JEE Main 4 April 2024 Shift 1 question paper with solutions
JEE Main 4 April 2024 Shift 1: all 90 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Continuity and Differentiability · Single correct
Let $f : \mathbb{R} \to \mathbb{R}$ be a function given by $$f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2}, & x 0 \end{cases}$$ where $\alpha, \beta \in \mathbb{R}$. If $f$ is continuous at $x = 0$, then $\alpha^2 + \beta^2$ is equal to :
3
12
48
6
Answer: (b)
Solution
Given $f(0^-) = \lim_{x \to 0^-} \frac{2 \sin^2 x}{x^2} = 2 = \alpha$. For $f(0^+)$, we have $f(0^+) = \lim_{x \to 0^+} \beta \times \sqrt{2} \frac{\sin \frac{x}{2}}{\frac{x}{2}} = \frac{\beta}{\sqrt{2}} = 2$. This implies $\beta = 2\sqrt{2}$. Therefore, $\alpha^2 + \beta^2 = 4 + 8 = 12$.
Question 2
Maths · Probability · Single correct
Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :
$\frac{5}{18}$
$\frac{5}{16}$
$\frac{4}{17}$
$\frac{7}{18}$
Answer: (a)
Solution
Given the bags A, B, and C with contents as follows: A has 7 red and 5 blue, B has 5 red and 7 blue, C has 6 red and 6 blue. The probability of selecting a blue ball, P(B), is calculated as: $$P(B) = \frac{1}{3} \cdot \frac{5}{12} + \frac{1}{3} \cdot \frac{7}{12} + \frac{1}{3} \cdot \frac{6}{12}$$ The required probability is: required probability = $$\frac{\frac{1}{3} \cdot \frac{5}{12}}{\frac{1}{3} \cdot \left[ \frac{5}{12} + \frac{7}{12} + \frac{6}{12} \right]} = \frac{5}{18}$$
Question 3
Maths · Straight Lines and Pair of Straight Lines · Single correct
The vertices of a triangle are A(-1, 3), B(-2, 2) and C(3, -1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :
$x + y + \left(2 - \sqrt{2}\right) = 0$
$-x + y - \left(2 - \sqrt{2}\right) = 0$
$x + y - \left(2 - \sqrt{2}\right) = 0$
$x - y - \left(2 + \sqrt{2}\right) = 0$
Answer: (c)
Solution
Equation of AC is $x + y = 2$. Equation of line parallel to AC is $x + y = d$. $$\left| \frac{d - 2}{\sqrt{2}} \right| = 1$$ $$d = 2 - \sqrt{2}$$ Equation of new required line is $x + y = 2 - \sqrt{2}$.
Question 4
Maths · Differential Equations · Single correct
If the solution $y = y(x)$ of the differential equation $(x^4 + 2x^3 + 3x^2 + 2x + 2) \, dy - (2x^2 + 2x + 3) \, dx = 0$ satisfies $y(-1) = -\frac{\pi}{4}$, then $y(0)$ is equal to:
$\frac{\pi}{2}$
$-\frac{\pi}{2}$
0
$\frac{\pi}{4}$
Answer: (d)
Solution
Given $$\int dy = \int \frac{(2x^2 + 2x + 3)}{x^4 + 2x^3 + 3x^2 + 2x + 2} \, dx$$ we have $$y = \int \frac{(2x^2 + 2x + 3)}{(x^2 + 1)(x^2 + 2x + 2)} \, dx$$ which simplifies to $$y = \int \frac{dx}{x^2 + 2x + 2} + \int \frac{dx}{x^2 + 1}$$ resulting in $$y = \tan^{-1}(x + 1) + \tan^{-1} x + C$$ Given $$y(-1) = -\frac{\pi}{4}$$ we find $$-\frac{\pi}{4} = 0 - \frac{\pi}{4} + C \Rightarrow C = 0$$ Therefore, $$y = \tan^{-1}(x + 1) + \tan^{-1} x$$ Finally, $$y(0) = \tan^{-1} 1 = \frac{\pi}{4}$$
Question 5
Maths · Applications of Derivatives · Single correct
Let the sum of the maximum and the minimum values of the function $f(x) = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8}$ be $\frac{m}{n}$, where $\gcd(m, n) = 1$. Then $m + n$ is equal to:
Maths · Applications of Integrals · Single correct
One of the points of intersection of the curves $y = 1 + 3x - 2x^2$ and $y = \frac{1}{x}$ is $\left( \frac{1}{2}, 2 \right)$. Let the area of the region enclosed by these curves be $\frac{1}{24} \left( l \sqrt{5} + m \right) - n \log_e \left( 1 + \sqrt{5} \right)$, where $l, m, n \in \mathbb{N}$. Then $l + m + n$ is equal to
If the system of equations $$x + (\sqrt{2} \sin \alpha) y + (\sqrt{2} \cos \alpha) z = 0$$ $$x + (\cos \alpha) y + (\sin \alpha) z = 0$$ $$x + (\sin \alpha) y - (\cos \alpha) z = 0$$ has a non-trivial solution, then $\alpha \in \left(0, \frac{\pi}{2}\right)$ is equal to:
$\frac{11\pi}{24}$
$\frac{5\pi}{24}$
$\frac{7\pi}{24}$
$\frac{3\pi}{4}$
Answer: (b)
Solution
The determinant of the matrix is set to zero: $$\begin{vmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \sin \alpha & -\cos \alpha \\ 1 & \cos \alpha & \sin \alpha \end{vmatrix} = 0$$ Expanding the determinant, we have: $$1 - \sqrt{2} \sin \alpha (\sin \alpha + \cos \alpha) + \sqrt{2} \cos \alpha (\cos \alpha - \sin \alpha) = 0$$ Simplifying, we get: $$1 + \sqrt{2} \cos 2\alpha - \sqrt{2} \sin 2\alpha = 0$$ This implies: $$\cos 2\alpha - \sin 2\alpha = -\frac{1}{\sqrt{2}}$$ Rewriting, we have: $$\cos \left(2\alpha + \frac{\pi}{4}\right) = -\frac{1}{2}$$ Solving for $2\alpha$, we get: $$2\alpha + \frac{\pi}{4} = 2n\pi \pm \frac{2\pi}{3}$$ Thus: $$\alpha + \frac{\pi}{8} = n\pi \pm \frac{\pi}{3}$$ For $n = 0$, we find: $$x = \frac{\pi}{3} - \frac{\pi}{8} = \frac{5\pi}{24}$$
Question 8
Maths · Permutations and Combinations · Single correct
There are 5 points $P_1, P_2, P_3, P_4, P_5$ on the side $AB$, excluding $A$ and $B$, of a triangle $ABC$. Similarly there are 6 points $P_6, P_7, \ldots, P_{11}$ on the side $BC$ and 7 points $P_{12}, P_{13}, \ldots, P_{18}$ on the side $CA$ of the triangle. The number of triangles, that can be formed using the points $P_1, P_2, \ldots, P_{18}$ as vertices, is:
776
796
751
771
Answer: (c)
Solution
The expression is given by: $$^{18}C_3 - ^{5}C_3 - ^{6}C_3 - ^{7}C_3$$ The result is: $$= 751$$
Question 9
Maths · Integrals · Single correct
Let $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \\ x - 2, & 0 < x \leq 2 \end{cases}$ and $h(x) = f(|x|) + |f(x)|$. Then $\int_{-2}^{2} h(x) \, dx$ is equal to :
1
6
4
2
Answer: (d)
Solution
Given the function $h(x)$ defined as follows: $$h(x) = \begin{cases} x - 2 + 2 - x = 0, & 0 \leq x \leq 2 \\ -x - 2 + 2 = -x, & -2 \leq x < 0 \end{cases}$$ We find that $$\int_0^2 h(x) \, dx = 0 and \int_{-2}^0 h(x) \, dx = 2$$
Question 10
Maths · Binomial Theorem · Single correct
The sum of all rational terms in the expansion of $\left(2^{\frac{1}{5}} + 5^{\frac{1}{3}}\right)^{15}$ is equal to:
Let a unit vector which makes an angle of $60^\circ$ with $2\hat{i} + 2\hat{j} - \hat{k}$ and angle $45^\circ$ with $\hat{i} - \hat{k}$ be $\vec{C}$. Then $\vec{C} + \left( -\frac{1}{2} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k} \right)$ is :
Let the first three terms $2, p$ and $q$, with $q \neq 2$, of a G.P. be respectively the $7^{th}$, $8^{th}$ and $13^{th}$ terms of an A.P. If the $5^{th}$ term of the G.P. is the $n^{th}$ term of the A.P., then $n$ is equal to:
Let $\alpha, \beta \in \mathbb{R}$. Let the mean and the variance of 6 observations $-3, 4, 7, -6, \alpha, \beta$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is:
Maths · Complex Numbers and Quadratic Equations · Single correct
If 2 and 6 are the roots of the equation $ax^2 + bx + 1 = 0$, then the quadratic equation, whose roots are $\frac{1}{2a+b}$ and $\frac{1}{6a+b}$, is:
$2x^2 + 11x + 12 = 0$
$x^2 + 8x + 12 = 0$
$4x^2 + 14x + 12 = 0$
$x^2 + 10x + 16 = 0$
Answer: (b)
Solution
Sum $= 8 = -\frac{b}{a}$ Product $= 12 = \frac{1}{a} \implies a = \frac{1}{12}$ $b = -\frac{2}{3}$ $2a + b = \frac{2}{12} - \frac{2}{3} = -\frac{1}{2}$ $6a + b = \frac{6}{12} - \frac{2}{3} = -\frac{1}{6}$ sum $= -8$ $P = 12$ $x^2 + 8x + 12 = 0$
Question 15
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $\left(\bar{z}\right)^2 + |z| = 0, z \in \mathbb{C}$. Then $4 \left(\alpha^2 + \beta^2\right)$ is equal to:
6
8
2
4
Answer: (d)
Solution
Given $z = x + iy$ and $\bar{z} = x - iy$. Then $\bar{z}^2 = x^2 - y^2 - 2ixy$. This implies $$x^2 - y^2 - 2ixy + \sqrt{x^2 + y^2} = 0.$$ Therefore, $x = 0$ or $y = 0$. If $-y^2 + |y| = 0$, then $|y| = |y|^2$, which gives $y = 0, \pm 1$. Thus, $i, -i$ are roots. If $x^2 + |x| = 0$, then $x = 0$. This implies $\alpha = i - i = 0$ and $\beta = i(-i) = 1$. Therefore, $4(0 + 1) = 4$.
Question 16
Maths · Three Dimensional Geometry · Single correct
Let the point, on the line passing through the points $P(1, -2, 3)$ and $Q(5, -4, 7)$, farther from the origin and at distance of 9 units from the point $P$, be $(\alpha, \beta, \gamma)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:
A square is inscribed in the circle $x^2 + y^2 - 10x - 6y + 30 = 0$. One side of this square is parallel to $y = x + 3$. If $(x_i, y_i)$ are the vertices of the square, then $\Sigma (x_i^2 + y_i^2)$ is equal to:
148
152
160
156
Answer: (b)
Solution
Given the equations $y = x + c$ and $x + y + d = 0$. The conditions are: $$\left| \frac{5 - 3 + c}{\sqrt{2}} \right| = \sqrt{2}$$ $$\left| \frac{8 + d}{\sqrt{2}} \right| = \sqrt{2}$$ Solving these, we get: $$|c + 2| = 2$$ $$8 + d = \pm 2$$ Thus, $c = 0, -4$ and $d = -10, -6$. The points are $(5, 5), (3, 3), (7, 3), (5, 1)$. The sum is: $$\sum (x_i^2 + y_i^2) = 25 + 25 + 9 + 9 + 49 + 9 + 25 + 1 = 152$$
Question 18
Maths · Inverse Trigonometric Functions · Single correct
If the domain of the function $\sin^{-1}\left(\frac{3x-22}{2x-19}\right) + \log_e\left(\frac{3x^2-8x+5}{x^2-3x-10}\right)$ is $(\alpha, \beta]$, then $3\alpha + 10\beta$ is equal to:
100
95
97
98
Answer: (c)
Solution
Given the inequality $$-1 \leq \frac{3x - 22}{2x - 19} \leq 1$$ and $$\frac{3x^2 - 8x + 5}{x^2 - 3x - 10} > 0$$. The solution for $x$ is $$x \in \left(5, \frac{41}{5}\right]$$. Additionally, $$3\alpha + 10\beta = 97$$.
Question 19
Maths · Continuity and Differentiability · Single correct
Let $f(x) = x^5 + 2e^{x/4}$ for all $x \in \mathbb{R}$. Consider a function $g(x)$ such that $(g \circ f)(x) = x$ for all $x \in \mathbb{R}$. Then the value of $8g'(2)$ is:
If $\lim_{x \to 1} \frac{(5x+1)^{1/3}-(x+5)^{1/3}}{(2x+3)^{1/2}-(x+4)^{1/2}} = \frac{m\sqrt{5}}{n(2n)^{2/3}}$, where gcd$(m, n) = 1$, then $8m + 12n$ is equal to
Answer: 100
Solution
The limit is given by $$ \lim_{x \to 1} \frac{\frac{1}{3}(5x + 1)^{-2/3}5 - \frac{1}{3}(x + 5)^{-2/3}}{\frac{1}{2}(2x + 3)^{-1/2} \cdot 2 - \frac{1}{2}(x + 4)^{-1/2}} $$ This simplifies to $$ \frac{\frac{8}{3} \sqrt{5}}{6^{2/3}} m = 8 n = 3 $$ Given the equation $$ 8m + 12n = 100 $$
Question 22
Maths · Sets · Numerical
In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these subjects. Let $m$ and $n$ respectively be the least and the most number of students who studied all the three subjects. Then $m + n$ is equal to
Answer: 45
Solution
Given the Venn diagram, we have the following inequalities: $$125 \leq m + 90 - x \leq 130$$ $$85 \leq P + 70 - x \leq 95$$ $$75 \leq C + 80 - x \leq 90$$ The equation for the total is: $$m + P + C + 120 - 2x = 210$$ Solving these, we find: $$\Rightarrow 15 \leq x \leq 45 and 30 - x \geq 0$$ This implies: $$\Rightarrow 15 \leq x \leq 30$$ Finally, we have: $$30 + 15 = 45$$
Question 23
Maths · Differential Equations · Numerical
Let the solution $y = y(x)$ of the differential equation $\frac{dy}{dx} - y = 1 + 4 \sin x$ satisfy $y(\pi) = 1$. Then $y\left(\frac{\pi}{2}\right) + 10$ is equal to
Answer: 7
Solution
Given $y e^{-x} = \int \left( e^{-x} + 4 e^{-x} \sin x \right) \, dx$. $y e^{-x} = -e^{-x} - 2 \left( e^{-x} \sin x e^{-x} \cos x \right) + C$. $y = -1 - 2(\sin x + \cos x) + c e^x$. Therefore, $y(\pi) = 1 \Rightarrow c = 0$. $y'(\pi/2) = -1 - 2 = -3$. Ans = 10 - 3 = 7
Question 24
Maths · Three Dimensional Geometry · Numerical
If the shortest distance between the lines $\frac{x+2}{2} = \frac{y+3}{3} = \frac{z-5}{4}$ and $\frac{x-3}{1} = \frac{y-2}{-3} = \frac{z+4}{2}$ is $\frac{38}{3\sqrt{5}}k$, and $\int_0^k [x^2] \, dx = \alpha - \sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6\alpha^3$ is equal to .
Let $A$ be a square matrix of order $2$ such that $|A|=2$ and the sum of its diagonal elements is $-3$. If the points $(x,y)$ satisfying \[ A^2+xA+yI=0 \] lie on a hyperbola, whose transverse axis is parallel to the $x$-axis, eccentricity is $e$ and the length of the latus rectum is $\ell$, then $e^4+\ell^4$ is equal to ____
Answer: 233
Solution
Question 26
Maths · Binomial Theorem · Numerical
Let $a = 1 + \frac{{^2C_2}}{{3!}} + \frac{{^3C_2}}{{4!}} + \frac{{^4C_2}}{{5!}} + \cdots,$ $b = 1 + \frac{{^1C_0 + ^1C_1}}{{1!}} + \frac{{^2C_0 + ^2C_1 + ^2C_2}}{{2!}} + \frac{{^3C_0 + ^3C_1 + ^3C_2 + ^3C_3}}{{3!}} + \cdots$ Then $\frac{{2b}}{{a^2}}$ is equal to
Answer: 8
Solution
Given $$f(x) = 1 + \frac{(1+x)}{1!} + \frac{(1+x)^2}{2!} + \frac{(1+x)^3}{3!} + \ldots$$ $$\frac{e^{(1+x)}}{1+x} = \frac{1}{1+x} + 1 + \frac{(1+x)}{2!} + \frac{(1+x)^2}{3!} + \frac{(1+x)^2}{4!}$$ Coefficient of $x^2$ in RHS: $$1 + \frac{2C_2}{3} + \frac{3C_2}{4} + \ldots = a$$ Coefficient of $x^2$ in LHS. $$e \left(1 + x + \frac{x^2}{2!}\right) \cdots \left(1 - x + \frac{x^2}{2!} \cdots \right)$$ is $e - e + \frac{e}{2!} = a$ $$b = 1 + \frac{2}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \ldots = e^2$$ $$\frac{2}{a^2} b = 8$$
Question 27
Maths · Matrices · Numerical
Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$. Then the maximum value of $\det(A)$ is .
Let the length of the focal chord PQ of the parabola $y^2 = 12x$ be 15 units. If the distance of PQ from the origin is $p$, then $10p^2$ is equal to
Answer: 72
Solution
The length of the focal chord is given by $4a \csc^2 \theta = 15$. Thus, $12 \csc^2 \theta = 15$. We have $\sin^2 \theta = \frac{4}{5}$. Therefore, $\tan^2 \theta = 4$ and $\tan \theta = 2$. The equation is $\frac{y - 0}{x - 3} = 2$. This simplifies to $y = 2x - 6$. Rearranging gives $2x - y - 6 = 0$. We find $P = \frac{6}{\sqrt{5}}$. Finally, $10p^2 = 10 \cdot \frac{36}{5} = 72$.
Question 29
Maths · Vector Algebra · Numerical
Let ABC be a triangle of area $15\sqrt{2}$ and the vectors $\overrightarrow{AB} = \hat{i} + 2\hat{j} - 7\hat{k}$, $\overrightarrow{BC} = a\hat{i} + b\hat{j} + c\hat{k}$ and $\overrightarrow{AC} = 6\hat{i} + d\hat{j} - 2\hat{k}$, $d > 0$. Then the square of the length of the largest side of the triangle ABC is
Answer: 54
Solution
Given the triangle with vectors, we calculate the area using the determinant method. $$Area = \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{vmatrix} = 15 \sqrt{2}$$ Expanding the determinant, we have: $$(-4 + 7d) \hat{i} - \hat{j}(-2 + 42) + \hat{k}(d - 12)$$ Simplifying, we get: $$(7d - 4)^2 + (40)^2 + (d - 12)^2 = 1800$$ Expanding and simplifying: $$50d^2 - 80d - 40 = 0$$ Dividing by 10: $$5d^2 - 8d - 4 = 0$$ Factoring gives: $$5d(d - 2) + 2(d - 2) = 0$$ Thus, $d = 2$ or $d = -\frac{2}{5}$. Since $d > 0$, $d = 2$. Substituting back: $$(a + 1) \hat{i} + (b + 2) \hat{j} + (c - 7) \hat{k} = 6 \hat{i} + 2 \hat{j} - 2 \hat{k}$$ Solving the equations: $$a + 1 = 6, b + 2 = 2, c - 7 = -2$$ We find: $$a = 5, b = 0, c = 5$$ Calculating the magnitudes: $$|AB| = \sqrt{1 + 4 + 49} = \sqrt{54}$$ $$|BC| = \sqrt{25 + 25} = \sqrt{50}$$ $$|AC| = \sqrt{86 + 4 + 4} = \sqrt{44}$$ The answer is 54.
Question 30
Maths · Integrals · Fill in the blank
If $\displaystyle\int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x\cos x}\,dx = \frac{1}{a}\log_e\left(\frac{a}{3}\right) + \frac{\pi}{b\sqrt{3}}$, where $a, b \in \mathbb{N}$, then $a+b$ is equal to _____
Answer: 8
Solution
Given the integral $$\int_0^{\pi/2} \frac{\sin^2 x}{1 + \frac{1}{2} \sin 2x} \, dx = \int_0^{\pi/4} \frac{1 - \cos 2x}{2 + \sin 2x} \, dx$$ we have $$\int \frac{1}{2 + \sin 2x} - \int \frac{\cos 2x}{2 + \sin 2x}$$ which is $$(I_1) - (I_2).$$ For $$(I_1) = \int \frac{dx}{2 + \frac{2 \tan x}{1 + \tan^2 x}}$$ we have $$\int_0^{\pi/4} \frac{\sec^2 x \, dx}{2 \tan^2 x + 2 \tan x + 2}.$$ Let $$\tan x = t.$$ Then $$\frac{1}{2} \int_0^1 \frac{dt}{\left(t + \frac{1}{2}\right)^2 + \frac{3}{4}} = \frac{\pi}{6\sqrt{3}}.$$ For $$I_2 = \int_0^{\pi/4} \frac{\cos 2x}{2 + \sin 2x} \, dx = \frac{1}{2} \left(\ln \frac{3}{2}\right).$$ Therefore, $$I_1 - I_2 = \frac{1}{\sqrt{3}} \frac{\pi}{6} + \frac{1}{2} \ln \frac{2}{3}.$$ Thus, $$a = 2, \ b = 6.$$
Physics
Question 31
Physics · Moving Charges and Magnetism · Single correct
An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then :
the electron will continue to move with uniform velocity along the axis of the solenoid.
the electron will be accelerated along the axis.
the electron path will be circular about the axis.
the electron will experience a force at 45^$\circ$ to the axis and execute a helical path.
Answer: (a)
Solution
Since $\vec{v} \parallel \vec{B}$ so force on electron due to magnetic field is zero. So it will move along axis with uniform velocity.
Question 32
Physics · Electromagnetic Waves · Single correct
The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos \omega (t - z/c) \mathrm{NC}^{-1}$. The magnetic field induction of this wave is (in SI unit):
Given $\vec{E} = \hat{i} 40 \cos \omega \left( t - \frac{z}{c} \right)$. $\vec{E}$ is along $+x$ direction. $\vec{v}$ is along $+z$ direction. So direction of $\vec{B}$ will be along $+y$ and magnitude of $B$ will be $\frac{E}{c}$. So answer is $\frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right) \hat{j}$.
Question 33
Physics · Nuclei · Single correct
Which of the following nuclear fragments corresponding to nuclear fission between neutron $\left( ^1_0 \mathrm{n} \right)$ and uranium isotope $\left( ^{235}_{92} \mathrm{U} \right)$ is correct:
Balancing mass number and atomic number $$^{235}_{92} \mathrm{U} + ^{1}_{0} \mathrm{n} \rightarrow ^{144}_{56} \mathrm{Ba} + ^{89}_{36} \mathrm{Kr} + 3 \, ^{1}_{0} \mathrm{n}$$
Question 34
Physics · Mathematics in Physics · Single correct
In an experiment to measure focal length $(f)$ of convex lens, the least counts of the measuring scales for the position of object $(u)$ and for the position of image $(v)$ are $\Delta u$ and $\Delta v$, respectively. The error in the measurement of the focal length of the convex lens will be:
Given $f^{-1} = v^{-1} - u^{-1}$. Differentiating, we have $-f^{-2} \, df = -v^{-2} \, dv - u^{-2} \, du$. Therefore, $$\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}.$$ Thus, $$df = f^2 \left[ \frac{dv}{v^2} + \frac{du}{u^2} \right].$$
Question 35
Physics · Mechanical Properties of Fluids · Single correct
Given below are two statements: Statement I : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation $P_1 - P_2 = \rho g \left( h_2 - h_1 \right)$. Statement II : In ventury tube shown $2gh = v_1^2 - v_2^2$ 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.
Statement I is correct but Statement II is incorrect.
Statement I is incorrect but Statement II is correct.
Both Statement I and Statement II are incorrect.
Answer: (b)
Solution
Applying Bernoulli's equation $$P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$$ $\($[h_1 $\&$ h_2 are height of point from any reference level]$\)$ Given $V_1 = V_2 = 0$ (for statement-1) $$\therefore P_1 - P_2 = g (h_2 - h_1)$$ For statement-2 $$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$$ $$P_1 - P_2 = \rho g h$$ $$P_1 - P_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$$ $$\rho g h = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$$ $$2 g h = v_2^2 - v_1^2$$
Question 36
Physics · Thermal Properties of Matter · Single correct
The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $8\,\Omega$ and $10\,\Omega$ respectively. After inserting in a hot bath of temperature $400^\circ\mathrm{C}$, the resistance of platinum wire is:
A metal wire of uniform mass density having length $L$ and mass $M$ is bent to form a semicircular arc and a particle of mass $m$ is placed at the centre of the arc. The gravitational force on the particle by the wire is :
Physics · Thermal Properties of Matter · Single correct
On celcius scale the temperature of body increases by $40^\circ \mathrm{C}$. The increase in temperature on Fahrenheit scale is:
$68^\circ \mathrm{F}$
$75^\circ \mathrm{F}$
$72^\circ \mathrm{F}$
$70^\circ \mathrm{F}$
Answer: (c)
Solution
We know that per $1^\circ \mathrm{C}$ change is equivalent to $1.8^\circ$ change in $^\circ \mathrm{F}$. Therefore, $40^\circ$ change on Celsius scale will correspond to $72^\circ$ change on Fahrenheit scale. Hence option (3)
Question 39
Physics · Ray Optics and Optical Instruments · Single correct
An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is 25D. Focal length of each of the convex lens is :
20 $\mathrm{\ cm}$
50 $\mathrm{\ cm}$
500 $\mathrm{\ cm}$
25 $\mathrm{\ cm}$
Answer: (a)
Solution
We know that $P_{eq} = \Sigma P_i$. Given all lenses are identical, $5P = 25D$. Therefore, $P = 5D$. Thus, $\frac{1}{f} = 5 \Rightarrow f = \frac{1}{5} \, \mathrm{m} = 20 \, \mathrm{cm}$. Hence option (1).
Question 40
Physics · Dual Nature of Radiation and Matter · Single correct
Which figure shows the correct variation of applied potential difference (V) with photoelectric current (I) at two different intensities of light ($I_1 < I_2$) of same wavelengths :
Answer: (b)
Solution
Given lights are of same wavelength. Hence stopping potential will remain same. Since $I_2 > I_1$, hence saturation current corresponding to $I_2$ will be greater than that corresponding to $I_1$.
Question 41
Physics · Laws of Motion · Single correct
A wooden block, initially at rest on the ground, is pushed by a force which increases linearly with time $t$. Which of the following curve best describes acceleration of the block with time:
Answer: (d)
Solution
Given $F = ma \Rightarrow a = \frac{F}{m} = \frac{kt}{m}$. $a$ vs $t$ will be a straight line passing through the origin.
Question 42
Physics · Work, Energy and Power · Single correct
If a rubber ball falls from a height $h$ and rebounds upto the height of $h/2$. The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are :
50$\%$, $\sqrt{2gh}$
50$\%$, $\sqrt{gh}$
40$\%$, $\sqrt{2gh}$
50$\%$, $\sqrt{\frac{gh}{2}}$
Answer: (a)
Solution
Velocity just before collision is $\sqrt{2gh}$. Velocity just after collision is $\sqrt{2g \left( \frac{h}{2} \right)}$. Therefore, $\Delta KE = \frac{1}{2} m(2gh) - \frac{1}{2} mgh = \frac{1}{2} mgh$. Therefore, the percentage loss in energy is $$\frac{\Delta KE}{KE_i} \times 100 = \frac{\frac{1}{2} mgh}{\frac{1}{2} mg2h} \times 100 = 50\%$$
Question 43
Physics · Physical World, Units and Measurements · Single correct
The equation of stationary wave is : $$y = 2a \sin\left(\frac{2\pi nt}{\lambda}\right) \cos\left(\frac{2\pi x}{\lambda}\right).$$ Which of the following is NOT correct :
The dimensions of $n/\lambda$ is $[T]$
The dimensions of $n$ is $[LT^{-1}]$
The dimensions of $x$ is $[L]$
The dimensions of $nt$ is $[L]$
Answer: (a)
Solution
Comparing the given equation with standard equation of standing $$\frac{2\pi n}{\lambda} = \omega$$ and $$\frac{2\pi}{\lambda} = k$$ $$\left[ \frac{n}{\lambda} \right] = [\omega] = \mathrm{T}^{-1}$$ $$[nt] = [\lambda] = \mathrm{L}$$ $$[n] = [\lambda \omega] = \mathrm{LT}^{-1}$$ $$[x] = [\lambda] = \mathrm{L}$$
Question 44
Physics · Motion in a Straight Line · Single correct
A body travels $102.5\,\mathrm{m}$ in $n^{\text{th}}$ second and $115.0\,\mathrm{m}$ in $(n+2)^{\text{th}}$ second. The acceleration is:
$9\,\mathrm{m/s^2}$
$6.25\,\mathrm{m/s^2}$
$12.5\,\mathrm{m/s^2}$
$5\,\mathrm{m/s^2}$
Answer: (b)
Solution
Question 45
Physics · Current Electricity · Single correct
To measure the internal resistance of a battery, potentiometer is used. For $R = 10\Omega$, the balance point is observed at $l = 500 \, \mathrm{cm}$ and for $R = 1\Omega$ the balance point is observed at $l = 400 \, \mathrm{cm}$. The internal resistance of the battery is approximately :
Physics · Electric Charges and Fields · Single correct
An infinitely long positively charged straight thread has a linear charge density $\lambda \mathrm{Cm}^{-1}$. An electron revolves along a circular path having axis along the length of the wire. The graph that correctly represents the variation of the kinetic energy of electron as a function of radius of circular path from the wire is:
Answer: (c)
Solution
Question 47
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The value of net resistance of the network as shown in the given figure is :
6$\Omega$
($\frac{5}{2}$)$\Omega$
($\frac{15}{4}$)$\Omega$
($\frac{30}{11}$)$\Omega$
Answer: (a)
Solution
Diode 2 is in reverse bias. So current will not flow in the branch of the 2nd diode, so we can assume it to be a broken wire. Diode 1 is in forward bias, so it will behave like a conducting wire. So the new circuit will be $$R_{eq} = \frac{15 \times 10}{15 + 10} = \frac{15 \times 10}{25} = 6 \, \Omega$$
Question 48
Physics · Thermodynamics · Single correct
P-T diagram of an ideal gas having three different densities $\rho_1, \rho_2, \rho_3$ (in three different cases) is shown in the figure. Which of the following is correct:
$\rho_1 > \rho_2$
$\rho_2 < \rho_3$
$\rho_1 = \rho_2 = \rho_3$
$\rho_1 < \rho_2$
Answer: (a)
Solution
For ideal gas $$PV = nRT$$ $$PV = \frac{m}{M} RT$$ $$P = \left( \frac{M}{V} \right) \frac{RT}{M}$$ $$P = \frac{\rho RT}{M}$$ (Where $m$ is mass of gas and $M$ is molecular mass of gas) For same temperature $P_1 > P_2 > P_3$ So $\rho_1 > \rho_2 > \rho_3$
Question 49
Physics · Motion in a Plane · Single correct
The co-ordinates of a particle moving in $x - y$ plane are given by : $x = 2 + 4t, y = 3t + 8t^2$. The motion of the particle is :
uniformly accelerated having motion along a parabolic path.
uniform motion along a straight line.
uniformly accelerated having motion along a straight line.
non-uniformly accelerated.
Answer: (a)
Solution
Given $$x = 2 + 4t$$ $$\frac{dx}{dt} = v_x = 4$$ $$\frac{dv_x}{dt} = a_x = 0$$ $$y = 3t + 8t^2$$ $$\frac{dy}{dt} = v_y = 3 + 16t$$ $$\frac{dv_y}{dt} = a_y = 16$$ The motion will be uniformly accelerated motion. For path $$x = 2 + 4t$$ $$\frac{(x - 2)}{4} = t$$ Put this value of $t$ in equation of $y$ $$y = 3 \left(\frac{x-2}{4}\right) + 8 \left(\frac{x-2}{4}\right)^2$$ This is a quadratic equation so path will be parabola.
Question 50
Physics · Alternating Current · Single correct
In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to: \begin{enumerate} \item[(A)] pure inductor. \item[(B)] pure capacitor. \item[(C)] pure resistor. \item[(D)] combination of an inductor and capacitor. \end{enumerate} Choose the correct answer from the options given below:
A, B and C only
A and B only
B, C and D only
A, B and D only
Answer: (d)
Solution
This is possible when phase difference is $\frac{\pi}{2}$ between current and voltage so correct answer will be (4).
Question 51
Physics · Electric Charges and Fields · Numerical
An infinite plane sheet of charge having uniform surface charge density $+\sigma_s \, \mathrm{C/m^2}$ is placed on $x-y$ plane. Another infinitely long line charge having uniform linear charge density $+\lambda_e \, \mathrm{C/m}$ is placed at $z = 4 \, \mathrm{m}$ plane and parallel to $y$-axis. If the magnitude values $|\sigma_s| = 2 \, |\lambda_e|$ then at point $(0, 0, 2)$, the ratio of magnitudes of electric field values due to sheet charge to that of line charge is $\pi \sqrt{n} : 1$. The value of $n$ is _______.
Answer: 16
Solution
The ratio of $E_S$ to $E_\ell$ is given by $$\frac{E_S}{E_\ell} = \frac{\sigma}{2\epsilon_0} \times \frac{2\pi \epsilon_0 r}{\lambda}$$ Simplifying, we have $$= \frac{\pi \times \sigma r}{\lambda}$$ Further simplifying, $$= \frac{\pi \times 2 \lambda \times 2}{\lambda} = \frac{4\pi}{1}$$ Therefore, $n = 16$.
Question 52
Physics · Atoms · Numerical
A hydrogen atom changes its state from $n = 3$ to $n = 2$. Due to recoil, the percentage change in the wavelength of emitted light is approximately $1 \times 10^{-n}$. The value of $n$ is_____. [Given $\mathrm{Rhc}$ = 13.6 $\mathrm{eV}$, $\mathrm{hc}$ = 1242 $\mathrm{eVnm}$, $\mathrm{h}$ = 6.6 $\times$ $10^{-34}$ $\mathrm{Js}$ mass of the hydrogen atom = 1.6 $\times$ $10^{-27}$ $\mathrm{kg}$ ]
Physics · Moving Charges and Magnetism · Numerical
The magnetic field existing in a region is given by $\vec{B} = 0.2(1 + 2x)\hat{k}$ T. A square loop of edge 50 cm carrying 0.5 A current is placed in $x-y$ plane with its edges parallel to the $x-y$ axes, as shown in figure. The magnitude of the net magnetic force experienced by the loop is_____ mN.
Answer: 50
Solution
Force on segment parallel to x-axis will cancel each other. Hence $F_{net}$ will be due to portion parallel to y-axis. $$F = 0.5 \times 0.5 \times 6 \times 0.2 - 0.5 \times 0.5 \times 0.2 \times 5$$ $$= 0.5 \times 0.5 \times 0.2$$ $$= 0.25 \times 0.2$$ $$= 50 \times 10^{-3} \, \mathrm{N}$$ $$= 50 \, \mathrm{mN}$$
Question 54
Physics · Alternating Current · Numerical
A alternating current at any instant is given by $i = [6 + \sqrt{56} \sin(100 \pi t + \pi/3)] \, \mathrm{A}$. The $rms$ value of the current is ________ A.
Answer: 8
Solution
Given $$I_{DNs} = \sqrt{\frac{\int i^2 \, dt}{\int dt}}$$ We have $$I_{Dns} = \sqrt{\frac{(6)^2 + (\sqrt{56})^2}{2}}$$ Simplifying, $$= \sqrt{36 + 28}$$ $$= \sqrt{64}$$ $$= 8 \, \mathrm{A}$$
Question 55
Physics · Current Electricity · Numerical
Twelve wires each having resistance $2\,\Omega$ are joined to form a cube. A battery of $6\,\mathrm{V}$ emf is joined across point $a$ and $c$. The voltage difference between $e$ and $f$ is _____ $\mathrm{V}$.
Answer: 1
Solution
From symmetry, current through e-b and g-d is 0. Therefore, $R_{eq} = \frac{3}{4} \times R = \frac{3}{2} \, \Omega$. Therefore, current through battery $= \frac{6 \times 2}{3} = 4 \, A$. $i_2 = \frac{4}{8} \times 2 = 1 \, A$. Therefore, $\Delta V$ across e-f $= \frac{i_2}{2} \times R = \frac{1}{2} \times 2 = 1 \, V$.
Question 56
Physics · Mechanical Properties of Fluids · Numerical
A soap bubble is blown to a diameter of 7 cm. 36960 $\mathrm{erg}$ of work is done in blowing it further. If surface tension of soap solution is 40 $\mathrm{dyne/cm}$ then the new radius is ________ cm. Take ( $\pi$ = $\frac{22}{7}$ ) .
Two wavelengths $\lambda_1$ and $\lambda_2$ are used in Young's double slit experiment. $\lambda_1 = 450 \, \mathrm{nm}$ and $\lambda_2 = 650 \, \mathrm{nm}$. The minimum order of fringe produced by $\lambda_2$ which overlaps with the fringe produced by $\lambda_1$ is $n$. The value of $n$ is _____.
Physics · Mechanical Properties of Solids · Numerical
An elastic spring under tension of 3 N has a length $a$. Its length is $b$ under tension 2 N. For its length $(3a - 2b)$, the value of tension will be ______N.
Two forces $\vec{F}_1$ and $\vec{F}_2$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between $\vec{F}_1$ and $\vec{F}_2$ is $\cos^{-1}\left(\frac{1}{n}\right)$. The value of $|n|$ is .
Answer: 6
Solution
Given $|\vec{F}_1| = F$ and $|\vec{F}_R| = |\vec{F}_2| = 3F$. $$F_R^2 = F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta$$ $$9F^2 = F^2 + 9F^2 + 6F^2 \cos \theta$$ Solving for $\cos \theta$, we get: $$\cos \theta = -\frac{1}{6}$$ Therefore, $$\theta = \cos^{-1}\left(-\frac{1}{6}\right)$$ Given $n = -6$, we have $$|n| = 6$$
Question 60
Physics · System of Particles and Rotational Motion · Numerical
A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed $v$. The sphere and the cylinder reaches upto maximum heights $h_1$ and $h_2$, respectively, above the initial level. The ratio $h_1 : h_2$ is $\frac{n}{10}$. The value of $n$ is _____.
Answer: 7
Solution
Gain in P.E. = Loss in K.E. $$mgh = \frac{1}{2} mv^2 \left( 1 + \frac{K^2}{R^2} \right)$$ $$h \propto 1 + \frac{K^2}{R^2}$$ $$\frac{h_1}{h_2} = \frac{1 + \frac{2}{5}}{1 + 1} = \frac{7}{5 \times 2} = \frac{7}{10}$$ $$n = 7$$
Chemistry
Question 61
Chemistry · Electrochemistry · Single correct
What pressure (bar) of $\mathrm{H}_2$ would be required to make emf of hydrogen electrode zero in pure water at $25^\circ \mathrm{C}$?
$10^{-7}$
$0.5$
$1$
$10^{-14}$
Answer: (d)
Solution
The reaction is given by $$2e^- + 2\mathrm{H}^+ (\mathrm{aq}) \rightarrow \mathrm{H}_2 (\mathrm{g})$$ The Nernst equation is $$E = E^\circ - \frac{0.059}{n} \log \frac{P_{\mathrm{H}_2}}{[\mathrm{H}^+]^2}$$ Substituting the values, we have $$0 = 0 - \frac{0.059}{2} \log \frac{P_{\mathrm{H}_2}}{(10^{-7})^2}$$ Simplifying, $$\log \frac{P_{\mathrm{H}_2}}{(10^{-7})^2} = 0$$ Therefore, $$\frac{P_{\mathrm{H}_2}}{10^{-14}} = 1$$ Thus, $$P_{\mathrm{H}_2} = 10^{-14} bar$$
Question 62
Chemistry · Co-ordination Compounds · Single correct
The correct sequence of ligands in the order of decreasing field strength is :
According to spectrochemical series, ligand field strength is $\mathrm{CO} > \mathrm{H_2O} > \mathrm{F^-} > \mathrm{S^{2-}}$.
Question 63
Chemistry · Co-ordination Compounds · Single correct
Match List I with List II : Choose the correct answer from the options given below :
(A)- (IV), (B) - (III), $(C)$ - (I), (D) - (II)
(A)- (I), (B) - (II), $(C)$ - (IV), (D) - (III)
(A)- (III), (B) - (I), $(C)$ - (II), (D) - (IV)
(A)- (II), (B) - (IV), $(C)$ - (III), (D) - (I)
Answer: (a)
Solution
The first reaction shows a +R effect with $\mathrm{NH_2}$ as an electron donating group. The second reaction shows a -R effect with $\mathrm{NO_2}$ as an electron withdrawing group. The third reaction shows a +E effect with $\mathrm{H^+}$ as an electron deficient species. The fourth reaction shows a -E effect with $\mathrm{CN}$ as an electron deficient species.
Question 64
Chemistry · Equilibrium · Single correct
What will be the decreasing order of basic strength of the following conjugate bases? $\mathrm{OH^-},\ \mathrm{RO^-},\ \mathrm{CH_3COO^-},\ \mathrm{Cl^-}$
Strong acids have weak conjugate bases. Acidic strength: $$H - Cl > CH_3COOH > H_2O > R - OH$$ Conjugate base strength: $$Cl^- < CH_3COO^- < OH^- < RO^-$$
Question 65
Chemistry · Analytical Chemistry · Single correct
In the precipitation of the iron group (III) in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to :
increase concentration of $\mathrm{Cl}^-$ ions
increase concentration of $\mathrm{NH}_4^+$ ions
prevent interference by phosphate ions
decrease concentration of $\mathrm{OH}^-$ ions
Answer: (d)
Solution
Given the reactions: $$\mathrm{NH_4OH} \rightleftharpoons \mathrm{NH_4^+} + \mathrm{OH^-}$$ $$\mathrm{NH_4Cl} \rightarrow \mathrm{NH_4^+} + \mathrm{Cl^-}$$ Due to the common ion effect of $\mathrm{NH_4^+}$, $[\mathrm{OH^-}]$ decreases to such an extent that only group-III cation can be precipitated, due to their very low $K_{sp}$ in the range of $10^{-38}$.
Question 66
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Identify (B) and $(C)$ and how are (A) and $(C)$ related?
Answer: (c)
Solution
Compound (A) undergoes an elimination reaction (E2) with alcoholic NaOH to form compound (B). Compound (B) then reacts with HBr in ether through an electrophilic addition reaction to form compound (C). A and C are position isomers.
Question 67
Chemistry · Electrochemistry · Single correct
One of the commonly used electrode is calomel electrode. Under which of the following categories, calomel electrode comes?
Oxidation - Reduction electrodes
Metal ion - Metal electrodes
Gas - Ion electrodes
Metal - Insoluble Salt - Anion electrodes
Answer: (d)
Solution
Theory based
Question 68
Chemistry · Co-ordination Compounds · Single correct
Number of complexes from the following with even number of unpaired " d " electrons is $$[\mathrm{V(H_2O)_6}]^{3+}, \ [\mathrm{Cr(H_2O)_6}]^{2+}, \ [\mathrm{Fe(H_2O)_6}]^{3+}, \ [\mathrm{Ni(H_2O)_6}]^{3+}, \ [\mathrm{Cu(H_2O)_6}]^{2+}$$ [Given atomic numbers : $V = 23$, $Cr = 24$, $Fe = 26$, $Ni = 28$, $Cu = 29$]
2
1
4
5
Answer: (a)
Solution
$[\mathrm{V(H_2O)_6}]^{3+} \rightarrow d^2sp^3$ $_{23}\mathrm{V} : -[\mathrm{Ar}]\,3d^3\,4s^2$ $\mathrm{V}^{3+} : -[\mathrm{Ar}]\,3d^2,\; n=2$ (even number of unpaired $e^-$) --- $[\mathrm{Cr(H_2O)_6}]^{2+} \rightarrow sp^3d^2$ $_{24}\mathrm{Cr} : -[\mathrm{Ar}]\,3d^5\,4s^1$ $\mathrm{Cr}^{2+} : -[\mathrm{Ar}]\,3d^4,\; n=4$ (even number of unpaired $e^-$) \[ \begin{array}{c} e_g\\ \boxed{\uparrow}\boxed{\phantom{\uparrow}} \end{array} \] \[ \begin{array}{c} t_{2g}\\ \boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \end{array} \] --- $[\mathrm{Fe(H_2O)_6}]^{3+} \rightarrow sp^3d^2$ $\mathrm{Fe}^{3+} : -[\mathrm{Ar}]\,3d^5\,4s^0$ $n=5$ (odd number of unpaired $e^-$) --- $[\mathrm{Ni(H_2O)_6}]^{3+} \rightarrow sp^3d^2$ $\mathrm{Ni} : -[\mathrm{Ar}]\,3d^8\,4s^2$ $\mathrm{Ni}^{3+} : -[\mathrm{Ar}]\,3d^7,\; n=3$ (odd number of unpaired $e^-$) --- $[\mathrm{Cu(H_2O)_6}]^{2+} \rightarrow sp^3d^2$ $\mathrm{Cu} : -[\mathrm{Ar}]\,3d^9\,4s^0$ $n=1$ (odd number of unpaired $e^-$)
Question 69
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Which one of the following molecules has maximum dipole moment?
$\mathrm{NF}_3$
$\mathrm{CH}_4$
$\mathrm{PF}_5$
$\mathrm{NH}_3$
Answer: (d)
Solution
For $\mathrm{CH_4}$ and $\mathrm{PF_5}$, $\mu_{net} = 0$ (non polar). $$|\mu|_{\mathrm{NH_3}} > |\mu|_{\mathrm{NF_3}}$$ Vector addition of bond moment and lone pair moment is greater than vector subtraction of bond moment and lone pair moment.
Question 70
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Number of molecules/ions from the following in which the central atom is involved in $sp^3$ hybridization is _____ $\mathrm{NO_3^-}$, $\mathrm{BCl_3}$, $\mathrm{ClO_2^-}$, $\mathrm{ClO_3}$
4
3
2
1
Answer: (c)
Solution
Question 71
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Which among the following is incorrect statement?
Electromeric effect dominates over inductive effect
The electromeric effect is, temporary effect
Hydrogen ion ($H^+$) shows negative electromeric effect
The organic compound shows electromeric effect in the presence of the reagent only.
Answer: (c)
Solution
Hydrogen ion ($\mathrm{H}^+$) shows positive electromeric effect.
Question 72
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Given below are two statements: Statements I: Acidity of $\alpha$-hydrogens of aldehydes and ketones is responsible for Aldol reaction. Statement II: Reaction between benzaldehyde and ethanal will NOT give Cross - Aldol product. In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I is correct but Statement II is incorrect
Both Statement I and Statement II are correct
Both Statement I and Statement II are incorrect
Statement I is incorrect but Statement II is correct
Answer: (a)
Solution
Aldehyde and ketones having acidic $\alpha$-hydrogen show aldol reaction. Benzaldehyde reacts with ethanal in the presence of a base to form a cross aldol product.
Question 73
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Which of the following nitrogen containing compound does not give Lassaigne's test?
Urea
Phenyl hydrazine
Glycene
Hydrazine
Answer: (d)
Solution
Hydrazine ($\mathrm{NH_2 - NH_2}$) have no carbon so does not show Lassaigne's test.
Question 74
Chemistry · Biomolecules · Single correct
Which of the following is the correct structure of L-Glucose?
Answer: (a)
Solution
The structure of L-Glucose is shown with the following configuration: At the top, there is $\mathrm{CHO}$. The first hydroxyl group ($\mathrm{HO}$) is on the left. The second hydroxyl group ($\mathrm{HO}$) is also on the left. The third hydroxyl group ($\mathrm{HO}$) is on the left. The fourth hydroxyl group ($\mathrm{OH}$) is on the right. At the bottom, there is $\mathrm{CH_2OH}$.
Question 75
Chemistry · The d-and f-Block Elements · Single correct
The element which shows only one oxidation state other than its elemental form is :
Cobalt
Titanium
Nickel
Scandium
Answer: (d)
Solution
Co, Ti, Ni can show $+2$, $+3$ and $+4$ oxidation states. But 'Sc' only shows $+3$ stable oxidation state.
Question 76
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Identify the product in the following reaction:
Answer: (a)
Solution
The reaction shown is a Clemmensen reduction. It involves the reduction of a carbonyl group to a methylene group using zinc amalgam (Zn-Hg) and hydrochloric acid (HCl).
Question 77
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Number of elements from the following that CANNOT form compounds with valencies which match with their respective group valencies is _____. B, C, N, S, O, F, P, Al, Si
7
3
5
6
Answer: (b)
Solution
N, O, F can't extend their valencies up to their group number due to the non-availability of vacant 2 d like orbital.
Question 78
Chemistry · Solutions · Single correct
The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is : (Given : Molar Mass Na : 23 and Cl : 35.5 $\mathrm{g/mol}$ )
2
20
4
0.2
Answer: (d)
Solution
Molarity $M$ is given by the formula: $$M = \frac{n_{\mathrm{NaCl}}}{V_{\mathrm{sol}} (in L)}$$ Substituting the given values: $$M = \frac{5.85}{58.5}$$ $$M = \frac{1}{0.5} = 0.2 \, \mathrm{M}$$
Question 79
Chemistry · Haloalkanes and Haloarenes · Single correct
Identify the correct set of reagents or reaction conditions 'X' and 'Y' in the following set of transformation
X = dil.aq. NaOH, 20°C, Y = Br₂/CHCl₃
X = conc.alc. NaOH, 80°C, Y = Br₂/CHCl₃
X = dil.aq. NaOH, 20°C, Y = HBr /acetic acid
X = conc.alc. NaOH, 80°C, Y = HBr/ acetic acid
Answer: (d)
Solution
The reaction involves the conversion of $\mathrm{CH_3 - CH_2 - CH_2 - Br}$ to $\mathrm{CH_3 - CH = CH_2}$ using concentrated alcoholic $\mathrm{NaOH}$ (X). Then, $\mathrm{CH_3 - CH = CH_2}$ is converted to $\mathrm{CH_3 - CHBr - CH_3}$ using $\mathrm{HBr}$ in acetic acid (Y).
Question 80
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The correct order of first ionization enthalpy values of the following elements is : (A) O (B) N (C) Be (D) F (E) B Choose the correct answer from the options given below :
E < C < A < B < D
C < E < A < B < D
B < D < C < E < A
A < B < D < C < E
Answer: (a)
Solution
The correct order of first ionization energy (IE) is: $$\mathrm{Li < B < Be < C < O < N < F < Ne}$$ This corresponds to the order: $$\mathrm{E < C < A < B < D}$$
Question 81
Chemistry · Thermodynamics · Numerical
The enthalpy of formation of ethane ($C_2H_6$) from ethylene by addition of hydrogen where the bond-energies of $\mathrm{C-H}$, $\mathrm{C-C}$, $\mathrm{C=C}$, $\mathrm{H-H}$ are $414 \, \mathrm{kJ}$, $347 \, \mathrm{kJ}$, $615 \, \mathrm{kJ}$ and $435 \, \mathrm{kJ}$ respectively is kJ
Answer: 125
Solution
For the reaction $\mathrm{C_2H_4 (g) + H_2 (g) \rightarrow C_2H_6 (g)}$, the change in enthalpy $\Delta H$ is calculated as follows. $\n$ $\n$ $\Delta$ H = \mathrm{BE(C=C) + 4BE(C-H) + BE(H-H)}$$ $\n$$$- \mathrm{BE(C-C) - 6BE(C-H)}$$ $\n$$\nSimplifying$, $\n$ $\n$ $\Delta$ H = \mathrm{BE(C=C) + BE(H-H) - BE(C-C)}$$ $\n$$$- 2\mathrm{BE(C-H)}$$ $\n$$\nSubstituting$ the bond energies, $\n$ $\n$= 615 + 435 - 347 - 2 $\times$ 414 $\n$= -125 \, \mathrm{kJ}
Question 82
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical
The number of the correct reaction(s) among the following is
Answer: (a)
Solution
Option (A) involves the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl3, leading to the formation of a benzyl ketone, which is incorrect. Option (B) involves the reduction of benzoyl chloride to benzoic acid using hydrogen and Pd-BaSO4, which is incorrect. Option (C) involves the reaction of benzene with carbon monoxide and HCl in the presence of anhydrous AlCl3 and CuCl, leading to the formation of benzaldehyde, which is correct. Option (D) involves the hydrolysis of benzamide to aniline using $\mathrm{H_3O^+}$ and heat, which is incorrect.
Question 83
Chemistry · Amines · Numerical
X $\mathrm{g}$ of ethylamine is subjected to reaction with $\mathrm{NaNO_2/HCl}$ followed by water; evolved dinitrogen gas which occupied 2.24 $\mathrm{L}$ volume at STP. X is __ $\times 10^{-1}$ $\mathrm{g}$.
Answer: 45
Solution
Given: $\mathrm{N_2}$ evolved is $2.24 \, \mathrm{L}$ i.e. $0.1 \, mole$. i.e. $\mathrm{CH_3CH_2NH_2}$ (ethyl amine) will be $4.5 \, \mathrm{g}$ ($= 0.1 \, mole$). Hence the answer $= 45 \times 10^{-1} \, \mathrm{g}$.
Question 84
Chemistry · Structure of Atom · Numerical
The de-Broglie's wavelength of an electron in the 4th orbit is _______$\pi a_0 \cdot (a_0 = Bohr's radius)$
Answer: 8
Solution
Given $$2 \pi r_n = n \lambda_d$$ Substituting $$2 \pi a_0 \frac{n^2}{Z} = n \lambda_d$$ For $$2 \pi a_0 \frac{4^2}{1} = 4 \lambda_d$$ We find $$\lambda_d = 8 \pi a_0$$
Question 85
Chemistry · Redox Reactions · Numerical
Only 2 mL of $KMnO_4$ solution of unknown molarity is required to reach the end point of a titration of 20 $\mathrm{\, mL}$ of oxalic acid (2M) in acidic medium. The molarity of $KMnO_4$ solution should be____ $\mathrm{M.}$
Chemistry · The d-and f-Block Elements · Numerical
Consider the following reaction $$\mathrm{MnO_2 + KOH + O_2 \rightarrow A + H_2O}$$. Product ' A ' in neutral or acidic medium disproportionate to give products ' B ' and ' C ' along with water. The sum of spin-only magnetic moment values of B and C is ______$\mathrm{BM}$ . (nearest integer) (Given atomic number of Mn is 25)
Answer: 4
Solution
Given the reaction $\mathrm{MnO_2 + KOH + O_2 \rightarrow K_2MnO_4 + H_2O}$. In neutral or acidic solution, $\mathrm{K_2MnO_4}$ converts to $\mathrm{KMnO_4 + MnO_2}$. For $\mathrm{Mn^{+4}}$, the electronic configuration is $[\mathrm{Ar}] 3d^3$. The number of unpaired electrons $n = 3$, and the magnetic moment $\mu = \sqrt{3(3+2)} = 3.87 \, \mathrm{B.M.}$ The nearest integer is (4).
Question 87
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below. \[ A+B \xrightleftharpoons[\text{Step 3}]{\text{Step 1}} C \xrightarrow{\text{Step 2}} P \] Some details of the above reactions are listed below. \[ \begin{array}{ccc} \text{Step} & \text{Rate constant }(\mathrm{sec}^{-1}) & \text{Activation energy }(\mathrm{kJ\,mol}^{-1})\\ 1 & k_1 & 300\\ 2 & k_2 & 200\\ 3 & k_3 & E_{a3} \end{array} \] If the overall rate constant of the above transformation $(k)$ is given as \[ k=\frac{k_1k_2}{k_3} \] and the overall activation energy $(E_a)$ is $400\,\mathrm{kJ\,mol^{-1}}$, then the value of $E_{a3}$ is \[ \underline{\hspace{2cm}}\,\mathrm{kJ\,mol^{-1}} \] (nearest integer).
2.5 $\mathrm{g}$ of a non-volatile, non-electrolyte is dissolved in 100 $\mathrm{g}$ of water at $25^\circ$ $\mathrm{C}$. The solution showed a boiling point elevation by $2^\circ$ $\mathrm{C}$. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is _____ mm of Hg (nearest integer) [Given : Molal boiling point elevation constant of water ($K_b$) = 0.52 $\mathrm{K}$ $\cdot$ $\mathrm{kgmol}^{-1}$, 1 atm pressure = 760 mm of Hg, molar mass of water = 18 $\mathrm{g}$ $\cdot$ $\mathrm{mol}^{-1}$]
Answer: 707
Solution
Given $2 = 0.52 \times \mathrm{m}$. $$\mathrm{m} = \frac{2}{0.52}$$ According to the question, the solution is much diluted, so $$\frac{\Delta P}{P^\circ} = \frac{n_{solute}}{n_{solvent}}$$ $$\frac{\Delta P}{P^\circ} = \frac{\mathrm{m}}{1000} \times M_{solvent}$$ $$\Delta P = P^\circ \times \frac{\mathrm{m}}{1000} \times M_{solvent}$$ $$= 760 \times \frac{\frac{2}{0.52}}{1000} \times 18 = 52.615$$ $$P_5 = 760 - 52.615 = 707.385 \, mm of Hg$$
The number of different chain isomers for $\mathrm{C_7H_{16}}$ is
Answer: 9
Solution
Question 90
Chemistry · Chemical Bonding and Molecular Structure · Numerical
Number of molecules/species from the following having one unpaired electron is ____$\mathrm{O_2}$, $\mathrm{O_2^{-1}}$, $\mathrm{NO}$, $\mathrm{CN^{-1}}$, $\mathrm{O_2^{2-}}$
Answer: 2
Solution
According to M.O.T. $\mathrm{O_2} \rightarrow$ no. of unpaired electrons $= 2$ $\mathrm{O_2^-} \rightarrow$ no. of unpaired electron $= 1$ $\mathrm{NO} \rightarrow$ no. of unpaired electron $= 1$ $\mathrm{CN^-} \rightarrow$ no. of unpaired electron $= 0$ $\mathrm{O_2^{2-}} \rightarrow$ no. of unpaired electron $= 0$