JEE Advanced 3 October 2021 Paper 2 question paper with solutions
JEE Advanced 3 October 2021 Paper 2: all 57 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 · Permutations and Combinations · Multiple correct
Let $$S_1 = \{(i, j, k) : i, j, k \in \{1, 2, \ldots, 10\}\}$$ $$S_2 = \{(i, j) : 1 \leq i < j + 2 \leq 10, \ i, j \in \{1, 2, \ldots, 10\}\},$$ $$S_3 = \{(i, j, k, l) : 1 \leq i < j < k < l, \ i, j, k, l \in \{1, 2, \ldots, 10\}\}.$$ and $$S_4 = \{(i, j, k, l) : i, j,$$ k and l are distinct elements in $$\{1, 2, \ldots, 10\}\}.$$ If the total number of elements in the set $S_r$ is $n_r$, $r = 1, 2, 3, 4$, then which of the following statements is (are) TRUE?
Maths · Properties of Triangles · Multiple correct
Consider a triangle PQR having sides of lengths p, q and r opposite to the angles P, Q and R, respectively. Then which of the following statements is (are) TRUE?
$\cos P \geq 1 - \frac{p^2}{2qr}$
$\cos R \geq \left(\frac{q-r}{p+q}\right) \cos P + \left(\frac{p-r}{p+q}\right) \cos Q$
Given the triangle with sides $p$, $q$, and $r$, and angles $P$, $Q$, and $R$, we analyze the options: (A) $\($ $\cos$ P = $\frac{q^2 + r^2 - p^2}{2qr}$ = $\frac{q^2 + r^2}{2qr}$ - $\frac{p^2}{2qr}$ $\geq$ 1 - $\frac{p^2}{2qr}$ $\)$ (as $p^2 + q^2 \geq 2qr$ (AM ≥ GM)), so (A) is correct. (B) $\($(p + q) $\cos$ R $\geq$ (q - r) $\cos$ P + (p - r) $\cos$ Q $\)$ $\($$\Rightarrow$ (p $\cos$ R + r $\cos$ P) + (q $\cos$ R + r $\cos$ Q) $\geq$ q $\cos$ P + p $\cos$ Q $\)$ $\($$\Rightarrow$ q + p $\geq$ r $\)$ So (B) is correct. (C) $\($ $\frac{q + r}{p}$ = $\frac{\sin Q + \sin R}{\sin P}$ $\geq$ $\frac{2 \sqrt{\sin Q \times \sin R}}{\sin P}$ $\)$ so (C) is incorrect. (D) $\($ $\cos$ Q > $\frac{p}{r}$ $\Rightarrow$ $\sin$ R $\cos$ Q > $\sin$ P $\)$ $\($$\Rightarrow$ $\sin$ P + $\sin$ (R - Q) > 2 $\sin$ P $\)$ $\($$\Rightarrow$ $\sin$ (R - Q) > $\sin$ P $\)$ need not necessarily hold true if $R < Q$ Hence (A), (B)
Question 3
Maths · Integrals · Multiple correct
Let $f: \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \to \mathbb{R}$ be a continuous function such that \[ f(0) = 1 \quad \text{and} \quad \int_0^{\frac{\pi}{3}} f(t)\,dt = 0. \] Then which of the following statements is (are) \textbf{TRUE}?
The equation $f(x) - 3 \cos 3x = 0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
The equation $f(x) - 3 \sin 3x = -\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
For any real numbers $\alpha$ and $\beta$, let $y_{\alpha, \beta}(x)$, $x \in \mathbb{R}$, be the solution of the differential equation \[\frac{dy}{dx} + \alpha y = x e^{\beta x}, y(1) = 1 \] Let $S = \{ y_{\alpha, \beta}(x) : \alpha, \beta \in \mathbb{R} \}$. Then which of the following functions belong(s) to the set $S$?
Let $\mathrm{O}$ be the origin and $\overrightarrow{\mathrm{OA}}$ = 2$\hat{i}$ + 2$\hat{j}$ + $\hat{k}$, $\overrightarrow{\mathrm{OB}}$ = $\hat{i}$ - 2$\hat{j}$ + 2$\hat{k}$ and $\overrightarrow{\mathrm{OC}}$ = $\frac{1}{2}$ ($\overrightarrow{\mathrm{OB}}$ - $\lambda$ $\overrightarrow{\mathrm{OA}}$) for some $\lambda$ > 0. If | $\overrightarrow{\mathrm{OB}}$ $\times$ $\overrightarrow{\mathrm{OC}}$ | = $\frac{9}{2}$, then which of the following statements is (are) $\textbf{TRUE}$?
Projection of $\overrightarrow{\mathrm{OC}}$ on $\overrightarrow{\mathrm{OA}}$ is -$\frac{3}{2}$
Area of the triangle OAB is $\frac{9}{2}$
Area of the triangle ABC is $\frac{9}{2}$
The acute angle between the diagonals of the parallelogram with adjacent sides $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OC}}$ is $\frac{\pi}{3}$
Answer: (a), (b), (c)
Solution
Given $\overline{OB} \times \overline{OC} = \frac{1}{2} \overline{OB} \times (\overline{OB} - \lambda \overline{OA})$. This equals $\frac{\lambda}{2} (\overline{OA} \times \overline{OB})$. We have $|\overline{OB}| \times |\overline{OC}| = \frac{|\lambda|}{2} |\overline{OA}| \times |\overline{OB}|$ (Note $\overline{OA}$ and $\overline{OB}$ are perpendicular). Thus, $$\frac{9 \lambda}{2} = \frac{9}{2} \Rightarrow \lambda = 1 (given \lambda > 0)$$ So $\overline{OC} = \frac{\overline{OB} - \overline{OA}}{2} = \frac{\overline{AB}}{2}$. M is the midpoint of $AB$. Note the projection of $\overline{OC}$ on $\overline{OA} = -\frac{3}{2}$. We have $\tan \theta = \frac{1}{3}$. The area of $\Delta ABC = \frac{9}{2}$. The acute angle between diagonals is $$\tan^{-1} \left( \frac{1 + \frac{1}{3}}{1 - \frac{1}{3}} \right) = \tan^{-1} 2.$$
Question 6
Maths · Conic Sections · Multiple correct
Let E denote the parabola $y^2 = 8x$. Let $P = (-2, 4)$, and let $Q$ and $Q'$ be two distinct points on E such that the lines $PQ$ and $PQ'$ are tangents to E. Let $F$ be the focus of E. Then which of the following statements is (are) TRUE?
The triangle $PFQ$ is a right-angled triangle
The triangle $QPQ'$ is a right-angled triangle
The distance between $P$ and $F$ is $5\sqrt{2}$
$F$ lies on the line joining $Q$ and $Q'$
Answer: (a), (b), (d)
Solution
Note that P lies on directrix so triangle PQQ' is right angled, hence QQ' passes through focus F. PF = 4$\sqrt{2}$. Equation of QF is y = x - 2 and PF is x + y = 2. Hence A, B, D.
Question 7
Maths · Conic Sections · Fill in the blank
Consider the region $R = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x \geq 0$ and $y^2 \leq 4 - x\}$. Let $F$ be the family of all circles that are contained in $R$ and have centers on the $x$-axis. Let $C$ be the circle that has largest radius among the circles in $F$. Let $(\alpha, \beta)$ be a point where the circle $C$ meets the curve $y^2 = 4 - x$. The radius of the circle $C$ is _____.
Answer: 1.5
Solution
Let the circle be $x^2 + y^2 + \lambda x = 0$. For point of intersection of circle and parabola $y^2 = 4 - x$. $$x^2 + 4 - x + \lambda x = 0 \implies x^2 + x(\lambda - 1) + 4 = 0$$ For tangency: $\Delta = 0 \implies (\lambda - 1)^2 - 16 = 0 \implies \lambda = 5$ (rejected) or $\lambda = -3$. Circle: $x^2 + y^2 - 3x = 0$. Radius $= \frac{3}{2} = 1.5$.
Question 8
Maths · Conic Sections · Fill in the blank
Consider the region $R = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x \geq 0$ and $y^2 \leq 4 - x\}$. Let $F$ be the family of all circles that are contained in $R$ and have centers on the $x$-axis. Let $C$ be the circle that has largest radius among the circles in $F$. Let $(\alpha, \beta)$ be a point where the circle $C$ meets the curve $y^2 = 4 - x$. The value of $\alpha$ is _____.
Answer: 2
Solution
For point of intersection: $$x^2 - 4x + 4 = 0 \implies x = 2 so \alpha = 2$$
Question 9
Maths · Sets · Numerical
Let $f_1 : (0, \infty) \to \mathbb{R}$ and $f_2 : (0, \infty) \to \mathbb{R}$ be defined by $$f_1(x) = \int_0^x \prod_{j=1}^{21} (t-j) \, dt, x > 0$$ and $$f_2(x) = 98(x-1)^{50} - 600(x-1)^{49} + 2450, x > 0,$$ where, for any positive integer $n$ and real numbers $a_1, a_2, \ldots, a_n$, $\prod_{i=1}^n a_i$ denotes the product of $a_1, a_2, \ldots, a_n$. Let $m_i$ and $n_i$, respectively, denote the number of points of local minima and the number of points of local maxima of function $f_i$, $i = 1, 2$, in the interval $(0, \infty)$. The value of $2m_1 + 3n_1 + m_1n_1$ is .
Answer: 57
Solution
Given $f_1(x) = \int_0^x \prod_{j=1}^{21} (t-j)^j \, dt$. Differentiating, we have $$f_1'(x) = \prod_{j=1}^{21} (x-j)^j = (x-1)(x-2)^2(x-3)^3 \ldots (x-21)^{21}$$ The graph of $f_1'(x)$ shows points of minima at $4m + 1$ where $m = 0, 1, \ldots, 5 \Rightarrow m_1 = 6$. Points of maxima are at $4m - 1$ where $m = 1, 2, \ldots, 5 \Rightarrow n_1 = 5$. Thus, $2m_1 + 3n_1 + m_1n_1 = 57$.
Question 10
Maths · Binomial Theorem · Numerical
Let $f_1 : (0, \infty) \to \mathbb{R}$ and $f_2 : (0, \infty) \to \mathbb{R}$ be defined by $$f_1(x) = \int_0^x \prod_{j=1}^{21} (t-j)^j \, dt, x > 0$$ and $$f_2(x) = 98(x-1)^{50} - 600(x-1)^{49} + 2450, x > 0,$$ where, for any positive integer $n$ and real numbers $a_1, a_2, \ldots, a_n$, $\prod_{i=1}^n a_i$ denotes the product of $a_1, a_2, \ldots, a_n$. Let $m_i$ and $n_i$, respectively, denote the number of points of local minima and the number of points of local maxima of function $f_i$, $i = 1, 2$, in the interval $(0, \infty)$ The value of $6m_2 + 4n_2 + 8m_2n_2$ is ______.
Answer: 6
Solution
Given $f_2'(x) = 98(x-1)^{50} - 600(x-1)^{49} + 2450$. Therefore, $f_2''(x) = 2 \times 49 \times 50(x-1)^{49} - 50 \times 12 \times 49(x-1)^{48}$. This simplifies to $50 \times 49 \times 2(x-1)^{48}(x-1-6)$. Further simplifying gives $50 \times 49 \times 2(x-1)^{48}(x-7)$. The sign chart for $f_2'(x)$ shows a change from negative to positive at $x = 7$. Point of minima is at $x = 7$. Thus, $m_2 = 1$. There is no point of maxima, so $n_2 = 0$. Finally, $6m_2 + 4n_2 + 8m_2n_2 = 6$.
Question 11
Maths · Applications of Integrals · Fill in the blank
Let $g_i: \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}$, $i = 1, 2$, and $f: \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}$ be functions such that $g_1(x) = 1$, $g_2(x) = |4x - \pi|$ and $f(x) = \sin^2 x$, for all $x \in \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right]$ Define $S_i = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_i(x) \, dx$, $i = 1, 2$ The value of $\frac{16S_1}{\pi}$ is _____.
Let $g_i: \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}$, $i = 1, 2$, and $f: \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right] \to \mathbb{R}$ be functions such that $g_1(x) = 1$, $g_2(x) = |4x - \pi|$ and $f(x) = \sin^2 x$, for all $x \in \left[ \frac{\pi}{8}, \frac{3\pi}{8} \right]$ Define $S_i = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_i(x) \, dx$, $i = 1, 2$ The value of $\frac{48S_2}{\pi^2}$ is _____.
Answer: 1.5
Solution
Let $S_2 = \int_{\pi/8}^{3\pi/8} f(x)g_2(x) \, dx = \int_{\pi/8}^{3\pi/8} \sin^2 x |4x - \pi| \, dx$. This equals $\int_{\pi/8}^{3\pi/8} \sin^2 \left( \frac{\pi}{2} - x \right) \left| 4 \left( \frac{\pi}{2} - x \right) \right| \, dx - \pi \, dx$. This simplifies to $\int_{\pi/8}^{3\pi/8} (\cos^2 x) |\pi - 4x| \, dx$. Thus, $\Rightarrow 2S_2 = \int_{\pi/8}^{3\pi/8} |4x - \pi| (\sin^2 x + \cos^2 x) \, dx = \int_{\pi/8}^{3\pi/8} |4x - \pi| \, dx$. This equals $2 \times \frac{1}{2} \times \frac{\pi}{8} \times \frac{\pi}{2} = \frac{\pi^2}{16}$. Therefore, $\Rightarrow \frac{48S_2}{\pi^2} = \frac{3}{2} = 1.5$.
Question 13
Maths · Conic Sections · Single correct
Let $M = \{(x,y)\in\mathbb{R}\times\mathbb{R}:x^2+y^2\leq r^2\}$, where $r>0$. Consider the geometric progression $a_n=\frac{1}{2^{n-1}},\ n=1,2,3,\ldots$. Let $S_0=0$ and, for $n\geq1$, let $S_n$ denote the sum of the first $n$ terms of this progression. For $n\geq1$, let $C_n$ denote the circle with center $(S_{n-1},0)$ and radius $a_n$, and denote the circle with center $(S_n,1)$ and radius $a_n$. Consider M with $r = \frac{1025}{513}$. Let $k$ be the number of all those circles $C_n$ that are inside M. Let $l$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect. Then
$k + 2l = 22$
$2k + l = 26$
$2k + 3l = 34$
$3k + 2l = 40$
Answer: (d)
Solution
Given $S_n = 1 + \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n-1}}$. $$= 2 \left(1 - \frac{1}{2^n}\right) = 2 - \frac{1}{2^{n-1}}$$ Centre of $C_n$ is $\left(2 - \frac{1}{2^{n-2}}, 0\right)$ and radius of $C_n$ is $\frac{1}{2^{n-1}}$. When $r = \frac{1025}{S_{13}} < 2$, $C_n$ will lie inside $m$ when $$2 - \frac{1}{2^{n-2}} + \frac{1}{2^{n-1}} < \frac{1025}{S_{13}}$$ Therefore, $k = 10$. Also $\ell = 5$. $$3k + 2\ell = 30 + 10 = 40$$
Question 14
Maths · Conic Sections · Single correct
Let $M = \{(x,y)\in\mathbb{R}\times\mathbb{R}:x^2+y^2\leq r^2\}$, where $r>0$. Consider the geometric progression $a_n=\frac{1}{2^{n-1}},\ n=1,2,3,\ldots$. Let $S_0=0$ and, for $n\geq1$, let $S_n$ denote the sum of the first $n$ terms of this progression. For $n\geq1$, let $C_n$ denote the circle with center $(S_{n-1},0)$ and radius $a_n$, and denote the circle with center $(S_n,1)$ and radius $a_n$. Consider M with $r = \frac{(2^{199} - 1) \sqrt{2}}{2^{198}}$. The number of all those circles $D_n$ that are inside M is
198
199
200
201
Answer: (b)
Solution
Center of $D_n$ is $(S_{n-1}, S_{n-1})$. $$r = \frac{1}{2^{n-1}}$$ $D_n$ will lie inside when $\sqrt{2}(S_{n-1}) \frac{\sqrt{2}}{2^{198}} + \frac{1}{2^{n-1}}$$ $$\Rightarrow n = 199$$
Question 15
Maths · Applications of Integrals · Single correct
Let \[ \psi_1:[0,\infty)\to\mathbb{R},\qquad \psi_2:[0,\infty)\to\mathbb{R}, \] \[ f:[0,\infty)\to\mathbb{R} \] and \[ g:[0,\infty)\to\mathbb{R} \] be functions such that \[ f(0)=g(0)=0, \] \[ \psi_1(x)=e^{-x}+x,\qquad x\geq0, \] \[ \psi_2(x)=x^2-2x-2e^{-x}+2,\qquad x\geq0, \] \[ f(x)=\int_{-x}^{x}\left(|t|-t^2\right)e^{-t^2}\,dt, \qquad x>0 \] and \[ g(x)=\int_{0}^{x^2}\sqrt{t}\,e^{-t}\,dt, \qquad x>0. \] Which of the following statements is TRUE ?
A number is chosen at random from the set {1, 2, 3, $\ldots$ , 2000$\}$. Let p be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of 500p is ___.
Answer: 214
Solution
A = set of numbers divisible by 3 A = $\{$3, 6, 9, 12, $\ldots$, 1998$\}$ $\therefore$ $\;$ n(A) = 666 B = set of numbers divisible by 7 B = $\{$7, 14, 21, $\ldots$, 1995$\}$ $\therefore$ $\;$ n(B) = 285 A $\cap$ B = $\{$21, 42, $\ldots$, 1995$\}$ $\therefore$ $\;$ n(A $\cup$ B) = 606 + 285 - 95 = 856 required $\;$ probability = $\frac{856}{2000}$ = P so, $\;$ 500 $\;$ P = $\frac{856}{2000}$ $\times$ 500 = 214
Question 18
Maths · Conic Sections · Numerical
Let E be the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$. For any three distinct points P, Q and Q' on E, let M (P, Q) be the mid-point of the line segment joining P and Q, and M (P, Q') be the mid-point of the line segment joining P and Q'. Then the maximum possible value of the distance between M(P, Q) and M(P, Q'), as P, Q and Q' vary on E, is ____.
Answer: 4
Solution
A and B be midpoints of segment PQ and PQ' respectively. AB = distance between M(P, Q) and M(P, Q') = $\frac{1}{2}$ $\cdot$ QQ'. Since, Q, Q' must be on E, so, maximum of QQ' = 8. Therefore, Maximum of AB = $\frac{8}{2}$ = 4.
Question 19
Maths · Integrals · Numerical
For any real number x, let [x] denote the largest integer less than or equal to x. If \[ I=\int_{0}^{10}\sqrt{\frac{10x}{x+1}}\;dx \] then the value of 9I is ____.
Physics · System of Particles and Rotational Motion · Multiple correct
One end of a horizontal uniform beam of weight $W$ and length $L$ is hinged on a vertical wall at point $O$ and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point $Q$, at a height $L$ above the hinge at point $O$. A block of weight $\alpha W$ is attached at the point $P$ of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of $(2\sqrt{2})W$. Which of the following statement(s) is(are) correct?
The vertical component of reaction force at $O$ does not depend on $\alpha$
The horizontal component of reaction force at $O$ is equal to $W$ for $\alpha = 0.5$
The tension in the rope is $2W$ for $\alpha = 0.5$
The rope breaks if $\alpha > 1.5$
Answer: (a), (b), (d)
Solution
Given the diagram, we have the following equations: $$R_y + \frac{T}{\sqrt{2}} = W + \alpha W ...(i)$$ $$R_x = \frac{T}{\sqrt{2}} ...(ii)$$ Taking torque about 'O': $$W \frac{\ell}{2} + \alpha W \ell = \frac{T}{\sqrt{2}} \ell$$ $$T = \sqrt{2} \left( \frac{W}{2} + \alpha W \right) ...(iii)$$ Substituting into equation for $R_x$: $$R_x = \frac{T}{\sqrt{2}} = \left( \frac{W}{2} + \alpha W \right)$$ Taking torque about P: $$R_y \ell = W \frac{\ell}{2}$$ $$R_y = \frac{W}{2}$$ When $T = T_{max}$: $$2 \sqrt{2} W = \sqrt{2} \left( \frac{W}{2} + \alpha W \right)$$ We get $\alpha = \frac{3}{2}$
Question 21
Physics · Waves · Multiple correct
A source, approaching with speed $u$ towards the open end of a stationary pipe of length $L$, is emitting a sound of frequency $f_s$. The farther end of the pipe is closed. The speed of sound in air is $v$ and $f_0$ is the fundamental frequency of the pipe. For which of the following combination(s) of $u$ and $f_s$, will the sound reaching the pipe lead to a resonance?
$u = 0.8 \, v$ and $f_s = f_0$
$u = 0.8 \, v$ and $f_s = 2f_0$
$u = 0.8 \, v$ and $f_s = 0.5f_0$
$u = 0.5 \, v$ and $f_s = 1.5f_0$
Answer: (a), (d)
Solution
Given $f = f_s \left( \frac{v}{v-u} \right)$. (A) $f = f_0 \left( \frac{v}{v-0.8v} \right) = 5f_0$ (B) $f = 2f_0 \left( \frac{v}{v-0.8v} \right) = 10f_0$ (C) $f = 0.5f_0 \left( \frac{v}{v-0.8v} \right) = 2.5f_0$ (D) $f = 1.5f_0 \left( \frac{v}{v-0.5v} \right) = 3f_0$ All odd harmonics are available in closed pipe therefore correct Ans (A,D)
Question 22
Physics · Ray Optics and Optical Instruments · Multiple correct
For a prism of prism angle $\theta = 60^\circ$, the refractive indices of the left half and the right half are, respectively, $n_1$ and $n_2$ ($n_2 \geq n_1$) as shown in the figure. The angle of incidence $i$ is chosen such that the incident light rays will have minimum deviation if $n_1 = n_2 = n = 1.5$. For the case of unequal refractive indices, $n_1 = n$ and $n_2 = n + \Delta n$ (where $\Delta n \ll n$), the angle of emergence $e = i + \Delta e$. Which of the following statement(s) is (are) correct?
The value of $\Delta e$ (in radians) is greater than that of $\Delta n$
$\Delta e$ is proportional to $\Delta n$
$\Delta e$ lies between $2.0$ and $3.0$ milliradians, if $\Delta n = 2.8 \times 10^{-3}$
$\Delta e$ lies between $1.0$ and $1.6$ milliradians, if $\Delta n = 2.8 \times 10^{-3}$
Answer: (b), (c)
Solution
Given $1 \times \sin i = \mu \sin \left( \frac{A}{2} \right)$. $\sin i = \frac{3}{4}$. $n_1 \sin 30^\circ = 1 \sin(e)$. On differentiating both sides, $$dn \sin 30^\circ = de \cos(e)$$ $$de = \frac{dn}{2 \cos(e)}$$ $$= \frac{dn}{2 \times \frac{\sqrt{7}}{4}}$$ $$de = \frac{2}{\sqrt{7}} dn \Rightarrow de < dn$$ $$de = \frac{2.8 \times 10^{-3} \times 2}{\sqrt{7}} = 2.11 \, \mathrm{mrad}$$
A physical quantity $\vec{S}$ is defined as $\vec{S} = (\vec{E} \times \vec{B}) / \mu_0$, where $\vec{E}$ is electric field, $\vec{B}$ is magnetic field and $\mu_0$ is the permeability of free space. The dimensions of $\vec{S}$ are the same as the dimensions of which of the following quantity (ies)?
$\frac{Energy}{charge \times current}$
$\frac{Force}{Length \times Time}$
$\frac{Energy}{Volume}$
$\frac{Power}{Area}$
Answer: (b), (d)
Solution
The pointing vector $\vec{S}$ is given by $$\vec{S} = [\vec{E} \times \vec{B}] \frac{1}{\mu_0}$$ $S$ is the pointing vector that denotes the flow of energy per unit area per unit time. $$\vec{S} = \frac{watt}{m^2}$$ Hence B, D are correct.
Question 24
Physics · Nuclei · Multiple correct
A heavy nucleus $N$, at rest, undergoes fission $N \rightarrow P + Q$, where $P$ and $Q$ are two lighter nuclei. Let $\delta = M_N - M_P - M_Q$, where $M_P$, $M_Q$ and $M_N$ are the masses of $P$, $Q$ and $N$, respectively. $E_P$ and $E_Q$ are the kinetic energies of $P$ and $Q$, respectively. The speed of $P$ and $Q$ are $v_P$ and $v_Q$, respectively. If $c$ is the speed of light, which of the following statement(s) is(are) correct?
The magnitude of momentum for $P$ as well as $Q$ is $c \sqrt{2 \mu \delta}$, where $\mu = \frac{M_P M_Q}{(M_P + M_Q)}$
Answer: (a), (c), (d)
Solution
Question 25
Physics · Moving Charges and Magnetism · Multiple correct
Two concentric circular loops, one of radius $R$ and the other of radius $2R$, lie in the $xy$-plane with the origin as their common center, as shown in the figure. The smaller loop carries current $I_1$ in the anti-clockwise direction and the larger loop carries current $I_2$ in the clockwise direction, with $I_2 > 2I_1$. $\vec{B}(x, y)$ denotes the magnetic field at a point $(x, y)$ in the $xy$-plane. Which of the following statement(s) is(are) current?
$\vec{B}(x, y)$ is perpendicular to the $xy$-plane at any point in the plane
$|\vec{B}(x, y)|$ depends on $x$ and $y$ only through the radial distance $r = \sqrt{x^2 + y^2}$
$|\vec{B}(x, y)|$ is non-zero at all points for $r < R$
$\vec{B}(x, y)$ points normally outward from the $xy$-plane for all the points between the two loops
Answer: (a), (b)
Solution
(A) $d\vec{B} = \frac{\mu_0 i d\vec{\ell} \times \vec{r}}{4 \pi r^3}$ $d\vec{\ell}$ is in the xy plane and $\vec{r}$ is also in the xy plane, so $d\vec{B}$ is perpendicular to the xy plane. (B) Due to symmetry, it depends only on $r = \sqrt{x^2 + y^2}$. (C) At the center $B_1 = \frac{\mu_0 I_1}{2R}$; $B_2 = \frac{\mu_0 I_2}{4R} \Rightarrow B_2 > B_1$ But as we approach towards the first loop, $B_1$ increases to infinity, hence $B_1$ dominates. So it would be zero at some point between the inner loops and the center. Ans. (A, B)
Question 26
Physics · Current Electricity · Numerical
A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle is sealed at atmospheric pressure $p_0 = 10^5 \, \mathrm{Pa}$ so that the volume of the trapped air is $v_0 = 3.3 \, \mathrm{cc}$. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure $P_0 + \Delta p$ without changing its orientation. At this pressure, the volume of the trapped air is $v_0 - \Delta v$. Let $\Delta v = X \, \mathrm{cc}$ and $\Delta p = Y \times 10^3 \, \mathrm{Pa}$. The value of $X$ is ____.
A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle is sealed at atmospheric pressure $p_0 = 10^5 \, \mathrm{Pa}$ so that the volume of the trapped air is $v_0 = 3.3 \, \mathrm{cc}$. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure $p_0 + \Delta p$ without changing its orientation. At this pressure, the volume of the trapped air is $v_0 - \Delta v$. Let $\Delta v = X \, \mathrm{cc}$ and $\Delta p = Y \times 10^3 \, \mathrm{Pa}$. The value of $Y$ is ____.
Answer: 10.0
Solution
Isothermal process for air $P_1 V_1 = P_2 V_2$ $10^5 \times 3.3 = P_2 \times 3$ $P_2 = 1.1 \times 10^5$ $\Delta P = P_2 - P_1 = 1.1 \times 10^5 - 10^5$ $= 0.1 \times 10^5$ $= 10 \times 10^3$ Pascal $= Y \times 10^3$ Pascal So $Y = 10$
Question 28
Physics · Alternating Current · Numerical
A pendulum consists of a bob of mass $m = 0.1 \, \mathrm{kg}$ and a massless inextensible string of length $L = 1.0 \, \mathrm{m}$. It is suspended from a fixed point at height $H = 0.9 \, \mathrm{m}$ above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse $P = 0.2 \, \mathrm{kg\cdot m/s}$ is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is $J \, \mathrm{kg\cdot m^2/s}$. The kinetic energy of the pendulum just after the lift-off is $K$ Joules. The value of $J$ is _____.
Answer: 0.18
Solution
Given $L = P \times 0.9 = 0.18 \, \mathrm{kgm^2/s}$. Just after the string becomes taut, there will be no velocity along the string. Therefore, $$V_\perp = \frac{P \cos \theta}{m} = \frac{0.2 \times 0.9}{1 \times 0.1} = 1.8 \, \mathrm{m/s}$$ $$\therefore K = \frac{1}{2} m V_\perp^2 = \frac{1}{2} \times 0.1 \times 1.8^2$$ $$= 0.162 \, \mathrm{J}$$
Question 29
Physics · Current Electricity · Numerical
A pendulum consists of a bob of mass $m = 0.1 \, \mathrm{kg}$ and a massless inextensible string of length $L = 1.0 \, \mathrm{m}$. It is suspended from a fixed point at height $H = 0.9 \, \mathrm{m}$ above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse $P = 0.2 \, \mathrm{kg\cdot m/s}$ is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is $J \, \mathrm{kg\cdot m^2/s}$. The kinetic energy of the pendulum just after the lift-off is $K$ Joules. The value of $K$ is ____.
Answer: 0.16
Solution
Given $L = P \times 0.9 = 0.18 \, \mathrm{kgm^2/s}$. Just after the string becomes taut, there will be no velocity along the string. Therefore, $$V_\perp = \frac{P \cos \theta}{m} = \frac{0.2 \times 0.9}{1 \times 0.1} = 1.8 \, \mathrm{m/s}$$ $$\therefore K = \frac{1}{2} m V_\perp^2 = \frac{1}{2} \times 0.1 \times 1.8^2$$ $$= 0.162 \, \mathrm{J}$$
Question 30
Physics · Electrostatic Potential and Capacitance · Numerical
In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance $C \, \mu F$ across a $200 \, V$, $50 \, \mathrm{Hz}$ supply. The power consumed by the lamp is $500 \, W$ while the voltage drop across it is $100 \, V$. Assume that there is no inductive load in the circuit. Take $rms$ values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is $\phi$. Assume, $\pi \sqrt{3} \approx 5$. The value of C is ____.
Answer: 100.0
Solution
Question 31
Physics · Oscillations · Numerical
In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance $C \, \mu \mathrm{F}$ across a $200 \, \mathrm{V}$, $50 \, \mathrm{Hz}$ supply. The power consumed by the lamp is $500 \, \mathrm{W}$ while the voltage drop across it is $100 \, \mathrm{V}$. Assume that there is no inductive load in the circuit. Take $rms$ values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is $\phi$. Assume, $\pi \sqrt{3} \approx 5$. The value of $\phi$ is ____.
Physics · Electromagnetic Induction · Single correct
A special metal \[ S \] conducts electricity without any resistance. A closed wire loop, made of \[ S, \] does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius \[ a, \] with its center at the origin. A magnetic dipole of moment \[ m \] is brought along the axis of this loop from infinity to a point at distance \[ r\ (r\gg a) \] from the center of the loop with its north pole always facing the loop, as shown in the figure below. The magnitude of magnetic field of a dipole \[ m, \] at a point on its axis at distance \[ r, \] is \[ \frac{\mu_0}{2\pi}\frac{m}{r^3}, \] where \[ \mu_0 \] is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, \[ m_1 \text{ and } m_2, \] separated by a distance \[ r \] on the common axis, with their north poles facing each other, is \[ \frac{k m_1 m_2}{r^4}, \] where \[ k \] is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles. When the dipole $m$ is placed at a distance $r$ from the center of the loop (as shown in the figure), the current induced in the loop will be proportional to
$\frac{m}{r^3}$
$\frac{m^2}{r^2}$
$\frac{m}{r^2}$
$\frac{m^2}{r}$
Answer: (a)
Question 33
Physics · Electric Charges and Fields · Single correct
A special metal \[ S \] conducts electricity without any resistance. A closed wire loop, made of \[ S, \] does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius \[ a, \] with its center at the origin. A magnetic dipole of moment \[ m \] is brought along the axis of this loop from infinity to a point at distance \[ r\ (r\gg a) \] from the center of the loop with its north pole always facing the loop, as shown in the figure below. The magnitude of magnetic field of a dipole \[ m, \] at a point on its axis at distance \[ r, \] is \[ \frac{\mu_0}{2\pi}\frac{m}{r^3}, \] where \[ \mu_0 \] is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, \[ m_1 \text{ and } m_2, \] separated by a distance \[ r \] on the common axis, with their north poles facing each other, is \[ \frac{k m_1 m_2}{r^4}, \] where \[ k \] is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles. The work done in bringing the dipole from infinity to a distance $r$ from the center of the loop by the given process is proportional to
$\frac{m}{r^5}$
$\frac{m^2}{r^5}$
$\frac{m^2}{r^6}$
$\frac{m^2}{r^7}$
Answer: (c)
Solution
The solution starts with the expression for $\phi$ as $L i = \frac{\mu_0 m}{2 \pi r^3} \times \pi a^2$. This implies $i = \frac{\mu_0 m \pi a^2}{2 \pi r^3 L}$. Therefore, $i \propto \frac{m}{r^3}$. The expression for $m'$ is $\pi a^2 i = \frac{\mu_0 m \pi^2 a^4}{2 \pi r^3 L}$. The force $F$ is given by $\frac{k m^2 \pi^2 a^4}{2 \pi^7 L}$. The work $W$ is $\int F dr \propto \int \frac{m^2 dr}{r^7}$. Finally, $W \propto \frac{m^2}{r^6}$.
Question 34
Physics · Thermodynamics · Single correct
A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume, $C_v = 2R$. Here, $R$ is the gas constant. Initially, each side has a volume $V_0$ and temperature $T_0$. The left side has an electric heater, which is turned on at very low power to transfer heat $Q$ to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to $V_0/2$. Consequently, the gas temperatures on the left and the right sides become $T_L$ and $T_R$, respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition. The value of $\frac{T_R}{T_0}$ is
A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume, $C_v = 2R$. Here, $R$ is the gas constant. Initially, each side has a volume $V_0$ and temperature $T_0$. The left side has an electric heater, which is turned on at very low power to transfer heat $Q$ to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to $V_0/2$. Consequently, the gas temperatures on the left and the right sides become $T_L$ and $T_R$, respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition. The value of $\frac{Q}{RT_0}$ is
In order to measure the internal resistance $r_1$ of a cell of emf $E$, a meter bridge of wire resistance $R_0 = 50 \, \Omega$, a resistance $R_0/2$, another cell of emf $E/2$ (internal resistance $r$) and a galvanometer $G$ are used in a circuit, as shown in the figure. If the null point is found at $l = 72 \, \mathrm{cm}$, then the value of $r_1 = \, \Omega$.
Answer: 3
Solution
Given the circuit, we have the equation for current $i$ as follows: $$i \left( \frac{R_0}{2} + 0.28 R_0 \right) = \frac{E_0}{2}$$ Simplifying, we get: $$i \times 0.78 R_0 = \frac{E_0}{2}$$ Solving for $i$, we have: $$i = \frac{E_0}{2 \times 0.78 R_0} = \frac{E_0}{r_1 + \frac{3}{2} R_0}$$ From this, we find: $$r_1 + 1.5 R_0 = 1.56 R_0$$ Therefore, $$r_1 = 0.06 R_0$$ Converting to ohms: $$= 0.06 \times 50 = 3 \, \Omega$$
Question 37
Physics · Gravitation · Numerical
The distance between two stars of masses $3M_S$ and $6M_S$ is $9R$. Here $R$ is the mean distance between the centers of the Earth and the Sun, and $M_S$ is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period $nT$, where $T$ is the period of Earth’s revolution around the Sun. The value of $n$ is ___.
Physics · Dual Nature of Radiation and Matter · Numerical
In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals $P$, $Q$ and $R$ are $E_P$, $E_Q$ and $E_R$, respectively, and they are related by $E_P = 2E_Q = 2E_R$. In this experiment, the same source of monochromatic light is used for metals $P$ and $Q$ while a different source of monochromatic light is used for the metal $R$. The work functions for metals $P$, $Q$ and $R$ are $4.0 \, \mathrm{eV}$, $4.5 \, \mathrm{eV}$ and $5.5 \, \mathrm{eV}$, respectively. The energy of the incident photon used for metal $R$, in eV, is ___.
The reaction sequence(s) that would lead to \[ o\text{-xylene} \] as the major product is (are)
Answer: (a), (b)
Question 40
Chemistry · Amines · Multiple correct
Correct option(s) for the following sequence of reactions is(are)
Q = $KNO_2$, W = $LiAlH_4$
R = benzenamine, V = KCN
Q = $AgNO_2$, R = phenylmethanamine
W = $LiAlH_4$, V = AgCN
Answer: (c), (d)
Solution
The solution involves a series of chemical reactions starting with PhCH_3. 1. PhCH_3 is treated with Br_2 in the presence of light to form PhCH_2Br (P). 2. PhCH_3 is also oxidized using (i) KMnO_4, KOH, $\Delta$ and (ii) H_3O^+ to form PhCOOH (T). 3. PhCOOH is converted to PhCONH_2 (U) using (i) NH_3 and (ii) $\Delta$. 4. PhCH_2Br (P) is reacted with 1. AgNO_2 and 2. H_2, Pd/C to form PhCH_2NH_2 (R). 5. PhCH_2NH_2 (R) is treated with CHCl_3 and KOH to form PhCH_2NC, which is foul smelling. 6. PhCH_2Br (P) can also be converted to PhCH_2NC using V(AgCN). 7. PhCONH_2 (U) can be reduced to PhCH_2NH_2 (R) using W(LiAlH_4).
Question 41
Chemistry · Chemical Kinetics and Nuclear Chemistry · Multiple correct
For the following reaction $$2\mathrm{X} + \mathrm{Y} \xrightarrow{k} \mathrm{P}$$ the rate of reaction is $$\frac{\mathrm{d} [\mathrm{P}]}{\mathrm{dt}} = k[\mathrm{X}]$$. Two moles of $$\mathrm{X}$$ are mixed with one mole of $$\mathrm{Y}$$ to make 1.0 L of solution. At 50 s, 0.5 mole of $$\mathrm{Y}$$ is left in the reaction mixture. The correct statement(s) about the reaction is(are) (Use: $$\ln 2 = 0.693$$)
The rate constant, $k$, of the reaction is $13.86 \times 10^{-4} \, \mathrm{s}^{-1}$.
Some standard electrode potentials at 298 K are given below: $Pb^{2+}/Pb \qquad -0.13$ V \\ $Ni^{2+}/Ni \qquad -0.24$ V \\ $Cd^{2+}/Cd \qquad -0.40$ V \\ $Fe^{2+}/Fe \qquad -0.44$ V To a solution containing $0.001$ M of $\mathbf{X}^{2+}$ and $0.1$ M of $\mathbf{Y}^{2+}$, the metal rods $\mathbf{X}$ and $\mathbf{Y}$ are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of $\mathbf{X}$. The correct combination(s) of $\mathbf{X}$ and $\mathbf{Y}$, respectively, is (are) (Given: Gas constant, $R = 8.314$ J K$^{-1}$ mol$^{-1}$, Faraday constant, $F = 96500$ C mol$^{-1}$)
Cd and Ni
Cd and Fe
Ni and Pb
Ni and Fe
Answer: (a), (b), (c)
Solution
x(s) $\rightarrow$ x^{+2} (0.001 $\,$ $\mathrm{M}$) + 2e^- (anode) $\newline$ y^{+2} (0.1 $\,$ $\mathrm{M}$) + 2e^- $\rightarrow$ y $\,$ (s) (cathode) $\newline$ $\newline$ E_{cell} = E^$\circ$_{cell} - $\frac{0.06}{2}$ $\log$ $\frac{x^{+2}}{y^{+2}}$ $\newline$ $\newline$ E_{cell} = E^$\circ$_{cell} + 0.06 $\newline$ $\newline$ (A) $\;$ Cd and Ni $\;$ E^$\circ$_{cell} = +0.4 - 0.24 ; $\;$ E_{cell} = 0.22 $\newline$ (B) $\;$ Cd and Fe $\;$ E^$\circ$_{cell} = -0.04 ; $\;$ E_{cell} = 0.02 $\newline$ (C) $\;$ Ni and Pb $\;$ E^$\circ$_{cell} = 0.11 ; $\;$ E_{cell} = 0.17 $\newline$ (D) $\;$ Ni and Fe $\;$ E^$\circ$_{cell} = -0.2 ; $\;$ E_{cell} = -0.14 $\newline$ since in (A) (B) (C) E_{cell} is positive hence answer is (A) (B) (C).
The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are) (Note: py = pyridine Given: Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively)
[$\mathrm{FeCl}$_4]^- and [$\mathrm{Fe(CO)}$_4]^{2-}
[$\mathrm{Co(CO)}$_4]^- and [$\mathrm{CoCl}$_4]^{2-}
[$\mathrm{Ni(CO)}$_4] and [$\mathrm{Ni(CN)}$_4]^{2-}
[$\mathrm{Cu(py)}$_4]^+ and [$\mathrm{Cu(CN)}$_4]^{3-}
Answer: (a), (b), (d)
Solution
Sol.(A) $[\mathrm{FeCl}_4]^-$ Fe $\rightarrow [\mathrm{Ar}] \, 3d^6 4s^2$ Fe$^{+3}$ $\rightarrow [\mathrm{Ar}] \, 3d^5 4s^0$ Cl$^-$ is W.F.L. and does not pair up the unpaired electron of central metal atom. $$\therefore \mathrm{Fe}^{3+} \, (d^5) in [\mathrm{FeCl}_4]^-$$ $$\begin{array}{cccccc} \uparrow & \uparrow & \uparrow & \uparrow & \uparrow & \uparrow \\ \end{array}$$ $$\begin{array}{cccc} 3d & 4s & 4p & sp^3 \\ \end{array}$$ Tetrahedral $[\mathrm{Fe(CO)}_4]^{2-}$ Fe $\rightarrow [\mathrm{Ar}] \, 3d^6 4s^2$ Fe$^{2-}$ $\rightarrow [\mathrm{Ar}] \, 3d^8 4s^2$ $$\therefore \mathrm{Fe}^{2-} \, (d^{10}) in [\mathrm{Fe(CO)}_4]^{2-}$$ $$\begin{array}{cccccc} \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow \\ \end{array}$$ $$\begin{array}{cccc} 3d & 4s & 4p & sp^3 \\ \end{array}$$ Tetrahedral (B) $[\mathrm{Co(CO)}_4]^-$ Co $\rightarrow [\mathrm{Ar}] \, 3d^7 4s^2$ Co$^{-1}$ $\rightarrow [\mathrm{Ar}] \, 3d^8 4s^2$ $$\therefore \mathrm{Co} \, (d^{10}) in [\mathrm{Co(CO)}_4]^-$$ $$\begin{array}{cccccc} \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow \\ \end{array}$$ $$\begin{array}{cccc} 3d & 4s & 4p & sp^3 \\ \end{array}$$ Tetrahedral $[\mathrm{CoCl}_4]^{2-}$ Co $\rightarrow [\mathrm{Ar}] \, 3d^7 4s^2$ Co$^{+2}$ $\rightarrow [\mathrm{Ar}] \, 3d^7 4s^0$ Cl$^-$ is W.F.L. and does not pair up the unpaired electron of central metal atom. $$\therefore \mathrm{Co}^{2+} \, (d^7) in [\mathrm{CoCl}_4]^{2-}$$ $$\begin{array}{cccccc} \uparrow\downarrow & \uparrow\downarrow & \uparrow & \uparrow & \uparrow & \uparrow \\ \end{array}$$ $$\begin{array}{cccc} 3d & 4s & 4p & sp^3 \\ \end{array}$$ Tetrahedral (C) $[\mathrm{Ni(CO)}_4]$ Ni $\rightarrow [\mathrm{Ar}] \, 3d^8 4s^2$ Ni$^0$ $\rightarrow [\mathrm{Ar}] \, 3d^{10} 4s^0$ $$\therefore \mathrm{Ni} \, (d^{10}) in [\mathrm{Ni(CO)}_4]$$ $$\begin{array}{cccccc} \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow \\ \end{array}$$ $$\begin{array}{cccc} 3d & 4s & 4p & sp^3 \\ \end{array}$$ Tetrahedral $[\mathrm{Ni(CN)}_4]^{2-}$ Ni $\rightarrow [\mathrm{Ar}] \, 3d^8 4s^2$ Ni$^{+2}$ $\rightarrow [\mathrm{Ar}] \, 3d^8 4s^0$ CN$^-$ is S.F.L. and pair up the unpaired electron of central metal atom. $$\therefore \mathrm{Ni} \, (d^8) in [\mathrm{Ni(CN)}_4]^{2-}$$ $$\begin{array}{cccccc} \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow \\ \end{array}$$ $$\begin{array}{cccc} 3d & 4s & 4p & dsp^2 \\ \end{array}$$ Square planar (D) $[\mathrm{Cu(py)}_4]^+$ Cu $\rightarrow [\mathrm{Ar}] \, 3d^{10} 4s^1$ Cu$^{+1}$ $\rightarrow [\mathrm{Ar}] \, 3d^{10} 4s^0$ $$\therefore \mathrm{Cu}^{+1} \, (d^{10}) in [\mathrm{Cu(py)}_4]^+$$ $$\begin{array}{cccccc} \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow \\ \end{array}$$ $$\begin{array}{cccc} 3d & 4s & 4p & sp^3 \\ \end{array}$$ Tetrahedral $[\mathrm{Cu(CN)}_4]^{3-}$ Cu $\rightarrow [\mathrm{Ar}] \, 3d^{10} 4s^1$ Cu$^{+1}$ $\rightarrow [\mathrm{Ar}] \, 3d^{10} 4s^0$ CN$^-$ is S.F.L. and pair up the unpaired electron of central metal atom. $$\therefore \mathrm{Cu}^{+1} \, (d^{10}) in [\mathrm{Cu(CN)}_4]^{3-}$$ $$\begin{array}{cccccc} \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow \\ \end{array}$$ $$\begin{array}{cccc} 3d & 4s & 4p & sp^3 \\ \end{array}$$ Tetrahedral
Question 44
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Multiple correct
The correct statement(s) related to oxoacids of phosphorous is(are):
Upon heating, $\mathrm{H_3PO_3}$ undergoes disproportionation reaction to produce $\mathrm{H_3PO_4}$ and $\mathrm{PH_3}$.
While $\mathrm{H_3PO_3}$ can act as reducing agent, $\mathrm{H_3PO_4}$ cannot.
$\mathrm{H_3PO_3}$ is a monobasic acid.
The H atom of P–H bond in $\mathrm{H_3PO_3}$ is not ionizable in water.
Answer: (a), (b), (d)
Solution
(A) $4\mathrm{H_3PO_3} \xrightarrow{\Delta} 3\mathrm{H_3PO_4} + \mathrm{PH_3}$ (correct) (B) $\mathrm{H_3PO_4}$ has "P" in its highest oxidation state, hence cannot act as a reducing agent (correct) (C) Dibasic acid (incorrect) Two OH group present in $\mathrm{H_3PO_3}$ (D) Non-ionizable Ionizable (Correct) The hydrogen which is directly attached to phosphorous does not ionized in water.
Question 45
Chemistry · Some Basic Concepts of Chemistry · Numerical
At 298 K, the limiting molar conductivity of a weak monobasic acid is $4 \times 10^2 \, \mathrm{S \, cm^2 \, mol^{-1}}$. At 298 K, for an aqueous solution of the acid the degree of dissociation of $\alpha$ and the molar conductivity is $y \times 10^2 \, \mathrm{S \, cm^2 \, mol^{-1}}$. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes $3y \times 10^2 \, \mathrm{S \, cm^2 \, mol^{-1}}$. The value of $\alpha$ is ______.
At 298 K, the limiting molar conductivity of a weak monobasic acid is $4 \times 10^2 \, \mathrm{S \, cm^2 \, mol^{-1}}$. At 298 K, for an aqueous solution of the acid the degree of dissociation of $\alpha$ and the molar conductivity is $y \times 10^2 \, \mathrm{S \, cm^2 \, mol^{-1}}$. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes $3y \times 10^2 \, \mathrm{S \, cm^2 \, mol^{-1}}$. The value of $y$ is
Chemistry · Alcohols, Phenols and Ethers · Numerical
Reaction of $x$ g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with $y$ g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively). (Use Molar masses (in g mol$^{-1}$) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively). The value of $x$ is ______.
Answer: 3.57
Solution
The value of x is So to get 1.29 gm organic salt. We have to form 0.01 mole salt. So 0.01 mole nitrobenzene is required. 0.03 mole Sn is required. So the amount of nitrobenzene = 0.01 $\times$ 123 = 1.23 $\,$ $\mathrm{gm}$ the amount of Sn required = 0.01 $\times$ 357 = 3.57 $\,$ $\mathrm{gm}$ Ans. 3.57 $\&$ 1.23
Question 48
Chemistry · Some Basic Concepts of Chemistry · Numerical
Reaction of $x$ g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with $y$ g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively). (Use Molar masses (in g mol$^{-1}$) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively). The value of $y$ is ______.
Answer: 1.23
Solution
The value of $x$ is $$3\mathrm{Sn} + 6\mathrm{HCl} + \mathrm{NO_2} \rightarrow \mathrm{NH_2} + 3\mathrm{SnCl_2} + 2\mathrm{H_2O}$$ $$357 \, \mathrm{gm} \,(3 \, \mathrm{mole}) + 123 \, \mathrm{gm} \,(1 \, \mathrm{mole}) \rightarrow 1 \, \mathrm{mole}$$ $$(72 + 8 + 35) + 14 = 129 \, \mathrm{gm} \,(molecular weight of organic salt)$$ So to get 1.29 gm organic salt. We have to form 0.01 mole salt. So 0.01 mole nitrobenzene is required. 0.03 mole Sn is required. So the amount of nitrobenzene $= 0.01 \times 123 = 1.23 \, \mathrm{gm}$ The amount of Sn required $= 0.01 \times 357 = 3.57 \, \mathrm{gm}$ Ans. 3.57 $\&$ 1.23
Question 49
Chemistry · Redox Reactions · Numerical
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M $KMnO_4$ solution to reach the end point. Number of moles of Fe^{2+} present in 250 mL solution is x $\times$ 10^{-2} (consider complete dissolution of $FeCl_2$). The amount of iron present in the sample of y$\%$ by weight. (Assume : $KMnO_4$ reacts only with Fe^{2+} in the solution Use : Molar mass of iron as 56 $\mathrm{g \, mol^{-1}}$) The value of x is ______.
Answer: 1.87,1.88
Solution
Fe + 2HCl $\rightarrow$ $\mathrm{FeCl_2}$ + $\mathrm{H_2}$ $\,$ (x mole) x mole $\mathrm{Fe^{+2}}$ + $\mathrm{MnO_4^-}$ $\frac{x}{10}$ $\,$ mole 12.5 $\,$ ml 0.03 $\,$ M n_f = 1 n_f = 5 $\frac{x}{10}$ = $\frac{12.5 \times 0.03 \times 5}{1000}$ x = 0.01875 (x = 1.88 or 1.87) wt of Fe = 1.05 $\,$ g $\%$ $\mathrm{Fe}$ = $\frac{1.05}{5.6}$ $\times$ 100 = 18.75
Question 50
Chemistry · Redox Reactions · Numerical
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M $KMnO_4$ solution to reach the end point. Number of moles of Fe^{2+} present in 250 mL solution is x $\times$ 10^{-2} (consider complete dissolution of $FeCl_2$). The amount of iron present in the sample of y$\%$ by weight. (Assume : $KMnO_4$ reacts only with Fe^{2+} in the solution Use : Molar mass of iron as 56 $\mathrm{g \, mol^{-1}}$) The value of $y$ is
Answer: 18.75
Solution
Fe + 2HCl $\rightarrow$ FeCl_2 + H_2 x mole x mole Fe^{+2} + MnO_4^- $\frac{x}{10}$ mole 12.5 ml 0.03 M n_f = 1 n_f = 5 $\frac{x}{10}$ = $\frac{12.5 \times 0.03 \times 5}{1000}$ x = 0.01875 (x = 1.88 or 1.87) wt of Fe = 1.05 g $\%$ Fe = $\frac{1.05}{5.6}$ $\times$ 100 = 18.75
Question 51
Chemistry · Chemical Bonding and Molecular Structure · Single correct
The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below : Correct match of the C–H bonds (shown in bold) in Column J with their BDE in Column K is
P – iii, Q – iv, R – ii, S – i
P – i, Q – ii, R – iii, S – iv
P – iii, Q – ii, R –i, S – iv
P – ii, Q – i, R – iv, S – iii
Answer: (a)
Solution
Most stability of radical, less is the bond energy. (P) $H \rightarrow \cdot C \cdot + H^\cdot$ 2° Carbon Free radical (Q) $Ph--CH_2--H \rightarrow Ph--CH_2^\cdot + H^\cdot$ Most stable due to resonance (R) $CH_2=CH--H \rightarrow CH_2=CH^\cdot + H^\cdot$ (less stable) (S) $CH \equiv C--H \rightarrow CH \equiv C^\cdot + H^\cdot$ More % S-Character decreases stability of free radical Q require least BDE and S Required maximum BDE So, Order of BDE Q < P < R < S
Question 52
Chemistry · Haloalkanes and Haloarenes · Single correct
The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below : For the following reaction \[ \mathrm{CH_4(g)+Cl_2(g)\xrightarrow{light}} \] $CH_3Cl(g) + HCl(g)$ the correct statement is
Initiation step is exothermic with $\Delta H^\circ = -58 \, \mathrm{kcal \, mol^{-1}}$
Propagation step involving $\cdot \mathrm{CH_3}$ formation is exothermic with $\Delta H^\circ = -2 \, \mathrm{kcal \, mol^{-1}}$.
Propagation step involving $\mathrm{CH_3Cl}$ formation is endothermic with $\Delta H^\circ = +27 \, \mathrm{kcal \, mol^{-1}}$.
The reaction is exothermic with $\Delta H^\circ = -25 \, \mathrm{kcal \, mol^{-1}}$.
Answer: (d)
Solution
Initiation step is endothermic hence option (A) is wrong. Propagation step involving $\cdot \mathrm{CH_3}$ formation is endothermic hence option (B) is wrong. Propagation step involving $\mathrm{CH_3Cl}$ formation is exothermic hence option (C) is wrong. Reaction $$\mathrm{CH_4 + Cl_2 \longrightarrow CH_3 - Cl + HCl}$$ $$\mathrm{CH_4 \longrightarrow \cdot CH_3 + \cdot H} \Delta H = 105 \, \mathrm{KCal/mol}$$ $$\mathrm{Cl_2 \longrightarrow \cdot Cl + \cdot Cl} \Delta H = 58 \, \mathrm{KCal/mol}$$ $$\mathrm{\cdot Cl + \cdot CH_3 \longrightarrow CH_3 - Cl} \Delta H = -85 \, \mathrm{KCal/mol}$$ $$\mathrm{\cdot Cl + \cdot H \longrightarrow HCl} \Delta H = -103 \, \mathrm{KCal/mol}$$ $$\mathrm{CH_4 + Cl_2 \longrightarrow CH_3 - Cl + HCl} \Delta H = -25 \, \mathrm{KCal/mol}$$ Overall reaction is exothermic with $\Delta H^\circ = -25 \, \mathrm{KCal/mol}$, hence option (D) is correct.
Question 53
Chemistry · Co-ordination Compounds · Single correct
The reaction of $K_3[Fe(CN)_6]$ with freshly prepared $FeSO_4$ solution produces a dark blue precipitate called Turnbull’s blue. Reaction of $K_4[Fe(CN)_6]$ with the $FeSO_4$ solution in complete absence of air produces a white precipitate $X$, which turns blue in air. Mixing the $FeSO_4$ solution with $NaNO_3$, followed by a slow addition of concentrated $H_2SO_4$ through the side of the test tube produces a brown ring. Precipitate $X$ is
Fe_4[Fe(CN)_6]_3
Fe[Fe(CN)_6]
K_2Fe[Fe(CN)_6]
KFe[Fe(CN)_6]
Answer: (c)
Solution
In the absence of air, $\mathrm{K_4[Fe(CN)_6]}$ reacts with $\mathrm{FeSO_4}$ to form $\mathrm{K_2Fe[Fe(CN)_6]}$, which is a white precipitate. In the presence of air, this further reacts to form $\mathrm{Fe_4[Fe(CN)_6]_3}$, known as Prussian Blue.
Question 54
Chemistry · Co-ordination Compounds · Single correct
The reaction of $K_3[Fe(CN)_6]$ with freshly prepared $FeSO_4$ solution produces a dark blue precipitate called Turnbull’s blue. Reaction of $K_4[Fe(CN)_6]$ with the $FeSO_4$ solution in complete absence of air produces a white precipitate $X$, which turns blue in air. Mixing the $FeSO_4$ solution with $NaNO_3$, followed by a slow addition of concentrated $H_2SO_4$ through the side of the test tube produces a brown ring. Among the following, the brown ring is due to the formation of
[$\mathrm{Fe(NO)_2(SO_4)_2}$]^{2-}
[$\mathrm{Fe(NO)_2(H_2O)_4}$]^{3+}
[$\mathrm{Fe(NO)_4(SO_4)_2}$]
[$\mathrm{Fe(NO)(H_2O)_5}$]^{2+}
Answer: (d)
Solution
FeSO$_4$ with slow addition of conc. H$_2$SO$_4$ and NaNO$_3$ forms [Fe(H$_2$O)$_5$NO]SO$_4$ (Brown Ring Complex).
Question 55
Chemistry · Thermodynamics · Numerical
One mole of an ideal gas at 900 $\,$ $\mathrm{K}$, undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are same, the value of $\ln$ $\frac{V_3}{V_2}$ is . \[ (U:\ \text{internal energy},\ S:\ \text{entropy},\ p:\ \text{pressure},\ V:\ \text{volume},\ R:\ \text{gas constant}) \] \[ (\text{Given: molar heat capacity at constant volume, }C_{V,m}\ \text{of the gas is}\ \frac{5}{2}R) \]
Answer: 10
Solution
Given $\Delta U_I = n C_{v,m} \Delta T = W_I \{ q_I = 0 \}$. $$-1800 \, R = 1 \times \frac{5R}{2} \times \Delta T = -720 \, \mathrm{K}$$ $T_2 = 180 \, \mathrm{K}$ $$W_{II} = W_I = -1800 \, R = -1 \times R \times 180 \ln \left( \frac{V_3}{V_2} \right)$$ $$\ln \left( \frac{V_3}{V_2} \right) = 10 \Rightarrow 10$$
Question 56
Chemistry · Structure of Atom · Numerical
Consider a helium (He) atom that absorbs a photon of wavelength 330 $\mathrm{nm}$. The change in the velocity (in $\mathrm{cm}$ $\mathrm{s}^{-1}$) of He atom after the photon absorption is ___. (Assume: Momentum is conserved when photon is absorbed. Use: Planck constant = 6.6 $\times$ 10^{-34} $\mathrm{J}$ $\mathrm{s}$, Avogadro number = 6 $\times$ 10^{23} $\mathrm{mol}^{-1}$, Molar mass of He = 4 $\mathrm{g}$ $\mathrm{mol}^{-1}$)
Answer: 30
Solution
Given $\lambda = \frac{h}{p}$, we have $p = \frac{6.6 \times 10^{-34}}{330 \times 10^{-9}} = \frac{4 \times 10^{-3}}{6 \times 10^{23}} \times v$ where $(p = m \times v)$. $v = 0.3 \, \mathrm{m/s} = 30 \, \mathrm{cm/s}$.
Question 57
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Numerical
Ozonolysis of $\mathrm{ClO_2}$ produces an oxide of chlorine. The average oxidation state of chlorine in this oxide is ___.
Answer: 6
Solution
The reaction is $2\mathrm{ClO_2} + 2\mathrm{O_3} \rightarrow \mathrm{Cl_2O_6} + 2\mathrm{O_2}$. For $\mathrm{Cl_2O_6}$: $$2x + 6(-2) = 0$$ Solving for $x$ gives $x = +6$. The average oxidation state of Cl in $\mathrm{Cl_2O_6}$ is 6.