JEE Main 4 April 2025 Shift 1 question paper with solutions
JEE Main 4 April 2025 Shift 1: all 75 questions in paper order (Maths, Physics, Chemistry) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Maths
Question 1
Maths · Relations and Functions · Single correct
Let $f, g : (1, \infty) \to \mathbb{R}$ be defined as $f(x) = \frac{2x+3}{5x+2}$ and $g(x) = \frac{2-3x}{1-x}$. If the range of the function $f \circ g : [2, 4] \to \mathbb{R}$ is $[\alpha, \beta]$, then $\frac{1}{\beta - \alpha}$ is equal to
Consider the sets $A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 = 25\}$, $B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + 9y^2 = 144\}$, $C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \leq 4\}$, and $D = A \cap B$. The total number of one-one functions from the set $D$ to the set $C$ is:
15120
19320
17160
18290
Answer: (c)
Solution
A: $x^2 + y^2 = 25$...(i) B: $\frac{x^2}{144} + \frac{y^2}{16} = 1$...(ii) C: $x^2 + y^2 \leq 4$...(iii) Solve (1) & (2) $$x^2 + 9(25 - x^2) = 144$$ $$-8x^2 = 144 - 225 = -81$$ $$x = \pm \frac{9}{2\sqrt{2}}$$ By (1) $\Rightarrow y = \pm \sqrt{25 - x^2}$ $$= \pm \sqrt{25 - \frac{81}{8}} = \pm \frac{\sqrt{119}}{2\sqrt{2}}$$ Therefore, $D = A \cap B =$ $$\left\{ \left( \frac{9}{2\sqrt{2}}, \frac{\sqrt{119}}{2\sqrt{2}} \right), \left( \frac{9}{2\sqrt{2}}, -\frac{\sqrt{119}}{2\sqrt{2}} \right), \left( -\frac{9}{2\sqrt{2}}, \frac{\sqrt{119}}{2\sqrt{2}} \right), \left( -\frac{9}{2\sqrt{2}}, -\frac{\sqrt{119}}{2\sqrt{2}} \right) \right\}$$ Number of elements in set $D = 4$ Therefore, $C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \leq 4\}$ $$= \{(0, 2), (2, 0), (0, -2), (-2, 0), (1, 1), (-1, -1), (1, -1), (-1, 1), (1, 0), (0, 1), (-1, 0), (0, -1), (0, 0)\}$$ Number of elements in set $C = 13$ Total number of one-one function from Set $D$ to set $C$ $\Rightarrow 13 \times 12 \times 11 \times 10 = 17160$
Question 3
Maths · Sets · Single correct
Let $A = \{1, 6, 11, 16, \ldots\}$ and $B = \{9, 16, 23, 30, \ldots\}$ be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n(A \cup B)$ is
For an integer $n \geq 2$, if the arithmetic mean of all coefficients in the binomial expansion of $(x+y)^{2n-3}$ is $16$, then the distance of the point $P\,(2n-1, n^2 - 4n)$ from the line $x+y=8$ is:
$\sqrt{2}$
$2\sqrt{2}$
$5\sqrt{2}$
$3\sqrt{2}$
Answer: (d)
Solution
Number of terms in $(x+y)^{(2n-3)} \implies \left[ \frac{(2n-3+1)}{(2n-2)} \right]$. Therefore, sum of all coefficients is $2^{2n-3}$. (Put $x = y = 1$) Therefore, arithmetic mean of all coefficients $$= \binom{2^{2n-3}}{2n-2} = 16$$ $$\Rightarrow 2^{2n-3} = 2^5(n-1) \Rightarrow n = 5$$ Therefore, $P(2n-1, n^2 - 4n) = (9, 5)$. Given $x+y=8$ Therefore, $$PM = \left| \frac{9+5-8}{\sqrt{2}} \right| = \frac{6}{\sqrt{2}} = \frac{3 \times 2}{\sqrt{2}} = 3\sqrt{2}$$
Question 5
Maths · Probability · Single correct
The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is:
Maths · Three Dimensional Geometry · Single correct
Let the shortest distance between the lines $\frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}$ and $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}$ be $3\sqrt{30}$. Then the positive value of $5\alpha + \beta$ is
If $\lim_{x \to 1^+} \frac{(x-1)(6 + \lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1$, where $\lambda, \mu \in \mathbb{R}$, then $\lambda + \mu$ is equal to
Let $f : [0, \infty) \to \mathbb{R}$ be differentiable function such that $f(x) = 1 - 2x + \int_0^x e^{x-t} f(t) \mathrm{d}t$ for all $x \in [0, \infty)$. Then the area of the region bounded by $y = f(x)$ and the coordinate axes is
Maths · Three Dimensional Geometry · Single correct
Let A and B be two distinct points on the line L: $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both A and B are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1,2,3)$ on the line L. If O is the origin, then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to:
Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function satisfying $f(0) = 1$ and $f(2x) - f(x) = x$ for all $x \in \mathbb{R}$. If $\lim_{n \to \infty} \left\{ f(x) - f\left( \frac{x}{2^n} \right) \right\} = G(x)$, then $\sum_{r=1}^{10} G\left( r^2 \right)$ is equal to
\[ 1 + 3 + 5^2 - 7 + 9^2 + \ldots \text{ up to 40 terms is equal to} \]
43890
41880
33980
40870
Answer: (b)
Solution
The given expression is $\left(1^2 + 5^2 + 9^2 + \ldots up to 20 terms \right) + \left(3 + 7 + 11 + \ldots up to 20 terms \right)$. We start by rewriting the expression as: $$= \sum_{r=1}^{20} (4r - 3)^2 + \sum_{r=1}^{20} (4r - 1)$$ Simplifying further: $$= \sum_{r=1}^{20} (4r - 3)^2 + (4r - 1)$$ Expanding the square: $$= 4 \sum_{r=1}^{20} (4r^2 - 5r + 2)$$ Breaking it down: $$= 16 \sum_{r=1}^{20} r^2 - 20 \sum_{r=1}^{20} r + 8 \sum_{r=1}^{20} 1 = 41880$$
Question 12
Maths · Binomial Theorem · Single correct
In the expansion of $\left( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \right)^n$, $n \in \mathbb{N}$, if the ratio of $15^{th}$ term from the beginning to the $15^{th}$ term from the end is $\frac{1}{6}$, then the value of $^nC_3$ is:
4060
1040
2300
4960
Answer: (c)
Solution
Given $T_{r+1} = \binom{n}{r} \left(2^{1/3}\right)^{n-r} \left(\frac{1}{3^{1/3}}\right)^r$. Let $r = 14$. Then, $$T_{15} = \binom{n}{14} \left(2^{1/3}\right)^{n-14} \left(\frac{1}{3^{1/3}}\right)^{14}$$ $T'_{15} = 15^{th}$ term from last is $(n-13)^{th}$ term from beginning. $$T'_{15} = \binom{n}{n-14} \left(2^{1/3}\right)^{14} \left(\frac{1}{3^{1/3}}\right)^{n-14}$$ Therefore, $$\frac{T_{15}}{T'_{15}} = \frac{\binom{n}{14} \left(2^{1/3}\right)^{n-14} \left(\frac{1}{3^{1/3}}\right)^{14}}{\binom{n}{n-14} \left(2^{1/3}\right)^{14} \left(\frac{1}{3^{1/3}}\right)^{n-14}} = \frac{1}{6}$$ $$= \left(2^{1/3}\right)^{n-28} \left(3^{1/3}\right)^{n-28} = \frac{1}{6}$$ $$= 6^{\frac{n-28}{3}} = 6^{-1}$$ So, $nC_3 = \binom{25}{3} = 2300$
Question 13
Maths · Inverse Trigonometric Functions · Single correct
Considering the principal values of the inverse trigonometric functions, $$\sin^{-1}\left(\frac{\sqrt{3}}{2}x + \frac{1}{2}\sqrt{1-x^2}\right), -\frac{1}{2} < x < \frac{1}{\sqrt{2}},$$ is equal to
$\frac{\pi}{4} + \sin^{-1} x$
$\frac{\pi}{6} + \sin^{-1} x$
$-\frac{5\pi}{6} - \sin^{-1} x$
$\frac{5\pi}{6} - \sin^{-1} x$
Answer: (b)
Solution
Given $$\sin^{-1}\left( \frac{\sqrt{3}}{2}x + \frac{1}{2}\sqrt{1-x^2} \right), -1 < x < \frac{1}{\sqrt{2}}$$. Let $$\sin^{-1}(x) = \theta$$ where $$-\frac{\pi}{6} < \theta < \frac{\pi}{4}$$. Then $$x = \sin \theta$$. Therefore, $$\sin^{-1}\left( \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta \right)$$. This simplifies to $$\sin^{-1}\left( \sin \left( \theta + \frac{\pi}{6} \right) \right) = \theta + \frac{\pi}{6}$$. Thus, $$\sin^{-1}(x) + \frac{\pi}{6}$$.
Question 14
Maths · Vector Algebra · Single correct
Consider two vectors $\vec{u} = 3\hat{i} - \hat{j}$ and $\vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k}, \lambda > 0$. The angle between them is given by $\cos^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$. Let $\vec{v} = \vec{v}_1 + \vec{v}_2$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$. Then the value $\left|\vec{v}_1\right|^2 + \left|\vec{v}_2\right|^2$ is equal to
$\frac{23}{2}$
14
$\frac{25}{2}$
10
Answer: (b)
Solution
Given $\vec{u} = 3\hat{i} - \hat{j}$, $\vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k}$. The dot product $\vec{u} \cdot \vec{v}$ is given by $$\frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|} = \cos \theta$$ This implies $$\frac{5}{\sqrt{10} \sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}}$$ Solving gives $\lambda^2 = 9$ which implies $\lambda = 3$ (since $\lambda > 0$). We can express $\vec{v}$ as $\vec{v} = \vec{v}_1 + \vec{v}_2$. Then $$\|\vec{v}\|^2 = \|\vec{v}_1\|^2 + \|\vec{v}_2\|^2 + 2 \vec{v}_1 \cdot \vec{v}_2$$ This simplifies to $$14 = \|\vec{v}_1\|^2 + \|\vec{v}_2\|^2 + 0 (\because \vec{v}_1 \perp \vec{v}_2)$$ Thus, $$\|\vec{v}_1\|^2 + \|\vec{v}_2\|^2 = 14$$
Question 15
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let the three sides of a triangle are on the lines $4x - 7y + 10 = 0$, $x + y = 5$ and $7x + 4y = 15$. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $x = 0$, $y = 0$ and $x + y = 1$ is
5
$\sqrt{5}$
$\sqrt{20}$
20
Answer: (b)
Solution
The given equations are $7x + 4y = 15$ and $4x - 7y + 10 = 0$. The point of intersection (P.O.I) is $B = (1, 2)$. The line $x + y = 1$ intersects the axes at $P$ and $Q$. The distance between $P$ and $B$ is $\sqrt{5}$.
Question 16
Maths · Integrals · Single correct
The value of $$\int_{-1}^{1} \frac{(1+\sqrt{|x|-x})e^x+(\sqrt{|x|-x})e^{-x}}{e^x+e^{-x}} \, dx$$ is equal to
The length of the latus-rectum of the ellipse, whose foci are (2, 5) and (2, -3) and eccentricity is $\frac{4}{5}$, is
$\frac{6}{5}$
$\frac{50}{3}$
$\frac{10}{3}$
$\frac{18}{5}$
Answer: (d)
Solution
Given $2be = 8$, we have $be = 4$. The coordinates of the foci are $F_1(2, 5)$ and $F_2(2, -3)$. Solving for $b$, we have: $$b \left( \frac{4}{5} \right) = 4 \implies b = 5$$ Thus, $c^2 = b^2 - a^2$. Solving for $a$: $$16 = 25 - a^2 \implies a = 3$$ The length of the latus rectum (L.R.) is given by: $$L.R. = \frac{2a^2}{b} = \frac{18}{5}$$
Question 18
Maths · Complex Numbers and Quadratic Equations · Single correct
Consider the equation $x^2 + 4x - n = 0$, where $n \in [20, 100]$ is a natural number. Then the number of all distinct values of $n$, for which the given equation has integral roots, is equal to
7
8
6
5
Answer: (c)
Solution
Given $x^2 + 4x + 4 = n + 4$. $(x + 2)^2 = n + 4$ $x = -2 \pm \sqrt{n + 4}$ Therefore, $20 \leq n \leq 100$ $\sqrt{24} \leq \sqrt{n + 4} \leq \sqrt{104}$ $\Rightarrow \sqrt{n + 4} \in \{5, 6, 7, 8, 9, 10\}$ Therefore, $6$ integral values of $n$ are possible.
Question 19
Maths · Probability · Single correct
A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let $X$ denote the number of defective pens. Then the variance of $X$ is
$\frac{11}{15}$
$\frac{28}{75}$
$\frac{2}{15}$
$\frac{3}{5}$
Answer: (b)
Solution
Given the probabilities for different values of $x$: For $x = 0$, $P(x) = \frac{{^7C_2}}{{^{10}C_2}}$. For $x = 1$, $P(x) = \frac{{^7C_1 \cdot ^3C_1}}{{^{10}C_2}}$. For $x = 2$, $P(x) = \frac{{^3C_2}}{{^{10}C_2}}$. The mean $\mu$ is calculated as $\mu = \Sigma x_i P(x_i) = 0 + \frac{7}{15} + \frac{2}{15} = \frac{3}{5}$. The variance is calculated as $Variance (x) = \Sigma P_i (x_i - \mu)^2 = \frac{28}{75}$.
Question 20
Maths · Trigonometric Functions · Single correct
If 10 $\sin^4 \theta$ + 15 $\cos^4 \theta$ = 6, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:
If the area of the region $\{(x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to ______.
Answer: 368
Solution
The area $A$ is given by the integral: $$A = \int_{1}^{25} 4\sqrt{x} \, dx - \frac{1}{2} \times 4 \times 4 - \frac{1}{2} \times 20 \times 20$$ Evaluating the integral, we have: $$A = \left[ \frac{4x^{3/2}}{3/2} \right]_{1}^{25} - 8 - 200$$ Simplifying, we get: $$A = \frac{8}{3} (125 - 1) - 208$$ This simplifies to: $$A = \frac{368}{3} \implies 3A = 368$$
Question 22
Maths · Matrices · Fill in the blank
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to .
Answer: 6
Solution
Since $A$ is an orthogonal matrix, $A^T = A^{-1}$. Therefore, $A^2 = A^{-1} \left( \because A^2 = A^T \right)$. This implies $A^3 = I$. Let $B = (A + I)^3 + (A - I)^3 - 6A$. Then, $B = 2 \left( A^3 + 3A \right) - 6A = 2A^3$. Therefore, $B = 2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$. Now, the sum of diagonal elements is $2 + 2 + 2 = 6$.
Question 23
Maths · Complex Numbers and Quadratic Equations · Fill in the blank
Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}, B = \{ z \in \mathbb{C} : \mathrm{Re}(z - iz) = 2 \}$ and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to ________.
Let C be the circle $x^2 + (y - 1)^2 = 2$, $E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on x-axis and y-axis respectively. Let the straight line $x + y = 3$ touch the curves $C$, $E_1$ and $E_2$ at $P(x_1, y_1)$, $Q(x_2, y_2)$ and $R(x_3, y_3)$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$, the value of $9(x_1y_1 + x_2y_2 + x_3y_3)$ is equal to .
Answer: 46
Solution
Let $E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $(a > b)$. $E_2 : \frac{x^2}{c^2} + \frac{y^2}{d^2} = 1$, $(c < d)$. $C : x^2 + (y - 1)^2 = 2$. Equation of tangent at $P(x_1, y_1)$ is $xx_1 + y(y_1 - 1) = (y_1 + 1)$. Comparing with $x + y = 3$, we get $P(1, 2)$. Therefore, the parametric equation of $x + y = 3$ is $$\begin{pmatrix} x - 1 \\ -1/\sqrt{2} \end{pmatrix} = \begin{pmatrix} y - 2 \\ 1/\sqrt{2} \end{pmatrix} = \pm \frac{2\sqrt{2}}{3} \left( \because PQ = \frac{2\sqrt{2}}{3} \right).$$ On solving, we get $Q \left( \frac{5}{3}, \frac{4}{3} \right)$, $R \left( \frac{1}{3}, \frac{8}{3} \right)$. So, $9(x_1y_1 + x_2y_2 + x_3y_3)$ is $$9 \left( 2 + \frac{5}{3} \times \frac{4}{3} + \frac{1}{3} \times \frac{8}{3} \right) \Rightarrow 46.$$
Question 25
Maths · Continuity and Differentiability · Numerical
Let $m$ and $n$ be the number of points at which the function $f(x) = \max \{x, x^3, x^5, \ldots, x^{21}\}$, $x \in \mathbb{R}$, is not differentiable and not continuous, respectively. Then $m + n$ is equal to .
Answer: 3
Solution
Given $$f(x) = \begin{cases} x, & x < -1 \\ x^{21}, & -1 \leq x < 0 \\ x, & 0 \leq x < 1 \\ x^{21}, & x \geq 1 \end{cases}$$ $f(x)$ is continuous everywhere. Therefore, $n = 0$. Differentiating, we have $$f'(x) = \begin{cases} 1, & x < -1 \\ 21x^{20}, & -1 \leq x < 0 \\ 1, & 0 < x < 1 \\ 21x^{20}, & x \geq 1 \end{cases}$$ Therefore, $f(x)$ is non-differentiable at $x = -1, 0, 1$. Thus, $m = 3$. Finally, $m + n = 3$.
Physics
Question 26
Physics · Kinetic Theory · Single correct
The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are $3 \times 10^{-7} \, \mathrm{m}$ and $600 \, \mathrm{m/s}$ respectively. Find the frequency of its collisions.
$2 \times 10^{10} \, /\mathrm{s}$
$9 \times 10^{5} \, /\mathrm{s}$
$2 \times 10^{9} \, /\mathrm{s}$
$5 \times 10^{8} \, /\mathrm{s}$
Answer: (c)
Solution
Frequency is given by the formula $$Frequency = \frac{1}{T} = \frac{V_{avg}}{\lambda}$$ where $T$ is the period, $V_{avg}$ is the average velocity, and $\lambda$ is the wavelength. Substituting the given values: $$= \frac{600}{3 \times 157} = 2 \times 10^9 \, sec^{-1}$$
Question 27
Physics · Dual Nature of Radiation and Matter · Single correct
A small mirror of mass $m$ is suspended by a massless thread of length $l$. Then the small angle through which the thread will be deflected when a short pulse of laser of energy $E$ falls normal on the mirror ($c =$ speed of light in vacuum and $g =$ acceleration due to gravity)
$\theta = \frac{3E}{4mc\sqrt{gl}}$
$\theta = \frac{E}{mc\sqrt{gl}}$
$\theta = \frac{E}{2mc\sqrt{gl}}$
$\theta = \frac{2E}{mc\sqrt{gl}}$
Answer: (d)
Solution
Force due to beam assuming complete reflection $F = \frac{2P}{C} = \frac{2}{C} \frac{dE}{dt}$; $P$ is power. So change in momentum of mirror. $$m(V - 0) = \int F dt = \frac{2}{C} \int dE = \frac{2E}{C}$$ Now using work energy theorem ...(1) $$W_g = \Delta k$$ $$-mg\ell(1 - \cos \theta) = 0 - \frac{1}{2} mv^2$$ $$g\ell \left(2 \sin^2 \frac{\theta}{2}\right) = \frac{v^2}{2}$$ as $\theta$ is small $$g\ell^2 \left(\frac{\theta}{2}\right)^2 = \frac{1}{2} \frac{4E^2}{m^2c^2}$$ $$g\ell\theta^2 = \frac{4E^2}{m^2c^2}$$ $$\theta = \frac{m c \sqrt{g\ell}}{2E}$$
Question 28
Physics · Mechanical Properties of Fluids · Single correct
Two liquids A and B have $\theta_A$ and $\theta_B$ as contact angles in a capillary tube. If $K = \cos \theta_A / \cos \theta_B$, then identify the correct statement:
$K$ is negative, then liquid $A$ and liquid $B$ have convex meniscus.
$K$ is negative, then liquid $A$ and liquid $B$ have concave meniscus.
$K$ is negative, then liquid $A$ has concave meniscus and liquid $B$ has convex meniscus.
$K$ is zero, then liquid $A$ has convex meniscus and liquid $B$ has concave meniscus.
Answer: (c)
Solution
Given $k = \frac{\cos \theta_A}{\cos \theta_B}$. It is negative when $\cos \theta_A$ and $\cos \theta_B$ are of opposite sign, so option (3).
Question 29
Physics · System of Particles and Rotational Motion · Multiple correct
Which of the following are correct expression for torque acting on a body? ( $\vec{r} = position vector; \vec{p} = linear momentum; \vec{L} = angular momentum; \vec{\alpha} = angular acceleration; I = moment of inertia; \vec{F} = force; t = time$ ) Choose the correct answer from the options given below :
B, D and E Only
C and D Only
B, C, D and E Only
A, B, D and E Only
Answer: (c)
Solution
Conceptual
Question 30
Physics · Wave Optics · Single correct
In a Young's double slit experiment, the slits are separated by 0.2 mm. If the slits separation is increased to 0.4 mm, the percentage change of the fringe width is:
0$\%$
100$\%$
50$\%$
25$\%$
Answer: (c)
Solution
Given $\beta = \frac{D \lambda}{d} \propto \frac{1}{d}$. If $d$ is doubled then $\beta$ is half so 50$\%$ decrement.
Question 31
Physics · Alternating Current · Single correct
An alternating current is represented by the equation, $i = 100\sqrt{2} \sin(100\pi t)$ ampere. The RMS value of current and the frequency of the given alternating current are
Given $i_r = \frac{i_0}{\sqrt{2}} = 100 \, \mathrm{A}$. Frequency $f = \frac{w}{2\pi} = \frac{100\pi}{2\pi} = 50 \, \mathrm{Hz}$.
Question 32
Physics · Waves · Single correct
Consider the sound wave travelling in ideal gases of He, CH$_4$, and CO$_2$. All the gases have the same ratio $\frac{P}{\rho}$, where $P$ is the pressure and $\rho$ is the density. The ratio of the speed of sound through the gases $v_{\text{He}} : v_{\text{CH}_4} : v_{\text{CO}_2}$ is given by
Physics · Physical World, Units and Measurements · Single correct
In an electromagnetic system, the quantity representing the ratio of electric flux and magnetic flux has dimension of $M^P L^Q T^R A^S$, where value of ' $Q$ ' and ' $R$ ' are
$(3, -5)$
$(-2, 2)$
$(-2, 1)$
$(1, -1)$
Answer: (d)
Solution
Given $\($ $\phi$_E = $\frac{EA}{BA}$ = $\frac{E}{B}$ $\)$ and $\($ $\phi$_M $\)$. $\($ B = M $\ell$ T^{-2} = ATLT^{-1} $\)$ So, $\[$ $\left$[ $\frac{E}{B}$ $\right$] = $\frac{ML^{-3}A^{-1}}{MT^{-2}A^{-1}}$ = LT^{-1} $\]$ Or $\($ E = c $\cdot$ B $\)$ $\[$ $\left$[ $\frac{E}{B}$ $\right$] = LT^{-1} $\]$ Hence, $\($ (4) $\)$
Question 34
Physics · Ray Optics and Optical Instruments · Single correct
When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$, the object is to be moved:
40 cm away from the mirror.
80 cm away from the mirror.
20 cm towards the mirror.
20 cm away from the mirror.
Answer: (a)
Solution
Given $m = \frac{1}{2} = \frac{f}{f-u}$. $$\frac{1}{2} = \frac{f}{f - (-40)}$$ $$f + 40 = 2f \Rightarrow f = 40 \, \mathrm{cm}$$ Now $m = \frac{1}{3} = \frac{40}{40-u}$. $$40 - u = 120 \Rightarrow u = -80$$
Question 35
Physics · Dual Nature of Radiation and Matter · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R : Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency. In the light of the above statements, choose the correct answer from the options given below
Both A and R are true but R is NOT the correct explanation of A
A is false but R is true
A is true but R is false
Both A and R are true and R is the correct explanation of A
Answer: (b)
Solution
Q6. $V_S = \frac{h\nu - \phi}{e}$ (2) so stopping potential doesn't depend on intensity $I = \frac{\eta h \nu}{A}$ On increasing intensity, number of photons per second $n$ increases, so the number of electrons.
Question 36
Physics · System of Particles and Rotational Motion · Single correct
If $\vec{L}$ and $\vec{P}$ represent the angular momentum and linear momentum respectively of a particle of mass 'm' having position vector $\vec{r} = a(\hat{i} \cos \omega t + \hat{j} \sin \omega t)$. The direction of force is
Opposite to the direction of $\vec{r}$
Opposite to the direction of $\vec{L}$
Opposite to the direction of $\vec{P}$
Opposite to the direction of $\vec{L} \times \vec{P}$
Answer: (a)
Solution
Given $\vec{a} = -\omega^2 \vec{r}$. Therefore, $\vec{F}$ is opposite to $\vec{r}$.
Question 37
Physics · Laws of Motion · Single correct
A body of mass $m$ is suspended by two strings making angles $\theta_1$ and $\theta_2$ with the horizontal ceiling with tensions $T_1$ and $T_2$ simultaneously. $T_1$ and $T_2$ are related by $T_1 = \sqrt{3} T_2$. The angles $\theta_1$ and $\theta_2$ are
$\theta_1 = 30^\circ$, $\theta_2 = 60^\circ$ with $T_2 = \frac{3mg}{4}$
$\theta_1 = 60^\circ$, $\theta_2 = 30^\circ$ with $T_2 = \frac{mg}{2}$
$\theta_1 = 45^\circ$, $\theta_2 = 45^\circ$ with $T_2 = \frac{3mg}{4}$
$\theta_1 = 30^\circ$, $\theta_2 = 60^\circ$ with $T_2 = \frac{4mg}{5}$
Answer: (b)
Solution
Given $T_1 \sin \theta_1 + T_2 \sin \theta_2 = mg$ and $T_1 = \sqrt{3} \, T_2$. Therefore, $$T_2 \left[ \sqrt{3} \sin \theta_1 + \sin \theta_2 \right] = mg.$$ For $\theta_1 = 60^\circ$ and $\theta_2 = 30^\circ$, $$T_2 = \frac{mg}{2}.$$
Question 38
Physics · Current Electricity · Single correct
Current passing through a wire as function of time is given as $I(t) = 0.02t + 0.01 \, \mathrm{A}$. The charge that will flow through the wire from $t = 1 \, \mathrm{s}$ to $t = 2 \, \mathrm{s}$ is:
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The kinetic energy needed to project a body of mass $m$ from earth surface to infinity is $\frac{1}{2} mgR$, where $R$ is the radius of earth. Reason R: The maximum potential energy of a body is zero when it is projected to infinity from earth surface. In the light of the above statements, choose the correct answer from the option given below
A False but R is true
Both A and R are true and R is the correct explanation of A
A is true but R is false
Both A and R are true but R is NOT the correct explanation of A
Answer: (a)
Solution
Given $KE = \frac{1}{2} m \left( \frac{2Gm}{R} \right) = mgR$. Assertion wrong at $\infty$. $U = 0$. Therefore, reason correct.
Question 40
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The Boolean expression $Y = A\overline{B}C + \overline{AC}$ can be realised with which of the following gate configurations. A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate, B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate C. 3-input OR gate, 3 NOT gates and one 2-input AND gate Choose the correct answer from the options given below
In an experiment with a closed organ pipe, it is filled with water by $\left( \frac{1}{5} \right)$th of its volume. The frequency of the fundamental note will change by
25$\%$
20$\%$
-20$\%$
-25$\%$
Answer: (a)
Solution
Given $\lambda_1 = 4\ell$. For $f_1$, we have: $$f_1 = \frac{v}{4\ell}$$ Given $\lambda_2 = \frac{16\ell}{5}$. For $f_2$, we have: $$f_2 = \frac{5}{5}V = \frac{v}{16\ell}$$ The change in frequency $\Delta f$ is given by: $$\frac{\Delta f}{f} = \left( \frac{1}{16} \right) \frac{v}{\ell} \times \frac{4\ell}{v} \times 100 = 25\%$$
Question 42
Physics · Oscillations · Single correct
Two simple pendulums having lengths $l_1$ and $l_2$ with negligible string mass undergo angular displacements $\theta_1$ and $\theta_2$, from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?
Physics · Electric Charges and Fields · Single correct
Two infinite identical charged sheets and a charged spherical body of charge density ' $\rho$ ' are arranged as shown in figure. Then the correct relation between the electrical fields at A, B, C and D points is:
Physics · Electric Charges and Fields · Single correct
Two small spherical balls of mass 10 g each with charges $-2\mu C$ and $2\mu C$, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite nonconducting charge sheet with uniform charge density of $100\mu C/m^2$ such that length of rod makes an angle of $30^\circ$ with electric field generated by charge sheet. Net torque acting on the rod is: (Take $\varepsilon_0 : 8.85 \times 10^{-12} C^2/Nm^2$)
112 Nm
1.12 Nm
2.24 Nm
11.2 Nm
Answer: (b)
Solution
Given $E = \frac{\sigma}{2\varepsilon_0}$. The torque $\tau$ is given by $\tau = PE \sin \theta$. Substituting the values, we have: $$\tau = \left[ \left(2 \times 10^{-6}\right) \left(\frac{2}{10}\right)\right] \left[ \frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \right] \left(\frac{1}{2}\right)$$ Simplifying, we get: $$\tau = \frac{10}{8.85} = 1.12 \, \mathrm{Nm}$$
Question 45
Physics · Atoms · Single correct
Considering the Bohr model of hydrogen like atoms, the ratio of the ratio of the radius $5^{th}$ orbit of the electron in $\mathrm{Li}^{2+}$ and $\mathrm{He}^{+}$ is
$\frac{3}{2}$
$\frac{4}{9}$
$\frac{9}{4}$
$\frac{2}{3}$
Answer: (d)
Solution
Given $r = r \cdot \frac{n^2}{Z}$. For $\mathrm{Li}^{2+}$, $$r_5 = r \cdot \frac{25}{3}$$ For $\mathrm{He}^+$, $$r_5 = r \cdot \frac{25}{2}$$ Therefore, $$\frac{r_{\mathrm{Li}^{2+}}}{r_{\mathrm{He}^+}} = \frac{2}{3}$$
Question 46
Physics · System of Particles and Rotational Motion · Numerical
A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is $\sqrt{\frac{x}{5}}$ where x =
Answer: 4
Solution
Applying Mechanical Energy conservation: $$k_i + U_i = k_f + U_f$$ $$\Rightarrow 0 + \mathrm{Mgh} = \frac{1}{2} mv^2 \left(1 + \frac{k^2}{R^2}\right) + 0$$ $$\Rightarrow V = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$$ So Ratio of velocities $$\frac{V_{Ring}}{V_{solids sphere}} = \sqrt{\frac{1 + \frac{2}{5}}{1 + 1}} = \sqrt{\frac{7}{10}}$$ $$x = 3.5$$ Rounding off $$x = 4$$
Question 47
Physics · Mechanical Properties of Solids · Numerical
Two slabs with square cross section of different materials (1, 2) with equal sides (l) and thickness $d_1$ and $d_2$ such that $d_2 = 2 \, d_1$ and $l > d_2$. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is $\theta_2 = 2 \theta_1$. If the shear moduli of material 1 is $4 \times 10^9 \, \mathrm{N/m^2}$, then shear moduli of material 2 is $x \times 10^9 \, \mathrm{N/m^2}$, where value of $x$ is ________.
Physics · Ray Optics and Optical Instruments · Numerical
Distance between object and its image (magnified by $-\frac{1}{3}$) is 30 cm. The focal length of the mirror used is $\left(\frac{x}{4}\right)$ cm, where magnitude of value of $x$ is _______.
Answer: 45
Solution
Given $M = -\frac{1}{3}$. $$\frac{-V}{-U} = \frac{-1}{3} \Rightarrow V = \frac{U}{3}$$ Distance between object and image: $$U - V = 30$$ $$U - \frac{U}{3} = 30$$ Solving gives $U = 45$ and $V = 15$. Using the lens formula: $$\frac{1}{f} = \frac{1}{V} + \frac{1}{U} = \frac{-1}{15} - \frac{1}{45}$$ This simplifies to: $$\Rightarrow F = \frac{45}{4}$$ Finally, $x = 45$.
Question 49
Physics · Electrostatic Potential and Capacitance · Fill in the blank
Four capacitor each of capacitance $16\mu \mathrm{F}$ are connected as shown in the figure. The capacitance between points A and B is : _____ (in $\mu \mathrm{F}$).
Answer: 64
Solution
Redrawing $$C_{eq} = 4C = 64$$
Question 50
Physics · Electromagnetic Induction · Fill in the blank
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of 10 cm/s, induced emf between points A and E is _____ mV.
Answer: 10
Solution
As field is uniform we can replace the bent wire with straight wire from A to B. So EMF: $$\varepsilon = Bv\ell_{AB}$$ $$= \frac{1}{\sqrt{2}} \times \frac{10 \, \mathrm{cm}}{5} \times 2 \left(10 \sin 45^\circ \right) \, \mathrm{cm}$$ $$\varepsilon = 10 \, \mathrm{mV}$$
Chemistry
Question 51
Chemistry · Solutions · Single correct
$XY$ is the membrane / partition between two chambers $1$ and $2$ containing sugar solutions of concentration $c_1$ and $c_2$ $(c_1 > c_2)$ moLL$^{-1}$. For the reverse osmosis to take place identify the correct condition (Here $p_1$ and $p_2$ are pressures applied on chamber $1$ and $2$) $(A)$ Membrane/Partition; Cellophane, $p_1 > \pi$ \newline $(B)$ Membrane/Partition; Porous, $p_2 > \pi$ \newline $(C)$ Membrane/Partition; Parchment paper, $p_1 > \pi$ \newline $(D)$ Membrane/Partition; Cellophane, $p_2 > \pi$ Choose the correct answer from the option given below:
$B$ and $D$ only
$A$ and $D$ only
$A$ and $C$ only
$C$ only
Answer: (c)
Solution
Normal osmosis occurs from $(2)$ to $(1)$ For reverse osmosis from $(1)$ to $(2)$ Pressure: $P_1 > \pi$ $\therefore$ Answer $[A \& C]$ only
Question 52
Chemistry · Thermodynamics · Single correct
Let us consider a reversible reaction at temperature, T. In this reaction, both $\Delta H$ and $\Delta S$ were observed to have positive values. If the equilibrium temperature is $T_e$, then the reaction becomes spontaneous at:
$T = T_e$
$T_e > T$
$T > T_e$
$T_e = 5T$
Answer: (c)
Solution
For reaction to be spontaneous according to 2nd law: $$\Delta G \left( \frac{\Delta H}{\Delta S} \right) = T_e$$ $$\Rightarrow T > T_e$$
Question 53
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Which of the following molecules(s) show/s paramagnetic behavior? (A) $\mathrm{O_2}$ (B) $\mathrm{N_2}$ $(C)$ $\mathrm{F_2}$ (D) $\mathrm{S_2}$ (E) $\mathrm{Cl_2}$ Choose the correct answer from the options given below:
B only
A & C only
A & E only
A & D only
Answer: (d)
Solution
\begin{tabular}{|c|c|c|} \hline & \textbf{ } & \textbf{No. of unpaired $e^-$} \\ \hline (A) & O$_2$ & 2 \\ \hline (B) & N$_2$ & 0 \\ \hline (C) & F$_2$ & 0 \\ \hline (D) & S$_2$ & 2 \\ \hline (E) & Cl$_2$ & 0 \\ \hline \end{tabular} If species contain unpaired electron than it is paramagnetic. So A & D are paramagnetic.
Question 54
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Aldol condensation is a popular and classical method to prepare $\alpha, \beta$-unsaturated carbonyl compounds. This reaction can be both intermolecular and intramolecular. Predict which one of the following is not a product of intramolecular aldol condensation?
Answer: (d)
Solution
The reactions shown are examples of aldol reactions. 1. The first reaction is an intramolecular aldol reaction where the compound undergoes cyclization in the presence of an aqueous base and heat to form a bicyclic compound. 2. The second reaction is also an intramolecular aldol reaction. The compound cyclizes to form a six-membered ring in the presence of an aqueous base and heat. 3. The third reaction is another example of an intramolecular aldol reaction, resulting in the formation of a bicyclic compound. 4. The fourth reaction is an intermolecular aldol reaction. Two different molecules react in the presence of an aqueous base and heat to form a new compound with a methyl group attached.
Question 55
Chemistry · Thermodynamics · Single correct
One mole of an ideal gas expands isothermally and reversibly from $10 \, \mathrm{dm}^3$ to $20 \, \mathrm{dm}^3$ at $300 \, \mathrm{K}$. $\Delta U$, $q$ and work done in the process respectively are: Given: $R = 8.3 \, \mathrm{JK}^{-1} \, \mathrm{mol}^{-1}$ $\ln 10 = 2.3$ $\log 2 = 0.30$ $\log 3 = 0.48$
$0, 21.84 \, \mathrm{kJ}, -1.26 \, \mathrm{kJ}$
$0, -17.18 \, \mathrm{kJ}, 1.718 \, \mathrm{J}$
$0, 21.84 \, \mathrm{kJ}, 21, 84 \, \mathrm{kJ}$
$0, 178 \, \mathrm{kJ}, -1.718 \, \mathrm{kJ}$
Answer: (d)
Solution
Given the process from $(10 \, \mathrm{L}, 300 \, \mathrm{K})$ to $(20 \, \mathrm{L}, 300 \, \mathrm{K})$ with $n=1$. The equation is $-q = w = -nRT \ln \frac{V_2}{V_1}$. Substituting the values, we have: $$-q = -8.3 \times 300 \times \ln \left( \frac{20}{10} \right)$$ This simplifies to: $$= -1.718 \, \mathrm{kJ}$$ Therefore, $q = 1.718 \, \mathrm{kJ}$ and $w = -1.718 \, \mathrm{kJ}$. The change in internal energy is $\Delta U = 0$ because $\Delta T = 0$.
Question 56
Chemistry · Co-ordination Compounds · Single correct
Which one of the following complexes will have $\Delta_0 = 0$ and $\mu = 5.96 \, \mathrm{B.M.}$?
$[\mathrm{Fe(CN)}_6]^{4-}$
$[\mathrm{Co(NH}_3)_6]^{3+}$
$[\mathrm{FeF}_6]^{4-}$
$[\mathrm{Mn(SCN)}_6]^{4-}$
Answer: (d)
Solution
For (1) $[\mathrm{Co(NH_3)_6}]^{3+}$, $\mathrm{Co}^{3+} \Rightarrow 3d^6 4s^0$. $\mathrm{NH_3}$ is a strong field ligand, $\mu = 0$. $$= [-0.4 \times 6 + 0.6 \times (0)] \Delta_0 = -2.4 \Delta_0$$ For (2) $[\mathrm{Mn(SCN)_6}]^{4-}$, $\mathrm{Mn}^{2+} \Rightarrow 3d^5 4s^0$. $\mathrm{SCN}$ is a weak field ligand. $$\mu = \sqrt{35} \, \mathrm{B.M.} = 5.96 \, \mathrm{B.M.}$$ $$\mathrm{CFSE} = (-0.4 \times 3 + 0.6 \times 2) \Delta_0$$ So $\Delta_0 = 0$. For (3) $[\mathrm{Fe(CN)_6}]^{4-}$, $\mathrm{Fe}^{2+} \Rightarrow 3d^6 4s^0$. $\mathrm{CN}$ is a strong field ligand (SFL), $\mu = 0$. $$\mathrm{CFSE} = -2.4 \Delta_0$$ For (4) $[\mathrm{FeF_6}]^{4-}$, $\mathrm{Fe}^{2+} \Rightarrow 3d^6 4s^0$. $$\mu = \sqrt{24} \, \mathrm{B.M.} = 4.89 \, \mathrm{B.M.}$$ $$\mathrm{CFSE} = (-0.4 \times 4 + 0.6 \times 2) \Delta_0 = -1.2 \Delta_0$$
Question 57
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
For $\mathrm{A_2} + \mathrm{B_2} \rightleftharpoons 2\mathrm{AB}$ $E_a$ for forward and backward reaction are $180$ and $200 \, \mathrm{kJ \, mol^{-1}}$ respectively If catalyst lowers $E_a$ for both reaction by $100 \, \mathrm{kJ \, mol^{-1}}$. Which of the following statement is correct?
Catalyst does not alter the Gibbs energy change of a reaction.
Catalyst can cause non-spontaneous reactions to occur.
The enthalpy change for the reaction is $+20 \, \mathrm{kJ \, mol^{-1}}$.
The enthalpy change for the catalysed reaction is different from that of uncatalysed reaction.
Answer: (a)
Solution
For the reaction $\mathrm{A_2 + B_2 \rightleftharpoons 2AB}$: $E_f = 180 \, \mathrm{kJ \, mol^{-1}}$ $E_b = 200 \, \mathrm{kJ \, mol^{-1}}$ $\Delta H = E_f - E_b = -20 \, \mathrm{kJ \, mol^{-1}}$ In presence of catalyst: $E_f = 180 - 100 = 80 \, \mathrm{kJ \, mol^{-1}}$ $E_b = 200 - 100 = 100 \, \mathrm{kJ \, mol^{-1}}$ Catalyst does not change $\Delta H$ or $\Delta G$ of a reaction.
Question 58
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
Rate law for a reaction between A and B is given by $$ R = k [\mathrm{A}]^n [\mathrm{B}]^m $$ If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $$ \left( \frac{r_2}{r_1} \right) $$ is
$2^{(n-m)}$
$(n-m)$
$(m+n)$
$\frac{1}{2^{m+n}}$
Answer: (a)
Solution
Given $r_1 = k[A]^n[B]^m$. Now A is doubled and B is halved in concentration. Therefore, $r_2 = k2^n[A]^n \cdot \frac{[B]^m}{2^m}$. Now $\frac{r_2}{r_1} = 2^{(n-m)}$.
Question 59
Chemistry · Co-ordination Compounds · Single correct
Number of stereoisomers possible for the complexes, $[\mathrm{CrCl}_3(\mathrm{py})_3]$ and $[\mathrm{CrCl}_2(\mathrm{ox})_2]^{3-}$ are respectively (py = pyridine, ox = oxalate)
3 $\&$ 3
2 $\&$ 2
2 $\&$ 3
1 $\&$ 2
Answer: (c)
Solution
So total stereo isomer = 2. Geometrical isomer = 2 (1 cis + 1 trans). Optical isomer = 3 (2 optically active + 1 optically inactive). Stereoisomer = 3.
Question 60
Chemistry · Amines · Single correct
The major product (A) formed in the following reaction sequence is
Answer: (a)
Solution
The reaction sequence starts with the reduction of nitrobenzene to aniline using $\mathrm{Sn, HCl}$. Aniline is then acetylated with acetic anhydride in the presence of pyridine to form acetanilide. Bromination of acetanilide with $\mathrm{Br_2, CH_3COOH}$ leads to the formation of bromoacetanilide. Finally, hydrolysis with $\mathrm{NaOH (aq.)}$ yields $p-bromo aniline$.
Question 61
Chemistry · Electrochemistry · Single correct
On charging the lead storage battery, the oxidation state of lead changes from $x_1$ to $y_1$ at the anode and from $x_2$ to $y_2$ at the cathode. The values of $x_1, y_1, x_2, y_2$ are respectively:
$+4, +2, 0, +2$
$+2, 0, +2, +4$
$0, +2, +4, +2$
$+2, 0, 0, +4$
Answer: (b)
Solution
For charging of lead storage battery cell reaction is $$2\mathrm{PbSO_4} (s) + 2\mathrm{H_2O}(l) \rightarrow \mathrm{Pb}(s) + \mathrm{PbO_2}(s) + 2\mathrm{H_2SO_4}(aq)$$ At anode $\mathrm{PbSO_4}$ reduced back to $\mathrm{Pb}$ and at cathode $\mathrm{PbSO_4}$ oxidised back to $\mathrm{PbO_2}$. Therefore, $x_1 = +2, y_1 = 0$ $x_2 = +2, y_2 = 4$
Question 62
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Given below are two statements: Statement I: Nitrogen forms oxides with $+1$ to $+5$ oxidation states due to the formation of $p\pi - p\pi$ bond with oxygen. Statement II: Nitrogen does not form halides with $+5$ oxidation state due to the absence of d-orbital in it. In the light of given statements, choose the correct answer from the options given below.
Statement I is true but Statement II is false
Both Statement I and Statement II are false
Statement I is false but Statement II is true
Both Statement I and Statement II are true
Answer: (d)
Solution
In oxide of nitrogen it can achieve +5 oxidation state because it can form $p\pi - p\pi$ bond with oxygen e.g. $\mathrm{N_2O_5}$. Nitrogen cannot form halide in +5 oxidation state because it does not contain d-orbital. e.g. $\mathrm{NX_5}$ does not exist. $X = halide$
Question 63
Chemistry · Hydrocarbons · Single correct
Benzene is treated with oleum to produce compound (X) which when further heated with molten sodium hydroxide followed by acidification produces compound (Y). The compound Y is treated with zinc metal to produce compound (Z). Identify the structure of compound (Z) from the following option.
Answer: (b)
Solution
Benzene reacts with oleum to form benzene sulphonic acid. This is then treated with NaOH to produce sodium phenoxide. Acidification with $\mathrm{H^+}$ yields phenol. Finally, phenol is reduced with Zn to regenerate benzene.
Question 64
Chemistry · Biomolecules · Single correct
Identify the pair of reactants that upon reaction, with elimination of HCl will give rise to the dipeptide Gly-Ala.
NH$_2$--CH$_2$--COCl and NH$_2$--CH--COOH CH$_3$
NH$_2$--CH$_2$--COCl and NH$_3^+$--CH--COCl CH$_3$
NH$_2$--CH$_2$--COOH and NH$_2$--CH--COCl CH$_3$
NH$_2$--CH$_2$--COOH and NH$_2$--CH--COOH CH$_3$
Answer: (a)
Solution
The reaction involves the nucleophilic substitution of the chlorine atom in $\mathrm{NH_2{-}CH_2{-}C(=O)Cl}$ by the amine group in $\mathrm{NH_2{-}CH{-}(CH_3){-}COOH}$. This forms an intermediate with a positive charge on the nitrogen atom. The chloride ion leaves, resulting in the formation of a peptide bond between glycine and alanine, producing $\mathrm{NH_2{-}CH_2{-}C(=O){-}NH{-}CH{-}(CH_3){-}COOH}$ and $\mathrm{HCl}$. The final product is Gly-Ala.
Question 65
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Given below are the pairs of group 13 element showing their relation in terms of atomic radius. $(\mathrm{B} < \mathrm{Al})$, $(\mathrm{Al} < \mathrm{Ga})$, $(\mathrm{Ga} < \mathrm{In})$ and $(\mathrm{In} < \mathrm{Tl})$ Identify the elements present in the incorrect pair and in that pair find out the element $(\mathrm{X})$ that has higher ionic radius $(\mathrm{M}^{3+})$ that the other one. The atomic number of the element $(\mathrm{X})$ is
31
49
13
81
Answer: (a)
Solution
Size order $\mathrm{Al} > \mathrm{Ga}$ $\mathrm{Al}^{3+} < \mathrm{Ga}^{3+}$ Atomic number of Ga is 31
Question 66
Chemistry · Alcohols, Phenols and Ethers · Single correct
An organic compound (X) with molecular formula $C_3H_6O$ is not readily oxidised. On reduction it gives $C_3H_8O$ (Y) which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are:
CH_3COCH_3, CH_3CH_2CH_2OH, CH_3CH(Br)CH_3
CH_3COCH_3, CH_3CH(OH)CH_3, CH_3CH(Br)CH_3
CH_3CH_2CHO, CH_3CH_2CH_2OH, CH_3CH_2CH_2Br
CH_3CH_2CHO, CH_3CH = CH_2, CH_3CH(Br)CH_3
Answer: (b)
Solution
The compound $\mathrm{CH_3C(O)CH_3}$ (X) undergoes reduction to form $\mathrm{CH_3CH(OH)CH_3}$ (Y). Upon reaction with HBr, it forms $\mathrm{CH_3CH(Br)CH_3}$ (Z). This compound reacts with magnesium in ether to form the Grignard reagent $\mathrm{CH_3CH(MgBr)CH_3}$. This Grignard reagent reacts with $\mathrm{CH_3C(O)CH_3}$ and undergoes hydrolysis to form $\mathrm{CH_3C(OH)(CH_3)_2CH_3}$, which is 2,3-dimethyl butan-2-ol.
Question 67
Chemistry · Haloalkanes and Haloarenes · Single correct
Predict the major product of the following reaction sequence :-
Answer: (b)
Solution
The reaction sequence starts with the bromination of methylcyclohexane using $\mathrm{Br_2/h\nu}$ to form bromomethylcyclohexane. This is followed by dehydrohalogenation using alcoholic $\mathrm{KOH}$ and heat $\Delta$ to form methylenecyclohexane. Finally, the addition of $\mathrm{HBr}$ in the presence of a peroxide $\mathrm{R-O-O-R}$ leads to the formation of 1-bromo-1-methylcyclohexane.
Question 68
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Given below are two statements. Statement I : The dipole moment of $\overset{4}{\mathrm{CH_3}}-\overset{3}{\mathrm{CH}}=\overset{2}{\mathrm{C}}-\overset{1}{\mathrm{H}}-\mathrm{O}=\mathrm{O}$ is greater than $\overset{4}{\mathrm{CH_3}}-\overset{3}{\mathrm{CH_2}}-\overset{2}{\mathrm{C}}-\overset{1}{\mathrm{CH}}=\mathrm{O}$ Statement II : $C_1$--$C_2$ bond length of $\underset{4}{\mathrm{CH_3}}-\underset{3}{\mathrm{CH}}=\underset{2}{\mathrm{CH}}-\underset{1}{\mathrm{CH}}=\mathrm{O}$ is greater than $C_1$--$C_2$ bond length of $\underset{4}{\mathrm{CH_3}}-\underset{3}{\mathrm{CH_2}}-\underset{2}{\mathrm{CH_2}}-\underset{1}{\mathrm{CH}}=\mathrm{O}$ In the light of the above statements, choose the correct answer from the options given below:
Statement I is false but Statement II is true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Both Statement I and Statement II are true
Answer: (c)
Solution
Statement-I: $$\mu = q \times d$$ More charges and more distance between charges than other compound so more dipole moment. Statement-I is true. Statement-II: $C_1 - C_2$ bond has partial double bond character that means lesser bond length than $C_1 - C_2$ bond of other compound. Statement-II is false.
Question 69
Chemistry · The d-and f-Block Elements · Single correct
Pair of transition metal ions having the same number of unpaired electrons is :
Which one of the following about an electron occupying the 1 s orbital in a hydrogen atom is incorrect ? (Bohr's radius is represented by $a_0$)
The probability density of finding the electron is maximum at the nucleus
The electron can be found at a distance $2a_0$ from the nucleus
The 1 s orbital is spherically symmetrical
The total energy of the electron is maximum when it is at a distance $a_0$ from the nucleus
Answer: (d)
Solution
The graph shows $\Psi^2$ as a function of $r$. 1. $\Psi^2$ = Probability density is maximum at nucleus. 2. Electron can exist up to infinity from nucleus. 3. True 4. Energy of electron is maximum at infinite distance from nucleus.
In Dumas' method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is _____ % (nearest integer). (Aqueous tension at 300 K = 15 $\mathrm{mmHg}$)
Chemistry · The d-and f-Block Elements · Numerical
$KMnO_4$ acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of X + Y is _____.
Answer: 10
Solution
$\overset{+7}{KMnO_4} \xrightarrow{\text{Acidic medium}} Mn^{2+}$ (O.A.) X is difference in oxidation state. $$6 \mathrm{CH_3COO^-} + \mathrm{Fe^{3+}} + \mathrm{H_2O} \rightarrow [\mathrm{Fe_3(OH)_2(CH_3COO)_6}]^{\oplus} + 2\mathrm{H^\oplus}$$ $$[\mathrm{Fe_3(OH)_2(CH_3COO)_6}]^{\oplus} (Brown red ppt) + 4\mathrm{H_2O} \rightarrow$$ $$[\mathrm{Fe(OH)_2(CH_3COO)}] + \mathrm{CH_3COOH} + \mathrm{H^\oplus}$$ $\mathrm{Fe^{3+}} \Rightarrow 3d^5 4s^0$ contains 5 d electrons. So $Y = 5$. $$X + Y = 5 + 5 = 10$$
Question 73
Chemistry · Some Basic Concepts of Chemistry · Numerical
Fortification of food with iron is done using $\mathrm{FeSO_4 \cdot 7H_2O}$. The mass in grams of the $\mathrm{FeSO_4 \cdot 7H_2O}$ required to achieve $12\text{ ppm}$ of iron in $150\text{ kg}$ of wheat is _________ (Nearest integer) \\ [\textbf{Given:} Molar mass of $\mathrm{Fe}$, $\mathrm{S}$, and $\mathrm{O}$ respectively are $56$, $32$, and $16\text{ g mol}^{-1}$]
Answer: 8.935
Solution
Let mass of iron be $w \, \mathrm{g}$. $$\frac{w}{150 \times 10^3} \times 10^6 = 12$$ $$\Rightarrow w = 150 \times 12 \times 10^{-3} = 1.8 \, \mathrm{gm}$$ Let mass of $\mathrm{FeSO_4 \cdot 7H_2O} = w_1 \, \mathrm{gm}$. $$\Rightarrow Moles of Fe = \frac{1.8}{56} = \left( \frac{w_1}{56 + 96 + 7 \times 18} \right)$$ $$\Rightarrow w_1 = 8.935 \, \mathrm{gm}$$
Question 74
Chemistry · Equilibrium · Numerical
The pH of a 0.01 M weak acid HX $(K_a = 4 \times 10^{-10})$ is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6. The new concentration of the diluted weak acid is given as $x \times 10^{-4}$ M. The value of $x$ is ________ (nearest integer)
Answer: 25
Solution
Given $\mathrm{HX_{(aq)}} \rightleftharpoons \mathrm{H^+_{(aq)}} + \mathrm{X^-_{(aq)}}$, $K_a = 4 \times 10^{-10}$. $0.01(1 - \alpha)$, $0.01\alpha$, $0.01\alpha$. Not justified. $\Rightarrow 0.01\alpha = 10^{-5} \Rightarrow \alpha = 10^{-3}$. $K_a = 0.01\alpha^2 = 10^{-8}$. On dilution let final concentration of HX = $cM$. $\mathrm{HX_{(an)}} \rightleftharpoons \mathrm{H^+_{(aq)}} + \mathrm{X^-_{(an)}}$. $C(1 - \alpha)$, $C\alpha$, $C\alpha$. $\Rightarrow C\alpha = 10^{-6} \ldots (1)$ $$\frac{C\alpha^2}{1 - \alpha} = K_a = 10^{-8} \ldots (2)$$ $$\Rightarrow \frac{10^{-6}\alpha}{1 - \alpha} = 10^{-8}$$ Data given is inconsistent $\&$ contradictory. This should be bonus.
Question 75
Chemistry · Biomolecules · Numerical
The total number of hydrogen bonds of a DNA-double Helix strand whose one strand has the following sequence of bases is_______. $5'$ - G - G - C - A - A - A - T - C - G - G - C - T - A - $3'$
Answer: 33
Solution
Two nucleic acid chains are wound about each other and held together by H bonds between pair of bases. Adenine from two hydrogen bonds with thymine and Guanine form three hydrogen bond with cytosine. 5'G - G - C - A - A - A - T - C - G - G - C - T - A - 3' In given DNA strand total seven guanine and cytosine bases which form total 21 H-bonds and six adenine and thymine base which will form total 12 H-bonds with other DNA strand. Total no. of H bonds = 7 $\times$ 3 + 6 $\times$ 2 = 33