JEE Main 20 July 2021 Shift 1: all 90 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 · Mathematical Reasoning · Single correct
The Boolean expression $(p \land \sim q) \Rightarrow (q \lor \sim p)$ is equivalent to:
Answer: (b)
The truth table is given as follows: Therefore, $(p \land \sim q) \Rightarrow (q \lor \sim p)$ is equivalent to $p \Rightarrow q$. So, option (2) is correct.
Maths · Integrals · Single correct
Let a be a positive real number such that $\int_0^a e^{x - \lfloor x \rfloor} dx = 10e - 9$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. Then $a$ is equal to:
Answer: (b)
Given $a > 0$. Let $n \leq a < n + 1$, $n \in \mathbb{W}$. Therefore, $a = [a] + \{a\}$. Here $[a] = n$. Now, $$\int_0^a e^{x - [x]} \, dx = 10e - 9$$ implies $$\int_0^n e^{x} \, dx + \int_n^a e^{x - [x]} \, dx = 10e - 9$$ Therefore, $$n \int_0^1 e^x \, dx + \int_n^a e^{x - n} \, dx = 10e - 9$$ implies $$n(e - 1) + (e^{a-n} - 1) = 10e - 9$$ Therefore, $n = 0$ and $\{a\} = \log_e 2$. So, $a = [a] + \{a\} = (10 + \log_e 2)$. Option (2) is correct.
Maths · Statistics · Single correct
The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are:
Answer: (a)
Let other two numbers be $a$, $(21-a)$. Now, $$10.25 = \frac{4+16+25+49+a^2+(21-a)^2}{6} - (6.5)^2$$ (Using formula for variance) $$\Rightarrow 6(10.25) + 6(6.5)^2 = 94 + a^2 + (21-a)^2$$ $$\Rightarrow a^2 + (21-a)^2 = 221$$ $$\therefore a = 10 and (21-a) = 21 - 10 = 11$$ So, remaining two observations are 10, 11. $$\Rightarrow$$ Option (1) is correct.
Maths · Integrals · Single correct
The value of the integral $\int_{-1}^{1} \log_e (\sqrt{1-x} + \sqrt{1+x}) \, dx$ is equal to:
Answer: (b)
Let I = 2 $\int$_0^1 $\ln$($\sqrt{1-x}$ + $\sqrt{1+x}$) $\cdot$ $\frac{1}{1}$ $\,$ dx (I.B.P.) $\therefore$ $\ $I = 2 $\left$[ $\left$( x $\cdot$ $\ln$($\sqrt{1-x}$ + $\sqrt{1-x}$) $\right$)_0^1 $\right$. - $\left$. $\int$_0^1 x $\cdot$ $\left$( $\frac{1}{\sqrt{1-x}+\sqrt{1+x}}$ $\right$) $\cdot$ $\left$( $\frac{1}{2\sqrt{1+x}}$ - $\frac{1}{2\sqrt{1-x}}$ $\right$) $\,$ dx $\right$] = 2($\ln$ $\sqrt{2}$ - 0) - $\frac{2}{2}$ $\int$_0^1 $\frac{x \sqrt{1-x} - \sqrt{1+x} \, dx}{(\sqrt{1-x}+\sqrt{1+x})\sqrt{1-x^2}}$ = ($\log$_e 2) - $\int$_0^1 $\frac{x \cdot (2 - 2\sqrt{1-x^2})}{-2x\sqrt{1-x^2}}$ $\,$ dx (After rationalisation) = ($\log$_e 2) + $\int$_0^1 $\left$( $\frac{1 - \sqrt{1-x^2}}{\sqrt{1-x^2}}$ $\right$) $\,$ dx = ($\log$_e 2) + ($\sin$^{-1} x)_0^1 - 1 = $\log$_e 2 + $\left$( $\frac{\pi}{2}$ - 0 $\right$) - 1 $\therefore$ $\ $I = ($\log$_e 2) + $\frac{\pi}{2}$ - 1 $\Rightarrow$ Option (3) is correct.
Maths · Complex Numbers and Quadratic Equations · Single correct
If $\alpha$ and $\beta$ are the distinct roots of the equation $x^2 + (3)^{1/4}x + 3^{1/2} = 0$, then the value of $\alpha^{96} \left(\alpha^{12} - 1\right) + \beta^{96} \left(\beta^{12} - 1\right)$ is equal to:
Answer: (c)
As, $\left(\alpha^2 + \sqrt{3}\right)^3 = -(3)^{1/4} \cdot \alpha$. $$\Rightarrow \left(\alpha^4 + 2\sqrt{3}\alpha^2 + 3\right) = \sqrt{3}\alpha^2 (On squaring)$$ $$\therefore (\alpha^4 + 3) = (-)\sqrt{3}\alpha^2$$ $$\Rightarrow \alpha^8 + 6\alpha^4 + 9 = 3\alpha^4 (Again squaring)$$ $$\therefore \alpha^8 + 3\alpha^4 + 9 = 0$$ $$\Rightarrow \alpha^8 = -9 - 3\alpha^4$$ (Multiply by $\alpha^4$) So, $\alpha^{12} = -9\alpha^4 - 3\alpha^8$. $$\therefore \alpha^{12} = -9\alpha^4 - 3(-9 - 3\alpha^4)$$ $$\Rightarrow \alpha^{12} = -9\alpha^4 + 27 + 9\alpha^4$$ Hence, $\alpha^{12} = (27)^2$. $$\Rightarrow (\alpha^{12})^8 = (27)^8$$ $$\Rightarrow \alpha^{96} = (3)^{24}$$ Similarly $\beta^{96} = (3)^{24}$ $$\therefore \alpha^{96}(\alpha^{12} - 1) + \beta^{96}(\beta^{12} - 1) = (3)^{24} \times 52$$ Option (3) is correct.
Maths · Determinants · Single correct
Let $A=\begin{bmatrix}2&3\\a&0\end{bmatrix}$, $a\in\mathbb{R}$, be written as $P+Q$, where $P$ is a symmetric matrix and $Q$ is a skew-symmetric matrix. If $\det(Q)=9$, then the modulus of the sum of all possible values of $\det(P)$ is equal to:
Answer: (a)
Given $$A = \begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix}, \quad a \in \mathbb{R}$$ and $$P = \frac{A + A^T}{2} = \begin{bmatrix} 2 & \frac{3+a}{2} \\ \frac{a+3}{2} & 0 \end{bmatrix}$$ and $$Q = \frac{A - A^T}{2} = \begin{bmatrix} 0 & \frac{3-a}{2} \\ \frac{a-3}{2} & 0 \end{bmatrix}$$ As, $\det(Q) = 9$ \[ \Rightarrow (a - 3)^2 = 36 \] \[ \Rightarrow a = 3 \pm 6 \] \[ \therefore a = 9, -3 \] \[ = 0 - \frac{(a+3)^2}{4} = 0, \text{ for } a = -3 \] \[ = 0 - \frac{(a+3)^2}{4} = -\frac{1}{4}(12)(12), \text{ for } a = 9 \] \[ \therefore \text{Modulus of the sum of all possible values of } \det(P) = | -36 | + | 0 | = 36 \text{ Ans.} \] \[ \Rightarrow \text{Option (1) is correct.} \]
Maths · Complex Numbers and Quadratic Equations · Single correct
If $z$ and $\omega$ are two complex numbers such that $|z\omega| = 1$ and $arg(z) - arg(\omega) = \frac{3\pi}{2}$, then $arg\left(\frac{1 - 2\bar{z}\bar{\omega}}{1 + 3\bar{z}\bar{\omega}}\right)$ is: (Here $arg(z)$ denotes the principal argument of complex number $z$)
Answer: (c)
As $|zw| = 1$. If $|z| = r$, then $|w| = \frac{1}{r}$. Let $\arg(z) = \theta$. Therefore, $\arg(\omega) = \left( \theta - \frac{3\pi}{2} \right)$. So, $z = re^{i\theta}$. Thus, $\overline{z} = re^{-i\theta}$. $\omega = \frac{1}{r} e^{i \left( \theta - \frac{3\pi}{2} \right)}$. Now, consider $$\frac{1 - 2z\omega}{1 + 3z\omega} = \frac{1 - 2e^{i \left( -\frac{3\pi}{2} \right)}}{1 + 3e^{i \left( -\frac{3\pi}{2} \right)}} = \left( \frac{1 - 2i}{1 + 3i} \right)$$ $$= \frac{(1 - 2i)(1 - 3i)}{(1 + 3i)(1 - 3i)} = -\frac{1}{2} (1 + i)$$ Therefore, $\operatorname{prin} \arg \left( \frac{1 - 2z\omega}{1 + 3z\omega} \right)$ $$= \operatorname{prin} \arg \left( \frac{1 - 2z\omega}{1 + 3z\omega} \right)$$ $$= \left( -\frac{1}{2} (1 + i) \right)$$ $$= -\left( \pi - \frac{\pi}{4} \right) = -\frac{3\pi}{4}$$ So, option (2) is correct.
Maths · Properties of Triangles · Single correct
If in a triangle ABC, $AB = 5$ units, $\angle B = \cos^{-1}\left(\frac{3}{5}\right)$ and radius of circumcircle of $\triangle ABC$ is 5 units, then the area (in sq. units) of $\triangle ABC$ is:
Answer: (c)
As, $\cos B = \frac{3}{5}$ $$\Rightarrow B = 53^\circ$$ As, $R = 5 \Rightarrow \frac{c}{\sin c} = 2R$ $$\Rightarrow \frac{5}{10} = \sin c$$ $$\Rightarrow C = 30^\circ$$ Now, $\frac{b}{\sin B} = 2R$ $$\Rightarrow b = 2(5) \left( \frac{4}{5} \right) = 8$$ Now, by cosine formula $$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$$ $$\Rightarrow \frac{3}{5} = \frac{a^2 + 25 - 64}{2(5)a}$$ $$\Rightarrow a^2 - 6a - 3g = 0$$ $$\therefore a = \frac{6 \pm \sqrt{192}}{2} = \frac{6 \pm 8\sqrt{3}}{2}$$ $$\Rightarrow 3 + 4\sqrt{3} (Reject a = 3 - 4\sqrt{3})$$ Now, $\Delta = \frac{abc}{4R} = \frac{(3 + 4\sqrt{3})(8)(5)}{4(5)} = 2(3 + 4\sqrt{3})$ $$\Rightarrow \Delta = (6 + 8\sqrt{3})$$ Option (3) is correct.
Maths · Relations and Functions · Single correct
\quad Let $[x]$ denote the greatest integer $\le x$, where $x\in\mathbb{R}$. If the domain of the real valued function $f(x)=\sqrt{\dfrac{[x]|-2}{||x||-3}}$ is $(-\infty,a)\cup[b,c)\cup[4,\infty)$, $a<b<c$, then the value of $a+b+c$ is:
Answer: (c)
For domain, $$\frac{|x| - 2}{|x| - 3} \geq 0$$ Case I: When $$|x| - 2 \geq 0$$ and $$|x| - 3 > 0$$ Therefore, $$x \in (-\infty, -3) \cup [4, \infty)$$ Case II: When $$|x| - 2 \leq 0$$ and $$|x| - 3 < 0$$ Therefore, $$x \in [-2, 3)$$ So, from (1) and (2) we get Domain of function $$= (-\infty, -3) \cup [-2, 3) \cup [4, \infty)$$ Therefore, $$a + b + c = -3 + (-2) + 3 = -2 (a < b < c)$$ Thus, Option (3) is correct.
Maths · Differential Equations · Single correct
Let $y = y(x)$ be the solution of the differential equation $x \tan\left(\frac{y}{x}\right) dy = \left(y \tan\left(\frac{y}{x}\right) - x\right) dx$ $-1 \leq x \leq 1, y\left(\frac{1}{2}\right) = \frac{\pi}{6}$. Then the area of the region bounded by the curves $x = 0, x = \frac{1}{\sqrt{2}}$ and $y = y(x)$ in the upper half plane is:
Answer: (a)
We have $$\frac{dy}{dx} = \frac{x \left( \frac{y}{x} \cdot \tan \frac{y}{x} - 1 \right)}{x \tan \frac{y}{x}}$$ Therefore, $$\frac{dy}{dx} = \frac{y}{x} - \cot \left( \frac{y}{x} \right)$$ Put $$\frac{y}{x} = v$$ Thus, $$y = vx$$ Therefore, $$\frac{dy}{dx} = v + x \frac{dv}{dx}$$ Now, we get $$v + x \frac{dv}{dx} = v - \cot(v)$$ Thus, $$\int \left( \tan \right) dv = - \int \frac{dx}{x}$$ Therefore, $$\ln \left| \sec \left( \frac{y}{x} \right) \right| = -\ln |x| + c$$ As $$\left( \frac{1}{2} \right) = \left( \frac{y}{x} \right) \Rightarrow C = 0$$ Therefore, $$\sec \left( \frac{y}{x} \right) = \frac{1}{x}$$ Thus, $$\cos \left( \frac{y}{x} \right) = x$$ Therefore, $$y = x \cos^{-1}(x)$$ So, required bounded area is $$\int_{0}^{1/\sqrt{2}} x \left( \cos^{-1} x \right) \, dx = \left( \frac{\pi - 1}{8} \right)$$ (II) (I) (I.B.P.) Therefore, option (1) is correct.
Maths · Binomial Theorem · Single correct
The coefficient of $x^{256}$ in the expansion of $(1x)^{101} \left(x^2 + x + 1\right)^{100}$ is:
Answer: (b)
$(1-x)^{160}\cdot(x^{2}+x+1)^{100}\cdot(1-x)$ $=((1-x)(x^{2}+x+1))^{100}(1-x)$ $=(1-x^{3})^{100}(1-x)$ $=(1-x^{3})^{100}-x(1-x^{3})^{100}$ No term of $x^{256}$ We find coefficient of $x^{255}$ Required coefficient $=(-1)\times(-1)^{85}\times{}^{100}C_{85}$ $={}^{100}C_{85}$ $={}^{100}C_{15}$
Maths · Determinants · Single correct
Let $$ A=[a_{ij}] $$ be a $3\times 3$ matrix, where $$ a_{ij}= \begin{cases} 1, & \text{if } i=j,\\ -x, & \text{if } |i-j|=1,\\ 2x+1, & \text{otherwise}. \end{cases} $$ Let a function $$ f:\mathbb{R}\rightarrow\mathbb{R} $$ be defined as $$ f(x)=\det(A). $$ Then the sum of maximum and minimum values of $f$ on $\mathbb{R}$ is equal to:
Answer: (d)
Given $$A = \begin{bmatrix} 1 & -x & 2x + 1 \\ -x & 1 & -x \\ 2x + 1 & -x & 1 \end{bmatrix}$$ The determinant is $$|A| = 4x^3 - 4x^2 - 4x = f(x)$$ Differentiating, we have $$f'(x) = 4 \left(3x^2 - 2x - 1\right) = 0$$ This implies $$x = 1; x = \frac{-1}{3}$$ Therefore, $$f(1) = -4; f\left(\frac{-1}{3}\right) = \frac{20}{27}$$ The sum is $$-4 + \frac{20}{27} = \frac{-88}{27}$$
Maths · Vector Algebra · Single correct
Let $\vec{a}$ = 2$\hat{i}$ + $\hat{j}$ - 2$\hat{k}$ and $\vec{b}$ = $\hat{i}$ + $\hat{j}$. If $\vec{c}$ is a vector such that $\vec{a}$ $\cdot$ $\vec{c}$ = |$\vec{c}$|, |$\vec{c}$ - $\vec{a}$| = 2$\sqrt{2}$ and the angle between ($\vec{a}$ $\times$ $\vec{b}$) and $\vec{c}$ is $\frac{\pi}{6}$, then the value of |($\vec{a}$ $\times$ $\vec{b}$) $\times$ $\vec{c}$| is:
Answer: (d)
Given $|\vec{a}| = 3 = a; \vec{a} \cdot \vec{c} = c$. Now $|\vec{c} - \vec{a}| = 2\sqrt{2}$. $$c^2 + a^2 - 2c \cdot \vec{a} = 8$$ $$c^2 + 9 - 2(c) = 8$$ $$c^2 - 2c + 1 = 0 \Rightarrow c = 1 = |\vec{c}|$$ Also, $\vec{a} \times \vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$. Given $$(\vec{a} \times \vec{b}) = |\vec{a} \times \vec{b}||\vec{c}| \sin \frac{\pi}{6}$$ $$= (3)(1)(1/2)$$ $$= 3/2$$
Maths · Inverse Trigonometric Functions · Single correct
The number of real roots of the equation $$\tan^{-1} \sqrt{x(x+1)} + \sin^{-1} \sqrt{x^2 + x + 1} = \frac{\pi}{4}$$ is:
Answer: (d)
$\tan^{-1}\left(\sqrt{x^2+x}\right)+\sin^{-1}\left(\sqrt{x^2+x+1}\right)=\frac{\pi}{4}$. For the equation to be defined, $x^2+x\geq 0$. Also, $x^2+x+1\leq 1$ $\Rightarrow x^2+x\leq 0$. Therefore, the only possibility for the equation to be defined is $x^2+x=0$ $\Rightarrow x=0$ or $x=-1$. None of these values satisfy the given equation. Therefore, the number of roots is $0$.
Maths · Differential Equations · Single correct
Let y = y(x) be the solution of the differential equation $e^x \sqrt{1-y^2} \, dx + \left( \frac{y}{x} \right) \, dy = 0, y(1) = -1.$ Then the value of $(y(3))^2$ is equal to:
Answer: (b)
Given $e^x \sqrt{1-y^2} \, dx + \frac{y}{x} \, dy = 0$. Therefore, $e^x \sqrt{1-y^2} \, dx + \frac{-y}{x} \, dy$. Integrating, $$\int \frac{-y}{\sqrt{1-y^2}} \, dy = \int e^x \, dx$$ which implies $$\sqrt{1-y^2} = e^x(x-1) + c.$$ Given: At $x = 1$, $y = -1$, $$0 = 0 + c \Rightarrow c = 0.$$ Therefore, $$\sqrt{1-y^2} = e^x(x-1).$$ At $x = 3$, $$1-y^2 = (e^3 2)^2 \Rightarrow y^2 = 1 - 4e^6.$$
Maths · Applications of Derivatives · Single correct
Let 'a' be a real number such that the function $$f(x) = ax^2 + 6x - 15, x \in \mathbb{R}$$ is increasing in $$\left( -\infty, \frac{3}{4} \right)$$ and decreasing in $$\left( \frac{3}{4}, \infty \right)$$. Then the function $$g(x) = ax^2 - 6x + 15, x \in \mathbb{R}$$ has a:
Answer: (a)
Given $\left(-\frac{B}{2A}=\frac{3}{4}\right)$ $\Rightarrow -\frac{-6}{2a}=\frac{3}{4}$ $\Rightarrow a=\frac{(-6)\times4}{-6}$ $\Rightarrow a=-4$ Therefore, $g(x)=-4x^{2}-6x+15$ Local maximum at $x=-\frac{B}{2A}$ $=-\frac{-6}{2(-4)}$ $=-\frac{3}{4}$
Maths · Continuity and Differentiability · Single correct
Let a function $f : \mathbb{R} \to \mathbb{R}$ be defined as $$f(x) = \begin{cases} \sin x - e^x & if x \leq 0 \\ a + \lfloor -x \rfloor & if 0 < x < 1 \\ 2x - b & if x \geq 1 \end{cases}$$ Where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. If $f$ is continuous on $\mathbb{R}$, then $(a + b)$ is equal to:
Answer: (b)
Continuous at $x = 0$ $f(0^+) = f(0^-)$ implies $a - 1 = 0 - e^0$ implies $a = 0$ Continuous at $x = 1$ $f(1^+) = f(1^-)$ implies $2(1) - b = a + (-1)$ implies $b = 2 - a + 1$ implies $b = 3$ Therefore, $a + b = 3$
Maths · Probability · Single correct
Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter M appears at the fourth position in any such word is:
Answer: (b)
Total words with M at fourth place = $\frac{10!}{2!2!2!}$ Total words = $\frac{11!}{2!2!2!}$ Required probability = $\frac{10!}{11!}$ = $\frac{1}{11}$
Maths · Probability · Single correct
The probability of selecting integers $a \in [-5, 30]$ such that $x^2 + 2(a + 4)x - 5a + 64 > 0$, for all $x \in \mathbb{R}$, is:
Answer: (b)
Given $D < 0$. $$4(a + 4)^2 - 4(-5a + 64) < 0$$ This simplifies to: $$a^2 + 16 + 8a + 5a - 64 < 0$$ Which further simplifies to: $$a^2 + 13a - 48 < 0$$ Factoring gives: $$(a + 16)(a - 3) < 0$$ Thus, $a \in (-16, 3)$. Therefore, possible $a$ values are: $\{-5, -4, \ldots, 3\}$. The required probability is: $$\frac{8}{36}$$ Which simplifies to: $$\frac{2}{9}$$
Maths · Conic Sections · Single correct
Let the tangent to the parabola $S : y^2 = 2x$ at the point $P(2,2)$ meet the $x$-axis at $Q$ and normal at it meet the parabola $S$ at the point $R$. Then the area (in sq. units) of the triangle $PQR$ is equal to:
Answer: (a)
Tangent at $P$: $y(2) = 2(1/2)(x + 2)$ $$\Rightarrow 2y = x + 2$$ $$\therefore \; Q = (-2, 0)$$ Normal at $P$: $y - 2 = -\frac{(2)}{2^{1/2}}(x - 2)$ $$\Rightarrow y - 2 = -2(x - 2)$$ $$\Rightarrow y = 6 - 2x$$ $$\therefore Solving with y^2 = 2x \Rightarrow R \left( \frac{9}{2} - 3 \right)$$ $$\therefore Ar(\Delta PQR) = \frac{1}{2} \begin{vmatrix} 2 & 2 & 1 \\ -2 & 1 & 1 \\ \frac{9}{2} & 3 & -1 \end{vmatrix}$$ $$= \frac{25}{2} sq. units$$
Maths · Vector Algebra · Numerical
Let $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$, with the vector $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}$. Then $36 \cos^2 2\theta$ is equal to ___
Answer: 4
Given $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c})$. This equals 3. Therefore, $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}$. The dot product $\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c}) = |\vec{a}| + |\vec{a} + \vec{b} + \vec{c}| \cos \theta$. Thus, $1 = \sqrt{3} \cos \theta$. This implies $\cos 2\theta = -\frac{1}{3}$. Finally, $36 \cos^2 2\theta = 4$.
Maths · Matrices · Numerical
Let $A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}$ and $B = 7 A^{20} - 20 A^7 + 2I$, where $I$ is an identity matrix of order $3 \times 3$. If $B = [b_{ij}]$, then $b_{13}$ is equal to
Answer: 910
Let $A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix} = I + C$ where $I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$, $C = \begin{pmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}$. $$C^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},$$ $$C^3 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = C^4 = C^5 = \ldots$$ $B = 7A^{20} - 20A^7 + 2I$ $$= 7(I + C)^{20} - 20(I + C)^7 + 2I$$ $$= 7 \left( I + 20C + \binom{20}{2}C^2 \right) - 20 \left( I + 7C + \binom{7}{2}C^2 \right) + 2I$$ So $$b_{13} = 7 \times \binom{20}{2} - 20 \times \binom{7}{2} = 910$$
Maths · Three Dimensional Geometry · Numerical
Let P be a plane passing through the points $(1, 0, 1)$, $(1, -2, 1)$ and $(0, 1, -2)$. Let a vector $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ be such that $\vec{a}$ is parallel to the plane P, perpendicular to $(\hat{i} + 2\hat{j} + 3\hat{k})$ and $\vec{a} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 2$, then $(\alpha - \beta + \gamma)^2$ equals
Answer: 81
Equation of plane: $$\begin{vmatrix} x-1 & y-0 & z-1 \\ 1-1 & 2 & 1-1 \\ 1-0 & 0-1 & 1+2 \end{vmatrix} = 0$$ This simplifies to: $$3x - z - 2 = 0$$ Let $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ be parallel to $3x - z - 2 = 0$. This gives: $$3\alpha - 8 = 0 ...(1)$$ Also, $\vec{a} \perp \hat{i} + 2\hat{j} + 3\hat{k}$ This gives: $$\alpha + 2\beta + 38 = 0 ...(2)$$ And: $$\vec{a} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 0$$ This gives: $$\alpha + \beta + 28 = 2 ...(3)$$ On solving equations 1, 2, and 3, we get: $$\alpha = 1, \beta = -5, \gamma = 3$$ So, $\alpha - \beta + 8 = 81$
Maths · Binomial Theorem · Numerical
The number of rational terms in the binomial expansion of $\left(4^{\frac{1}{4}} + 5^{\frac{1}{6}}\right)^{120}$ is
Answer: 21
Given $\($ $\left$( 4^{1/4} + 5^{1/6} $\right$)^{120} $\)$. $\[$ T_{r+1} = $\binom{120}{r}$ $\left$( 2^{1/2} $\right$)^{120-r} $\left$( 5 $\right$)^{r/6} $\]$ For rational terms $\($ r = 6$\lambda$ $\)$. $\($ 0 $\leq$ r $\leq$ 120 $\)$ so total number of forms are 21.
Maths · Three Dimensional Geometry · Numerical
If the shortest distance between the lines $\vec{r}_1 = \alpha \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda (\hat{i} - 2 \hat{j} + \hat{k}), \lambda \in \mathbb{R}, \alpha > 0$ and $\vec{r}_2 = -4 \hat{i} - \hat{k} + \mu (3 \hat{i} - 2 \hat{j} - 2 \hat{k}), \mu \in \mathbb{R}$ is 9, then $\alpha$ is equal to
Answer: 6
If $\vec{r} = \vec{a} + \lambda \vec{b}$ and $\vec{r} = \vec{c} + \lambda \vec{d}$ then shortest distance between two lines is $L = \frac{(\vec{a} - \vec{c}) \cdot (\vec{b} \times \vec{d})}{|\vec{b} \times \vec{d}|}$. Therefore, $\vec{a} - \vec{c} = ((\alpha + 4) \hat{i} + 2 \hat{j} + 3 \hat{k})$. $\frac{\vec{b} \times \vec{d}}{|\vec{b} \times \vec{d}|} = \frac{(2 \hat{i} + 2 \hat{j} + \hat{k})}{3}$. Therefore, $((\alpha + 4) \hat{i} + 2 \hat{j} + 3 \hat{k}) \cdot \frac{(2 \hat{i} + 2 \hat{j} + \hat{k})}{3} = 9$. or $\alpha = 6$
Maths · Conic Sections · Numerical
Let $T$ be the tangent to the ellipse $E: x^2 + 4y^2 = 5$ at the point $P(1,1)$. If the area of the region bounded by the tangent $T$, ellipse $E$, lines $x=1$ and $x=\sqrt{5}$ is $$ \alpha\sqrt{5}+\beta+\gamma\cos^{-1}\!\left(\frac{1}{\sqrt{5}}\right), $$ then $|\alpha+\beta+\gamma|$ is equal to
Answer: 1
Tangent at P: $x + 4y = 5$ Required Area = $$\int_{1}^{\sqrt{5}} \left( \frac{5-x}{4} - \frac{\sqrt{5-x^2}}{2} \right) \, dx$$ $$= \left[ \frac{5x}{4} - \frac{x^2}{8} - \frac{x}{4} \sqrt{5-x^2} - \frac{5}{2} \sin^{-1} \frac{x}{\sqrt{5}} \right]_{1}^{\sqrt{5}}$$ $$= \frac{5}{4} \sqrt{5} - \frac{5}{4} - \frac{5}{4} \cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$$ If we assume $\alpha, \beta, \gamma, \in \mathbb{Q}$ (Not given in question) then $\alpha = \frac{5}{4}, \beta = -\frac{5}{4} \& \gamma = -\frac{5}{4}$ $$|\alpha + \beta + \gamma| = 1.25$$
Maths · Determinants · Numerical
Let a, b, c, d be in arithmetic progression with common difference $\lambda$. If $$\begin{vmatrix} x + a - c & x + b & x + a \\ x - 1 & x + c & x + b \\ x - b + d & x + d & x + c \end{vmatrix} = 2$$ then value of $\lambda^2$ is equal to ___
Answer: 1
Given the determinant equation: $$\begin{vmatrix} x + a - c & x + b & x + a \\ x - 1 & x + c & x + b \\ x - b + d & x + d & x + c \end{vmatrix} = 2$$ Perform the column operation $C_2 \rightarrow C_2 - C_3$: $$\begin{vmatrix} x - 2\lambda & \lambda & x + a \\ x - 1 & \lambda & x + b \\ x + 2\lambda & \lambda & x + c \end{vmatrix} = 2$$ Perform the row operations $R_2 \rightarrow R_2 - R_1$, $R_3 \rightarrow R_3 - R_1$: $$\begin{vmatrix} x - 2\lambda & 1 & x + a \\ \lambda & 2\lambda - 1 & 0 \\ 4\lambda & 0 & 2\lambda \end{vmatrix} = 2$$ Simplifying gives: $$1 (4\lambda^2 - 4\lambda^2 + 2\lambda) = 2$$ Thus, $\lambda^2 = 1$.
Maths · Permutations and Combinations · Numerical
There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicketkeeper, is ___
Answer: 777
15 : Players 6 : Bowlers 7 : Batsmen 2 : Wicket keepers Total number of ways for at least $4$ bowlers, $5$ batsmen \& $1$ wicket keeper $={}^{6}C_{4}\times{}^{7}C_{5}\times{}^{2}C_{1}+{}^{6}C_{5}\times{}^{7}C_{5}\times{}^{2}C_{1}$ $=630+147$ $=777$
Maths · Conic Sections · Numerical
Let $y = mx + c$, $m > 0$ be the focal chord of $y^2 = -64x$, which is tangent to $(x + 10)^2 + y^2 = 4$ Then, the value of $4\sqrt{2}(m + c)$ is equal to
Answer: 34
Given $y^2 = -64x$. The focus is $(-16, 0)$. The equation $y = mx + c$ is a focal chord. Therefore, $c = 16m$. The line $y = mx + c$ is tangent to $(x + 10)^2 + y^2 = 4$. Thus, $$y = m(x + 10) \pm 2\sqrt{1 + m^2}$$ which implies $$c = 10m \pm 2\sqrt{1 + m^2}$$ Therefore, $$16m = 10m \pm 2\sqrt{1 + m^2}$$ Solving gives $$6m = 2\sqrt{1 + m^2} (m > 0)$$ which leads to $$9m^2 = 1 + m^2$$ Solving for $m$, we get $$m = \frac{1}{2\sqrt{2}} \&c = \frac{8}{\sqrt{2}}$$ Finally, $$4\sqrt{2}(m + c) = 4\sqrt{2}\left(\frac{17}{2\sqrt{2}}\right) = 34$$
Maths · Limits and Derivatives · Fill in the blank
If the value of $\displaystyle \lim_{x\to0}\left(2-\cos x\sqrt{\cos 2x}\right)^{\left(\frac{x+2}{x^2}\right)}$ is equal to $e^a$, then $a$ is equal to ________.
Answer: 3
Given the limit $$\lim_{x \to 0} \left(2 - \cos x \sqrt{\cos x}\right)^{\frac{x+2}{x^2}}$$ form: $1^\infty$. We have: $$= e^{\lim_{x \to 0} \left(\frac{1 - \cos x \sqrt{\cos 2x}}{x^2}\right) \times (x+2)}$$ Now consider the limit: $$\lim_{x \to 0} \frac{1 - \cos x \sqrt{\cos 2x}}{x^2}$$ Using L'Hospital's Rule: $$= \lim_{x \to 0} \frac{\sin x \sqrt{\cos 2x} - \cos x \times \frac{1}{2 \sqrt{\cos 2x}} \times (-2 \sin 2x)}{2x}$$ (by L'Hospital's Rule) $$= \lim_{x \to 0} \frac{\sin x \cos 2x + \sin 2x \cdot \cos x}{2x}$$ $$= \frac{1}{2} + 1 = \frac{3}{2}$$ So, $$e^{\lim_{x \to 0} \left(\frac{1 - \cos x \sqrt{\cos 2x}}{x^2}\right) (x+2)}$$ $$= e^{\frac{3}{2} \times 2} = e^3$$ Thus, $$a = 3$$
Physics · Current Electricity · Single correct
The value of current in the $6\,\Omega$ resistance is :
Answer: (c)
Applying KCL at point P, $$\frac{V - 0}{6} + \frac{V - 90}{5} + \frac{V - 140}{20} = 0$$ which simplifies to $$10 \, \mathrm{V} + 12 \, \mathrm{V} - 1080 + 3 \, \mathrm{V} - 420 = 0$$ Therefore, $$V = 60$$ Thus, the current in $6\, \Omega$ is $$\frac{V - 0}{6} = 10 \, \mathrm{A}$$ Hence option 3.
Physics · Laws of Motion · Single correct
The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a $30^\circ$ banked road at maximum possible speed without skidding is ___ $\times 10^3$ kg m/s$^2$
Answer: (a)
At $v_{max}$, $f$ will be limiting in nature. Therefore, balancing force in vertical direction, $N \cos 30^\circ - mg - \mu N \cos 60^\circ = 0$ $$\Rightarrow N [\cos 30^\circ - \mu \cos 60^\circ] = mg$$ Therefore, $$N = \frac{800 \times 10}{(0.87 - 0.1)} \approx 10.2 \times 10^3 \, \mathrm{kg \, m/s^2}$$ Hence option 1.
Physics · Nuclei · Single correct
A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years respectively. What will be the time after the which one third of the material remains ? (Take $\ln 3 = 1.1$ )
Answer: (d)
Given $\lambda_1 = \frac{\ln 2}{700} year^{-1}$, $\lambda_2 = \frac{\ln 2}{1400} year^{-1}$. Therefore, $\lambda_{met} = \lambda_1 + \lambda_2 = \ln 2 \left[ \frac{1}{700} + \frac{1}{1400} \right]$. $$= \frac{3 \ln 2}{1400} year^{-1}$$ Now, let the initial number of radioactive nuclei be No. Thus, $\frac{N_0}{3} = N_0 e^{-\lambda_{mat} t}$. $$\Rightarrow \ln \frac{1}{3} = -\lambda_{net} t$$ $$\Rightarrow 1.1 = \frac{3 \times 0.693}{1400} t \Rightarrow t \approx 740 years$$ Hence option 4.
Physics · Laws of Motion · Single correct
A steel block of 10 kg rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration 0.2 $\mathrm{m/s^2}$. The normal reaction $R'$ by the floor if mass of the iron cylinders are equal and of 20 kg each, is N. [Take $g = 10 \, \mathrm{m/s^2}$ and $\mu_s = 0.2$]
Answer: (b)
Writing force equation in vertical direction $$Mg - N = Ma$$ $$\Rightarrow 70 \, g - N = 70 \times 0.2$$ $$\Rightarrow N = 70 [g - 0.2] = 70 \times 9.8$$ $$\therefore N = 686 \, Newton$$ Note: Since there is no compressive normal from the sides, hence friction will not act. Hence option 2.
Physics · Alternating Current · Single correct
AC voltage $V(t) = 20 \sin \Omega t$ of frequency $50 \, \mathrm{Hz}$ is applied to a parallel plate capacitor. The separation between the plates is $2 \, \mathrm{mm}$ and the area is $1 \, \mathrm{m}^2$. The amplitude of the oscillating displacement current for the applied AC voltage is ___ [ Take $\varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{F/m}$ ]
Answer: (c)
From the given information, $$C = \frac{\varepsilon_0 \, A}{d} = \frac{\varepsilon_0 \times 1}{2 \times 10^{-3}} \, \mathrm{F}$$ $$\therefore \; X_C = \frac{1}{\omega C} = \frac{2 \times 10^{-3}}{2 \times 50 \pi \times \varepsilon_0} = \frac{2 \times 10^{-3}}{25 \times 4 \pi \varepsilon_0} \, \Omega$$ $$\therefore \; X_C = \frac{2 \times 10^{-3}}{25} \times 9 \times 10^9 = \frac{18}{25} \times 10^6 \, \Omega$$ $$\therefore \; i_0 = \frac{V_0}{X_C} = \frac{20 \times 25}{18} \times 10^{-6} \, \mathrm{A} = 27.47 \, \mu \mathrm{A}$$ The value of amplitude of displacement current will be same as value of amplitude of conventional current. Hence option 3.
Physics · Ray Optics and Optical Instruments · Single correct
Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is:
Answer: (d)
Given \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}. \] Substituting the given values, \[ \frac{1.4}{v} - \frac{1.25}{-40} = \frac{1.4 - 1.25}{-25}. \] Therefore, \[ \frac{1.4}{v} = -\frac{0.15}{25} - \frac{1.25}{40}. \] Solving, we get \[ v = -37.58\,\mathrm{cm}. \] Hence, the correct option is **(4)**.
Physics · Gravitation · Single correct
A person whose mass is 100 kg travels from Earth to Mars in a spaceship. Neglect all other objects in sky and take acceleration due to gravity on the surface of the Earth and Mars as 10 $\mathrm{m/s^2}$ and 4 $\mathrm{m/s^2}$ respectively. Identify from the below figures, the curve that fits best for the weight of the passenger as a function of time.
Answer: (a)
At neutral point $g = 0$ so graph $(C)$ is correct. Hence option (1).
Physics · Thermodynamics · Single correct
The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from $0^\circ \mathrm{C}$ to $50^\circ \mathrm{C}$ when no work is done is ___ (R is the universal gas constant)
Answer: (d)
Physics · Mathematics in Physics · Single correct
If $\vec{A}$ and $\vec{B}$ are two vectors satisfying the relation $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$. Then the value of $|\vec{A} - \vec{B}|$ will be
Answer: (d)
Given $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$. $AB \cos \theta = AB \sin \theta \Rightarrow \theta = 45^\circ$. $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos 45^\circ}$ $$= \sqrt{A^2 + B^2 - \sqrt{2}AB}$$ Hence option (4).
Physics · Moving Charges and Magnetism · Single correct
A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_d$ and $r_\alpha$ be their respective radii of circular path. The value of $\frac{r_d}{r_\alpha}$ is equal to:
Answer: (b)
Given $r = \frac{mv}{qB} = \frac{\sqrt{2mk}}{qB}$. $$\frac{r_d}{r_\alpha} = \sqrt{\frac{m_d}{m_\alpha}} \frac{q_\alpha}{q_d} = \sqrt{\frac{2}{4}} \left(\frac{2}{1}\right) = \sqrt{2}$$ Hence option (2).
Physics · Nuclei · Single correct
A nucleus of mass M emits $\gamma$-ray photon of frequency '$\nu$'. The loss of internal energy by the nucleus is: [Take 'c' as the speed of electromagnetic wave]
Answer: (d)
Energy of $\gamma$ ray $[E_\gamma] = h\nu$ Momentum of $\gamma$ ray $[P_\gamma] = \frac{h}{\lambda} = \frac{h\nu}{c}$ Total momentum is conserved. $\vec{P}_\gamma + \vec{P}_{Nu} = 0$ Where $\vec{P}_{Nu} =$ Momentum of decayed nuclei $\Rightarrow P_\gamma = P_{Nu}$ $\Rightarrow \frac{h\nu}{c} = P_{Nu}$ $\Rightarrow$ K. E. of nuclei $$= \frac{1}{2} Mv^2 = \frac{(P_{Nu})^2}{2M} = \frac{1}{2M} \left[ \frac{h\nu}{c} \right]^2$$ Loss in internal energy $= E_\gamma + K \cdot E_{Nu}$ $$= h\nu + \frac{1}{2M} \left[ \frac{h\nu}{c} \right]^2$$ $$= h\nu \left[ 1 + \frac{h\nu}{2Mc^2} \right]$$
Physics · Electric Charges and Fields · Single correct
A certain charge $Q$ is divided into two parts $q$ and $(Q-q)$. How should the charges $Q$ and $q$ be divided so that $q$ and $(Q-q)$ placed at a certain distance apart experience maximum electrostatic repulsion?
Answer: (b)
The force $F_q$ is given by $$F_q = \frac{kq(Q-q)}{L^2} = \frac{k}{L^2} (qQ - q^2)$$ The derivative of $F$ with respect to $q$ is zero when the force is maximum. $$\frac{dF}{dq} = 0$$ Differentiating, we have $$\frac{dF}{dq} = \frac{k}{L^2} [Q - 2q] = 0$$ This implies $$Q - 2q = 0 \Rightarrow Q = 2q$$
Physics · Current Electricity · Single correct
A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m$^2$. At every point the direction of current density is at an angle of 60$^\circ$ with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is: (Resistivity of magnesium $\rho = 44 \times 10^{-8} \, \Omega \mathrm{m}$)
Answer: (c)
Given $I = \vec{J} \cdot \vec{A} = JA \cos(\theta)$. $$5 = J \left( \frac{4}{100} \right) \times \cos(60)$$ $$J = 5 \times 50 = 250 \, \mathrm{A/m^2}$$ Now, $\vec{E} = \rho \cdot \vec{J}$ $$= 44 \times 10^{-8} \times 250 = 11 \times 10^{-5} \, \mathrm{V/m}$$
Physics · Kinetic Theory · Single correct
Consider a mixture of gas molecule of types A, B and C having masses $m_A < m_B < m_C$. The ratio of their root mean square speeds at normal temperature and pressure is :
Answer: (d)
The root mean square velocity $V_{RMS}$ is given by the equation $$V_{RMS} = \sqrt{\frac{3RT}{M}}.$$ Given $m_A V_B > V_C.$$ Therefore, $$\frac{1}{V_A} < \frac{1}{V_B} < \frac{1}{V_C}.$$
Physics · Motion in a Plane · Single correct
A butterfly is flying with a velocity $4\sqrt{2} \, \mathrm{m/s}$ in North-East direction. Wind is slowly blowing at $1 \, \mathrm{m/s}$ from North to South. The resultant displacement of the butterfly in $3$ seconds is:
Answer: (d)
The velocity of the butterfly in the frame of the wind is given by: $$\vec{V}_{BW} = 4\sqrt{2} \cos 45^\circ \hat{i} + 4\sqrt{2} \sin 45^\circ \hat{j}$$ This simplifies to: $$= 4\hat{i} + 4\hat{j}$$ The velocity of the wind is: $$\vec{V}_{w} = -\hat{j}$$ The velocity of the butterfly is: $$\vec{V}_{B} = \vec{V}_{BW} + \vec{V}_{w} = 4\hat{i} + 3\hat{j}$$ The displacement of the butterfly is: $$\vec{S}_{B} = \vec{V}_{B} \times t = (4\hat{i} + 3\hat{j}) \times 3 = 12\hat{i} + 9\hat{j}$$ The magnitude of the displacement is: $$|\vec{S}_{B}| = \sqrt{(12)^2 + (9)^2} = 15 \, m$$
Physics · Mechanical Properties of Solids · Single correct
The value of tension in a long thin metal wire has been changed from $T_1$ to $T_2$. The lengths of the metal wire at two different values of tension $T_1$ and $T_2$ are $\ell_1$ and $\ell_2$ respectively. The actual length of the metal wire is :
Answer: (a)
Given $$Y = \frac{FL}{A \Delta L}$$ Therefore, $$Y = \frac{T_1 \ell_0}{A(\ell_1 - \ell_0)} = \frac{T_2 \ell_0}{A(\ell_2 - \ell_0)}$$ Thus, $$1 = \frac{T_1 (\ell_2 - \ell_0)}{T_2 (\ell_1 - \ell_0)}$$ Expanding, $$T_2 \ell_1 - T_2 \ell_0 = T_1 \ell_2 - T_1 \ell_0$$ Rearranging gives, $$(T_1 - T_2) \ell_0 = T_1 \ell_2 - T_2 \ell_1$$ Finally, $$\ell_0 = \left( \frac{T_1 \ell_2 - T_2 \ell_1}{T_1 - T_2} \right)$$
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
For the circuit shown below, calculate the value of $I_Z$:
Answer: (a)
Given $I = \frac{50}{1000} = 50 \, \mathrm{mA}$. The resistance $R = 1000 \, \Omega$. For the second part, $I = \frac{50}{2000} = 25 \, \mathrm{mA}$. The current $I_Z = I_{1000} - I_{2000} = 50 - 25 = 25 \, \mathrm{mA}$.
Physics · Electromagnetic Induction · Single correct
The arm PQ of a rectangular conductor is moving from $x = 0$ to $x = 2b$ outwards and then inwards from $x = 2b$ to $x = 0$ as shown in the figure. A uniform magnetic field perpendicular to the plane is acting from $x = 0$ to $x = b$. Identify the graph showing the variation of different quantities with distance:
Answer: (c)
As rod moves in field area increases up to $\bar{x} = b$ then field is absent and again flux is generated on return journey from $x = b$ to $x = 0$. Thus plot $A$ for flux. $\Rightarrow \; e = - \frac{d\phi}{dt} \Rightarrow$ curve $B$ for emf $\Rightarrow$ Power dissipated $= vi \Rightarrow$ curve $C$ for power dissipated
Physics · Physical World, Units and Measurements · Single correct
The entropy of a system is given by $S=\alpha^2\beta\ln\left[\frac{\mu kR}{J\beta^2}+3\right]$, where $\alpha$ and $\beta$ are constants. $\mu$, $J$, $k$, and $R$ represent the number of moles, mechanical equivalent of heat, Boltzmann constant, and gas constant, respectively. Given that $S=\frac{\mathrm{d}Q}{T}$, choose the incorrect option from the following:
Answer: (d)
Physics · Atoms · Single correct
The radiation corresponding to $3 \to 2$ transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of $5 \times 10^{-4} \, \mathrm{T}$. Assume that the radius of the largest circular path followed by these electrons is $7 \, \mathrm{mm}$, the work function of the metal is: (Mass of electron $= 9.1 \times 10^{-31} \, \mathrm{kg}$)
Answer: (d)
Given the transition from $3 \to 2$, the energy is $1.89 \, \mathrm{eV}$. The magnetic field is $5 \times 10^{-4} \, \mathrm{T}$ and the radius is $r = 7 \, \mathrm{mm}$. Using the relation $r = \frac{mv}{qB}$, we have $mv = qrB$. Thus, the energy $E$ is given by $$E = \frac{p^2}{2m} = \frac{(qRB)^2}{2m}$$ Substituting the values, $$= \frac{(1.6 \times 10^{-19} \times 7 \times 10^{-3} \times 5 \times 10^{-4})^2}{2 \times 9.1 \times 10^{-31}} \, \mathrm{Joule}$$ $$= \frac{3136 \times 10^{-52}}{18.2 \times 10^{-31} \times 1.6 \times 10^{-19}} \, \mathrm{eV}$$ $$= 1.077 \, \mathrm{eV}$$ We know the work function $= energy incident - (\mathrm{KE})_{electron}$. Therefore, $\phi = 1.89 - 1.077 = 0.813 \, \mathrm{eV}$.
Physics · Motion in a Plane · Numerical
In a spring gun having spring constant $100 \, \mathrm{N/m}$ a small ball 'B' of mass $100 \, \mathrm{g}$ is put in its barrel (as shown in figure) by compressing the spring through $0.05 \, \mathrm{m}$. There should be a box placed at a distance 'd' on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of $2 \, \mathrm{m}$ above the ground. The value of $d$ is ___ m. $(g = 10 \, \mathrm{m/s^2})$
Answer: 1
Given $\($ $\frac{1}{2}$ kx^2 = $\frac{1}{2}$ mv^2 $\)$. Therefore, $\($ Kx^2 = mv^2 $\)$. The velocity $\($ v $\)$ is given by $\($ v = x $\sqrt{\frac{k}{m}}$ = 0.05 $\sqrt{\frac{100}{0.1}}$ = 0.05 $\times$ 10 $\sqrt{10}$ $\)$. Thus, $\($ v = 0.5 $\sqrt{10}$ $\)$. From $\($ h = $\frac{1}{2}$ gt^2 $\)$, we have $\($ t = $\sqrt{\frac{2h}{g}}$ = $\sqrt{\frac{2 \times 2}{10}}$ = $\frac{2}{\sqrt{10}}$ $\)$. Therefore, $\($ d = vt = 0.5 $\sqrt{10}$ $\times$ $\frac{2}{\sqrt{10}}$ = 1 $\mathrm{m}$ $\)$.
Physics · Alternating Current · Numerical
In an LCR series circuit, an inductor $30 \, \mathrm{mH}$ and a resistor $1 \, \Omega$ are connected to an AC source of angular frequency $300 \, \mathrm{rad/s}$. The value of capacitance for which, the current leads the voltage by $45^\circ$ is $\frac{1}{x} \times 10^{-3} \, \mathrm{F}$. Then the value of $x$ is ___
Answer: 3
Given $\($ $\tan$ $\phi$ = $\frac{x_C - x_L}{R}$ $\)$. Since $\($ $\tan$ 45 = $\frac{x_C - x_L}{R}$ $\)$, we have $\($ x_C - x_L = R $\)$. Therefore, $\($ $\frac{1}{\omega C}$ - $\omega$ L = R $\)$. Solving $\($ $\frac{1}{\omega C}$ - 300 $\times$ 0.03 = 1 $\)$, we find $\($ $\frac{1}{\omega C}$ = 10 $\)$. Thus, $\($ C = $\frac{1}{10 \omega}$ = $\frac{1}{10 \times 300}$ $\)$. Therefore, C = $\frac{1}{3} \times 10^{-3}$ \. Finally, $\($ X = 3 $\)$.
Physics · Waves · Numerical
The amplitude of wave disturbance propagating in the positive $x$-direction is given by $y = \frac{1}{(1+x)^2}$ at time $t = 0$ and $y = \frac{1}{1+(x-2)^2}$ at $t = 1\, \mathrm{s}$, where $x$ and $y$ are in meters. The shape of wave does not change during the propagation. The velocity of the wave will be ___ m/s
Answer: 2
At $t = 0$, $y = \frac{1}{1 + x^2}$. At time $t = t$, $y = \frac{1}{1 + (x - vt)^2}$. At $t = 1$, $y = \frac{1}{1 + (x - v)^2}$ $\ldots$ (i) At $t = 1$, $y = \frac{1}{1 + (x - 2)^2}$ $\ldots$ (ii) Comparing (i) $\&$ (ii), $v = 2 \, \mathrm{m/s}$
Physics · System of Particles and Rotational Motion · Numerical
A body having specific charge $8 \mu \mathrm{C/g}$ is resting on a frictionless plane at a distance $10 \, \mathrm{cm}$ from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of $100 \, \mathrm{V/m}$ is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be ___ s.
Answer: 1
Given $F = ma$. $qE = ma$. $a = \frac{qE}{m}$. Now $d = \frac{1}{2}at^2$. $$t = \sqrt{\frac{2d}{a}}$$ $$t = \sqrt{\frac{2d}{\frac{qE}{m}}}$$ $$t = \sqrt{\frac{2 \times 0.1}{\left(\frac{8 \times 10^{-6}}{10^{-3}}\right) \times 100}} = \frac{1}{2}$$ Therefore, the time period $= 2t = 1 sec$. Ans. = 1.00
Physics · Thermodynamics · Fill in the blank
In the reported figure, heat energy absorbed by a system in going through a cyclic process is ___ $\pi$ J
Answer: 100
Physics · System of Particles and Rotational Motion · Fill in the blank
A circular disc moves from the top to the bottom of an inclined plane of length $L$. When it slips down the plane, it takes time $t_1$. When it rolls down the plane, it takes time $t_2$. The value of $\frac{t_2}{t_1}$ is $\sqrt{\frac{3}{x}}$. The value of $x$ is ________.
Answer: 2
If the disk slips on the inclined plane, then its acceleration $a_1 = g \sin \theta$. The distance $L = \frac{1}{2} a_1 t_1^2$. Therefore, $$t_1 = \sqrt{\frac{2L}{a_1}} ...(i)$$ If the disk rolls on the inclined plane, its acceleration $a_2 = \frac{g \sin \theta}{1 + \frac{I}{mR^2}}$. Simplifying, $a_2 = \frac{g \sin \theta}{1 + \frac{mR^2}{2mR^2}} = \frac{2}{3} g \sin \theta$. Now $L = \frac{1}{2} a_2 \cdot t_2^2$. Therefore, $$t_2 = \sqrt{\frac{2L}{a_2}} ...(ii)$$ Now, $$\frac{t_2}{t_1} = \sqrt{\frac{a_1}{a_2}} = \sqrt{\frac{3}{2}}$$ Thus, $x = 2$.
Physics · System of Particles and Rotational Motion · Numerical
A rod of mass $M$ and length $L$ is lying on a horizontal frictionless surface. A particle of mass $m$ travelling along the surface hits at one end of the rod with a velocity $u$ in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $\left( \frac{m}{M} \right)$ is $\frac{1}{x}$. The value of $x$ will be ___
Answer: 4
Just after collision. From momentum conservation, $P_i^0 = P_f$. $$mu = Mv (i)$$ From angular momentum conservation about $O$, $$mu \cdot \frac{L}{2} = \frac{ML^2}{12} \omega$$ $$\Rightarrow \omega = \frac{6mu}{ML} (ii)$$ From $e = \frac{R.V.S}{R.V.A}$, $$1 = \frac{v + \frac{\omega L}{2}}{u}$$ $$v + \frac{\omega L}{2} = u$$ $$v + \frac{3mu}{M} = u$$ $$\frac{mu}{M} + \frac{3mu}{M} = u$$ $$\frac{4mu}{M} = u$$ $$\frac{m}{M} = \frac{1}{4}$$ $$X = 4$$
Physics · Ray Optics and Optical Instruments · Numerical
An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 \,$\mathrm{m}$, if the focal length of the eyepiece is equal to ___ cm.
Answer: 25
For simple microscope, $$m = 1 + \frac{D}{f_0}$$ $$6 = 1 + \frac{D}{f_0}$$ $$5 = \frac{25}{f_0}$$ $$f_0 = 5 \, \mathrm{cm}$$ For compound microscope, $$m = \frac{\ell \cdot D}{f_0 \cdot f_e}$$ $$12 = \frac{60 \times 25}{5 \cdot f_e}$$ $$f_e = 25 \, \mathrm{cm}$$
Physics · Waves · Numerical
The frequency of a car horn encountered a change from 400 $\mathrm{Hz}$ to 500 $\mathrm{Hz}$. When the car approaches a vertical wall. If the speed of sound is 330 $\,$ $\mathrm{m/s}$. Then the speed of car is ___ $\mathrm{km/h}$.
Answer: 132
Wall as an observer. Frequency received by wall $$f_1 = f_0 \left( \frac{C}{C-V} \right)$$ Again wall as a source. Frequency received by observer on car $$f_2 = f_1 \left( \frac{C+V}{C} \right)$$ $$f_2 = f_0 \left( \frac{C+V}{C-V} \right)$$ $$500 = 400 \left( \frac{C+V}{C-V} \right)$$ $$\frac{5}{4} = \frac{C+V}{C-V}$$ $$C = 9 \, V$$ $$V = \frac{C}{9} = \frac{330}{9} \, \mathrm{m/s}$$ $$V = \frac{330}{9} \times \frac{18}{5} = 132 \, \mathrm{km/h}$$
Physics · Communication Systems · Numerical
A carrier wave $V_c(t) = 160 \sin (2 \pi \times 10^6 t)$ volts is made to vary between $V_{max} = 200 \, \mathrm{V}$ and $V_{min} = 120 \, \mathrm{V}$ by a message signal $V_m(t) = A_m \sin (2 \pi \times 10^3 t)$ volts. The peak voltage $A_m$ of the modulating signal is ____.
Answer: 40
Maximum amplitude $$A_{max} = A_m + A_C$$ $$\Rightarrow V_{max} = V_m + V_C$$ $$200 = V_m + 160$$ $$V_m = 40$$ Therefore, peak voltage $A_m = 40$ Ans. 40
Chemistry · Co-ordination Compounds · Single correct
According to the valence bond theory the hybridization of central metal atom is $\mathrm{dsp}^2$ for which one of the following compounds?
Answer: (b)
Chemistry · Biomolecules · Single correct
The correct structure of Rhumann's Purple, the compound formed in the reaction of ninhydrin with proteins is:
Answer: (d)
The reaction shown is the Ninhydrin Test. Ninhydrin reacts with an amino acid to form a colored compound. The reaction involves the decarboxylation of the amino acid and the formation of a Schiff base, resulting in the release of $\mathrm{RCHO}$, $\mathrm{CO_2}$, and $4\mathrm{H_2O}$.
Chemistry · Environmental Chemistry · Single correct
Green chemistry in day-to-day life is in the use of:
Answer: (d)
Chlorine gas was used earlier for bleaching paper. These days, hydrogen peroxide ($\mathrm{H_2O_2}$) with suitable catalyst. Tetra chloroethene ($\mathrm{Cl_2C = CCl_2}$) was earlier used as solvent for dry cleaning. The compound contaminates the ground water and is also a suspected carcinogen. Replacement of halogenated solvent by liquid $\mathrm{CO_2}$ will result in less harm to groundwater. Hence given statement (4) is correct.
Chemistry · Co-ordination Compounds · Single correct
The correct order of intensity of colors of the compounds is:
Answer: (c)
Given $[\mathrm{NiCl}_4]^{2-} > [\mathrm{Ni(H_2O)}_6]^{2+} > [\mathrm{Ni(CN)}_4]^{2-}$. Splitting $\Delta_t [\mathrm{Ni(H_2O)}_6]^{2+} > [\mathrm{Ni(CN)}_4]^{2-}$. Colour of compound.
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The set in which compounds have different nature is:
Answer: (b)
(1) $\mathrm{B(OH)_3}$ acidic and $\mathrm{H_3PO_3}$ acidic (2) $\mathrm{B(OH)_3}$ acidic and $\mathrm{Al(OH)_3}$ amphoteric (3) $\mathrm{NaOH}$ basic and $\mathrm{Ca(OH)_2}$ basic (4) $\mathrm{Be(OH)_2}$ amphoteric and $\mathrm{Al(OH)_3}$ amphoteric
Chemistry · Redox Reactions · Single correct
The species given below that does NOT show disproportionation reaction is:
Answer: (a)
In $\mathrm{BrO}_4^-$, Br is in highest oxidation state $(+7)$. So it cannot oxidise further hence it cannot show disproportionation reaction.
Chemistry · The Solid State · Single correct
Given below are two statements. One is labelled as Assertion $\textbf{A}$ and the other is labelled as Reason $\textbf{R}$. Assertion $\textbf{A}$ : Sharp glass edge becomes smooth on heating it upto its melting point. Reason $\textbf{R}$ : The viscosity of glass decreases on melting. Choose the most appropriate answer from the options given below.
Answer: (b)
Hence given statement (A) is not correct. But statement (B) is correct.
Chemistry · Polymers · Single correct
Orlon fibres are made up of:
Answer: (a)
Orlon fibers are made up of Polyacrylonitrile.
Chemistry · Hydrogen · Single correct
Given below are two statements: One is labelled as Assertion A and other is labelled as Reason R. Assertion A : The dihedral angles in $\mathrm{H_2O_2}$ in gaseous phase is $90.2^\circ$ and in solid phase is $111.5^\circ$. Reason R : The change in dihedral angle in solid and gaseous phase is due to the difference in the intermolecular forces. Choose the most appropriate answer from the options given below for A and R.
Answer: (d)
(a) $\mathrm{H_2O_2}$ structure in gas phase, dihedral angle is $111.5^\circ$. (b) $\mathrm{H_2O_2}$ structure in solid phase at $110 \, \mathrm{K}$, dihedral angle is $90.2^\circ$. Hence given statement (A) is not correct but statement (B) is correct.
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Chemical nature of the nitrogen oxide compound obtained from a reaction of concentrated nitric acid and $\mathrm{P_4O_{10}}$ (in 4 : 1 ratio) is:
Answer: (a)
Given the reaction: $$4 \mathrm{HNO_3} + \mathrm{P_4O_{10}} \rightarrow 2 \mathrm{N_2O_5} + (\mathrm{HPO_3})_4$$ Answer: $\mathrm{N_2O_5}$ is acidic in nature.
Chemistry · Some Basic Concepts of Chemistry · Single correct
An inorganic Compound 'X' on treatment with concentrated $\mathrm{H_2SO_4}$ produces brown fumes and gives dark brown ring with $\mathrm{FeSO_4}$ in presence of concentrated $\mathrm{H_2SO_4}$. Also Compound 'X' gives precipitate 'Y', when its solution in dilute $\mathrm{HCl}$ is treated with $\mathrm{H_2S}$ gas. The precipitate 'Y' on treatment with concentrated $\mathrm{HNO_3}$ followed by excess of $\mathrm{NH_4OH}$ further gives deep blue coloured solution, Compound 'X' is:
Answer: (c)
Given $\mathrm{NO_3^-} + \mathrm{H_2SO_4} \rightarrow \mathrm{NO_2} \uparrow + \mathrm{H_2O}$ with brown fumes. $\mathrm{FeSO_4} + \mathrm{H_2SO_4} + \mathrm{NO_3^-}$ Sol $^n$ conc. X $$\downarrow$$ $[\mathrm{Fe(H_2O)_5(NO)}] \mathrm{SO_4}$ (Dark brown ring) $\mathrm{Cu^{2+}} + (dilHCl + \mathrm{H_2S})$ X (cation) (Group-II reagent) $$\downarrow$$ $\mathrm{CuS} \downarrow$ (Black ppt) (Y) $\mathrm{CuS} \xrightarrow{conc. \mathrm{HNO_3}} \mathrm{Cu(NO_3)_2} + \mathrm{NO_2} + \mathrm{S} + \mathrm{H_2O}$ Excess $\mathrm{NH_4OH}$ Sol $^n$ $[\mathrm{Cu(NH_3)_4}]^{2+}$ Deep blue colour solution. Therefore, $X \rightarrow \mathrm{Cu(NO_3)_2}$
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Among the given species the Resonance stabilised carbocations are:
Answer: (c)
The question asks about resonance in structures (A) and (B). Structure (A) shows resonance involving the movement of the positive charge around the benzene ring. Structure (B) shows resonance involving the movement of the positive charge along the allylic system.
Chemistry · The s-Block Elements · Single correct
A s-block element (M) reacts with oxygen to form an oxide of the formula $\mathrm{MO_2}$. The oxide is pale yellow in colour and paramagnetic. The element (M) is:
Answer: (d)
(A) $2\mathrm{Mg} + \mathrm{O}_2 \rightarrow 2\mathrm{MgO}$ (Diamagnetic) (B) $2\mathrm{Na} + \mathrm{O}_2 \rightarrow \mathrm{Na}_2\mathrm{O}$ (Diamagnetic) $2\mathrm{Na} + \mathrm{O}_2\underset{\text{(excess)}}{} \rightarrow \mathrm{Na}_2\mathrm{O}_2$ (Diamagnetic) (C) $2\mathrm{Ca} + \mathrm{O}_2 \rightarrow 2\mathrm{CaO}$ (Diamagnetic) $\mathrm{Ca} + \mathrm{O}_2 \rightarrow \mathrm{CaO}_2$ (Diamagnetic) (D) $\mathrm{K} + \mathrm{O}_2\underset{\text{(excess)}}{} \rightarrow \mathrm{KO}_2$ (Paramagnetic)
Chemistry · Alcohols, Phenols and Ethers · Single correct
In the given reaction 3-Bromo-2,2-dimethylbutane $\xrightarrow{\mathrm{C_2H_5OH}}$ 'A' (Major Product) Product A is:
Answer: (c)
The reaction starts with 3-Bromo-2,2-dimethyl butane. In the presence of $\mathrm{C_2H_5OH}$, $\mathrm{Br^-}$ is removed, forming a carbocation. A 1,2-methyl shift occurs, resulting in a more stable carbocation. The $\mathrm{C_2H_5OH}$ then attacks the carbocation, forming an ether linkage. The final product is 2-Ethoxy-2,3-dimethyl butane.
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The metal that can be purified economically by fractional distillation method is:
Answer: (b)
Zinc can be purified economically by fractional distillation.
Chemistry · Amines · Single correct
Compound A is converted to B on reaction with $CHCl_3$ and $KOH$. The compound B is toxic and can be decomposed by C. A, B and C respectively are :
Answer: (c)
The reaction sequence involves the conversion of a primary amine $\mathrm{R-NH_2}$ to an isonitrile $\mathrm{R-N\equiv C}$ using chloroform $\mathrm{CHCl_3}$ and then hydrolysis with $\mathrm{H_3O^+}$ to regenerate the primary amine $\mathrm{R-NH_2}$ and formic acid $\mathrm{HCOOH}$.
Chemistry · Surface Chemistry · Single correct
The conditions given below are in the context of observing Tyndall effect in colloidal solutions: (A) The diameter of the colloidal particles is comparable to the wavelength of light used. (B) The diameter of the colloidal particles is much smaller than the wavelength of light used. (C) The diameter of the colloidal particles is much larger than the wavelength of light used. (D) The refractive indices of the dispersed phase and the dispersion medium are comparable. (E) The dispersed phase has a very different refractive index from the dispersion medium. Choose the most appropriate conditions from the options given below:
Answer: (a)
The phenomenon of scattering of light by colloidal particles as a result of which the path of the beam becomes visible is called a Tyndall effect. Smaller the diameter and similar the magnitude of refractive indices, lesser is the scattering and hence the Tyndall effect and vice-versa. The diameter of the dispersed phase particle should not be smaller than the wavelength of light used because they won't be able to scatter the light so, therefore, the diameter of the dispersed particles should be equal or not much smaller than the wavelength of the light used. The refractive indices (i.e. the ratio of the velocity of light in vacuum to the velocity of light in any medium) of the dispersed phase and the dispersion medium should differ greatly in magnitude than only the particles will be able to scatter the light and Tyndall effect will be observed. On the other hand, if the refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude, then there will be no scattering of light and hence, therefore, no Tyndall effect is observed. Hence answer A and E are correct.
Chemistry · Biomolecules · Single correct
Identify the incorrect statement from the following
Answer: (a)
Amylose is a linear chain polymer of $\alpha$-D-glucose while amylopectine is branched chain polymer of $\alpha$-D-glucose.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Which among the above compound/s does/do not form Silver mirror when treated with Tollen's reagent?
Answer: (c)
Aldehydes give positive Tollen's Test (Silver mirror test). (I) Positive (II) Negative (III) Positive (IV) Positive
Chemistry · Hydrocarbons · Single correct
For above chemical reactions, identify the correct statement from the following
Answer: (d)
Chemistry · Chemical Bonding and Molecular Structure · Numerical
The number of lone pairs of electrons on the central I atom in $\mathrm{I}_3^-$ is ____
Answer: 3
The number of lone pairs of electron on the central atom is 3.
Chemistry · Redox Reactions · Numerical
$250\,\mathrm{mL}$ of $0.5\,\mathrm{M}$ $\mathrm{NaOH}$ was added to $500\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\mathrm{HCl}$. The number of unreacted $\mathrm{HCl}$ molecules in the solution after complete reaction is \_\_\_\_ $\times 10^{21}$ (Nearest integer) $(N_A = 6.022 \times 10^{23})$
Answer: 226
We know that the number of moles = $V_{litre} \times Molarity$. And the number of millimoles = $V_{ml} \times Molarity$. So millimoles of NaOH = $250 \times 0.5 = 125$. Millimoles of HCl = $500 \times 1 = 500$. Now the reaction is NaOH + HCl $\rightarrow$ NaCl + H$_2$O. At $t = 0$: $$125 500 0 0$$ At $t = t$: $$0 375 125 125$$ So millimoles of HCl left = 375. Moles of HCl = $375 \times 10^{-3}$. Number of HCl molecules = $6.022 \times 10^{23} \times 375 \times 10^{-3}$. $$= 225.8 \times 10^{21}$$ $$\approx 226 \times 10^{21} = 226$$
Chemistry · Structure of Atom · Numerical
The Azimuthal quantum number for the valence electrons of $\mathrm{Ga}^{+}$ ion is ___ (Atomic number of Ga = 31)
Answer: 0
The azimuthal quantum number for the valence electrons (4s-subshell) of $\mathrm{Ga^+}$ ion is zero (0).
Chemistry · Co-ordination Compounds · Fill in the blank
The spin-only magnetic moment value for the complex $[\mathrm{Co(CN)}_6]^{4-}$ is $\_$$\_$ BM. [At. no. of Co = 27]
Answer: 2
Given $[\mathrm{Co(CN)}_6]^{4-}$, we have the equation $x + 6 \times (-1) = -4$. Solving for $x$, we get $x = +2$. Therefore, $\mathrm{Co}^{2+} : [\mathrm{Ar}] 3d^7$. CN$^-$ is a strong field ligand which can pair the electron of the central atom. It has one unpaired electron $(n)$ in the 4d-subshell. So the spin-only magnetic moment $(\mu)$ is given by $\mu = \sqrt{n(n+2)} \, \mathrm{B.M}$ where $n$ is the number of unpaired electrons. $$\mu = \sqrt{3} \, \mathrm{B.M}$$ $$\mu = 1.73 \, \mathrm{BM}$$
Chemistry · Equilibrium · Numerical
$2\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{SO_3(g)}$ In an equilibrium mixture, the partial pressures are $P_{\mathrm{SO_3}} = 43\,\mathrm{kPa}$, $P_{\mathrm{O_2}} = 530\,\mathrm{Pa}$ and $P_{\mathrm{SO_2}} = 45\,\mathrm{kPa}$. The equilibrium constant $K_P =$ \underline{\hspace{1cm}} $\times 10^{-2}$. (Round off to the nearest integer.)
Answer: 172
The reaction is given by $$2\mathrm{SO_2}(g) + \mathrm{O_2}(g) = 2\mathrm{SO_3}(g)$$ The equilibrium constant $K_p$ is given by $$K_p = \frac{(p\mathrm{SO_3}(g))^2}{p\mathrm{SO_2}(g)} \times p\mathrm{O_2}(g)$$ Substituting the values, $$= \frac{43 \times 43}{45 \times 45} \times 530 \, \mathrm{Pa^{-1}}$$ $$= 172.28 \times 10^{-5} \, \mathrm{Pa^{-1}}$$ $$= 172.28 \, \mathrm{atm}$$ $$= 17228 \times 10^{-2} \, \mathrm{atm}$$ The answer is 17228.
Chemistry · Amines · Numerical
The number of nitrogen atoms in a semicarbazone molecule of acetone is __
Answer: 3
Semicarbazone molecule of acetone
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical
To synthesise 1.0 mole of 2-methylpropan-2-ol from Ethylethanoate ___ equivalents of $\mathrm{CH_3MgBr}$ reagent will be required. (Integer value)
Answer: 2
The reaction involves the addition of $\mathrm{CH_3MgBr}$ to an ester, resulting in the formation of a tertiary alcohol. The ethyl group is replaced by the $\mathrm{CH_3}$ group from the Grignard reagent. The intermediate is then hydrolyzed with water to form 2-methylpropan-2-ol.
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, $10\%$ of the virus is inactivated. The rate constant for viral inactivation is ___ $\times 10^{-3}\,\mathrm{min}^{-1}$. (Nearest integer) [Use: $\ln 10=2.303$; $\log_{10}3=0.477$; property of logarithms: $\log x^y=y\log x$]
Answer: 106
As the unit of rate constant is $\mathrm{min}^{-1}$ so it must be a first order reaction $K \times t = 2.303 \log \frac{A_0}{A_t}$. In 1 min, 10% is inactivated so taking $A_0 = 100$, $A_1 = 90$ in 1 min. So $$K \times 1 = 2.303 \times \log \frac{100}{90}$$ $$= 2.303 \times (\log 10 - 2 \log 3)$$ $$= 2.303 \times (1 - 2 \times 0.477)$$ $$= 0.10593$$ $$= 105.93 \times 10^{-3}$$ $$\approx = 106$$
Chemistry · Thermodynamics · Numerical
An average person needs about 10000 kJ energy per day. The amount of glucose (molar mass $= 180.0 \, \mathrm{g \, mol^{-1}}$) needed to meet this energy requirement is ___ g. ( Use : $\Delta_C H(glucose) = -2700 \, \mathrm{kJ \, mol^{-1}}$ )
Answer: 667
1 mole glucose gives 2700 kJ energy so moles of glucose needed for $10^5$ kJ energy is $$\frac{100000}{2700} = 370 moles$$ wt. of glucose is $3.10 \times 180$ $$= 666.666$$ $$\approx 667 gm$$ $$\frac{Y_{Benzene}}{Y_{M.B}} = \frac{P^0_B X_B}{P^0_{MB} X_{MB}} = \frac{70 \times 1}{20 \times 1} = \frac{7}{2}$$ $$Y_{Benzene} = \frac{7}{9} = 77.77 \times 10^{-2}$$ $$= 78 \times 10^{-12}$$
Chemistry · Solutions · Numerical
At $20^\circ\mathrm{C}$, the vapour pressure of benzene is $70\,\mathrm{torr}$ and that of methylbenzene is $20\,\mathrm{torr}$. The mole fraction of benzene in the vapour phase at $20^\circ\mathrm{C}$ above an equimolar mixture of benzene and methylbenzene is ___ $\times 10^{-2}$. (Nearest integer)
Answer: 78
Given $P_B^\circ = 40$, $P_T^\circ = 20$, $K_B = 0.5 = K_T$. Now, $$y_B = \frac{K_B P_B^\circ}{K_B P_B^\circ + K_T P_T^\circ}$$ $$= \frac{70 \times 0.5}{70 \times 0.5 + 20 \times 0.5}$$