JEE Advanced 4 June 2023 Paper 2 question paper with solutions
JEE Advanced 4 June 2023 Paper 2: 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.
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Maths
Question 1
Maths · Integrals · Single correct
Let $f : [1, \infty) \to \mathbb{R}$ be a differentiable function such that $f(1) = \frac{1}{3}$ and $3 \int_{1}^{x} f(t) \, dt = x f(x) - \frac{x}{3}, \, x \in [1, \infty)$. Let $e$ denote the base of the natural logarithm. Then the value of $f(e)$ is
$\frac{e^2 + 4}{3}$
$\frac{\log_e 4 + e}{3}$
$\frac{4e^2}{3}$
$\frac{e^2 - 4}{3}$
Answer: (c)
Solution
Differentiate with respect to 'x'. $$3f(x) = f(x) + x f'(x) - x^2$$ $$\frac{dy}{dx} - \left( \frac{2}{x} \right) y = x$$ If $IF = e^{-2 \ln x} = \frac{1}{x^2}$ $$y \left( \frac{1}{x^2} \right) = \int x \cdot \frac{1}{x^2} \, dx$$ $$y = x^2 \ln x + cx^2$$ Therefore, $y(1) = \frac{1}{3} \Rightarrow c = \frac{1}{3}$ $$y(e) = \frac{4e^2}{3}$$
Question 2
Maths · Probability · Single correct
Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is $\frac{1}{3}$, then the probability that the experiment stops with head is.
$\frac{1}{3}$
$\frac{5}{21}$
$\frac{4}{21}$
$\frac{2}{7}$
Answer: (b)
Solution
Given $\mathrm{P(H)} = \frac{1}{3}$; $\mathrm{P(T)} = \frac{2}{3}$. Required probability $= \mathrm{P(HH or HTHH or HTHTHH or \ldots)}$ $$+ \mathrm{P(THH or THTHH or THTHTHH or \ldots)}$$ $$= \frac{1}{3} \cdot \frac{1}{3} + \frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}$$ $$= \frac{1}{1 - \frac{2}{3} \cdot \frac{1}{3}} + \frac{2}{3} \cdot \frac{1}{1 - \frac{2}{3} \cdot \frac{1}{3}} = \frac{5}{21}$$
Question 3
Maths · Inverse Trigonometric Functions · Single correct
For any $y \in \mathbb{R}$, let $\cot^{-1}(y) \in (0, \pi)$ and $\tan^{-1}(y) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the sum of all the solutions of the equation $\tan^{-1}\left(\frac{6y}{9-y^2}\right) + \cot^{-1}\left(\frac{9-y^2}{6y}\right) = \frac{2\pi}{3}$ for $0 < |y| < 3$, is equal to
$2\sqrt{3} - 3$
$3 - 2\sqrt{3}$
$4\sqrt{3} - 6$
$6 - 4\sqrt{3}$
Answer: (c)
Question 4
Maths · Vector Algebra · Single correct
Let the position vectors of the points P, Q, R and S be $\vec{a} = \hat{i} + 2\hat{j} - 5\hat{k}$, $\vec{b} = 3\hat{i} + 6\hat{j} + 3\hat{k}$, $\vec{c} = \frac{17}{5} \hat{i} + \frac{16}{5} \hat{j} + 7\hat{k}$ and $\vec{d} = 2\hat{i} + \hat{j} + \hat{k}$, respectively. Then which of the following statements is true?
The points P, Q, R and S are NOT coplanar
$\frac{\vec{b} + 2\vec{d}}{3}$ is the position vector of a point which divides PR internally in the ratio 5 : 4
$\frac{\vec{b} + 2\vec{d}}{3}$ is the position vector of a point which divides PR externally in the ratio 5 : 4
The square of the magnitude of the vector $\vec{b} \times \vec{d}$ is 95
Let $M = (a_{ij})$, $i, j \in \{1, 2, 3\}$, be the $3 \times 3$ matrix such that $a_{ij} = 1$ if $j+1$ is divisible by $i$, otherwise $a_{ij} = 0$. Then which of the following statements is (are) true?
M is invertible
There exists a nonzero column matrix $\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ such that $M \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} = \begin{pmatrix} -a_1 \\ -a_2 \\ -a_3 \end{pmatrix}$
The set $\{X \in \mathbb{R}^3 : MX = 0 \} \neq \{0\}$, where $0 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$
The matrix $(M - 2I)$ is invertible, where $I$ is the $3 \times 3$ identity matrix
Maths · Continuity and Differentiability · Multiple correct
Let $f : (0,1) \to \mathbb{R}$ be the function defined as $f(x) = [4x] \left( x - \frac{1}{4} \right) \left( x - \frac{1}{2} \right)$, where $[x]$ denotes the greatest integer less than or equal to $x$. Then which of the following statements is(are) true?
The function $f$ is discontinuous exactly at one point in $(0,1)$
There is exactly one point in $(0,1)$ at which the function $f$ is continuous but NOT differentiable
The function $f$ is NOT differentiable at more than three points in $(0,1)$
The minimum value of the function $f$ is $-\frac{1}{512}$
Answer: (a), (b)
Solution
Given $$f(x) = \begin{cases} 0 & ; \; 0 < x < \frac{1}{4} \\ \left( x - \frac{1}{4} \right)^2 \left( x - \frac{1}{2} \right) & ; \; \frac{1}{4} \leq x < \frac{1}{2} \\ 2 \left( x - \frac{1}{4} \right)^2 \left( x - \frac{1}{2} \right) & ; \; \frac{1}{2} \leq x < \frac{3}{4} \\ 3 \left( x - \frac{1}{4} \right)^2 \left( x - \frac{1}{2} \right) & ; \; \frac{3}{4} \leq x < 1 \end{cases}$$ $f(x)$ is discontinuous at $x = \frac{3}{4}$ only. Differentiating, we have $$f'(x) = \begin{cases} 0 & ; \; 0 < x < \frac{1}{4} \\ 2 \left( x - \frac{1}{4} \right) \left( x - \frac{1}{2} \right) + \left( x - \frac{1}{4} \right)^2 & ; \; \frac{1}{4} < x < \frac{1}{2} \\ 4 \left( x - \frac{1}{4} \right) \left( x - \frac{1}{2} \right) + 2 \left( x - \frac{1}{4} \right)^2 & ; \; \frac{1}{2} < x < \frac{3}{4} \\ 6 \left( x - \frac{1}{4} \right) \left( x - \frac{1}{2} \right) + 3 \left( x - \frac{1}{4} \right)^2 & ; \; \frac{3}{4} < x < 1 \end{cases}$$ $f(x)$ is non-differentiable at $x = \frac{1}{2}$ and $\frac{3}{4}$. Minimum values of $f(x)$ occur at $x = \frac{5}{12}$ whose value is $-\frac{1}{432}$.
Question 7
Maths · Continuity and Differentiability · Multiple correct
Let S be the set of all twice differentiable functions f from $\mathbb{R}$ to $\mathbb{R}$ such that $\frac{d^2 f}{dx^2}(x) > 0$ for all $x \in (-1, 1)$. For $f \in S$, let $X_f$ be the number of points $x \in (-1, 1)$ for which $f(x) = x$. Then which of the following statements is(are) true?
There exists a function $f \in S$ such that $X_f = 0$
For every function $f \in S$, we have $X_f \leq 2$
There exists a function $f \in S$ such that $X_f = 2$
There does NOT exist any function $f$ in $S$ such that $X_f = 1$
Answer: (a), (b), (c)
Solution
S = Set of all twice differentiable functions $f : \mathbb{R} \to \mathbb{R}$ $$\frac{d^2 f}{dx^2} > 0 in (-1, 1)$$ Graph $f$ is concave upward. Number of solutions of $f(x) = x \to x_f$ Graph of $y = f(x)$ can intersect graph of $y = x$ at at most two points $\Rightarrow 0 \leq x_f \leq 2$ Aliter $$\frac{d^2 f(x)}{dx^2} > 0$$ Let $\phi(x) = f(x) - x$ $\phi''(x) > 0$ $\therefore \phi(x) = 0$ has at most 1 root in $x \in (-1, 1)$ $\therefore \phi(x) = 0$ has at most 2 roots in $x \in (-1, 1)$ $\therefore x_f \leq 2$
Question 8
Maths · Applications of Integrals · Numerical
For $x \in \mathbb{R}$, let $\tan^{-1}(x) \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$. Then the minimum value of the function $f : \mathbb{R} \to \mathbb{R}$ defined by $$f(x) = \int_{0}^{x \tan^{-1} x} \frac{e^{(t - \cos t)}}{1 + t^{2023}} \, dt$$ is
Answer: 0
Solution
Given $$f(x) = \int_0^{\tan^{-1} x} \frac{e^{-t - \cos t}}{1 + t^{2023}} \, dt$$ Differentiating, we have $$f'(x) = \frac{e^{x \tan^{-1} x - \cos(x \tan^{-1} x)}}{1 + (x \tan^{-1} x)^{2023}} \cdot \left( \frac{x}{1 + x^2} + \tan^{-1} x \right)$$ For $x 0 & For x > 0 \\ < 0 & For x < 0 \\ 0 & For x = 0 \end{cases}$$ This indicates a point of minima. Hence, the minimum value is $f(0) = \int_0^0 = 0$.
Question 9
Maths · Differential Equations · Numerical
For $x\in\mathbb{R}$, let $y(x)$ be a solution of the differential equation $$ (x^2-5)\frac{\mathrm{d}y}{\mathrm{d}x}-2xy=-2x(x^2-5)^2 $$ such that $y(2)=7$. Then the maximum value of the function $y(x)$ is:
Answer: 16
Solution
Given $\($ $\frac{dy}{dx}$ - $\frac{2x}{x^2 - 5}$y = -2x(x^2 - 5) $\)$. IF $\($ e^{$\int$ $\frac{2x}{x^2 - 5}$ $\,$ dx} = $\frac{1}{(x^2 - 5)}$ $\)$ $\($ y $\cdot$ $\frac{1}{x^2 - 5}$ = $\int$ -2x $\,$ dx + c $\)$ $\($ $\Rightarrow$ $\frac{y}{x^2 - 5}$ = -x^2 + c $\)$ $\($ x = 2, $\ $y = 7 $\)$ $\($ $\frac{7}{-1}$ = -4 + c $\Rightarrow$ c = -3 $\)$ $\($ y = -(x^2 - 5)(x^2 + 3) $\)$ Put $\($ x^2 = t > 0 $\)$ $\($ y = -(t - 5)(t + 3) $\)$ $\($ y_{max} = 16 $\)$ when $\($ x^2 = 1 $\)$ $\($ y_{max} = 16 $\)$
Question 10
Maths · Probability · Numerical
Let X be the set of all five digit numbers formed using 1,2,2,2,4,4,0. For example, 22240 is in X while 02244 and 44422 are not in X. Suppose that each element of X has an equal chance of being chosen. Let p be the conditional probability that an element chosen at random is a multiple of 20 given that it is a multiple of 5. Then the value of 38p is equal to
Answer: 31
Solution
Number of elements in X which are multiple of 5 $$\frac{4}{3} = 4$$ $$\frac{4}{2} = 12$$ $$\frac{4}{3} = 4$$ Total = 38 Among these 38 elements, let us calculate when element is not divisible by 20 $$\frac{3}{3} = 1$$ $$\frac{3}{2} = 3$$ Total = 7 Therefore, $p = \frac{38 - 7}{38}$, so $38p = 31$
Question 11
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $A_1, A_2, A_3, \ldots, A_8$ be the vertices of a regular octagon that lie on a circle of radius $2$. Let $P$ be a point on the circle and let $PA_i$ denote the distance between the points $P$ and $A_i$ for $i = 1, 2, \ldots, 8$. If $P$ varies over the circle, then the maximum value of the product $PA_1 \cdot PA_2 \cdot \ldots \cdot PA_8$, is
Let $R = \left\{ \begin{vmatrix} c & 2 & d \\ 0 & 5 & 0 \end{vmatrix} : a, b, c, d \in \{0, 3, 5, 7, 11, 13, 17, 19\} \right\}$. Then the number of invertible matrices in $R$ is
Answer: 3780
Solution
Let us calculate when $|R| = 0$. Case-I $ad = bc = 0$. Now $ad = 0$. Therefore, Total – (When none of $a$ and $d$ is 0) is $$8^2 - 1 = 15$$ ways. Similarly $bc = 0 \Rightarrow 15$ ways. Thus, $15 \times 15 = 225$ ways of $ad = bc = 0$. Case-II $ad = bc \neq 0$. Either $a = d = b = c$ or $a \neq d$, $b \neq d$ but $ad = bc$. $$^7C_1 = 7$$ ways. $$^7C_2 \times 2 \times 2 = 84$$ ways. Total 91 ways. Thus, $|R| = 0$ in $225 + 91 = 316$ ways. $|R| \neq 0$ in $8^4 - 316 = 3780$.
Question 13
Maths · Conic Sections · Numerical
Let $C_1$ be the circle of radius 1 with center at the origin. Let $C_2$ be the circle of radius $r$ with center at the point $A = (4,1)$, where $1 < r < 3$. Two distinct common tangents $PQ$ and $ST$ of $C_1$ and $C_2$ are drawn. The tangent $PQ$ touches $C_1$ at $P$ and $C_2$ at $Q$. The tangent $ST$ touches $C_1$ at $S$ and $C_2$ at $T$. Mid points of the line segments $PQ$ and $ST$ are joined to form a line which meets the $x$-axis at a point $B$. If $AB = \sqrt{5}$, then the value of $r^2$ is
Let $p_i$ be the probability that a randomly chosen point has $i$ many friends, $i = 0, 1, 2, 3, 4$. Let $X$ be a random variable such that for $i = 0, 1, 2, 3, 4$, the probability $P(X = i) = p_i$. Then the value of $7E(X)$ is
Two distinct points are chosen randomly out of the points $A_1, A_2, \ldots, A_{49}$. Let $p$ be the probability that they are friends. Then the value of $7p$ is
Answer: 0.5
Solution
Total number of ways of selecting 2 persons = $\binom{49}{2}$. Number of ways in which 2 friends are selected = 6 $\times$ 7 $\times$ 2 = 84. $$7P = \frac{84 \times 2}{49 \times 48} \times 7 = \frac{1}{2}$$
Physics
Question 18
Physics · Electric Charges and Fields · Single correct
An electric dipole is formed by two charges $+q$ and $-q$ located in xy-plane at $(0, 2) \, \mathrm{mm}$ and $(0, -2) \, \mathrm{mm}$, respectively, as shown in the figure. The electric potential at point $P(100, 100) \, \mathrm{mm}$ due to the dipole is $V_0$. The charges $+q$ and $-q$ are then moved to the points $(-1, 2) \, \mathrm{mm}$ and $(1, -2) \, \mathrm{mm}$, respectively. What is the value of electric potential at $P$ due to the new dipole?
$V_0/4$
$V_0/2$
$V_0/\sqrt{2}$
$3V_0/4$
Answer: (b)
Solution
Given $P_1 = q(4)$ and $\vec{P_1} = P_1 \hat{j}$. The position vector is $\vec{r} = 100(\hat{i} + \hat{j}) \, \mathrm{mm}$. The initial velocity is given by $$v_0 = \frac{K P_1 \cdot \vec{r}}{r^3} = \frac{K (100 P_1)}{(100 \sqrt{2})^3}.$$ For the second part, $\tan \theta = 2$, and $\vec{P_2} = P_2 \left[ -\cos \theta \hat{i} + \sin \theta \hat{j} \right]$. The position vector remains $\vec{r} = 100(\hat{i} + \hat{j}) \, \mathrm{mm}$. We have $P_2 = q \ell$. The velocity is $$v = \frac{K \vec{P_2} \cdot \vec{r}}{r^3} = \frac{K (100 P_2)(-\cos \theta + \sin \theta)}{(100 \sqrt{2})^3}.$$ The ratio of velocities is $$\frac{v_0 P_2}{P_1} = (-\cos \theta + \sin \theta).$$ Therefore, $$v = v_0 \frac{q \ell}{q(4)} [-\cos \theta + \sin \theta] = \frac{v_0}{4} [-2 + 4] = \frac{v_0}{2}.$$
Question 19
Physics · Physical World, Units and Measurements · Single correct
Young's modulus of elasticity $Y$ is expressed in terms of three derived quantities, namely, the gravitational constant $G$, Planck's constant $h$ and the speed of light $c$, as $Y = c^\alpha h^\beta G^\gamma$. Which of the following is the correct option?
$\alpha = 7, \beta = -1, \gamma = -2$
$\alpha = -7, \beta = -1, \gamma = -2$
$\alpha = 7, \beta = -1, \gamma = 2$
$\alpha = -7, \beta = 1, \gamma = -2$
Answer: (a)
Solution
Given $Y = c^\alpha h^\beta G^\gamma$. $$\mathrm{ML^{-1}T^{-2} = (LT^{-1})^\alpha (ML^2T^{-1})^\beta (M^{-1}L^3T^{-2})^\gamma}$$ From the equations, we have: 1. $1 = \beta - \gamma$ ...(1) 2. $-1 = \alpha + 2\beta + 3\gamma$ ...(2) 3. $-2 = -\alpha - \beta - 2\gamma$ ...(3) Subtracting equation (3) from equation (2): $$-3 = \beta + \gamma$$ Using equation (1): $$1 = \beta - \gamma$$ Subtracting these two equations: $$-2 = 2\beta \implies \beta = -1, \gamma = -2$$ Substituting $\beta$ and $\gamma$ into equation (2): $$-1 = \alpha - 2 - 6 \implies \alpha = 7$$
Question 20
Physics · Laws of Motion · Single correct
A particle of mass $m$ is moving in the $xy$-plane such that its velocity at a point $(x, y)$ is given as $\vec{v} = \alpha (y\hat{x} + 2x\hat{y})$, where $\alpha$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle?
An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas in $n$. The internal energy of one mole of the gas is $U_n$ and the speed of sound in the gas is $v_n$. At a fixed temperature and pressure, which of the following is the correct option?
$v_3 U_6$
$v_5 > v_3$ and $U_3 > U_5$
$v_5 > v_7$ and $U_5 < U_7$
$v_6 < v_7$ and $U_6 < U_7$
Answer: (c)
Solution
U = $\frac{1}{2}$ fnrT = $\frac{frT}{2}$ Therefore, A and B are wrong. v_{sound} = $\sqrt{\frac{\gamma RT}{M}}$ = $\sqrt{\left(\frac{2}{f} + 1\right) \frac{RT}{M}}$ $\Rightarrow$ more 'f', less 'v' Therefore, v_5 > v_7
Question 22
Physics · Ray Optics and Optical Instruments · Multiple correct
A monochromatic light wave is incident normally on a glass slab of thickness $d$, as shown in the figure. The refractive index of the slab increases linearly from $n_1$ to $n_2$ over the height $h$. Which of the following statement(s) is (are) true about the light wave emerging out of the slab?
It will deflect up by an angle $\tan^{-1}\left[\frac{(n_2^2 - n_1^2)d}{2h}\right]$
It will deflect up by an angle $\tan^{-1}\left[\frac{(n_2 - n_1)d}{h}\right]$
It will not deflect.
The deflection angle depends only on $(n_2 - n_1)$ and not on the individual values of $n_1$ and $n_2$.
Answer: (b), (d)
Solution
Given the diagram, we have the equation $n_1 d + \ell = n_2 d$. The tangent of the angle $\theta$ is given by $$\tan \theta = \frac{\ell}{h} = \frac{(n_2 - n_1) d}{h}.$$
Question 23
Physics · System of Particles and Rotational Motion · Multiple correct
An annular disk of mass $M$, inner radius $a$ and outer radius $b$ is placed on a horizontal surface with coefficient of friction $\mu$, as shown in the figure. At some time, an impulse $J_0 \hat{x}$ is applied at a height $h$ above the center of the disk. If $h = h_m$ then the disk rolls without slipping along the $x$-axis. Which of the following statement(s) is(are) correct?
For $\mu \neq 0$ and $a \to 0$, $h_m = b/2$
For $\mu \neq 0$ and $a \to b$, $h_m = b$
For $h = h_m$, the initial angular velocity does not depend on the inner radius $a$.
For $\mu = 0$ and $h = 0$, the wheel always slides without rolling.
Answer: (a), (b), (c), (d)
Solution
Given $J_0 = mv$ (1) $J_0 h_m = I_c \omega$ (2) $v = \omega R$ (3) Therefore, $h_m = \frac{I_c}{mR}$ (A) If $a = 0$, $I_c = \frac{1}{2} mb^2$ and $R = b$, therefore $h_m = \frac{b}{2}$ (B) If $a = b$, $I_c = mb^2$ and $R = b$, therefore $h_m = b$ (C) $v = \frac{J_0}{m}$ therefore $100 = \frac{V}{R} = \frac{J_0}{mR}$ (D) Force is acting on COM. Therefore, no rotation.
The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by $\vec{E} = 30(2\hat{x} + \hat{y}) \sin \left[ 2\pi \left( 5 \times 10^{14} \, t - \frac{10^7}{3} \, z \right) \right] \, \mathrm{V \, m^{-1}}$. Which of the following option(s) is(are) correct? [Given: The speed of light in vacuum, $c = 3 \times 10^8 \, \mathrm{ms^{-1}}$]
The wave is polarized in the $xy$-plane with polarization angle $30^\circ$ with respect to the $x$-axis.
The refractive index of the medium is 2.
Answer: (a), (d)
Solution
The speed of light in the medium is calculated as follows: $$C_{medium} = \frac{5 \times 10^{14}}{10^7 / 3} = 1.5 \times 10^8 \, m/s \therefore \mu = 2$$ Given $$C_{medium} = \frac{E}{B} \implies B = \frac{E}{C_m} = \frac{30 \sqrt{5}}{1.5 \times 10^8} = 2 \sqrt{5} \times 10^{-7}$$ The direction of $\vec{B}$ is given by: $$\vec{B}_{direction} \equiv \hat{k} \times (2 \hat{i} + \hat{j}) \equiv \frac{2 \hat{j} - \hat{i}}{\sqrt{5}}$$ Therefore, $$\vec{B} = 2 \times 10^{-7} (-\hat{i} + 2 \hat{j}) \sin \left[ 27 \left( 5 \times 10^{17} t - \frac{10^7}{3} z \right) \right]$$ From the diagram, $$\tan \theta = \frac{1}{2}$$
Question 25
Physics · System of Particles and Rotational Motion · Numerical
A thin circular coin of mass $5 \, \mathrm{gm}$ and radius $4/3 \, \mathrm{cm}$ is initially in a horizontal $xy$-plane. The coin is tossed vertically up (+$z$ direction) by applying an impulse of $\frac{\sqrt{\pi}}{2} \times 10^{-2} \, \mathrm{N\cdot s}$ at a distance $2/3 \, \mathrm{cm}$ from its center. The coin spins about its diameter and moves along the +$z$ direction. By the time the coin reaches back to its initial position, it completes $n$ rotations. The value of $n$ is ____. [Given: The acceleration due to gravity $g = 10 \, \mathrm{m/s^2}$]
Answer: 30
Solution
Given the equations: $$J = mv ......(1)$$ $$Jr = I_c \omega ......(2)$$ $$J_c = \frac{1}{4} m R^2 ......(3)$$ $$t = \frac{2v}{g} ......(4)$$ $$\theta = 2 \pi N = \omega t ......(5) : N = 30$$
Question 26
Physics · Electromagnetic Induction · Numerical
A rectangular conducting loop of length 4 cm and width 2 cm is in the $xy$-plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction $\frac{\sqrt{3}}{2} \hat{x} + \frac{1}{2} \hat{y}$ with a constant speed $v$. The wire is carrying a steady current $I = 10 \, \mathrm{A}$ in the positive $x$-direction. A current of $10 \, \mu\mathrm{A}$ flows through the loop when it is at a distance $d = 4 \, \mathrm{cm}$ from the wire. If the resistance of the loop is $0.1 \, \Omega$, then the value of $v$ is _________ $\mathrm{ms}^{-1}$. [Given: The permeability of free space $\mu_0 = 4\pi \times 10^{-7} \, \mathrm{NA}^{-2}$]
A string of length 1 m and mass $2 \times 10^{-5}$ kg is under tension $T$. when the string vibrates, two successive harmonics are found to occur at frequencies $750$ Hz and $1000$ Hz. The value of tension $T$ is _______ Newton.
Answer: 5
Solution
Given the equation for frequency: $$f = \frac{P}{2\ell} \sqrt{\frac{T}{\mu}}$$ From equation (1): $$750 = \frac{P}{2} \sqrt{\frac{T}{\mu}} (1)$$ From equation (2): $$1000 = \frac{P+1}{2} \sqrt{\frac{T}{\mu}} (2)$$ Equating the ratios: $$\frac{4}{3} = \frac{P+1}{P} \therefore P = 3$$ Substituting back: $$1000 = \frac{4}{2} \sqrt{\frac{T}{2 \times 10^{-5}}} \therefore T = 5 \, N$$
Question 28
Physics · Mechanical Properties of Fluids · Numerical
An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius 0.1 mm is dipped vertically into the liquid through the airtight piston hole, as shown in the figure. The air in the container is isothermally compressed from its original volume $V_0$ to $\frac{100}{101} V_0$ with the movable piston. Considering air as an ideal gas, the height $(h)$ of the liquid column in the capillary above the liquid level in cm is______. [Given: Surface tension of the liquid is $0.075 \, \mathrm{Nm}^{-1}$, atmospheric pressure is $10^5 \, \mathrm{N} \, \mathrm{m}^{-2}$, acceleration due to gravity $(g)$ is $10 \, \mathrm{m} \, \mathrm{s}^{-2}$, density of the liquid is $10^3 \, \mathrm{kg} \, \mathrm{m}^{-3}$ and contact angle of capillary surface with the liquid is zero]
Answer: 25
Solution
Given $$h_0 = \frac{2T \cos \theta}{\rho g r} = \frac{2 \times 0.075 \times 1}{10^3 \times 10 \times 10^{-4}} = 15 \, \mathrm{cm}$$ $$P_0 V_0 = P \frac{100 V_0}{101} \Rightarrow P = \frac{101}{100} P_0$$ $$P_0 - \frac{2T \cos \theta}{r} + \rho g h = P = \frac{101}{100} P_0$$ $$\Rightarrow -\rho g h_0 + \rho g h = \frac{P_0}{100}$$ $$\Rightarrow h = h_0 + \frac{P_0}{100 \rho g}$$ $$= 15 \, \mathrm{cm} + \frac{10^5}{100 \times 10^3 \times 10} = 25 \, \mathrm{cm}$$
Question 29
Physics · Nuclei · Fill in the blank
In a radioactive decay process, the activity is defined as $A = -\frac{dN}{dt}$, where $N(t)$ is the number of radioactive nuclei at time $t$. Two radioactive sources, $S_1$ and $S_2$ have same activity at time $t = 0$. At a later time, the activities of $S_1$ and $S_2$ are $A_1$ and $A_2$, respectively. When $S_1$ and $S_2$ have just completed their $3^{rd}$ and $7^{th}$ half-lives, respectively, the ratio $A_1/A_2$ is .
Answer: 16
Solution
At time $t = 0$, the amounts are $A_0$ for both $S_1$ and $S_2$. At time $t = \tau$, the amounts are $A_1$ and $A_2$. The ratio is given by: $$\frac{A_1}{A_2} = \frac{A_0 (0.5)^{t/(t_{1/2})_1}}{A_0 (0.5)^{t/(t_{1/2})_2}} = \frac{(0.5)^3}{(0.5)^7} = 2^4 = 16$$
Question 30
Physics · Thermodynamics · Numerical
One mole of an ideal gas undergoes two different cyclic processes I and II, as shown in the $P-V$ diagrams below. In cycle I, processes $a$, $b$, $c$ and $d$ are isobaric, isothermal, isobaric and isochoric, respectively. In cycle II, processes $a'$, $b'$, $c'$ and $d'$ are isothermal, isochoric, isobaric and isochoric, respectively. The total work done during cycle I is $W_I$ and that during cycle II is $W_{II}$. The ratio $W_I/W_{II}$ is _______.
$S_1$ and $S_2$ are two identical sound sources of frequency $656\,\text{Hz}$. The source $S_1$ is located at $O$ and $S_2$ moves anti-clockwise with a uniform speed $4\sqrt{2}\,\text{m s}^{-1}$ on a circular path around $O$, as shown in the figure. There are three points $P$, $Q$ and $R$ on this path such that $P$ and $R$ are diametrically opposite while $Q$ is equidistant from them. A sound detector is placed at point $P$. The source $S_1$ can move along direction $OP$. [Given: The speed of sound in air is $324\,\text{m s}^{-1}$] When only $S_2$ is emitting sound and it is $Q$, the frequency of sound measured by the detector in Hz is
Answer: 648
Solution
The frequency observed $f'$ is given by the formula: $$f' = \frac{C}{C + v \cos 45^\circ} f$$ Substituting the given values: $$f' = \frac{324}{324 + 4\sqrt{2} \times \frac{1}{\sqrt{2}}} \times 656 = 648 \, Hz$$
Question 32
Physics · Waves · Numerical
$S_1$ and $S_2$ are two identical sound sources of frequency $656\,\text{Hz}$. The source $S_1$ is located at $O$ and $S_2$ moves anti-clockwise with a uniform speed $4\sqrt{2}\,\text{m s}^{-1}$ on a circular path around $O$, as shown in the figure. There are three points $P$, $Q$ and $R$ on this path such that $P$ and $R$ are diametrically opposite while $Q$ is equidistant from them. A sound detector is placed at point $P$. The source $S_1$ can move along direction $OP$. [Given: The speed of sound in air is $324\,\text{m s}^{-1}$] Consider both sources emitting sound. When $S_2$ is at $R$ and $S_1$ approaches the detector with a speed $4 \, \mathrm{ms}^{-1}$, the beat frequency measured by the detector is $\mathrm{Hz}$.
Answer: 8.2
Solution
Given $f_{P from S_2} = 656 \, Hz$. $f_{P from S_1} = \frac{C}{C-V} f = \frac{656 \times 324}{324 - 4} = 664.2$. $\Delta f = 664.2 - 656 = 8.2 \, Hz$.
Question 33
Physics · Mechanical Properties of Fluids · Fill in the blank
A cylindrical furnace has height $(H)$ and diameter $(D)$ both $1\,\text{m}$. It is maintained at temperature $360\,\text{K}$. The air gets heated inside the furnace at constant pressure $P_a$ and its temperature becomes $T = 360\,\text{K}$. The hot air with density $\rho$ rises up a vertical chimney of diameter $d = 0.1\,\text{m}$ and height $h = 9\,\text{m}$ above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density $\rho_a = 1.2\,\text{kg m}^{-3}$, pressure $P_a$ and temperature $T_a = 300\,\text{K}$ enters the furnace. Assume air as an ideal gas, neglect the variations in $\rho$ and $T$ inside the chimney and the furnace. Also ignore the viscous effects. [Given: The acceleration due to gravity $g = 10\,\text{m s}^{-2}$ and $\pi = 3.14$] Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is ________ gm $s^{-1}$.
Answer: 60.80,60.81
Solution
Given $\rho_0 T_0 = \rho T$. $$1.2 \times 300 = \rho (360) \therefore \rho = 1$$ Between A and B: $$P_0 + \frac{1}{2} \rho V_0^2 = P + \frac{1}{2} \rho V^2 + \rho gh \ldots (1)$$ $$\frac{\pi D^2}{4} V_0 = \frac{\pi d^2}{4} V \ldots (2)$$ Between B and C: $$P + \frac{1}{2} \rho V^2 = P_0 - \rho_0 g(H + h) + \frac{1}{2} \rho V^2 \ldots (3)$$ From (1) and (2): $$\Rightarrow P_0 + \frac{1}{2} \rho \left( \frac{V d^2}{D^2} \right)^2 = P + \frac{1}{2} \rho V^2 + \rho gh$$ $$\Rightarrow \rho_0 g(H + h) = \frac{1}{2} \rho V^2 \left[ 1 - \frac{d^4}{D^4} \right] + \rho gh$$ $$\Rightarrow V^2 \simeq \frac{2 \rho_0}{\rho} g(H + h) - 2gh$$ $$= 2 \times 1.2 \times 10 \times 10^{-2} \times 10 \times 9$$ $$= 240 - 180 = 60 \therefore V = \sqrt{60} \mathrm{m/s}$$ $$Q_m = \frac{\pi d^2}{4} V = 1 \times \frac{\pi}{4} \times 10^{-2} \times \sqrt{60} \approx 60.80$$ Ans. 60.80 to 60.81
Question 34
Physics · States of Matter · Numerical
A cylindrical furnace has height $(H)$ and diameter $(D)$ both $1\,\text{m}$. It is maintained at temperature $360\,\text{K}$. The air gets heated inside the furnace at constant pressure $P_a$ and its temperature becomes $T = 360\,\text{K}$. The hot air with density $\rho$ rises up a vertical chimney of diameter $d = 0.1\,\text{m}$ and height $h = 9\,\text{m}$ above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density $\rho_a = 1.2\,\text{kg m}^{-3}$, pressure $P_a$ and temperature $T_a = 300\,\text{K}$ enters the furnace. Assume air as an ideal gas. Neglect the variations in $\rho$ and $T$ inside the chimney and the furnace. Also ignore the viscous effects. [Given: The acceleration due to gravity $g = 10\,\text{m s}^{-2}$ and $\pi = 3.14$] When the chimney is closed using a cap at the top, a pressure difference $\Delta P$ develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of $\Delta P$ is ______ $\mathrm{Nm}^{-2}$.
Answer: 30
Solution
Given $P = constant$, we have $\rho_a T_a = \rho T$. Substituting the values, $1.2 \times 300 = \rho \times 360$. Solving for $\rho$, we get $\rho = 1 \, \mathrm{kg/m^3}$. The pressure difference $\Delta P$ is given by $\rho_a g (h + H) - \rho g h$. Calculating, $1.2 \times 10 \times 10 - 1 \times 10 \times 9$. This simplifies to $120 - 90 = 30 \, \mathrm{N/m^2}$.
Chemistry
Question 35
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The correct molecular orbital diagram for $\mathrm{F}_2$ molecule in the ground state is
Answer: (c)
Solution
Solution for $\mathrm{F_2}$ (18 electrons). Naming of molecular orbitals are as per preference of formation of $\sigma$ and $\pi$ bonds respectively.
Question 36
Chemistry · Surface Chemistry · Single correct
Consider the following statements related to colloids. (I) Lyophobic colloids are not formed by simple mixing of dispersed phase and dispersion medium. (II) For emulsions, both the dispersed phase and the dispersion medium are liquid. (III) Micelles are produced by dissolving a surfactant in any solvent at any temperature. (IV) Tyndall effect can be observed from a colloidal solution with dispersed phase having the same refractive index as that of the dispersion medium. The option with the correct set of statements is
(I) and (II)
(II) and (III)
(III) and (IV)
(II) and (IV)
Answer: (a)
Solution
(I) As in Lyophobic colloids there is no interaction between dispersed phase and dispersion medium, special methods are used for preparation, simple mixing will not form colloid. (II) Emulsions are liquid in liquid type colloids. (III) Dissolving surfactant in a proper solvent will only form micelles at temperature above Kraft's temperature. (IV) For Tyndall effect there must be a large difference in refractive index between dispersed phase and dispersion medium in order to have diffraction of light. Hence ans (I) & (II) are correct.
Question 37
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
In the following reactions, $P$, $Q$, $R$, and $S$ are the major products. The correct statement about $P$, $Q$, $R$, and $S$ is
$P$ is a primary alcohol with four carbons.
$Q$ undergoes Kolbe’s electrolysis to give an eight-carbon product.
$R$ has six carbons and it undergoes Cannizzaro reaction.
$S$ is a primary amine with six carbons.
Answer: (b)
Solution
The solution involves several steps of organic reactions. First, the reaction of the compound with $Mg$ in dry ether forms $CH_2MgCl$. Then, the addition of $H_2O$ leads to the formation of $CH_3CH_2OH$ and $Mg(OH)Cl$. Next, the compound reacts with $CO_2$ to form $CH_2C(OMgCl)OH$, which upon hydrolysis gives $CH_2COOH$. Kolbe electrolysis of $CH_2COONa$ results in the formation of an 8-carbon compound. Further reaction with $NaOH$ gives $CH_2COOH$ and $Mg(OH)Cl$. The compound is then reacted with $CH_3C=O$ to form $CH_3CH(OH)CH_3$. Oxidation with $CrO_3$ yields $CH_3COCH_3$, which does not undergo the Cannizzaro reaction. The reaction with $NaCN$ followed by hydrogenation with $H_2, Ni$ gives $CH_2CH_2NH_2$. Further reaction with $CHCl_3, KOH$ and heating results in $CH_2CH_2NC$. Reduction with $LiAlH_4$ and hydrolysis gives $CH_2CH_2NHCH_3$, which is a secondary amine.
Question 38
Chemistry · Biomolecules · Single correct
A disaccharide $X$ cannot be oxidised by bromine water. The acid hydrolysis of $X$ leads to a laevorotatory solution. The disaccharide $X$ is
Answer: (a)
Solution
Sucrose in the presence of $\mathrm{H_3O^+}$ converts to glucose and fructose. The specific rotation changes from $+52.5^\circ$ to $-92^\circ$, indicating that the mixture of products is laevorotatory. Sucrose does not react with $\mathrm{Br_2 + H_2O}$. BCD are reducing sugars and will get oxidized by $\mathrm{Br_2 + H_2O}$.
The complex(es), which can exhibit the type of isomerism shown by $[Pt(NH_3)_2Br_2]$, is(are) $[en = H_2NCH_2CH_2NH_2]$
$[Pt(en)(SCN)_2]$
$[Zn(NH_3)_2Cl_2]$
$[Pt(NH_3)_2Cl_4]$
$[Cr(en)_2(H_2O)(SO_4)]^+$
Answer: (c), (d)
Solution
Given $[Pt(NH_3)_2Br_2]^+$. Hybridisation: $dsp^2$, geometry: square planar. (A) $[Pt(en)(SCN)_2]$: square planar, cis–trans not possible. (B) $[Zn(NH_3)_2Cl_2]$: tetrahedral, cis–trans not possible. (C) $[Pt(NH_3)_2Cl_4]$: octahedral, cis–trans possible. (D) $[Cr(en)_2(H_2O)SO_4]^+$: Octahedral.
Question 40
Chemistry · The Solid State · Multiple correct
Atoms of metals x, y, and z form face-centred cubic (fcc) unit cell of edge length $L_x$, body-centred cubic (bcc) unit cell of edge length $L_y$, and simple cubic unit cell of edge length $L_z$, respectively. If $r_z = \frac{\sqrt{3}}{2} r_y; r_y = \frac{8}{\sqrt{3}} r_x; M_z = \frac{3}{2} M_y$ and $M_z = 3 M_x$, then the correct statement (s) is (are) [Given : $M_x, M_y,$ and $M_z$ are molar masses of metals x, y, and z, respectively. $r_x, r_y,$ and $r_z$ are atomic radii of metals x, y, and z, respectively.]
Packing efficiency of unit cell of x > Packing efficiency of unit cell of y > Packing efficiency of unit cell of z
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Multiple correct
In the following reactions, $\textbf{P, Q, R, and S}$ are the major products. The correct statement (S) about P, Q, R, and S is (are)
$\textbf{P}$ and $\textbf{Q}$ are monomers of polymers dacron and glyptal, respectively.
$\textbf{P, Q, and R}$ are dicarboxylic acids.
Compounds $\textbf{Q}$ and $\textbf{R}$ are the same.
$\textbf{R}$ does $\textit{not}$ undergo aldol condensation and $\textbf{S}$ does $\textit{not}$ undergo Cannizzaro reaction.
Answer: (c), (d)
Solution
The first reaction involves the oxidation of the ethyl group on the benzene ring using $KMnO_4$, $KOH$, and heat, followed by acidification with $H_3O^+$ to form compound (P). Compound (P) is then reacted with $NaOH$ and $H_2O$ in a $S_N^AE$ reaction, followed by acidification to form compound (Q), a $\beta$-keto acid. Upon heating, compound (Q) undergoes decarboxylation to form a carboxylic acid, which is then oxidized using $H_2CrO_4$ to form compound (R). In the second reaction sequence, the brominated compound is reacted with magnesium in dry ether to form a Grignard reagent. This Grignard reagent reacts with carbon dioxide to form a carboxylate, which upon acidification with $H_3O^+$ gives a carboxylic acid. This carboxylic acid is then reacted with ammoniacal $AgNO_3$ to form the final product.
Question 42
Chemistry · Redox Reactions · Numerical
\[ \mathrm{H_2S} \] \[ (5\ \text{moles}) \] reacts completely with acidified aqueous potassium permanganate solution. In this reaction, the number of moles of water produced is \[ x, \] and the number of moles of electrons involved is \[ y. \] The value of \[ (x+y) \] is
Answer: 18
Solution
The reaction is given by: $$2\mathrm{KMnO_4} + 5\mathrm{H_2S} + 3\mathrm{H_2SO_4} \rightarrow \mathrm{K_2SO_4} + 2\mathrm{MnSO_4} + 5\mathrm{S} + 8\mathrm{H_2O}$$ Let $x = 8$ (moles of $\mathrm{H_2O}$ produced). Let $y = 14 - 4 = 10$ (number of electrons involved). Thus, $x + y = 10 + 8 = 18$.
Question 43
Chemistry · Chemical Bonding and Molecular Structure · Numerical
Among $[I_3]^+$, $[SiO_4]^{4-}$, $SO_2Cl_2$, $XeF_2$, $SF_4$, $ClF_3$, $Ni(CO)_4$, $XeO_2F_2$, $[PtCl_4]^{2-}$, $XeF_4$, and $SOCl_2$, the total number of species having $sp^3$ hybridised central atom is _____.
Answer: 5
Solution
The hybridization states for the given molecules are as follows: $I_3^-$: $sp^3$ $\mathrm{SiO_4^{4-}}$: $sp^3$ $\mathrm{SO_2Cl_2}$: $sp^3$ $\mathrm{XeF_2}$: $sp^3d$ $\mathrm{SF_4}$: $sp^3d$ $\mathrm{ClF_3}$: $sp^3d$ $\mathrm{[Ni(CO)_4]}$: $sp^3$ $\mathrm{XeO_2F_2}$: $sp^3d$ $\mathrm{[PtCl_4]^{2-}}$: $dsp^2$ $\mathrm{XeF_4}$: $sp^3d^2$ $\mathrm{SOCl_2}$: $sp^3$
Question 44
Chemistry · Redox Reactions · Numerical
Consider the following molecules: $\mathrm{Br_3O_8}$, $\mathrm{F_2O}$, $\mathrm{H_2S_4O_6}$, $\mathrm{H_2S_5O_6}$, and $\mathrm{C_3O_2}$. Count the number of atoms existing in their zero oxidation state in each molecule. Their sum is ____.
Answer: 6
Solution
For $\mathrm{Br_3O_8}$, the number of atoms with zero oxidation state is 0. For $\mathrm{F_2O}$, the number of atoms with zero oxidation state is 0. For $\mathrm{H_2S_4O_6}$, the number of atoms with zero oxidation state is 2. For $\mathrm{H_2S_5O_6}$, the number of atoms with zero oxidation state is 3. For $\mathrm{C_3O_2}$, the number of atoms with zero oxidation state is 1.
Question 45
Chemistry · Structure of Atom · Numerical
For $\mathrm{He}^+$, a transition takes place from the orbit of radius $105.8 \, \mathrm{pm}$ to the orbit of radius $26.45 \, \mathrm{pm}$. The wavelength (in nm) of the emitted photon during the transition is ___.
Answer: 30
Solution
For single electron system $r = 52.9 \times \frac{n^2}{Z} \ \mathrm{pm}$ Given $Z = 2$ for $\mathrm{He}^+$ $r_2 = 105.8 \ \mathrm{pm}$ So $105.8 = 52.9 \times \frac{n_2^2}{2}$ $n_2 = 2$ $r_1 = 26.45$ So $26.45 = 52.9 \times \frac{n_1^2}{2}$ $n_1 = 1$ So transition is from 2 to 1. Now $\frac{hc}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ So $\lambda = 30 \times 10^{-9} \ \mathrm{m} = 30 \ \mathrm{nanometer}$. Here '$R_H$' is given in terms of energy value.
Question 46
Chemistry · Solutions · Numerical
50 $\mathrm{mL}$ of 0.2 molal urea solution (density = 1.012 $\mathrm{g}$ $\mathrm{mL}^{-1}$ at 300 $\mathrm{K}$) is mixed with 250 $\mathrm{mL}$ of a solution containing 0.06 $\mathrm{g}$ of urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in Torr) of the resulting solution at 300 $\mathrm{K}$ is . [Use : Molar mass of urea = 60 $\mathrm{g}$ $\mathrm{mol}^{-1}$; gas constant, R = 62 $\mathrm{L}$ $\,$ $\mathrm{Torr}$ $\mathrm{K}^{-1}$ $mol^{-1}$; Assume, $\Delta_{mix}$ H = 0, $\Delta_{mix}$ V = 0]
Answer: 682
Solution
Weight of 50 ml 0.2 molal urea = V $\times$ d = 50 $\times$ 1.012 = 50.6 $\,$ $\mathrm{gm}$. Given 0.2 molal implies 1000 $\,$ $\mathrm{gm}$ $\,$ solvent has $\,$ 0.2 $\,$ moles urea So weight of solution = 1000 + 0.2 $\times$ 60 = 1012 $\,$ $\mathrm{gm}$. So wt. of urea in 50.6 $\,$ $\mathrm{gm}$ $\,$ solution = $\frac{12 \times 50.6}{1012}$ = 0.6 $\,$ $\mathrm{gm}$ Total urea = 0.6 + 0.06 = 0.66 $\,$ $\mathrm{gm}$ Total volume = 300 $\,$ $\mathrm{ml}$ Now, osmotic pressure $\pi = C \times R \times T = \frac{0.66 \times 62 \times 300}{60 \times 0.3} = 682 \, Torr.$
Question 47
Chemistry · Haloalkanes and Haloarenes · Numerical
The reaction of \[ 4\text{-methyloct-}1\text{-ene}\ (P,\ 2.52\ \mathrm{g}) \] with \[ \mathrm{HBr} \] in the presence of \[ (C_6H_5CO)_2O_2 \] gives two isomeric bromides in a \[ 9:1 \] ratio, with combined yield of \[ 50\%. \] Of these, the entire amount of the primary alkyl bromide was reacted with appropriate amount of diethylamine followed by treatment with excess \[ K_2CO_3 \] to give a non-ionic product \[ S \] in \[ 100\% \] yield. The mass in mg of \[ S \] obtained is Use molar mass \[ [\text{ (in g mol}^{-1} H=1,\ C=12,\ N=14,\ Br=80] \]The reaction of 4-methyloct-ene (P, 2.52 $\mathrm{g}$) with $\mathrm{HBr}$ in the presence of ($\mathrm{C_6H_5CO}$)_2$\mathrm{O_2}$ gives two isomeric bromides in a 9 : 1 ratio, with combined yield of 50. Of these, the entire amount of the primary alkyl bromide was reacted with an appropriate amount of diethylamine followed by treatment with eq. $\mathrm{K_2CO_3}$ to give a non-ionic product S in 100$\%$ yield. The mass (in mg) of S obtained is __. [Use molar mass (in g $\mathrm{mol}^{-1}$) : $\mathrm{H}$ = 1, $\mathrm{C}$ = 12, $\mathrm{N}$ = 14, $\mathrm{Br}$ = 80]
Answer: 1791
Solution
The compound $\mathrm{H_2C = CH - CH_2 - CH - CH_2 - CH_2 - CH_2 - CH_3}$ (P) is given. It reacts to form $\mathrm{H_2C - CH_2 - CH - CH_2 - CH_2 - CH_3}$ with $\mathrm{Br}$ and $\mathrm{H_3C - CH - CH_2 - CH_2 - CH_2 - CH_3}$. The reaction involves $2.52 \, \mathrm{gm}$ or $0.02 \, \mathrm{mole}$. The primary alkyl bromide is formed with $0.009 \, \mathrm{mole}$. The reaction proceeds with $\mathrm{HN(CH_2CH_3)_2}$ in a $9:1$ ratio. The intermediate formed is $\mathrm{H_3C - H_2C - N^+ - CH_2 - CH_2 - CH - CH_2 - CH_2 - CH_2 - CH_3}$ with $\mathrm{Br^-}$. Upon treatment with aqueous $\mathrm{K_2CO_3}$, the final product is $\mathrm{H_3C - H_2C - N - CH_2 - CH_2 - CH - CH_2 - CH_2 - CH_3}$ along with $\mathrm{KBr}$ and $\mathrm{KHCO_3}$. The molecular mass is $199 \, \mathrm{gm}$. For $S = 0.009 \, \mathrm{mole}$, the mass is calculated as $199 \times 0.009 = 1791 \, \mathrm{mg}$.
Question 48
Chemistry · Thermodynamics · Numerical
The entropy versus temperature plot for phases $\alpha$ and $\beta$ at $1\ \mathrm{bar}$ pressure is given. $S_T$ and $S_0$ are the entropies of the phases at temperatures $T$ and $0\ \mathrm{K}$, respectively. The transition temperature for $\alpha$ to $\beta$ phase is $600\ \mathrm{K}$ and $C_{P,\beta}-C_{P,\alpha}$ is independent of temperature in the range of $200\ \mathrm{K}$ to $700\ \mathrm{K}$. If $C_{P,\alpha}$ and $C_{P,\beta}$ are heat capacities of $\alpha$ and $\beta$ phases, respectively, the value of entropy change, $S_\beta-S_\alpha$ (in $\mathrm{J\,mol^{-1}\,K^{-1}}$), at $300\ \mathrm{K}$ is _____. [Use: $\ln 2=0.69$. Given: $S_\beta-S_\alpha=0$ at $0\ \mathrm{K}$]
The entropy versus temperature plot for phases $\alpha$ and $\beta$ at $1\ \mathrm{bar}$ pressure is given. $S_T$ and $S_0$ are the entropies of the phases at temperatures $T$ and $0\ \mathrm{K}$, respectively. The transition temperature for $\alpha$ to $\beta$ phase is $600\ \mathrm{K}$, and $C_{P,\beta}-C_{P,\alpha}$ is independent of temperature in the range of $200\ \mathrm{K}$ to $700\ \mathrm{K}$. If $C_{P,\alpha}$ and $C_{P,\beta}$ are the heat capacities of $\alpha$ and $\beta$ phases, respectively, the value of the enthalpy change, $H_\beta-H_\alpha$ (in $\mathrm{J\,mol^{-1}}$), at $300\ \mathrm{K}$ is ______.
Answer: 300
Solution
As the phase transition temperature is 600 K. So at 600 K $\Delta G^\circ_{rxn} = 0$. So $\Delta H^\circ_{reaction (600)} = T \Delta S^\circ_{reaction (600)}$. $$\Delta H^\circ_{(600)} = 600 \times 1 = 600 Joule/mole$$ So $\Delta H_{600} - \Delta H_{300} = \Delta C_P (T_2 - T_1)$. $$\Delta H_{600} - \Delta H_{300} = 1 \times 300$$ $$\Delta H_{300} = \Delta H_{600} - 300 = 600 - 300 = 300 Joule/mole.$$
Question 50
Chemistry · Amines · Numerical
A trinitro compound, $1,3,5\text{-tris-(4-nitrophenyl)benzene}$, on complete reaction with an excess of $\mathrm{Sn/HCl}$ gives a major product, which on treatment with an excess of $\mathrm{NaNO_2/HCl}$ at $0^\circ\mathrm{C}$ provides $P$ as the product. $P$ upon treatment with excess of $\mathrm{H_2O}$ at room temperature gives the compound $Q$. Bromination of $Q$ in aqueous medium furnishes the product $R$. The compound $P$ upon treatment with an excess of phenol under basic conditions gives the product $S$. The molar mass difference between compounds $Q$ and $R$ is $474\ \mathrm{g\,mol^{-1}}$ and between compounds $P$ and $S$ is $172.5\ \mathrm{g\,mol^{-1}}$. The number of heteroatoms present in one molecule of $R$ is _____. [Use: Molar mass (in $\mathrm{g\,mol^{-1}}$): H = 1, C = 12, N = 14, O = 16, Br = 80, Cl = 35.5. Atoms other than C and H are considered as heteroatoms.]
Answer: 9
Solution
The reaction starts with the reduction of the nitro groups using $\mathrm{Sn/HCl}$ in excess, converting them to amino groups. The resulting compound is then treated with $\mathrm{NaNO_2/HCl}$ to form the diazonium salt $(P)$. In a basic medium, this diazonium salt reacts with phenol to form the azo compound $(S)$. The number of carbons in $(S)$ is 42, and the number of heteroatoms is 9, giving a total of 51 atoms. The compound $(S)$ is then treated with $\mathrm{H_2O}$ in excess, followed by $\mathrm{Br_2/H_2O}$, to form compound $(R)$. The number of heteroatoms in $(R)$ is 9.
Question 51
Chemistry · Amines · Numerical
A trinitro compound, $1,3,5\text{-tris-(4-nitrophenyl)benzene}$, on complete reaction with an excess of $\mathrm{Sn/HCl}$ gives a major product, which on treatment with an excess of $\mathrm{NaNO_2/HCl}$ at $0^\circ\mathrm{C}$ provides $P$ as the product. $P$ upon treatment with excess of $\mathrm{H_2O}$ at room temperature gives the product $Q$. Bromination of $Q$ in aqueous medium furnishes the product $R$. The compound $P$ upon treatment with an excess of phenol under basic conditions gives the product $S$. The molar mass difference between compounds $Q$ and $R$ is $474\ \mathrm{g\,mol^{-1}}$ and between compounds $P$ and $S$ is $172.5\ \mathrm{g\,mol^{-1}}$. The total number of carbon atoms and heteroatoms present in one molecule of $S$ is ____. [Use: Molar mass in $\mathrm{g\,mol^{-1}}$: H = 1, C = 12, N = 14, O = 16, Br = 80, Cl = 35.5. Atoms other than C and H are considered as heteroatoms.]
Answer: 51
Solution
The reaction starts with the reduction of the nitro groups to amino groups using $\mathrm{Sn/HCl}$ in excess. The resulting compound is then treated with $\mathrm{NaNO_2/HCl}$ to form the diazonium salt $(P)$. This diazonium salt undergoes a coupling reaction with phenol in a basic medium to form compound $(S)$. The number of carbons in $(S)$ is 42, and the number of heteroatoms is 9, giving a total of 51 atoms. Compound $(S)$ is then treated with $\mathrm{H_2O}$ in excess to form compound $(R)$. The number of heteroatoms in $(R)$ is 9.