JEE Main 16 March 2021 Shift 2 question paper with solutions

Perform the column operation $C_1 + C_2 \rightarrow C_1$ on the matrix: $$\begin{vmatrix} 2 & 1 + \cos^2 x & \cos 2x \\ 2 & \cos^2 x & \cos 2x \\ 1 & \cos^2 x & \sin 2x \end{vmatrix}$$ Next, perform the row operation $R_1 - R_2 \rightarrow R_1$: $$\begin{vmatrix} 0 & 1 & 0 \\ 2 & \cos^2 x & \cos 2x \\ 1 & \cos^2 x & \sin 2x \end{vmatrix}$$ Open with respect to $R_1$: $$-(2 \sin 2x - \cos 2x)$$ Let $\cos 2x - 2 \sin 2x = f(x)$. The maximum value of $f(x)$ is: $$f(x)\big|_{\max} = \sqrt{1 + 4} = \sqrt{5}$$

Total cases: $$6 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2$$ $$n(s) = 6 \cdot 6!$$ Favourable cases: Number divisible by 3 Sum of digits must be divisible by 3 Case-I 1, 2, 3, 4, 5, 6 Number of ways = $6!$ Case-II 0, 1, 2, 4, 5, 6 Number of ways = $5 \cdot 5!$ Case-III 0, 1, 2, 3, 4, 5 Number of ways = $5 \cdot 5!$ $$n(favourable) = 6! + 2 \cdot 5 \cdot 5!$$ $$P = \frac{6! + 2 \cdot 5 \cdot 5!}{6 \cdot 6!} = \frac{4}{9}$$

Given $\lim_{x \to 0^+} f(x) = f(0) = \lim_{x \to 0^-} (x)$. $$\lim_{x \to 0^+} \frac{\cos^{-1}(1-x^2) \cdot \sin^{-1}(1-x)}{x(1-x)(1+x)}$$ $$\lim_{x \to 0^+} \frac{\cos^{-1}(1-x^2)}{x \cdot 1 \cdot 1} \cdot \frac{\pi}{2}$$ Let $1 - x^2 = \cos \theta$. $$\frac{\pi}{2} \lim_{x \to 0^+} \frac{\theta}{\sqrt{1-\cos \theta}}$$ $$\frac{\pi}{2} \lim_{\theta \to 0^+} \frac{\theta}{\sqrt{2} \sin \frac{\theta}{2}} = \frac{\pi}{\sqrt{2}}$$ Now, $\lim_{x \to 0^-} \frac{\cos^{-1}(1-(1+x)^2) \sin^{-1}(-x)}{(1+x)-(1+x)^3}$ $$\lim_{x \to 0^-} \frac{\frac{\pi}{2} (-\sin^{-1} x)}{(1+x)(2+x)(-x)}$$ $$\lim_{x \to 0^-} \frac{\frac{\pi}{2}}{1 \cdot 2} \cdot \frac{\sin^{-1} x}{x} = \frac{\pi}{4}$$ Therefore, $\Rightarrow RHL \neq LHL$. Function can't be continuous. $\Rightarrow$ No value of $\alpha$ exist.

Plane passing through $(42, 0, 0)$, $(0, 42, 0)$, $(0, 0, 42)$. From intercept form, equation of plane is $x + y + z = 42$. $$\Rightarrow (x - 11) + (y - 19) + (z - 12) = 0$$ Let $a = x - 11$, $b = y - 19$, $c = z - 12$. $a + b + c = 0$. Now, given expression is $$3 + \frac{a}{b^2c^2} + \frac{b}{a^2c^2} + \frac{c}{a^2b^2} - \frac{42}{14abc}$$ $$3 + \frac{a^3 + b^3 + c^3 - 3abc}{a^2b^2c^2}$$ If $a + b + c = 0$ $$\Rightarrow a^3 + b^3 + c^3 = 3abc$$ $$\Rightarrow 3$$

Given $$I = \int_0^{10} [x] \cdot e^{[x] - x + 1}$$ We can write $$I = \int_0^1 0 \, dx + \int_1^2 1 \cdot e^{2-x} + \int_2^3 2 \cdot e^{3-x} + \ldots + \int_9^{10} 9 \cdot e^{10-x} \, dx$$ This implies $$I = \sum_{n=0}^{9} \int_n^{n+1} n \cdot e^{n+1-x} \, dx$$ Evaluating the integral, we have $$= -\sum_{n=0}^{9} n \left( e^{n+1-x} \right)_{n}^{n+1}$$ Simplifying, $$= (e-1) \sum_{n=0}^{9} n$$ Calculating the sum, $$= -\sum_{n=0}^{9} n \cdot \left( e^0 - e^1 \right)$$ Finally, $$= 45(e-1)$$

Given $y^2 = 4x$. Mirror image on $y = x \Rightarrow C : x^2 = 4y$. $2x = 4 \cdot \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{2}$. $$\left. \frac{dy}{dx} \right|_{P(2,1)} = \frac{2}{2} = 1$$ Equation of tangent at $(2,1)$ $$\Rightarrow y - 1 = 1(x - 2)$$ $$\Rightarrow x - y = 1$$

Given $\($ $\frac{dy}{dx}$ + ($\tan$ x)y = $\sin$ x; $\ $0 $\leq$ x $\leq$ $\frac{\pi}{3}$ $\)$. Integrating factor (I.F.) is $\($ e^{$\int$ $\tan$ x $\,$ dx} = e^{$\ln$ $\sec$ x} = $\sec$ x $\)$. $\($ y $\sec$ x = $\int$ $\tan$ x $\,$ dx $\)$ $\($ y $\sec$ x = $\ln$ |$\sec$ x| + C $\)$ Given $\($ x = 0, y = 0 $\)$ implies $\($ $\therefore$ c = 0 $\)$. Thus, $\($ y $\sec$ x = $\ln$ |$\sec$ x| $\)$. $\($ y = $\cos$ x $\cdot$ $\ln$ |$\sec$ x| $\)$. $\($ y|_{x = $\frac{\pi}{4}$} = $\left$( $\frac{1}{\sqrt{2}}$ $\right$) $\cdot$ $\ln$ $\sqrt{2}$ $\)$ $\($ y|_{x = $\frac{\pi}{4}$} = $\frac{1}{2\sqrt{2}}$ $\log$_e 2 $\)$

Given the set $A = \{2, 3, 4, 5, \ldots, 30\}$. The condition $(a, b) \simeq (c, d)$ implies $ad = bc$. For $(4, 3) \simeq (c, d)$, we have $4d = 3c$. This implies $$\frac{4}{3} = \frac{c}{d}$$ and $c, d \in \{2, 3, \ldots, 30\}$. Therefore, $$\frac{c}{d} = \frac{4}{3}$$ and the pairs $(c, d)$ are $\{(4, 3), (8, 6), (12, 9), (16, 12), (20, 15), (24, 18), (28, 21)\}$. The number of ordered pairs is $7$.

Given $x^2 + y^2 + ax + 2ay + c = 0$ $2\sqrt{g^2-c} = 2\sqrt{\dfrac{a^2}{4}-c} = 2\sqrt{2}$ $\Rightarrow \dfrac{a^2}{4}-c = 2 \qquad ...(1)$ $2\sqrt{f^2-c} = 2\sqrt{a^2-c} = 2\sqrt{5}$ $\Rightarrow a^2-c = 5 \qquad ...(2)$ From (1) and (2), $\dfrac{3a^2}{4} = 3$ $\Rightarrow a = -2 \; (a<0)$ $\therefore c = -1$ Circle: $\Rightarrow x^2 + y^2 - 2x - 4y - 1 = 0$ $\Rightarrow (x-1)^2 + (y-2)^2 = 6$ Given $x + 2y = 0$ $\Rightarrow m = -\dfrac{1}{2}$ $m_{\text{tangent}} = 2$ Equation of tangent: $\Rightarrow (y-2) = 2(x-1) \pm \sqrt{6}\sqrt{1+4}$ $\Rightarrow 2x - y \pm \sqrt{30} = 0$ Perpendicular distance from $(0,0)$ $= \left|\dfrac{\pm\sqrt{30}}{\sqrt{4+1}}\right|$ $= \sqrt{6}$

Given $$\exp\left(\frac{(|z|+3)(|z|-1)}{|z|+1} \ln 2\right) \geq \log_{\sqrt{2}} |5\sqrt{7} + 9i|$$ This implies $$\frac{(|z|+3)(|z|-1)}{|z|+1} \geq \log_{\sqrt{2}}(16)$$ Therefore, $$\frac{(|z|+3)(|z|-1)}{|z|+1} \geq 2^3$$ This simplifies to $$\frac{(|z|+3)(|z|-1)}{|z|+1} \geq 3$$ Thus, $$(|z|+3)(|z|-1) \geq 3(|z|+1)$$ Expanding gives $$|z|^2 + 2|z| - 3 \geq 3|z| + 3$$ Simplifying, $$|z|^2 + |z| - 6 \geq 0$$ Factoring, $$(|z|-3)(|z|+2) \geq 0 \Rightarrow |z| - 3 \geq 0$$ Therefore, $$|z| \geq 3 \Rightarrow |z|_{\min} = 3$$

Let $\alpha$ be the number of triangles. $$\alpha = 5 \cdot 6 \cdot 7 + 5 \cdot 7 \cdot 9 + 5 \cdot 6 \cdot 9 + 6 \cdot 7 \cdot 9$$ $$= 210 + 315 + 270 + 378$$ $$= 1173$$ Let $\beta$ be the number of quadrilaterals. $$\beta = 5 \cdot 6 \cdot 7 \cdot 9 = 1890$$ The difference is given by $$\beta - \alpha = 1890 - 1173 = 717$$

Given the equations $y^2 = 3x^2$ and $x^2 + y^2 = 4b$. Solve both we get so $x^2 = b$. $$\frac{x^2}{16} + \frac{3x^2}{b^2} = 1$$ $$\frac{b}{16} + \frac{3}{b} = 1$$ $$b^2 - 16b + 48 = 0$$ $$(b - 12)(b - 4) = 0$$ $b = 12, b > 4$

Given $\sin^{-1} \frac{3x}{5} + \sin^{-1} \frac{4x}{5} = \sin^{-1} x$. $$\sin^{-1} \left( \frac{3x}{5} \sqrt{1 - \frac{16x^2}{25}} + \frac{4x}{5} \sqrt{1 - \frac{9x^2}{25}} \right) = \sin^{-1} x$$ $$\frac{3x}{5} \sqrt{1 - \frac{16x^2}{25}} + \frac{4x}{5} \sqrt{1 - \frac{9x^2}{25}} = x$$ $$x = 0, 3\sqrt{25 - 16x^2} + 4\sqrt{25 - 9x^2} = 25$$ $$4\sqrt{25 - 9x^2} = 25 - 3\sqrt{25 - 16x^2}$$ squaring we get $$16 \left( 25 - 9x^2 \right) = 625 + 9 \left( 25 - 16x^2 \right) - 150\sqrt{25 - 16x^2}$$ $$400 = 625 + 225 - 150\sqrt{25 - 16x^2} - 16x^2$$ $$\sqrt{25 - 16x^2} = 3 \Rightarrow 25 - 16x^2 = 9$$ $$\Rightarrow x^2 = 1$$ Put $x = 0, 1, -1$ in the original equation. We see that all values satisfy the original equation. Number of solution = 3

P $\equiv$ (x_1, mx_1) Q $\equiv$ (x_2, mx_2) A_1 = $\frac{1}{2}$ $\begin{vmatrix}$ 3 & 4 & 1 $\\$ 2 & 0 & 1 $\\$ -1 & 1 & 1 $\end{vmatrix}$ = $\frac{13}{2}$ A_2 = $\frac{1}{2}$ $\begin{vmatrix}$ x_1 & mx_1 & 1 $\\$ x_2 & mx_2 & 1 $\\$ 2 & 0 & 1 $\end{vmatrix}$ A_2 = $\frac{1}{2}$ |2(mx_1 - mx_2)| = m |x_1 - x_2| A_1 = 3 A_2 $\Rightarrow$ $\frac{13}{2}$ = 3m |x_1 - x_2| AC : x + 3y = 2 BC : y = 4x - 8 P : x + 3y = 2 & y = mx $\Rightarrow$ x_1 = $\frac{2}{1+3m}$ Q : y = 4x - 8 & xy = mx $\Rightarrow$ x_2 = $\frac{8}{4-m}$ |x_1 - x_2| = $\left$| $\frac{2}{1+3m}$ - $\frac{8}{4-m}$ $\right$| = $\left$| $\frac{-26m}{(1+3m)(4-m)}$ $\right$| = $\frac{26m}{(3m+1)|m-4|}$ |x_1 - x_2| = $\frac{13}{6m}$ $\Rightarrow$ $\frac{26m}{(3m+1)(4-m)}$ = $\frac{13}{6m}$ $\Rightarrow$ 12m^2 = -(3m+1)(m-4) $\Rightarrow$ 12m^2 = -(3m^2 - 11m - 4) $\Rightarrow$ 15m^2 - 11m - 4 = 0 $\Rightarrow$ 15m^2 - 15m + 4m - 4 = 0 $\Rightarrow$ (15m + 4)(m - 1) = 0 $\Rightarrow$ m = 1

Given $f(x) = 3 \ln(x-1) - 3 \ln(x+1) - \frac{2}{x-1}$. Differentiating, we have: $$f'(x) = \frac{3}{x-1} - \frac{3}{x+1} + \frac{2}{(x-1)^2}$$ Simplifying, $$f'(x) = \frac{4(2x-1)}{(x-1)^2(x+1)}$$ We need $f'(x) \geq 0$. Therefore, $$x \in (-\infty, -1) \cup \left[ \frac{1}{2}, 1 \right) \cup (1, \infty)$$

Given $\ln f(x+1) = \ln(x f(x))$. $\ln f(x+1) = \ln x + \ln f(x)$ Therefore, $g(x+1) = \ln x + g(x)$ This implies $g(x+1) - g(x) = \ln x$ Thus, $g''(x+1) - g''(x) = -\frac{1}{x^2}$ Put $x = 1, 2, 3, 4$ $$g''(2) - g''(1) = -\frac{1}{1^2} \cdots (1)$$ $$g''(3) - g''(2) = -\frac{1}{2^2} \cdots (2)$$ $$g''(4) - g''(3) = -\frac{1}{3^2} \cdots (3)$$ $$g''(5) - g''(4) = -\frac{1}{4^2} \cdots (4)$$ Add all the equations we get $$g''(5) - g''(1) = -\frac{1}{1^2} - \frac{1}{2^2} - \frac{1}{3^2} - \frac{1}{4^2}$$ $$|g''(5) - g''(1)| = \frac{205}{144}$$

Given $\int_0^1 (x^2 + bx + c), dx = 1$. $\frac{1}{3} + \frac{b}{2} + c = 1$ $\frac{b}{2} + c = \frac{2}{3}$ $3b + 6c = 4$ ...(1) Given $P(2) = 5$. $4 + 2b + c = 5$ $2b + c = 1$ ...(2) From (1) and (2): $b = \frac{2}{9}$ $c = \frac{5}{9}$ $9(b + c) = 7$

(3,5,7) satisfy the line $L_1$ $$\frac{3-a}{\ell} = \frac{5-2}{3} = \frac{7-b}{4}$$ $$\frac{3-a}{\ell} = 1 \& \frac{7-b}{4} = 1$$ $$a + \ell = 3 \ldots (1)$$ $$b = 3 \ldots (2)$$ $$\vec{v_1} = \langle 4, 3, 8 \rangle - \langle 3, 5, 7 \rangle$$ $$\vec{v_1} = \langle 1, -2, 1 \rangle$$ $$\vec{v_2} = \langle \ell, 3, 4 \rangle$$ $$\vec{v_1} \cdot \vec{v_2} = 0 \Rightarrow \ell - 6 + 4 = 0 \Rightarrow \ell = 2$$ $$a + \ell = 3 \Rightarrow a = 1$$ Line $L_1$: $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ Line $L_2$: $$\frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$ $$A = \langle 1, 2, 3 \rangle$$ $$B = \langle 2, 4, 5 \rangle$$ $$\overrightarrow{AB} = \langle 1, 2, 2 \rangle$$ $$\vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$ $$\vec{q} = 3\hat{i} + 4\hat{j} + 5\hat{k}$$ $$\vec{p} \times \vec{q} = -\hat{i} + 2\hat{j} - \hat{k}$$ Shortest distance = $$\frac{\left| \overrightarrow{AB} \cdot (\vec{p} \times \vec{q}) \right|}{\left| \vec{p} \times \vec{q} \right|} = \frac{1}{\sqrt{6}}$$

Given $\($ $\frac{dy}{dx}$ = $\frac{y^2 - x^2}{2xy}$ $\)$, $\($ x $\in$ (0, $\infty$) $\)$. Put $\($ y = vx $\)$. $\[$ x $\frac{dv}{dx}$ + v = $\frac{v^2 - 1}{2v}$ $\]$ $\[$ $\frac{2v}{v^2 + 1}$ dv = -$\frac{dx}{x}$ $\]$ Integrate, $\[$ $\ln$(v^2 + 1) = -$\ln$ x + C $\]$ $\[$ $\ln$$\left$( $\frac{y^2}{x^2}$ + 1 $\right$) = -$\ln$ x + C $\]$ Put $\($ x = 1, y = 1, C = $\ln$ 2 $\)$ $\[$ $\ln$$\left$( $\frac{y^2}{x^2}$ + 1 $\right$) = -$\ln$ x + $\ln$ 2 $\]$ $\[$ $\Rightarrow$ x^2 + y^2 - 2x = 0 (Curve C_1) $\]$ Similarly, $\($ $\frac{dy}{dx}$ = $\frac{2xy}{x^2 - y^2}$ $\)$ Put $\($ y = vx $\)$ $\[$ x^2 + y^2 - 2y = 0 $\]$ Required area = $\($ 2 $\int$_0^1 $\left$( $\sqrt{2x - x^2}$ - x $\right$) $\,$ dx = $\frac{\pi}{2}$ - 1 $\)$

Given $\vec{r} \times \vec{a} = \vec{b} \times \vec{r} = \vec{r} \times (\vec{a} + \vec{b}) = 0$. $\vec{r} = \lambda (\vec{a} + \vec{b}) \Rightarrow \vec{r} = \lambda (\hat{i} + 2\hat{j} - 3\hat{k} + 2\hat{i} - 3\hat{j} + 5\hat{k})$. $\vec{r} = \lambda (3\hat{i} - \hat{j} + 2\hat{k}) \ldots (1)$. $\vec{r} \cdot (\alpha \hat{i} + 2\hat{j} + \hat{k}) = 3$. Put $\vec{r}$ from (1) $\alpha \lambda = 1 \ldots (2)$. $\vec{r} \cdot (2\hat{i} + 5\hat{j} - \alpha \hat{k}) = -1$. Put $\vec{r}$ from (1) $2\lambda \alpha - \lambda = 1 \ldots (3)$. Solve (2) and (3). $\alpha = 1, \lambda = 1$. $\Rightarrow \vec{r} = 3\hat{i} - \hat{j} + 2\hat{k}$. $|\vec{r}|^2 = 14$ and $\alpha = 1$. $\alpha + |\vec{r}|^2 = 15$.

DC of line $\equiv \left( \frac{3}{\sqrt{m^2+10}}, \frac{-m}{\sqrt{m^2+10}}, \frac{1}{\sqrt{m^2+10}} \right)$ Q lies on $x + 2y - 3z + 10 = 0$ $$1 + \frac{3r}{\sqrt{m^2+10}} - 4 - \frac{2mr}{\sqrt{m^2+10}} - 9 - \frac{3r}{\sqrt{m^2+10}} + 10 = 0$$ $$\Rightarrow \frac{r}{\sqrt{m^2+10}} (3 - 2m - 3) = 2$$ $$\Rightarrow \frac{r}{\sqrt{m^2+10}} (-2m) = 2$$ $$r^2 m^2 = m^2 + 10$$ $$\frac{7}{2} m^2 = m^2 + 10 \Rightarrow \frac{5}{2} m^2 = 10 \Rightarrow m^2 = 4$$ $$|m| = 2$$

Given $$ \sigma^2 = \frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2} + \frac{n_1 n_2}{(n_1 + n_2)} \left( \overline{x_1} - \overline{x_2} \right)^2 $$ Let $n_1 = 10$, $n_2 = n$, $\sigma_1^2 = 2$, $\sigma_2^2 = 1$. $$ \overline{x_1} = 2, \overline{x_2} = 3, \sigma^2 = \frac{17}{9} $$ $$ \frac{17}{9} = \frac{10 \times 2 + n}{n + 10} + \frac{10n}{(n + 10)^2} (3 - 2)^2 $$ $$ \Rightarrow \frac{17}{9} = \frac{(n + 20)(n + 10) + 10n}{(n + 10)^2} $$ $$ \Rightarrow 17n^2 + 1700 + 340n = 90n + 9 \left(n^2 + 30n + 200\right) $$ $$ \Rightarrow 8n^2 - 20n - 100 = 0 $$ $$ 2n^2 - 5n - 25 = 0 $$ $$ \Rightarrow (2n + 5)(n - 5) = 0 \Rightarrow n = \frac{-5}{2}, 5 $$ (Rejected) Hence $n = 5$

Given $A = XB$ $$\begin{bmatrix} a_1 \\ a_2 \\ \sqrt{3}a_1 \\ \sqrt{3}a_2 \end{bmatrix} = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 & -1 \\ 1 & k \\ \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} = \begin{bmatrix} b_1 - b_2 \\ b_1 + kb_2 \end{bmatrix}$$ $b_1 - b_2 = \sqrt{3}a_1 \ldots (1)$ $b_1 + kb_2 = \sqrt{3}a_2 \ldots (2)$ Given, $a_1^2 + a_2^2 = \frac{2}{3} \left(b_1^2 + b_2^2\right)$ $(1)^2 + (2)^2$ $\left(b_1 + b_2\right)^2 + \left(b_1 + kb_2\right)^2 = 3 \left(a_1^2 + a_2^2\right)$ $a_1^2 + a_2^2 = \frac{2}{3}b_1^2 + \frac{(1+k^2)}{3}b_2^2 + \frac{2}{3}b_1b_2(k-1)$ Given, $a_1^2 + a_2^2 = \frac{2}{3}b_1^2 + \frac{2}{3}b_2^2$ On comparing we get $$\frac{k^2+1}{3} = \frac{2}{3} \Rightarrow k^2 + 1 = 2$$ $$\Rightarrow k = \pm 1$$ & $\frac{2}{3}(k-1) = 0 \Rightarrow k = 1$ From both we get $k = 1$

Given $$\int \frac{(x^2-1)dx}{(x^4+3x^2+1)\tan^{-1}\left(x+\frac{1}{x}\right)} + \int \frac{dx}{x^4+3x^2+1}$$ $$\frac{\left(1-\frac{1}{x^2}\right)dx}{\left(\left(x+\frac{1}{x}\right)^2+1\right)\tan^{-1}\left(x+\frac{1}{x}\right)} + \frac{1}{2}\int \frac{(x^2+1)-(x^2-1)dx}{x^4+3x^2+1}$$ Put $\tan^{-1}\left(x+\frac{1}{x}\right) = t$ $$\frac{dt}{t} + \frac{1}{2}\int \frac{\left(1+\frac{1}{x^2}\right)dx}{\left(x-\frac{1}{x}\right)^2+5} - \frac{1}{2}\int \frac{\left(1-\frac{1}{x^2}\right)dx}{\left(x+\frac{1}{x}\right)^2+1}$$ Put $x-\frac{1}{x} = y, x+\frac{1}{x} = z$ $$\log_e t + \frac{1}{2}\int \frac{dy}{y^2+5} - \frac{1}{2}\int \frac{dz}{z^2+1}$$ $$-\frac{1}{2}\tan^{-1}\left(\frac{x^2+1}{x}\right) + C$$ $$= \log_e \tan^{-1}\left(x+\frac{1}{x}\right) + \frac{1}{2\sqrt{5}}\tan^{-1}\left(\frac{x^2-1}{\sqrt{5}x}\right)$$ $$\alpha = 1, \beta = \frac{1}{2\sqrt{5}}, \gamma = \frac{1}{\sqrt{5}}, \delta = -\frac{1}{2}$$ or $$\alpha = 1, \beta = -\frac{1}{2\sqrt{5}}, \gamma = \frac{1}{\sqrt{5}}, \delta = -\frac{1}{2}$$ $$10(\alpha + \beta \gamma + \delta) = 10\left(1 + \frac{1}{10} - \frac{1}{2}\right) = 6$$

Given $$g[f(x)] = \begin{cases} f(x) + 1 & f(x) < 0 \\ (f(x) - 1)^2 + b & f(x) \geq 0 \end{cases}$$ $$g[f(x)] = \begin{cases} x + a + 1 & x + a < 0 \& x < 0 \\ |x - 1| + 1 & |x - 1| < 0 \& x \geq 0 \end{cases}$$ $$g[f(x)] = \begin{cases} (x + a - 1)^2 + b & x + a \geq 0 \& x < 0 \\ (|x - 1| - 1)^2 + b & |x - 1| \geq 0 \& x \geq 0 \end{cases}$$ $$g[f(x)] = \begin{cases} x + a + 1 & x \in (-\infty, -a) \& x \in (-\infty, 0) \\ |x - 1| + 1 & x \in \phi \\ (x + a - 1)^2 + b & x \in [-a, 0) \& x \in (-\infty, 0) \\ (|x - 1| - 1)^2 + b & x \in \mathbb{R} \& x \in [0, \infty) \end{cases}$$ $$g[f(x)] = \begin{cases} x + a + 1 & x \in (-\infty, -a) \\ (x + a - 1)^2 + b & x \in [-a, 0) \\ (|x - 1| - 1)^2 + b & x \in [0, \infty) \end{cases}$$ g(f(x)) is continuous at $x = -a$ and at $x = 0$ $$1 = b + 1 \& (a - 1)^2 + b = b$$ $$b = 0 \& a = 1$$ Therefore, $a + b = 1$

Given $a^2 = \frac{b}{16}$, we have $\frac{1}{b} = \frac{1}{16a^2}$. Then, $\frac{2}{b} = \frac{1}{a} + 6$. This implies $\frac{1}{8a^2} = \frac{1}{a} + 6$. Rearranging gives $\frac{1}{a^2} - \frac{8}{a} - 48 = 0$. Solving $\frac{1}{a} = 12, -4$ gives $a = \frac{1}{12}, -\frac{1}{4}$. Since $a > 0$, we take $a = \frac{1}{12}$. Then $b = 16a^2 = \frac{1}{9}$. Thus, $72(a + b) = 6 + 8 = 14$.

Given $\Delta = \frac{1}{2} \cdot 5 \cdot 12 \cdot \sin A = 30$. Therefore, $\sin A = 1$. Since $A = 90^\circ$, it follows that $BC = 13$. Thus, $BC = 2R = 13$. The inradius $r$ is given by $r = \frac{\Delta}{S} = \frac{30}{15} = 2$. Therefore, $2R + r = 15$.

Given $$A = \sum_{k=0}^{n} \binom{n}{k} \left[ \left( -\frac{1}{2} \right)^2 + \left( -\frac{3}{4} \right)^2 + \left( -\frac{7}{8} \right)^2 + \left( -\frac{15}{16} \right)^2 + \left( -\frac{37}{32} \right)^2 \right]$$ We have $$A = \left( 1 - \frac{1}{2} \right)^2 + \left( 1 - \frac{3}{4} \right)^n + \left( 1 - \frac{7}{8} \right)^n + \left( 1 - \frac{15}{16} \right)^n + \left( 1 - \frac{31}{32} \right)^n$$ This simplifies to $$A = \frac{1}{2^n} + \frac{1}{4^n} + \frac{1}{8^n} + \frac{1}{16^n} + \frac{1}{32^n}$$ Further simplification gives $$A = \frac{1}{2^n} \left( \frac{1 - \left( \frac{1}{2^n} \right)^5}{1 - \frac{1}{2^n}} \right) \implies A = \frac{\left( 1 - \frac{1}{25n} \right)}{(2^n - 1)}$$ Thus, $$(2^n - 1) A = 1 - \frac{1}{2^{5n}}, Given 63 A = 1 - \frac{1}{2^{30}}$$ Clearly, $5n = 30$ Therefore, $n = 6$

Given $\vec{c} = \lambda (\vec{a} \times \vec{b})$. Calculate $\vec{a} \times \vec{b}$ using the determinant: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix}$$ This results in $\vec{a} \times \vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$. Now, compute the dot product: $$\vec{c} \cdot (\hat{i} + \hat{j} + 3\hat{k}) = \lambda (3\hat{i} - 2\hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} + 3\hat{k})$$ This implies $\lambda (4) = 8 \Rightarrow \lambda = 2$. Therefore, $\vec{c} = 2(\vec{a} \times \vec{b})$. Finally, compute: $$\vec{c} \cdot (\vec{a} \times \vec{b}) = 2|\vec{a} \times \vec{b}|^2 = 28$$

Given $S_n(x) = (2 + 3 + 6 + 11 + 18 + 27 + \ldots + n - terms) \log_a x$. Let $S_1 = 2 + 3 + 6 + 11 + 18 + 27 + \ldots + T_n$. $S_1 = 2 + 3 + 6 + \ldots + T_n$. $T_n = 2 + 1 + 3 + 5 + \ldots + n terms$. $T_n = 2 + (n - 1)^2$. $S_1 = \Sigma T_n = 2n + \frac{(n-1)n(2n-1)}{6}$. Therefore, $S_n(x) = \left(2n + \frac{n(n-1)(2n-1)}{6}\right) \log_a x$. $S_{24}(x) = 1093 (Given)$. $\log_a x \left(48 + \frac{23 \cdot 24 \cdot 47}{6}\right) = 1093$. $\log_a x = \frac{1}{4} \ldots (1)$. $S_{12}(2x) = 265$. $S_{12}(2x) = 265$. $\log_a 2x = \frac{1}{2} \ldots (2)$. (2) - (1) $\log_a 2x - \log_a x = \frac{1}{4}$. $\log_a 2 = \frac{1}{4} \Rightarrow a = 16$

The truth table for the given logic gate is: $$\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \hline \end{array}$$ The truth table is similar to that of a NOR gate.

From energy conservation, after bullet gets embedded till the system comes momentarily at rest, $$(M + m)gh = \frac{1}{2}(M + m)v_1^2$$ $v_1$ is velocity after collision. Therefore, $$v_1 = \sqrt{2gh}$$ Applying momentum conservation, just before and just after collision, $$mv = (M + m)v_1$$ $$v = \left(\frac{M+m}{m}\right)v_1 = \frac{6}{10 \times 10^{-3}} \times \sqrt{2 \times 9.8 \times 9.8 \times 10^{-2}}$$ $$\approx 831.55 \, \mathrm{m/s}$$

Since force on a point charge by magnetic field is always perpendicular to $\vec{V} \left[ \vec{F} = q \vec{V} \times \vec{B} \right]$. Therefore, work by magnetic force on the point charge is zero.

The nature of flow is determined by Reynolds Number. $$\mathrm{Re} = \frac{\rho v D}{\eta}$$ $$\begin{aligned} &\left[ \begin{array}{l} \rho \rightarrow density of fluid \\ \eta \rightarrow coefficient of viscosity \\ v \rightarrow velocity of flow \\ D \rightarrow Diameter of pipe \end{array} \right] \end{aligned}$$ From NCERT If $\mathrm{Re} 2000$ $\rightarrow$ flow is turbulent $$\mathrm{Re}_{initial} = 10^3 \times \frac{0.18 \times 10^{-3}}{\pi \times (0.5 \times 10^{-2})^2 \times 60} \times \frac{1 \times 10^{-2}}{10^{-3}}$$ $$= 382.16$$ $$\mathrm{Re}_{final} = 10^3 \times \frac{0.48 \times 10^{-3}}{\pi \times (0.5 \times 10^{-2})^2 \times 60} \times \frac{1 \times 10^{-2}}{10^{-3}}$$ $$= 1019.09$$

Given: $$\vec{v} = 0.5t^2 \hat{i} + 3t \hat{j} + 9 \hat{k}$$ $$\vec{v}_{at t=2} = 2 \hat{i} + 6 \hat{j} + 9 \hat{k}$$ Therefore, the angle made by the direction of motion of the mosquito will be, $$\cos^{-1} \frac{2}{11} (from x-axis) = \tan^{-1} \frac{\sqrt{117}}{2}$$ $$\cos^{-1} \frac{6}{11} (from y-axis) = \tan^{-1} \frac{\sqrt{85}}{6}$$ $$\cos^{-1} \frac{9}{11} (from z-axis) = \tan^{-1} \frac{\sqrt{40}}{9}$$

Given $x = 3\,\mathrm{m}$, the equation is $\dfrac{2K\lambda}{r} = \dfrac{\sigma}{\varepsilon_0}$. Solving for $\sigma$, we find $\sigma = 0.424 \times 10^{-9}\,\mathrm{C/m^2}$.

The de Broglie wavelength is given by $$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}.$$ The ratio of wavelengths is $$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}}.$$ For electrons and protons, $$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{1831.4} = 42.79.$$

Given $V_S = \frac{P}{i} = \frac{60}{0.11} = 545.45$. $V_P = 220$. Since $V_S > V_P$, this is a step up transformer.

Given $A = \lambda N$. $N = n N_A$. $N = \left( \frac{1.5 \times 10^{-3}}{198} \right) N_A$. $A = \left( \frac{\ln 2}{t_{1/2}} \right) N$. $1 Curie = 3.7 \times 10^{10} Bq$. $A = 365 Bq$.

Given the formula for lambda: $$\lambda = \frac{RT}{\sqrt{2 \pi d^2} \, N_A P}$$ We have: $$\lambda = 102 \, \mathrm{nm}$$

Given $\($ $\frac{1}{F}$ = $\left$[ $\frac{\mu_L}{\mu_S}$ - 1 $\right$] $\left$[ $\frac{1}{R_1}$ - $\frac{1}{R_2}$ $\right$] $\)$. If $\($ $\mu$_L = $\mu$_S $\Rightarrow$ $\frac{1}{F}$ = 0 $\Rightarrow$ F = $\infty$ $\)$.

Given $$Y = \frac{Stress}{Strain} = \frac{FL}{Al} = \frac{mg \cdot L}{\pi R^2 \cdot \ell}$$ $$\frac{\Delta Y}{Y} = \frac{\Delta m}{m} + \frac{\Delta L}{L} + 2 \cdot \frac{\Delta R}{R} + \frac{\Delta \ell}{\ell}$$ Multiplying by 100, we have: $$\frac{\Delta Y}{Y} \times 100 = 100 \left[ \frac{1}{1000} + \frac{1}{1000} + 2 \left( \frac{0.001}{0.2} \right) + \frac{0.001}{0.5} \right]$$ Simplifying, we get: $$= \frac{1}{10} + \frac{1}{10} + 1 + \frac{1}{5} = \frac{14}{10} = 1.4\%$$

Given $\alpha_A > \alpha_B$. Length of both strips will decrease $\Delta L_A > \Delta L_B$.

Given the equation: $$500 = (1.5)^2 \times R \times 20$$ We need to find the value of $E$ using the equation: $$E = (3)^2 \times R \times 20$$ Substituting the value of $R$ from the first equation, we find: $$E = 2000 \, \mathrm{J}$$

Statement I: $v_{max} = \sqrt{\mu R g} = \sqrt{(0.2) \times 2 \times 9.8}$ $v_{max} = 1.97 \, \mathrm{m/s}$ $7 \, \mathrm{km/h} = 1.944 \, \mathrm{m/s}$ Speed is lower than $v_{max}$, hence it can take safe turn. Statement II $v_{max} = \sqrt{R g \left[ \frac{\tan \theta + \mu}{1 - \mu \tan \theta} \right]}$ $$= \sqrt{2 \times 9.8 \left[ \frac{1 + 0.2}{1 - 0.2} \right]} = 5.42 \, \mathrm{m/s}$$ $18.5 \, \mathrm{km/h} = 5.14 \, \mathrm{m/s}$ Speed is lower than $v_{max}$, hence it can take safe turn.

Given $D = 2\sqrt{2Rh}$. $$h = \frac{D^2}{8R} = \frac{45^2}{8 \times 6400} km \cong 39.55 m$$

Given: $$E_1 = \frac{B_0 (x + d)}{a} v_0 d$$ $$E_2 = \frac{B_0 (x)}{a} v_0 d$$ The net electric field is: $$E_{net} = E_1 - E_2$$ Substituting the values, we get: $$E_{net} = \frac{B_0 v_0 d^2}{a}$$

Given $A = A_0 e^{-\gamma t} = A_0 e^{-\frac{bt}{2m}}$. $$\frac{A_0}{2} = A_0 e^{-\frac{bt}{2m}}$$ $$\frac{bt}{2m} = \ln 2$$ $$t = \frac{2m}{b} \ln 2 = \frac{2 \times 500 \times 0.693}{20}$$ $$t = 34.65 second.$$

Given $N_1 = N_0 e^{-\lambda t_1}$, we have $$\frac{N_1}{N_0} = e^{-\lambda t_1}$$ $$0.67 = e^{-\lambda t_1}$$ Taking the natural logarithm, $$\ln(0.67) = -\lambda t_1$$ Similarly, for $N_2 = N_0 e^{-\lambda t_2}$, we have $$\frac{N_2}{N_0} = e^{-\lambda t_2}$$ $$0.33 = e^{-\lambda t_2}$$ Taking the natural logarithm, $$\ln(0.33) = -\lambda t_2$$ Subtracting these equations, $$\ln(0.67) - \ln(0.33) = \lambda t_1 - \lambda t_2$$ $$\lambda (t_1 - t_2) = \ln \left( \frac{0.67}{0.33} \right)$$ Approximating, $$\lambda (t_1 - t_2) \cong \ln 2$$ Therefore, $$t_1 - t_2 \cong \frac{\ln 2}{\lambda} = t_{1/2}$$ The half-life is $t_{1/2} = 20$ minutes.

Red light and blue light have different wavelength and different frequency.

Given $Q = i^2 R T$. $$R = \frac{Q}{i^2 t} = \frac{10 \times 10^{-3}}{4 \times 10^{-6} \times 1} = 2500 \Omega$$

Given $$C = \frac{\varepsilon_0 \, A}{\frac{d}{2K} + \frac{d}{2}} = \frac{2\varepsilon_0 \, A}{\frac{d}{K} + d}$$ Simplifying, $$= \frac{2 \times 2\varepsilon_0}{\frac{1}{3.2} + 1} = \frac{4 \times 3.2}{4.2} \varepsilon_0$$ Finally, $$= 3.04\varepsilon_0$$

Given $\vec{\tau} = \vec{r} \times \vec{F}$. $\vec{r} = (2\hat{i}) - (2\hat{i} + 3\hat{j} + 4\hat{k}) = -3\hat{j} - 4\hat{k}$ $\vec{F} = 4\hat{i} + 3\hat{j} + 4\hat{k}$ $\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -4 \\ 4 & 3 & 4 \end{vmatrix}$ $= \hat{i}(-12 + 12) - \hat{j}(0 + 16) + \hat{k}(0 + 12)$ $= -16\hat{i} + 12\hat{k}$ Therefore, $|\vec{\tau}| = \sqrt{16^2 + 12^2} = 20$

Energy given = $U_f - U_i$ $$= 0 - \left( -\frac{3GM^2}{5} \frac{R}{R} \right)$$ $$= \frac{3}{5} \frac{GM^2}{R}$$ $x = 3$

Given $\omega_1 = 0.02; \mu_1 = 1.5; \omega_2 = 0.03; \mu_2 = 1.6$. Achromatic combination. Therefore, $\theta_{net} = 0$. $\theta_1 - \theta_2 = 0$. $\theta_1 = \theta_2$. $\omega_1 \delta_1 = \omega_2 \delta_2$. $\delta_{net} = \delta_1 - \delta_2 = 2^\circ$. $$\delta_1 - \frac{\omega_1 \delta_1}{\omega_2} = 2^\circ$$ $$\delta_1 \left(1 - \frac{\omega_1}{\omega_2}\right) = 2^\circ$$ $$\delta_1 \left(1 - \frac{2}{3}\right) = 2^\circ$$ $$\delta_1 = 6^\circ$$ $$\delta_1 = (\mu_1 - 1) A_1$$ $$6^\circ = (1.5 - 1) A_1$$ $$A_1 = 12^\circ$$

Given $\($ $\vec{a}$ = $\frac{\vec{F}}{m}$ = $\frac{2\hat{i} + 3\hat{j} + 5\hat{k}}{2}$ $\)$. This simplifies to $\($ $\hat{i}$ + 1.5$\hat{j}$ + 2.5$\hat{k}$ $\)$. The position vector $\($ $\vec{r}$ = $\vec{u}$t + $\frac{1}{2}$ $\vec{a}$ t^2 $\)$. Substituting the values, we have: $\[$ = 0 + $\frac{1}{2}$ ($\hat{i}$ + 1.5$\hat{j}$ + 2.5$\hat{k}$)(16) $\]$ $\[$ = 8$\hat{i}$ + 12$\hat{j}$ + 20$\hat{k}$ $\]$ Thus, $\($ b = 12 $\)$.

Given $12 \sin \theta = v_r$. Since $\sin \theta = \frac{1}{2}$, we have $\theta = 30^\circ$. Therefore, $\alpha = 120^\circ$.

Given $f_c = f_0$. $$\frac{3 \, V_c}{4 \, L} = \frac{2 \, V_0}{2 \, L'}$$ $$\frac{3 \, V_c}{4 \, L} = \frac{V_0}{L'}$$ $$L' = \frac{4 \, L}{3} \frac{V_0}{V_c} = \frac{4 \, L}{3} \sqrt{\frac{B \cdot \rho_1}{\rho_2 \cdot B}} (B is bulk modulus)$$ $$= \frac{4 \, L}{3} \sqrt{\frac{\rho_1}{\rho_2}}$$ Given $x = 4$.

Given $$a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{2}{3} g \sin \theta$$ $$b = 3$$

Given $n = 0.60 = 1 - \frac{T_L}{T_H}$. Solving for $\frac{T_L}{T_H=0.4 \Rightarrow T_L}$ gives $0.4 \times 400$. This equals $160 \, \mathrm{K}$, which is $-113^\circ \mathrm{C}$.

The greenhouse gases are $\mathrm{CO_2}$, $\mathrm{H_2O_{(vapour)}}$ and $\mathrm{CH_4}$.

The given reaction involves the conversion of a methyl group to a carboxylic acid group. This is typically achieved through oxidation. The reagent "A" is likely an oxidizing agent. In the second step, alkaline $\mathrm{KMnO_4}$ is used, which is a strong oxidizing agent. Therefore, "A" is likely $\mathrm{KMnO_4}$ as well.

Reduction of $\mathrm{Al_2O_3} \rightarrow \mathrm{Al}$ is carried out by electrolytic reduction of its fused salts. $\mathrm{ZnO}$, $\mathrm{Fe_2O_3}$, and $\mathrm{Cu_2O}$ can be reduced by carbon.

Na $\rightarrow [\mathrm{Ne}] 3s^1$ $\mathrm{IE}_1$ is very low but $\mathrm{IE}_2$ is very high due to stable noble gas configuration of $\mathrm{Na}^+$. Mg $\rightarrow [\mathrm{Ne}] 3s^2$ $\mathrm{IE}_1$ and $\mathrm{IE}_2$ are low. $\mathrm{IE}_3$ is very high.

For substitution at allylic position in the given compound, the reagent used is $\mathrm{Cl_2}$/ uv light. The reaction is free radical halogenation.

The secondary structure of protein includes two types: (a) $\alpha$-Helix (b) $\beta$-pleated sheet. In $\alpha$-Helix structure, the polypeptide chain is coiled around due to the presence of intramolecular H-Bonding.

Urea–HCHO resin is used in manufacture of wood laminates.

(1) NaH (sodium Hydride) is used as a reducing reagent. (2) In pyridine, due to free electron on N atom, it is basic in nature. Hence statement I is false & II is true.

The structure of $\mathrm{H_2O_2}$ is open book type, which is non-planar. $\mathrm{H_2O_2}$ is used in the treatment of effluents. It can act as both an oxidizing agent (O.A) and a reducing agent (R.A). $\mathrm{H_2O_2}$ is miscible in water due to hydrogen bonding.

The reaction starts with an alkyl halide $\mathrm{R-X}$ reacting with ammonia $\mathrm{NH_3}$ to form $[\mathrm{R-NH_3}]^+ \mathrm{X}^-$. This intermediate reacts with $\mathrm{OH}^-$ to form a primary amine $\mathrm{R-NH_2}$, sodium halide $\mathrm{NaX}$, and water $\mathrm{H_2O}$. The primary amine can further react with another molecule of alkyl halide $\mathrm{R-X}$ to form a secondary amine $\mathrm{R-NHR}$, sodium halide $\mathrm{NaX}$, and water $\mathrm{H_2O}$. The secondary amine can react with another molecule of alkyl halide $\mathrm{R-X}$ in the presence of $\mathrm{NaOH}$ to form a tertiary amine $\mathrm{R_3N}$, sodium halide $\mathrm{NaX}$, and water $\mathrm{H_2O}$. So the purpose of $\mathrm{NaOH}$ in the above reactions is to remove acidic impurities.

For the given compounds: (a) $\mathrm{CH_3-C-NH-C_2H_5}$; L.P. on Nitrogen is delocalised. (b) $\mathrm{CH_2CH_2-N-CH_2CH_3}$; L.P. on Nitrogen is delocalised. (c) $\mathrm{CH_3-C-NH-C-CH_3}$; L.P. on Nitrogen is delocalised due to conjugation with both $\mathrm{C=O}$ (Hence least basic) (d) $\mathrm{CH_3-CH_2-NH-CH_2-CH_3}$; L.P. on Nitrogen is localised.

$X={}_{33}\mathrm{As}\rightarrow$ Metalloid $Y={}_{53}\mathrm{I}\rightarrow$ Nonmetal $Z={}_{83}\mathrm{Bi}\rightarrow$ Metal

Match list: \begin{tabular}{|c|p{7cm}|c|p{4cm}|} \hline (a) & Lassaigne's Test & (iii) & N, S, P and Halogen \\ \hline (b) & Cu(II) Oxide & (i) & Carbon \\ \hline (c) & AgNO$_3$ & (iv) & Halogen specifically \\ \hline (d) & Sodium fusion extract gives black precipitate with acetic acid and lead acetate (CH$_3$COOH/(CH$_3$COO)$_2$Pb) & (ii) & Sulphur \\ \hline \end{tabular} Option-(a)-(iii); (b)-(i); $(c)$-(iv); (d)-(ii)

1. Colloidal solution exhibits colligative properties. 2. An ordinary filter cannot stop the flow of colloidal particles. 3. Flocculating power increases with increase in the opposite charge of electrolyte. 4. Colloidal particles show Brownian motion.

Q2 (4) (a) Ce $\rightarrow [\mathrm{Xe}]4f^25d^06s^2$ In complex $\mathrm{Ce^{4+}} \rightarrow [\mathrm{Xe}]4f^05d^06s^0$ there is no unpaired electron so $\mu_m = 0$ (b) $\mathrm{^{64}Gd^{3+}} \rightarrow [\mathrm{Xe}]4f^75d^06s^0$ contain seven unpaired electrons so, $\mu_m = \sqrt{7(7+2)} = \sqrt{63} B.M.$ (c) $\mathrm{^{63}Eu^{3+}} \rightarrow [^{54}\mathrm{Xe}] 4f^65d^06s^0$ contain six unpaired electron so, $\mu_m = \sqrt{6(6+2)} = \sqrt{48} B.M$ Hence, order of spin only magnetic movement $b > c > a$

The reaction is given by: $$\mathrm{H_3PO_3 + 2NaOH \rightarrow Na_2HPO_3 + 2H_2O}$$ Given 50 ml, 1 M solution. We need to find the volume $V$ for 1 M solution. The stoichiometry gives: $$\frac{n_{\mathrm{NaOH}}}{n_{\mathrm{H_3PO_3}}} = \frac{2}{1}$$ Thus, $$\frac{1 \times V}{50 \times 1} = \frac{2}{1} \implies V_{\mathrm{NaOH}} = 100 \, \mathrm{ml}$$ For the second reaction: $$\mathrm{H_3PO_2 + 2NaOH \rightarrow NaH_2PO_3 + H_2O}$$ Given 100 ml, 2 M solution. We need to find the volume $V$. The stoichiometry gives: $$\frac{n_{\mathrm{NaOH}}}{n_{\mathrm{H_3PO_2}}} = \frac{1}{1}$$ Thus, $$\frac{1 \times V}{2 \times 100} = \frac{1}{1} \implies V_{\mathrm{NaOH}} = 200 \, \mathrm{ml}$$

The reaction begins with the addition of phenylmagnesium bromide (PhMgBr) to the imine group in dry ether. This forms a Grignard addition product with a magnesium bromide ion. Upon hydrolysis with $\mathrm{H_3O^+}$, the imine is converted to a ketone. The final product is a phenyl ketone with a methoxy group.

HCP structure: Per atom, there will be one octahedral void (OV) and two tetrahedral voids (TV). Therefore total three voids per atom are present in HCP structure. Therefore total number of atoms of Ga will be: $$\frac{Mass}{Molar Mass} \times N_A = \frac{0.58 \, g}{70 \, g/mol} \times 6.023 \times 10^{23}$$ Now, total number of voids = 3 $\times$ total number of atoms $$= 3 \times \frac{0.581}{70} \times 6.023 \times 10^{23} = 14.99 \times 10^{21}$$ $$\simeq 15 \times 10^{21}$$

Given conc $n$ of KCl $= \frac{\mathrm{m} \cdot \mathrm{mol}}{\mathrm{L}}$. Therefore, Conductance $(G) = 0.55 \mathrm{mS}$. Cell constant $\left( \frac{\ell}{A} \right) = 1.3 \, \mathrm{cm}^{-1}$. To Calculate: Molar conductivity $(\lambda_m)$ of solution. $\rightarrow$ $\rightarrow$ Molarity $= 5 \times 10^{-3} \frac{\mathrm{mol}}{\mathrm{L}}$ $\rightarrow$ Conductivity $= G \times \left( \frac{\ell}{A} \right) = 0.55 \mathrm{mS} \times \frac{1.3}{\frac{1}{100}} \, \mathrm{m}^{-1}$ $= 55 \times 1.3 \, \mathrm{mSm}^{-1}$ $\mathrm{eqn}^{n}(1) \lambda_m = \frac{1}{1000} \times \frac{55 \times 1.3}{\left( \frac{5}{1000} \right)} \, \mathrm{mSm}^2 \mathrm{mol}^{-1}$ $\Rightarrow \lambda_m = 14.3 \, \mathrm{mSm}^2 \mathrm{mol}^{-1}$

Given $t_2 = 54 \, min$ for A and $T_{1/2} = 18 \, min$ for B. At $t = 0$, concentration is $x \, M$. To calculate: $[A]_t = 16 \times [B]_t$ at time $t = ?$. For first order kinetics: $$[A]_t = \frac{A_0}{(2)^n}$$ where $n$ is the number of half-lives. Now from the relation (1), $[A]_t = 16 \times [B]_t$. $$\frac{x}{(2)^{n_1}} = \frac{x}{(2)^{n_2}} \times 16$$ implies $$(2)^{n_2} = (2)^{n_1} \times (2)^4$$ Therefore, $n_2 = n_1 + 4$. $$\frac{t}{(t_{1/2})_2} = \frac{t}{(t_{1/2})_1} + 4$$ $$t \left( \frac{1}{18} - \frac{1}{54} \right) = 4 \Rightarrow t = \frac{4 \times 18 \times 54}{36}$$ Therefore, $$t = 108 \, min$$

In Duma's method of estimation of Nitrogen. 0.1840 $\mathrm{gm}$ of organic compound gave 30 $\mathrm{mL}$ of nitrogen which is collected at 287 $\mathrm{K}$ $\&$ 758 $\mathrm{mm}$ of Hg. Given; Aqueous tension at 287 $\mathrm{K}$ = 14 $\mathrm{mm}$ of Hg. Hence actual pressure = (758 - 14) = 744 $\mathrm{mm}$ of Hg. Volume of nitrogen at STP = $\frac{273 \times 744 \times 30}{287 \times 760}$ $$V = 27.935 \mathrm{mL}$$ $\therefore$ 22400 $\mathrm{mL}$ of $\mathrm{N_2}$ at STP weighs = 28 $\mathrm{gm}$. $\therefore$ 27.94 $\mathrm{mL}$ of $\mathrm{N_2}$ at STP weighs = $\left$( $\frac{28}{22400}$ $\times$ 27.94 $\right$) $\mathrm{gm}$ = 0.0349 $\mathrm{gm}$ Hence $\%$ of Nitrogen = $\left$( $\frac{0.0349}{0.1840}$ $\times$ 100 $\right$) = 18.97$\%$ Rond off. Answer = 19$\%$

For, $n = 5$ $\ell = (0, 1, 2, 3, 4)$ If $\ell = 0$, $m = 0$ $\ell = 1$, $m = \{-1, 0, +1\}$ $\ell = 2$, $m = \{-2, -1, 0, +1, +2\}$ $\ell = 3$, $m = \{-3, -2, -1, 0, +1, +2, +3\}$ $\ell = 4$, $m = \{-4, -3, -2, -1, 0, +1, +2, +3, +4\}$ 5 d, 5f and 5 g subshell contain one-one orbital having $m_c = +2$

Given $P_A^0 = 21 kPa$ and $P_B^0 = 18 kPa$. An ideal solution is prepared by mixing 1 mol A and 2 mol B. Therefore, $X_A = \frac{1}{3}$ and $X_B = \frac{2}{3}$. According to Raoult's law, $$P_T = X_A P_A^0 + X_B P_B^0$$ $$\Rightarrow P_T = \left( \frac{1}{3} \times 21 \right) + \left( \frac{2}{3} \times 18 \right)$$ $$\Rightarrow P_T = 7 + 12 = 19 kPa$$

Given $\mathrm{H_2SO_3}$ [Dibasic acid] $c = 0.588 \, \mathrm{M}$. The pH of the solution is due to the first dissociation only since $K_{a_1} \gg K_{a_2}$. First dissociation of $\mathrm{H_2SO_3}$: $$\mathrm{H_2SO_3(aq)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{HSO_3^-(aq)} : k_{a_1} = 1.7 \times 10^{-2}$$ At $t = 0$: $C$ At $t$: $C - x x x$ $$\Rightarrow K_{a_1} = \frac{1.7}{100} = \frac{[\mathrm{H^+}][\mathrm{HSO_3^-}]}{[\mathrm{H_2SO_3}]}$$ $$\frac{1.7}{100} = \frac{x^2}{(0.58 - x)}$$ $$\Rightarrow 1.7 \times 0.588 - 1.7x = 100x^2$$ $$\Rightarrow 100x^2 + 1.7x - 1 = 0$$ $$\Rightarrow [\mathrm{H^+}] = x = \frac{-1.7 + \sqrt{(1.7)^2 + 4 \times 100 \times 1}}{2 \times 100} = 0.09186$$ Therefore, the pH of the solution is: $\mathrm{pH} = -\log[\mathrm{H^+}]$ $$\Rightarrow \mathrm{pH} = -\log(0.09186) = 1.036 \simeq 1$$

Given the reaction: $$3 \mathrm{Pb(NO_3)_2} + \mathrm{Cr_2(SO_4)_3} \rightarrow 3\mathrm{PbSO_4} + 2\mathrm{Cr(NO_3)_3}$$ Volumes and concentrations are given as: 35 ml of 0.15 M and 20 ml of 0.12 M. Calculating moles: $$= 5.25 \, \mathrm{m.mol} = 2.4 \, \mathrm{m \cdot mol} 5.25 \, \mathrm{m \cdot mol}$$ $$= 5.25 \times 10^{-3} \, \mathrm{mol}$$ Therefore, moles of $\mathrm{PbSO_4}$ formed: $$= 5.25 \times 10^{-3} = 525 \times 10^{-5}$$

Given $T = 298 \, \mathrm{K}$, $R = 8.314 \, \mathrm{\frac{J}{mol \cdot K}}$. The chemical reaction is $\mathrm{Fe} + 2\mathrm{HCl} \rightarrow \mathrm{FeCl_2} + \mathrm{H_2(g)}$. $50 \, \mathrm{g}$, $P = 1 \, \mathrm{bar}$. $$= \frac{50}{55.85} \, \mathrm{mol}$$ Work done for $1 \, \mathrm{mol}$ gas $= -P_{ext} \times \Delta V$. $$= \Delta n g R T$$ $$= -1 \times 8.314 \times 298 \, \mathrm{J}$$ Work done for $\frac{50}{55.85} \, \mathrm{mol}$ of gas $$= -1.8314 \times 298 \times \frac{50}{55.85} \, \mathrm{J}$$ $$= -2218.059 \, \mathrm{J}$$ $$\simeq -2218 \, \mathrm{J}$$

Given $\lambda_{absorbed} = 498 \, \mathrm{nm}$ (given). The octahedral splitting energy $\Delta_0$ or $E$ is calculated as follows: $$\Delta_0 or E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{498 \times 10^{-9}}$$ $$= 0.0399 \times 10^{-17} \, \mathrm{J}$$ $$= 3.99 \times 10^{-19} \, \mathrm{J}$$ $$= 4.00 \times 10^{-19} \, \mathrm{J} (round off)$$