JEE Main 29 July 2022 Shift 2 question paper with solutions

Given $|z - \frac{1}{z}| = 2$. Let $|z| = r$. We have: $$\left| z - \frac{1}{|z|} \right| \leq |z - \frac{1}{z}| \leq |z| + \frac{1}{|z|}$$ This implies: $$\left| r - \frac{1}{r} \right| \leq 2 \leq r + \frac{1}{r}$$ This is always true. Therefore: $$r - \frac{1}{r} \geq -2 and r - \frac{1}{r} \leq 2$$ Solving these inequalities: $$r^2 - 1 \leq 2r$$ $$r^2 - 2r \leq 1$$ $$(r - 1)^2 \leq 2$$ $$r - 1 \leq \sqrt{2}$$ Thus, $|z|_{\max} = 1 + \sqrt{2}$.

Given the matrix $$A = \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}$$. (1) $R_1 \rightarrow R_1 + R_2$; $$\begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}$$ possible. (2) $R_1 \leftrightarrow R_2$; $$\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$$ possible. (3) Option is not possible. (4) $R_2 \rightarrow R_2 + 2R_1$; $$\begin{bmatrix} -1 & 2 \\ -1 & 3 \end{bmatrix}$$ possible.

Given the equations: $$x + y + z = 6 (1)$$ $$2x + 5y + \alpha z = \beta (2)$$ $$x + 2y + 3z = 14 (3)$$ From equation (1), we have $$x + y = 6 - z$$ From equation (3), we have $$x + 2y = 14 - 3z$$ On solving, $$x = z - 2 \implies y = 8 - 2z in (2)$$ Substitute in equation (2): $$2(z - 2) + 5(8 - 2z) + \alpha z = \beta$$ Simplifying, $$(\alpha - 8)z = \beta - 36$$ For having infinite solutions, $$\alpha - 8 = 0 \& \beta - 36 = 0$$ Thus, $$\alpha = 8, \beta = 36 (\alpha + \beta = 44)$$

Given $$f(x) = \begin{cases} \frac{\ln(1+5x) - \ln(1+\alpha x)}{x} & ; x \neq 0 \\ 10 & ; x = 0 \end{cases}$$ We find $$\lim_{x \to 0} \frac{\ln(1+5x) - \ln(1+\alpha x)}{x} = 10$$ Using expansion, $$\lim_{x \to 0} \frac{(5x + \ldots) - (\alpha x + \ldots)}{x} = 10$$ This gives $$5 - \alpha = 10 \implies \alpha = -5$$

Given $$I = \int_0^1 [2x - 13x^2 - 3x - 2x + 2 + 1] \, dx$$ Simplifying, we have $$I = \int_0^1 [2x - 1(3x - 2)(x - 1)] \, dx + \int_0^1 1 \, dx$$ Further simplifying, $$I = \int_0^{2/3} \left[ (2x - (3x^2 - 5x + 2)) \right] \, dx + \int_{2/3}^1 (2x + (3x^2 - 5x + 2)) \, dx + 1$$ Using the graph, we have $$\alpha = \frac{7 - \sqrt{37}}{6}, \beta = \frac{7 - \sqrt{13}}{6}$$ The integral becomes $$\int_0^\alpha (-2) \, dx + \int_\alpha^{1/3} (-1) \, dx + \int_{1/3}^\beta 0 \, dx + \int_\beta^{2/3} 1 \, dx$$ This simplifies to $$= -2\alpha - \left( \frac{1}{3} - \alpha \right) + \frac{2}{3} - \beta = -\alpha - \beta + \frac{1}{3}$$ For $$x \in \left( \frac{2}{3}, 1 \right)$$ We have $$3x^2 - 3x + 2 \in \left( \frac{4}{3}, 2 \right)$$ Thus, $$[3x^2 - 3x + 2] = 1$$ Therefore, $$\int_{2/3}^1 [3x^2 - 3x + 2] \, dx = 1 \left( 1 - \frac{2}{3} \right) = \frac{1}{3}$$ Hence, $$I = \left( \frac{1}{3} - (\alpha + \beta) \right) + \left( \frac{1}{3} \right) + 1$$ This simplifies to $$= \frac{5}{3} - \left( \frac{7 - \sqrt{37}}{6} + \frac{7 - \sqrt{13}}{6} \right)$$

Given $a_0 = 0$, $a_1 = 0$. $$a_{n+2} = 3 \, a_{n+1} - 2 \, a_n : n \geq 0$$ $$a_{n+2} - a_{n+1} = 2 \left( a_{n+1} - a_n \right) + 1$$ For $n = 0$, $a_2 - a_1 = 2 \left( a_1 - a_0 \right) + 1$. For $n = 1$, $a_3 - a_2 = 2 \left( a_2 - a_1 \right) + 1$. For $n = 2$, $a_4 - a_3 = 2 \left( a_3 - a_2 \right) + 1$. For $n = n$, $a_{n+2} - a_{n+1} = 2 \left( a_{n+1} - a_n \right) + 1$. $$(a_{n+2} - a_1) - 2 \left( a_{n+1} - a_0 \right) - (n + 1) = 0$$ $$a_{n+2} = 2 a_{n+1} + (n + 1)$$ Let $n \to n - 2$. $$a_n - 2 a_{n-1} = n - 1$$ Now $a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}$ $$= (a_{25} - 2 a_{24}) (a_{23} - 2 a_{22}) = (24)(22) = 528$$

Given the expression: $$\sum_{x=1}^{20} (r^2 + 1) r!$$ This can be rewritten as: $$\sum_{x=1}^{20} ((r+1)^2 - 2r) r!$$ Which simplifies to: $$\sum_{x=1}^{20} ((r+1)(r+1)! - r \cdot r!) - \sum_{r=1}^{20} r \cdot r!$$ Further simplifying gives: $$\sum_{x=1}^{20} ((r+1)(r+1)! - r \cdot r!) - \sum_{r=1}^{20} ((r+1)! - r!)$$ This results in: $$(21 \cdot 21! - 1) - (21! - 1)$$ Finally, we have: $$= 20 \cdot 21! = 22! - 2 \cdot 21!$$

$I(x)=\int \sec^2x\cdot\sin^{-2022}x\,dx$ Let $I=\sin^{-2022}x,\qquad II=\int \sec^2x\,dx=\tan x$ $= \tan x\cdot\sin^{-2022}x$ $-\int (2022)\tan x\, (\sin x)^{-2023}\cos x\,dx$ $=\tan x\cdot\sin^{-2022}x$ $-2022\int (\sin x)^{-2022}\,dx$ Hence, $I(x)=\tan x\cdot\sin^{-2022}x+C$ At $x=\dfrac{\pi}{4}$, $2^{1011} =\left(\dfrac1{\sqrt2}\right)^{-2022}+C$ $\therefore C=0$ Hence, $I(x)=\dfrac{\tan x}{(\sin x)^{2022}}$ $I\!\left(\dfrac{\pi}{6}\right) =\dfrac1{\sqrt3} \left(\dfrac12\right)^{-2022} =\dfrac{2^{2022}}{\sqrt3}$ $I\!\left(\dfrac{\pi}{3}\right) =\dfrac{\sqrt3} {\left(\dfrac{\sqrt3}{2}\right)^{2022}}$ $=\dfrac{2^{2022}} {(\sqrt3)^{2021}}$ $=\dfrac1{3^{1010}} \left(\dfrac{2^{2022}}{\sqrt3}\right)$ $\therefore 3^{1010}\,I\!\left(\dfrac{\pi}{3}\right) =I\!\left(\dfrac{\pi}{6}\right)$

\[ \frac{dy}{dx} = \frac{x+y-2}{x-y} = \frac{(x-1)+(y-1)}{(x-1)-(y-1)} \] \[ x - 1 = X, \quad y - 1 = Y \] \[ \frac{dY}{dX} = \frac{X+Y}{X-Y} \] \[ Y = VX \qquad \frac{dY}{dX} = V + X\frac{dV}{dX} \] \[ V + X\frac{dV}{dX} = \frac{1+V}{1-V} \qquad X\frac{dV}{dX} = \frac{V^2+1}{1-V} \] \[ \int \frac{1-V}{1+V^2}\,dV = \int \frac{dX}{X} \] \[ \int \frac{dV}{1+V^2} - \frac{1}{2}\int \frac{2V\,dV}{1+V^2} = \int \frac{dX}{X} \] \[ \tan^{-1}V - \frac{1}{2}\ln\left(1+V^2\right) = \ln X + c \] \[ \tan^{-1}\left(\frac{Y}{X}\right) - \frac{1}{2}\ln\left(1+\frac{Y^2}{X^2}\right) = \ln(X) + c \] \[ \tan^{-1}\left(\frac{y-1}{x-1}\right) - \frac{1}{2}\ln\left(1+\frac{(y-1)^2}{(x-1)^2}\right) = \ln(x-1) + c \] Passes through $(2,1)$ \[ 0 - \frac{1}{2}\ln 1 = \ln 1 + c \quad \therefore c = 0 \] Passes through $(k+1, 2)$ \[ \therefore \tan^{-1}\left(\frac{1}{k}\right) - \frac{1}{2}\ln\left(1+\frac{1}{k^2}\right) = \ln k \] \[ \Rightarrow 2\tan^{-1}\left(\frac{1}{k}\right) = \log_e\left(k^2+1\right) \]

Given $\($ $\frac{dy}{dx}$ + $\left$( $\frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11x + 6}$ $\right$) y = $\frac{x+3}{x+1}$ $\)$. $\($ $\int$ p(x) dx $\)$ I.F. = $\($ e^{$\int$ p(x) dx} $\)$ $\($ $\int$ p(x) dx = $\int$ $\left$( $\frac{2x^2 + 11x + 13}{(x+1)(x+2)(x+3)}$ $\right$) dx $\)$ Using partial fraction $\($ $\frac{2x^2 + 11x + 13}{(x+1)(x+2)(x+3)}$ = $\frac{A}{x+1}$ + $\frac{B}{x+2}$ + $\frac{C}{x+3}$ $\)$ $\($ A = $\frac{4}{2}$ = 2 $\)$ $\($ B = 1 $\)$ $\($ C = -1 $\)$ Therefore, $\($ $\int$ p(x) dx = A $\ln$(x+1) + B $\ln$(x+2) + C $\ln$(x+3) $\)$ $\($ = $\ln$ $\left$( $\frac{(x+1)^2(x+2)}{x+3}$ $\right$) $\)$ I.F. = $\($ e^{$\int$ p(x) dx} = $\frac{(x+1)^2(x+2)}{x+3}$ $\)$ Solution $\($ y(IF) = $\int$ Q(IF) dx $\)$ $\($ y $\left$( $\frac{(x+1)^2(x+2)}{x+3}$ $\right$) = $\int$ $\left$( $\frac{x+3}{x+1}$ $\right$) $\left$( $\frac{(x+1)^2(x+2)}{x+3}$ $\right$) dx $\)$ $\($ y $\left$( $\frac{(x+1)^2(x+2)}{x+3}$ $\right$) = $\frac{x^3}{3}$ + $\frac{3x^2}{2}$ + 2x + c $\)$ Passes through $\($ (0, 1) $\)$ $\($ C = $\frac{2}{3}$ $\)$ Now put $\($ x = 1 $\)$ $\($ $\Rightarrow$ y(1) = $\frac{3}{2}$ $\)$

Given $m_1 m_2 = -1$. $$a^2 + 11a + 3 \left( m_1^2 + \frac{1}{m_1^2} \right) = 220$$ Equation of AC: $$AC = (\cos \alpha - \sin \alpha) + (\sin \alpha + \cos \alpha) \, y = 10$$ $$BD = (\sin \alpha - \cos \alpha) \, x + (\sin \alpha - \cos \alpha) \, y = 0$$ $$(10 \, (\cos \alpha - \sin \alpha), 10 \, (\sin \alpha - \cos \alpha))$$ Slope of AC: $$\left( \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} \right) = \tan \theta = M$$ Equation of line making an angle $\frac{\pi}{4}$ with AC: $$m_1, m_2 = \frac{m \pm \tan \frac{\pi}{4}}{1 \mp m \tan \frac{\pi}{4}}$$ $$= \frac{m + 1}{1 - m} or \frac{m - 1}{1 + m}$$ $$= \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}, \frac{\sin \alpha - \cos \alpha - 1}{\sin \alpha + \cos \alpha}$$ $$1 - \left( \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} \right), \frac{1 + \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}}{5(\cos \alpha + \sin \alpha)}$$ $$B \, (10(\cos \alpha - \sin \alpha), 10(\cos \alpha + \sin \alpha))$$ $$a = AB = \sqrt{2} \, BM = \sqrt{2} \, (5 \sqrt{2}) = 10$$ $$a = 10$$ Therefore, $$a^2 + 11a + 3 \left( m_1^2 + \frac{1}{m_1^2} \right) = 220$$ $$100 + 110 + 3 \left( \tan^2 \alpha + \cot^2 \alpha \right) = 220$$ Hence $\tan^2 \alpha = 3$, $\tan^2 \alpha = \frac{1}{3}$ $\implies \alpha = \frac{\pi}{3} or \frac{\pi}{6}$ Now $72 \left( \sin^4 \alpha + \cos^4 \alpha \right) + a^2 - 3a + 13$ $$= 72 \left( \frac{9}{16} + \frac{1}{16} \right) + 100 - 30 + 13$$ $$= 72 \left( \frac{5}{8} \right) + 83 = 45 + 83 = 128$$

Given $$2 \cos \left( \frac{x^2 + x}{6} \right) = 4^x + 4^{-x}$$ L.H.S $\leq$ 2 and R.H.S $\geq$ 2. Hence L.H.S = 2 and R.H.S = 2. $$2 \cos \left( \frac{x^2 + x}{6} \right) = 2 4^x + 4^{-x} = 2$$ Check $x = 0$. Possible, hence only one solution.

$A(\alpha, -2)$, $B(\alpha, 6)$, $C\left(\dfrac{\alpha}{4}, -2\right)$ Since $AC$ is perpendicular to $AB$, $\triangle ABC$ is right angled at $A$. Circumcentre $=$ midpoint of $BC = \left(\dfrac{5\alpha}{8},\ 2\right)$ $$\therefore\ \frac{5\alpha}{8} = 5 \quad \text{and} \quad \frac{\alpha}{4} = 2$$ $\alpha = 8$ $$B = (8, 6)$$ $\text{Area} = \dfrac{1}{2}(6)(8) = 24$ $\text{Perimeter} = 24$ $\text{Circumradius} = 5$ $\text{Inradius} = \dfrac{\Delta}{s} = \dfrac{24}{12} = 2$

Length of perpendicular $$PQ = \frac{|1 + 4 + 3 - 14|}{\sqrt{6}} = \sqrt{6}$$ $$QR = (PQ) \cot 60^\circ = \sqrt{2}$$ Therefore, the area of $\triangle PQR = \frac{1}{2} (PQ)(QR) = \sqrt{3}$

Given points A(2, 3, 9), B(5, 2, 1), C(1, $\lambda$, 8), and D($\lambda$, 2, 3). The condition is that the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$ are coplanar, which implies: $$\begin{vmatrix} 3 & -1 & -8 \\ -1 & \lambda - 3 & -1 \\ \lambda - 2 & -1 & -6 \end{vmatrix} = 0$$ Expanding the determinant, we have: $$[-6(\lambda - 3) - 1] - 8(1 - (\lambda - 3)(\lambda - 2)) + (6 + (\lambda - 2)) = 0$$ Simplifying, we get: $$3(-6\lambda + 17) - 8(-\lambda^2 + 5\lambda - 5) + (\lambda + 4) = 8$$ This simplifies to: $$8\lambda^2 - 57\lambda + 95 = 0$$ The product of the roots is given by: $$\lambda_1 \lambda_2 = \frac{95}{8}$$

Given: A: Drown ball from boy II is black B: Red ball transferred The probability is given by: $$ \mathrm{P} \left( \frac{B}{A} \right) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(A)} $$ Calculating: $$ = \frac{\frac{3}{9} \times \frac{5}{10}}{\frac{3}{9} \times \frac{5}{10} + \frac{4}{9} \times \frac{6}{10} + \frac{3}{9} \times \frac{5}{10}} $$ Simplifying: $$ = \frac{15}{15 + 24 + 15} = \frac{15}{54} = \frac{5}{18} $$

Given $|z - 1 + i| \geq |z|$; $|z| < 2$; $|z + i| = |z - 1|$. Hence $w = 2x + iy \in S$. $2x \leq \frac{1}{2} \therefore x \leq \frac{1}{4}$. Now $$(2x)^2 + (2x)^2 < 4$$ $$x^2 < \frac{1}{2} \implies x \in \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$$ $$\therefore x \in \left( -\frac{1}{\sqrt{2}}, \frac{1}{4} \right]$$

Given $|\vec{a} \parallel \vec{b} \parallel \vec{c}| = 14$. $\vec{a} \wedge \vec{b} = \vec{b} \wedge \vec{c} = \vec{c} \wedge \vec{a} = \theta = \frac{2\pi}{3}$. So, $\vec{a} \cdot \vec{b} = -\frac{1}{2} ab$, $\vec{b} \cdot \vec{c} = -\frac{1}{2} bc$, $\vec{a} \cdot \vec{c} = -\frac{1}{2} ac$. Let $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b})$. $$= \frac{1}{4} ab^2 c + \frac{1}{2} ab^2 c = \frac{3}{4} ab^2 c$$ Similarly, $$(\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = \frac{3}{4} abc^2$$ $$(\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = \frac{3}{4} a^2 bc$$ $$168 = \frac{3}{4} abc(a + b + c)$$ So, $(a + b + c) = 16$.

Given $f(x) = \sin^{-1}\left( \frac{x^2 - 3x + 2}{x^2 + 2x + 7} \right)$. Domain $$\frac{x^2 - 3x + 2}{x^2 + 2x + 7} \geq -1 and \frac{x^2 - 3x + 2}{x^2 + 2x + 7} \leq 1$$ $$2x^2 - x + 9 \geq 0 and 5x - 5 \Rightarrow x \geq -1$$ $x \in \mathbb{R}$ Hence Domain $x \in [-1, \infty)$

(p $\rightarrow$ q) $\lor$ (p $\rightarrow$ r) ($\sim$ p $\lor$ q) $\lor$ ($\sim$ p $\lor$ r) = $\sim$ p $\lor$ (q $\lor$ r) = p $\rightarrow$ (q $\lor$ r) $\equiv$ (3) is true. Now (1) (p $\land$ $\sim$ r) $\rightarrow$ q $\sim$ (p $\land$ $\sim$ r) $\lor$ q = ($\sim$ p $\lor$ r) $\lor$ q = $\sim$ p $\lor$ (r $\lor$ q) = p $\rightarrow$ (q $\lor$ r) (4) (p $\land$ $\sim$ q) $\rightarrow$ r = p $\rightarrow$ (q $\lor$ r)

Let, mean = m = np and variance = v = npq, p + q = 1. Sum = m + v = $\frac{165}{2}$. Product = mv = 1350. On solving, m = np = 60 and v = npq = $\frac{45}{2}$. Therefore, q = $\frac{3}{8}$ and P = $\frac{5}{8}$. Hence n = 96.

Given $P_n = \alpha^n - \beta^n$ and $x^2 - x - 4 = 0$. $$\frac{P_{15}P_{16} - P_{14}P_{16} - P_{15}^2 + P_{14}P_{15}}{P_{13}P_{14}} (1)$$ As $P_n - P_{n-1} = (\alpha^n - \beta^n) - (\alpha^{n-1} - \beta^{n-1})$ $$= \alpha^{n-2}(\alpha^2 - \alpha) - \beta^{n-2}(\beta^2 - \beta)$$ $$= 4(\alpha^{n-2} - \beta^{n-2})$$ $$P_n - P_{n-1} = 4 \, P_{n-2}$$ Hence Expression (1) $$\frac{P_{16}(P_{15} - P_{14}) - P_{15}(P_{15} - P_{14})}{P_{13}P_{14}}$$ $$= \frac{(P_{15} - P_{14})(P_{16} - P_{15})}{P_{13}P_{14}} = \frac{(4P_{13})(4P_{14})}{P_{13}P_{14}} = 16$$

Given the matrix equation: $\begin{bmatrix} 1&1&1 \end{bmatrix} \begin{bmatrix} -1&2&3\\ 0&1&6\\ 0&0&-1 \end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} =33$ As $A^2$ $= \begin{bmatrix} -1&2&3\\ 0&1&6\\ 0&0&-1 \end{bmatrix} \begin{bmatrix} -1&2&3\\ 0&1&6\\ 0&0&-1 \end{bmatrix}$ $= \begin{bmatrix} 1&0&6\\ 0&1&0\\ 0&0&1 \end{bmatrix}$ $A^4$ $=A^2A^2$ $= \begin{bmatrix} 1&0&6\\ 0&1&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0&6\\ 0&1&0\\ 0&0&1 \end{bmatrix}$ $= \begin{bmatrix} 1&0&12\\ 0&1&0\\ 0&0&1 \end{bmatrix}$ $A^8$ $=A^4A^4$ $= \begin{bmatrix} 1&0&12\\ 0&1&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0&12\\ 0&1&0\\ 0&0&1 \end{bmatrix}$ $= \begin{bmatrix} 1&0&24\\ 0&1&0\\ 0&0&1 \end{bmatrix}$ $A^{10}$ $=A^2A^8$ $= \begin{bmatrix} 1&0&6\\ 0&1&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0&24\\ 0&1&0\\ 0&0&1 \end{bmatrix}$ $= \begin{bmatrix} 1&0&30\\ 0&1&0\\ 0&0&1 \end{bmatrix}$ For $K$ Even $A^K= \begin{bmatrix} 1&0&3K\\ 0&1&0\\ 0&0&1 \end{bmatrix}$ $X^TA^KX=33$ (This is not correct) $\begin{bmatrix} 1&1&1 \end{bmatrix} \begin{bmatrix} 1&0&3K\\ 0&1&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ Therefore, $3K+3=33$ $\therefore\ K=10$ But it should be dropped as $33$ is not a matrix. If $K$ is odd $X^TA^KX=33$ $X^TAA^{K-1}X=33$ $\begin{bmatrix} 1&1&1 \end{bmatrix} \begin{bmatrix} -1&2&3\\ 0&1&6\\ 0&0&-1 \end{bmatrix} \begin{bmatrix} 1&0&3K-3\\ 0&1&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} =33$ $\begin{bmatrix} -1&3&8 \end{bmatrix} \begin{bmatrix} 3K-2\\ 1\\ 1 \end{bmatrix} =[33]$ $[-3K+13]=[33]$ $K=\dfrac{20}{3}$ (not possible)

4 digit numbers For divisibility by 55, the number should be divisible by 5 and 11 both. Also, for divisibility by 11: $$a + c = b + 5$$ For $b = 1$: $a = 2$, $c = 4$ $a = 4$, $c = 2$ For $b = 2$: $a = 3$, $c = 4$ $a = 4$, $c = 3$ For $b = 3$: $a = 6$, $c = 2$ $a = 2$, $c = 6$ Therefore, 6 possible four digit numbers are divisible by 55. (II) 5 digit number is not possible. (Not possible)

Given $$\sum_{K=1}^{10} K^2 \left( \binom{10}{K} \right)^2$$ This is equal to $$\sum_{K=1}^{10} \left( K \cdot \binom{10}{K} \right)^2 = \sum_{K=1}^{10} \left( 10 \cdot \binom{9}{K-1} \right)^2$$ This simplifies to $$= 100 \sum_{K=1}^{10} \binom{9}{K-1} \cdot \binom{9}{10-K}$$ Which is equal to $$= 100 \binom{18}{9} = 100 \left( \frac{18!}{9!9!} \right)$$ Thus, $$\Rightarrow 4862000 = 22000L$$ Hence, $$L = 221$$

Given $f(x) = 4|2x + 3| + 9\left[x + \frac{1}{2}\right] - 12[x + 20]$. $x \in (-20, 20)$. $f(x)$ is not differentiable at $x = I \in \{-19, -18, \ldots, 0, \ldots, 19\} = 39$ points. At $x = I + \frac{1}{2}$, $f(x)$ is non-differentiable at 39 points. Check at $x = \frac{-3}{2}$. Discount at $x = \frac{-3}{2}$. Therefore, $N. R(1)$. Number of points of non-differentiability: $$= 39 + 39 + 1 = 79$$

Given $y = x^3 - x^2 + x$. At point $P(2, -1)$, the derivative $\frac{dy}{dx}$ is $22$. Therefore, the tangent to the curve at $P$ is given by: $$y + 1 = 22(x - 2)$$ Simplifying, we get: $$y = 22x - 45$$ For the curve $y = x^3 - x^2 + x$, the derivative is: $$\frac{dy}{dx} \bigg|_{C_2} = 3x^2 - 2x + 1$$ At point $Q(a, b)$, the derivative is: $$\frac{dy}{dx} \bigg|_Q = 3a^2 - 2a + 1$$ Setting this equal to $22$, we have: $$3a^2 - 2a + 1 = 22$$ Simplifying, we get: $$3a^2 - 2a - 21 = 0$$ Factoring, we find: $$3a^2 - 9a + 7a - 21 = 0$$ Solving, we find $a = 3$ and $b = 21$. The equation $2a + 9b = 195$ is satisfied. At $a = -7/3$, the tangent will be parallel, hence it is rejected.

Given the diagram, we have: $$\cos \theta = \frac{6}{\frac{13}{2}} = \frac{12}{13}$$ $$\sin \theta = \frac{5}{13}$$ The relation for PM is given by: $$\mathrm{PM} = \mathrm{AM} \cot \theta$$ Substituting the values, we get: $$\mathrm{PM} = 6 \left( \frac{12}{5} \right) \therefore 5(\mathrm{PM}) = 72$$

Given $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2$; $\vec{a} \cdot \vec{b} = 3$. As $|\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} = |\vec{a}|^2 + 2|\vec{b}|^2$. $$|\vec{b}|^2 = 2 \vec{a} \cdot \vec{b} = 6$$ $$|\vec{a} \times \vec{b}|^2 = 75$$ $$|\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = 75$$ $$6 |\vec{a}|^2 - 9 = 75 \implies |\vec{a}|^2 = 14$$

Given the set S: $\($ $\frac{(x-3)^2}{16}$ + $\frac{(y-4)^2}{9}$ $\leq$ 1 $\)$; $\($ x, y $\in$ $\{$1, 2, 3, $\ldots$$\}$ $\)$. Set T: $\($ (x-7)^2 + (y-4)^2 $\leq$ 36 $\)$; $\($ x, y $\in$ $\mathbb{R}$ $\)$. Let $\($ x-3 = x : y-4 = y $\)$. The set S: $\($ $\frac{x^2}{16}$ + $\frac{y^2}{9}$ $\leq$ 1 $\)$; $\($ x $\in$ $\{$-2, -1, 0, 1, $\ldots$$\}$ $\)$. The set T: $\($ (x-4)^2 + y^2 $\leq$ 36 $\)$; $\($ y $\in$ $\{$-3, -2, -1, 0, $\ldots$$\}$ $\)$. The intersection $\($ S $\cap$ T = (-2, 0), (-1, 0), $\ldots$, (4, 0) $\rightarrow$ (7) $\)$. $\($(-1, 1), (0, 1), $\ldots$, (3, 1) $\rightarrow$ (5) $\)$. $\($(-1, -1), (0, -1), $\ldots$, (3, -1) $\rightarrow$ (5) $\)$. $\($(-1, 2), (0, 2), (1, 2), (2, 2) $\rightarrow$ (4) $\)$. $\($(-1, -2), (0, -2), (1, -2), (2, -2) $\rightarrow$ (4) $\)$. $\($(0, 3) (0, -3) $\rightarrow$ (2) $\)$.

Let $q_A = q_B = q$. $$F = \frac{Kq^2}{r^2}$$ When C is placed in contact with A, charge on A and C will be $\frac{q}{2}$. Now C is placed in contact with B, charge on B and C will be $\frac{q + \frac{q}{2}}{2} = \frac{3q}{4}$. Now, $$\frac{q}{2} \overset{F_1}{\longleftarrow} A \overset{F_2}{\longrightarrow} C \overset{F_2}{\longrightarrow} B \frac{3q}{4}$$ $$F' = F_2 - F_1 = \left( \frac{K \frac{3q}{4} - K \frac{q}{2}}{\frac{r^2}{4}} \right) \cdot \frac{3q}{4}$$ $$= \frac{3Kq^2}{4r^2} = \frac{3F}{4} (B)$$

Torque is given by $Torque = F \times r_\perp$ and is measured in Nm. Stress is given by $Stress = \frac{Force}{Area}$ and is measured in N/m$^2$. Latent heat is given by $Latent heat = \frac{Energy}{Mass}$ and is measured in J Kg$^{-1}$. Power is given by $Power = \frac{Work}{Time}$ and is measured in N ms$^{-1}$. The correct matching is A-III, B-IV, C-II, D-I.

Electric field between plates is given by $E = \frac{q_1 - q_2}{2A \epsilon_0}$. The potential $V$ is given by $V = Ed = \frac{q_1 - q_2}{2A \epsilon_0} d$. Thus, $V = \frac{q_1 - q_2}{2C}$.

Theory based

Given $\($ $\frac{R_X}{R_Y}$ = $\frac{\ell_X}{\ell_Y}$ $\)$. When wire is stretched to double of its length, then resistance becomes 4 times. $\($ R_W = 4R_X = 2R_Y $\)$. $\($ $\frac{R_X}{R_Y}$ = $\frac{1}{2}$ $\)$. So, $\($ $\frac{\ell_X}{\ell_Y}$ = $\frac{1}{2}$ $\)$.

Force of interaction = $I_1 \, \ell_1 \, B_{12}$ $$= \frac{\mu_0 I_1 I_2}{2 \pi r} \ell_1$$ $$= \frac{4 \pi \times 10^{-7} \times 6 \times 0.5}{2 \pi \times 5 \times 10^{-2}}$$ $$= 1.2 \times 10^{-5} towards X$$

Time taken by ball to reach highest point is $\frac{u}{g}$. Frequency of throw is $\frac{g}{u} = n$. Therefore, $u = \frac{g}{n}$. The maximum height $H_{max}$ is given by $$H_{max} = \frac{u^2}{2g} = \frac{\left(\frac{g}{n}\right)^2}{2g}$$ which simplifies to $$\frac{g}{2n^2}.$$

Element X should be resistive with $R = 20 \, \Omega$. Element Y should be inductive with $X_L = 20 \, \Omega$. When X and Y are connected in series, $$Z = \sqrt{X_L^2 + R^2} = 20 \sqrt{2}$$ $$I_0 = \frac{E_0}{Z} = \frac{100}{20 \sqrt{2}} = \frac{5}{\sqrt{2}} \, \mathrm{A}$$ $$I_{\mathrm{rms}} = \frac{I_0}{\sqrt{2}} = \frac{5}{2} \, \mathrm{A}$$

Given $I_1 = \frac{1}{2}(2I_0) = I_0$. $I_2 = I_1 \cos^2 30^\circ$. Therefore, $I_2 = I_0 \cdot \frac{3}{4} = \frac{3I_0}{4}$.

Given $p = \sqrt{2mE} = \sqrt{2mqV}$. The ratio $\frac{p_\alpha}{p_p}$ is given by $$\frac{p_\alpha}{p_p} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\frac{4}{1} \times \frac{2}{1}}$$ This simplifies to $$= \frac{2\sqrt{2}}{1}$$

Given $R \propto A^{1/3}$. $$V = \frac{4}{3} \pi R^3 \propto A$$ Mass $\propto A$. So density is independent of $A$.

The initial potential energy is given by $$U_i = -\frac{GMm}{R}$$ The final potential energy is $$U_f = -\frac{GMm}{4R}$$ The change in potential energy is $$\Delta U = U_f - U_i = \frac{3GMm}{4R}$$ This is equal to $$= \frac{3}{4} mgR$$ Substituting the values, $$= \frac{3}{4} \times 1 \times 10 \times 64 \times 10^5$$ This results in $$= 48 \, MJ$$

For first $\frac{h}{2}$ $$\frac{h}{2} = \frac{1}{2} g t_1^2$$ For total height $h$ $$h = \frac{1}{2} g (t_1 + t_2)^2$$ $$\frac{1}{\sqrt{2}} = \frac{t_1}{t_1 + t_2}$$ $$1 + \frac{t_2}{t_1} = \sqrt{2}$$ $$\frac{t_1}{t_2} = \frac{1}{\sqrt{2} - 1}$$

For equilibrium $m_2 g = m_1 g \sin \theta$. $$\sin \theta = \frac{m_2}{m_1} = \frac{3}{5}$$ $$\cos \theta = \frac{4}{5}$$ Normal force on $m_1 = 5g \cos \theta$ $$= 5 \times 10 \times \frac{4}{5} = 40 \, \mathrm{N}$$

P' = P + $\frac{20}{100}$ P = 1.2 P $\%$ change in KE = $\frac{K' - K}{K}$ $\times$ 100 = $\left$( $\frac{\frac{P'^2}{2m} - \frac{P^2}{2m}}{\frac{P^2}{2m}}$ $\right$) $\times$ 100 = [(1.2)^2 - 1] $\times$ 100 = 44 $\%$

Given $\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 5 & 3 & -7 \end{vmatrix}$. This expands to $= \hat{i}(2(-7) - 1 \cdot 3) - \hat{j}(2(-7) - 1 \cdot 5) + \hat{k}(2 \cdot 3 - 2 \cdot 5)$. Simplifying, we get $= \hat{i}(-14 - 3) - \hat{j}(-14 - 5) + \hat{k}(6 - 10)$. Thus, $\vec{\tau} = -17\hat{i} + 19\hat{j} - 4\hat{k}$.

Given $\delta = 37^\circ$. The vertical component $B_V = B \sin \delta$. $$6 \times 10^{-5} = B \frac{3}{5}$$ Solving for $B$, we have: $$B = 10 \times 10^{-5} \, \mathrm{T}$$ Thus, $B = 10^{-4} \, \mathrm{T}$.

The root mean square velocity $V_{rms}$ is given by the equation: $$V_{rms} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 293}{5 \times 10^{-17}}}$$ This is approximately equal to $15 \, mm/s$.

Given $i = 45^\circ$. The deviation $D = i - r$. So, $15^\circ = 45 - r \Rightarrow r = 30^\circ$. Using Snell's law, $n_1 \sin i = n_2 \sin r$. Therefore, $1 \cdot \sin 45^\circ = \mu \cdot \sin 30^\circ$. This gives $\frac{1}{\sqrt{2}} = \mu \cdot \frac{1}{2}$. Solving for $\mu$, we get $\mu = \sqrt{2} = 1.414$.

The force is given by the integral of the differential mass times the square of the angular velocity times the distance. $$F = \int (dm) \omega^2 x$$ Substituting the limits from 0 to $L$, we have: $$= \frac{m}{L} \omega^2 \frac{L^2}{2}$$ Simplifying, $$= \frac{m \omega^2 L}{2}$$ Solving for $\omega$, $$\omega = \sqrt{\frac{2}{mL}} \sqrt{F}$$ Substituting the given values, $$= \sqrt{\frac{2}{0.25 \times 0.5}} \sqrt{F}$$ Simplifying further, $$= \sqrt{16} \sqrt{F}$$ Finally, $$= 4 \sqrt{F}$$

Given $P' = 10\%$ of $110 \, \mathrm{W}$. $$= \frac{10}{100} \times 110 \, \mathrm{W}$$ $$= 11 \, \mathrm{W}$$ $I_1 - I_2 = \frac{P'}{4 \pi r_1^2} - \frac{P'}{4 \pi r_2^2}$ $$= \frac{11}{4 \pi} \left[ \frac{1}{1} - \frac{1}{25} \right]$$ $$= \frac{11}{4 \pi} \times \frac{24}{25}$$ $$= \frac{264}{\pi} \times 10^{-2} = 84 \times 10^{-2} \, \mathrm{W/m^2}$$

Given $T = \frac{mv^2}{\ell} = \frac{10 \times v^2}{0.5} = 20v^2$. $T_{max} = Breaking stress \times Area$ $= 5 \times 10^8 \times 10^{-4} = 5 \times 10^4$. $20V^2 = 5 \times 10^4$ $V = \sqrt{\frac{1}{4} \times 10^4} = 50 \, \mathrm{m/s}$

Given $F_V + F_B = mg$ (v = constant). $F_V = mg - F_B$ $= \rho_B V g - \rho_L V g$ $= (\rho_B - \rho_L) V g$ $= (8 - 1.3) \times 10^3 \times \frac{0.3 \times 10^{-3}}{8 \times 10^3} \times 10$ $= \frac{6.7 \times 0.3}{8} \times 10^{-2}$ (g = 10) $= \frac{67 \times 3}{8} \times 10^{-4} = 25.125 \times 10^{-4}$ Ans. 25.125

Frequencies present in output of square law device $2f_c$, $f_c + f_m$, $f_c$, $f_c - f_m$, $2f_m$, $f_m$. After passing through band bass filter: $f_c + f_m$, $f_c$, $f_c - f_m$. Band width $= 2f_m$ $$= \frac{2\omega_m}{2\pi} = \frac{6.28 \times 10^6}{3.14}$$ $$= 2 \, MHz$$

Given $V_w = \sqrt{\frac{T}{\mu}}$. $$60 = \sqrt{\frac{T}{10 \times 10^{-3}}} \times 0.5$$ $$T = \frac{(60)^2 \times 10^{-2}}{0.5} = 72 \, \mathrm{N}$$ $$\Delta \ell = \frac{F \ell}{AY} = \frac{72 \times 0.5}{2 \times 10^{-6} \times 1.2 \times 10^{11}}$$ $$= \frac{72 \times 5}{24} \times 10^{-5} = 15 \times 10^{-5}$$ Ans. 15

The equation for the effective weight is given by: $$mg' = mg - F_B$$ where $F_B$ is the buoyant force. Therefore, the effective acceleration $g'$ is: $$g' = \frac{mg - F_B}{m}$$ Substituting the expression for the buoyant force, we have: $$g' = \frac{\rho_B V g - \rho_w V g}{\rho_B V}$$ Simplifying, we get: $$g' = \left( \frac{\rho_B - \rho_w}{\rho_B} \right) g$$ Substituting the given values: $$g' = \frac{5 - 1}{5} \times g$$ $$g' = \frac{4}{5} g$$ The period $T$ is given by: $$T = 2\pi \sqrt{\frac{\ell}{g}}$$ The new period $T'$ is: $$\frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{\frac{g}{\frac{4}{5}g}} = \sqrt{\frac{5}{4}}$$ Therefore: $$T' = 5\sqrt{5}$$

Given the equation $\varepsilon - IR - V_z = 0$. Substituting the values, we have: $$20 - IR - 6 = 0$$ Solving for $IR$, we get: $$IR = 12$$ Given $I = 25 \times 10^{-3}$, we substitute to find $R$: $$25 \times 10^{-3} R = 12$$ Solving for $R$, we have: $$R = \frac{12}{25 \times 10^{-3}} = 480 \, \Omega$$

\[ N=N_0e^{-\lambda t} \] $\frac{N_B}{N_A}$ = $\frac{e^{-\lambda_2 t}}{e^{-\lambda_1 t}}$ = $e^{-\lambda_2 t}e^{\lambda_1 t}$ $e^1$ = $e^{(\lambda_1-\lambda_2)t}$ $(\lambda_1-\lambda_2)$t=1 t = $\frac{1}{\lambda_1-\lambda_2}$ = $\frac{1}{25\lambda-16\lambda}$ = $\frac{1}{9\lambda}$

Energy stored in capacitor $$= \frac{1}{2} CV^2 = \frac{1}{2} \times 500 \times 10^{-6} \times 10^4$$ $$= \frac{5}{2} \, \mathrm{J}$$ Current will be maximum when whole energy of capacitor becomes energy of inductor. $$\frac{1}{2} LI^2 = \frac{5}{2}$$ $$I = \sqrt{\frac{5}{L}} = \sqrt{\frac{5}{50 \times 10^{-3}}} = 10 \, \mathrm{A}.$$

$4\mathrm{HNO_3}(l)$+$3\mathrm{KCl}(s)$ $\rightarrow$ $\mathrm{Cl_2}(g)$+$\mathrm{NOCl}(g)$+$2\mathrm{H_2O}(g)$+$3\mathrm{KNO_3}(s)$

Energy order of subshell decided by $(n+\lambda)$ rule. A $\Rightarrow$ 3d $\Rightarrow n + 1 = 5$ B $\Rightarrow$ 4p $\Rightarrow n + \lambda = 5$ C $\Rightarrow$ 4d $\Rightarrow n + \ell = 6$ D $\Rightarrow$ 3s $\Rightarrow (n+\ell) = 4$ D < A < B < C

Given the reactions: $$\mathrm{C(S) + O_2(g) \rightarrow CO_2(g) + 400 \, kJ}$$ 1 g mole $$\mathrm{C(s) + \frac{1}{2} O_2 (g) \rightarrow CO(g) + 100 \, kJ} (II)$$ Calculating the mass: $$0.6 \times 1000 = 600 \, \mathrm{gm}$$ $$600 \times \frac{60}{100} (Pure Carbon)$$ $$= 360 \, \mathrm{gm} = \frac{360}{12} = 30 \, mole (Pure Carbon)$$ Carbon converted into $\mathrm{CO_2}$: $$= \left( 30 - 30 \times \frac{60}{100} \right)$$ $$= 12 \, mole$$ And carbon converted into $\mathrm{CO}$: $$= 30 \times \frac{60}{100} = 18 \, mole$$ Energy generated during II equation: $$= 18 \times 100$$ $$= 1800 \, \mathrm{kJ}$$ Energy generated during I reaction: $$= 12 \times 400$$ $$= 4800$$ Total: $$= 1800 + 4800 = 6600 \, \mathrm{kJ}$$

For the reaction of HCl and $\mathrm{H_2SO_4}$, the concentration of $[\mathrm{H^+}]$ is calculated as follows: $$[\mathrm{H^+}] = \frac{(0.01 \times 200) + (0.01 \times 2 \times 400)}{600}$$ Simplifying the expression: $$= \frac{2 + 8}{600} = \frac{10}{600} = \frac{1}{60}$$ The pH is given by: $$\mathrm{pH} = -\log \left[ \frac{1}{60} \right]$$ Calculating the pH: $$= 1.78$$

More the critical temp. of gas greater is the ease of liquefaction hence greater is the adsorption.

Liquation process is used for metal having low melting point such as tin in which they are heated and brought to molten state and made to flow down the slope while impurities with higher melting point left on the top.

SnH$_4$ is non planar molecular hydride. Tetrahedral shape, sp$^3$ hybridisation.

Gypsum ($\mathrm{CaSO_4.2H_2O}$) is used to enhance setting time in portland cement.

$Na_2B_4O_7\cdot10H_2O \xrightarrow{\Delta} Na_2B_4O_7+10H_2O$ $Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2\ \text{(sodium metaborate)}+B_2O_3$ $B_2O_3+CoO \longrightarrow Co(BO_2)_2\ \text{(cobalt (II) metaborate)}$ Blue Bead

Enthalpy of hydration increases with increase in charge density. $\mathrm{Mn^{+2}}$ has least charge density (as Mn has highest size among the given options) so it will have least enthalpy of hydration.

There is unsymmetric filling of $e_g$ subset of $\mathrm{Cu}^{+2}$ ion, while there is symmetrical distribution in $t_{2g}$ set, if the complex has same ligand there will be equal repulsion which leads to symmetrical bond length along $t_{2g}$, but due to uneven filling of electron in $e_g$ subset, either octahedral will be elongated or compressed.

The reaction is given by: $$\mathrm{N_2(g) + O_2(g) \rightarrow 2NO(g)}$$ Then: $$\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)}$$ $\mathrm{NO_2}$ damages plant leaves.

The compound shown is γ-methyl cyclohexane carbaldehyde.

Having same configuration.

The reaction starts with $\mathrm{CH_3CH_2OH}$ which is treated with concentrated $\mathrm{H_2SO_4}$ and heat $\Delta$ to form $\mathrm{CH_2=CH_2}$. This is then treated with Bayer's Reagent to form glycol, $\mathrm{CH_2(OH)CH_2(OH)}$.

B.S.C (Benzene sulphonyl chloride) is known as Hinsberg Reagent.

Rayon is semisynthetic polymer.

Amylose is water soluble.

Phenolphthalein is a weak acid and gives color in a basic medium.

Given $\mathrm{O_2 + Ne}$ with $x \, \mathrm{gm}$ and $200 \, \mathrm{gm}$. The total pressure $P_{total} = 25 \, \mathrm{bar}$ and $P_{\mathrm{Ne}} = 20 \, \mathrm{bar}$. The pressure of $\mathrm{O_2}$ is given by: $$P_{\mathrm{O_2}} + P_{\mathrm{Ne}} = 25$$ $$P_{\mathrm{O_2}} = 25 - 20 = 5 \, \mathrm{bar}$$ Using the equation: $$\frac{x}{32} + \frac{200}{20}$$ We have: $$\frac{1}{5} = \frac{\frac{x}{32}}{\frac{x}{32} + 10}$$ Simplifying: $$1 = \frac{x \times 32}{32(x + 320)}$$ Solving for $x$: $$5x = x + 320$$ $$4x = 320$$ $$x = \frac{320}{4} = 80 \, \mathrm{gm}$$

Mass of $\mathrm{C_2H_5OH} = 62.5 \times 0.8 = 50 \, \mathrm{g}$. $\Delta T_f = K_f \times m$ $$0.9 = 2 \times \frac{1.8 \times 1000}{M_w \times 50}$$ $$M_w = \frac{2 \times 1.8 \times 1000}{0.9 \times 50} = 80$$

At anode $$\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2e^-$$ At cathode $$2\mathrm{Ag}^+ + 2e^- \rightarrow 2\mathrm{Ag}$$ Cell reaction $$\mathrm{Cu} + 2\mathrm{Ag}^+ \rightarrow \mathrm{Cu}^{2+} + 2\mathrm{Ag}$$ $$E_{cell} = E^0_{cell} - \frac{0.06}{2} \log \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^+]^2}$$ $$0.43 = E^0_{cell} - \frac{0.06}{2} \log \frac{(0.001)}{(0.01)^2}$$ $$E^0_{cell} = 0.46$$ $$E^0_{cell} = E^0_{\mathrm{Ag}^+/\mathrm{Ag}} - E^0_{\mathrm{Cu}^{2+}/\mathrm{Cu}}$$ $$0.46 = 0.80 - E^0_{\mathrm{Cu}^{2+}/\mathrm{Cu}}$$ $$E^0_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.34 \, volt$$ $$E^0_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 34 \times 10^{-2}$$

Given the equation $$t = \frac{1}{\lambda} \ln \left( \frac{a}{a-x} \right)$$. We have $$100 = \frac{30}{\ln 2} \ln \left( \frac{1}{w} \right)$$. Solving for $$w$$, we get $$\frac{1}{w} = 10$$. Therefore, $$W = 0.1 \times \mu g$$. The answer is $$1 \times 10^{-1} \mu g$$.

Coordination no. = 6 Oxidation state = 3 $$6 + 3 = 9$$

R(OH)_x $\rightarrow$ $\mathrm{H}$_2 PoAC on H: $$x \left( \frac{1.84 \times 10^{-3}}{92} \right) = \frac{1.344}{22.4} \times 2$$ $$x = \frac{1.344 \times 2 \times 92 \times 1000}{1.84 \times 22400} = 6$$ x = 6

The reaction of the given compound with HCN results in the formation of a compound with 3 stereoisomers.

Bithinol contains two benzene rings connected by a sulfur atom. Each benzene ring has two chlorine atoms attached. Therefore, the total number of chlorine atoms is 4.