JEE Main 10 April 2023 Shift 2 question paper with solutions

Given equation is $\int_0^{t^2} \left( f(x) + x^2 \right) dx = \frac{4}{3} t, \forall t > 0$. According to Newton Leibnitz theorem we have $$\frac{d}{dx} \left( \int_{u(x)}^{v(x)} f(t) dt \right) = f(v(x)) \times v'(x) - f(u(x)) \times u'(x).$$ Apply Newton's Leibnitz theorem in the given equation. $$\Rightarrow \left( f(t^2) + t^4 \right) 2t - 0 = 4t^2$$ $$\Rightarrow f(t^2) + t^4 = 2t$$ $$\Rightarrow f(x^2) = -x^4 + 2x$$ $$\Rightarrow f(x) = -x^2 + 2 \sqrt{x}$$ $$\Rightarrow f\left( \frac{\pi^2}{4} \right) = -\frac{\pi^4}{4} + 2 \times \frac{\pi}{2}$$ $$= -\frac{\pi^4}{16} + \pi$$ $$= \pi \left( 1 - \frac{\pi^3}{16} \right)$$ Hence this is the correct option.

Let us find the different ways so that 8 persons can travel in 3 cars. $$\begin{array}{|c|c|c|} \hline C_1 & C_2 & C_3 \\ \hline 3 & 3 & 2 \\ 2 & 3 & 3 \\ 3 & 2 & 3 \\ \hline \end{array}$$ Hence we have 3 ways. Now the number of ways to distribute 8 persons such that they can travel in 3 cars with any car carrying maximum of 3 persons is $$\left( \frac{8!}{3!3!2!} \right) \times 3 = 1680$$ Therefore, the required answer is 1680.

Given, $$\int \left( \left( \frac{x}{e} \right)^{2x} + \left( \frac{e}{x} \right)^{2x} \right) \ln x \, dx = \frac{1}{\alpha} \left( \frac{x}{e} \right)^{\beta x} - \frac{1}{\gamma} \left( \frac{e}{x} \right)^{\delta x} + C$$ Now let $$I = \int \left( \left( \frac{x}{e} \right)^{2x} + \left( \frac{e}{x} \right)^{2x} \right) \ln x \, dx$$ Now let $$\left( \frac{x}{e} \right)^{2x} = t$$ $$\Rightarrow 2x(\ln x - 1) = \ln t$$ $$\Rightarrow \ln x \, dx = \frac{1}{2t} \, dt$$ So, $$I = \frac{1}{2} \int \left( t + \frac{1}{t} \right) \frac{dt}{t}$$ $$\Rightarrow I = \frac{1}{2} \int \left( 1 + \frac{1}{t^2} \right) dt$$ $$\Rightarrow I = \frac{1}{2} \left( t - \frac{1}{t} \right) + C$$ $$\Rightarrow I = \frac{1}{2} \left( \left( \frac{x}{e} \right)^{2x} - \left( \frac{e}{x} \right)^{2x} \right) + C$$ Now on comparing with $$I = \frac{1}{\alpha} \left( \frac{x}{e} \right)^{\beta x} - \frac{1}{\gamma} \left( \frac{e}{x} \right)^{\delta x} + C$$, we get $$\alpha = 2, \beta = 2, \gamma = 2, \delta = 2$$ Hence, $$\alpha + 2\beta + 3\gamma - 4\delta = 2 + 4 + 6 - 8 = 4$$ Hence this is the required option.

Plane passes through the points A(1, 2, 0), B(1, 4, 1) and C(0, 5, 1). We know that the equation of a plane passing through three non-collinear points $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ is $$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0$$ $$\begin{vmatrix} x - 1 & y - 2 & z - 0 \\ 1 - 1 & 4 - 2 & 1 - 0 \\ 0 - 1 & 5 - 2 & 1 - 0 \end{vmatrix} = 0$$ $$\begin{vmatrix} x - 1 & y - 2 & z - 0 \\ 0 & 2 & 1 \\ -1 & 3 & 1 \end{vmatrix} = (x - 1)(2 - 3) + (y - 2)(0 + 1) + z(0 + 2) = 0$$ $$-x + 1 - y + 2 + 2z = 0$$ $$x + y - 2z = 3 \ldots (i)$$ Now image of P(1, 2, 6) in (i) $$\frac{x - 1}{1} = \frac{y - 2}{1} = \frac{z - 6}{-2} = \frac{-2(-9 - 3)}{6} = 4$$ $$x = 5, \; y = 6, \; z = -2$$ $$\alpha = 5, \; \beta = 6, \; \gamma = -2$$ $$\therefore \alpha^2 + \beta^2 + \gamma^2 = 65$$ Hence this is the required option.

Given, $A = \{2, 3, 4\}$, $B = \{8, 9, 12\}$. And $a_1 \in A$, $b_2 \in B$. Now finding elements where $a_1$ divides $b_2$, we get, $$(a_1, b_2) \in \{(2, 8), (2, 12), (3, 9), (3, 12), (4, 8), (4, 12)\}$$ Also given, $a_2 \in A$, $b_1 \in B$. Now finding elements when $a_2$ divides $b_1$, we get, $$(a_2, b_1) \in \{(2, 8), (2, 12), (3, 9), (3, 12), (4, 8), (4, 12)\}$$ Hence, Number of relations $= 6 \times 6 = 36$

Given, $$A = \frac{1}{5!6!7!} \begin{bmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}$$ $\($$\Rightarrow$ |A| = $\frac{1}{5!6!7!}$ $\begin{vmatrix}$ 5! & 6! & 7! $\\$ 6! & 7! & 8! $\\$ 7! & 8! & 9! $\end{vmatrix}$$\)$ $\($$\Rightarrow$ |A| = $\frac{1}{5!6!7!}$ $\begin{vmatrix}$ 1 & 6 & 42 $\\$ 1 & 7 & 56 $\\$ 1 & 8 & 72 $\end{vmatrix}$$\)$ $\($$\Rightarrow$ |A| = 2$\)$ Now using the property of adjoint of matrix we get, $$|adj(adj(A))| = |A|^{(n-1)^2} \& |kA| = k^n |A|,$$ where $\($n$\)$ is order of square matrix, Now using the above two formula we get, $$|adj(adj(2A))| = |2A|^{(2-1)^2} |A|^4$$ $\($$\Rightarrow$ |adj(adj(2A))| = $2^{12}$ $\cdot$ $2^4$\)$ $\($$\Rightarrow$ |adj(adj(2A))| = $2^{16}$\)

Given, let $A$ be the point $(1, 2)$ and $B$ be any point on the curve $x^2 + y^2 = 16$. And the centre of the locus of the point $P$, which divides the line segment $AB$ in the ratio $3 : 2$ is the point $C(\alpha, \beta)$. Now let the point on the circle $x^2 + y^2 = 16$ be $B(4 \cos \theta, 4 \sin \theta)$. Now using section formula in $A(1, 2)$ and $B(4 \cos \theta, 4 \sin \theta)$ we get, $$P \left( \frac{12 \cos \theta + 2}{5}, \frac{12 \sin \theta + 4}{5} \right) \equiv (h, k)$$ $$\Rightarrow \cos \theta = \frac{5h - 2}{12} \& \sin \theta = \frac{5k - 4}{12}$$ Now squaring and adding we get, $$\left( \frac{5h - 2}{12} \right)^2 + \left( \frac{5k - 4}{12} \right)^2 = 1$$ $$\Rightarrow \left( h - \frac{2}{5} \right)^2 + \left( k - \frac{4}{5} \right)^2 = \left( \frac{12}{5} \right)^2$$ So, the centre of the locus is $C \left( \frac{2}{5}, \frac{4}{5} \right)$. Hence, by distance formula we get, $$AC = \sqrt{ \left( \frac{3}{5} \right)^2 + \left( \frac{6}{5} \right)^2 } = \frac{3 \sqrt{5}}{5}$$

Given that $P(getting odd 7 times) = P(getting odd 9 times)$. This is the case of binomial distribution with $n$ successes and probability of success $= p = \frac{3}{6}$ and failure is $q = 1 - p = \frac{1}{2}$. We know that $P(X = k) = \binom{n}{k} (p)^k (q)^{n-k}$. Now $P(X = 7) = P(X = 9)$ $$\Rightarrow \binom{n}{7} \left( \frac{1}{2} \right)^7 \left( \frac{1}{2} \right)^{n-7} = \binom{n}{9} \left( \frac{1}{2} \right)^9 \left( \frac{1}{2} \right)^{n-9}$$ $$\Rightarrow \binom{n}{7} = \binom{n}{9}$$ $$\Rightarrow \binom{n}{n-7} = \binom{n}{9}$$ $$\Rightarrow n = 16$$ $$\Rightarrow P\left(2 times even\right) = \binom{16}{2} \left( \frac{1}{2} \right)^{14} \left( \frac{1}{2} \right)^2$$ $$= \frac{15 \times 4}{2^{15}} = \frac{60}{2^{15}} = \frac{k}{2^{15}}$$ $$\Rightarrow k = 60$$ Hence this is the required option.

Given, $g(x) = f(x) + f(1-x)$ and $f''(x) > 0$, $x \in (0, 1)$. If $g$ is decreasing in the interval $(0, \alpha)$ and increasing in the interval $(\alpha, 1)$, Now solving, $$g(x) = f(x) + f(1-x)$$ Now differentiating both sides we get, $$g'(x) = f'(x) - f'(1-x)$$ Differentiating again we get, $$g''(x) = f''(x) + f''(1-x) > 0$$ So, $g'(x)$ is increasing as given $f''(x) > 0$ $$\Rightarrow g'(0) < g'(1)$$ $$\Rightarrow f'(0) - f'(1) < f'(1) - f'(0)$$ $$\Rightarrow f'(0) < f'(1)$$ Now finding $g'(x) = 0$ we get, $$\Rightarrow f'(x) = f'(1-x)$$ $$\Rightarrow x = 1 - x$$ $$\Rightarrow x = \frac{1}{2}$$ Now $g'(x)$ is positive for $x \in \left(0, \frac{1}{2}\right)$ And $g'(x)$ is negative for $x \in \left(\frac{1}{2}, 1\right)$ $$\therefore \alpha = \frac{1}{2}$$ So, $$\tan^{-1} 2\alpha + \tan^{-1} \left(\frac{1}{\alpha}\right) + \tan^{-1} \left(\frac{\alpha + 1}{\alpha}\right)$$ $$= \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3$$ $$= \frac{\pi}{4} + \tan^{-1} \left(\frac{2 + 3}{1 - 6}\right)$$ $$= \frac{\pi}{4} + \tan^{-1} (-1)$$ $$= \frac{\pi}{4} + \frac{3\pi}{4} = \pi$$

Given, $$15x^2 + 19y^2 = 285$$ $\($$\Rightarrow$ $\frac{x^2}{19}$ + $\frac{y^2}{15}$ = 1$\)$ Now plotting the diagram with given circle of radius 4 we get, $$\frac{x^2}{19} + \frac{y^2}{15} = 1$$ Now, let the equation of tangent of ellipse be $\($y = mx $\pm$ $\sqrt{19m^2 + 15}$$\)$ $\{$as $\($c^2 = a^2m^2 + b^2$\)$$\}$ If it is tangent to circle $\($x^2 + y^2 = 16$\)$ then using the formula, $\($c^2 = r^2(1 + m^2)$\)$ we get, $$\frac{\sqrt{19m^2 + 15}}{\sqrt{1 + m^2}} = 4$$ $\($$\Rightarrow$ 19m^2 + 15 = 16 + 16m^2$\)$ $\($$\Rightarrow$ 3m^2 = 1$\)$ $\($$\Rightarrow$ m = $\pm$ $\frac{1}{\sqrt{3}}$$\)$ $\($$\Rightarrow$ $\theta$ = 30^$\circ$$\)$ with $\($x-axis$\)$, Hence, Angle made by tangent with minor axis i.e., with $\($y-axis$\)$ is $\($$\frac{\pi}{3}$$\)$

Given, $\($ $\vec{a}$ = 2$\hat{i}$ + 7$\hat{j}$ - $\hat{k}$, $\;$ $\vec{b}$ = 3$\hat{i}$ + 5$\hat{k}$ $\;$ and $\;$ $\vec{c}$ = $\hat{i}$ - $\hat{j}$ + 2$\hat{k}$ $\)$ And $\($ $\vec{d}$ $\)$ be a vector which is perpendicular to both $\($ $\vec{a}$ $\)$ and $\($ $\vec{b}$ $\)$, So, $\($ $\vec{d}$ = $\lambda$ $\left$( $\vec{a}$ $\times$ $\vec{b}$ $\right$) $\)$ Now finding, $\[$ $\vec{a}$ $\times$ $\vec{b}$ = $\begin{vmatrix}$ $\hat{i}$ & $\hat{j}$ & $\hat{k}$ $\\$ 2 & 7 & -1 $\\$ 3 & 0 & 5 $\end{vmatrix}$ $\]$ $\($ $\Rightarrow$ $\vec{a}$ $\times$ $\vec{b}$ = 35$\hat{i}$ - 13$\hat{j}$ - 21$\hat{k}$ $\)$ So, $\($ $\vec{d}$ = $\lambda$ $\left$( 35$\hat{i}$ - 13$\hat{j}$ - 21$\hat{k}$ $\right$) $\)$ Also given, $\($ $\vec{c}$ $\cdot$ $\vec{d}$ = 12 $\)$ $\($ $\Rightarrow$ $\vec{c}$ $\cdot$ $\vec{d}$ = ($\hat{i}$ - $\hat{j}$ + 2$\hat{k}$) $\cdot$ $\lambda$ (35$\hat{i}$ - 13$\hat{j}$ - 21$\hat{k}$) $\)$ $\($ $\Rightarrow$ $\lambda$ (35 + 13 - 42) = 12 $\)$ $\($ $\Rightarrow$ $\lambda$ = 2 $\)$ So, $\($ $\vec{c}$ $\times$ $\vec{d}$ = 2 $\times$ $\begin{vmatrix}$ $\hat{i}$ & $\hat{j}$ & $\hat{k}$ $\\$ 1 & -1 & 2 $\\$ 35 & -13 & -21 $\end{vmatrix}$ $\)$ $\($ $\Rightarrow$ $\vec{c}$ $\times$ $\vec{d}$ = 2 $\left$( 47$\hat{i}$ + 91$\hat{j}$ + 22$\hat{k}$ $\right$) $\)$ Now finding, $\[$ $\left$( -$\hat{i}$ + $\hat{j}$ - $\hat{k}$ $\right$) $\cdot$ $\left$( $\vec{c}$ $\times$ $\vec{d}$ $\right$) = 2 $\left$( -$\hat{i}$ + $\hat{j}$ - $\hat{k}$ $\right$) $\cdot$ $\left$( 47$\hat{i}$ + 91$\hat{j}$ + 22$\hat{k}$ $\right$) $\]$ $\($ $\Rightarrow$ $\left$( -$\hat{i}$ + $\hat{j}$ - $\hat{k}$ $\right$) $\cdot$ $\left$( $\vec{c}$ $\times$ $\vec{d}$ $\right$) = 2(-47 + 91 - 22) = 44 $\)$

Given, $S_n=4+11+21+34+\cdots+T_n$ $S_n=4+11+21+34+\cdots+T_{n-1}+T_n$ Subtracting above equations, we get $0=4+7+10+13+\cdots-T_n$ $\Rightarrow\ T_n=4+7+10+13+\cdots$ The above series is in AP. $\Rightarrow\ T_n=\frac{n}{2}\left(2\times4+(n-1)3\right)$ $\Rightarrow\ T_n=\frac{n}{2}(3n+5)$ So, $S_n=\sum_{n=1}^{n}\frac{3n^2+5n}{2}$ $\Rightarrow\ S_n=\frac{1}{2}\left(\frac{3n(n+1)(2n+1)}{6}+\frac{5n(n+1)}{2}\right)$ $\Rightarrow\ S_n=\frac{n(n+1)}{2}\left(\frac{2n+1}{2}+\frac{5}{2}\right)$ $\Rightarrow\ S_{29}=\frac{29\times30}{2}\left(\frac{59}{2}+\frac{5}{2}\right)$ $\Rightarrow\ S_{29}=29\times15\times32=13920$ $\Rightarrow\ S_{9}=\frac{9\times10}{2}\left(\frac{19}{2}+\frac{5}{2}\right)$ $\Rightarrow\ S_{9}=9\times5\times12=540$ $\Rightarrow\ \frac{S_{29}-S_{9}}{60}=\frac{13920-540}{60}=223$ Therefore, the required answer is $223$.

Let $P$ be origin. Now from the above triangle we can write $\overrightarrow{PA} = \overrightarrow{a}$, $\overrightarrow{PB} = \overrightarrow{b}$, $\overrightarrow{PC} = \overrightarrow{c}$. Also, we know that centroid $G$ divides orthocentre $Q$ and circumcentre $P$ in the ratio $2 : 1$. $$\overrightarrow{PG} = \frac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}}{3}$$ Now $\overrightarrow{PQ} = 3 \left( \frac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}}{3} \right)$ $$\Rightarrow \overrightarrow{PQ} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$$ Therefore, the required answer is $PQ$.

Given, $\sim [p \lor (\sim (p \land q))]$ Now we know that, By demorgan law, $\sim (A \lor B) \equiv \sim A \land \sim B$ Now using above formula we get, $\sim [p \lor (\sim (p \land q))]$ $\equiv \sim p \land (p \land q)$ [as $\sim (\sim A) \equiv A$] $\equiv (p \land q) \land (\sim p)$

Given, $9^{1 - \tan^2 x} + 9^{\tan^2 x} = 10$. Now let $9^{\tan^2 x} = t$, then the above equation will be, $$\frac{9}{t} + t = 10$$ $$\Rightarrow t^2 - 10t + 9 = 0$$ $$\Rightarrow t = 9 or t = 1$$ So, when $9^{\tan^2 x} = 9 \Rightarrow \tan^2 x = 1$ $$\Rightarrow \tan x = \pm 1 \Rightarrow x = \pm \frac{\pi}{4}, as given x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$$ Now when $9^{\tan^2 x} = 1$ $$\Rightarrow \tan^2 x = 0 \Rightarrow x = 0$$ Hence, $$\beta = \sum_{x \in S} \left( \frac{x}{3} \right) = \tan^2 \left( \frac{0}{3} \right) + \tan^2 \left( \frac{\pi}{12} \right) + \tan^2 \left( -\frac{\pi}{12} \right)$$ $$= 0 + 2 \left( 2 - \sqrt{3} \right)^2$$ $$= 14 - 8\sqrt{3}$$ So, the value of $\frac{1}{6} \left( \beta - 14 \right)^2 = \frac{64 \times 3}{32} = 32$. Hence this is the correct option.

Given that the coefficient of $x$ and $x^2$ in $(1 + x)^p (1 - x)^q$ are 4 and $-5$. $$\Rightarrow \left(1 + x\right)^p (1 - x)^q = \left(1 + px + \frac{p(p-1)}{2!} x^2 + \ldots \right) \left(1 - qx + \frac{q(q-1)}{2!} x^2 - \ldots \right)$$ Now coefficient of $x$ from the above expansion will be $p - q$ which is equal to 4 $$\Rightarrow p - q = 4.$$ Similarly coefficient of $x^2$ is $-5$. $$\Rightarrow \frac{p(p-1)}{2} + \frac{q(q-1)}{2} - pq = -5$$ $$\Rightarrow \frac{p^2 - 2pq + q^2}{2} - \frac{(p+q)}{2} = -5$$ $$\Rightarrow \frac{(p-q)^2}{2} - \frac{(p+q)}{2} = -5$$ $$\Rightarrow \frac{16}{2} + 5 = \frac{(p+q)}{2}$$ $$\Rightarrow p + q = 26 and p - q = 4$$ On solving the above equations we get, $$p = 15 and q = 11.$$ $$\Rightarrow 2p + 3q = 2(15) + 3(11) = 63$$ Therefore, the required answer is 63.

Given, the line $\frac{x}{1} = \frac{6-y}{2} = \frac{z+8}{5}$ intersect the lines $\frac{x-5}{4} = \frac{y-7}{3} = \frac{z+2}{1}$ and $\frac{x+3}{6} = \frac{3-y}{3} = \frac{z-6}{1}$ at the points $A$ and $B$ respectively. Now plotting the diagram, we get, $$\frac{x}{1} = \frac{y-6}{-2} = \frac{z+8}{5}$$ $$\frac{x-5}{4} = \frac{y-7}{3} = \frac{z+2}{1}$$ $$\frac{x+3}{6} = \frac{y-3}{-3} = \frac{z-6}{1}$$ Now solving, $\frac{x}{1} = \frac{y-6}{-2} = \frac{z+8}{5} = \lambda$ and $\frac{x-5}{4} = \frac{y-7}{3} = \frac{z+2}{1} = t$ we get, $$A \equiv (\lambda, -2\lambda + 6, 5\lambda - 8) \equiv (4t + 5, 3t + 7, t - 2)$$ Now comparing both side and solving we get, $t = -1, \lambda = 1$. Hence, point $A(1, 4, -3)$. Now solving, $\frac{x}{1} = \frac{y-6}{-2} = \frac{z+8}{5} = \mu$ and $\frac{x+3}{6} = \frac{y-3}{-3} = \frac{z-6}{1} = k$ we get, $$B \equiv (\mu, -2\mu + 6, 5\mu - 8) \equiv (6k - 3, -3k + 3, k + 6)$$ Now comparing both side and solving we get, $k = 1, \mu = 3$. Hence, point $B(3, 0, 7)$. So, mid-point of $AB = M(2, 2, 2)$. Now finding, distance of $M$ from the plane $2x - 2y + z - 14 = 0$ we get, $$\left| \frac{4 - 4 + 2 - 14}{\sqrt{2^2 + 2^2 + 1^2}} \right| = \left| \frac{-12}{3} \right| = 4$$

We have, $$Z = \frac{2x - 3i}{4x + 2i}$$ Therefore, $$Z = \frac{2x + i(2y - 3)}{4x + i(4y + 2)}$$ Multiply by the conjugate, $$Z = \frac{(2x + i(2y - 3))(4x - i(4y + 2))}{(4x + i(4y + 2))(4x - i(4y + 2))}$$ Simplifying, $$Z = \frac{8x^2 + (4y + 2)(2y - 3) + i[4x(2y - 3) - (8xy + 4x)]}{16x^2 + (4y + 2)^2}$$ Since, $Z$ is purely real, $$\frac{4x(2y - 3) - (8xy + 4x)}{16x^2 + (4y + 2)^2} = 0$$ This implies, $$-12x + 8xy - (8xy + 4x) = 0$$ Therefore, $$x = 0$$ And, $$4x + i(4y + 2) \neq 0$$ Thus, $$i(4y + 2) \neq 0$$ Therefore, $$y \neq -\frac{1}{2}$$ Hence this is the required option.

Given that $\alpha$ be the remainder when $(22)^{2022} + (2022)^{22}$ is divided by 3 and $\beta$ be the remainder when the same is divided by 7. $$\Rightarrow (22)^{2022} + (2022)^{22} = (21 + 1)^{2022} + (2022)^{22}.$$ Here $(2022)^{22}$ is divisible by 3 as 2022 is divisible by 3. So on expanding $(21 + 1)^{2022}$, we get $$\Rightarrow (21 + 1)^{2022} = {}^{2022}C_{0}(21)^{2022} + {}^{2022}C_{1}(21)^{2021} + \ldots + {}^{2022}C_{2022}(1)^{2022}$$ $$= 3(3^{2021} \times 7^{2022} + {}^{2022}C_{1} \times 3^{2020} \times 7^{2021} + \ldots) + 1$$ $$= 3k_1 + 1$$ In this case the remainder is 1. Hence, $\alpha = 1$. Now, $$\Rightarrow (22)^{2022} + (2022)^{22} = (21 + 1)^{2022} + (2023 - 1)^{22}$$ Take $(2023 - 1)^{22}$ $$\Rightarrow (2023 - 1)^{22} = {}^{22}C_{0}(2023)^{22} - {}^{22}C_{1}(2023)^{21} + \ldots + {}^{22}C_{22}(-1)^{22}$$ $$= 7\left({}^{22}C_{0}(7)^{21}(289)^{22} - {}^{22}C_{1}(7)^{20}(289) + \ldots\right) + 1$$ $$= 7k_2 + 1$$ $$\Rightarrow (21 + 1)^{2022} + (2023 - 1)^{22} = 7k_1 + 1 + 7k_2 + 1$$ $$= 7\mu + 2$$ $$\Rightarrow \beta = 2.$$ Hence, $\alpha + \beta^2 = 1^2 + 2^2 = 5.$ Therefore, the required answer is 5.

Given, Now we know that, Mean = $\frac{\sum X_i f_i}{\sum f_i}$ $\Rightarrow$ $\mu$ = $\frac{2k + 2k^2 - 2 + 3k^2 - 3 + 4k^2 + 4 - 5k - 15}{62}$ $\Rightarrow$ $\mu$ = $\frac{9k^2 + 7k - 16}{62}$ $\ldots$ $\ldots$ (1) And $\sum$ f_i = 62 $\Rightarrow$ 3k^2 + 4k - 2 = 62 $\Rightarrow$ 3k^2 + 4k - 64 = 0 $\Rightarrow$ (3k + 16)(k - 4) = 0 $\Rightarrow$ k = 4 Now putting the value of k in above equation (1) we get, $\mu$ = $\frac{156}{62}$ Now finding variance we get, $\sigma$^2 = $\frac{\sum (X_i)^2 f_i}{\sum f_i}$ - (Mean)^2 $\Rightarrow$ $\sigma$^2 = $\frac{1 \cdot (2k) + 4(k^2 - 1) + 9(k^2 - 1) + 16(k^2 + 1) + 25(k - 3)}{62}$ - $\left$($\frac{156}{62}$$\right$)^2 $\Rightarrow$ $\sigma$^2 = $\frac{29k^2 + 27k - 72}{62}$ - $\mu$^2 Now putting the value of k in above equation we get, $\Rightarrow$ $\sigma$^2 = $\frac{500}{62}$ - $\mu$^2 $\Rightarrow$ $\sigma$^2 + $\mu$^2 = $\frac{500}{62}$ $\Rightarrow$ [$\sigma$^2 + $\mu$^2] = 8

Given, the equations of two adjacent sides of a parallelogram $ABCD$ be $2x - 3y = -23$ and $5x + 4y = 23$. So, $AB \equiv 2x - 3y = -23$ and $BC \equiv 5x + 4y = 23$. Also given, $AC \equiv 3x + 7y = 23$. Solving the above lines we get, $A(-4, 5)$, $B(-1, 7)$, $C(3, 2)$. We know that, diagonals of a parallelogram have the same midpoint, so $AC$ and $BD$ have the same midpoint and let point $D$ be $(x, y)$. So midpoint formula we get, $$\frac{x - 1}{2} = \frac{-4 + 3}{2} \Rightarrow x = 0 and \frac{y + 7}{2} = \frac{2 + 5}{2} \Rightarrow y = 0$$ Hence, point $D$ is $(0, 0)$. Now equation of $BD$ will be $7x + y = 0$. Now finding the distance of $A(-4, 5)$ from $7x + y = 0$ we get, $$d = \left| \frac{7(-4) + 5}{\sqrt{7^2 + 1^2}} \right| = \frac{23}{\sqrt{50}}$$ Hence, $50d^2 = 23^2 = 529$

Given, system of equations, $$6 \lambda x - 3y + 3z = 4 \lambda^2$$ $$2x + 6 \lambda y + 4z = 1$$ $$3x + 2y + 3 \lambda z = \lambda$$ For no solution $\Delta = 0$, $$\begin{vmatrix} 6 \lambda & -3 & 3 \\ 2 & 6 \lambda & 4 \\ 3 & 2 & 3 \lambda \end{vmatrix} = 0$$ So, $$9 \lambda^3 - 7 \lambda - 2 = 0$$ $$\Rightarrow (\lambda - 1)(3 \lambda + 1)(3 \lambda + 2) = 0$$ Hence, $$\lambda \in \left\{ 1, -\frac{1}{3}, -\frac{2}{3} \right\}$$ Now putting the value of $\lambda$ in $12 \sum_{\lambda \in S} |\lambda|$ we get, $$12 \sum_{\lambda \in S} |\lambda| = 12 \times \left( 1 + \frac{1}{3} + \frac{2}{3} \right) = 24$$

Given, the foot of perpendicular from the point $A(4, 3, 1)$ on the plane $P: x - y + 2z + 3 = 0$ be $N$. So, by foot of perpendicular formula we get, $$\frac{x_1 - 4}{1} = \frac{y_1 - 3}{-1} = \frac{z_1 - 1}{2} = \frac{-4 - 3 + 2 + 3}{1^2 + 1^2 + 2^2}$$ $$\Rightarrow N \equiv (x_1, y_1, z_1) = (3, 4, -1)$$ And $B(5, \alpha, \beta), \alpha, \beta \in \mathbb{Z}$ is a point on plane $P$ such that the area of the triangle $ABN$ is $3\sqrt{2}$. So putting the value of point $B(5, \alpha, \beta)$ in plane we get, $$5 - \alpha + 2\beta + 3 = 0$$ $$\Rightarrow \alpha - 2\beta = 8$$ And given area of triangle $ABN$ is $3\sqrt{2}$, we get $$\frac{1}{2} AN \times BN = 3\sqrt{2}$$ $$\Rightarrow AN \times BN = 6\sqrt{2}$$ $$\Rightarrow \frac{6}{\sqrt{6}} \times BN = 6\sqrt{2}$$ $$\Rightarrow BN = 2\sqrt{3}$$ $$\Rightarrow 4 + (\alpha - 4)^2 + (\beta + 1)^2 = 12$$ $$\Rightarrow (\alpha - 4)^2 + (\beta + 1)^2 = 8$$ $$\Rightarrow 5\beta^2 + 18\beta + 9 = 0 \{as \alpha - 2\beta = 8\}$$ $$\Rightarrow \beta = -3 \left\{as \beta \neq -\frac{3}{5}\right\}$$ So, $\beta = -3$, $\alpha = 2$. Hence, $\alpha^2 + \beta^2 + \alpha\beta = 3^2 + 2^2 - 6 = 7$

Given, the quadratic curve passing through the point $(-1, 0)$ and touching the line $y = x$ at $(1, 1)$ be $y = f(x)$. Now let the curve be $y = ax^2 + bx + c$. Passes through $(-1, 0)$ $$\Rightarrow a - b + c = 0 \ldots (i)$$ Passes through $(1, 1)$ $$\Rightarrow a + b + c = 1 \ldots (ii)$$ Now given tangent to $y = x$ at $(1, 1)$, so, $$\frac{dy}{dx} = 1$$ $$\Rightarrow 2a + b = 1 \ldots (iii)$$ Solving all above three equations we get, $$a = \frac{1}{4}, \; b = \frac{1}{2} \; and \; c = \frac{1}{4}$$ Hence, the equation of curve is $y = \frac{x^2}{4} + \frac{x}{2} + \frac{1}{4}$. Now $(\alpha, \alpha + 1)$ lies on curve $$\alpha + 1 = \frac{\alpha^2}{4} + \frac{\alpha}{2} + \frac{1}{4}$$ $$\Rightarrow (\alpha - 3)(\alpha + 1) = 0$$ $$\Rightarrow \alpha = 3$$ Point in first quadrant is $(3, 4)$. Now finding slope of tangent at $(3, 4)$ we get, $$\left( \frac{dy}{dx} \right)_{(3,4)} = \left( \frac{x}{2} + \frac{1}{2} \right)_{(3,4)} = 2$$ $$\Rightarrow Slope of normal = -\frac{1}{2}$$ Hence, equation of normal will be, $$y - 4 = -\frac{1}{2}(x - 3)$$ So, $x$-intercept $= 11$

Given, the tangent at any point P on a curve passing through the points $(1, 1)$ and $\left( \frac{1}{10}, 100 \right)$, intersect positive x-axis and y-axis at the points A and B respectively. And $PA : PB = 1 : 1$. $k$ and $y = y(x)$ is the solution of the differential equation $\frac{dy}{dx} = kx + \frac{k}{2}$, $y(0) = k$. Now on plotting the diagram we get, Equation of tangent at $P(x, y)$ is: $$Y - y = \frac{dy}{dx} (X - x)$$ Coordinate of $A = \left( x - y \frac{dy}{dx}, 0 \right)$ Coordinate of $B = \left( 0, y - x \frac{dy}{dx} \right)$ $$\therefore (x, y) = \left( \frac{kx - ky}{k + 1}, \frac{y - x \frac{dy}{dx}}{k + 1} \right)$$ So, $y = \frac{y - x \frac{dy}{dx}}{k + 1}$ $$\Rightarrow y(k + 1) = y - x \frac{dy}{dx}$$ $$\Rightarrow ky = -x \frac{dy}{dx}$$ $$\Rightarrow k \frac{dz}{z} = - \frac{dx}{x}$$ Now integrating both sides, we get $$k \ln |x| + \ln |y| = \ln C \ldots (i)$$ Now given equation (i) passes through $(1, 1)$ and $\left( \frac{1}{10}, 100 \right)$ So, $C = 1$ and $k = 2$ Now putting the value of $k$, we get $$e^{\frac{dy}{dx}} = kx + \frac{k}{2}$$ $$\Rightarrow e^{\frac{dy}{dx}} = 2x + 1 \{as k = 2\}$$ $$\Rightarrow \frac{dy}{dx} = \ln(2x + 1)$$ Integrating the above equation we get, $$y = \frac{1}{2}(2x + 1)[\ln(2x + 1) - 1] + C$$ And the above equation passes through $(0, 2)$ So, $C = \frac{5}{2}$ So, $2y = (2x + 1)(\ln(2x + 1) - 1) + 5$ $$\Rightarrow 2y(1) = 3(\ln 3 - 1) + 5$$ $$\Rightarrow 2y(1) = 3 \ln(3) + 2$$ Hence, $4y(1) - 5 \ln 3 = 3$, $2 = 3(2 \ln 3 - 1) - 5 \ln 3$ $$\Rightarrow 4y(1) - 5 \ln 3 = 6 \ln 3 + 4 - 5 \ln 3 = 13$$ $$\Rightarrow 4y(1) - 5 \ln 3 = 4 - \ln 3 \approx 5$$ Note: This question was found in JEE Mains 2023 April session.

Given, $a_1, a_2, 2, a_3, a_4$ be in an arithmetico-geometric progression. And the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is $\frac{49}{2}$. So, common ratio here $r = 2$ and let common difference $= d$. And we know that, $$a_2 = (a_1 + d) \times 2$$ $$2 = (a_1 + 2d) \times 2^2 \ldots (i)$$ $$a_3 = (a_1 + 3d) \times 2^3$$ $$a_4 = (a_1 + 4d) \times 2^4$$ Now adding all above equations we get, $$31a_1 + 98d = \frac{49}{2} \ldots (ii)$$ Now from $(i)$ and $(ii)$ $$a_1 = 2, d = \frac{1}{4}$$ Hence, $a_4 = 16(0 + 1) = 16$

We know that, Domain of $\sec^{-1} x$ is $(-\infty, -1] \cup [1, \infty)$. Now, $f(x) = \sec^{-1} \left( \frac{2x}{5x+3} \right)$. So, $$\frac{2x}{5x+3} \leq -1 or \frac{2x}{5x+3} \geq 1$$ $$\Rightarrow \frac{2x}{5x+3} + 1 \leq 0 or \frac{2x}{5x+3} - 1 \geq 0$$ $$\Rightarrow \frac{7x+3}{5x+3} \leq 0 or \frac{-3x-3}{5x+3} \geq 0$$ $$\Rightarrow \frac{7x+3}{5x+3} \leq 0 or \frac{x+1}{5x+3} \leq 0$$ $$\Rightarrow x \in \left[ -1, -\frac{3}{5} \right) \cup \left( -\frac{3}{5}, -\frac{3}{7} \right]$$ So, $\alpha = -1$, $\beta = -\frac{3}{5}$, $\gamma = -\frac{3}{5}$, $\delta = -\frac{3}{7}$. Now, $$|3\alpha + 10(\beta + \gamma) + 21\delta|$$ $$= |-3 - 6 - 6 - 9|$$ $$= 24$$ Hence this is the required answer.

Total 4 digit numbers using the digits 1, 2, 2, 3 is $$\frac{4!}{2!} = 12$$ Numbers are _ _ _ 3 _ _ _ 3 _ _ _ 3 _ _ _ 2 _ _ _ 2 _ _ _ 2 _ _ _ 2 _ _ _ 2 _ _ _ 2 _ _ _ 1 _ _ _ 1 _ _ _ 1 Required sum is $$= [3 \times 3 + 2 \times 6 + 1 \times 3] + [(3 \times 3 + 2 \times 6 + 1 \times 3) \times 10] + [(3 \times 3 + 2 \times 6 + 1 \times 3) \times 100] + [(3 \times 3 + 2 \times 6 + 1 \times 3) \times 1000]$$ $$= [24] + [24 \times 10] + [24 \times 100] + [24 \times 1000]$$ $$= 24 + 240 + 2400 + 24000 = 26664$$

Given, And $\theta_1 + \theta_2 = \frac{\pi}{2}$. Now, Let $AB = x$, $BD = y$. Also given, $\sqrt{3}BE = 4AB$ $$\Rightarrow \sqrt{3}(y + DE) = 4x$$ $$\Rightarrow DE = \frac{4x}{\sqrt{3}} - y$$ Now given area of triangle $\Delta CAB = 2\sqrt{3} - 3$ $$\Rightarrow \frac{1}{2} xy = 2\sqrt{3} - 3$$ $$\Rightarrow y = \frac{4\sqrt{3} - 6}{x}$$ Now finding, $\tan \theta_2 = \frac{x}{y} = \frac{x^2}{4\sqrt{3}} = \frac{4}{\sqrt{3}} \left( \frac{4\sqrt{3} - 6}{x^2} \right)$ $\tan \theta_1 = \frac{y}{x} = \frac{4\sqrt{3} - 6}{x^2}$ Now taking tan both of $\theta_1 + \theta_2 = \frac{\pi}{2}$ we get, $\tan \theta_1 \cdot \tan \theta_2 = 1$ $$\Rightarrow \left( \frac{4}{\sqrt{3}} \right) \left( \frac{4\sqrt{3} - 6}{x^2} \right) = 1$$ $$\Rightarrow \frac{4\sqrt{3} - 6}{x^2} = \sqrt{3} or \frac{4\sqrt{3} - 6}{x^2} = \frac{1}{\sqrt{3}}$$ So, $\theta_1$ is maximum when $\frac{4\sqrt{3} - 6}{x^2} = \frac{1}{\sqrt{3}}$ $$\Rightarrow x^2 = \sqrt{3} (4\sqrt{3} - 6)$$ $$\Rightarrow x^2 = 12 - 6\sqrt{3}$$ $$\Rightarrow x^2 = (3 - \sqrt{3})$$ $$\Rightarrow x = 3 - \sqrt{3} and \angle C = 60^\circ$$ $$\Rightarrow \tan 60^\circ = \frac{DE}{CD} = \frac{DE}{CE} = x : \sqrt{3} = 3\sqrt{3} - 3$$ And $\cos 60^\circ = \frac{CD}{CE} = \frac{2x - 6}{2\sqrt{3}}$ So, the Perimeter of $\Delta AEC = CD + DE + CE = 6$

If the area of the region $\{(x, y) : |x^2 - 2| \leq y \leq x \}$ is $A$, then $6A + 16\sqrt{2}$ is equal to $_$ Given, $$|x^2 - 2| \leq y \leq x$$ Now on finding the intersection point we get, $(x, y) \equiv (1, 1)$ and $(2, 2)$ Now plotting the curve of the above function we get, Now from above diagram required bounded area will be, A = ∫[1 to √2] (x - (2 - x²)) dx + ∫[√2 to 2] (x - (x² - 2)) dx A = [x²/2 - 2x + x³/3]₁^√2 + [x²/2 - x³/3 + 2x]√2^2 A = [1 - 2√2/3] - [-7/6] + [10/3] - [1 - 2√2/3 + 2√2] A = 9/2 - 8√2/3 ⇒ 6A = 27 - 16√2 ⇒ 6A + 16√2 = 27

The time ($t_1$) taken by the person to travel distance $x$ with velocity $v_1$ is given by $$t_1 = \frac{x}{v_1} \ldots(1)$$ The time ($t_2$) taken by the person to travel distance $x$ with velocity $v_2$ is given by $$t_2 = \frac{x}{v_2} \ldots(2)$$ The formula to calculate the average speed ($v$) of the person is given by $$v = \frac{2x}{t_1 + t_2} \ldots(3)$$ Substitute the expressions from equations (1) and (2) into equation (3) to obtain the required relation. $$v = \frac{2x}{\left( \frac{x}{v_1} + \frac{x}{v_2} \right)}$$ $$\Rightarrow \frac{x}{v_1} + \frac{x}{v_2} = \frac{2x}{v}$$ $$\Rightarrow \frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}$$

It is given that $t_{\frac{1}{2}} = T$. After disintegration of $\left( \frac{7}{8} \right)^{th}$ part, $\left( \frac{1}{8} \right)^{th}$ part is only left. Using the decay formula, $$N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{\frac{1}{2}}}}$$ $$\Rightarrow \frac{N_0}{8} = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T}}$$ $$\Rightarrow -\ln 8 = -\frac{t}{T} \ln 2$$ $$\Rightarrow 3 \ln 2 = \frac{t}{T} \ln 2$$ $$\Rightarrow t = 3T$$

The internal energy of a gas is given by $U = nC_V T$. A diatomic gas has 3 translational and 2 rotational degrees of freedom. A monoatomic gas has 3 translational degrees of freedom. Hence, the internal energy of the mixture $= n_1 C_{V_1} T + n_2 C_{V_2} T$ $$= 2 \times \left( \frac{5R}{2} \right) T + 4 \times \frac{3R}{2} T$$ $$= 5RT + 6RT$$ $$= 11RT$$

The least count is $LC = 0.1 \, \mathrm{mm}$. Zero error $(ZE) = 6 \times LC = 6 \times 0.1 \, \mathrm{mm} = 0.6 \, \mathrm{mm}$. The total reading is $$TR = MSR + (VSR \times LC) - ZE$$ $$\Rightarrow TR = 32 \, \mathrm{mm} + (0.1 \times 4) \, \mathrm{mm} - 0.6 \, \mathrm{mm}$$ $$\Rightarrow TR = 31.8 \, \mathrm{mm} = 3.18 \, \mathrm{cm}$$

There are no atomic dipoles in diamagnetic materials because the resultant magnetic moment of each atom is zero due to paired electrons. Diamagnetic materials are repelled by a magnet. These substances are weakly repelled by the field, so in a nonuniform field, these substances have a tendency to move from a strong to a weak part of the external magnetic field. For diamagnetic substance $(-1 < \chi < 0)$ and a diamagnetic substance is repelled by the magnetic field.

The formula of capacitance is $C_1 = \left( \frac{\varepsilon_0 A}{d} \right)$. The dielectric constant of the metal sheet is $K_{metal} = \infty$. The new value of capacitance is $$C_2 = \frac{\varepsilon_0 A}{d + \frac{t}{K_{metal}}} = \frac{\varepsilon_0 A}{\frac{2d}{3} + 0} = 3C_1.$$ Clearly, the ratio, $$\frac{C_2}{C_1} = \frac{3C_1}{C_1} = 3 : 1.$$

The variation of acceleration due to gravity with the rotation of Earth can be written as $$g' = g - \omega^2 R \cos^2 \lambda \ldots (1)$$ Substitute $0^\circ$ for $\lambda$ for equatorial position to obtain the acceleration due to gravity. $$g' = g - \omega^2 R \cos^2 0^\circ$$ $$= g - \omega^2 R$$ Substitute $90^\circ$ for $\lambda$ for polar position to obtain the acceleration due to gravity. $$g' = g - \omega^2 R \cos^2 90^\circ$$ $$= g$$ The above two equations show that the effect on acceleration due to gravity is maximum at the equator and minimum at pole.

The velocity of the satellite revolving round the Earth at a height $r$ from the centre of the Earth is given by $$v = \sqrt{\left( \frac{GM}{r} \right)} \ldots (1)$$ The formula to calculate the time period in the circular orbit is given by $$T = \frac{2 \pi r}{v} \ldots (2)$$ In this case: $r = R + R = 2R$ Substitute the expression for the velocity from equation (1) into equation (2) and simplify to obtain the required time period. $$T = \frac{2 \pi r}{\sqrt{\frac{GM}{r}}}$$ $$= \frac{2 \pi \times 2R}{\sqrt{\frac{gR}{2}}}$$ $$= \frac{4 \pi R \sqrt{2}}{\pi \sqrt{R}}$$ $$= 4 \sqrt{2R}$$ $$= \sqrt{32R}$$

When a fan is rotating, it acquires some kinetic energy. When the switch is off, because of its motion, it continues rotating for some time until it stops because of the air friction. So it is due to inertia of motion that fan continues to rotate. Hence, both A and R are correct and R is the correct explanation of A.

The relation between the electric field and magnetic field is given by $E_0 = c B_0$. The given data is $B_0 = 6 \times 10^{-7} \, \mathrm{T}$ and $c = 3 \times 10^8 \, \mathrm{m \, s^{-1}}$. Thus, the maximum value of the electric field is $$E_0 = (3 \times 10^8) \times (6 \times 10^{-7}) = 180 \, \mathrm{V \, m^{-1}}.$$

In adiabatic process, $\Delta Q = 0$ and $\Delta U \neq 0$. Using the first law of thermodynamics, $\Delta Q = \Delta U + W$. Therefore, $\Delta U = -W$ ($W$ is the work done by the gas, so $-W$ will be the work done on the gas). As the gas is compressed $W 0$ and hence $\Delta T > 0$.

The formula to calculate the intensity of the interference pattern in Young's double slit experiment can be written as $$I = 4I_0 \cos^2 \left( \frac{\phi}{2} \right) \cdots (1)$$ Hence, the ratio of the required intensities can be calculated as follows- $$\frac{I_1}{I_2} = \frac{4I_0 \cos^2 \left( \frac{\pi}{6} \right)}{4I_0 \cos^2 \left( \frac{\pi}{4} \right)}$$ $$= \frac{\frac{3}{4}}{\frac{1}{2}}$$ $$= \frac{3}{2}$$

The relation between the stopping potential and the frequency of the incident light is given by $$eV_0 = h\nu - \phi \ldots (1)$$ When $V_0 = 0$, $$\phi = h\nu \ldots (2)$$ Substitute the value of the frequency corresponding to zero stopping potential into equation (2) to calculate the required work function of the material. $$\phi = \frac{6.626 \times 10^{-34} \, \mathrm{J\cdot s} \times 5 \times 10^{14} \, \mathrm{Hz}}{1.6 \times 10^{-19} \, \mathrm{J/eV}}$$ $$= \frac{6.626 \times 5}{16} \approx 2.07 \, \mathrm{eV}$$

The electrons inside the conductor move with an average velocity due to the applied field in a direction opposite to the field, this velocity is known as drift velocity. Electric field are applied from high to low potential region of the conductor. Also, electrons are negatively charged. Electron drifts from lower potential to higher potential on curved path.

The formula to calculate the elongation of a metallic wire because of the application of a load can be written as $$\Delta L = \left( \frac{FL}{AY} \right) \ldots (1)$$ The ratio of the elongations for both the wires is, then, calculated as follows- $$\frac{\Delta L_1}{\Delta L_2} = \frac{\frac{FL}{A_1Y_1}}{\frac{FL}{A_2Y_2}}$$ $$= \frac{A_2Y_2}{A_1Y_1}$$ $$= \frac{3}{1} \times \frac{4}{1}$$ $$= \frac{12}{1}$$

The maximum height is given by, $H = \frac{u^2 \sin^2 \theta}{2g}$. The given data is $\theta_1 = 30^\circ$, $\theta_2 = 60^\circ$, $u_1 = u_2$. Therefore, $$\frac{(H_1)_{\max}}{(H_2)_{\max}} = \frac{u_1^2 \sin^2 \theta_1}{u_2^2 \sin^2 \theta_2}$$ $$= \left(\frac{\sin 30^\circ}{\sin 60^\circ}\right)^2$$ $$= \left(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)^2$$ $$= \left(\frac{1}{\sqrt{3}}\right)^2$$ $$= \frac{1}{3}$$

The relation between bandwidth and the signal frequency is given by $$Bandwidth = 2 \times signal frequency (1)$$ Substitute the values of the known parameter into equation (1) to calculate the required bandwidth. $$Bandwidth = 2 \times 3 \, \mathrm{kHz}$$ $$= 6 \, \mathrm{kHz}$$

As the diode $D_2$ is connected in reverse order, there is no current through it. The effective resistances of the circuit can, then, be calculated as follows: $$R_{eff} = 25 + \frac{(25 + 125)^2}{(25 + 125) + (25 + 125)} \, \Omega$$ $$= 100 \, \Omega$$ Each current through the diodes can be calculated as follows: $$I_1 = \left( \frac{5 \, \mathrm{V}}{100 \, \Omega} \right)$$ $$= 0.05 \, \mathrm{A}$$ As the diode $D_2$ is non-conducting, the current $I_1$ will be divided equally among the other two diodes $D_1$ and $D_3$. Hence, the currents in the diodes can be written as $$I_2 = \frac{0.05 \, \mathrm{A}}{2}$$ $$= 0.025 \, \mathrm{A}$$ $$I_4 = 0.025 \, \mathrm{A}$$ Thus, $$\frac{I_1}{I_2} = \frac{0.05 \, \mathrm{A}}{0.025 \, \mathrm{A}}$$ $$= 2$$

The flux connected to the copper tube will fluctuate when the magnet moves under the influence of gravity. Eddy currents will be created in the copper tube's body as a result, these induced currents oppose the collapse of the magnet, in accordance with Lenz's law. A retarding force will consequently act on the magnet. This force will keep rising with the magnet's rising velocity until it equals the force of gravity. The net force acting on the magnet will then vanish. The magnet will therefore reach a steady speed.

It is given that the current is varying sinusoidally. The current can be written as $I = I_0 \sin \omega t$. The magnetic field through the solenoid $B = \mu_0 n I$. The flux through the square is $\phi = \mu_0 n I A$. The emf is $$\varepsilon = \mu_0 n A \times \frac{d(I_0 \sin \omega t)}{dt}$$ $$\Rightarrow \varepsilon = \mu_0 n A I_0 \omega \cos \omega t$$ The amplitude of the emf is, $$\varepsilon = \mu_0 n A I_0 \omega = 4 \pi \times 10^{-7} \times \frac{50}{10^{-2}} \times 4 \times 10^{-4} \times 2.5 \times 700$$ $$\Rightarrow \varepsilon = 44 \times 10^{-4} \, \mathrm{V}$$

By Newton's second law, $F = ma_{max}$. The maximum acceleration for a SHM is given by $a_{max} = \omega^2 A \cdots (i)$. The time period is given by $$T = \frac{2\pi}{\omega} = 3.14 \, \mathrm{s}$$ $$\Rightarrow \omega = \frac{2\pi}{T} = \frac{2\pi}{3.14} = 2 \, \mathrm{rad} \, \mathrm{s}^{-1}$$ Thus, the maximum acceleration from equation (i) $$a_{max} = 4 \times 1 = 4 \, \mathrm{m} \, \mathrm{s}^{-2}$$ The maximum force is $$F = 5 \, \mathrm{kg} \times 4 \, \mathrm{m} \, \mathrm{s}^{-2} = 20 \, \mathrm{N}$$

Using Rydberg's formula, $$\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ For the lowest wavelength in Lyman series, $n_1 = 1$, $n_2 = \infty$. $$\frac{1}{\lambda} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = RZ^2$$ For Balmer series, $n_1 = 2$, $n_2 = \infty$ $$\frac{1}{\lambda'} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{RZ^2}{4}$$ So, $$\frac{1}{\lambda'} = \frac{1}{4\lambda}$$ $$\Rightarrow \lambda' = 4 \times 917 \, \mathrm{\AA} = 3668 \, \mathrm{\AA}$$

The given data is $A_1 = 2 \times 10^{-4} \, \mathrm{m^2}$ $A_2 = 10 \times 10^{-6} \, \mathrm{m^2}$ $v_1 = 4 \, \mathrm{cm \, s^{-1}}$ Using equation of continuity which is a conservation of mass principle, $$A_1 v_1 = A_2 v_2$$ $$\Rightarrow \; v_2 = \frac{A_1 v_1}{A_2} = \frac{2 \times 10^{-4} \times 4}{10 \times 10^{-6}}$$ $$= 80 \, \mathrm{cm \, s^{-1}}$$

It is given that $i = 14 \, \mathrm{A}$ and $R = 2.2 \times 10^{-2} \, \mathrm{m}$. The magnetic field at the centre $O$ is given by $$B = \frac{\mu_0 i}{4R}$$ $$= \frac{4\pi \times 10^{-7} \times 14}{4 \times 2.2 \times 10^{-2}}$$ $$= 2 \times 10^{-4} \, \mathrm{T}$$

For $L_1$: $$\frac{1}{v_1} + \frac{1}{6} = \frac{1}{24}$$ $$\Rightarrow \frac{1}{v_1} = \frac{1}{24} - \frac{1}{6}$$ $$= \frac{1 - 4}{24}$$ $$\Rightarrow \frac{1}{v_1} = \left( \frac{-3}{24} \right)$$ $$\Rightarrow v_1 = -8 \, cm$$ For $L_2$: $$u = -18 \, cm$$ $$\Rightarrow \frac{1}{v} + \frac{1}{18} = \frac{1}{9}$$ $$\Rightarrow \frac{1}{v} = \frac{1}{9} - \frac{1}{18}$$ $$= \frac{2 - 1}{18}$$ $$\Rightarrow \frac{1}{v} = \frac{1}{18}$$ $$\Rightarrow v = 18 \, cm$$ Distance between object and distance $= 16 + 18 = 34 \, cm$.

It is given that $l = 1 \, \mathrm{cm} = 0.01 \, \mathrm{m}$ $A = 100 \, \mathrm{cm} \times 1 \, \mathrm{cm} = 100 \, \mathrm{cm}^2$ $\Rightarrow A = 0.01 \, \mathrm{m}^2$ $\rho = 3 \times 10^{-7} \, \Omega \, \mathrm{m}$ Using the formula of resistance, $$R = \frac{\rho l}{A} = \frac{3 \times 10^{-7} \times 0.01}{0.01} = 3 \times 10^{-7} \, \Omega$$

It is given that the force $\vec{F} = -P \hat{k}$. The coordinates of the origin is $(0, 0)$. The coordinates of the point given is $(2, -3)$. The displacement vector is $$\vec{r} = (0 - 2) \hat{i} + (0 + 3) \hat{j} = -2 \hat{i} + 3 \hat{j}$$ Hence, the torque is $$\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ 0 & 0 & -P \end{vmatrix}$$ $$\Rightarrow \vec{\tau} = -2P \hat{j} - 3P \hat{i} = P \left( a \hat{i} + b \hat{j} \right)$$ So, it can be written $a = -3, b = -2$ $$\therefore \frac{a}{b} = \left( \frac{3}{2} \right) \Rightarrow so x = 3$$

The maximum force acting is, $F_{max} = (14000 + 2000) \, \mathrm{N}$ $$= 16000 \, \mathrm{N}$$ The formula for power is given by, $P = F_{max} v$. It is given, $v = 3 \, \mathrm{m \, s^{-1}}$. $$\Rightarrow P = 16000 \, \mathrm{N} \times 3 \, \mathrm{m \, s^{-1}} = 48000 \, \mathrm{W} = 48 \, \mathrm{kW}$$ $$\Rightarrow W = 48 \, \mathrm{kW}$$

The electric field is given by $$E = \left( \frac{2k\lambda}{r} \right)$$ The required centripetal force is $$F = \left( \frac{mv^2}{r} \right)$$ The force due to the field is equal to the required centripetal force, $$\Rightarrow \left( \frac{2k\lambda}{r} \right) e = \frac{mv^2}{r}$$ $$\Rightarrow v = \sqrt{\frac{2k\lambda e}{m}}$$ $$= \sqrt{\frac{2 \times 9 \times 10^9 \times 2 \times 10^{-8} \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}}$$ $$= \sqrt{1.6 \times 4 \times 10^{13}}$$ $$= 4 \times 2 \times 10^6 \, \mathrm{m} \, \mathrm{s}^{-1}$$ $$= 8 \times 10^6 \, \mathrm{m} \, \mathrm{s}^{-1}$$

Reacting a Grignard reagent with a ketone gives a tertiary alcohol. When alkyl halide reacts with aqueous sodium hydroxide to give alcohol, the reaction is called nucleophilic substitution. The Hydroboration Oxidation reaction is an organic chemical reaction which is employed for the conversion of alkenes into alcohols that are neutral. Reductive ozonolysis of alkenes will lead to formation of aldehyde or ketones, oxidative ozonolysis of alkenes will lead to formation of carboxylic acids or ketones. So, alcohol is not formed by ozonolysis of alkenes.

Mg have 2 electrons in valence shell and after donation of 2 electrons it converts into $\mathrm{Mg^{2+}}$ ion and attains inert gas configuration, now to ionize one more electron from inert gas configuration is very difficult, hence it is easy to ionize Mg to $\mathrm{Mg^{2+}}$. $\mathrm{IE_2}$ is always greater than $\mathrm{IE_1}$ because on removal of $e^-$, size decreases and effective nuclear charge increases. So Assertion is correct. $\mathrm{Mg^{2+}}$ is a smaller ion than $\mathrm{Mg^+}$ and has a greater charge density. This means that $\mathrm{Mg^{2+}}$ has a stronger attraction to its electrons than $\mathrm{Mg^+}$ and requires more energy to remove an electron from it. So reason is also correct.

Barium chloride is used to estimate sulphur content in the molecule. In Carius tube test, a known mass of sulphur present in an organic compound with Silver Nitrate is heated into a sealed tube known as Carius tube with silver. The organic compound is then converted to Sulphuric acid which then reacts with excess Barium Chloride solution and produces a precipitate of Barium Sulphate. $Z$ is $\mathrm{BaSO_4}$ and $Y$ is $\mathrm{H_2SO_4}$. It is clear that "X" must contain sulphur. Methionine is $\mathrm{C_5H_{11}NO_2S}$, a sulphur containing amino acid.

Molarity $\left( M \right) = \frac{wt}{mwt \times V} \times 1000$ Using this formula, we can calculate the molarity of the oxalic acid solution as follows: 1. Mass of hydrated oxalic acid $= 3.1500 \, \mathrm{g}$ 2. Molar mass of hydrated oxalic acid $(\mathrm{H_2C_2O_4 \cdot 2H_2O}) = 126.064 \, \mathrm{g/mol}$ 3. Volume of the solution $= 250.0 \, \mathrm{mL}$ Molarity $\left( M \right) = \frac{3.15 \times 1000}{126 \times 250} = 0.10$ So both assertion and reason are correct.

Buna-S is a copolymer of 1,3-butadiene and styrene. It is prepared by copolymerisation of 1,3 butadiene and styrene along with sodium. In this process peroxide is used as a catalyst at $5^\circ \mathrm{C}$ therefore the formed product is also known as cold rubber. $$n \mathrm{CH_2} = \mathrm{HC} - \mathrm{CH} = \mathrm{CH_2} + n \rightarrow -(\mathrm{CH_2} - \mathrm{CH} = \mathrm{CH} - \mathrm{CH_2} - \mathrm{CH} - \mathrm{CH_2})_n$$

Lithium aluminium hydride, is a reducing agent that is commonly used in modern organic synthesis. It is a nucleophilic reducing agent that is best suited for reducing multiple polar bonds such as $\mathrm{C=O}$. It can reduce carbonyl compounds to alcohols and amides to amines. $\mathrm{C=C}$ is not effected during this reaction.

Ferric chloride is a hemostatic agent, which means that it can be used to stop bleeding by promoting the coagulation of blood. Blood is a colloidal solution of an albuminoid substance and can be coagulated to form clot by applying $\mathrm{Fe^{3+}}$ ion on the open wound. Thus, bleeding caused by a razor during shaving can be treated by rubbing potash alum against the cut which helps in the coagulation of the blood by sealing the blood vessels.

The reaction used for preparation of soap from fat is not alkaline hydrolysis, but rather saponification. In saponification, a triglyceride (fat or oil) is reacted with a strong base, such as sodium hydroxide (or potassium hydroxide), in the presence of heat to produce soap molecules and glycerol as a byproduct. The reaction involves the hydrolysis of the ester bond in the triglyceride by the base to form carboxylate ions (the soap molecules) and glycerol.

The stability of a carbocation increases with the number of alkyl groups attached to the carbon atom, due to the inductive and hyperconjugation effects of the alkyl groups. The more the number of alkyl groups, the greater is the electron density on the positively charged carbon, and hence the more stable is the carbocation. The above carbocation can be stabilised by resonance and hyperconjugation. The above carbocation stabilised by resonance caused by three benzene rings. This carbocation is more stable than the previous carbocation. The above carbocation stabilised by hyperconjugation. It is the least stable carbocation among the four carbocations given. The above carbocation is the most stable carbocation among the four due to the resonance caused by bent orbitals. From the given options Stability of carbocation: D > B > A > C The least stable carbocation will have the highest hydride affinity, because it is the most electron deficient and has the highest positive charge density. As a result, it will have the strongest attraction for a negatively charged hydride ion. Hydride affinity order: C > A > B > D

A face-centered cubic (fcc) unit cell contains a component in the center of each face in addition to those at the corners of the cube. The atoms touch each other along the face diagonal. For FCC: $a \sqrt{2} = 4r$ $$a = 2 \sqrt{2} r$$ A body-centered cubic (bcc) unit cell contains a component in the center of the cube in addition to those at the corners of the cube. The atoms touch each other along the body diagonal. For BCC: $a \sqrt{3} = 4r$

The chemical formula for washing soda is $\mathrm{Na_2 \, CO_3 \, 10H_2O}$, which means that each molecule of washing soda contains 10 water molecules. The chemical formula for soda ash is $\mathrm{Na_2 \, CO_3}$, which does not contain any water molecules. Therefore, there are 10 water molecules in washing soda and 0 water molecules in soda ash.

Plants use up carbon dioxide during photosynthesis and release oxygen into the atmosphere. This maintains the delicate balance between oxygen and carbon dioxide in the atmosphere. The burning of fuels and cutting of trees is disturbing this balance in nature. As a result of such activities, the amount of carbon dioxide in the air is going up and the level of oxygen is coming down. Continuous and excessive use of fuels and deforestation would increase this imbalance. This is bringing about a warming of the atmosphere, called global warming. Respiration, is a natural process, so balance of $\mathrm{CO_2}$ and $\mathrm{O_2}$ not disturbed by respiration.

The number of unpaired electrons in coordination complexes is determined by the nature of the central metal ion, the coordination number, and the identity and arrangement of the ligands around the metal ion. Based on these factors, the correct order of the number of unpaired electrons in the given complexes is: A $[\mathrm{Fe(CN)_6}]^{3-}$: $\mathrm{Fe^{3+}}$ $(3d^5)$ (Pairing of electrons takes place) $1$ B $[\mathrm{FeF_6}]^{3-}$: $\mathrm{Fe^{3+}}$ $(3d^5)$ $5$ C $[\mathrm{CoF_6}]^{3-}$: $\mathrm{Co^{3+}}$ $(3d^6)$ $4$ D $[\mathrm{Cr(oxalate)}_3]^{3-}$: $\mathrm{Cr^{3+}}$ $(3d^3)$ $3$ E $[\mathrm{Ni(CO)_4}]$: $\mathrm{Ni(3d^{10})}$ (Pairing of electrons takes place) $0$ No. of unpaired electrons order: E < A < D < C < B

The compounds A & B are alcohols, and rest of the compounds are phenols. Phenols are more acidic than alcohols. Among the A & B, methyl alcohol is more acidic than tertiary butyl alcohol as alkyl groups decreases the acidic nature by electron releasing tendency. Among the phenols, the phenol with electron withdrawing group is more acidic. $\mathrm{NO_2}$ (Electron withdrawing group) group will increase the acidity, $\mathrm{OCH_3}$ (Electron releasing group) group at para position will decrease the acidity. So, Acidic strength $\mathrm{E > C > D > A > B}$

Friedel-Crafts Alkylation refers to the replacement of an aromatic proton with an alkyl group. This is done through an electrophilic attack on the aromatic ring with the help of a carbocation generated from the alkyl halide and Lewis acid aluminium chloride. In the given case ring closure reaction takes place and $OCH_3$ is ortho/para directing, major product will be obtained from para attack.

d-orbital is degenerate which split into two levels $e_g$ and $t_{2g}$ in the presence of ligands. This splitting is due to the presence of ligands. This is called the crystal-field splitting and the energy difference between the two levels is called the crystal-field splitting energy, $\Delta_0$. (A) $\mathrm{Ti^{2+} : 4s^03d^2}$ $t_{2g} \ e^- = 2$ $e_g \ e^- = 0$ $$\Delta_0 = 2(-.4\Delta_0) = -.8\Delta_0$$ (B) $\mathrm{V^{2+} : 4s^03d^3}$ $t_{2g} \ e^- = 3$ $e_g \ e^- = 0$ $$\Delta_0 = 3(-.4\Delta_0)$$ $$= -1.2\Delta_0$$ (C) $\mathrm{Mn^{3+} = 4s^03d^4}$ $t_{2g} \ e^- = 3$ $e_g \ e^- = 1$ $$\Delta_0 = 3(-.4\Delta_0) + 1(.6\Delta_0)$$ $$= -1.2\Delta_0 + 0.6\Delta_0$$ $$= -0.6\Delta_0$$ (D) $\mathrm{Fe^{3+} = 4s^03d^5}$ $t_{2g} \ e^- = 3$ $e_g \ e^- = 2$ $$\Delta_0 = 3(-.4\Delta_0) + 2(.6\Delta_0)$$ $$= -1.2\Delta_0 + 1.2\Delta_0 = 0$$

The isotopes of hydrogen are protium, deuterium, and tritium, which have different physical properties. For example, deuterium has a higher boiling point and a lower vapor pressure than protium, while tritium is radioactive and has a much shorter half-life than the other two isotopes. The mass difference between the isotopes of hydrogen is indeed significant, with protium having a mass of 1 atomic mass unit (amu), deuterium having a mass of 2 amu, and tritium having a mass of 3 amu. So both A and R are true and R is the correct explanation of A.

The number of moles $= \frac{\text{Mass}}{\text{Molar mass}}$ The number of molecules $= \text{moles} \times N_A$ One mole of any gas occupies $22.7 \, \mathrm{L}$ at STP. $(A)$ Moles of $\mathrm{CH_4} = \frac{16}{16} = 1$ Molecules $= 6.02 \times 10^{23}$ Electrons $= 10 \times 6.02 \times 10^{23} = 60.2 \times 10^{23}$ $(B)$ Moles of $\mathrm{H_2} = \frac{1}{2} = 0.5$ mole Volume at STP $= \frac{1}{2} \times 22.7 = 11.35$ lit. $(C)$ $1$ mole $\mathrm{N_2} = 28$ gm $(D)$ $0.5$ mole $\mathrm{SO_2} = \frac{1}{2} \times 64 = 32$ gm

The metallic character of an element is defined as the easiness of its atom in losing electrons. According to the modern periodic table, the metallic character of an element decreases while moving from left to right across a period. K is an alkali metal. For alkaline earth metals, metallic character increases down the group. Hence, the correct order of metallic character is $\mathrm{K} > \mathrm{Ca} > \mathrm{Be}$.

Ellingham diagrams are a particular graphical form of the principle that the thermodynamic feasibility of a reaction depends on the sign of $\Delta G$, the Gibbs free energy change, which is equal to $\Delta H - T \Delta S$, where $\Delta H$ is the enthalpy change and $\Delta S$ is the entropy change. Hence, element below in Ellingham diagram can reduce oxide of element above it. C can reduce FeO at $600 \, ^\circ \mathrm{C}$.

$\mathrm{A(g)} \rightleftharpoons 2\mathrm{B(g)} + \mathrm{C(g)}$ $$ \begin{array}{cccc} \text{Initial:} & 450 & 0 & 0 \\ \text{At time} = t: & 450 - p & 2p & p \end{array} $$ $$(450 - p) + (2p) + (p) = 720$$ $$450 + 2p = 720$$ $$2p = 270$$ $$p = 135$$ Fraction of A(g) decomposed $= \frac{\text{Amount of A decomposed}}{\text{Initial amount}} = \frac{135}{450} = 0.3$ The fraction A(g) decomposed $= 3 \times 10^{-1} \Rightarrow x = 3$

In faintly alkaline medium potassium permanganate changes to manganese dioxide in its oxidising action. $$\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$$ The oxidation state of manganese in $\mathrm{MnO_4^-}$ is $+7$ and in $\mathrm{MnO_2}$ is $+4$. Hence, it transfers 3 electrons during the oxidising action in alkaline medium.

The decomposition reactions are endothermic reactions. The reactions $\mathrm{I_2(g) \rightarrow 2I(g)}$ and $\mathrm{HCl(g) \rightarrow H(g) + Cl(g)}$ are decomposition reactions. The heat vaporisation is an endothermic reaction. Hence, the conversion of $\mathrm{H_2O(l) \rightarrow H_2O(g)}$ is endothermic process. The heat of combustion is an exothermic process. The heat involved in $\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$ is heat of combustion. The dissolution of ammonium chloride in water is an endothermic process. $\mathrm{NH_4Cl(s) \rightarrow NH_4^+(aq) + Cl^-(aq)}; \Delta H = +15.1 \, \mathrm{kJ/mol}.$

The lone pair of electrons can be identified by drawing structure of the given molecules as follows, $\mathrm{H_2O}$ has 2 lone pairs. $\mathrm{N_2}$ has 2 lone pairs. $\mathrm{CO}$ has 2 lone pairs. $\mathrm{XeF_4}$ has $2 + (4 \times 3) = 14$ lone pairs. $\mathrm{NH_3}$ has 1 lone pair. $\mathrm{NO}$ has 3 lone pairs. Reference for structure NCERT - XI pg 105 $\mathrm{CO_2}$ has 4 lone pairs. $\mathrm{F_2}$ has 6 lone pairs. The number of molecules having only 2 lone pair of electrons $= 3$ Which are $\mathrm{H_2O}$, $\mathrm{N_2}$ and $\mathrm{CO}$. $\mathrm{XeF_4}$ have 2 lone pairs on central atom, but we are asked about lone pair in molecule.

Hydrolysis of $\mathrm{XeF_4}$ will give $\mathrm{XeO_3}$. This hydrolysis reaction is a redox reaction. This redox reaction is an example of disproportionation reaction. $$6\, \mathrm{XeF_4} + 12\, \mathrm{H_2O} \rightarrow 4\, \mathrm{Xe} + 2\, \mathrm{XeO_3} + 24\, \mathrm{HF} + 3\, \mathrm{O_2}$$ The oxidation states of Xenon in the reactant and product is $$\mathrm{XeO_3} : \mathrm{Xe}(+6)$$ $$\mathrm{XeF_4} : \mathrm{Xe}(+4)$$ Hence, the difference in oxidation state is 2.

The osmotic pressure of a non-electrolytic solution can be calculated as follows, $$\pi = CRT$$ where $\pi$ is the osmotic pressure. $C$ is the molarity, given by $$C = \frac{Number of moles}{Volume of solution in L}$$ Calculating the molarity: $$1.29 \times 10^{-3} = \frac{0.63}{(\mathrm{MW})(0.3)} \times (0.083) \times 300$$ Solving for MW: $$\mathrm{MW} = \frac{0.63 \times 0.083 \times 300}{(0.3) \times (1.29 \times 10^{-3})}$$ This gives: $$= 40534.88$$ Rounding to the nearest integer: $$\approx 40535$$ (Nearest integer)

The magnetic moment of the metal ion can be calculated using $$\mu = \sqrt{n(n+2)} \, BM$$ where $n$ is the number of unpaired electrons. Given $\mu = 4.90 \, BM$, $$4.90 \, BM = \sqrt{n(n+2)} \, BM$$ $$24 = n(n+2)$$ Solving for $n$, we find $$n = 4$$

The difference energy between two energy levels is $$\Delta E = 2.18 \times 10^{-18} \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \, \mathrm{J}$$ $$1.47 \times 10^{-17} = 2.18 \times 10^{-18} \times (9) \left( \frac{1}{(n)^2} - \frac{1}{(n+1)^2} \right)$$ $$0.749 = \frac{1}{n^2} - \frac{1}{(n+1)^2}$$ $$n = 1$$ Hence, the transition is taking place between $1 \rightarrow 2$, and it comes under Lyman series. If we take $0.749 \simeq 0.75 = \frac{3}{4}$

The relation between Molar conductance $\lambda_m$ and specific conductance $\kappa$ is $$\lambda_m = \frac{k \times 1000}{M}$$ $M$ = molarity $$\lambda_m = \frac{\left(5 \times 10^{-5}\right) \times \left(10^3\right)}{2.5 \times 10^{-3}} = 20 \, \mathrm{S \, cm^2 \, mol^{-1}}$$ Now, the degree of dissociation, $\alpha = \frac{\lambda_m}{\lambda_\infty}$ $$\alpha = \frac{20}{400} = \frac{1}{20}$$ The acid dissociation constant, $K_a = \frac{C \alpha^2}{(1-\alpha)} = \frac{\left(2.5 \times 10^{-3}\right) \left(\frac{1}{20} \times \frac{1}{20}\right)}{\left(\frac{19}{20}\right)}$$ $$= 65.789 \times 10^{-7}$$ $$\approx 66 \times 10^{-7}$$

A. $C_0 \xrightarrow{t = \frac{2k}{C_0}} \frac{C_0}{2} \xrightarrow{t = \frac{4k}{C_0}} \frac{C_0}{4} \xrightarrow{t = \frac{8k}{C_0}} \frac{C_0}{8}$ Successive half lives of zero order reactions decreases with time. Note: successive half lives of first order reactions remains same. B. For zero order reactions, the rate of reaction is independent on the concentration of reactants. C. In some reaction molecularity (theoretical value) may be equal to the order of a reaction (experimental value). The order of a reaction may be fractional, zero, positive or negative. Molecularity can never be fractional. (D) Unit of $k = \frac{Units of rate}{[units of concentration]^{order}}$ Zero order $= mollit^{-1} sec^{-1}$ Second order $= mol^{-1} lit^{-1}$