{8} \pi^3$$ This implies $$= \frac{\pi^3}{32} \leq K \pi^3 < \frac{7}{8} \pi^3$$ Therefore, $$\frac{1}{32} \leq K < \frac{7}{8}$$
Question 12
Maths · Inverse Trigonometric Functions · Single correct
The domain of the function $$f(x) = \frac{\cos^{-1}\left(\frac{x^2 - 5x + 6}{x^2 - 9}\right)}{\log_e(x^2 - 3x + 2)}$$ is
Given the inequality $$-1 \leq \frac{x^2 - 5x + 6}{x^2 - 9} \leq 1$$ we split it into two parts: 1. $$\frac{x^2 - 5x + 6}{x^2 - 9} - 1 \leq 0$$ simplifies to $$\frac{1}{x+3} \geq 0$$ which gives $$x \in (-3, \infty) \ldots (1)$$ 2. $$\frac{x^2 - 5x + 6}{x^2 - 9} + 1 \geq 0$$ simplifies to $$\frac{2x + 1}{x+3} \geq 0$$ which gives $$x \in (-\infty, -3) \cup \left[-\frac{1}{2}, \infty\right) \ldots (2)$$ After taking the intersection, we have $$x \in \left[-\frac{1}{2}, \infty\right)$$. For the inequality $$x^2 - 3x + 2 > 0$$, we find $$x \in (-\infty, 1) \cup (2, \infty)$$. Since $$x^2 - 3x + 2 \neq 1$$, we find $$x \neq \frac{3 \pm \sqrt{5}}{2}$$. After taking the intersection of each solution, we have $$\left[-\frac{1}{2}, 1\right) \cup (2, \infty) - \left(\frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}\right)$$.
Question 13
Maths · Applications of Derivatives · Single correct
$f(x)=4\log_e(x-1)-2x^2+4x+5,\quad x>1,$ which one of the following is NOT correct?
f is increasing in (1, 2) and decreasing in (2, $\infty$)
f(x) = -1 has exactly two solutions
f'(e) - f''(2) < 0
f(x) = 0 has a root in the interval (e, e+1)
Answer: (c)
Solution
Given $f(x) = 4 \log_e(x - 1) - 2x^2 + 4x + 5$, $x > 1$. $$f'(x) = \frac{4}{x-1} - 4(x-1)$$ For $1 0$. For $x > 2 \Rightarrow f'(x) 0$. $f(e + 1) 0$$ (option C is incorrect)
Question 14
Maths · Applications of Derivatives · Single correct
The tangent at the point $(x_1, y_1)$ on the curve $y = x^3 + 3x^2 + 5$ passes through the origin, then $(x_1, y_1)$ does NOT lie on the curve:
$x^2 + \frac{y^2}{81} = 2$
$\frac{y^2}{9} - x^2 = 8$
$y = 4x^2 + 5$
$\frac{x}{3} - y^2 = 2$
Answer: (d)
Solution
The tangent at $(x_1, y_1)$ to the curve $$y = x^3 + 3x^2 + 5$$ $$y - y_1 = \left(3x_1^2 + 6x_1\right)(x - x_1)$$ passing through origin $$-y_1 = \left(3x_1^3 + 6x_1\right)(-x_1)$$ $$y_1 = \left(3x_1^3 + 6x_1^2\right) -------(1)$$ And $(x_1, y_1)$ lies on the curve $$y = x^3 + 3x^2 + 5$$ $$y_1 = x_1^3 + 3x_1^2 + 5 ----(2)$$ From equation (1) and (2) $$2y_1 = 3x_1^2 + \frac{15}{2}$$ Hence the equation of curve $y = \frac{3}{2}x^2 + \frac{15}{2}$ This curve does not intersect $\frac{x}{3} - y^2 = 2$
Question 15
Maths · Applications of Derivatives · Single correct
The sum of absolute maximum and absolute minimum values of the function f(x) = |2x^2 + 3x - 2| + $\sin$ x $\cos$ x in the interval [0, 1] is:
3 + $\frac{\sin(1) \cos^2(\frac{1}{2})}{2}$
3 + $\frac{1}{2}$ (1 + 2$\cos$(1)) $\sin$(1)
5 + $\frac{1}{2}$ ($\sin$(1) + $\sin$(2))
2 + $\sin$($\frac{1}{2}$) $\cos$($\frac{1}{2}$)
Answer: (b)
Question 16
Maths · Applications of Derivatives · Single correct
The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :
9
10
11
12
Answer: (a)
Solution
Let $r$ be the radius of the spherical balloon. $S =$ Surface area $S = 4\pi r^2$ $$\frac{dS}{dt} = 8\pi r \times \frac{dr}{dt} = k (constant)$$ $$4\pi r^2 = kt + C (C is constant of integration)$$ For $t = 0$, $r = 3$ $\Rightarrow 36\pi = C$ For $t = 5$, $r = 7$ $\Rightarrow K = 32\pi$ $$4\pi r^2 = 32\pi t + 36\pi$$ $$r^2 = 8t + 9$$ for $t = 9$ $$r^2 = 81$$ $r = 9$
Question 17
Maths · Differential Equations · Single correct
If $x = x(y)$ is the solution of the differential equation $y \frac{dx}{dy} = 2x + y^3(y+1)e^y, x(1) = 0$; then $x(e)$ is equal to :
Let $\hat{a}, \hat{b}$ be unit vectors. If $\vec{c}$ be a vector such that the angle between $\hat{a}$ and $\vec{c}$ is $\frac{\pi}{12}$, and $\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})$, then $|6\vec{c}|^2$ is equal to
Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $\frac{6}{11}$, then n is equal to _____.
13
6
4
3
Answer: (c)
Solution
Let $E_1$ denote the selection for the 1st bag and $E_2$ denote the selection for the 2nd bag. The probabilities are $\mathrm{P}(E_1) = \frac{1}{2}$ and $\mathrm{P}(E_2) = \frac{1}{2}$. Let $A$ be the event that the selected balls are 1 red and 1 black. The probability $\mathrm{P}\left( \frac{A}{E_1} \right) = \frac{{^3C_1 \times ^1C_1}}{{^6C_2}} = \frac{1}{5}$. The probability $\mathrm{P}\left( \frac{A}{E_1} \right) = \frac{{^3C_1 \times ^2C_1}}{{(n+5)C_2}} = \frac{12}{(n+5)(n+4)}$. The probability $\mathrm{P}\left( \frac{E_1}{A} \right) = \frac{\mathrm{P}(E_1) \times \mathrm{P}\left( \frac{A}{E_1} \right)}{\mathrm{P}(E_1) \times \mathrm{P}\left( \frac{A}{E_1} \right) + \mathrm{P}(E_2) \times \mathrm{P}\left( \frac{A}{E_2} \right)}$. This simplifies to $\frac{1}{10} = \frac{6}{11}$. Thus, $\Rightarrow n = 4$.
Question 20
Maths · Probability · Single correct
If a random variable X follows the Binomial distribution B (33, p) such that $3P(X = 0) = P(X = 1)$, then the value of $\frac{P(X = 15)}{P(X = 18)} - \frac{P(X = 16)}{P(X = 17)}$ is equal to
In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, -2 marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is _.
Answer: 40
Solution
Given $x_1 + x_2 + x_3 + x_4 + x_5 = 5$. Only one possibility is $3, 3, 3, -2, -2$. Number of ways is $$\frac{5!}{3!2!} \times 2 = 40$$
Question 22
Maths · Applications of Derivatives · Fill in the blank
Let A $( $\frac{3}{\sqrt{a}}$, $\sqrt{a}$) a > 0, be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If D(3 $\cos\theta$, a $\sin\theta$) is a point in the fourth quadrant such that the maximum area of $\Delta$ ACD is 12 square units, then a is equal to _______.
Answer: 8
Solution
Given $$A = \left( \frac{3}{\sqrt{a}}, \sqrt{a} \right)$$ $$B = \left( \frac{-3}{\sqrt{a}}, \sqrt{a} \right)$$ $$C = \left( \frac{-3}{\sqrt{a}}, -\sqrt{a} \right)$$ Area of ACD is $$\frac{1}{2} \begin{vmatrix} \frac{3}{\sqrt{a}} & \sqrt{a} \\ \frac{-3}{\sqrt{a}} & -\sqrt{a} \\ 3 \cos \theta & a \sin \theta \end{vmatrix}$$ This simplifies to $$\frac{1}{2} 6 \sqrt{a} (\cos \theta - \sin \theta)$$ Thus, $$3 \sqrt{a} (\cos \theta - \sin \theta)$$ The maximum value of the function is $$3 \sqrt{a} \sqrt{2}$$ Given $$3 \sqrt{a} \sqrt{2} = 12$$ Then $$2a = 16$$ So $$a = 8$$
Question 23
Maths · Conic Sections · Numerical
If two tangents drawn from a point $(\alpha, \beta)$ lying on the ellipse $25x^2 + 4y^2 = 1$ to the parabola $y^2 = 4x$ are such that the slope of one tangent is four times the other, then the value of $(10\alpha + 5)^2 + (16\beta^2 + 50)^2$ equals __________
Answer: 2929
Solution
Given $\alpha = \frac{1}{5} \cos \theta$, $\beta = \frac{1}{2} \sin \theta$. The equation of the tangent to $y^2 = 4x$ is $y = mx + \frac{1}{m}$. It passes through $(\alpha, \beta)$. $$\frac{1}{2} \sin \theta = m \frac{1}{5} \cos \theta + \frac{1}{m}$$ $$m^2 \left( \frac{\cos \theta}{5} \right) - m \left( \frac{1}{2} \sin \theta \right) + 1 = 0$$ It has two roots $m_1$ and $m_2$, where $m_1 = 4m_2$. $$m_1 + m_2 = \frac{1}{2} \sin \theta = \frac{\cos \theta}{5}$$ $$m_1 m_2 = \frac{5}{\cos \theta}$$ After eliminating $m_1$ and $m_2$, $$\cos \theta = \frac{-5 \pm \sqrt{29}}{2}$$ $$\alpha = \frac{-5 \pm \sqrt{29}}{10} \implies 10\alpha + 5 = \pm \sqrt{29}$$ $$\beta^2 = \frac{1}{4} \sin^2 \theta \implies 16\beta^2 = -50 + 10\sqrt{29}$$ $$(10\alpha + 5)^2 + (16\beta^2 + 50)^2 = 2929$$
Question 24
Maths · Permutations and Combinations · Fill in the blank
The number of one-one function $f : \{a, b, c, d\} \rightarrow \{0, 1, 2, \ldots, 10\}$ such that $2f(a) - f(b) + 3f(c) + f(d) = 0$ is $\_$$\_$$\_$$\_$ .
Answer: 31
Solution
Given $2f(a) + 3f(c) = f(d) - f(b)$. Using fundamental principle of counting. Number of one-one function is 31.
Question 25
Maths · Continuity and Differentiability · Numerical
The number of points where the function $$f(x) = \begin{cases} |2x^2 - 3x - 7| & if x \leq -1 \\ [4x^2 - 1] & if -1 < x < 1 \\ |x + 1| + |x - 2| & if x \geq 1 \end{cases}$$ $[t]$ denotes the greatest integer $\leq t$, is discontinuous is .
Answer: 7
Solution
Question 26
Maths · Integrals · Numerical
Let $f(\theta) = \sin \theta + \int_{-\pi/2}^{\pi/2} (\sin \theta + t \cos \theta) f(t) \, dt$. Then the value of $$\left| \int_{0}^{\pi/2} f(\theta) \, d\theta \right|$$ is .
Answer: 1
Solution
Given $f(\theta) = \sin \theta + \int_{-\pi/2}^{\pi/2} (\sin \theta + \cos \theta) f(t) \, dt$. $f(\theta) = \sin \theta + \sin \theta \int_{-\pi/2}^{\pi/2} f(t) \, dt + \cos \theta \int_{-\pi/2}^{\pi/2} tf(t) \, dt$. Let $A = \int_{-\pi/2}^{\pi/2} f(t) \, dt$, $B = \int_{-\pi/2}^{\pi/2} tf(t) \, dt$. $f(\theta) = \sin \theta + A \sin \theta + B \cos \theta$. $f(\theta) = (A + 1) \sin \theta + B \cos \theta$. $A = \int_{-\pi/2}^{\pi/2} (A + 1) \sin t + B \cos t \, dt$. $A = 2B \ldots (1)$. $B = \int_{-\pi/2}^{\pi/2} t((A + 1) \sin t + B \cos t) \, dt$. $B = \int_{-\pi/2}^{\pi/2} t(A + 1) \sin t \, dt$. $B = (A + 1) 2.1$. $B = (A + 1) 2.1$. $2A + 2 - B = 0 \ldots (2)$. After solving $B = -\frac{2}{3}, A = -\frac{4}{3}$. $\int_{0}^{\pi/2} f(\theta) d\theta = \int_{0}^{\pi/2} \frac{1}{3} \sin \theta - \frac{2}{3} \cos \theta$. $= 1$
Question 27
Maths · Applications of Integrals · Fill in the blank
Let $$\max_{0\le x\le2}\left\{\frac{9-x^2}{5-x}\right\}=\alpha \quad \min_{0\le x\le2}\left\{\frac{9-x^2}{5-x}\right\}=\beta$$ If $$\int_{\beta-\frac{8}{3}}^{2\alpha-1}\max\left\{\frac{9-x^2}{5-x},x\right\}dx=\alpha_1+\alpha_2\log_e\left(\frac{8}{15}\right)$$ $$\alpha_1+\alpha_2=\underline{\hspace{8cm}}$$
Answer: 34
Solution
Given $y = \frac{9 - x^2}{5 - x} = 5 + x + \frac{16}{x - 5}$. The derivative is $\frac{dy}{dx} = 1 - \frac{16}{(x - 5)^2}$. So the critical point is $x = 1$ in $[0, 2]$. We have $y(0) = \frac{9}{5}$, $y(1) = 2$, $y(2) = \frac{5}{3}$. So $\alpha = 2$ and $\beta = \frac{5}{3}$. The integral $I = \int_{-1}^{3} \max \left( \frac{9 - x^2}{5 - x}, x \right) \, dx$. This becomes $I = \int_{-1}^{9/5} \frac{9 - x^2}{5 - x} \, dx + \int_{9/5}^{3} x \, dx$. Further, $I = \int_{-1}^{9/5} \left( 5 + x + \frac{16}{x - 5} \right) \, dx + \int_{9/5}^{3} x \, dx$. After solving, $I = 14 + \frac{28}{25} + 16 \ln \left( \frac{8}{15} \right) + \frac{72}{25}$. Thus, $\alpha_1 = 18$ and $\alpha_2 = 16$.
Question 28
Maths · Applications of Integrals · Numerical
Let S be the region bounded by the curves $y = x^3$ and $y^2 = x$. The curve $y = 2|x|$ divides S into two regions of areas $R_1$ and $R_2$. If $\max \{R_1, R_2\} = R_2$, then $\frac{R_2}{R_1}$ is equal to .
Answer: 19
Solution
The solution starts with calculating the integral $S = \int_0^1 \sqrt{x} - x^3 \, dx$. This evaluates to: $$\left[ \frac{2x^{3/2}}{3} - \frac{x^4}{4} \right]_0^1$$ which simplifies to $\frac{5}{12}$. Next, $R_1 = \int_0^{1/4} (\sqrt{x} - 2x) \, dx$. This evaluates to: $$\left[ \frac{2x^{3/2}}{3} - x^2 \right]_0^{1/4} = \frac{1}{48}$$ Therefore, $R_2 = \frac{19}{48}$. So, the ratio $\frac{R_2}{R_1} = 19$.
Question 29
Maths · Three Dimensional Geometry · Numerical
Let a line having direction ratios 1, -4, 2 intersect the lines $\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1}$ and $\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1}$ at the point A and B. Then $(AB)^2$ is equal to ____.
If the shortest distance between the line $$\vec{r} = (-\hat{i} + 3\hat{k}) + \lambda (\hat{i} - a\hat{j})$$ and $$\vec{r} = (-\hat{j} + 2\hat{k}) + \mu (\hat{i} - \hat{j} + \hat{k})$$ is $\frac{\sqrt{2}}{3}$, then the integral value of $a$ is equal to
Answer: 2
Solution
Given $\mathbf{a}_1 = (-1, 0, 3)$ and $\mathbf{a}_2 = (0, -1, 2)$. The direction ratios of line (1) are $\mathbf{b}_1 = (1, -a, 0)$ and for line (2) are $\mathbf{b}_2 = (1, -1, 1)$. The vector $\mathbf{a}_2 - \mathbf{a}_1 = (1, -1, -1)$. The cross product $\mathbf{b}_1 \times \mathbf{b}_2$ is given by: $$\mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 1 & -1 & 1 \end{vmatrix}$$ Calculating the determinant, we have: $$\mathbf{b}_1 \times \mathbf{b}_2 = \hat{i}(-a) - \hat{j} + \hat{k}(a-1)$$ The magnitude of the cross product is: $$|\mathbf{b}_1 \times \mathbf{b}_2| = \sqrt{a^2 + 1 + (a-1)^2}$$ The dot product $\mathbf{a}_2 - \mathbf{a}_1 \cdot \mathbf{b}_1 \times \mathbf{b}_2 = 2 - 2a$. Therefore, $$\frac{2(1-a)}{\sqrt{a^2 + 1 + (a-1)^2}} = \frac{\sqrt{2}}{3}$$ Squaring on both sides and solving for $a = 2$, we get $\frac{1}{2}$.
Physics
Question 31
Physics · Physical World, Units and Measurements · Single correct
Identify the pair of physical quantities which have different dimensions:
Wave number and Rydberg's constant
Stress and Coefficient of elasticity
Coercivity and Magnetisation
Specific heat capacity and Latent heat
Answer: (d)
Solution
S = $\frac{Q}{m \Delta T}$ = $\frac{J}{\mathrm{Kg}^\circ \mathrm{C}}$ L = $\frac{Q}{m}$ = $\frac{J}{\mathrm{Kg}}$
Question 32
Physics · Motion in a Plane · Single correct
A projectile is projected with velocity of 25 m/s at an angle $\theta$ with the horizontal. After $t$ seconds its inclination with horizontal becomes zero. If $R$ represents horizontal range of the projectile, the value of $\theta$ will be: [use $g = 10 \, \mathrm{m/s^2}$]
A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is $\frac{K}{\pi}$ rev. / min. The value of K is : (Assume the string is massless and unstretchable)
A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 $\mathrm{ms^{-1}}$. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is: [use g = 9.8 $\mathrm{ms^{-2}}$]
A particle experiences a variable force $\vec{F} = (4x\hat{i} + 3y^2\hat{j})$ in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane, the Kinetic Energy changes by
50.0 J
12.5 J
25.0 J
0 J
Answer: (c)
Solution
Given $\mathbf{F} = 4x \hat{i} + 3y^2 \hat{j}$. Work done $\mathrm{WD} = \Delta \mathrm{KE}$. The work done $W$ is given by $$W = \int \mathbf{F} \cdot (dx \hat{i} + dy \hat{j})$$ This becomes $$= \int_1^2 4x \, dx + \int_2^3 3y^2 \, dy$$ Evaluating the integrals, we have $$= (2x^2)_1^2 + (y^3)_2^3$$ Substituting the limits, $$= (8 - 2) + (27 - 8)$$ Simplifying, $$= 6 + 19 = 25 \, \mathrm{J}$$
Question 36
Physics · Gravitation · Single correct
The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is: [Radius of earth $R = 6400 \, \mathrm{km}$ and $\sqrt{3} = 1.732$]
3840 km
4685 km
2133 km
4267 km
Answer: (b)
Solution
Given $Mg' = \frac{M}{3} g$. Therefore, $g' = \frac{g}{3}$. We have $g' = g \left( \frac{R}{R + h} \right)^2 = \frac{g}{3}$. Thus, $\frac{R}{R + h} = \frac{1}{\sqrt{3}}$. Solving for $h$, we get $h = (\sqrt{3} - 1)R$. Substituting $R = 6400$, we find $h = (1.732 - 1)6400$. Therefore, $h = 4685 \, \mathrm{km}$.
Question 37
Physics · Mechanical Properties of Solids · Single correct
The bulk modulus of a liquid is $3 \times 10^{10} \, \mathrm{Nm}^{-2}$. The pressure required to reduce the volume of liquid by $2\%$ is:
$3 \times 10^8 \, \mathrm{Nm}^{-2}$
$9 \times 10^8 \, \mathrm{Nm}^{-2}$
$6 \times 10^8 \, \mathrm{Nm}^{-2}$
$12 \times 10^8 \, \mathrm{Nm}^{-2}$
Answer: (c)
Solution
Given $B = 3 \times 10^{10}$. The fractional change in volume is $-\frac{\Delta V}{V} = 0.02$. Using the formula for bulk modulus, $B = \frac{\Delta P}{-\frac{\Delta V}{V}}$, we have $\Delta P = -B \left( \frac{\Delta V}{V} \right)$. Substituting the values, we get: $$\Delta P = (3 \times 10^{10})(0.02)$$ $$= 6 \times 10^8 \, \mathrm{N/m^2}$$
Question 38
Physics · Thermal Properties of Matter · Single correct
Two metallic blocks $M_1$ and $M_2$ of same area of cross-section are connected to each other (as shown in figure). If the thermal conductivity of $M_2$ is $K$ then the thermal conductivity of $M_1$ will be: [Assume steady state heat conduction]
10 K
8 K
12.5 K
2 K
Answer: (b)
Solution
Given that $\Delta T \propto R \propto \frac{\ell}{k}$, we have: $$\frac{\Delta T_1}{\Delta T_2} = \frac{\ell_1}{k_1} \times \frac{k_2}{\ell_2} = \frac{16}{k_1} \times \frac{k}{8}$$ $$\frac{20}{80} = \frac{16}{k_1} \times \frac{k}{8} \rightarrow k_1 = 8k$$
Question 39
Physics · Thermodynamics · Single correct
A Carnot engine whose heat sinks at $27^{\circ}\mathrm{C}$, has an efficiency of $25\%$. By how many degrees should the temperature of the source be changed to increase the efficiency by $100\%$ of the original efficiency?
Increases by $18^{\circ}\mathrm{C}$
Increase by $200^{\circ}\mathrm{C}$
Increase by $120^{\circ}\mathrm{C}$
Increase by $73^{\circ}\mathrm{C}$
Answer: (b)
Solution
Question 40
Physics · Waves · Single correct
The equations of two waves are given by : $$ y_1 = 5 \sin 2\pi(x - vt) cm $$ $$ y_2 = 3 \sin 2\pi(x - vt + 1.5) cm $$ These waves are simultaneously passing through a string. The amplitude of the resulting wave is
2 cm
4 cm
5.8 cm
8 cm
Answer: (a)
Solution
Given $A_1 = 5$ and $A_2 = 3$. The change in angle is $\Delta \theta = 2\pi(1.5) = 3\pi$. The net amplitude $A_{net}$ is given by $$A_{net} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(3\pi)}$$ This simplifies to $$= |A_1 - A_2|$$ Therefore, $A_{net} = 2 \, cm$.
Question 41
Physics · Electric Charges and Fields · Single correct
A vertical electric field of magnitude $4.9 \times 10^5 \, \mathrm{N/C}$ just prevents a water droplet of a mass $0.1 \, \mathrm{g}$ from falling. The value of charge on the droplet will be: (Given $g = 9.8 \, \mathrm{m/s^2}$)
$1.6 \times 10^{-9} \, \mathrm{C}$
$2.0 \times 10^{-9} \, \mathrm{C}$
$3.2 \times 10^{-9} \, \mathrm{C}$
$0.5 \times 10^{-9} \, \mathrm{C}$
Answer: (b)
Solution
Given the equation $Mg = qE$. Substitute the values: $$(0.1 \times 10^{-3})(9.8) = 4.9 \times 10^5 q$$ Solving for $q$: $$\frac{2 \times 10^{-4}}{10^5} = q$$ Therefore, $$q = 2 \times 10^{-9} \, \mathrm{C}$$
Question 42
Physics · Electrostatic Potential and Capacitance · Single correct
A parallel plate capacitor is formed by two plates each of area $30\pi\,\mathrm{cm}^2$ separated by $1\,\mathrm{mm}$. A material of dielectric strength $3.6 \times 10^7\,\mathrm{Vm}^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6}\,\mathrm{C}$, the value of dielectric constant of the material is: $$\left\{\text{Use: } \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,\mathrm{Nm^2C^{-2}}\right\}$$
Two identical cells each of emf 1.5 $\mathrm{V}$ are connected in parallel across a parallel combination of two resistors each of resistance 20$\Omega$. A voltmeter connected in the circuit measures 1.2 $\mathrm{V}$. The internal resistance of each cell is
2.5$\Omega$
4$\Omega$
5$\Omega$
10$\Omega$
Answer: (c)
Solution
Given the circuit, we have the equation for voltage as $V = E - ir/2$. Substituting the values, we get: $$1.2 = 1.5 - i \left( \frac{r}{2} \right)$$ Rearranging gives: $$i \frac{r}{2} = 0.3$$ The current $i$ is given by: $$i = \frac{1.5}{10 + \frac{r}{2}} \Rightarrow 10i + \frac{ir}{2} = 1.5$$ Substituting the known values: $$10i = 1.5 - 0.3$$ Solving for $i$ gives: $$i = 0.12 \, \mathrm{A}$$ Therefore, solving for $r$: $$\Rightarrow \frac{0.6}{0.12} = 5 \, \Omega$$
Question 44
Physics · Moving Charges and Magnetism · Single correct
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In an uniform magnetic field, speed and energy remains the same for a moving charged particle. Reason (R) : Moving charged particle experiences magnetic force perpendicular to its direction of motion.
Both (A) and (R) are true and (R) is the correct explanation of (A)
Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
is true but (R) is false
is false but (R) is true.
Answer: (a)
Solution
Given $\vec{F} = q(\vec{v} \times \vec{B})$. Since $\vec{F} \perp \vec{v}$, the work done is $\vec{F} \cdot \vec{S}$. Therefore, the work done is $0$.
Question 45
Physics · Moving Charges and Magnetism · Single correct
The magnetic field at the centre of a circular coil of radius $r$, due to current $I$ flowing through it, is $B$. The magnetic field at a point along the axis at a distance $\frac{r}{2}$ from the centre is:
$\frac{B}{2}$
$2B$
$\left(\frac{2}{\sqrt{5}}\right)^3 B$
$\left(\frac{2}{\sqrt{3}}\right)^3 B$
Answer: (c)
Solution
Given $B_C = \frac{\mu_0 I}{2r}$, $B_a = \frac{\mu_0 I r^2}{2(x^2 + r^2)^{3/2}}$. At $x = \frac{r}{2}$, $$B_a = \frac{\mu_0 I r^2}{2 \left( \frac{r^2}{4} + r^2 \right)^{3/2}}$$ $$= \frac{\mu_0 I r^2}{2 \left( \frac{5}{4} r^2 \right)^{3/2}} = \frac{\mu_0 I}{2r} \left( \frac{4}{5} \right)^{3/2}$$ $$= \frac{\mu_0 I}{2r} \left( \frac{2}{\sqrt{5}} \right)^3$$
Question 46
Physics · Alternating Current · Single correct
A resistance of 40 $\Omega$ is connected to a source of alternating current rated 220 V, 50 Hz. Find the time taken by the current to change from its maximum value to rms value:
2.5 ms
1.25 ms
2.5 s
0.25 s
Answer: (a)
Solution
Considering sinusoidal AC. Phase at maximum value $= \frac{\pi}{2}$. Phase at rms value $= \frac{3\pi}{4}$. Thus phase change $= \frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$. Now $\omega = 2\pi f$. $= 2\pi \times 50$. $= 100\pi$. Time taken $t = \frac{\theta}{\omega} = \frac{\pi/4}{100\pi} = \frac{1}{400} s$. $t = 2.5 \times 10^{-3} = 2.5 ms$.
Question 47
Physics · Electromagnetic Waves · Single correct
A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is $4.5 \times 10^{-2} \, \mathrm{Am}^{-1}$ at a point, what will be the approximate magnitude of electric field intensity at that point? (Given: permeability of free space $\mu_0 = 4\pi \times 10^{-7} \, \mathrm{NA}^{-2}$, speed of light in vacuum $c = 3 \times 10^8 \, \mathrm{ms}^{-1}$)
Choose the correct option from the following options given below:
In the ground state of Rutherford's model electrons are in stable equilibrium. While in Thomson's model electrons always experience a net-force.
An atom has a nearly continuous mass distribution in a Rutherford's model but has a highly non-uniform mass distribution in Thomson's model
A classical atom based on Rutherford's model is doomed to collapse.
The positively charged part of the atom possesses most of the mass in Rutherford's model but not in Thomson's model.
Answer: (c)
Solution
According to Rutherford, $e^-$ revolves around the nucleus in a circular orbit. Thus $e^-$ is always accelerating (centripetal acceleration). An accelerating charge emits EM radiation and thus $e^-$ should lose energy and finally should collapse in the nucleus.
Question 49
Physics · Nuclei · Single correct
Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments 'B' and 'C' of mass numbers 105 and 115. The binding energy of nucleons in 'B' and 'C' is 6.4 MeV per nucleon. The energy Q released per fission will be:
A baseband signal of 3.5 $\mathrm{MHz}$ frequency is modulated with a carrier signal of 3.5 $\mathrm{GHz}$ frequency using amplitude modulation method. What should be the minimum size of antenna required to transmit the modulated signal?
42.8 $\mathrm{m}$
42.8 $\mathrm{mm}$
21.4 $\mathrm{mm}$
21.4 $\mathrm{m}$
Answer: (c)
Solution
Given $f_c = 3.5 \, \mathrm{GHz}$ and $f_m = 3.5 \, \mathrm{MHz}$. Side band frequencies are $f_c - f_m$ and $f_c + f_m$, which are almost $f_c$. $$\lambda = \frac{c}{f_c}$$ Minimum length of antenna = $$\frac{c}{f_c \cdot 4} = \frac{\lambda}{4} = \frac{3 \times 10^8}{3.5 \times 10^9 \times 4}$$ $$= 21.4 \, \mathrm{mm}$$
Question 51
Physics · Motion in a Straight Line · Numerical
From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 $\mathrm{s}$. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 $\mathrm{s}$. A third ball released, from the rest from the same location, will reach the ground in $\mathrm{s}$.
Answer: 3
Solution
Let height of tower be $h$ and speed of projection in first two cases be $u$. For case-I: 2nd equation is $s = ut + \frac{1}{2} at^2$ $$h = -u(6) + \frac{1}{2} g (6)^2$$ $$H = -6u + 18 \, g \ldots (i)$$ For case-II: $h = u(1.5) + \frac{1}{2} g (1.5)^2$ $$h = 1.5u + \frac{2.25g}{2} \ldots (ii)$$ Multiplying equation (ii) by 4 we get $$4h = 6u + 4.5 \, g \ldots (iii)$$ Equation (i) + equation (iii) we get $5h = 22.5g$ $$h = 4.5g \ldots (iv)$$ For case-III: $$h = 0 + \frac{1}{2} gt^2 \ldots (v)$$ Using equation (4) $\&$ equation (5) $$4.5g = \frac{1}{2} gt^2$$ $$t^2 = 9 \Rightarrow t = 3s$$
Question 52
Physics · Work, Energy and Power · Numerical
A ball of mass 100 g is dropped from a height $h = 10 \, \mathrm{cm}$ on a platform fixed at the top of vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\frac{h}{2}$. The spring constant is ________ $\mathrm{Nm^{-1}}$. (Use $g = 10 \, \mathrm{ms^{-2}}$)
Answer: 120
Solution
By energy conservation, PE = KE. $$mg \left( H + \frac{H}{2} \right) = \frac{1}{2} k x^2 \left( x = \frac{H}{2} \right)$$ $$0.100 \times 10 \times \frac{3}{2} (0.10) = \frac{1}{2} k (0.05 \times 0.05)$$ $$k = \frac{3 \times 0.10}{0.05 \times 0.05}$$ $$= \frac{3 \times 1000}{25} = 120 \, \mathrm{N/m}$$
Question 53
Physics · System of Particles and Rotational Motion · Numerical
A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10 \, $\mathrm{g}$ are put one on the top of the other at the 10.0 \, $\mathrm{cm}$ mark the scale is found to be balanced at 40.0 \, $\mathrm{cm}$ mark. The mass of the metre scale is found to be $x \times 10^{-2}\,\mathrm{kg}$. The value of $x$ is
Answer: 6
Solution
Let mass of meter scale be $m$. Balancing torque about knife edge: $$(0.02 \, \mathrm{g}) \times (30 \times 10^{-2}) = mg \times (10 \times 10^{-2})$$ $$m = 0.06 \, \mathrm{kg} = 6 \times 10^{-2} \, \mathrm{kg}$$
Question 54
Physics · Kinetic Theory · Numerical
0.056 kg of Nitrogen is enclosed in a vessel at a temperature of $127^\circ \mathrm{C}$. The amount of heat required to double the speed of its molecules is _____ k cal. (Take $R = 2 \, \mathrm{cal \, mole^{-1} \, }K^{-1}$)
In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3 : 2, the difference in the balancing length of the potentiometer wire in above two cases will be ______ cm.
As shown in the figure, an inductor of inductance $200\,\mathrm{mH}$ is connected to an AC source of emf $220\,\mathrm{V}$ and frequency $50\,\mathrm{Hz}$. The instantaneous voltage of the source is $0\,\mathrm{V}$ when the peak value of current is $\dfrac{\sqrt{a}}{\pi}\,\mathrm{A}$. The value of $a$ is _______.
Physics · Ray Optics and Optical Instruments · Numerical
Two identical thin biconvex lenses of focal length $15 \, \mathrm{cm}$ and refractive index $1.5$ are in contact with each other. The space between the lenses is filled with a liquid of refractive index $1.25$. The focal length of the combination is _____ cm.
Answer: 10
Solution
Given the lens system, we have the following equations: $$\frac{1}{f_1} = \frac{1}{15} = \left(\frac{3}{2} - 1\right)\left[\frac{2}{R}\right]$$ Thus, $$\frac{1}{R} = \frac{1}{15}$$ For the equivalent focal length: $$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$$ Substituting the values: $$= \frac{1}{15} + \left(\frac{5}{4} - 1\right)\left[\frac{-2}{R}\right] + \frac{1}{15}$$ Simplifying further: $$= \frac{1}{15} - \frac{1}{30} + \frac{1}{15}$$ Finally: $$= \frac{2 - 1 + 2}{30}$$ $$= \frac{3}{30} = \frac{1}{10}$$
Question 58
Physics · Wave Optics · Numerical
Sodium light of wavelengths 650 nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is _____ $\times$ $10^{-5}$ $\mathrm{m}$.
Physics · Dual Nature of Radiation and Matter · Numerical
When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $v_1$. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $v_2$. If $v_2 = x \, v_1$, the value of $x$ will be _____.
Answer: 2
Solution
Given $h\nu = h\nu_{th} + \frac{1}{2}mv^2$. Let $v = 2v_{th}$. Then, $2h\nu_{th} = h\nu_{th} + \frac{1}{2}mv_1^2 \ldots (1)$. Now, let $v = 5v_{th}$. Then, $5h\nu_{th} = h\nu_{th} + \frac{1}{2}mv_2^2 \ldots (2)$. Dividing equation (1) by equation (2), we have $$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{h\nu_{th}}{4h\nu_{th}}.$$ Simplifying, $$\left(\frac{v_1}{v_2}\right)^2 = \frac{1}{4} \implies v_2 = 2v_1.$$
A transistor is used in common-emitter mode in an amplifier circuit. When a signal of $10\,\mathrm{mV}$ is added to the base-emitter voltage, the base current changes by $10\,\mathrm{µA}$ and the collector current changes by $1.5\,\mathrm{mA}$. The load resistance is $5\,\mathrm{k\Omega}$. The voltage gain of the transistor will be _____.
Answer: 750
Solution
Given $r_i = \frac{10 \, \mathrm{mV}}{10 \, \mu \mathrm{A}} = 10^3 \, \Omega$. $\beta = \frac{1.5 \, \mathrm{mA}}{10 \, \mu \mathrm{A}} = 150$. The voltage gain $A_V$ is given by $$A_V = \left( \frac{R_0}{r_i} \right) \beta = \left( \frac{5000}{1000} \right) \times 150 = 750.$$
Chemistry
Question 61
Chemistry · Some Basic Concepts of Chemistry · Single correct
If a rocket runs on a fuel $(C_{15} H_{30})$ and liquid oxygen, the weight of oxygen required and $CO_2$ released for every litre of fuel respectively are: (Given: density of the fuel is $0.756 \, \mathrm{g/mL}$)
1188 $\mathrm{g}$ and 1296 $\mathrm{g}$
2376 $\mathrm{g}$ and 2592 $\mathrm{g}$
2592 $\mathrm{g}$ and 2376 $\mathrm{g}$
3429 $\mathrm{g}$ and 3142 $\mathrm{g}$
Answer: (c)
Solution
The reaction is given by: $$\mathrm{C_{15}H_{30} + \frac{45}{2}O_2 \rightarrow 15CO_2 + 15H_2O}$$ Mass of fuel = $0.756 \times 1000 \, \mathrm{g}$ Number of moles of fuel = $$\frac{0.756 \times 1000}{210}$$ Weight of oxygen = $$\frac{0.756 \times 1000}{210} \times \frac{45}{2} \times 32 = 2592 \, \mathrm{g}$$ Weight of $\mathrm{CO_2}$ = $$\frac{0.756 \times 1000}{210} \times 15 \times 44 = 2376 \, \mathrm{g}$$
Question 62
Chemistry · Structure of Atom · Single correct
Consider the following pairs of electrons The pairs of electron present in degenerate orbitals is/are:
Only A
Only B
Only C
$(B) and (C)$
Answer: (b)
Solution
Based on "n + l" rule only (B) has pair of electron in degenerate orbitals
Question 63
Chemistry · Equilibrium · Single correct
For a reaction at equilibrium A(g) $\rightleftharpoons$ B(g) + $\frac{1}{2}$C(g) the relation between dissociation constant (K), degree of dissociation ($\alpha$) and equilibrium pressure P is given by:
K = $\frac{\frac{1}{\alpha^2} p^{\frac{3}{2}}}{\left(1 + \frac{3}{2} \alpha\right)^{\frac{1}{2}} (1 - \alpha)}$
K = $\frac{\frac{3}{\alpha^2} p^{\frac{1}{2}}}{(2 + \alpha)^{\frac{1}{2}} (1 - \alpha)}$
K = $\frac{(\alpha p)^{\frac{3}{2}}}{\left(1 + \frac{3}{2} \alpha\right)^{\frac{1}{2}} (1 - \alpha)}$
K = $\frac{(\alpha p)^{\frac{3}{2}}}{(1 + \alpha)(1 - \alpha)^{\frac{1}{2}}}$
The highest industrial consumption of molecular hydrogen is to produce compounds of element:
Carbon
Nitrogen
Oxygen
Chlorine
Answer: (b)
Solution
Nitrogen. Around 55$\%$ of hydrogen around would go to ammonia production.
Question 65
Chemistry · The s-Block Elements · Single correct
Which of the following statements are correct? (A) Both $\mathrm{LiCl}$ and $\mathrm{MgCl_2}$ are soluble in ethanol. (B) The oxides $\mathrm{Li_2O}$ and $\mathrm{MgO}$ combine with excess of oxygen to give superoxide. (C ) $\mathrm{LiF}$ is less soluble in water than other alkali metal fluorides. (D) $\mathrm{Li_2O}$ is more soluble in water than other alkali metal oxides. Choose the most appropriate answer from the options given below:
(A) and ( C) only
(A), ( C) and (D) only
(B)and (C ) only
(A)and ( C) only
Answer: (a)
Solution
(A) Both $\mathrm{LiCl}$ and $\mathrm{MgCl_2}$ are soluble in ethanol. (B) Li and Mg do not form superoxide. (C) $\mathrm{LiF}$ has high lattice energy. (D) $\mathrm{Li_2O}$ is least soluble in water than other alkali metal oxides.
Question 66
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Identify the correct statement for $\mathrm{B_2H_6}$ from those given below. (A) In $\mathrm{B_2H_6}$, all B-H bonds are equivalent. (B) In $\mathrm{B_2H_6}$ there are four 3-centre-2-electron bonds. $(C)$ $\mathrm{B_2H_6}$ is a Lewis acid. (D) $\mathrm{B_2H_6}$ can be synthesized form both $\mathrm{BF_3}$ and $\mathrm{NaBH_4}$. (E) $\mathrm{B_2H_6}$ is a planar molecule. Choose the most appropriate answer from the options given below :
(A) and (E) only
(A), $(C)$ and (E) only
(A) and (D) only
(A) and (E) only
Answer: (c)
Solution
Two 3 centre – 2 – electron bonds. $\mathrm{B_2H_6}$ is electron deficient species. $\mathrm{B_2H_6}$ is non-planar molecule. $\mathrm{BF_3 + LiAlH_4 \rightarrow 2B_2H_6 + 3LiF + 3AlF_3}$. $\mathrm{NaBH_4 + I_2 \rightarrow B_2H_6 + 2NaI + H_2}$.
Question 67
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Which of the following is an example of conjugated diketone?
Answer: (c)
Solution
The compound shown is a conjugated diketone.
Question 68
Chemistry · Amines · Single correct
In the given reactions sequence, the major product 'C' is: $C_8H_{10} \xrightarrow[\mathrm{H_2SO_4}]{\mathrm{HNO_3}} A \xrightarrow[\Delta]{\mathrm{Br_2}} B \xrightarrow{\mathrm{alcoholic\ KOH}} C$
Answer: (b)
Solution
Question 69
Chemistry · Surface Chemistry · Single correct
Given below are two statements : Statement I : Emulsions of oil in water are unstable and sometimes they separate into two layers on standing. Statement II : For stabilisation of an emulsion, excess of electrolyte is added. In the light of the above statements, choose the most appropriate answer from the options given below :
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
Statement I: Fact Statement II: The principle emulsifying agents for O/W emulsions are proteins, gums natural and synthetic soaps etc...
Question 70
Chemistry · Co-ordination Compounds · Single correct
Match List - I with List - II Choose the most appropriate answer from the options given below:
(A) - (IV), (B) - (III), (C) - (II), (D) - (I)
(A) - (IV), (B) - (I), (C) - (II), (D) - (III)
(A) - (II), (B) - (III), (C) - (I), (D) - (IV)
(A) - (III), (B) - (IV), (C) - (II), (D) - (I)
Answer: (a)
Solution
List - I and List - II are matched as follows: (A) Sphalerite matches with (IV) $\mathrm{ZnS}$ (B) Calamine matches with (III) $\mathrm{ZnCO_3}$ (C) Galena matches with (II) $\mathrm{PbS}$ (D) Siderite matches with (I) $\mathrm{FeCO_3}$
Question 71
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Given below are the oxides: $\mathrm{Na_2O, As_2O_3, N_2O, NO}$ and $\mathrm{Cl_2O_7}$ Number of amphoteric oxides is:
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
The most stable trihalide of nitrogen is:
$\mathrm{NF}_3$
$\mathrm{NCl}_3$
$\mathrm{NBr}_3$
$\mathrm{NI}_3$
Answer: (a)
Solution
Order of stability: $$\mathrm{NF_3 > NCl_3 > NBr_3 > NI_3}$$
Question 73
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Which one of the following elemental forms is not present in the enamel of the teeth?
$\mathrm{Ca}^{2+}$
$\mathrm{P}^{3+}$
$\mathrm{F}^{-}$
$\mathrm{P}^{5+}$
Answer: (b)
Solution
Calcium and phosphate are the major components of teeth enamel.
Question 74
Chemistry · Co-ordination Compounds · Single correct
Match List – I with List - II
(A) →(II), (B)→(IV), (C)→(I), (D)→(III)
(A) →(III), (B)→(IV), (C)→(I), (D)→(II)
(A) →(III), (B)→(I), (C)→(IV), (D)→(II)
(A) →(II), (B)→(I), (C)→(IV), (D)→(III)
Answer: (b)
Solution
Match the items in List I with those in List II based on their hybridization. (A) $[\mathrm{PtCl}_4]^{2-}$ corresponds to (III) $\mathrm{dsp}^2$. (B) $\mathrm{BrF}_5$ corresponds to (IV) $\mathrm{sp}^3\mathrm{d}^2$. (C) $\mathrm{PCl}_5$ corresponds to (I) $\mathrm{sp}^3\mathrm{d}$. (D) $[\mathrm{Co(NH}_3)_6]^{3+}$ corresponds to (II) $\mathrm{d}^2\mathrm{sp}^3$.
Question 75
Chemistry · Haloalkanes and Haloarenes · Single correct
The major product of the above reaction is
Answer: (d)
Solution
The reaction sequence starts with the substitution of the bromine atom by a cyano group using $NaCN$. This is followed by a reaction with $OH^-$ to form a cyclic compound. The intermediate is then reduced using $H_2/Ni$ to form the final product.
Question 76
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Two statements are given below : Statement I: The melting point of monocarboxylic acid with even number of carbon atoms is higher than that of with odd number of carbon atoms acid immediately below and above it in the series. Statement II : The solubility of monocarboxylic acids in water decreases with increase in molar mass. Choose the most appropriate 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: (a)
Solution
I. Better packing efficiency of monocarboxylic acids with even number of carbon atoms results in higher M.P. II. As molar mass increases hydrophobic part size increase hence solubility decreases.
Question 77
Chemistry · Polymers · Single correct
Which of the following is an example of polyester?
Chemistry · Chemistry in Everyday Life · Single correct
Which of the following is not a broad spectrum antibiotic?
Vancomycin
Ampicillin
Ofloxacin
Penicillin G
Answer: (d)
Solution
Penicillin G following is a narrow spectrum antibiotic.
Question 79
Chemistry · Analytical Chemistry · Single correct
During the qualitative analysis of salt with cation $y^{2+}$, addition of a reagent (X) to alkaline solution of the salt gives a bright red precipitate. The reagent (X) and the cation $(y^{2+})$ present respectively are:
Dimethylglyoxime and $\mathrm{Ni}^{2+}$
Dimethylglyoxime and $\mathrm{Co}^{2+}$
Nessler’s reagent and $\mathrm{Hg}^{2+}$
Nessler’s reagent and $\mathrm{Ni}^{2+}$
Answer: (a)
Solution
$Ni^{2+} + \mathrm{DMG} \rightarrow [Ni(\mathrm{DMG})_2]\downarrow$ (Bright red precipitate)
Question 80
Chemistry · Biomolecules · Single correct
A polysaccharide 'X' on boiling with dil $\mathrm{H_2SO_4}$ at 393 K under 2-3 atm pressure yields 'Y'. 'Y' on treatment with bromine water gives gluconic acid. 'X' contains $\beta$-glycosidic linkages only. Compound 'X' is :
$2\mathrm{O}_3(g) \rightleftharpoons 3\mathrm{O}_2(g)$ At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (−) __ J mol⁻¹ (Nearest integer) [Given: $\ln$ 1.35 = 0.3 and R = 8.3 $\mathrm{J}$ $\mathrm{K}^{-1}$ $\mathrm{mol}^{-1}$]
Answer: 747
Solution
The reaction is given by $2 \mathrm{O_3} \rightleftharpoons 3 \mathrm{O_2} (g)$. The initial pressures are $\frac{2}{5}$ and $\frac{3}{5}$. The equilibrium constant $k_p$ is given by $$k_p = \frac{P_{\mathrm{O_2}}^3}{P_{\mathrm{O_3}}^2}$$ where $k_p = 1.35$. The change in Gibbs free energy is calculated as $$\Delta G^\circ = -RT \ln k_p$$ $$= -8.3 \times 300 \times \ln 1.35$$ $$= -747 \, \mathrm{J/mol}$$
Question 82
Chemistry · Redox Reactions · Numerical
A $0.166\,\mathrm{g}$ sample of an organic compound was digested with conc. $\mathrm{H_2SO_4}$ and then distilled with $\mathrm{NaOH}$. The ammonia gas evolved was passed through $50.0\,\mathrm{mL}$ of $0.5\,\mathrm{N}$ $\mathrm{H_2SO_4}$. The used acid required $30.0\,\mathrm{mL}$ of $0.25\,\mathrm{N}$ $\mathrm{NaOH}$ for complete neutralization. The mass percentage of nitrogen in the organic compound is \_\_\_\_.
Answer: 63
Solution
Given $m_{eq}$ of NaOH used $= 30 \times 0.25$. $m_{eq}$ of H$_2$SO$_4$ taken $= 50 \times 0.5$. Therefore, $m_{eq}$ of H$_2$SO$_4$ used $= 50 \times 0.25 \times 30 \times 0.25 = 17.5$ m mol of NH$_3$. Therefore, $\% N = \frac{17.5 \times 10^{-3} \times 14}{0.166} \times 100 = 147.59\%$. (Not possible)
Number of electrophilic centre in the given compound is ____
Answer: 3
Solution
Question 84
Chemistry · Hydrocarbons · Numerical
The major product 'A' of the following given reaction has $\underline{\qquad} sp^2$ hybridized carbon atoms. 2,7-Dimethyl-2, 6-octadien $\xrightarrow{H^+}$ Major Product $A$
Answer: 2
Solution
Question 85
Chemistry · The Solid State · Numerical
Atoms of element X form hcp lattice and those of element Y occupy $\frac{2}{3}$ of its tetrahedral voids. The percentage of element X in the lattice is ________ (Nearest integer)
Answer: 43
Solution
Given $X \to 6$ and $Y \to \frac{2}{3} \times 6 = 8$. The percentage of $X$ is calculated as $$\% X = \frac{6}{14} \times 100 = 42.8 \approx 43\%$$
Question 86
Chemistry · Solutions · Numerical
The osmotic pressure of blood is $7.47\,\mathrm{bar}$ at $300\,\mathrm{K}$. To inject glucose to a patient intravenously, it has to be isotonic with blood. The concentration of glucose solution in $\mathrm{g\,L^{-1}}$ is \_\_\_\_ (Molar mass of glucose $= 180\,\mathrm{g\,mol^{-1}}$, $R = 0.083\,\mathrm{L\,bar\,K^{-1}\,mol^{-1}}$) (Nearest integer)
Answer: 54
Solution
Given the equation $\pi = C \cdot R \cdot T$. Substitute the values: $$7.47 = C \times 0.083 \times 300$$ Solve for $C$: $$C = 0.3 \, \mathrm{M}$$ Calculate the concentration in grams per liter: $$= 0.3 \times 180 \, \mathrm{gL^{-1}}$$ Result: $$= 54 \, \mathrm{gL^{-1}}$$
Question 87
Chemistry · Electrochemistry · Numerical
The cell potential for the following cell Pt | $\mathrm{H}_2(g)$ | $\mathrm{H}^+(aq)$ || $\mathrm{Cu}^{2+}$ (0.01 $\mathrm{M}$) | $\mathrm{Cu}(s)$ is $0.576\, \mathrm{V}$ at $298\, \mathrm{K}$. The pH of the solution is ___. (Nearest integer)
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700 - 1000 K. The data has been analysed by plotting $\ln k$ vs $\frac{10^3}{T}$ graph. The value of activation energy for the reaction is ___ kJ mol$^{-1}$. (Nearest integer) (Given : $R = 8.31 \, \mathrm{J \, K}^{-1} \, \mathrm{mol}^{-1}$)
Answer: 154
Solution
Given $\ln k = \ln A - \frac{E_a}{10^3 RT} \times 10^3 = \ln A + \frac{10^3}{T} \left[ \frac{-E_a}{10^3 RT} \right]$. From the graph, $$\frac{-E_a}{10^3 \times R} = -18.5$$ $E_a = 153.735 \, \mathrm{kJ/mol}$, approximately $154$.
Question 89
Chemistry · The d-and f-Block Elements · Numerical
The difference in oxidation state of chromium in chromate and dichromate salts is
Answer: 0
Solution
For $\mathrm{CrO_4^{2-}}$ and $\mathrm{Cr_2O_7^{2-}}$, the difference is zero.
Question 90
Chemistry · Co-ordination Compounds · Numerical
In the cobalt-carbonyl complex: $[\mathrm{Co}_2(\mathrm{CO})_8]$, number of Co-Co bonds is "X" and terminal CO ligands is "Y". X + Y =