JEE Advanced 4 June 2023 Paper 1 question paper with solutions
JEE Advanced 4 June 2023 Paper 1: all 51 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.
Register free to attempt this shift as a timed 180-minute test with instant scoring and chapter-wise analysis.
Maths
Question 1
Maths · Relations and Functions · Multiple correct
Let $S = (0, 1) \cup (1, 2) \cup (3, 4)$ and $T = \{0, 1, 2, 3\}$. Then which of the following statements is(are) true?
There are infinitely many functions from $S$ to $T$
There are infinitely many strictly increasing functions from $S$ to $T$
The number of continuous functions from $S$ to $T$ is at most 120
Every continuous function from $S$ to $T$ is differentiable
Answer: (a), (c), (d)
Solution
Given $S = (0, 1) \cup (1, 2) \cup (3, 4)$ and $T = \{0, 1, 2, 3\}$. Number of functions: Each element of $S$ has 4 choices. Let $n$ be the number of elements in set $S$. Number of functions $= 4^n$. Here $n \to \infty$. Therefore, Option (A) is correct. Option (B) is incorrect (obvious). For continuous functions, each interval will have 4 choices. Number of continuous functions $= 4 \times 4 \times 4 = 64$. Therefore, Option (C) is correct. Every continuous function is piecewise constant functions, hence differentiable. Therefore, Option (D) is correct.
Question 2
Maths · Conic Sections · Multiple correct
Let $\mathrm{T}_1$ and $\mathrm{T}_2$ be two distinct common tangents to the ellipse $\mathrm{E} : \frac{x}{6} + \frac{y}{3} = 1$ and the parabola $\mathrm{P} : y^2 = 12x$. Suppose that the tangent $\mathrm{T}_1$ touches $\mathrm{P}$ and $\mathrm{E}$ at the point $A_1$ and $A_2$, respectively and the tangent $\mathrm{T}_2$ touches $\mathrm{P}$ and $\mathrm{E}$ at the points $A_4$ and $A_3$, respectively. Then which of the following statements is(are) true?
The area of the quadrilateral $A_1A_2A_3A_4$ is 35 square units
The area of the quadrilateral $A_1A_2A_3A_4$ is 36 square units
The tangents $\mathrm{T}_1$ and $\mathrm{T}_2$ meet the x-axis at the point $(-3, 0)$
The tangents $\mathrm{T}_1$ and $\mathrm{T}_2$ meet the x-axis at the point $(-6, 0)$
Answer: (a), (c)
Solution
The equation of the line is $y = mx + \frac{3}{m}$. The equation for $C^2$ is $C^2 = a^2 m^2 + b^2$. Solving $\frac{9}{m^2} = 6m^2 + 3$ gives $m^2 = 1$. The lines $T_1$ and $T_2$ are $y = x + 3$ and $y = -x - 3$. These lines cut the x-axis at $(-3, 0)$. The points are $A_1(3, 6)$, $A_4(3, -6)$, $A_2(-2, 1)$, and $A_3(-2, -1)$. The distances are $A_1A_4 = 12$, $A_2A_3 = 2$, and $MN = 5$. The area is $\frac{1}{2} (12 + 2) \times 5 = 35 sq. unit$.
Question 3
Maths · Applications of Integrals · Multiple correct
Let $f : [0, 1] \rightarrow [0, 1]$ be the function defined by $f(x) = \frac{x}{3} - x^2 + \frac{5}{9} x + \frac{17}{36}$. Consider the square region $S = [0, 1] \times [0, 1]$. Let $G = \{(x, y) \in S : y > f(x)\}$ be called the green region and $R = \{(x, y) \in S : y < f(x)\}$ be called the red region. Let $L_h = \{(x, h) \in S : x \in [0, 1]\}$ be the horizontal line drawn at a height $h \in [0, 1]$. Then which of the following statements is(are) true?
There exists an $h \in \left[ \frac{1}{4}, \frac{2}{3} \right]$ such that the area of the green region above the line $L_h$ equals the area of the green region below the line $L_h$
There exists an $h \in \left[ \frac{1}{4}, \frac{2}{3} \right]$ such that the area of the red region above the line $L_h$ equals the area of the red region below the line $L_h$
There exists an $h \in \left[ \frac{1}{4}, \frac{2}{3} \right]$ such that the area of the green region above the line $L_h$ equals the area of the red region below the line $L_h$
There exists an $h \in \left[ \frac{1}{4}, \frac{2}{3} \right]$ such that the area of the red region above the line $L_h$ equals the area of the green region below the line $L_h$
Answer: (b), (c), (d)
Solution
Given $f(x) = \frac{x^3}{3} - x^2 + \frac{5x}{9} + \frac{17}{36}$. $f'(x) = x^2 - 2x + \frac{5}{9}$. $f'(x) = 0$ at $x = \frac{1}{3}$ in $[0, 1]$. $A_R =$ Area of Red region $A_G =$ Area of Green region $A_R = \int_{0}^{\frac{1}{3}} f(x) \, dx = \frac{1}{2}$. Total area $= 1$ Therefore, $A_G = \frac{1}{2}$. $\int_{0}^{1} f(x) \, dx = \frac{1}{2}$. $A_G = A_R$. $f(0) = \frac{17}{36}$. $f(1) = \frac{13}{36} \approx 0.36$. $f\left(\frac{1}{3}\right) = \frac{181}{324} \approx 0.558$. (A) Correct when $h = \frac{3}{4}$ but $h \in \left[\frac{1}{4}, \frac{2}{3}\right]$. Therefore, (A) is incorrect. (B) Correct when $h = \frac{1}{4}$. Therefore, (B) is correct. (C) When $h = \frac{181}{324}$, $A_R = \frac{1}{2}$, $A_G < \frac{1}{2}$. $h = \frac{13}{36}$, $A_R < \frac{1}{2}$, $A_G = \frac{1}{2}$. Therefore, $A_R = A_G$ for some $h \in \left(\frac{13}{36}, \frac{181}{324}\right)$. Therefore, (C) is correct. (D) Option (D) is remaining coloured part of option (C), hence option (D) is also correct.
Question 4
Maths · Limits and Derivatives · Single correct
Let $f : (0, 1) \to \mathbb{R}$ be the functions defined as $f(x) = \sqrt{n}$ if $x \in \left[ \frac{1}{n+1}, \frac{1}{n} \right)$ where $n \in \mathbb{N}$. Let $g : (0, 1) \to \mathbb{R}$ be a function such that $\int_{x^2}^{x} \sqrt{\frac{1-t}{t}} \, dt < g(x) < 2\sqrt{x}$ for all $x \in (0, 1)$. Then $\lim_{x \to 0} f(x)g(x)$
Maths · Three Dimensional Geometry · Single correct
Let Q be the cube with the set of vertices $\{(x_1, x_2, x_3) \in \mathbb{R}^3 : x_1, x_2, x_3 \{0, 1\}\}$. Let F be the set of all twelve lines containing the diagonals of the six faces of the cube Q. Let S be the set of all four lines containing the main diagonals of the cube Q; for instance, the line passing through the vertices $(0, 0, 0)$ and $(1, 1, 1)$ is in S. For lines $\ell_1$ and $\ell_2$, let $d(\ell_1, \ell_2)$ denote the shortest distance between them. Then the maximum value of $d(\ell_1, \ell_2)$, as $\ell_1$ varies over F and $\ell_2$ varies over S, is
$\frac{1}{\sqrt{6}}$
$\frac{1}{\sqrt{8}}$
$\frac{1}{\sqrt{3}}$
$\frac{1}{\sqrt{12}}$
Answer: (a)
Solution
DR'S of OG = 1, 1, 1 DR'S of AF = -1, 1, 1 DR'S of CE = 1, 1, -1 DR'S of BD = 1, -1, 1 Equation of OG $\Rightarrow$ $\frac{x}{1}$ = $\frac{y}{1}$ = $\frac{z}{1}$ Equation of AB $\Rightarrow$ $\frac{x-1}{1}$ = $\frac{y}{-1}$ = $\frac{z}{0}$ Normal to both the line's $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 0 \end{vmatrix} = \hat{i} + \hat{j} - 2\hat{k}$$ OA = $\hat{i}$ S.D. = $\frac{\left| \hat{i} (\hat{i} + \hat{j} - 2\hat{k}) \right|}{\left| \hat{i} + \hat{j} - 2\hat{k} \right|}$ = $\frac{1}{\sqrt{6}}$ Ans. (A)
Question 6
Maths · Probability · Single correct
Let $X = \left\{ (x, y) \in \mathbb{Z} \times \mathbb{Z} : \frac{x}{8} + \frac{y}{20} < 1 and y^2 < 5x \right\}$. Three distinct points P, Q and R are randomly chosen from X. Then the probability that P, Q and R form a triangle whose area is a positive integer, is
$\frac{71}{220}$
$\frac{73}{220}$
$\frac{79}{220}$
$\frac{83}{220}$
Answer: (b)
Solution
Given $\frac{x^2}{8} + \frac{y^2}{20} < 1$ and $y^2 < 5x$. Solving corresponding equations $$\frac{x^2}{8} + \frac{y^2}{20} = 1 \& y^2 = 5x$$ $$\Rightarrow \begin{cases} x = 2 \\ y = \pm \sqrt{10} \end{cases}$$ $$X = \{(1,1), (1,0), (1,-1), (1,2), (1,-2), (2,3), (2,2), (2,1), (2,0), (2,-1), (2,-2), (2,-3)\}$$ Let $S$ be the sample space and $E$ be the event $n(S) = \binom{12}{3}$. For $E$, selecting 3 points in which 2 points are either $x = 1$ or $x = 2$ but the distance between them is even. Triangles with base 2: $$= 3 \times 7 + 5 \times 5 = 46$$ Triangles with base 4: $$= 1 \times 7 + 3 \times 5 = 22$$ Triangles with base 6: $$= 1 \times 5 = 5$$ $$P(E) = \frac{46 + 22 + 5}{\binom{12}{3}} = \frac{73}{220}$$ Ans. (B)
Question 7
Maths · Conic Sections · Single correct
Let P be a point on the parabola $y^2 = 4ax$, where $a > 0$. The normal to the parabola at P meets the x-axis at a point Q. The area of the triangle PFQ, where F is the focus of the parabola, is 120. If the slope $m$ of the normal and $a$ are both positive integers, then the pair $(a,m)$ is
(2, 3)
(1, 3)
(2, 4)
(3, 4)
Answer: (a)
Solution
Let point $P (at^2, 2at)$. The normal at $P$ is $y = -tx + 2at + at^3$. For $y = 0$, $x = 2a + at^2$. Therefore, $Q(2a + at^2, 0)$. The area of $\triangle PFQ = \frac{1}{2} (a + at^2)(2at) = 120$. Thus, $m = -t$. Therefore, $a^2 [1 + m^2] m = 120$. The pair $(a, m) = (2, 3)$ will satisfy.
Let $\tan^{-1}(x) \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, for $x \in \mathbb{R}$. Then the number of real solutions of the equation $$\sqrt{1 + \cos(2x)} = \sqrt{2} \tan^{-1}(\tan x)$$ in the set $\left( -\frac{3\pi}{2}, -\frac{\pi}{2} \right) \cup \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \cup \left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$ is equal to
Answer: 3
Solution
Given $\sqrt{2} |\cos x| = \sqrt{2} \cdot \tan^{-1}(\tan x)$. Therefore, $|\cos x| = \tan^{-1}(\tan x)$. The number of solutions is 3.
Question 9
Maths · Applications of Integrals · Numerical
Let $n \geq 2$ be a natural number and $f : [0,1] \to \mathbb{R}$ be the function defined by $$f(x) = \begin{cases} n(1-2nx) & if 0 \leq x \leq \frac{1}{2n} \\ 2n(2nx-1) & if \frac{1}{2n} \leq x \leq \frac{3}{4n} \\ 4n(1-nx) & if \frac{3}{4n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(nx-1) & if \frac{1}{n} \leq x \leq 1 \end{cases}$$ If $n$ is such that the area of the region bounded by the curves $x = 0$, $x = 1$, $y = 0$ and $y = f(x)$ is $4$, then the maximum value of the function $f$ is
Answer: 8
Solution
Area = Area of (I + II + III) = 4 $$= \frac{1}{2} \times \frac{1}{2n} \times n + \frac{1}{2} \times \frac{1}{2n} \times n + \frac{1}{2} \left(1 - \frac{1}{n}\right) \times n$$ $$= \frac{1}{4} + \frac{1}{4} + \frac{n-1}{2} = 4$$ Therefore, $n = 8$. Thus, the maximum value of $f(x) = 8$.
Question 10
Maths · Sequences and Series · Numerical
Let $75\ldots57$ denote the $(r + 2)$ digit number where the first and the last digits are 7 and the remaining $r$ digits are 5. Consider the sum $S = 77 + 757 + 7557 + \ldots + 75\ldots57$. If $S = \frac{75\ldots57 + m}{n}$, where $m$ and $n$ are natural numbers less than 3000, then the value of $m + n$ is
Answer: 1219
Solution
Given $S = 77 + 757 + 7557 + \ldots + 75\ldots57$ with 98 terms. Then $10S = 770 + 7570 + \ldots + 75\ldots570$. Subtracting these, we have: $$9S = -77 + \underbrace{13 + 13 + \ldots + 13}_{98 times} + 75\ldots570$$ This simplifies to: $$9S = -77 + 13 \times 98 + 75\ldots57 + 13$$ Therefore: $$S = \frac{75\ldots57 + 1210}{9}$$ Where $m = 1210$ and $n = 9$. Thus, $m + n = 1219$.
Question 11
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $A = \left\{ \frac{1967 + 1686i \sin \theta}{7 - 3i \cos \theta} : \theta \in \mathbb{R} \right\}$. If $A$ contains exactly one positive integer $n$, then the value of $n$ is
Let P be the plane $\sqrt{3}x + 2y + 3z = 16$ and let $$S = \left\{ \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} : \alpha^2 + \beta^2 + \gamma^2 = 1 and the distance of (\alpha, \beta, \gamma) from the plane P is \frac{7}{2} \right\}.$$ Let $\vec{u}, \vec{v}$ and $\vec{w}$ be three distinct vectors in $S$ such that $|\vec{u} - \vec{v}| = |\vec{v} - \vec{w}| = |\vec{w} - \vec{u}|$. Let $V$ be the volume of the parallelepiped determined by vectors $\vec{u}, \vec{v}$ and $\vec{w}$. Then the value of $\frac{80}{\sqrt{3}} V$ is
Answer: 45
Solution
Given $|\mathbf{u} - \mathbf{v}| = |\mathbf{v} - \mathbf{w}| = |\mathbf{w} - \mathbf{u}|$ implies $\($ $\Delta$ UVW $\)$ is an equilateral triangle. Now distances of $U$, $V$, $W$ from $P = \frac{7}{2}$. Therefore, $PQ = \frac{7}{2}$. Also, Distance of plane $P$ from origin implies $OQ = 4$. Thus, $OP = OQ - PQ \Rightarrow OP = \frac{1}{2}$. Hence, $PU = \sqrt{OU^2 - OP^2} \Rightarrow PU = \frac{\sqrt{3}}{2} = R$. Also, for $\Delta UVW$, $P$ is the circumcenter. Therefore, for $\Delta UVW$: $US = R \cos 30^\circ$ implies $UV = 2R \cos 30^\circ$. Thus, $UV = \frac{3}{2}$. Therefore, $Ar(\Delta UVW) = \frac{\sqrt{3}}{4} \left( \frac{3}{2} \right)^2 = \frac{9\sqrt{3}}{16}$. Thus, Volume of tetrahedron with coterminous edges $\mathbf{u}, \mathbf{v}, \mathbf{w}$ is $$= \frac{1}{3} (Ar. \Delta UVW) \times OP = \frac{1}{3} \times \frac{9\sqrt{3}}{16} \times \frac{1}{2} = \frac{3\sqrt{3}}{32}$$ Thus, parallelepiped with coterminous edges $\mathbf{u}, \mathbf{v}, \mathbf{w} = 6 \times \frac{3\sqrt{3}}{32} = \frac{9\sqrt{3}}{16} = V$. Therefore, $\frac{80}{\sqrt{3}} V = 45$.
Question 13
Maths · Binomial Theorem · Numerical
Let a and b be two nonzero real numbers. If the coefficient of $x^5$ in the expansion of $\left(ax^2 + \frac{70}{27bx}\right)^4$ is equal to the coefficient of $x^{-5}$ is equal to the coefficient of $\left(ax - \frac{1}{bx^2}\right)^7$, then the value of $2b$ is
\[ \text{Let } \alpha,\beta,\gamma \text{ be real numbers. Consider the system of linear equations} \] $$ \begin{cases} x+2y+z=7,\\ x+\alpha z=11,\\ 2x-3y+\beta z=\gamma \end{cases} $$ Match each entry in **List-I** to the correct entry in **List-II**. The correct option is :
($P$) $\rightarrow$ (3) (Q) $\rightarrow$ (2) $(R)\\rightarrow\(1)$ (S) $\rightarrow$ (4)
($P$) $\rightarrow$ (3) (Q) $\rightarrow$ (2) $(R)\\rightarrow\(5)$ (S) $\rightarrow$ (4)
($P$) $\rightarrow$ (2) (Q) $\rightarrow$ (1) $(R)\\rightarrow\(4)$ (S) $\rightarrow$ (5)
($P$) $\rightarrow$ (2) (Q) $\rightarrow$ (1) $(R)\\rightarrow\(1)$ (S) $\rightarrow$ (3)
Answer: (a)
Question 15
Maths · Statistics · Single correct
Consider the given data with frequency distribution \[ \begin{array}{c|cccccccc} x_i & 3 & 8 & 11 & 10 & 5 & 4 \\ f_i & 5 & 2 & 3 & 2 & 4 & 4 \end{array} \] Match each entry in List-I to the correct entries in List-II. The correct option is:
Maths · Three Dimensional Geometry · Single correct
Let $\ell_1$ and $\ell_2$ be the lines $\vec{r}_1 = \lambda (\hat{i} + \hat{j} + \hat{k})$ and $\vec{r}_2 = (\hat{j} - \hat{k}) + \mu (\hat{i} + \hat{k})$, respectively. Let $X$ be the set of all the planes $H$ that contain the line $\ell_1$. For a plane $H$, let $d(H)$ denote the smallest possible distance between the points of $\ell_2$ and $H$. Let $H_0$ be plane in $X$ for which $d(H_0)$ is the maximum value of $d(H)$ as $H$ varies over all planes in $X$. Match each entry in List-I to the correct entries in List-II.
Let the system of planes be $ax + by + cz = 0$. Since it contains $L_1$, we have $a + b + c = 0$. For the largest possible distance between plane (1) and $L_2$, the line $L_2$ must be parallel to plane (1). Thus, $a + c = 0$, which implies $b = 0$. Therefore, the plane $H_0$ is $x - z = 0$. Now, $d(H_0)$ is the perpendicular distance from the point $(0, 1, -1)$ on $L_2$ to the plane. Thus, $d(H_0) = \frac{|0 + 1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$. Therefore, $P \to 5$. For $Q$, the distance is $\left| \frac{2}{\sqrt{2}} \right| = \sqrt{2}$. Thus, $Q \to 4$. The point $(0, 0, 0)$ lies on the plane. Thus, $R \to 3$. For $S$, $x = z$, $y = z$, $x = 1$. The point of intersection is $p(1, 1, 1)$. Therefore, $OP = \sqrt{1 + 1 + 1} = \sqrt{3}$. Thus, $S \to 2$. Option [B] is correct.
Question 17
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $z$ be complex number satisfying $|z|^3 + 2z^2 + 4\bar{z} - 8 = 0$, where $\bar{z}$ denotes the complex conjugate of $z$. Let the imaginary part of $z$ be nonzero. Match each entry in List-I to the correct entries in List-II.
(P) → (1) (Q) → (3) (R) → (5) (S) → (4)
(P) → (2) (Q) → (1) (R) → (3) (S) → (5)
(P) → (2) (Q) → (4) (R) → (5) (S) → (1)
(P) → (2) (Q) → (3) (R) → (5) (S) → (4)
Answer: (b)
Solution
Given $|z|^3 + 2z^2 + 4z - 8 = 0$ (1). Take conjugate both sides. $\Rightarrow |z|^3 + 2\overline{z}^2 + 4\overline{z} - 8 = 0$ (2). By (1) - (2), $\Rightarrow 2(z^2 - \overline{z}^2) + 4(z - \overline{z}) = 0$. $\Rightarrow z + \overline{z} = 2$ (3). $\Rightarrow |z + \overline{z}| = 2$ (4). Let $z = x + iy$. $x = 1$. $\therefore z = 1 + iy$. Put in (1), $\Rightarrow (1 + y^2)^{3/2} + 2(1 - y^2 + 2iy) + 4(1 - iy) - 8 = 0$. $\Rightarrow (1 + y^2)^{3/2} = 2(1 + y^2)$. $\Rightarrow \sqrt{1 + y^2} = 2 = |z|$. Also $y = \pm \sqrt{3}$. $\therefore z = 1 \pm i\sqrt{3}$. $\Rightarrow z - \overline{z} = \pm 2i\sqrt{3}$. $\Rightarrow |z - \overline{z}| = 2\sqrt{3}$. $\Rightarrow |z - \overline{z}|^2 = 12$. Now $z + 1 = 2 + i\sqrt{3}$. $|z + 1|^2 = 4 + 3 = 7$. $\therefore P \rightarrow 2; \ Q \rightarrow 1; \ R \rightarrow 3; \ S \rightarrow 5$. $\therefore$ Option [B] is correct.
Physics
Question 18
Physics · Motion in a Plane · Multiple correct
A slide with a frictionless curved surface, which becomes horizontal at its lower end,, is fixed on the terrace of a building of height $3h$ from the ground, as shown in the figure. A spherical ball of mass $m$ is released on the slide from rest at a height $h$ from the top of the terrace. The ball leaves the slide with a velocity $\vec{u}_0 = u_0 \hat{x}$ and falls on the ground at a distance $d$ from the building making an angle $\theta$ with the horizontal. It bounces off with a velocity $\vec{v}$ and reaches a maximum height $h_1$. The acceleration due to gravity is $g$ and the coefficient of restitution of the ground is $1/\sqrt{3}$. Which of the following statement(s) is(are) correct?
$u_0 = \sqrt{2gh} \hat{x}$
$\vec{v} = \sqrt{2gh} (\hat{x} - \hat{z})$
$\theta = 60^\circ$
$d/h_1 = 2\sqrt{3}$
Answer: (a), (c), (d)
Solution
The initial velocity $\vec{v}_1 = \sqrt{2gh} \, \hat{i} - \sqrt{2g3h} \, \hat{k}$. The velocity $\vec{v} = \sqrt{2gh} \, \hat{i} + \sqrt{2g3h} \times \frac{1}{\sqrt{3}} \, \hat{k}$. This simplifies to $\sqrt{2gh} \, \hat{i} + \sqrt{2gh} \, \hat{k}$. The tangent of the angle $\theta$ is given by $\tan \theta = \frac{\sqrt{2g3h}}{\sqrt{2gh}} = \sqrt{3}$, which implies $\theta = 60^\circ$. The height $h_1$ is $h_1 = \frac{v_{1y}^2}{2g} = \frac{2gh}{2g} = h$. The distance $d$ is $d = v_x \, t = \sqrt{2gh} \times \sqrt{\frac{2 \times 3h}{g}}$. This simplifies to $\sqrt{2gh} \sqrt{\frac{6h}{g}} = 2\sqrt{3h}$. Therefore, $\frac{d}{h_1} = 2\sqrt{3}$.
Question 19
Physics · Ray Optics and Optical Instruments · Multiple correct
A plane polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is $\delta = 60^\circ$ (see Figure-1). The angle of minimum deviation for red light from the same prism is $\delta_{\min} = 30^\circ$ (see Figure-2). The refractive index of the prism material for blue light is $\sqrt{3}$. Which of the following statement(s) is(are) correct?
The blue light is polarized in the plane of incidence.
The angle of the prism is $45^\circ$.
The refractive index of the material of the prism for red light is $\sqrt{2}$.
The angle of refraction for blue light in air at the exit plane of the prism is $60^\circ$.
In a circuit shown in the figure, the capacitor $C$ is initially uncharged and the key $K$ is open. In this condition, a current of $1 \, \mathrm{A}$ flows through the $1 \, \Omega$ resistor. The key is closed at time $t = t_0$. Which of the following statement(s) is(are) correct? [Given: $e^{-1} = 0.36$]
The value of the resistance $R$ is $3 \, \Omega$.
For $t < t_0$, the value of current $I_1$ is $2 \, \mathrm{A}$.
At $t = t_0 + 7.2 \, \mu \mathrm{s}$, the current in the capacitor is $0.6 \, \mathrm{A}$.
For $t \to \infty$, the charge on the capacitor is $12 \, \mu \mathrm{C}$.
Physics · System of Particles and Rotational Motion · Single correct
A bar of mass $M = 1.00 \, \mathrm{kg}$ and length $L = 0.20 \, \mathrm{m}$ is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass $m = 0.10 \, \mathrm{kg}$ is moving on the same horizontal surface with $5.00 \, \mathrm{m} \, \mathrm{s}^{-1}$ speed on a path perpendicular to the bar. It hits the bar at a distance $L/2$ from the pivoted end and returns back on the same path with speed $v$. After this elastic collision, the bar rotates with an angular velocity $\omega$. Which of the following statement is correct?
Applying angular momentum conservation about hinge $$mv \frac{L}{2} + 0 = -mv \frac{L}{2} + \frac{ML^2}{3} \omega ....(i)$$ Also from eq. of restitution $$e = 1 = \frac{\omega \frac{L}{2} + V}{u} \implies u = \omega \frac{L}{2} + V ....(ii)$$ Solving (i) & (ii) $$\omega \approx 6.98 \, rad/sec \& v = 4.30 \, m/s$$ Hence option (A)
Question 22
Physics · Electrostatic Potential and Capacitance · Single correct
A container has a base of 50 cm $\times$ 5 cm and height 50 cm, as shown in the figure. It has two parallel electrically conducting walls each of area 50 cm $\times$ 50 cm. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant 3 at a uniform rate of 250 cm$^3$ s$^{-1}$. What is the value of the capacitance of the container after 10 seconds? [Given: Permittivity of free space $\varepsilon_0 = 9 \times 10^{-12}$ C$^2$ N$^{-1}$ m$^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]
One mole of an ideal gas expands adiabatically from an initial state $(T_A,V_0)$ to a final state $(T_f,5V_0)$. Another mole of the same gas expands isothermally from a different initial state $(T_B,V_0)$ to the same final state $(T_f,5V_0)$. The ratio of the specific heats at constant pressure and constant volume of this ideal gas is $\gamma$. What is the ratio $\frac{T_A}{T_B}$?
Two satellites P and Q are moving in different circular orbits around the Earth (radius $R$). The heights of P and Q from the Earth surface are $h_P$ and $h_Q$, respectively, where $h_P = R/3$. The accelerations of P and Q due to Earth's gravity are $g_P$ and $g_Q$, respectively. If $g_P/g_Q = 36/25$, what is the value of $h_Q$?
$3R/5$
$R/6$
$6R/5$
$5R/6$
Answer: (a)
Solution
Given $\($ $\frac{g_P}{g_Q}$ = $\frac{GM}{r_P^2}$ $\div$ $\frac{GM}{r_Q^2}$ = $\left$( $\frac{r_Q}{r_P}$ $\right$)^2 $\)$. $\[$ $\frac{36}{25}$ = $\left$( $\frac{r_Q}{r_P}$ $\right$)^2 $\]$ $\[$ $\frac{r_Q}{r_P}$ = $\frac{6}{5}$ $\]$ $\[$ r_Q = $\frac{6}{5}$ r_P $\]$ $\[$ R + h_Q = $\frac{6}{5}$ $\left$( R + $\frac{R}{3}$ $\right$) $\]$ $\[$ h_Q = $\frac{24}{15}$ R - R = $\frac{9}{15}$ R = $\frac{3}{5}$ R $\]$
Question 25
Physics · Atoms · Numerical
A Hydrogen-like atom has atomic number $Z$. Photons emitted in the electronic transitions from level $n = 4$ to level $n = 3$ in these atoms are used to perform photoelectric effect experiment on a target metal. The maximum kinetic energy of the photoelectrons generated is $1.95 \, \mathrm{eV}$. If the photoelectric threshold wavelength for the target metal is $310 \, \mathrm{nm}$, the value of $Z$ is _______. [Given: $hc = 1240 \, \mathrm{eV} \cdot \mathrm{nm}$ and $Rhc = 13.6 \, \mathrm{eV}$, where $R$ is the Rydberg constant, $h$ is the Planck's constant and $c$ is the speed of light in vacuum]
Physics · Ray Optics and Optical Instruments · Subjective
An optical arrangement consists of two concave mirrors $M_1$ and $M_2$, and a convex lens $L$ with a common principal axis, as shown in the figure. The focal length of $L$ is $10 \, \mathrm{cm}$. The radii of curvature of $M_1$ and $M_2$ are $20 \, \mathrm{cm}$ and $24 \, \mathrm{cm}$, respectively. The distance between $L$ and $M_2$ is $20 \, \mathrm{cm}$. A point object $S$ is placed at the mid-point between $L$ and $M_2$ on the axis. When the distance between $L$ and $M_1$ is $n/7 \, \mathrm{cm}$, one of the images coincides with $S$. The value of $n$ is
Answer: (80 or 150 or 220)
Solution
Two cases are possible if 1st refraction on lens: Since object is at focus, light will become parallel. 1st reflection at $M_1$: Light is parallel, image will be at focus. 2nd refraction from $L$: $u = -(d - 10)$ $f = 10 \, \mathrm{cm}$ $$\frac{1}{v} - \frac{1}{\mu f} = \frac{1}{f}$$ $$\frac{1}{v} + \frac{1}{d - 10} = \frac{1}{10}$$ $$\frac{1}{v} = \frac{1}{10} - \frac{1}{d - 10} \cdots (i)$$ This $v$ will be object for $M_2$, and image should be at $10 \, \mathrm{cm}$ $$\frac{1}{\mu} + \frac{1}{v_1} = \frac{1}{f}$$ $$-\frac{1}{20 - v} + \frac{1}{10} = -\frac{1}{12}$$ $$\frac{1}{12} - \frac{1}{10} = \frac{1}{20 - v}$$ $$-\frac{2}{120} = \frac{1}{20 - v}$$ $$20 - v = -60$$ $$v = 80 \, \mathrm{cm}$$ From equation (i) $$\frac{1}{80} = \frac{1}{10} - \frac{1}{d - 10}$$ $$\frac{1}{d - 10} = \frac{1}{10} - \frac{1}{80}$$ $$\frac{1}{d - 10} = \frac{80 - 10}{800} = \frac{70}{800}$$ $$d - 10 = \frac{80}{7} \Rightarrow d = 10 + \frac{80}{7} = \frac{150}{7}$$ $$n = 150$$ Case-2: If 1st reflection on mirror $m_2$ For $m_2$ $$\frac{1}{V_1} + \frac{1}{-10} = \frac{1}{-12}$$ $$V_1 = 60 \, \mathrm{cm}$$ Then refraction on lens $L$ $$u_2 = -80 \, \mathrm{cm}$$ $$\frac{1}{V_2} - \frac{1}{-60} = \frac{1}{10}$$ $$V_2 = \frac{80}{7}$$ Then reflection on $m_2$ Either $V_2$ is at centre (normal incidence) $$d - \frac{80}{7} = 20$$ $$d = \frac{220}{7}$$ $$\frac{n}{7} = \frac{220}{7},$$ $$n = 220$$ $V_2$ is at pole of $m_2$ $$d - \frac{80}{7} = 0$$ $$d = \frac{80}{7}$$ $$\frac{n}{7} = \frac{80}{7}$$ $$n = 80$$
Question 27
Physics · Ray Optics and Optical Instruments · Numerical
In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is $10 \pm 0.1 \, \mathrm{cm}$ and the distance of its real image from the lens is $20 \pm 0.2 \, \mathrm{cm}$. The error in the determination of focal length of the lens is $n \%$. The value of $n$ is .
A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas ($\gamma = 5/3$) and one mole of an ideal diatomic gas ($\gamma = 7/5$). Here, $\gamma$ is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of 66 Joule when heated at constant pressure. The change in its internal energy is ________ Joule.
Answer: 121
Solution
At constant pressure, $W = nR\Delta T = 66$. $\Delta U = n(C_V)_{mix} \Delta T$. $$(C_V)_{mix} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2}$$ $$(C_V)_{mix} = \frac{2 \times \frac{3}{2} R + 1 \times \frac{5}{2} R}{3}$$ $$(C_V)_{mix} = \frac{11}{6} R$$ $$\Delta U = \frac{11}{6} (nR\Delta T)$$ $$\Delta U = \frac{11}{6} \times 66 = 121 \, J$$
Question 29
Physics · Motion in a Straight Line · Numerical
A person of height 1.6 $\,$ $\mathrm{m}$ is walking away from a lamp post of height 4 $\,$ $\mathrm{m}$ along a straight path on the flat ground. The lamp post and the person are always perpendicular to the ground. If the speed of the person is 60 $\,$ $\mathrm{cm \, s^{-1}}$, the speed of the tip of the person's shadow on the ground with respect to the person is _______ $\,$ $\mathrm{cm \, s^{-1}}$.
Answer: 40
Solution
Given the similar triangles, we have the proportion: $$\frac{4}{y} = \frac{1.6}{y-x}$$ Solving for $x$, we get: $$4y - 4x = 1.6y$$ $$2.4y = 4x$$ $$x = 0.6y$$ Differentiating with respect to time $t$: $$\frac{dx}{dt} = 0.6 \times \frac{dy}{dt}$$ Given $\frac{dx}{dt} = 60 \, \mathrm{cm/s}$, we have: $$60 = 0.6 \times \frac{dy}{dt}$$ Therefore, $$\frac{dy}{dt} = 100 \, \mathrm{cm/s}$$ The speed of the tip of the person's shadow with respect to the person is: $$100 - 60 = 40 \, \mathrm{cm/s}$$
Question 30
Physics · Oscillations · Numerical
Two point-like objects of masses 20 gm and 30 gm are fixed at the two ends of a rigid massless rod of length 10 cm. This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is $1.2 \times 10^{-8} \, \mathrm{N} \, \mathrm{m} \, \mathrm{rad}^{-1}$. The angular frequency of the oscillations in $n \times 10^{-3} \, \mathrm{rad} \, \mathrm{s}^{-1}$. The value of $n$ is _____.
Answer: 10
Solution
The period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{C}}$. Therefore, $\omega = \sqrt{\frac{C}{I}}$. Where $I$ is the moment of inertia. $I = (30)(4)^2 + (20)(6)^2 = 1200 \, \mathrm{gm-cm^2} = 1.2 \times 10^{-4} \, \mathrm{kg-m^2}$. Thus, $\omega = \sqrt{\frac{1.2 \times 10^{-8}}{1.2 \times 10^{-4}}}$. This simplifies to $\omega = \sqrt{10^{-4}}$. Therefore, $\omega = (10^{-2})$. Finally, $n \times 10^{-3} = 10^{-2} \Rightarrow n = 10$.
Question 31
Physics · Nuclei · Single correct
List-I shows different radioactive decay processes and List-II provides possible emitted particles. Match each entry in List-I with an appropriate entry from List-II, and choose the correct option.
$$ P \rightarrow 4,\quad Q \rightarrow 3,\quad R \rightarrow 2,\quad S \rightarrow 1 $$
$$ P \rightarrow 5,\quad Q \rightarrow 3,\quad R \rightarrow 1,\quad S \rightarrow 4 $$
$$ P \rightarrow 5,\quad Q \rightarrow 3,\quad R \rightarrow 1,\quad S \rightarrow 4 $$
$$ P \rightarrow 5,\quad Q \rightarrow 1,\quad R \rightarrow 3,\quad S \rightarrow 2 $$
Physics · Thermal Properties of Matter · Single correct
Match the temperature of a black body given in List-I with an appropriate statement in List-II, and choose the correct option. [Given: Wien's constant as $2.9\times10^{-3}\,\mathrm{m\,K}$ and $\dfrac{hc}{e}=1.24\times10^{-6}\,\mathrm{V\,m}$]
A series LCR circuit is connected to a $45\sin(\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^{5}\,\mathrm{rad\,s^{-1}}$ and current amplitude at resonance is $I_0$. When the angular frequency of the source is $\omega=8\times10^{4}\,\mathrm{rad\,s^{-1}}$, the current amplitude in the circuit is $0.05\,I_0$. If $L=50\,\mathrm{mH}$, match each entry in List-I with an appropriate value from List-II and choose the correct option.
Physics · Electromagnetic Induction · Single correct
A thin conducting rod MN of mass 20 $\,$ $\mathrm{gm}$, length 25 $\,$ $\mathrm{cm}$ and resistance 10 $\,$ $\Omega$ is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field $B_0 = 4 \, \mathrm{T}$ directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time $t = 0$ and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II, and choose the correct option. [Given: The acceleration due to gravity $g = 10 \, \mathrm{ms^{-2}}$ and $e^{-1} = 0.4$]
P $\to$ 5, Q $\to$ 2, R $\to$ 3, S $\to$ 1
P $\to$ 3, Q $\to$ 1, R $\to$ 4, S $\to$ 5
P $\to$ 4, Q $\to$ 3, R $\to$ 1, S $\to$ 2
P $\to$ 3, Q $\to$ 4, R $\to$ 2, S $\to$ 5
Answer: (d)
Solution
From force equation $$mg - B i \ell = \frac{mdv}{dt}$$ $$mg - \frac{B B i \ell}{R} \times \ell = \frac{mdv}{dt}$$ $$\frac{mgR}{B^2 \ell^2} - v = \frac{mR}{B^2 \ell^2} \frac{dv}{dt}$$ $$\frac{B^2 \ell^2}{mR} \int_{t=0}^{t} dt = \int_{0}^{v} \frac{dv}{\frac{mgR}{B^2 \ell^2} - v}$$ Now $$\frac{mgR}{B^2 \ell^2} = \frac{20 \times 10^{-3} \times 10 \times 10}{16 \times \frac{1}{16}} = 2$$ And $$\frac{B^2 \ell^2}{mR} = \frac{16 \times \frac{1}{16}}{20 \times 10^{-3} \times 10} = \frac{1}{0.2} = 5$$ $$\therefore 5t = \left[-\ln(2-v)\right]_0^v$$ $$-5t = \ln \left[\frac{2-v}{v}\right]$$ $$\therefore v = 2 \left(1 - e^{-5t}\right)$$ At $t = 0.2$ sec $$v = 2 \left(1 - e^{-5 \times 0.2}\right)$$ $$v = 2 \left(1 - 0.4\right)$$ $$v = 1.2 \, \mathrm{m/s}$$ (P) Now at $t = 0.2$ sec The magnitude of the induced emf $= E = B v \ell$ $$= 4 \times 1.2 \times \frac{1}{4} = 1.2 \, \mathrm{Volt}$$ (Q) At $t = 0.2$ sec, the magnitude of magnetic force $= B I \ell \sin \theta$ $$= B \times \frac{B v}{R} \times \ell \times \sin 90^\circ$$ $$= 4 \times 4 \times \frac{1}{4} \times 1.3 \times \frac{1}{4}$$ $$= 0.12 \, \mathrm{Newton}$$ (R) At $t = 0.2$ sec, the power dissipated as heat $$P = i^2 R = \frac{v^2}{R} = \frac{1.2 \times 1.2}{10}$$ $$P = 0.144 \, \mathrm{watt}$$ (S) Magnitude of terminal velocity At terminal velocity, the net force become zero $$\therefore mg = B i \ell$$ $$mg = B \times \frac{B v_T}{R} \times \ell$$ $$\therefore v_T = \frac{mgR}{B^2 \ell^2} = \frac{20 \times 10^{-3} \times 10 \times 10}{16 \times \frac{1}{16}}$$ $$v_T = 2 \, \mathrm{m/s}$$ Hence, answer is (D)
Chemistry
Question 35
Chemistry · General Principles and Processes of Isolation of Elements · Multiple correct
The correct statement(s) related to processes involved in the extraction of metals is(are)
Roasting of Malachite produces Cuprite.
Calcination of Calamine produces Zincite.
Copper pyrites is heated with silica in a reverberatory furnace to remove iron.
Impure silver is treated with aqueous KCN in the presence of oxygen followed by reduction with zinc metal.
Answer: (b), (c), (d)
Solution
Question 36
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Multiple correct
In the following reactions, $P$, $Q$, $R$, and $S$ are the major products. The correct statement(s) about $P$, $Q$, $R$, and $S$ is(are)
Both $P$ and $Q$ have asymmetric carbon(s).
Both $Q$ and $R$ have asymmetric carbon(s).
Both $P$ and $R$ have asymmetric carbon(s).
$P$ has asymmetric carbon(s), $S$ does not have any asymmetric carbon.
Answer: (c), (d)
Solution
Formation of P: $\($ CH_3 - CH_2 - CH - CH_2 - CN $\xrightarrow{PhMgBr}$ Ph $\xrightarrow{H_3O^+}$ $\)$ leads to the formation of an asymmetric carbon compound (P). Formation of Q: $\($ C_6H_5 - C - CH_3 $\xrightarrow{anhy. AlCl_3}$ Ph - C - CH_3 $\xrightarrow{PhMgBr}$ Ph $\xrightarrow{H_3O^+}$ $\)$ results in a compound (Q) with no asymmetric carbon. Formation of R: $\($ C - Cl + $\frac{1}{2}$ (Ph - CH_2)_2Cd $\rightarrow$ Ph $\xrightarrow{PhMgBr}$ Ph $\xrightarrow{H_3O^+}$ $\)$ results in an asymmetric carbon compound (R). Formation of S: $\($ Ph - CH_2 - C - H $\xrightarrow{PhMgBr}$ Ph - CH_2 - CH - Ph $\xrightarrow{H_3O^+}$ $\)$ followed by $\($ CrO_3 with dil. H_2SO_4 $\rightarrow$ Ph - CH_2 - C - Ph $\xrightarrow{HCN}$ Ph - CH = C - Ph $\xrightarrow{\Delta}$ Ph - CH_2 - C - Ph $\)$ results in a compound (S) with no asymmetric carbon.
Question 37
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Multiple correct
Consider the following reaction scheme and choose the correct option(s) for the major products $Q$, $R$ and $S$. Styrene $\xrightarrow{\begin{array}{l} (i) \ \mathrm{B_2H_6} \\ (ii) \ \mathrm{NaOH, H_2O_2, H_2O} \end{array}}$ $\mathrm{P}$ $\xrightarrow{\begin{array}{l} (i) \ \mathrm{CrO_3, H_2SO_4} \\ (ii) \ \mathrm{Cl_2, Red \ Phosphorus} \\ (iii) \ \mathrm{H_2O} \end{array}}$ $\mathrm{Q}$ $\mathrm{P}$ $\xrightarrow{\begin{array}{l} (i) \ \mathrm{SOCl_2} \\ (ii) \ \mathrm{NaCN} \\ (iii) \ \mathrm{H_3O^+, \ \Delta} \end{array}}$ $\mathrm{R}$ $\xrightarrow{\mathrm{conc. \ H_2SO_4}}$ $\mathrm{S}$
Answer: (b)
Solution
The given reaction sequence starts with styrene, which is treated with $\mathrm{B_2H_6}$ followed by $\mathrm{NaOH}$, $\mathrm{H_2O_2}$, and $\mathrm{H_2O}$ to form compound (P) with an alcohol group. Compound (P) is then oxidized using $\mathrm{CrO_3}$ and $\mathrm{H_2SO_4}$ to form compound (O), which is a carboxylic acid. The compound undergoes the HVZ reaction with $\mathrm{Cl_2}$ and red phosphorus to form compound (Q), which is an alpha-chloro acid. Compound (Q) is then treated with $\mathrm{NaCN}$ to form a nitrile, which is hydrolyzed to form compound (R), a beta-hydroxy acid. Finally, compound (R) undergoes dehydration with concentrated $\mathrm{H_2SO_4}$ to form compound (S), which is a lactone.
Question 38
Chemistry · Analytical Chemistry · Single correct
In the scheme given below, X and Y, respectively, are
CrO$_4^{2-}$ and Br$_2$
MnO$_4^{2-}$ and Cl$_2$
MnO$_4^{-}$ and Cl$_2$
MnSO$_4$ and HOCl
Answer: (c)
Solution
The reaction of $\mathrm{MnCl_2}$ with $\mathrm{NaOH}$ produces $\mathrm{Mn(OH)_2}$ as a white precipitate (P) and $\mathrm{NaCl}$ as the filtrate (Q): $$\mathrm{MnCl_2 + NaOH \rightarrow Mn(OH)_2 \downarrow + NaCl}$$ When $\mathrm{Mn(OH)_2}$ is heated with $\mathrm{PbO_2}$ and $\mathrm{H^+}$ from $\mathrm{H_2SO_4}$, it forms $\mathrm{MnO_4^-}$ and $\mathrm{Pb^{2+}}$ with a purple color: $$\mathrm{Mn(OH)_2 \xrightarrow{PbO_2 + H^+ (H_2SO_4) heat} MnO_4^- + Pb^{2+}}$$ Chloride ions $\mathrm{Cl^-}$ react with $\mathrm{MnO(OH)_2}$ and concentrated $\mathrm{H_2SO_4}$ to produce $\mathrm{Cl_2}$, which then reacts with $\mathrm{I^-}$ to form $\mathrm{I_2}$: $$\mathrm{Cl^- \xrightarrow{MnO(OH)_2/conc.\ H_2SO_4} Cl_2}$$ $$\mathrm{Cl_2 + 2I^- \rightarrow I_2 + 2Cl^-}$$ The iodine $\mathrm{I_2}$ reacts with starch to give a blue coloration.
Question 39
Chemistry · Electrochemistry · Single correct
Plotting $1/\Lambda_m$ against $c \Lambda_m$ for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of $P$ and slope of $S$. The ratio $P/S$ is [$\Lambda_m$ = molar conductivity $\Lambda_m^\circ$ = limiting molar conductivity c = molar concentration$ $K_a$ = dissociation constant of HX]
On decreasing the $pH$ from 7 to 2, the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from $10^{-4} \, \mathrm{mol} \, \mathrm{L}^{-1}$ to $10^{-3} \, \mathrm{mol} \, \mathrm{L}^{-1}$. The $pK_a$ of HX is:
3
4
5
2
Answer: (b)
Solution
At pH = 7, pure water solubility $S_1 = \sqrt{K_{sp}}$. At pH = 2, $\[$ MX(s) + aq $\xrightleftharpoons$[K_{sp}]{} M^+(aq) + X^-(aq) $\]$ $\[$ X^-(aq) + H^+(aq) $\xrightleftharpoons$[1/K_a]{s-x} HX(aq) $\]$ Approximation: $s - x \approx 0$ [X is limiting reagent] $\[$ $\Rightarrow$ s $\approx$ x $\]$ $\[$ $\Rightarrow$ s(s-x) = K_{sp} $\ldots$ $\ldots$ (1) $\]$ $\[$ $\frac{s}{(s-x)(10^{-2})}$ = $\frac{1}{K_a}$ $\ldots$ $\ldots$ (2) $\]$ Multiply (1) and (2) $\[$ $\Rightarrow$ $\frac{s^2}{10^{-2}}$ = $\frac{K_{sp}}{K_a}$ $\]$ $\[$ $\Rightarrow$ s = $\frac{\sqrt{K_{sp}}}{10\sqrt{K_a}}$ $\]$ Now given: $\[$ $\frac{s}{s_1}$ = $\frac{10^{-3}}{10^{-4}}$ $\]$ $\[$ $\Rightarrow$ $\frac{\sqrt{K_{sp}}}{10\sqrt{K_a}}$ $\div$ $\sqrt{K_{sp}}$ = 10 $\]$ $\[$ $\Rightarrow$ $\frac{1}{10\sqrt{K_a}}$ = 10 $\]$ $\[$ $\Rightarrow$ $\sqrt{K_a}$ = 10^{-2} $\]$ $\[$ $\Rightarrow$ K_a = 10^{-4} $\]$ $\[$ $\Rightarrow$ pK_a = 4 $\]$
Question 41
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
In the given reaction scheme, $P$ is a phenyl alkyl ether, $Q$ is an aromatic compound; $R$ and $S$ are the major products. $$P \xrightarrow{HI} Q \xrightarrow{\begin{array}{c} (i) NaOH \\ (ii) CO_2 \\ (iii) H_3O^+ \end{array}} R \xrightarrow{\begin{array}{c} (i) (CH_3CO)_2O \\ (ii) H_3O^+ \end{array}} S$$ The correct statement about $S$ is
It primarily inhibits noradrenaline degrading enzymes.
It inhibits the synthesis of prostaglandin.
It is a narcotic drug.
It is ortho-acetylbenzoic acid.
Answer: (b)
Solution
P is phenyl alkyl ether. Q is aromatic compound. R and S are the major product. Phenyl alkyl ether reacts with HI to form Q and R-I. Q reacts with NaOH to form phenoxide ion. The phenoxide ion reacts with CO2 and then H3O+ to form salicylic acid. Salicylic acid reacts with acetic anhydride to form aspirin. Aspirin inhibits the synthesis of chemicals known as prostaglandins.
Question 42
Chemistry · The p-Block Elements (Group-13 and 14) · Numerical
The stoichiometric reaction of 516 g of dimethyldichlorosilane with water results in a tetrameric cyclic product $X$ in 75$\%$ yield. The weight (in g) of $X$ obtained is ___. [Use, molar mass (g mol$^{-1}$): H = 1, C = 12, O = 16, Si = 28, Cl = 35.5]
Answer: 222
Solution
The reaction is given by: $$4(\mathrm{CH_3})_2\mathrm{SiCl_2} + 4\mathrm{H_2O} \xrightarrow{75\%} (\mathrm{CH_3})_8\mathrm{Si_4O_4} + 8\mathrm{HCl}$$ Let $w = 516 \, \mathrm{g}$. The number of moles $n$ is calculated as: $$n = \frac{516}{129} = 4$$ The weight of the product is $296 \, \mathrm{g}$. The percentage yield is $75\%$. The weight of $X$ (in grams) is calculated as: $$296 \times \frac{75}{100} = 222 \, \mathrm{g}$$
Question 43
Chemistry · States of Matter · Fill in the blank
A gas has a compressibility factor of 0.5 and a molar volume of 0.4 $\mathrm{dm^3 \, mol^{-1}}$ at a temperature of 800 $\mathrm{K}$ and pressure $x$ $\mathrm{atm}$. If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be $y$ $\mathrm{dm^3 \, mol^{-1}}$. The value of $x/y$ is ___. [Use: Gas constant, $R = 8 \times 10^{-2}$ $\mathrm{L \, atm \, K^{-1} \, mol^{-1}}$]
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The plot of $\log k_f$ versus $\frac{1}{T}$ for a reversible reaction $\mathrm{A (g) \rightleftharpoons P (g)}$ is shown. Pre-exponential factors for the forward and backward reactions are $10^{15} \, \mathrm{s^{-1}}$ and $10^{11} \, \mathrm{s^{-1}}$, respectively. If the value of $\log K$ for the reaction at $500 \, \mathrm{K}$ is $6$, the value of $|\log k_b|$ at $250 \, \mathrm{K}$ is ______.
One mole of an ideal monoatomic gas undergoes two reversible processes (A $\rightarrow$ B and B $\rightarrow$ C) as shown in the given figure: A $\rightarrow$ B is an adiabatic process. If the total heat absorbed in the entire process (A $\rightarrow$ B and B $\rightarrow$ C) is $RT_2 \ln 10$, the value of $2 \log V_3$ is ________. [Use, molar heat capacity of the gas at constant pressure, $C_{p,m} = \frac{5}{2} R$]
Answer: 7
Solution
For A to B, (Reversible adiabatic) $$600 \left(V_1\right)^{2/3} = 60 \left(V_2\right)^{2/3}$$ $$\Rightarrow 10 = \left(\frac{V_2}{V_1}\right)^{2/3}$$ $$\Rightarrow 10 = \left(\frac{V_2}{10}\right)^{2/3}$$ $$\Rightarrow V_2 = 10(10)^{3/2} = 10^{5/2}$$ Now, $q_{net} = RT_2 \ln 10 = 60 R \ln 10 = q_{AB} + q_{BC}$ Therefore, $q_{AB} = 0$ $$\Rightarrow q_{BC} = 60 R \ln 10 = 60 R \ln \frac{V_3}{V_2}$$ [Therefore: B to C is reversible isothermal] $$\Rightarrow 60 R \ln 10 = 60 R \ln \left(\frac{V_3}{10^{5/2}}\right)$$ $$\Rightarrow \log 10 = \log V_3 - \frac{5}{2}$$ $$\Rightarrow \log V_3 = \frac{7}{2} \Rightarrow 2 \log V_3 = 7$$
Question 46
Chemistry · Equilibrium · Numerical
In a one-litre flask, $6$ moles of A undergoes the reaction $\mathrm{A(g)\rightleftharpoons P(g)}$. The progress of product formation at two temperatures (in Kelvin), $T_1$ and $T_2$, is shown in the figure: If $T_1=2T_2$ and $(\Delta G_2^\circ-\Delta G_1^\circ)=RT_2\ln x$, then the value of $x$ is _____. [$\Delta G_1^\circ$ and $\Delta G_2^\circ$ are standard Gibbs free energy change for the reaction at temperatures $T_1$ and $T_2$, respectively.]
The total number of $sp^2$ hybridised carbon atoms in the major product $\mathbf{P}$ (a non-heterocyclic compound) of the following reaction is ______.
Answer: 28
Solution
Total number of $sp^2$ hybridised C-atom in P = 28
Question 48
Chemistry · Some Basic Concepts of Chemistry · Single correct
Match the reactions (in the given stoichiometry of the reactants) in List-I with one of their products given in List-II and choose the correct option.
Chemistry · Co-ordination Compounds · Single correct
Match the electronic configurations in List-I with appropriate metal complex ions in List-II and choose the correct option. [Atomic Number: Fe = 26, Mn = 25, Co = 27]
P $\rightarrow$ 1; Q $\rightarrow$ 4; R $\rightarrow$ 2; S $\rightarrow$ 3
P $\rightarrow$ 1; Q $\rightarrow$ 2; R $\rightarrow$ 4; S $\rightarrow$ 5
P $\rightarrow$ 3; Q $\rightarrow$ 2; R $\rightarrow$ 5; S $\rightarrow$ 1
P $\rightarrow$ 3; Q $\rightarrow$ 2; R $\rightarrow$ 4; S $\rightarrow$ 1
Chemistry · Haloalkanes and Haloarenes · Single correct
Match the reactions in List-I with the features of their products in List-II and choose the correct option.
P $\to$ 1; Q $\to$ 2; R $\to$ 5; S $\to$ 3
P $\to$ 2; Q $\to$ 1; R $\to$ 3; S $\to$ 5
P $\to$ 1; Q $\to$ 2; R $\to$ 5; S $\to$ 4
P $\to$ 2; Q $\to$ 4; R $\to$ 3; S $\to$ 5
Answer: (b)
Solution
P undergoes $S_N2$ reaction with aqueous NaOH, resulting in retention of configuration. Q undergoes $S_N2$ reaction with aqueous NaOH, resulting in inversion of configuration. R undergoes $S_N1$ reaction with aqueous NaOH, resulting in a mixture of enantiomers. S undergoes $S_N1$ reaction with aqueous NaOH, resulting in a diastereomeric mixture.
Question 51
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match List-I with List-II and choose the correct option.
P $\rightarrow$ 2; Q $\rightarrow$ 4; R $\rightarrow$ 1; S $\rightarrow$ 3
P $\rightarrow$ 1; Q $\rightarrow$ 3; R $\rightarrow$ 5; S $\rightarrow$ 2
P $\rightarrow$ 3; Q $\rightarrow$ 2; R $\rightarrow$ 1; S $\rightarrow$ 4
P $\rightarrow$ 3; Q $\rightarrow$ 4; R $\rightarrow$ 5; S $\rightarrow$ 2
Answer: (d)
Solution
The solution involves several reactions: (i) Acetophenone is reduced using Zn-Hg/HCl to form ethylbenzene: $$Ph - C - CH_3 \xrightarrow{Zn-Hg/HCl} Ph - CH_2 - CH_3$$ (ii) Toluene is oxidized and then converted to benzoyl chloride, which is reduced by the Rosenmund reaction to form benzaldehyde (S): $$Toluene \xrightarrow{(i) KMnO_4, KOH / \Delta} \xrightarrow{(ii) SOCl_2} C - C - Cl \xrightarrow{H_2Pd - BaSO_4} CHO$$ (iii) Benzene is converted to benzyl chloride, which undergoes the Etard reaction to form benzaldehyde (P): $$Benzene \xrightarrow{CH_3 - Cl, Anhy. AlCl_3} \xrightarrow{CrO_2Cl_2, AcOAc / \Delta} CHO$$ (iv) Aniline is converted to a diazonium salt, which is then converted to chlorobenzene via the Gattermann reaction (Q): $$NH_2 \xrightarrow{NaNO_2 / HCl, 273-278 K} N_2Cl \xrightarrow{Cu / HCl} Cl$$ (v) Phenol is converted to benzaldehyde via the Gattermann Koch reaction (R): $$OH \xrightarrow{Zn / \Delta} \xrightarrow{CO / HCl, AlCl_3} CHO$$ The mapping is: P $\to$ 3, Q $\to$ 4, R $\to$ 5, S $\to$ 2.