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?

  1. $n_1 = 1000$
  2. $n_2 = 44$
  3. $n_3 = 220$
  4. $\frac{n_4}{12} = 420$

Answer: (a), (b), (d)

Solution

(A) $n_1 = 10 \times 10 \times 10 = 1000$ (B) As per given condition $1 \leq i < j + 2 \leq 10 \Rightarrow j \leq 8 and i \geq 1$ for $i = 1, 2$, $j = 1, 2, 3, \ldots, 8 \rightarrow (8 + 8)$ possibilities for $i = 3$, $j = 2, 3, \ldots, 8 \rightarrow 7$ possibilities $i = 4$, $j = 3, \ldots, 8 \rightarrow 6$ possibilities $i = 9$, $j = 1 \rightarrow 1$ possibility So $n_2 = (1 + 2 + 3 + \ldots + 8) + 8 = 44$ (C) $n_3 = \binom{10}{4}$ (Choose any four) $$= 210$$ (D) $n_4 = \binom{10}{4} \cdot 4! = (210) (24)$ $$\Rightarrow \frac{n_4}{12} = 420$$ So correct Ans. (A), (B), (D)

Question 2

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?

  1. $\cos P \geq 1 - \frac{p^2}{2qr}$
  2. $\cos R \geq \left(\frac{q-r}{p+q}\right) \cos P + \left(\frac{p-r}{p+q}\right) \cos Q$
  3. $\frac{q+r}{p} < 2 \sqrt{\frac{\sin Q \sin R}{\sin P}}$
  4. If $p \frac{p}{r}$ and $\cos R > \frac{p}{q}$

Answer: (a), (b)

Solution

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}?

  1. The equation $f(x) - 3 \cos 3x = 0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
  2. The equation $f(x) - 3 \sin 3x = -\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
  3. $\lim_{x \to 0} \frac{x \int_{0}^{x} f(t) \, dt}{1 - e^{x^2}} = -1$
  4. $\lim_{x \to 0} \frac{\sin x \int_{0}^{x} f(t) \, dt}{x^2} = -1$

Answer: (a), (b), (c)

Solution

Let $g(x) = f(x) - 3 \cos 3x$. Now $$\int_0^{\pi/3} g(x) \, dx = \int_0^{\pi/3} f(x) \, dx - 3 \int_0^{\pi/3} \cos 3x \, dx = 0$$ Hence $g(x) = 0$ has a root in $\left(0, \frac{\pi}{3}\right)$. Let $h(x) = f(x) - 3 \sin 3x + \frac{6}{\pi}$. Now $$\int_0^{\pi/3} h(x) \, dx = \int_0^{\pi/3} f(x) \, dx - 3 \int_0^{\pi/3} \sin 3x \, dx + \int_0^{\pi/3} \frac{6}{\pi} \, dx = 0 - 2 + 2 = 0$$ Hence $h(x) = 0$ has a root in $\left(0, \frac{\pi}{3}\right)$. $$\lim_{x \to 0} \frac{x \int_0^x f(t) \, dt}{1 - e^{x^2}} = \lim_{x \to 0} \frac{x^2}{1 - e^{x^2}} \cdot \frac{\int_0^x f(t) \, dt}{x}$$ Apply L'Hopital's Rule: $$= -1 \lim_{x \to 0} \frac{f(x)}{1} = -1$$ $$\lim_{x \to 0} \frac{(\sin x) \int_0^x f(t) \, dt}{x^2} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{\int_0^x f(t) \, dt}{x}$$ Apply L'Hopital's Rule: $$= 1 \lim_{x \to 0} \frac{f(x)}{1} = 1$$

Question 4

Maths · Differential Equations · Multiple correct

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 $S$?

  1. $f(x) = \frac{x^2}{2} e^{-x} + \left( e - \frac{1}{2} \right) e^{-x}$
  2. $f(x) = -\frac{x^2}{2} e^{-x} + \left( e + \frac{1}{2} \right) e^{-x}$
  3. $f(x) = \frac{e^x}{2} \left( x - \frac{1}{2} \right) + \left( e - \frac{e^2}{4} \right) e^{-x}$
  4. $f(x) = \frac{e^x}{2} \left( \frac{1}{2} - x \right) + \left( e + \frac{e^2}{4} \right) e^{-x}$

Answer: (a), (c)

Solution

Integrating factor is $e^{\alpha x}$. So $y e^{\alpha x} = \int x e^{(\alpha + \beta)x} \, dx$. Case-I If $\alpha + \beta = 0$ $$y e^{\alpha x} = \frac{x^2}{2} + c$$ It passes through $(1, 1) \Rightarrow c = e^{\alpha} - \frac{1}{2}$. So $y e^{\alpha x} = \frac{x^2 - 1}{2} + e^{\alpha}$. For $\alpha = 1$ $$y = \frac{x^2}{2} e^{-x} + \left(e - \frac{1}{2}\right) e^{-x} \rightarrow (A)$$ Case-II If $\alpha + \beta \neq 0$ $$y e^{\alpha x} = \frac{x e^{(\alpha + \beta)x}}{\alpha + \beta} - \frac{1}{\alpha + \beta} e^{(\alpha + \beta)x} \, dx$$ $$\Rightarrow y e^{\alpha x} = \frac{x e^{(\alpha + \beta)x}}{\alpha + \beta} - \frac{e^{(\alpha + \beta)x}}{(\alpha + \beta)^2} + c$$ $$\Rightarrow So c = e^{\alpha} - \frac{e^{\alpha + \beta}}{\alpha + \beta} - \frac{e^{\alpha + \beta}}{(\alpha + \beta)^2}$$ $$y = \frac{e^{\beta x}}{(\alpha + \beta)^2} \left((\alpha + \beta)x - 1\right) e^{-\alpha x} \left(e^x - \frac{e^{-\alpha + \beta}}{\alpha + \beta} + \frac{e^{\alpha + \beta}}{(\alpha + \beta)^2}\right)$$ If $\alpha = \beta = 1$ $$y = \frac{e^x}{4} \left(2x - 1\right) e^{-x} \left(e - \frac{e^2}{4}\right)$$ $$y = \frac{e^x}{2} \left(x - \frac{1}{2}\right) e^{-x} \left(e - \frac{e^2}{4}\right) \rightarrow (c)$$

Question 5

Maths · Vector Algebra · Multiple correct

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) $\textbf{TRUE}$?

  1. Projection of $\overrightarrow{\mathrm{OC}}$ on $\overrightarrow{\mathrm{OA}}$ is -$\frac{3}{2}$
  2. Area of the triangle OAB is $\frac{9}{2}$
  3. Area of the triangle ABC is $\frac{9}{2}$
  4. 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?

  1. The triangle $PFQ$ is a right-angled triangle
  2. The triangle $QPQ'$ is a right-angled triangle
  3. The distance between $P$ and $F$ is $5\sqrt{2}$
  4. $F$ 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 _____.

Answer: 2.0

Solution

Given $S_1 = \int_{\pi/8}^{3\pi/8} f(x) \, dx = \int_{\pi/8}^{3\pi/8} \sin^2 x \, dx = \int_{\pi/8}^{3\pi/8} \sin^2 \left( \frac{\pi}{8} + \frac{3\pi}{8} - x \right) \, dx = \int_{\pi/8}^{3\pi/8} \cos^2 x \, dx$. Therefore, $$2S_1 = \int_{\pi/8}^{3\pi/8} (\sin^2 x + \cos^2 x) \, dx = \frac{3\pi}{8} - \frac{\pi}{8} = \frac{\pi}{4}$$ Thus, $$\frac{16S_1}{\pi} = 2$$

Question 12

Maths · Sequences and Series · Numerical

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

  1. $k + 2l = 22$
  2. $2k + l = 26$
  3. $2k + 3l = 34$
  4. $3k + 2l = 40$

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

  1. 198
  2. 199
  3. 200
  4. 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 ?

  1. $f\left(\sqrt{\ln 3}\right) + g\left(\sqrt{\ln 3}\right) = \frac{1}{3}$
  2. For every $x > 1$, there exists an $\alpha \in (1, x)$ such that $\psi_1(x) = 1 + \alpha x$
  3. For every $x > 0$, there exists a $\beta \in (0, x)$ such that $\psi_2(x) = 2x(\psi_1(\beta) - 1)$
  4. $f$ is an increasing function on the interval $\left[0, \frac{3}{2}\right]$

Answer: (c)

Solution

Given $f'(x) = (|x| - x^2)e^{-x^2} + (|x| - x^2)e^{-x^2}$, $x \geq 0$. $f' = 2(x - X^2)e^{-x^2}$. $$egin{array}{c|c|c|c|c} & + & + & + & - - - \\ 0 & & & & 1 \\ \end{array}$$ Hence option (D) is wrong. $g'(x) = xe^{-x^2} 2x$. $f'(x) + g'(x) = 2 \times e^{-x^2}$. $f(x) + g(x) = -e^{-x^2} + c$. $f(x) + g(x) = -e^{-x^2} + 1$. $F(\ln 3) + g(\sqrt{\ln 3}) = 1 - 1 = \frac{1}{3} = \frac{2}{3}$ (option (A) is wrong). $H(x) = \psi_1(x) - 1 - \alpha x = e^{-x} + x - 1 - \alpha x$, $x \geq 1$ and $\alpha \in (1, x)$. $H(1) = e^{-1} + 1 - 1 - \alpha 0 \implies H(x)$ is decreasing $\implies$ option (B) is wrong. $(C) \psi_2(x) = 2(\psi_1(\beta) - 1)$. Applying L.M.V.T to $\psi_2(x)$ in $[0, x]$. $\psi'_2(\beta) = \frac{\psi_2(x) - \psi_2(0)}{x}$. $\psi'_2(\beta) = \frac{\psi_2(x) - 0}{x}$. $2\beta - 2 + 2e^{-\beta} = \frac{\psi_2(x) - 0}{x}$. $\implies \psi_2(x) = 2x(\psi_1(\beta) - 1)$ has one solution. Option (C) is correct.

Question 16

Maths · Linear Inequalities · Single correct

Let $\psi_1 : [0, \infty) \to \mathbb{R}$, $\psi_2 : [0, \infty) \to \mathbb{R}$, $f : [0, \infty) \to \mathbb{R}$ and $g : [0, \infty) \to \mathbb{R}$ be functions such that $f(0) = g(0) = 0$, $\psi_1(x) = e^{-x} + x$, $x \geq 0$, $\psi_2(x) = x^2 - 2x - 2e^{-x} + 2$, $x \geq 0$, $f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} \, dt$, $x > 0$ and $g(x) = \int_{0}^{x^2} \sqrt{t} \, e^{-t} \, dt$, $x > 0$ Which of the following statements is $\textbf{TRUE}$?

  1. $\psi_1(x) \leq 1$, for all $x > 0$
  2. $\psi_2(x) \leq 0$, for all $x > 0$
  3. $f(x) \geq 1 - e^{-x^2} - \frac{2}{3}x^3 + \frac{2}{5}x^5$, for all $x \in \left(0, \frac{1}{2}\right)$
  4. $g(x) \leq \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7$, for all $x \in \left(0, \frac{1}{2}\right)$

Answer: (d)

Solution

Given $\psi_1(x) = e^{-x} + x$, $x \geq 0$. $\psi_1'(x) = 1 - e^{-x} > 0$ implies $\psi_1(x)$ is increasing. $\psi_1(x) \geq \psi_1(0)$ for all $x \geq 0 \implies \psi_1(x) \geq 1$. (B) $\psi_2(x) = x^2 - 2x + 2 - 2e^{-x}$, $x \geq 0$. $\psi_2'(x) = 2x - 2 + 2e^{-x} = 2\psi_1(x) - 2 \geq 0$ for all $x \geq 0$. Therefore, $\psi_2(x)$ is increasing and $\psi_2(x) \geq \psi_2(0) \implies \psi_2(x) \geq 0$. (C) $f(x) = 2 \int_0^x (t - t^2) e^{-t^2} \, dt$ and $x \in \left(0, \frac{1}{2}\right)$. $$= \int_0^x 2te^{-t^2} \, dt - \int_0^x 2t^2 e^{-t^2} \, dt$$ $$= -e^{-x^2} \bigg|_0^x -$$ Let $H(x) = f(x) - 1 + e^{-x^2} + \frac{2}{3} x^3 - \frac{2}{5} x^5$, $x \in \left(0, \frac{1}{2}\right)$. $H(0) = 0$. $H'(x) = 2(x - x^2) e^{-x^2} - 2xe^{-x^2} + 2x^2 - 2x^4$ $$= -2x^2 e^{-x^2} + 2x^2 - 2x^4$$ $$= 2x^2 (1 - x^2 - e^{-x^2})$$ Since $e^{-x^2} \geq 1 - x$ for all $x \geq 0$, $H'(x) \leq 0$. Therefore, $H(x)$ is decreasing $\implies 1 - 1(x) < 0$ for $x \in \left(0, \frac{1}{2}\right)$. Let $P(x) = g(x) - \frac{2}{3} x^3 + \frac{2}{5} x^5 - \frac{1}{7} x^7 x \in \left(0, \frac{1}{2}\right)$. $P'(x) = 2x e^{-x^2} - 2x^2 + 2x^4 - x^6$ $$= 2x^2 \left(1 - \frac{x^2}{1} + \frac{x^4}{2} - \frac{x^6}{3} + \ldots \right) - 2x^2 + 2x^4 - x^6$$ $$= -\frac{x^8}{3} + \frac{x^{10}}{12} \ldots$$ Therefore, $P'(x) \leq 0$ implies $P(x)$ is decreasing. Therefore, $P(x) \leq 0$ on the interval.

Question 17

Maths · Probability · Numerical

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 ____.

Answer: 182

Solution

Let $f(x) = \left( \frac{10x}{x+1} \right)$. So, $f'(x) = 10 \left( \frac{(x+1) - x}{(x+1)^2} \right) = \frac{10}{(x+1)^2} > 0 \; \forall \; x \in [0, 10]$, so $f(x)$ is an increasing function. So, the range of $f(x)$ is $\left[ 0, \sqrt{\frac{100}{11}} \right]$. $$I = \int_0^{1/9} \left[ \sqrt{\frac{10x}{x+1}} \right] \, dx + \int_{1/9}^{2/3} \left[ \sqrt{\frac{10x}{x+1}} \right] \, dx + \int_{2/3}^{9} \left[ \sqrt{\frac{10x}{x+1}} \right] \, dx + \int_{9}^{10} \left[ \sqrt{\frac{10x}{x+1}} \right] \, dx$$ $$= 0 + \int_{1/9}^{2/3} \, dx + 2 \int_{2/3}^{9} \, dx + 3 \int_{9}^{10} \, dx$$ $$= \frac{2}{3} - \frac{1}{9} + 2 \left( 9 - \frac{2}{3} \right) + 3(10 - 9)$$ $$= \frac{6 - 1}{9} + 2 \times \frac{25}{3} + 3 = \frac{5}{9} + \frac{50}{3} + 3$$ $$= \frac{5 + 150 + 27}{9} = \frac{182}{9} = 182$$

Physics

Question 20

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?

  1. The vertical component of reaction force at $O$ does not depend on $\alpha$
  2. The horizontal component of reaction force at $O$ is equal to $W$ for $\alpha = 0.5$
  3. The tension in the rope is $2W$ for $\alpha = 0.5$
  4. 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?

  1. $u = 0.8 \, v$ and $f_s = f_0$
  2. $u = 0.8 \, v$ and $f_s = 2f_0$
  3. $u = 0.8 \, v$ and $f_s = 0.5f_0$
  4. $u = 0.5 \, v$ 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?

  1. The value of $\Delta e$ (in radians) is greater than that of $\Delta n$
  2. $\Delta e$ is proportional to $\Delta n$
  3. $\Delta e$ lies between $2.0$ and $3.0$ milliradians, if $\Delta n = 2.8 \times 10^{-3}$
  4. $\Delta e$ 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, $$dn \sin 30^\circ = de \cos(e)$$ $$de = \frac{dn}{2 \cos(e)}$$ $$= \frac{dn}{2 \times \frac{\sqrt{7}}{4}}$$ $$de = \frac{2}{\sqrt{7}} dn \Rightarrow de < dn$$ $$de = \frac{2.8 \times 10^{-3} \times 2}{\sqrt{7}} = 2.11 \, \mathrm{mrad}$$

Question 23

Physics · Electromagnetic Waves · Multiple correct

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)?

  1. $\frac{Energy}{charge \times current}$
  2. $\frac{Force}{Length \times Time}$
  3. $\frac{Energy}{Volume}$
  4. $\frac{Power}{Area}$

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?

  1. $E_P + E_Q = c^2 \delta$
  2. $E_P = \left( \frac{M_P}{M_P + M_Q} \right) c^2 \delta$
  3. $\frac{v_P}{v_Q} = \frac{M_Q}{M_P}$
  4. 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?

  1. $\vec{B}(x, y)$ is perpendicular to the $xy$-plane at any point in the plane
  2. $|\vec{B}(x, y)|$ depends on $x$ and $y$ only through the radial distance $r = \sqrt{x^2 + y^2}$
  3. $|\vec{B}(x, y)|$ is non-zero at all points for $r < R$
  4. $\vec{B}(x, y)$ 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 ____.

Answer: 0.3

Solution

When it starts sinking, $F_B = mg$. $\rho_0 (V_{glass} + V_{gas}) = m$. $1(2 + V_{gas}) = 5 \Rightarrow V_{gas} = 3 \, cc$. Hence $\Delta V = 0.3 \, cc$.

Question 27

Physics · Alternating Current · 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 $Y$ 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 ____.

Answer: 60

Solution

Given $\sqrt{V_C^2 + V_R^2} = \varepsilon_{rms}$. Therefore, $V_C^2 + 100^2 = 200^2$. Thus, $V_C = 100\sqrt{3} \, V$ ....(i) $\tan \phi = \frac{V_C}{V_R} = \frac{100\sqrt{3}}{100}$. Therefore, $\phi = 60^\circ$ .....(ii) $P = I_{rms} \varepsilon_{rms} \cos \phi = \frac{\varepsilon_{rms}^2}{Z} \cdot \frac{1}{2}$. $500 = \frac{200}{Z} \cdot \frac{1}{2}$. Therefore, $Z = 40 \, \Omega$ ....(iii) $\cos \phi = \frac{R}{Z} \Rightarrow \frac{1}{2} = \frac{R}{40}$. Therefore, $R = 20$. And $x_C = \sqrt{Z^2 - R^2} = \sqrt{40^2 - 20^2} = 20\sqrt{3} \, \Omega$. Therefore, $\frac{1}{\omega C} = 20\sqrt{3}$. Thus, $C = 100$

Question 32

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

  1. $\frac{m}{r^3}$
  2. $\frac{m^2}{r^2}$
  3. $\frac{m}{r^2}$
  4. $\frac{m^2}{r}$

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

  1. $\frac{m}{r^5}$
  2. $\frac{m^2}{r^5}$
  3. $\frac{m^2}{r^6}$
  4. $\frac{m^2}{r^7}$

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

  1. $\sqrt{2}$
  2. $\sqrt{3}$
  3. 2
  4. 3

Answer: (a)

Solution

Finally $V_L = \frac{3V_0}{2}$, $V_R = \frac{V_0}{2}$. $C_V = \frac{R}{\gamma - 1} = 2R \Rightarrow \gamma - 1 = \frac{1}{2}$. $\gamma = \frac{3}{2}$. $T_0 V_0^{\gamma - 1} = T_R \left( \frac{V_0}{2} \right)^{\gamma - 1}$. $\frac{T_R}{T_0} = \sqrt{2}$. $\rho \left( \frac{V_0}{2} \right)^{\gamma} = P_0 V_0^{\gamma} \Rightarrow P = P_0 \times 2^{\frac{3}{2}}$. $\frac{PV}{T_L} = \frac{P_0 V_0}{T_0} \Rightarrow T_L = 2^{\frac{3}{2}} \times \frac{3}{2} T_0 = 3\sqrt{2} T_0$. $Q = nC_V \Delta T_1 + nC_V \Delta T_2$. $= 1 \times 2R \times \left( 3\sqrt{2} - 1 \right) T_0 + 1 \times 2R \times \left( \sqrt{2} - 1 \right) T_0$. $\frac{Q}{RT_0} = 2 \left( 3\sqrt{2} - 1 \right) + 2 \left( \sqrt{2} - 1 \right) = 8\sqrt{2} - 4$

Question 35

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 $\frac{Q}{RT_0}$ is

  1. 4(2$\sqrt{2}$+1)
  2. 4(2$\sqrt{2}$-1)
  3. (5$\sqrt{2}$+1)
  4. (5$\sqrt{2}$-1)

Answer: (b)

Solution

Finally $V_L = \frac{3V_0}{2}$, $V_R = \frac{V_0}{2}$. $C_V = \frac{R}{\gamma - 1} = 2R \Rightarrow \gamma - 1 = \frac{1}{2}$. $\gamma = \frac{3}{2}$. $T_0 V_0^{\gamma - 1} = T_R \left( \frac{V_0}{2} \right)^{\gamma - 1}$. $\frac{T_R}{T_0} = \sqrt{2}$. $\rho \left( \frac{V_0}{2} \right)^{\gamma} = P_0 V_0^{\gamma} \Rightarrow P = P_0 \times 2^{\frac{3}{2}}$. $\frac{PV}{T_L} = \frac{P_0 V_0}{T_0} \Rightarrow T_L = 2^{\frac{3}{2}} \times \frac{3}{2} T_0 = 3\sqrt{2} T_0$. $Q = nC_V \Delta T_1 + nC_V \Delta T_2$. $= 1 \times 2R \times \left( 3\sqrt{2} - 1 \right) T_0 + 1 \times 2R \times \left( \sqrt{2} - 1 \right) T_0$. $\frac{Q}{RT_0} = 2 \left( 3\sqrt{2} - 1 \right) + 2 \left( \sqrt{2} - 1 \right) = 8\sqrt{2} - 4$

Question 36

Physics · Current Electricity · Numerical

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 ___.

Answer: 9

Solution

Circular orbits $$T = 2\pi \sqrt{\frac{R^2}{GM_S}}$$ Binary stars $$nT = 2\pi \sqrt{\frac{(9R)^3}{G(3M_S + 6M_S)}}$$ $$n \times 2\pi \sqrt{\frac{R^3}{GM_S}} = 9 \times 2\pi \sqrt{\frac{R^3}{GM_S}}$$ $$n = 9$$

Question 38

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 ___.

Answer: 6

Solution

For P and Q $$E_1 - 4 = E_P$$ $$E_1 - 4.5 = E_Q$$ $$E_P = 2 E_Q$$ $$E_1 - 4 = 2 (E_1 - 4.5)$$ $$E_1 = 5 \, \mathrm{eV}$$ $$E_P = 1 \, \mathrm{eV}, \; E_Q = E_R = 0.5 \, \mathrm{eV}$$ For $E_2$ $$E_2 - 5.5 = 0.5$$ $$E_2 = 6 \, \mathrm{eV}$$

Chemistry

Question 39

Chemistry · Hydrocarbons · Multiple correct

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)

  1. Q = $KNO_2$, W = $LiAlH_4$
  2. R = benzenamine, V = KCN
  3. Q = $AgNO_2$, R = phenylmethanamine
  4. 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: $$\ln 2 = 0.693$$)

  1. The rate constant, $k$, of the reaction is $13.86 \times 10^{-4} \, \mathrm{s}^{-1}$.
  2. Half-life of $\mathrm{X}$ is 50 s.
  3. At 50 s, $-\frac{\mathrm{d} [\mathrm{X}]}{\mathrm{dt}} = 13.86 \times 10^{-3} \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}$.
  4. At 100 s, $-\frac{\mathrm{d} [\mathrm{Y}]}{\mathrm{dt}} = 3.46 \times 10^{-3} \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}$.

Answer: (b), (c), (d)

Solution

Given $\($ $\frac{dp}{dt}$ = k[x]^1 $\)$. $\($ 2x + y $\rightarrow$ p $\)$ At $\($ t = 0 $\)$, $\($ 2 $\)$ $\($ 1 $\)$ At $\($ t = 50 $\,$ $\mathrm{s}$ $\)$, $\($ (2-1) $\)$ $\($ (1-0.5) $\)$ $\($ 0.5 $\)$ $\[$ -$\frac{1}{2}$ $\frac{dx}{dt}$ = $\frac{dp}{dt}$ = k[x]^1 $\]$ $\[$ -$\frac{dx}{dt}$ = 2k[x]^1 $\]$ $\[$ 2k = $\frac{\ln 2}{50}$ $\implies$ k = $\frac{\ln 2}{100}$ $\]$ At $\($ 50 $\,$ $\mathrm{sec}$ $\)$ $\[$ -$\frac{dx}{dt}$ = 2k $\times$ (1)^1 = $\frac{\ln 2}{50}$ $\]$ At $\($ 100 $\,$ $\mathrm{sec}$ $\)$ $\[$ $\frac{1}{2}$ $\frac{dx}{dt}$ = -$\frac{dy}{dt}$ $\implies$ -$\frac{dy}{dt}$ = $\frac{\ln 2}{100}$ $\times$ $\frac{1}{2}$ $\left$$\{$ -$\frac{dy}{dt}$ = k[x]^1 $\right$$\}$ $\]$

Question 42

Chemistry · Electrochemistry · Multiple correct

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$^{-1}$)

  1. Cd and Ni
  2. Cd and Fe
  3. Ni and Pb
  4. 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).

Question 43

Chemistry · Co-ordination Compounds · Multiple correct

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)

  1. [$\mathrm{FeCl}$_4]^- and [$\mathrm{Fe(CO)}$_4]^{2-}
  2. [$\mathrm{Co(CO)}$_4]^- and [$\mathrm{CoCl}$_4]^{2-}
  3. [$\mathrm{Ni(CO)}$_4] and [$\mathrm{Ni(CN)}$_4]^{2-}
  4. [$\mathrm{Cu(py)}$_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):

  1. Upon heating, $\mathrm{H_3PO_3}$ undergoes disproportionation reaction to produce $\mathrm{H_3PO_4}$ and $\mathrm{PH_3}$.
  2. While $\mathrm{H_3PO_3}$ can act as reducing agent, $\mathrm{H_3PO_4}$ cannot.
  3. $\mathrm{H_3PO_3}$ is a monobasic acid.
  4. 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 ______.

Answer: 0.21,0.22

Solution

Given $K_a = \frac{\Lambda^2 C}{\Lambda_m^\circ (\Lambda_m^\circ - \Lambda_m)}$. $$K_a = \frac{(y \times 10^2)^2 \times C}{4 \times 10^2 (4 \times 10^2 - y \times 10^2)} = \frac{(3y \times 10^2)^2 \times \frac{C}{20}}{4 \times 10^2 (4 \times 10^2 - 3y \times 10^2)}$$ Therefore, $$\frac{1}{(4-y)} = \frac{9}{20(4-3y)} \Rightarrow y = \frac{44}{51}$$ $$\alpha = \frac{44}{51} \times \frac{10}{4 \times 10^2}$$ $$\alpha = 0.2156 \; (\alpha = 0.22 \; or \; 0.21)$$ $$y = 0.86$$

Question 46

Chemistry · Electrochemistry · 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 $y$ is

Answer: 0.86

Solution

Given $\($ K_a = $\frac{\Lambda^2 C}{\Lambda_m^\circ (\Lambda_m^\circ - \Lambda_m)}$ $\)$. $\[$ K_a = $\frac{(y \times 10^2)^2 \times C}{4 \times 10^2 (4 \times 10^2 - y \times 10^2)}$ = $\frac{(3y \times 10^2)^2 \times \frac{C}{20}}{4 \times 10^2 (4 \times 10^2 - 3y \times 10^2)}$ $\]$ $\[$ $\Rightarrow$ $\frac{1}{(4-y)}$ = $\frac{9}{20(4-3y)}$ $\Rightarrow$ y = $\frac{44}{51}$ $\]$ $\[$ $\alpha$ = $\frac{\frac{44}{51} \times 10}{4 \times 10^2}$ $\]$ $\($ $\alpha$ = 0.2156 $\)$ ($\($ $\alpha$ = 0.22 $\)$ or $\($ 0.21 $\)$) $\($ y = 0.86 $\)$

Question 47

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

  1. P – iii, Q – iv, R – ii, S – i
  2. P – i, Q – ii, R – iii, S – iv
  3. P – iii, Q – ii, R –i, S – iv
  4. 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

  1. Initiation step is exothermic with $\Delta H^\circ = -58 \, \mathrm{kcal \, mol^{-1}}$
  2. Propagation step involving $\cdot \mathrm{CH_3}$ formation is exothermic with $\Delta H^\circ = -2 \, \mathrm{kcal \, mol^{-1}}$.
  3. Propagation step involving $\mathrm{CH_3Cl}$ formation is endothermic with $\Delta H^\circ = +27 \, \mathrm{kcal \, mol^{-1}}$.
  4. 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

  1. Fe_4[Fe(CN)_6]_3
  2. Fe[Fe(CN)_6]
  3. K_2Fe[Fe(CN)_6]
  4. 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

  1. [$\mathrm{Fe(NO)_2(SO_4)_2}$]^{2-}
  2. [$\mathrm{Fe(NO)_2(H_2O)_4}$]^{3+}
  3. [$\mathrm{Fe(NO)_4(SO_4)_2}$]
  4. [$\mathrm{Fe(NO)(H_2O)_5}$]^{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.