JEE Main 10 April 2023 Shift 1 question paper with solutions

Given, $O$ be the origin and the position vector of the point $P$ be $-\hat{i} - 2\hat{j} + 3\hat{k}$, So, $\overrightarrow{OP} = -\hat{i} - 2\hat{j} + 3\hat{k}$. Also given the position vectors of the points $A$, $B$ and $C$ are $-2\hat{i} + \hat{j} - 3\hat{k}$, $2\hat{i} + 4\hat{j} - 2\hat{k}$ and $-4\hat{i} + 2\hat{j} - \hat{k}$. Now finding, $$\overrightarrow{AB} = 4\hat{i} + 3\hat{j} + \hat{k}$$ $$\overrightarrow{AC} = -2\hat{i} + \hat{j} + 2\hat{k}$$ And $\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{vmatrix}$. $$\Rightarrow \overrightarrow{AB} \times \overrightarrow{AC} = 5\hat{i} - 10\hat{j} + 10\hat{k}$$ Now finding the Projection of $\overrightarrow{OP}$ on vector perpendicular to $\overrightarrow{AB}$ and $\overrightarrow{AC}$ we get, $$Projection = \frac{\left( \overrightarrow{OP} \cdot \frac{\overrightarrow{AB} \times \overrightarrow{AC}}{|\overrightarrow{AB} \times \overrightarrow{AC}|} \right) \left( \frac{\overrightarrow{AB} \times \overrightarrow{AC}}{|\overrightarrow{AB} \times \overrightarrow{AC}|} \right)}{\sqrt{1+4+4}}$$ $$= \frac{5 \left( -\hat{i} - 2\hat{j} + 3\hat{k} \right) \left( \hat{i} - 2\hat{j} + 2\hat{k} \right)}{5\sqrt{1+4+4}} = 3$$

Given, Ellipse $E : x^2 + 9y^2 = 9 \ldots (i)$ Now point $A(3, 0)$ and $B(0, 1)$ which is the intersection of given ellipse with positive axis, So, equation of line passing through $A$ and $B$ is given by, $$L : \frac{x}{3} + \frac{y}{1} = 1 \Rightarrow x = 3 - 3y \ldots (ii)$$ Now equation of Circle with diametric point $(-3, 0)$ and $(3, 0)$ is given by, $$C : x^2 + y^2 = 9 \ldots (iii)$$ Let $Q$ be foot of perpendicular from $P$ upon major axis, So, from (ii) and (iii) we get, $$(3 - 3y)^2 + y^2 = 9$$ $$\Rightarrow y = \frac{9}{5}, 0$$ Hence, $PQ = \frac{9}{5}$ Now Area of triangle will be, $$\frac{1}{2} \times OA \times PQ = \frac{1}{2} \times 3 \times \frac{9}{5} = \frac{27}{10}$$ Hence, $m - n = 17$

Given, $$f(x) = \frac{\left(\tan 1^\circ\right)x + \log_e (123)}{x \log_e (1234) - \left(\tan 1^\circ\right)}, x > 0$$ Now, let $\tan 1^\circ = a$, $\log_e (123) = b$ and $\log_e (1234) = c$. So, $$f(x) = \frac{ax + b}{cx - a}$$ $$\Rightarrow f(f(x)) = \frac{af(x) + b}{cf(x) - a}$$ $$= \frac{a \left(\frac{ax + b}{cx - a}\right) + b}{c \left(\frac{ax + b}{cx - a}\right) - a}$$ $$= \frac{\frac{a^2 x + ab + bc (cx - a)}{cx - a} + b}{\frac{acx + bc - a (cx - a)}{cx - a}}$$ $$= \frac{a^2 x + ab + bcx - ac}{acx + bc - a (cx - a)} = x$$ $$\Rightarrow f(f(x)) + f\left(f\left(\frac{4}{x}\right)\right) = x + \frac{4}{x}$$ Therefore, $x > 0$, then least value $= \sqrt{x \cdot \frac{4}{x}} = 2$

Given that, the side of square is 30 cm and $x$ cm squares are cut off. The required diagram is, Now the dimensions of the cuboid formed will be $l(x) = 30 - 2x$, $b(x) = 30 - 2x$ and $h(x) = x$. The Volume of the cuboid will be $V(x) = (30 - 2x)^2(x)$. Now to get Maximum value, $$\frac{dV(x)}{dx} = 0$$ $$\Rightarrow 2(30 - 2x)(-2)x + (30 - 2x)^2(1) = 0$$ $$\Rightarrow (30 - 2x)(-4x + 30 - 2x) = 0$$ On simplifying we get, $$\Rightarrow x = 15 \, cm, \, 5 \, cm$$ But $x$ cannot be 15 cm as the volume becomes zero. Hence $x = 5$ cm. Now to find the surface area of the cuboid, Surface area will be $=(30 - 2x) \times x \times 4 + (30 - 2x)^2$ $$=(30 - 2 \times 5) \times 5 \times 4 + (30 - 2 \times 5)^2$$ $$= 800 \, cm^2.$$ Therefore, the required surface area will be 800 cm$^2$.

Given, a line segment $AB$ of length $\lambda$ moves such that the points $A$ and $B$ remain on the periphery of a circle of radius $\lambda$. Now taking the points on the circle of radius $\lambda$ as $B(\lambda \cos \theta_1, \lambda \sin \theta_1)$ and $A(\lambda \cos \theta_2, \lambda \sin \theta_2)$ and taking the point $P(h, k)$ which divides the line segment in $AB$ of length $\lambda$ in $2 : 3$. Now plotting the diagram we get, Now, let $O$ be the origin and radius of circle is $\lambda$ and $AB = \lambda$ and using distance formula we get, $$AB = \lambda = \sqrt{(\lambda \cos \theta_1 - \lambda \cos \theta_2)^2 + (\lambda \sin \theta_1 - \lambda \sin \theta_2)^2}$$ $$\Rightarrow 1 = 2 - 2 \cos(\theta_1 - \theta_2)$$ $$\Rightarrow \cos(\theta_1 - \theta_2) = \frac{1}{2}$$ Now using section formula we get, $$h = \frac{2\lambda \cos \theta_1 + 3\lambda \cos \theta_2}{5} and k = \frac{2\lambda \sin \theta_1 + 3\lambda \sin \theta_2}{5}$$ Now squaring and adding above two value we get, $$h^2 + k^2 = \frac{\lambda^2}{25} \left[ 4 + 9 + 12(\cos(\theta_1 - \theta_2)) \right]$$ $$\Rightarrow h^2 + k^2 = \frac{\lambda^2}{25} \cdot 19$$ Hence, Radius $= \frac{\lambda}{5} \sqrt{19}$

Given, the complex number $z = x + iy$ be such that $\frac{2z - 3i}{2z + i}$ is purely imaginary. Now putting the value of $z = x + iy$ in $\frac{2z - 3i}{2z + i}$ we get, $$\frac{2(x+iy) - 3i}{2(x+iy) + i} = \frac{(2x+i(2y-3))}{(2x+i(2y+1))} \cdot \frac{(2x-i(2y+1))}{(2x-i(2y+1))}$$ Now taking real part as, $$\frac{4x^2 + (2y-3)(2y+1)}{4x^2 + (2y+1)^2} = 0$$ $$\Rightarrow 4x^2 + 4y^2 - 4y - 3 = 0$$ $$\Rightarrow x^2 + y^2 - y - \frac{3}{4} = 0$$ Now using, $x + y^2 = 0 \Rightarrow x = -y^2$ we get, $$x^2 + y^2 - y - \frac{3}{4} = 0$$ $$\Rightarrow y^4 + y^2 - y - \frac{3}{4} = 0$$ $$\Rightarrow y^4 + y^2 = \frac{3}{4}$$

Given, Expression $96 \cdot \cos \frac{\pi}{33} \cdot \cos \frac{2\pi}{33} \cdot \cos \frac{4\pi}{33} \cdots \cos \frac{16\pi}{33}$. Now we know that, $$\cos A \cdot \cos 2A \cdot \cos 2^2 A \cdot \cos 2^3 A \cdots \cos 2^{n-1} A = \frac{\sin 2^n A}{2^n \sin A}$$ Now using the above formula in given expression we get, $$96 \cdot \cos \frac{\pi}{33} \cdot \cos \frac{2\pi}{33} \cdot \cos \frac{4\pi}{33} \cdots \cos \frac{16\pi}{33}$$ $$= 96 \times \frac{\sin \frac{32\pi}{33}}{2^5 \sin \frac{\pi}{33}}$$ $$= 96 \times \frac{\sin \left( \pi - \frac{\pi}{33} \right)}{2^5 \sin \frac{\pi}{33}}$$ $$= 96 \times \frac{\sin \left( \frac{\pi}{33} \right)}{2^5 \sin \frac{\pi}{33}} \{ as \sin(\pi - \alpha) = \sin \alpha \}$$ $$= 96 \times \frac{1}{32} = 3$$

We need to find the value of $|3 \operatorname{adj}(|3A|A^2)|$. We know that $|kA| = k^n |A|$ where $n$ is the order of the matrix and $k$ is a constant. $$|3A| = 3^3 (2)$$ $$|3 \operatorname{adj}(|3A|A^2)| = 3^3 |\operatorname{adj}((3^3 \cdot 2) A^2)|$$ We know that $\operatorname{adj}(kA) = k^{n-1} \operatorname{adj}(A)$ $$= 3^3 \left(3^3 \cdot 2\right)^2 \operatorname{adj}(A^2)$$ Now we know that $|\operatorname{adj}A| = |A|^{n-1}$ $$= 3^3 \left(3^3 \cdot 2\right)^2 3^3 |\operatorname{adj}(A^2)|$$ $$= 3^3 (3^3 \cdot 2)^6 |A^2|^{3-1}$$ $$= 3^3 (3^3 \cdot 2)^6 2^2$$ $$= 2^{10} \cdot 3^{21}$$ Hence, the required answer is $2^{10} \cdot 3^{11}$.

Given: $$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$$ Put $y = vx$ implies $$\frac{dy}{dx} = v + x \frac{dv}{dx}$$ $$v + x \frac{dv}{dx} = \frac{1 + v^2}{2v}$$ implies $$x \frac{dv}{dx} = \frac{1 - v^2}{2v}$$ implies $$\int \left( \frac{2v}{v^2 - 1} \right) dv = - \int \frac{dx}{x}$$ implies $$\log_e |v^2 - 1| = \log_e \left( \frac{C}{x} \right)$$ implies $$\frac{y^2 - x^2}{x^2} = \frac{C}{x}$$ implies $$y^2 - x^2 = Cx$$ Put $x = 2$ and $y = 0$ we get, $$0 - 2^2 = 2C \implies C = -2$$ implies $$y^2 = x^2 - 2x$$ $$y(8) = \sqrt{8^2 - 16}$$ $$y(8) = \sqrt{48} = 4\sqrt{3}$$

Given, for the system of linear equations $2x - y + 3z = 5$, $3x + 2y - z = 7$, $4x + 5y + \alpha z = \beta$. Now finding, $$D = \begin{vmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{vmatrix}$$ $$\Rightarrow D = 2(2\alpha + 5) + (3\alpha + 4) + 3(7)$$ $$\Rightarrow D = 7\alpha + 35$$ Now we know that, for unique solution $D \neq 0 \Rightarrow \alpha \neq -5$. So, in option (B) $\alpha = -6$ corresponds to infinitely many solution, which is completely wrong hence, this is answer.

Given that, $2^N < N!$ where $N$ is the sum of numbers of two dice. Therefore, $2 \leq N \leq 12$. Let us check the given condition $2^N < N!$. For $N = 1$ (not possible) $\rightarrow 0$. For $N = 2$ (not possible) $\rightarrow 1$. For $N = 3$ (not possible) $\rightarrow 2$. For $N = 4$ (possible). We need to find the probability for $N \geq 4$. Therefore, $$P(N \geq 4) = 1 - P(N = 2) - P(N = 3)$$ $$= 1 - \frac{1}{36} - \frac{2}{36}$$ $$= \frac{11}{12} = \frac{m}{n}$$ Now let us find $4m - 3n$. $$= 4 \times 11 - 3 \times 12 = 8.$$ Hence, the required answer is 8.

Given, $P$ be the point of intersection of the line $\frac{x+3}{3} = \frac{y+2}{1} = \frac{1-z}{2}$ and the plane $x + y + z = 2$. Now let, $\frac{x+3}{3} = \frac{y+2}{1} = \frac{1-z}{2} = \lambda$. Say, point $A(3\lambda - 3, \lambda - 2, 1 - 2\lambda)$. Since point lie on the plane also, So, $3\lambda - 3 + \lambda - 2 + 1 - 2\lambda = 2$ $$\Rightarrow 2\lambda = 6$$ $$\Rightarrow \lambda = 3$$ Hence, the point will be $P(6, 1, -5)$. Now finding the distance of the point $P(6, 1, -5)$ from the plane $3x - 4y + 12z = 32$ we get, $$q = \left| \frac{18 - 4 - 60 - 32}{\sqrt{9 + 16 + 144}} \right| = \frac{78}{13} = 6$$ Equation with roots $q$ and $2q$ will be, $$x^2 - 3qx + 2q^2 = 0$$ $$\Rightarrow x^2 - 18x + 72 = 0$$

Given, $ (p \lor q) \land (q \lor (\sim r)) $ Now negation of above expression will be, $ \sim ((p \lor q) \land (q \lor (\sim r))) $ $$ = \sim (p \lor q) \lor \sim (q \lor (\sim r)) \{ as (A \land B) \equiv \sim A \lor \sim B \} $$ $$ = (\sim p \land \sim q) \lor (\sim q \land r) $$ $$ = (\sim p \land \sim q) \lor (\sim q) \land (\sim p \land \sim q) \lor r $$ $$ = \sim q \land [(\sim p \lor r) \land (\sim q \lor r)] $$ $$ = (\sim q \land (\sim p \lor r)) \land (\sim q \land (\sim q \lor r)) $$ $$ = (\sim q \land (\sim p \lor r)) \land \sim q $$ $$ = ((\sim p) \lor r) \land (\sim q) $$

Given, the coefficient of $x^7$ in the expansion of $\left(ax^2 - \frac{1}{bx}\right)^{13}$ is equal to the coefficient of $x^{-5}$ in $\left(ax + \frac{1}{bx^2}\right)^{13}$. We know that, the general term $T_{r+1}$ in the expansion $(a + b)^n$ is $$T_{r+1} = {}^{n}C_{r} a^{n-r} b^r$$ Applying to $\left(ax^2 - \frac{1}{bx}\right)^{13}$, we get $$T_{r+1} = {}^{13}C_{r} (ax^2)^{13-r} \left(-\frac{1}{bx}\right)^r$$ $$\Rightarrow T_{r+1} = (-1)^r \times {}^{13}C_{r} (a)^{13-r} (x)^{26-3r} (b)^{-r}$$ Therefore, $26 - 3r = 7 \Rightarrow r = \frac{19}{3}$ for coefficient of $x^7$. Thus, $$T_3 = {}^{13}C_{2} \left(\frac{a}{b^2}\right)^{11}$$ Similarly, applying to $\left(ax + \frac{1}{bx^2}\right)^{13}$, we get $$T_{r+1} = {}^{13}C_{r} (ax)^{13-r} \left(\frac{1}{bx^2}\right)^r$$ $$\Rightarrow T_{r+1} = {}^{13}C_{r} (a)^{13-r} (x)^{13-3r} (b)^{-r}$$ Therefore, $13 - 3r = -5$ for coefficient of $x^{-5}$ $$\Rightarrow r = 6$$ So, $$T_7 = {}^{13}C_{6} (a)^7 (b)^{-6}$$ Hence, applying the given condition we get $${}^{13}C_{2} \left(\frac{a}{b^2}\right)^{11} = {}^{13}C_{6} (a)^7 (b)^{-6}$$ $$\Rightarrow a^4 b^{14} = \frac{{}^{13}C_{6}}{{}^{13}C_{2}}$$ $$\Rightarrow a^4 b^{14} = \frac{13!}{7!6!} \times \frac{2!11!}{13!}$$ $$\Rightarrow a^4 b^{14} = \frac{11 \times 10 \times 9 \times 8}{6 \times 5 \times 4 \times 3}$$ $$\Rightarrow a^4 b^{14} = 22$$

Given, two vertices of a triangle $ABC$ be $(2, 4, 6)$ and $(0, -2, -5)$, and its centroid be $(2, 1, -1)$. Now, let the vertex 'C' be $(a, b, c)$ and using centroid formula we get, $$2 = \frac{2+0+a}{3} \Rightarrow a = 4$$ $$1 = \frac{4-2+b}{3} \Rightarrow b = 1$$ $$-1 = \frac{6-5+c}{3} \Rightarrow c = -4$$ So, the third vertex will be, $C(4, 1, -4)$. Now finding image of $C$ in $x + 2y + 4z = 11$ we get, $$\frac{\alpha - 4}{1} = \frac{\beta - 1}{2} = \frac{\gamma + 4}{4} = -2 \Rightarrow \frac{-4 - 2 \times 1 - 4 \times (-4) - 11}{1^2 + 2^2 + 4^2} = -2 \Rightarrow \alpha - 4 = 2 \Rightarrow \alpha = 6, \; \beta - 1 = 2 \Rightarrow \beta = 5 \; and \; \frac{\gamma + 4}{4} = 2 \Rightarrow \gamma = 4$$ Now find the required answer we get, $$\alpha \beta + \beta \gamma + \alpha \gamma = 30 + 20 + 24 = 74$$

Given, $$\frac{x+2}{1} = \frac{y}{-2} = \frac{z-5}{2}$$ $$\frac{x-4}{1} = \frac{y-1}{2} = \frac{z+3}{0}$$ Now we know that, Shortest distance between two skew lines is given by, $$SD = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{n_1} \times \vec{n_2})}{|\vec{n_1} \times \vec{n_2}|} \right|$$ Now finding, $$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 1 & 2 & 0 \end{vmatrix}$$ $$\Rightarrow \vec{n_1} \times \vec{n_2} = -4\hat{i} + 2\hat{j} + 4\hat{k}$$ And $$\vec{a_2} - \vec{a_1} = \left( 4\hat{i} + \hat{j} - 3\hat{k} - (-2\hat{i} + 0\hat{j} + 5\hat{k}) \right) = 6\hat{i} + \hat{j} - 8\hat{k}$$ Now putting the value in formula we get, $$SD = \frac{(6\hat{i} + \hat{j} - 8\hat{k}) \cdot (-4\hat{i} + 2\hat{j} + 4\hat{k})}{\sqrt{16 + 16 + 4}}$$ $$\Rightarrow SD = \frac{-24 + 2 - 32}{6} = \frac{54}{6} = 9$$

Given, $$I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) \, dx$$ $$\Rightarrow I(x) = \int e^{\sin^2 x} \left( \cos x - \frac{1}{\cos x} \right) \sin 2x \, dx$$ Now let, $\sin^2 x = t \Rightarrow \sin 2x \, dx = dt$ $$I(x) = \int e^t \left( \sqrt{1-t} - \frac{1}{2\sqrt{1-t}} \right) \, dt$$ Now comparing with, $$I(x) = \int e^t (f(t) + f'(t)) \, dt = e^t f(t) + c$$ Here, $f(t) = \sqrt{1-t}$ and $f'(t) = -\frac{1}{2\sqrt{1-t}}$ $$\Rightarrow I(x) = e^t \sqrt{1-t} + c = e^{\sin^2 x} \cdot \cos x + c$$ Now using, $I(0) = 1 \Rightarrow c = 0$ Hence, the value of $I\left(\frac{\pi}{3}\right) = \frac{1}{2} e^{\frac{3}{4}}$$

Given that, $$a^2 + (ar)^2 + (ar^2)^2 = 33033, \ (a, r \in \mathbb{N})$$ $$\Rightarrow a^2 (1 + r^2 + r^4) = 11^2 \times 273$$ On comparing both sides of the equation we get, $$a = 11 and 1 + r^2 + r^4 = 273$$ $$\Rightarrow r^2 + r^4 = 272$$ $$\Rightarrow r^2 (1 + r^2) = 16 \times 17$$ $$\Rightarrow r = 4$$ Now $$a + ar + ar^2 = 11 + 11 \times 4 + 11 \times 16$$ $$= 11 + 44 + 176 = 231.$$ Therefore, the required value is 231.

Given, an arc $PQ$ of a circle subtends a right angle at its centre $O$. The midpoint of the arc $PQ$ is $R$, and $\overrightarrow{OP} = \vec{u}$, $\overrightarrow{OR} = \vec{v}$ and $\overrightarrow{OQ} = \alpha \vec{u} + \beta \vec{v}$. So, plotting the diagram of the given value we get, Now given, $\overrightarrow{OQ} = \alpha \vec{u} + \beta \vec{v}$. Where, $\overrightarrow{OR} = \vec{v}$ and $\overrightarrow{OP} = \vec{u}$. Now $\left| \overrightarrow{OP} \right| = \left| \overrightarrow{OQ} \right| = radius$, so $R$ will lie on angle bisector of $\overrightarrow{OQ}$ and $\overrightarrow{OP}$. Now using the perpendicular condition we get, $$\overrightarrow{OQ} \cdot \overrightarrow{OP} = 0$$ $$\Rightarrow \overrightarrow{OQ} \cdot \overrightarrow{OP} = \alpha \left| \vec{u} \right|^2 + \beta \cdot \left( \vec{v} \cdot \vec{u} \right) = 0$$ $$\Rightarrow \alpha \left| \vec{u} \right|^2 + \beta \cdot \left( \left| \vec{u} \right| \cos 45^\circ \right) = 0$$ $$\Rightarrow \alpha + \beta \cdot \cos 45^\circ = 0 \{ as \left| \vec{u} \right| = \left| \vec{v} \right| = radius \}$$ $$\Rightarrow \alpha = -\frac{\beta}{\sqrt{2}}$$ Now solving, $\overrightarrow{OQ} \cdot \overrightarrow{OR} = \left| \overrightarrow{OQ} \right| \left| \overrightarrow{OR} \right| \cos 45^\circ = \frac{r^2}{\sqrt{2}}$ $$\Rightarrow \left( \alpha \vec{u} + \beta \vec{v} \right) \cdot \vec{v} = \frac{r^2}{\sqrt{2}}$$ $$\Rightarrow \alpha \cdot \frac{r^2}{\sqrt{2}} + \beta \cdot r^2 = \frac{r^2}{\sqrt{2}}$$ $$\Rightarrow \alpha \cdot \frac{1}{\sqrt{2}} + \beta = \frac{1}{\sqrt{2}}$$ $$\Rightarrow \beta = \sqrt{2} \{ as \alpha = -\frac{\beta}{\sqrt{2}} \}$$ So, $\alpha = -1$ Now finding the equation with roots $\alpha = -1$, $\beta = \sqrt{2}$ we get, $x^2 - x - 2 = 0$

We need to find the coefficient of $x^7$ in $(1 - x + 2x^3)^{10}$. We know that for $(x + y + z)^n$, $$T_n = \frac{n!}{a!b!c!} (x)^a (y)^b (z)^c$$ such that $a + b + c = n$. Now for $(1 - x + 2x^3)^{10}$, $$T_n = \frac{10!}{a!b!c!} (1)^a (-x)^b (2x^3)^c$$ $$= \frac{10!}{a!b!c!} (-1)^b 2^c x^{b+3c}$$ with $a + b + c = 10$. Here we need the coefficient of $x^7$. Hence the combinations would be | a | b | c | | 3 | 7 | 0 | | 5 | 4 | 1 | | 7 | 1 | 2 | which satisfies both $a + b + c = 10$ and $b + 3c = 7$. Hence, the coefficient of $x^7$ will be, $$= \frac{10!}{2!1!7!} (2)^2 (-1)^1$$ $$+ \frac{10!}{1!4!5!} (2)^1 (-1)^4$$ $$+ \frac{10!}{0!7!3!} (2)^0 (-1)^7$$ $$= 960$$

Given, $f : (-2, 2) \to \mathbb{R}$ be defined by $f(x) = \begin{cases} x[x], & -2 < x < 0 \\ (x-1)[x], & 0 \leq x < 2 \end{cases}$ $$\Rightarrow f(x) = \begin{cases} -2x, & -2 < x < -1 \\ -x, & -1 \leq x < 0 \\ 0, & 0 < x < 1 \\ x-1, & 1 \leq x < 2 \end{cases}$$ Now plotting the diagram of the above function we get, Now from above diagram we can say that, $y = f(x)$ is same as $y = |f(x)|$ Hence, the function is not continuous at one point and non differentiable at three points, so $m = 1$, $n = 3$ Hence, $m + n = 4$

Given sequence is $3, 8, 13, \ldots, 373$. Here $a=3$, $d=5$ and $a_n=373$. Now, $a_n=a+(n-1)d$ $373=3+(n-1)5$ $n=75$ We know that $S_n=\frac{n}{2}(a+a_n)$ $S_{75}=\frac{75}{2}(3+373)$ $S_{75}=14100$ Now let us write the sequence of terms which are divisible by $3$. We get $3, 18, 33, \ldots, 363$ $363=3+(n-1)15$ $n=25$ Now let us find the sum of terms divisible by $3$. $S_{\text{div by }3}=\frac{25}{2}(3+363)=4575$ Required sum $=$ Sum of $75$ terms $-$ Sum of terms divisible by $3$ $14100-4575$ $=9525$ Therefore, the required sum is $9525$.

Given, common tangent to the curves $y^2 = 4x$ and $(x-4)^2 + y^2 = 16$ touch the curves at the points $P$ and $Q$. Let, the equation of tangent be, $y = mx + \frac{a}{m}$, where $a = 1$ is focus and $m$ is slope. So, equation of tangent to parabola will be, $y = mx + \frac{1}{m}$. Also line is tangent to circle, so, using perpendicular distance formula from centre $(4, 0)$ to tangent which will be radius $r = 4$, we get, $$\left| \frac{4m + \frac{1}{m}}{m^2 + 1} \right| = 4$$ $$\Rightarrow 16m^2 + \frac{1}{m^2} + 8 = 16m^2 + 16$$ $$\Rightarrow m^2 = \frac{1}{8}$$ $$\Rightarrow m = \pm \frac{1}{2\sqrt{2}}$$ Now if $m = \frac{1}{2\sqrt{2}}$ then $P = \left( \frac{a}{m^2}, \frac{2a}{m} \right) = \left( 8, 4\sqrt{2} \right)$. Now equation of tangent will be, $x - 2\sqrt{2}y + 8 = 0$. And $Q$ will be foot of perpendicular from $(4, 0)$ on tangent $x - 2\sqrt{2}y + 8 = 0$, so, using foot of perpendicular formula we get, $$\frac{x-4}{1} = \frac{y-0}{-2\sqrt{2}} = \frac{-4-0+8}{1^2 + \left( 2\sqrt{2} \right)^2}$$ $$\Rightarrow (x, y) = \left( \frac{8}{3}, \frac{8\sqrt{2}}{3} \right)$$ Hence, by distance formula we get, $$(PQ)^2 = \left( \frac{16}{3} \right)^2 + \left( \frac{4\sqrt{2}}{3} \right)^2$$ $$\Rightarrow (PQ)^2 = \frac{256 + 32}{9} = 32$$

Given, the number 1, 2, 3, $\ldots$, 7. Now total numbers of 7 digit number will be $7!$ without repetition. Now total numbers which contain string 154 will be $5!$. Total numbers which contain string 2367 will be $4!$. And number which contain both string 2367 $\&$ 154 will be 2. So, total number which does not contain string 154 or 2367 will be $7! - (5! + 4! - 2)$. $$= 5040 - (120 + 24 - 2)$$ $$= 5040 - 142 = 4898$$ Hence this is the required answer.

Given equation: $(2a)^{\ln(a)} = (bc)^{\ln(b)}$ $$\Rightarrow \ln(a) \cdot (\ln(2a)) = \ln(b) \cdot \ln(bc)$$ $$\Rightarrow \ln(a) \cdot (\ln(2) + \ln(a)) = \ln(b) \cdot (\ln(b) + \ln(c))$$ Let $\ln(a) = x$, $\ln(b) = y$, $\ln(c) = z$, $x \neq y \neq z$ $$\Rightarrow x(\ln(2) + x) = y(y + z)$$ $$\Rightarrow x \ln(2) = y^2 - x^2 + yz \ldots (i)$$ Similarly, from second equation $$\ln(2) \cdot \ln(b) = \ln(c) \cdot \ln(a)$$ $$\Rightarrow \ln(2) = \frac{zx}{y} \ldots (ii)$$ Substitute eq(ii) in eq(i), We get, $$\Rightarrow y^3 - yz^2 + y^2z = x^2z$$ $$\Rightarrow y^2(y + z) - x^2(y + z) = 0$$ $$\Rightarrow (y^2 - x^2)(y + z) = 0$$ $$\Rightarrow (x - y)(x + y)(y + z) = 0$$ Therefore, $x \neq y \Rightarrow (x + y)(y + z) = 0$ Now $x = -y$, $y = -z$ and $x = z$ But $\ln(2) = \frac{zx}{y}$ $$\Rightarrow \ln(2) = \frac{x \left( \frac{x}{-x} \right)}{-x}$$ $$\Rightarrow \ln(2) = -\ln(a)$$ $$= -\ln 2 \Rightarrow a = \frac{1}{2}$$ And $$y = -z$$ $$\Rightarrow \ln(b) = -\ln(c)$$ $$\Rightarrow bc = 1$$ Now $6a + 5bc = 6 \left( \frac{1}{2} \right) + 5(1) = 8$ Hence this is the required answer.

Let, the parabola be $y = ax^2 + bx + c$. Now given parabola is passing through $(-1, 0)$, $(0, 1)$ and $(1, 0)$. Now putting the value in the above equation of parabola we get, $$a - b + c = 0 \ldots (1), \ c = 1 \ldots (2) \ and \ a + b + c = 0 \ldots (3)$$ Now solving above three equations we get, Parabola as $y = 1 - x^2$. Now area of region between $y = 1 - x^2$ and $(x + 1)^2 + (y - 1)^2 \leq 1$ is given by, Now from above diagram, required area will be, $$A = \int_{-1}^{0} \left( -x^2 + \sqrt{1 - (x + 1)^2} \right) \, dx$$ $$\Rightarrow A = \frac{\pi}{4} - \frac{1}{3}$$ Hence, $12(\pi - 4A) = 16$

Given that the mean for the frequency distribution is 28. We know that Mean $\left( \bar{x} \right) = \frac{\sum f_i x_i}{\sum f_i}$. Now for the class interval 0 – 10 we get $x_1 = \frac{0+10}{2} = 5$. Similarly we can calculate $x_i$ for other class intervals. Now we know that mean $= 28$ $$\Rightarrow \frac{2 \times 5 + 3 \times 15 + x \times 25 + 5 \times 35 + 4 \times 45}{14 + x} = 28$$ $$\Rightarrow x = 6$$ Also we know that variance \[ \sigma^2=\frac{\sum f_i(\bar{x}-x_i)^2}{\sum f_i} \] \[ \sigma^2= \frac{ 2\times(28-5)^2 +3\times(28-15)^2 +6\times(28-25)^2 +5\times(28-35)^2 +4\times(28-45)^2 }{20} \] \[ \sigma^2= \frac{ 2\times(23)^2 +3\times(13)^2 +6\times(3)^2 +5\times(7)^2 +4\times(17)^2 }{20} \] \[ \sigma^2=151 \] Hence this is the required answer.

Given, some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840. Now let total number of persons be $2n$, so number of couples will be $n$. Now number of matches played, so that no couple played in a match is given by, $$\binom{n}{2} \cdot \binom{n-2}{2} \cdot 2 = 840$$ {as if we choose couple as $\{$A, B$\}$ and other couple as $\{$C, D$\}$ and taking other two couple as $\{$x, y$\}$ $\&$ $\{$w, z$\}$, so the possible combination of the match where no couple played in a match will be, $\{$A, x$\}$ $\rightarrow$ $\{$C, w$\}$ $\&$ $\{$C, x$\}$ $\rightarrow$ $\{$A, w$\}$, hence there are two possible combination after choosing } $$n(n-1)(n-2)(n-3) = 5 \cdot 6 \cdot 7 \cdot 8$$ So, $n = 8$. Hence, the number of persons will be, $2n = 16$

Given, $$|n^2 - 10n + 19| 0 and n^2 - 10n + 13 0 and \left(n - \left(5 - 2\sqrt{3}\right)\right)\left(n - \left(5 + 2\sqrt{3}\right)\right) < 0$$ $$\Rightarrow n \in \mathbb{Z} - \{5\} and n \in \{2, 3, 4, 5, 6, 7, 8\}$$ Hence, $n = 2, 3, 4, 6, 7, 8$.

The given quantity is $P = \frac{a^2 b^3}{c \sqrt{d}}$. The percentage error for a quantity can be given by the formula $\frac{\Delta x}{x} \times 100$. Relation of the relative errors will be, $$\left( \frac{\Delta P}{P} \right) = \left[ \left( \frac{2 \Delta a}{a} \right) + \left( \frac{3 \Delta b}{b} \right) + \left( \frac{\Delta c}{c} \right) + \left( \frac{1}{2} \frac{\Delta d}{d} \right) \right]$$ Hence, the percentage error is calculated by substituting the values, $$\left( \frac{\Delta P}{P} \right) \times 100 = (2 \times 1) + (3 \times 2) + 3 + \left( \frac{1}{2} \times 4 \right) = 13\%.$$

The acceleration due to gravity with depth is $$g' = g \left( 1 - \frac{d}{R} \right)$$ The given data is $$W = 200 \, \mathrm{N}$$ $$d = \frac{R}{2}$$ Let the weight be at the given depth, $W'$. Hence, the value is $$mg' = mg \left( 1 - \frac{R}{2R} \right) = m \left( \frac{g}{2} \right)$$ The Weight $W' = \frac{W}{2} = \frac{200}{2} = 100 \, \mathrm{N}$$

Power rating of zener diode = 1.6 W. From the diagram above, $V_A - V_B = 8 \, \mathrm{V}$. Power is given by $$P = Vi$$ $$\Rightarrow i = \frac{1.6 \, \mathrm{W}}{8 \, \mathrm{V}} = 0.2 \, \mathrm{A}$$ Also, $V_C - V_A = 10 \, \mathrm{V} - 8 \, \mathrm{V} = 2 \, \mathrm{V}$. Thus, using Ohm's law, $$R_s = \frac{2 \, \mathrm{V}}{0.2 \, \mathrm{A}} = 10 \, \Omega$$

The data given is $\theta_1 = 15^\circ$, $\theta_2 = 45^\circ$, $R_1 = 50 \, \mathrm{m}$. The formula for range of a projectile is given by $$R = \frac{u^2 \sin 2\theta}{g} \cdots (i)$$ Substituting the values in equation (i) $$R_1 = \frac{u^2 \sin 2\theta_1}{g}$$ $$\Rightarrow 50 = \frac{u^2 \sin 30^\circ}{g}$$ $$\Rightarrow \frac{u^2}{g} = 100$$ Let the new range be $R_2$. The magnitude of the range is $$R_2 = \frac{u^2 \sin 2\theta_2}{g}$$ $$\Rightarrow R_2 = \left( \frac{u^2}{g} \right) \sin 90^\circ$$ $$\Rightarrow R_2 = 100 \, \mathrm{m}$$

The given data is $A_c = 15 \, \mathrm{V}$, $A_m = 3 \, \mathrm{V}$. Thus, the maximum amplitude and minimum amplitude is $$A_{max} = A_c + A_m$$ $$A_{min} = A_c - A_m$$ Hence, the ratio is $$\frac{A_{max}}{A_{min}} = \frac{15 + 3}{15 - 3} = \frac{18}{12} = \left( \frac{3}{2} \right)$$

Bohr's formula for angular momentum of an electron in an orbit is given by $$L = \frac{nh}{2\pi}$$ For the first orbit, $n = 1$ and the angular momentum is $L_1 = \frac{h}{2\pi} = L$ For 2nd orbit $$L_2 = \frac{2h}{2\pi} = 2L$$ Hence, the change in angular momentum is $$\Delta L = 2L - L = L$$

The law of conservation of momentum is given as $$m_1 u + m_2 u' = (m_1 + m_2) v'$$ The given data is $u = v$ $u' = 0 \, \mathrm{m \, s^{-1}}$ $m_1 = m$ $m_2 = 2m$ Using the conservation of momentum equation, $$mv + 2m \left( 0 \right) = \left( m + 2m \right) v'$$ $$\Rightarrow v' = \frac{mv}{3m} = \frac{v}{3}$$

For a moving coil galvanometer, NiAB = kθ. The current sensitivity is given by S$_i$ = NAB/k. So, if N is doubled, the current sensitivity is also doubled. The voltage is given by V = (BNA/Rk)θ. The voltage sensitivity is given by S$_v$ = NAB/(kR). As N is increased, R also increases. So, the voltage sensitivity does not change as N/R remains constant.

The formula for degree of freedom is given by $$f = 3N - k$$ where $N$ is the number of particles, $k$ is the individual relationship in the system. For a monoatomic gas particle, $N = 1$, $k = 0$. So, $f = 3$. Hence, monoatomic gas has 3 translational degree of freedom. For a rigid diatomic gas molecule, there are 5 degrees of freedom, 3 translational and 2 rotational. A diatomic gas can have two extra degrees of freedom due to rotation along two independent axes. A non rigid diatomic molecule has one extra degree of freedom than a rigid one. This is the vibrational degree of freedom. Hence, it has 6 degrees of freedom. So, a non rigid diatomic gas has 3 translational, 2 rotational and 1 vibrational degrees of freedom. A polyatomic gas has more than two atoms per molecule. Hence, the total number of degrees of freedom is 3 translational, 3 rotational and more than one vibrational degree of freedom.

The resistance in $ab$ is parallel to all other resistances. Since all the resistances other than $R_b$ is $4 \, \Omega$ the potential is zero ($V_x = V_y$) as the upper branch will become a case of balanced Wheatstone bridge between $V_x$ and $V_y$. So, the simplified circuit looks like the one below. Hence, the equivalent resistance is $$\frac{1}{R_{eq}} = \frac{1}{8} + \frac{1}{8} + \frac{1}{16} = \frac{5}{16}$$ $$\Rightarrow R_{eq} = \frac{16}{5} \, \Omega = 3.2 \, \Omega$$

In an isothermal process the temperature is constant. $PV = nRT$ So, it can be written that $$PV = P_A \left( \frac{V}{8} \right)$$ $$\Rightarrow P_A = 8P$$ In an adiabatic process, $PV^\gamma = constant$ $\gamma$ for monoatomic gas is $\frac{5}{3}$ So, it can be written $$\frac{P_B}{P} = \left( \frac{V}{\frac{V}{8}} \right)^{\frac{5}{3}}$$ $$\Rightarrow P_B = 32P$$ Thus, the ratio of the pressures is $$\frac{P_B}{P_A} = \frac{32P}{8P} = 4$$

The formula for impedance is given by $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ At resonance, $X_L = X_C$. So, $Z_{\min} = R$ Power will be maximum when impedance is minimum. Thus the maximum power occurs in a purely resistive circuit.

The orbital velocity is given $$V = \sqrt{\left( \frac{GM}{r} \right)}$$ where $M$ is the mass of earth and $r$ is the radius of the orbit of satellite. The velocity of the satellite with mass $m$ is $$V_1 = \sqrt{\frac{GM}{r}}$$ and the velocity of the satellite with mass $3m$ is $$V_2 = \sqrt{\frac{GM}{3r}}$$ Hence, the ratio of the velocities is $$\frac{V_1}{V_2} = \sqrt{\frac{3r}{r}} = \frac{\sqrt{3}}{1}$$

Pressure in a static fluid will be same at each point on same horizontal level. $$P = P_0 + \rho gh$$ where $P_0$ is atmospheric pressure. Pascal's law states that pressure applied to enclosed water is transmitted equally in all directions. Hence, both the statements are true and this is the right option.

The potential difference across the branches $EF$ and $GH$ is zero. So, the equivalent capacitance will be only due to the combination of two capacitors in parallel. $$C_{eq} = C + C$$ $$\Rightarrow C_{eq} = 2C$$

The formula to calculate the energy confined within the electric component of an electromagnetic wave can be written as $u_E = \frac{1}{2} \varepsilon_0 E^2 dV \ldots (1)$ Let's assume that the equation of the electric component is given by $E = E_0 \sin(\omega t + \phi) \ldots (2)$ Substitute the expression for the electric field component from equation (2) into equation (1) and simplify to obtain the expression for the energy confined within the small volume $dV$. $$u_E = \frac{1}{2} \varepsilon_0 |E_0 \sin(\omega t + \phi)|^2 dV$$ $$= \frac{1}{2} \varepsilon_0 E_0^2 \sin^2(\omega t + \phi) dV$$ $$= \frac{1}{4} \varepsilon_0 E_0^2 [1 - \cos(2\omega t + 2\phi)] \ldots (3)$$ From equation (3), it can be concluded that the frequency of oscillation of the energy confined within a small volume is double the frequency of the wave.

When an object is placed in front of a plane mirror, object distance is equal to the image distance. In the first case, the image distance will be $12 \, \mathrm{cm}$ from the mirror. In the next case, as the mirror is moved towards the object, the object distance becomes $8 \, \mathrm{cm}$. Hence, image distance will also be $8 \, \mathrm{cm}$.

The root mean squared velocity of a gas is given by $$v = \sqrt{\frac{3RT}{M}}$$ Thus, $v \propto \sqrt{T}$. Let $$T_1 = 300 \, \mathrm{K}$$ $$T_2 = 600 \, \mathrm{K}$$ Taking velocity ratios at the given temperatures, $$\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{600}} = \sqrt{\frac{1}{2}}$$ The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$. So, $$\lambda \propto \frac{1}{v}$$ The ratio of the wavelengths is $$\frac{\lambda_1}{\lambda_2} = \left(\frac{v_2}{v_1}\right) = \frac{\sqrt{2}}{1}$$ Thus, $$\lambda_2 = \left(\frac{1}{\sqrt{2}} \lambda_1\right)$$

Given, the instantaneous equation of motion of the particle is $$x = A \sin(\omega t + \delta) \ldots (1)$$ As the particle is moving along positive x-axis, $v = \frac{dx}{dt} > 0$. Substitute 0 for $t$ and $\frac{A}{2}$ for $x$ into equation (1) and solve to calculate the required phase difference. $$\frac{A}{2} = A \sin(\omega \times 0 + \delta)$$ $$\Rightarrow \sin \delta = \frac{1}{2}$$ $$= \sin \frac{\pi}{6}$$ $$\Rightarrow \delta = \frac{\pi}{6}$$

From the position time graph the velocity is found out by the slope of the line. The slope is given by $$\frac{x_2 - x_1}{t_2 - t_1}$$. The slope of $B$ is greater than $A$. Hence, the velocity of $B$ is greater than $A$. As can be seen from the graph $X_A V_A$. Hence, this is the correct option.

The given data is $I_0 = 32 \, \mathrm{W \, m^{-2}}$ and $I_f = 3 \, \mathrm{W \, m^{-2}}$. Thus, by substituting the value, we get $$I_f = \left( \frac{I_0}{2} \cos^2 \theta \right) \times \cos^2 (90^\circ - \theta) \cdots (i)$$ $$\Rightarrow 3 = 16 (\cos^2 \theta \cdot \sin^2 \theta)$$ $$\Rightarrow \cos \theta \cdot \sin \theta = \frac{\sqrt{3}}{4}$$ $$\Rightarrow \sin 2\theta = \frac{\sqrt{3}}{2}$$ $$\Rightarrow \theta = 30^\circ$$

The data given is $r = 15 \, \mathrm{cm}$ $m = 1 \, \mathrm{kg}$ $v = 22 \, \mathrm{m} \, \mathrm{s}^{-1}$ Total work done by the tube on the block is equal to loss in mechanical energy, $$w = \frac{1}{2} mv^2 + mgh$$ $$\Rightarrow w = \frac{1}{2} \times 1 \times (22)^2 + 1 \times 10 \times 0.3$$ $$\Rightarrow w = 245 \, \mathrm{J}$$

Let $L$ be constant; $L = I \omega$. From the conservation of angular momentum, $$\frac{2}{5} MR^2 \omega_1 = \frac{2}{5} M \left( \frac{R}{4} \right)^2 \omega_2$$ $$\Rightarrow MR^2 \omega_1 = \frac{MR^2 \omega_2}{16}$$ $$\Rightarrow \frac{\omega_1}{\omega_2} = \frac{1}{16}$$ Using $\omega = \frac{2\pi}{T}$ for each case along with the respective angular frequencies, $$\frac{T_2}{T_1} = \frac{1}{16}$$ $$\Rightarrow \frac{T_1}{T_2} = \frac{16}{1}$$ Using $T_1 = 24 \, \mathrm{h}$, $x = 16$ by comparing with the given time period in the question.

The magnetic intensity is $H = \frac{B}{\mu_0} - M$. For $M = 0$, it can be written $H = \left( \frac{B}{\mu_0} \right) = ni$. The data given is $H = 2.4 \times 10^3 \, \mathrm{A \, m^{-1}}$, $l = 15 \times 10^{-2} \, \mathrm{m}$, $N = 60$. Using the relation, $n = \frac{N}{l}$ and $H = ni$, the value of the current is, $$i = \frac{H}{n} = \frac{2.4 \times 10^3 \times 15 \times 10^{-2}}{60} = 6 \, \mathrm{A}$$

The force on the rod is given by $F = i l B$. Therefore, $$F = \frac{B v l}{R} \times l B$$ $$\Rightarrow F = \frac{B^2 l^2 v}{R} \cdots (i)$$ The data given is $R = 5 \, \Omega$, $l = 1 \, \mathrm{m}$, $v = 4 \, \mathrm{m} \, \mathrm{s}^{-1}$, $B = 0.15 \, \mathrm{T}$. Substituting the values in equation (i), $$F = \frac{(0.15)^2 \times (1)^2 \times 4}{5}$$ $$= 18 \times 10^{-3} \, \mathrm{N}$$

The general equation for a wave can be written as $y = A \sin(\omega t + kx)$. Comparing it with the given equation, $y = 5 \sin(6t + 0.003x)$, $$\omega = 6 \, \mathrm{rad} \, \mathrm{s}^{-1}, k = 0.003 \, \mathrm{cm}^{-1}$$ Using the equation, $v = \frac{\omega}{k}$ $$v = \frac{6}{0.003 \times 10^2} = 20 \, \mathrm{m} \, \mathrm{s}^{-1}$$

It is given that $\lambda = 1.5 \times 10^{-5} \, \mathrm{s}^{-1}$. Number of moles is $\frac{1 \times 10^{-6}}{60} = \frac{10^{-7}}{6}$. Number of atoms = number of moles $\times \, N_A$. $$N_0 = \frac{10^{-7}}{6} \times 6 \times 10^{23} = 10^{16}$$ Hence, at $t = 0$, $A = A_0 = N_0 \lambda$, the activity can be given by $$A = N_0 \lambda e^{-\lambda t}$$ $$\Rightarrow A = 10^{16} \times 1.5 \times 10^{-5} \times 1 = 15 \times 10^{10} \, \mathrm{Bq}$$

The charges on the spheres are $$q_X = \sigma 4 \pi a^2$$ $$q_Y = -\sigma 4 \pi b^2$$ $$q_Z = \sigma 4 \pi c^2$$ It is given that $V_X = V_Z$. So, $$\frac{q_X}{4 \pi \varepsilon_0 a} + \frac{q_Y}{4 \pi \varepsilon_0 b} + \frac{q_Z}{4 \pi \varepsilon_0 c} = \frac{q_X}{4 \pi \varepsilon_0 c} + \frac{q_Y}{4 \pi \varepsilon_0 c} + \frac{q_Z}{4 \pi \varepsilon_0 c}$$ $$\Rightarrow \frac{\sigma 4 \pi a^2}{a} - \frac{\sigma 4 \pi b^2}{b} + \frac{\sigma 4 \pi c^2}{c} = \frac{\sigma 4 \pi (a^2 - b^2 + c^2)}{c}$$ $$\Rightarrow c \left( a - b + c \right) = \left( a^2 - b^2 + c^2 \right)$$ $$\Rightarrow c = a + b = 2 \, \mathrm{cm} + 3 \, \mathrm{cm} = 5 \, \mathrm{cm}$$

When all the resistors are connected in series, the total resistance is $R_{max} = 10R$. Therefore, $R_{max} = 10 \times 10 = 100 \, \Omega$. When all the resistors are connected in parallel, the total resistance is $R_{min} = \frac{R}{10}$. Therefore, $R_{min} = \frac{10}{10} = 1 \, \Omega$. Hence, the ratio, $$\frac{R_{max}}{R_{min}} = \frac{100 \, \Omega}{1 \, \Omega} = 100.$$

The force acting on steel wire $F$ is $(2 + 1.14)g = 3.14g$. From the relation of Young's modulus, it can be written $$\frac{F}{A} = Y \left( \frac{\Delta L}{L} \right)$$ $$\Rightarrow \Delta L = \left( \frac{FL}{AY} \right) = \frac{3.14 \times g \times 1.6}{\pi \times (0.2)^2 \times 2 \times 10^{11} \times 10^{-4}}$$ $$\Rightarrow \Delta L = 2 \times 10^{-5}$$ $$\Rightarrow \Delta L = 20 \times 10^{-6} \, \mathrm{m}$$

In chromatography, the retardation factor (Rf) is the fraction of an analyte in the mobile phase of a chromatographic system. In planar chromatography in particular, the retardation factor Rf is defined as the ratio of the distance travelled by the centre of a spot to the distance travelled by the solvent front. The more polar spot travels slower, and the less polar spot travels faster. Rf values, on the other hand, are directly related to the rate of movement. The fastest moving spot has the highest Rf value, least polar (fastest moving), and the spot with the lowest Rf value is the most polar (slowest moving). Hence, the compound B is strongly adsorbed and has low Rf value.

The formula for ferrous ammonium sulphate is $\mathrm{FeSO_4} \cdot (\mathrm{NH_4})_2 \mathrm{SO_4} \cdot 6\mathrm{H_2O}$. It is prepared by dissolving equimolar mixture of hydrated ferrous sulphate and ammonium sulphate in water containing a little sulphuric acid. The solution is subjected to crystallisation, ferrous ammonium sulphate separates out from the solution. The chemical reaction is given below. $$\mathrm{FeSO_4} + (\mathrm{NH_4})_2 \mathrm{SO_4} + 6\mathrm{H_2O} \rightarrow \mathrm{FeSO_4} \cdot (\mathrm{NH_4})_2 \mathrm{SO_4} \cdot 6\mathrm{H_2O} (Mohr's salt)$$ Avoid prolonged heating while preparing crystals of ferrous ammonium sulphate, as it may oxidise ferrous ions to ferric ions and change the stoichiometry of the crystals.

$\begin{array}{c} \text{Limited water} \\ \mathrm{CaO \rightarrow Ca(OH)_2} \\ \text{Lime} \end{array}$ The process is known as slaking of lime and the product is slaked lime. It is an exothermic process. $\mathrm{Ca(OH)_2}$ aq solution is known as lime water and used for test of $\mathrm{CO_2}$. $\mathrm{Ca(OH)_2(aq) + CO_2 \rightarrow CaCO_3 + H_2O}$ $\mathrm{CaCO_3 + CO_2\ (excess) + H_2O \rightarrow Ca(HCO_3)_2}$ Milkiness is due to the formation of calcium carbonate and disappearance of milkiness is due to the formation of calcium bicarbonate.

The dipole moment is the vector sum of all the bond moments. The $\mathrm{CH_2Cl_2}$ and $\mathrm{CHCl_3}$ both have net non-zero dipole moment because the bond moments are not cancelled each other due to their tetrahedral structure. The 1, 4-dichlorobenzene, trans-butene, benzene have zero dipole moment due to cancellation of bond moments. The 1, 3-dichlorobenzene, cis-butene, anisidine has non-zero dipole moment.

The various solid wastes in the form of slags and sludges that are emerged from steel plants are blast furnace slag. Fly ash is a fine by-product recovered from gases of burning coal in thermal power plants. Gypsum is the waste of fertiliser industry. Non-cellulosic part of plant is wastes of paper industry and it is biodegradable.

The isomers of $\mathrm{C_8H_{11}N}$ that can be prepared by Gabriel phthalimide synthesis are primary amines. These can be prepared by Gabriel phthalimide synthesis because $\mathrm{S_N2}$ is possible in the structure shown. Secondary amines react with Hinsberg's reagent to give solid insoluble in $\mathrm{NaOH}$. Hence, $Q$ is the secondary amine as shown. Aromatic amines like aniline give diazonium salts. The diazonium salt used in diazo coupling to get red dye with beta Naphthol in sodium hydroxide. Hence, $R$ is aromatic amine as shown below.

$\mathrm{K_2Cr_2O_7}$ is generally preferred over $\mathrm{Na_2Cr_2O_7}$ in volumetric analysis because $\mathrm{Na_2Cr_2O_7}$ is hygroscopic in nature. Therefore, it is difficult to prepare its standard solution accurately, as precise weighing is not possible under normal atmospheric conditions. A hygroscopic substance absorbs moisture from the atmosphere, which leads to inaccuracies in its measured mass.

Secondary structure of protein refers to local folded structures that form within a polypeptide due to interactions between atoms of the backbone. They are found to exist in two different types of structures $\alpha$ – helix and $\beta$ – pleated sheet structures. This structure arises due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between $- \mathrm{CO}$ group and $- \mathrm{NH}$ groups of the peptide bond. Tertiary Structure of Protein arises from further folding of the secondary structure of the protein. H-bonds, electrostatic forces, disulphide linkages, and Vander Waals forces stabilise this structure. The tertiary structure of proteins represents overall folding of the polypeptide chains, further folding of the secondary structure.

The reaction $\mathrm{C(s) + H_2O(g)} \xrightarrow{1270 \, \mathrm{K} \, (T_1)} \mathrm{CO(g) + H_2(g)}$ is known as coal gasification reaction. The mixture produced in this reaction is called water gas. The reaction $\mathrm{CO(g)+H_2O(g)}\xrightarrow[\mathrm{Catalyst}]{673\,\mathrm{K}\,(T_2)}\mathrm{CO_2(g)+H_2(g)}$ is known as water gas shift reaction. Using water gas shift reaction we can increase the concentration of hydrogen gas. Hence, $T_1 > T_2$.

The valence electronic configuration of $\mathrm{Mn^{3+}}$ is $3d^4$ and that of $\mathrm{Mn^{2+}}$ is stable $3d^5$. The valence electronic configuration of $\mathrm{Fe^{3+}}$ is $3d^5$ and that of $\mathrm{Fe^{2+}}$ is $3d^6$. The reduction electrode potential of $\mathrm{Mn^{3+}/Mn^{2+}}$ is $+1.57 \, \mathrm{V}$ while that of $\mathrm{Fe^{3+}/Fe^{2+}}$ is $+0.77 \, \mathrm{V}$, hence A is wrong. Higher oxidation state of smaller d-block elements is stabilised (or say form compounds) with smaller anion oxide that can be explained by stearic reason hence B is correct. The oxidation electrode potential of $\mathrm{Cr^{2+}/Cr^{3+}}$ is $+0.41 \, \mathrm{V}$ hence, it can reduce $\mathrm{H^+}$ and so liberate $\mathrm{H_2}$. The unpaired electrons in $\mathrm{V^{2+}}$ are 3 hence, the magnetic moment of $\mathrm{V^{2+}}$ will be lesser than $4.4 \, \mathrm{BM}$. Hence, only B and C are correct.

Froth flotation is a process for selectively separating hydrophobic materials from hydrophilic. The principle of froth flotation process is that sulphide ores are preferentially wetted by the pine oil, whereas the gangue particles are wetted by the water. Cresols and aniline are used as froth stabilisers.

In case of strong field ligand containing complexes, electrons get pair up against the Hund's rule to get required number of hybrid orbitals. Among the given complexes' ammonia is a strong-field ligand and other ligands are weak field ligands. $[\mathrm{Co(NH_3)_6}]^{3+}$ has $d^2 sp^3$ (inner orbital complex) with zero unpaired electrons. Hence, it is octahedral with diamagnetic character and low spin complex. $[\mathrm{CoF_6}]^{3-}$ and $[\mathrm{CoCl_6}]^{3-}$ are octahedral but having unpaired electrons. $[\mathrm{NiCl_4}]^{2-}$ is not octahedral.

(A) $2 \mathrm{CO}(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}_2(g) \Delta H_1^{\circ} = -x \ \mathrm{kJ} \ \mathrm{mol}^{-1}$ (B) $\mathrm{C_{graphite}} + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) \Delta H_2^{\circ} = -y \ \mathrm{kJ} \ \mathrm{mol}^{-1}$ If the equation is multiplied the value $n$, then the enthalpy change value also multiplied by factor $n$. If the enthalpy is positive in one direction, it is negative in other direction. Multiply equation B by 2 and subtract equation (A) from it $$2 \mathrm{C_{graphite}} + \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}(g) \Delta H^{\circ} = x - 2y$$ $$\mathrm{C_{graphite}} + \frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{CO}; \Delta H^{\circ} = \frac{x - 2y}{2}$$

Oxide and peroxides of sodium are stable ($\mathrm{Na_2O}$ and $\mathrm{Na_2O_2}$). $\mathrm{K}$, $\mathrm{Rb}$, $\mathrm{Cs}$ form superoxides. Beryllium forms a hydride with molecular formula $\mathrm{BeH_2}$. Organic lead (tetraethyl lead; TEL) is used as an antiknock agent in gasoline and jet fuels. The compound $\mathrm{(NH_4)_2BeF_4}$ on thermal decomposition produces $\mathrm{BeF_2}$ and $\mathrm{NH_4F}$.

Thermoset plastics retain their form and stay solid under heat once cured. Urea-formaldehyde is an example of thermosetting polymer. Polymers which disintegrate by themselves over a period of time due to environmental degradation by bacteria etc. are called biodegradable polymers. Nylon-2-nylon-6 is an example of biodegradable polymer. Any artificial elastomer is referred to as synthetic rubber. Buna-N is an example of synthetic rubber. Polyester is a synthetic or man-made fiber material. Dacron is an example polyester.

One mole of any ideal gas occupies 22.7 L at standard temperature and pressure. The number of moles is given by the formula: $$Moles = \frac{Given volume}{Volume at STP}$$ $$Moles = \frac{2.8375}{22.7} = 0.125$$ The number of molecules is given by: $$Molecules = moles \times N_A$$ $$Molecules = 0.125 \times 6.022 \times 10^{23}$$ $$= 7.527 \times 10^{22}$$

Adsorption is an exothermic process due to a decrease in surface energy. Micelle formation is endothermic. For any spontaneous process, entropy should always increase, hence, $\Delta S > 0$ for micelle formation. Micelle formation decreases the stability of the colloidal solution, so the energy of the mixture should increase, which means $\Delta H > 0$. The adsorption process $\Delta H$ is negative. For micelle formation, $\Delta H$ is positive and $\Delta S$ is positive.

Alkaline potassium permanganate is a strong oxidising agent. In the given molecule alkene part undergo oxidation with alk. $KMnO_4$. The alkene part present in the alkene, hence it converts to carboxylic acid group and carbon dioxide. Finally, ester part in the molecule undergo hydrolysis in acidic water.

The given methyl phenyl ether can be prepared from Williamson Ether Synthesis. Williamson Ether Synthesis usually takes place as an $S_N2$ reaction of a Alkyl halide with an alkoxide ion. Aryl halides cannot be used in this reaction. The reaction is given below. $$\chemfig{**6(------)-ONa} + \chemfig{CH_3Br} \rightarrow \chemfig{**6(------)-OCH_3} + NaBr$$

The rate of $S_N2$ reaction decreases with increase in steric crowding. In the given set of molecules the second molecule (Tertiary alkyl halide) is sterically crowded, hence, it is correct option. More stable carbocation formed, higher will be the reactivity of $S_N1$. Benzyl carbocation is more reactive than aliphatic carbocation. The deactivating groups decreases the rate of electrophilic substitution reaction. $NO_2$ is strong deactivating group for electrophilic substitution. In the aromatic nucleophilic substitution reactions, the intermediate formed is carbanion and, it is stabilised by $NO_2$ at ortho or para positions. Hence A, B, C, D all are correct.

The equilibrium is possible when there is no exchange of matter between system and surroundings. At equilibrium state the forward and backward processes occur with same rate (speed). When a system reaches equilibrium at a given temperature, the forward and reverse reactions are occurring at the same rate, and the concentrations (or partial pressures) of the reactants and products no longer change with time. At this point, the system has achieved a steady state, where the rate of the forward reaction is equal to the rate of the reverse reaction. So when equilibrium is attained, the value of all its parameters became constant. The solubility is function of solubility product, the value of which is constant at a given temperature.

For $\mathrm{N_3^-}$ (Azide): $\mathrm{N} \equiv \mathrm{N} = \mathrm{N} \leftrightarrow \mathrm{N} = \mathrm{N} \equiv \mathrm{N}$. It has $sp$ hybridised central atom. Hence it is linear in shape. For $\mathrm{I_3^-}$ (triiodide): It has linear geometry, $sp^3d$ hybridisation with three lone pairs at central atom. For $\mathrm{NO_2^-}$ (nitrite): It is (nonlinear) bent shaped as it has $sp^2$ hybridisation with one lone pair at central atom. For $\mathrm{O_3}$ (ozone): It is bent in shape with $sp^2$ hybridisation and one lone pair at central atom. For $\mathrm{SO_2}$ (sulphur dioxide): It is bent in shape with $sp^2$ hybridisation and one lone pair at central atom. So 3 among the given molecules is bent in shape.

For parallel reaction When adding the rate constants of the two reactions, we need to use the formula for combining rate constants of two reactions in parallel, which is: $$k = k_1 + k_2$$ where $k_1$ and $k_2$ are the rate constants of the individual reactions. In this case, the half-lives of the two reactions are 12 min and 3 min, respectively. The rate constants of the two reactions can be calculated using the formula for half-life of a first-order reaction: $$\frac{1}{t_{1/2}} = \frac{1}{12} + \frac{1}{3} = \frac{5}{12}$$ Net $t_{1/2} = \frac{12}{5} = 2.4 min \approx 2 min$ Hence time taken for 50$\%$ consumption of reactant will be close to 2 min.

A black body is an idealised object that absorbs all the radiation that falls on it and emits radiation over a broad range of frequencies. Black bodies emit and absorb energy in the form of electromagnetic radiation. The frequency distribution of the emitted radiation from a black body depends on its temperature and is given by Planck's law. At a given temperature, the intensity vs frequency curve for a black body radiation passes through a maximum value. This frequency is known as the peak frequency or the frequency of maximum emission. At higher temperatures, the peak frequency of the black body radiation shifts towards higher frequencies (shorter wavelengths). This means that the maximum of the intensity vs frequency curve is at a higher frequency at higher temperature compared to that at lower temperature.

Let's first balance the two given chemical equations: $$\mathrm{Na_2O + H_2O \longrightarrow 2 \, NaOH(X)}$$ In this equation, there are two $\mathrm{NaOH}$ molecules, so the total number of oxygen atoms in the products is 1. $$\mathrm{Cl_2O_7 + H_2O \longrightarrow 2 \, HClO_4(Y)}$$ In this equation, there are two $\mathrm{HClO_4}$ molecules, so the total number of oxygen atoms in the products is 4. So X has one O and Y has four O.

The relation between $\Delta G^\circ$ and $E^\circ_{\text{cell}}$ is $$\Delta G^\circ = -n F E^\circ_{\text{cell}}$$ $$\mathrm{FeO_4^{2-} \xrightarrow[+2.2\,\mathrm{V}]{} Fe^{3+}} \quad \Delta G^\circ_1 = -6.6 \, \mathrm{F} \quad \text{(3 electrons are involved)}$$ $$\mathrm{Fe^{3+} \xrightarrow[+0.70\,\mathrm{V}]{} Fe^{2+}} \quad \Delta G^\circ_2 = -0.7 \, \mathrm{F} \quad \text{(one electron is involved)}$$ Hence, for $$\mathrm{FeO_4^{2-} \rightarrow Fe^{2+}} \quad \Delta G^\circ_3 = -7.3 \, \mathrm{F}$$ $$= -n F E^\circ_{\text{cell}} \quad \text{(Four electrons are involved)}$$ $$E^\circ_{\mathrm{FeO_4^{2-}/Fe^{+2}}} = \frac{-7.3 \, \mathrm{F}}{-4 \, \mathrm{F}} = 1.825, \quad n = 4$$ $$= 1825 \times 10^{-3} \, \mathrm{V}$$ $n =$ Electron exchange of that half cell reaction.

The dissociation of HA is given by: $$\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}$$ with degrees of dissociation $1-\alpha$, $\alpha$, and $\alpha$ respectively. Van't Hoff factor, $i = 1 + \alpha(n - 1)$ $$= 1 + 0.3(2 - 1)$$ $$= 1 + 0.3$$ $$= 1.3$$ $\Delta T_f$ without considering dissociation $= k_f \times molality$ $\Delta T_f$ without considering dissociation $= ik_f \times molality$ $$= \Delta T_f'$$ $$\Delta T_f' - \Delta T_f = 0.3 \left(k_f \times molality\right)$$ $$\frac{\Delta T_f' - \Delta T_f}{\Delta T_f} = \frac{0.3 \left(k_f \times molality\right)}{k_f \times molality} = 0.3$$ % value = 30

Potassium ferrocyanide, $\mathrm{K_4[Fe(CN)_6]}$, has a central iron ion that is coordinated with six cyanide ligands in an octahedral geometry. The oxidation state of the iron ion in potassium ferrocyanide is $+2$, hence has $d^2 sp^3$ hybridised state. On applying C.F.T. it shows that all the 6 electrons of $d$ subshell are present in the form of three pairs in $t_{2g}$ orbitals. Hence answer is 3.

From Boyle's Law, $P_2 V_2 = P_1 V_1$. $P_1 = 940.3 \, \mathrm{mm \, Hg}$. $P_2 =$ Pressure at which volume is reduced by 40%. $V_2 =$ Volume decreased by 40% ($V_2 = V_1 - 0.4 \, V_1 = 0.6 \, V_1$). Then, $P_2 = \frac{940.3 \times V_1}{0.6 V_1}$. $$= 1567.16$$ $$\approx 1567 \, \mathrm{mm \, Hg}$$

In $\mathrm{IF_5}$, the central atom is iodine, which has 7 valence electrons. Iodine in this molecule has 5 bonding pairs and 1 lone pair. Therefore, the number of lone pairs on the central atom of $\mathrm{IF_5}$ is 1. In $\mathrm{IF_7}$, the central atom is also iodine, which has 7 valence electrons. Iodine in this molecule has 7 bond pairs and 0 lone pairs. Therefore, the number of lone pairs on the central atom of $\mathrm{IF_7}$ is 0. Therefore, the sum of the number of lone pairs in the central atom of $\mathrm{IF_5}$ and $\mathrm{IF_7}$ is $1 + 0 = 1$.