JEE Main 25 February 2021 Shift 2 question paper with solutions

Given $$A = \begin{bmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{bmatrix}$$ Then $$2A = \begin{bmatrix} 2R_{11} & 2R_{12} & 2R_{13} \\ 2R_{21} & 2R_{22} & 2R_{23} \\ 2R_{31} & 2R_{32} & 2R_{33} \end{bmatrix}$$ Perform the row operation $$R_2 \to 2R_2 + 5R_3$$ Thus, $$B = \begin{bmatrix} 2R_{11} & 2R_{12} & 2R_{13} \\ 4R_{21} + 10R_{31} & 4R_{22} + 10R_{32} & 4R_{23} + 10R_{33} \\ 2R_{31} & 2R_{32} & 2R_{33} \end{bmatrix}$$ Perform the row operation $$R_2 \to R_2 - 5R_3$$ Then $$B = \begin{bmatrix} 2R_{11} & 2R_{12} & 2R_{13} \\ 4R_{21} & 4R_{22} & 4R_{23} \\ 2R_{31} & 2R_{32} & 2R_{33} \end{bmatrix}$$ The determinant is $$|B| = \begin{vmatrix} 2R_{11} & 2R_{12} & 2R_{13} \\ 4R_{21} & 4R_{22} & 4R_{23} \\ 2R_{31} & 2R_{32} & 2R_{33} \end{vmatrix}$$ This can be simplified to $$|B| = 2 \times 2 \times 4 \begin{vmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{vmatrix}$$ Which equals $$= 16 \times 4$$ Thus, $$= 64$$

Given the integral $$\int \frac{e^{3 \log_e x} + 5 e^{2 \log_e x}}{e^{\log_e x} + 5 e^{\log_e x} - 7 e^{2 \log_e x}} \, dx$$ we simplify the expression. First, rewrite the exponents: $$= \int \frac{x^3 + 5x^2}{x + 5x - 7x^2} \, dx$$ Simplify the denominator: $$= \int \frac{x^3 + 5x^2}{6x - 7x^2} \, dx$$ Factor the denominator: $$= \int \frac{x^2(x + 5)}{x(6 - 7x)} \, dx$$ Cancel common terms: $$= \int \frac{x^2 + 5x}{6 - 7x} \, dx$$ Let $t = 6 - 7x$, then $dt = -7 \, dx$ or $dx = -\frac{1}{7} \, dt$. Substitute and integrate: $$= \int \frac{-4}{t} \, dt$$ Integrate: $$= -4 \ln |t| + C$$ Substitute back for $t$: $$= 4 \ln |6 - 7x| + C$$ Thus, the solution is: $$= 4 \ln |x^2 + 5x - 7| + c$$

Shortest distance must be along common normal. Given $m_1$ (slope of line $x - y = 1$) $= 1$ implies slope of perpendicular line $= -1$. $m_2 = \frac{2x}{2} = x \implies m_2 = h$ implies slope of normal $= -\frac{1}{h}$. $-\frac{1}{h} = -1 \implies h = 1$. So point is $\left(1, \frac{1}{2}\right)$. $$D = \left| \frac{1 - \frac{1}{2} - 1}{\sqrt{1+1}} \right| = \frac{1}{2\sqrt{2}}$$

$(1 - 2i)^2 + \alpha(1 - 2i) + \beta = 0$ $1 - 4 - 4i + \alpha - 2i\alpha + \beta = 0$ $(\alpha + \beta - 3) - i(4 + 2\alpha) = 0$ $\alpha + \beta - 3 = 0 \qquad \& \qquad 4 + 2\alpha = 0$ $\alpha = -2,\quad \beta = 5$ $\alpha - \beta = -7$

Given $e_1 = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$, foci $(\pm ae, 0)$. Foci $=(\pm 3, 0)$. Let the equation of the hyperbola be $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$. Passes through $(\pm 3, 0)$, $A^2 = 9$, $A = 3$, $e_2 = \frac{5}{3}$. $$e_2^2 = 1 + \frac{B^2}{A^2}$$ $$\frac{25}{9} = 1 + \frac{B^2}{9} \Rightarrow B^2 = 16$$ Answer: $$\frac{x^2}{9} - \frac{y^2}{16} = 1$$

Given $$2 \cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right) - \left[ 2 \cos^2 \left( \frac{x+y}{2} \right) - 1 \right] = \frac{3}{2}$$ $$2 \cos \left( \frac{x+y}{2} \right) \left[ \cos \left( \frac{x-y}{2} \right) - \cos \left( \frac{x+y}{2} \right) \right] = \frac{1}{2}$$ $$2 \cos \left( \frac{x+y}{2} \right) \left[ 2 \sin \left( \frac{x}{2} \right) \cdot \sin \left( \frac{y}{2} \right) \right] = \frac{1}{2}$$ $$\cos \left( \frac{x+y}{2} \right) \cdot \sin \left( \frac{x}{2} \right) \cdot \sin \left( \frac{y}{2} \right) = \frac{1}{8}$$ Possible when $\frac{x}{2} = 30^\circ$ and $\frac{y}{2} = 30^\circ$ $$x = y = 60^\circ$$ $$\sin x + \cos y = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3}+1}{2}$$

Equation of plane passing through A, B, C will be $$\begin{vmatrix} x-1 & y-2 & z-3 \\ 2-1 & 3-2 & 1-3 \\ 2-1 & 4-2 & 2-3 \end{vmatrix} = 0$$ $$\Rightarrow \begin{vmatrix} 1 & 1 & -2 \\ 1 & 2 & -1 \end{vmatrix} = 0$$ $$\Rightarrow (x-1)(-1+4) - (y-2)(-1+2) + (z-3)(2-1) = 0$$ $$\Rightarrow (x-1)(3) - (y-2)(1) + (z-3)(1) = 0$$ $$\Rightarrow 3x - 3 - y + 2 + z - 3 = 0$$ $$\Rightarrow 3x - y + z - 4 = 0,$$ is the required plane. Now, given O(0, 0, 0) & P(2, -1, 1) Plane is $3x - y + z - 4 = 0$ $O'$ & $P'$ are foot of perpendiculars. For $O'$ $$\frac{x-0}{3} = \frac{y-0}{-1} = \frac{z-0}{1} = \frac{-(-0+0+4)}{9+1+1}$$ $$\frac{x}{3} = \frac{y}{-1} = \frac{z}{1} = \frac{4}{11}$$ $$\Rightarrow O'\left(\frac{12}{11}, -\frac{4}{11}, \frac{4}{11}\right)$$ For $P'$ $$\frac{x-2}{3} = \frac{y+1}{-1} = \frac{z-1}{1} = \frac{-3(2)-(-1)+1-4}{9+1+1}$$ $$\frac{x-2}{3} = \frac{y+1}{-1} = \frac{z-1}{1} = \left(-\frac{4}{11}\right)$$ $$P'\left(-\frac{12}{11} + 2, \frac{4}{11} - 1, -\frac{4}{11} + 1\right)$$ $$\Rightarrow P'\left(\frac{10}{11}, -\frac{7}{11}, \frac{7}{11}\right)$$ $O'P' = \sqrt{\left(\frac{10}{11} - \frac{12}{11}\right)^2 + \left(-\frac{7}{11} + \frac{4}{11}\right)^2 + \left(\frac{7}{11} - \frac{4}{11}\right)^2}$ $$\Rightarrow O'P' = \frac{1}{11}\sqrt{4 + 9 + 9}$$ $$\Rightarrow O'P' = \frac{\sqrt{22}}{11}$$ $$\Rightarrow O'P' = \frac{\sqrt{2}\sqrt{11}}{11}$$ $$\Rightarrow O'P' = \frac{\sqrt{2}}{11}$$

Based on Baye's theorem Probability = $$\frac{\left(160 \times \frac{35}{100}\right)}{\left(160 \times \frac{35}{100}\right) + \left(100 \times \frac{20}{100}\right) + \left(140 \times \frac{10}{100}\right)}$$ = $$\frac{5600}{9000}$$ = $$\frac{28}{45}$$

Given $$\csc \left( 2 \cot^{-1}(5) + \cos^{-1} \left( \frac{4}{5} \right) \right)$$ $$\csc \left( 2 \tan^{-1} \left( \frac{1}{5} \right) + \cos^{-1} \left( \frac{4}{5} \right) \right)$$ $$= \csc \left( \tan^{-1} \left( \frac{2 \left( \frac{1}{5} \right)}{1 - \left( \frac{1}{5} \right)^2} \right) + \cos^{-1} \left( \frac{4}{5} \right) \right)$$ $$= \csc \left( \tan^{-1} \left( \frac{5}{12} \right) + \cos^{-1} \left( \frac{4}{5} \right) \right)$$ Let $\tan^{-1}(5/12) = \theta \Rightarrow \sin \theta = \frac{5}{13}, \cos \theta = \frac{12}{13}$ and $\cos^{-1} \left( \frac{4}{5} \right) = \phi \Rightarrow \cos \phi = \frac{4}{5}$ and $\sin \phi = \frac{3}{5}$ $$= \csc(\theta + \phi)$$ $$= \frac{1}{\sin \theta \cos \phi + \cos \theta \sin \phi}$$ $$= \frac{1}{\frac{5}{13} \cdot \frac{4}{5} + \frac{12}{13} \cdot \frac{3}{5}} = \frac{65}{56}$$

Ellipse $\frac{x^2}{2} + \frac{y^2}{1} = 1$ Line $x + y = 1$ Using homogenisation $x^2 + 2y^2 = 2(1)^2$ $$x^2 + 2y^2 = 2(x + y)^2$$ $$x^2 + 2y^2 = 2x^2 + 2y^2 + 4xy$$ $$x^2 + 4xy = 0$$ for $ax^2 + 2hxy + by^2 = 0$ $$\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$$ $$\tan \theta = \left| \frac{2\sqrt{(2)^2-0}}{1+0} \right|$$ $$\tan \theta = -4$$ $$\cot \theta = -\frac{1}{4}$$ $$\theta = \cot^{-1} \left( -\frac{1}{4} \right)$$ $$\theta = \pi - \cot^{-1} \left( \frac{1}{4} \right)$$ $$\theta = \pi - \left( \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{4} \right) \right)$$ $$\theta = \frac{\pi}{2} + \tan^{-1} \left( \frac{1}{4} \right)$$

Contrapositive of $p \rightarrow q$ is $\sim q \rightarrow \sim p$. $p \rightarrow$ you will work $q \rightarrow$ you will earn money $\sim q \rightarrow$ you will not earn money $\sim p \rightarrow$ you will not work $\sim q \rightarrow \sim p \Rightarrow$ if you will not earn money, you will not work.

Given $$f(x) = \frac{5^x}{5^x + 5} \ldots (i)$$ $$f(2-x) = \frac{5^{2-x}}{5^{2-x} + 5}$$ $$f(2-x) = \frac{5}{5^x + 5} \ldots$$ Adding equation (i) and (ii) $$f(x) + f(2-x) = 1$$ $$f\left(\frac{1}{20}\right) + f\left(\frac{39}{20}\right) = 1$$ $$f\left(\frac{2}{20}\right) + f\left(\frac{38}{20}\right) = 1$$ $$f\left(\frac{19}{20}\right) + f\left(\frac{21}{20}\right) = 1$$ and $$f\left(\frac{20}{20}\right) = f(1) = \frac{1}{2}$$ $$\Rightarrow 19 + \frac{1}{2} \Rightarrow \frac{39}{2}$$

Given the equation: $$ \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix} \begin{bmatrix} 1 & \alpha \\ -\alpha & \beta \end{bmatrix} = \begin{bmatrix} 1 + \alpha^2 & \alpha - \alpha \beta \\ \alpha - \alpha \beta & \alpha^2 + \beta^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$ From this, we have: $$ 1 + \alpha^2 = 1 $$ $$ \alpha^2 = 0 $$ $$ \alpha^2 + \beta^2 = 1 $$ $$ \beta^2 = 1 $$ $$ \alpha^4 = 0 $$ $$ \beta^4 = 1 $$ $$ \alpha^4 + \beta^4 = 1 $$

AM $\geq$ GM $$\frac{a^{ax} + \frac{a}{a^{ax}}}{2} \geq \left(a^{ax} \cdot \frac{a}{a^{ax}}\right)^{1/2} \Rightarrow a^{ax} + a^{1-ax} \geq 2\sqrt{a}$$

Given $$I_{n+2} + I_n = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^n x \cdot \cos e^2 x \, dx = \left[ \frac{-(\cot x)^{n+1}}{n+1} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$ We have $$I_{n+2} + I_n = \frac{1}{n+1}$$ Also, $$I_2 + I_4 = \frac{1}{3}, I_3 + I_5 = \frac{1}{4}, I_4 + I_6 = \frac{1}{5}$$

The limit is given by $$\lim_{x \to \infty} \sum_{r=0}^{n-1} \frac{n}{(n+r)^2} = \lim_{x \to \infty} \sum_{r=0}^{n-1} \frac{n^2}{n^2 \left(1 + \frac{r}{n}\right)^2} = \int_0^1 \frac{dx}{(1+x)^2}$$ This simplifies to $$= -\left[ \frac{1}{1+x} \right]_0^1 \Rightarrow -\left[ \frac{1}{2} - 1 \right] = \frac{1}{2}$$

Total cases $ (4 \times 9 \times 9 \times 9) - (3 \times 9 \times 9) $. Probability = $$ \frac{(3 \times 9 \times 9) - (2 \times 9) + (8 \times 9 \times 9)}{(4 \times 9^3) - (3 \times 9^2)} $$ $$ = \frac{97}{217} $$

$x^2-6x-2=0$ $\alpha+\beta=6$ $\alpha\beta=-2$ and $\alpha^2-6\alpha-2=0\Rightarrow\alpha^2-2=6\alpha$ Now $\beta^2-6\beta-2=0\Rightarrow\beta^2-2=6\beta$ $\dfrac{a_{10}-2a_8}{3a_9}=\dfrac{(\alpha^{10}-\beta^{10})-2(\alpha^8-\beta^8)}{3(\alpha^9-\beta^9)}$ $=\dfrac{(\alpha^{10}-2\alpha^8)-(\beta^{10}-2\beta^8)}{3(\alpha^9-\beta^9)}$ $=\dfrac{\alpha^8(\alpha^2-2)-\beta^8(\beta^2-2)}{3(\alpha^9-\beta^9)}$ $=\dfrac{\alpha^8(6\alpha)-\beta^8(6\beta)}{3(\alpha^9-\beta^9)}$ $=\dfrac{6(\alpha^9-\beta^9)}{3(\alpha^9-\beta^9)}$ $=\dfrac{6}{3}$ $=2$

Number of elements in $A = 3$. Number of elements in $B = 5$. Number of elements in $A \times B = 15$. Number of one-one function $x = 5 \times 4 \times 3$. $$x = 60$$ Number of one-one function $y = 15 \times 14 \times 13$. $$y = 15 \times 4 \times \frac{14}{4} \times 13$$ $$y = 60 \times \frac{7}{2} \times 13$$ $$2y = (13)(7x)$$ $$2y = 91x$$

Given $$\Delta = \begin{vmatrix} 2 & 3 & 2 \\ 3 & 2 & 2 \\ 1 & -1 & 4 \end{vmatrix} = -20 \neq 0$$ Therefore, there is a unique solution. Calculate $$\Delta_x = \begin{vmatrix} 9 & 3 & 2 \\ 9 & 2 & 2 \\ 8 & -1 & 4 \end{vmatrix} = 0$$ Calculate $$\Delta_y = \begin{vmatrix} 2 & 9 & 2 \\ 3 & 9 & 2 \\ 1 & 8 & 4 \end{vmatrix} = -20$$ Calculate $$\Delta_z = \begin{vmatrix} 2 & 3 & 9 \\ 3 & 2 & 9 \\ 1 & -1 & 8 \end{vmatrix} = -40$$ Therefore, $$x = \frac{\Delta_x}{\Delta} = 0$$ $$y = \frac{\Delta_y}{\Delta} = 1$$ $$z = \frac{\Delta_z}{\Delta} = 2$$ Unique solution: (0,1,2)

Given: $7^n = (10 - 3)^n = 10K + (-3)^n$. Therefore, $7^n + 3^n = 10K + (-3)^n + 3^n$. If $n$ is odd, $3^n = 3^{2t} = (10 - 1)^t = 10p + (-1)^t = 10p \pm 1$. If $n$ is even, then $7^n + 3^n$ will not be a multiple of 10. So if $n$ is odd, then only $7^n + 3^n$ will be a multiple of 10. Therefore, $n = 11, 13, 15, \ldots, 99$. Answer: 45

Points of non-differentiability in $(-3, 3)$ are at $x = -2, -1, 0, 1, 2$. i.e. 5 points.

Given $\vec{a} = \hat{i} + \alpha \hat{j} + 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \alpha \hat{j} + \hat{k}$. The area of the parallelogram is $|\vec{a} \times \vec{b}|$. This is equal to $| (\hat{i} + \alpha \hat{j} + 3 \hat{k}) \times (3 \hat{i} - \alpha \hat{j} + \hat{k}) |$. Calculating, we have $8 \sqrt{3} = |(4 \alpha) \hat{i} + 8 \hat{j} - (4 \alpha) \hat{k}|$. Then, $(64)(3) = 16 \alpha^2 + 64 + 16 \alpha^2$. Simplifying, $(64)(3) = 32 \alpha^2 + 64$. Thus, $6 = \alpha^2 + 2$. Solving for $\alpha$, $\alpha^2 = 4$. Therefore, $\vec{a} = \hat{i} + \alpha \hat{j} + 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \alpha \hat{j} + \hat{k}$. The dot product $\vec{a} \cdot \vec{b} = 3 - \alpha^2 + 3$. This simplifies to $6 - \alpha^2 = 6 - 4 = 2$.

Let $x = 4k + 3$. $$(2020 + x)^{2022}$$ $$= (2020 + 4k + 3)^{2022}$$ $$= (4(505) + 4k + 3)^{2022}$$ $$= (4P + 3)^{2022}$$ $$= (4P + 4 - 1)^{2022}$$ $$= (4A - 1)^{2022}$$ $$2022C_0 (4A)^0 (-1)^{2022} + 2022C_1 (4A)^1 (-1)^{2021} + \ldots$$ $$1 + 8\lambda$$ Reminder is 1

Given the equations $4y^3 \frac{dy}{dx} = 1$ and $x \frac{dy}{dx} + y = 0$. Let $m_1 = \frac{1}{4y^3} \frac{dy}{dx}$ and $m_2 = -\frac{y}{x}$. We have $m_1 m_2 = -1$. Thus, $\frac{1}{4y^3} \times \frac{-y}{x} = -1$. Therefore, $x = y^4$. Now, $\frac{1}{4y^6} = 1$ and $xy = k$. This implies $y^6 = \frac{1}{4}$, which gives $k^6 = y^{30}$. Thus, $k^6 = \left(\frac{1}{4}\right)^5$. Therefore, $(4k)^6 = 4^6 \times k^6 = 4$.

Circle: $(x - 3)^2 + y^2 = 9$ Parabola: $y^2 = 4x$ Let tangent $y = mx + \frac{a}{m}$ $$y = mx + \frac{1}{m}$$ $$m^2 x - my + 1 = 0$$ the above line is also tangent to circle $$(x - 3)^2 + y^2 = 9$$ $$\therefore l from (3, 0) = 3$$ $$\frac{\left| \frac{1}{m} \right|}{\sqrt{m^2 + 1}} = 3$$ $$(3m^2 + 1)^2 = 9 \left(m^2 + m^4\right)$$ $$6m^2 + 1 + 9m^4 = 9m^2 + 9m^4$$ $$3m^2 = 1$$ $$m = \frac{1}{\sqrt{3}}$$ , tangent is $$y = \frac{1}{\sqrt{3}} x + \sqrt{3}$$ or $$y = -\frac{1}{\sqrt{3}} x - \sqrt{3}$$ (it will be used) (rejected) $$m = \frac{1}{\sqrt{3}}$$ for Parabola $\left(\frac{a}{m^2}, \frac{2a}{m}\right) \equiv (3, 2\sqrt{3})$ $(c, d)$ for Circle $y = \frac{1}{\sqrt{3}} x + \sqrt{3}$ $\&$ $(x - 3)^2 + y^2 = 9$ $$y = \frac{1}{\sqrt{3}} x + \sqrt{3}$$ solving, $$(x - 3)^2 + \left(\frac{1}{\sqrt{3}} x - \sqrt{3}\right)^2 = 9$$ $$x^2 + 9 - 6x + \frac{1}{3} x^2 - 6x + 9 = 9$$ $$4x^2 - 12x + 9 = 0$$ $$4x^2 - 6x - 6x + 9 = 0$$ $$2x(2x - 3) - 3(2x - 3) = 0$$ $$(2x - 3)(2x - 3) = 0$$ $$x = \frac{3}{2}$$ $$y = \frac{1}{\sqrt{3}} \left(\frac{3}{2}\right) + \sqrt{3}$$ $$y = \frac{\sqrt{3}}{2} + \sqrt{3}$$ $$(a, b) = \left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$$ $$y = \frac{3\sqrt{3}}{2}$$ $$2(a + c) = 2 \left(\frac{3}{2} + 3\right)$$ $$= 9$$

Given $\($ $\lim$_{x $\to$ 0} $\frac{ax - (e^{4x} - 1)}{ax(e^{4x} - 1)}$ $\)$. Applying L'Hospital's Rule $\($ $\lim$_{x $\to$ 0} $\frac{a - 4e^{4x}}{a(e^{4x} - 1) + ax(4e^{4x})}$ $\)$. So $\($ a = 4 $\)$. Applying L'Hospital's Rule $\($ $\lim$_{x $\to$ 0} $\frac{-16e^{4x}}{a(4e^{4x}) + a(4e^{4x}) + ax(16e^{4x})}$ $\)$. $\($ $\frac{-16}{4a + 4a}$ = $\frac{-16}{32}$ = -$\frac{1}{2}$ = b $\)$. $\($ a - 2b = 4 - 2 $\left$( -$\frac{1}{2}$ $\right$) = 4 + 1 = 5 $\)$

Given, $$(2xy^2 - y) \, dx + x \, dx = 0$$ $$\Rightarrow \frac{dy}{dx} 2y^2 - \frac{y}{x} = 0$$ $$\Rightarrow -\frac{1}{y^2} \frac{dy}{dx} + \frac{1}{y} \left( \frac{1}{x} \right) = 2$$ Let $\frac{1}{y} = z$ $$-\frac{1}{y^2} \frac{dy}{dx} = \frac{dz}{dx}$$ $$\Rightarrow \frac{dz}{dx} + z \left( \frac{1}{x} \right) = 2$$ $$F_x = e^{\int \frac{1}{x} \, dx} = x$$ $$z(x) = \int 2(x) \, dx = x^2 + c$$ $$\Rightarrow \frac{x}{y} = x^2 + c$$ As it passes through $P(2, 1)$ [Point of intersection of $2x - 3y = 1$ and $3x + 2y = 8$] $$\therefore \frac{2}{1} = 4 + c$$ $$\Rightarrow c = -2$$ $$\Rightarrow \frac{x}{y} = x^2 - 2$$ Put $x = 1$ $$\frac{1}{y} = 1 - 2 = -1$$ $$\Rightarrow y(1) = -1$$ $$\Rightarrow |y(1)| = 1$$

Given the integral $3 \int_{-2}^{2} |x^2 - x - 2| \, dx$. The expression $x^2 - x - 2$ can be factored as $(x - 2)(x + 1)$. This changes sign at $x = -1$ and $x = 2$. Therefore, we split the integral as follows: $$= 3 \left\{ \int_{-2}^{-1} (x^2 - x - 2) \, dx + \int_{-1}^{2} (-x^2 + x + 2) \, dx \right\}$$ Evaluating the integrals, we have: $$= 3 \left[ \left( \frac{x^3}{3} - \frac{x^2}{2} - 2x \right)_{-2}^{-1} - \left( \frac{x^3}{3} - \frac{x^2}{2} - 2x \right)_{-1}^{2} \right]$$ Simplifying, we find: $$= 19$$

Given $\ell_1: \vec{r} = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$. For $\ell_1: \frac{x-3}{1} = \frac{y+1}{2} = \frac{z-4}{2}$, the direction ratios of $\ell_1$ are $1, 2, 2$. For $\ell_2: \vec{r} = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$. For $\ell_2: \frac{x-3}{2} = \frac{y-3}{2} = \frac{z-2}{1}$, the direction ratios of $\ell_2$ are $2, 2, 1$. The direction ratios of $\ell$ is $1$ to $\ell_1$ and $\ell_2$. Direction ratios of $\ell \parallel (\ell_1 \times \ell_2) \Rightarrow \langle -2, 3, -2 \rangle$. Equation of $\ell$: $\frac{x}{2} = \frac{y}{-3} = \frac{z}{2}$. Solving $\ell$ and $\ell_1$: $$(2\lambda, -3\lambda, 2\lambda) = (\mu + 3, 2\mu - 1, 2\mu + \mu)$$ $$\Rightarrow 2\lambda = \mu + 3$$ $$-3\lambda = 2\mu - 1$$ $$2\lambda = 2\mu + 4$$ $$\mu = -1$$ $$\lambda = -1$$ $P(2, -3, 2)$ is the intersection point. Let $Q(2v + 3, 2v + 3, v + 2)$ be a point on $\ell_2$. Now, $PQ = \sqrt{(2v + 3 - 2)^2 + (2v + 3 + 3)^2 + (v + 2 - 2)^2} = \sqrt{17}$. $$(2v + 1)^2 + (2v + 6)^2 + (v)^2 = 17$$ $$9v^2 + 28v + 36 + 1 - 17 = 0$$ $$9v^2 + 28v + 20 = 0$$ $$9v^2 + 18v + 10v + 20 = 0$$ $$(9v + 10)(v + 2) = 0$$ $$v = -2 (rejected), -\frac{10}{9} (accepted)$$ $Q\left(3 - \frac{20}{9}, 3 - \frac{20}{9}, 2 - \frac{10}{9}\right)$ $$\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$$ Therefore, $18(a + b + c)$: $$= 18\left(\frac{7}{9} + \frac{7}{9} + \frac{8}{9}\right)$$ $$= 44$$

In n-type semiconductor pentavalent impurity is added. Each pentavalent impurity donates a free electron. So the Fermi-level of n-type semiconductor will go upward. In p-type semiconductor trivalent impurity is added. Each trivalent impurity creates a hole in the valence band. So the Fermi-level of p-type semiconductor will go downward.

In a ferromagnetic material, below the Curie temperature a domain is defined as a macroscopic region with saturation magnetization.

From p-v diagram, given $P v^{1/2} = constant$ ...(i) We know that $$PV = nRT$$ $$P \propto \left( \frac{T}{v} \right)$$ Put in equation (i) $$\left( \frac{T}{v} \right) (v)^{1/2} = constant$$ $$T \propto v^{1/2}$$ $$\frac{T_2}{T_1} = \sqrt{\frac{v_2}{v_1}}$$ $$\frac{T_2}{T_1} = \sqrt{\frac{2v_1}{v_1}}$$ $$\frac{T_2}{T_1} = \sqrt{2}$$

For particle (1) $20 + h = 10t + \frac{1}{2}gt^2$ $\ldots$ (i) For particle (2) $h = \frac{1}{2}gt^2$ $\ldots$ (ii) Put equation (ii) in equation (i) $$20 + \frac{1}{2}gt^2 = 10t + \frac{1}{2}gt^2$$ $t = 2 sec$ Put in equation (ii) $$h = \frac{1}{2}gt^2$$ $$= \frac{1}{2} \times 10 \times 2^2$$ $$h = 20 m$$ The height of the building $= 25 + 20 = 45 m$

The translational kinetic energy and rotational kinetic energy both obey Maxwell's distribution independent of each other. T.K.E of diatomic molecules = $\frac{3}{2} kT$ R.K.E. of diatomic molecules = $\frac{2}{2} kT$ So statement I is true but statement II is false.

Due to parallel combination, $K_{eff} = 2k + 2k$. Therefore, $T = 2\pi \sqrt{\frac{m}{k_{eff}}}$. $$= 2\pi \sqrt{\frac{m}{4k}}$$ $$T = \pi \sqrt{\frac{m}{k}}$$

Given frequency of massage signal $= f_m$ frequency of carrier signal $= f_c$ the wavelength of the corresponding signal in air is $\Rightarrow \lambda = \frac{c}{f_c}$

Given $\phi = \frac{q}{24 \varepsilon_0}$. $$\phi_T = \left( \frac{q}{24 \varepsilon_0} + \frac{q}{24 \varepsilon_0} \right) \times \frac{1}{2}$$ Therefore, $\phi_T = \frac{q}{24 \varepsilon_0}$.

Given $\Delta E = 10.2 \, \mathrm{eV}$. $$\frac{hc}{\lambda} = 10.2 \, \mathrm{eV}$$ Therefore, $$\lambda = \frac{hc}{(10.2) \, e}$$ $$= \frac{12400}{10.2} \, \mathrm{\AA}$$ $$= 121.56 \, \mathrm{nm}$$ $$\approx 121.8 \, \mathrm{nm}$$

Since $\phi$ remain same, circuit is in resonance. $$I_{rms} = \frac{V_{rms}}{z}$$ $$= \frac{220}{110}$$ $$I_{rms} = 2 \, A$$

From energy conservation $$mgh = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega^2$$ $$mgh = \frac{1}{2}mv_0^2 + \frac{1}{2} \times \frac{2}{5}ma^2 \times \frac{v_0^2}{a^2}$$ $$gh = \frac{1}{2}v_0^2 + \frac{1}{5}v_0^2$$ $$gh = \frac{7}{10}v_0^2$$ $$h = \frac{7}{10} \frac{v_0^2}{g}$$ From triangle, $\sin \theta = \frac{h}{\ell}$ then $h = \ell \sin \theta$ $$\ell \sin \theta = \frac{7}{10} \frac{v_0^2}{g}$$ $$\ell = \frac{7}{10} \frac{v_0^2}{g \sin \theta}$$

Given mass of electron = $m_e$. Mass of proton = $m_p$. Therefore, given $m_p = 1836 \, m_e$. From de-Broglie wavelength $$\lambda = \frac{h}{p} = \frac{h}{mv}$$ $$\frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e}$$ $$= \frac{1836 \, m_e}{m_e}$$ $$\frac{\lambda_e}{\lambda_p} = 1836$$

The initial phase angle $\phi_0 = \pi - \frac{\pi}{6}$ is equal to $$\frac{5\pi}{6}$$

Given $e = electronic charge$, $c = speed of light in free space$, $h = planck's constant$, $\frac{1}{4\pi \varepsilon_0} \frac{e^2}{hc} = \frac{ke^2}{hc} \times \frac{\lambda^2}{\lambda^2}$. $$= \frac{F \times \lambda}{E}$$ $$= \frac{E}{E}$$ $$= dimensionless = \left[M^0 L^0 T^0\right]$$

Since $v_1 \cos \alpha = v_2 \cos \beta$, we have $$\frac{v_1}{v_2} = \frac{\cos \beta}{\cos \alpha}.$$ Then the ratio of kinetic energies is $$\frac{k_1}{k_2} = \frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \left( \frac{v_1}{v_2} \right)^2 = \left( \frac{\cos \beta}{\cos \alpha} \right)^2.$$ Therefore, $$\frac{k_1}{k_2} = \frac{\cos^2 \beta}{\cos^2 \alpha}.$$

The point a covers $30^\circ$ in $0.1$ sec. Means $\frac{\pi}{6} \rightarrow 0.1 sec$. 1 $\rightarrow \frac{0.1}{\tau}$ $$2\pi = \frac{0.1 \times 6}{\pi} \times 2\pi$$ $T = 1.2 sec$ We know that $\omega = \frac{2\pi}{T}$ $$\omega = \frac{2\pi}{1.2}$$ Restoration force $(F) = m \omega^2 A$ Then Restoration force per unit mass $\left( \frac{F}{m} \right) = \omega^2 A$ $$\left( \frac{F}{m} \right) = \left( \frac{2\pi}{1.2} \right)^2 \times 0.36$$ $\cong 9.87 N$

If $A = B = 0$ then output $y = 1$. If $A = B = 1$ then output $y = 1$.

From the photoelectric effect equation $$\frac{hc}{\lambda} = \phi + ev_s$$ so $$ev_{s_1} = \frac{hc}{\lambda_1} - \phi \ldots$$ $$ev_{s_2} = \frac{hc}{\lambda_2} - \phi \ldots (ii)$$ Subtract equation (i) from equation (ii) $$ev_{s_1} - ev_{s_2} = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2}$$ $$v_{s_1} - v_{s_2} = \frac{hc}{e} \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)$$ $$(0.710 - 1.43) = 1240 \left( \frac{1}{491} - \frac{1}{\lambda_2} \right)$$ $$\frac{-0.72}{1240} = \frac{1}{491} - \frac{1}{\lambda_2}$$ $$\frac{1}{\lambda_2} = \frac{1}{491} + \frac{0.72}{1240}$$ $$\frac{1}{\lambda_2} = 0.00203 + 0.00058$$ $$\frac{1}{\lambda_2} = 0.00261$$ $$\lambda_2 = 383.14$$ $$\lambda_2 \approx 382 nm$$

Q6 (2) (a) Rectifier: used to convert a.c voltage into d.c. Voltage. (b) Stabilizer: used for constant output voltage even when the input voltage or load current change. (c) Transformer: used either for stepping up or stepping down the a.c. voltage. (d) Filter: used to remove any ripple in the rectified output voltage.

Given $$\sin \theta = \frac{1.22 \lambda}{D}$$ If $D$ is increased, then $\sin \theta$ will decrease. Therefore, the size of the circular fringe will decrease but intensity increases.

Given $$I = \frac{1}{2} c \varepsilon_0 E_0^2$$ $$\frac{8}{4 \pi \times 10^2} = \frac{1}{2} \times c \times \frac{1}{\mu_0 c^2} \times E_0^2$$ $$E_0 = \frac{2}{10} \sqrt{\frac{\mu_0 c}{\pi}}$$ Therefore, $\($ x = 2 $\)$

Given the forces in the diagram, we have: $$T \sin \theta = \frac{kq^2}{r^2}$$ $$T \cos \theta = mg$$ Thus, $$\tan \theta = \frac{kq^2}{mgr^2} = \frac{\tan \theta mg r^2}{k}$$ Solving for $q^2$: $$q^2 = \frac{1}{5} \times \frac{10 \times 10^{-6} \times 10 \times 0.2 \times 0.2}{9 \times 10^9}$$ Simplifying gives: $$q = \frac{2 \sqrt{2}}{3} \times 10^{-8}$$ After comparison from the given equation, $a = 20$

Here $R = radius of the earth$. From energy conservation $$-\frac{Gm_e m}{R} + \frac{1}{2} mv_i^2 = -\frac{Gm_e m}{11R} + 0$$ $$\frac{1}{2} mv_i^2 = \frac{10}{11} \frac{Gm_e m}{R}$$ $$v_i = \sqrt{\frac{20}{11} \frac{Gm_e}{R}}$$ $$v_i = \sqrt{\frac{10}{11}} \, v_e \{ \cdot escape velocity v_e = \sqrt{\frac{2Gm_e}{R}} \}$$ Then the value of $x = 10$

Given wavelength of an $x$-ray beam $= 10 Å$. Therefore, $E = \frac{hc}{\lambda} = mc^2$. $m = \frac{h}{c\lambda}$. The mass of a fictitious particle having the same energy as that of the $x$-ray photons $= \frac{x}{3} hkg$. $$\frac{x}{3} h = \frac{h}{c\lambda}$$ $$x = \frac{3}{c\lambda}$$ $$= \frac{3}{3 \times 10^8 \times 10 \times 10^{-10}}$$ $$x = 10$$

Given: $n = \frac{W}{Q_{in}} = \frac{1}{4}$ $$\frac{1}{4} = 1 - \frac{T_1}{T_2}$$ $$\frac{T_1}{T_2} = \frac{3}{4}$$ When the temperature of the sink is reduced by $52\, k$ then its efficiency is doubled. $$\frac{1}{2} = 1 - \frac{(T_1 - 52)}{T_2}$$ $$\frac{T_1 - 52}{T_2} = \frac{1}{2}$$ $$\frac{T_1}{T_2} - \frac{52}{T_2} = \frac{1}{2}$$ $$\frac{3}{4} - \frac{52}{T_2} = \frac{1}{2}$$ $$\frac{52}{T_2} = \frac{1}{4}$$ $$T_2 = 208\, k$$

Speed of transverse wave is $V = \sqrt{\frac{T}{\mu}}$. $$\frac{\Delta n}{v} = \frac{1}{2} \ln T - \frac{1}{2} \ln \mu$$ $$\frac{\Delta v}{v} = \frac{1}{2} \frac{\Delta T}{T}$$ $$= \frac{1}{2} \times 4$$ $$\frac{\Delta v}{v} = 2\%$$

If $\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P}$ only if $\vec{P} = 0$ or $\vec{Q} = 0$. The angle between $\vec{P}$ and $\vec{Q}$ is $\theta$ ($0^\circ < \theta < 360^\circ$). So $\theta = 180^\circ$.

When they are brought into contact and then separated by a distance $= 0.5 \, \mathrm{m}$, the charge distribution will be: The electrostatic force acting between the spheres is $$F_e = \frac{k q_1 q_2}{r^2}$$ $$= \frac{9 \times 10^9 \times 1 \times 10^{-9} \times 1 \times 10^{-9}}{(0.5)^2}$$ $$= \frac{900}{25} \times 10^{-9}$$ $$F_e = 36 \times 10^{-9} \, \mathrm{N}$$

The current $i_1 = \left( \frac{R_2}{R_1 + R_2} \right) i$ is given by $$i_1 = \left( \frac{2}{4+2} \right) \times 6$$ $$i_1 = 2 \, \mathrm{A}$$

Since the relation between kinetic energy and momentum is given by $$P = \sqrt{2m \mathrm{KE}} (\because KE is the same)$$ $$\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$$ $$\frac{n}{2} = \sqrt{\frac{4}{16}}$$ $$n = 1$$

Sp^3 d Hybridisation Sea-saw shape & axial bond length is more than equitorial bond length

The reaction shown is the conversion of aniline to phenyl isocyanide using chloroform and alcoholic KOH. This is known as the carbylamine reaction.

The structure shown is the $\($ $\alpha$ $\)$-anomer of maltose.

The reaction begins with the addition of $H^+$ to the nitroalkene, resulting in the formation of a carbocation. The structure of the carbocation is $\mathrm{CH_3-CH-CH-CH_3}$ with a $\mathrm{NO_2}$ group attached to the second carbon. This carbocation then reacts with benzene under heat ($\Delta$) to form the final product, which is a nitroalkane with a benzene ring attached.

The reaction sequence involves the following steps: 1. Friedel-Crafts acylation of benzene with $\mathrm{CH_3COCl}$ in the presence of $\mathrm{AlCl_3}$ to form acetophenone ($\mathrm{COCH_3}$ group attached to benzene). 2. Clemmensen reduction of acetophenone using $\mathrm{Zn-Hg/HCl}$ to form ethylbenzene ($\mathrm{C_2H_5}$ group attached to benzene). 3. Bromination of ethylbenzene using $\mathrm{Br_2/Fe}$ to form bromoethylbenzene. 4. Nitration of bromoethylbenzene using concentrated $\mathrm{HNO_3}$ and $\mathrm{H_2SO_4}$ to form the final product with $\mathrm{NO_2}$ and $\mathrm{Br}$ groups on the benzene ring.

The reaction is given by: $$\mathrm{H_2O + CO_2 \rightarrow H_2CO_3}$$

Acidity of carboxylic acid is proportional to $-R > -H > -I$ and inversely proportional to $\frac{1}{+R > +H > +I}$. The order of acidity is: $$II > III > IV > I$$

The reaction involves nitration of aniline using $\mathrm{HNO_3}$ and $\mathrm{H_2SO_4}$ at $288 \, \mathrm{K}$. The products formed are ortho-nitroaniline (A), meta-nitroaniline (B), and para-nitroaniline (C). The percentages of each product are as follows: ortho-nitroaniline (A) is $51\%$, meta-nitroaniline (B) is $47\%$, and para-nitroaniline (C) is $2\%$.

Fact based $\mathrm{F_2}$ has $\mathrm{F} - \mathrm{F}$, $\mathrm{F_2}$ involves repulsion of non-bonding electrons and moreover its size is small and hence due to high repulsion its bond dissociation energy is very low.

Both statements are correct

Fact German silver is an alloy which does not have silver. Cu – 50%; Ni – 30%; Zn – 20%

$[\mathrm{FeF_6}]^{3-}$ $\mathrm{Fe^{3+}}$ $3\,d^5$ $\rightarrow$ 5 unpaired electrons as $\mathrm{F^-}$ is a weak field ligand. $[\mathrm{Co(NH_3)_6}]^{3+}$ $\mathrm{Co^{3+}}$ $3\,d^6$ $\rightarrow$ No unpaired electron as $\mathrm{NH_3}$ is a strong field ligand and causes pairing. $[\mathrm{NiCl_4}]^{2-}$ $\mathrm{Ni^{2+}}$ $3\,d^8$ $\rightarrow$ 2 unpaired electrons. $[\mathrm{Cu(NH_3)_4}]^{2+}$ $\mathrm{Cu^{2+}}$ $3\,d^9$ $\rightarrow$ 1 unpaired electron.

\[ \begin{aligned} \mathrm{NaCN + HNO_3} &\rightarrow \mathrm{NaNO_3 + HCN}\uparrow \\ \mathrm{Na_2S + 2HNO_3} &\rightarrow \mathrm{2NaNO_3 + H_2S}\uparrow \end{aligned} \] Nitric acid decomposes $\mathrm{NaCN}$ and $\mathrm{Na_2S}$; otherwise, they precipitate during the test and interfere with the result.

Fact base

The equilibrium reaction is given by: $$\mathrm{Ca(OH)_2 \rightleftharpoons Ca^{+2}_s + \frac{2\mathrm{OH}^-}{2s+10^{-7}}}$$ The equation for solubility is: $$s(2s + 10^{-7})^2 = 55 \times 10^{-7}$$ Simplifying, we have: $$4s^3 = 55 \times 10^{-7}$$ Solving for $s^3$: $$s^3 = \frac{5500}{4} \times 10^{-9}$$ Taking the cube root, we find: $$s = \left(\frac{2250}{2}\right)^{1/3} \times 10^{-3}$$ Simplifying further: $$s = (1125)^{1/3} \times 10^{-3}$$ Finally, we get: $$s = 1.11 \times 10^{-2}$$

The reaction involves $\mathrm{Ni^{2+}}$ with dimethylglyoxime in the presence of $\mathrm{NH_4OH}$. The dimethylglyoxime acts as a ligand, forming a complex with nickel, resulting in a rose red precipitate. Dimethylglyoxime is a negative bidentate ligand.

The reaction involves the hydroformylation of an alkene. The given alkene is $\mathrm{CH_3CH_2CH=CH_2}$. Under the influence of $\mathrm{H_2/CO}$ and a rhodium catalyst, the alkene undergoes hydroformylation to form an aldehyde. The product is $\mathrm{CH_3CH_2CH_2CHO}$.

Fact Ga, In, Si, Ge are refined by zone refining or vacuum refining.

The reaction is given as follows: $$\mathrm{CH_2 - OH \ CH_2 - OH + \ oxalic \ acid \ \xrightarrow{210^\circ C} \ CH_2 = CH_2}$$

The transformation from rhombic sulphur to monoclinic sulphur is given by the equation: $$S_{Rhombic} \rightleftharpoons S_{Monoclinic}$$ At $95.6^\circ C$, $\alpha$-sulphur converts to $\beta$-sulphur.

The reaction is given by $AB \rightarrow A^+ + B^-$. The concentrations are $1 - \alpha$, $\alpha$, and $\alpha$. Given $\alpha = 3/4$ and $N = 2$. The van't Hoff factor $i$ is calculated as $i = [1 + (2 - 1)\alpha]$. Substituting the values, we have: $$2.5 = [1 + (2 - 1)3/4] \times 0.52 \times m$$ Solving for $m$: $$m = \frac{2.5}{0.52 \times 7/4} = \frac{10}{3.64} = 2.747$$ Thus, $m = 2.747 \approx 3 \, \mathrm{mol/kg}$.

\textbf{Solution:} Sulphanilic acid contains the \(-\mathrm{SO_3H}\) group, not \(-\mathrm{COOH}\). Picric acid is \(2,4,6\)-trinitrophenol and contains a phenolic \(-\mathrm{OH}\) group. Aspirin (acetylsalicylic acid) contains one \(-\mathrm{COOH}\) group. Ascorbic acid (vitamin C) is a lactone and does not contain a \(-\mathrm{COOH}\) group. Therefore, only aspirin contains the \(-\mathrm{COOH}\) group. \[ \boxed{1} \]

$\dfrac{K_{52^\circ\mathrm{C}}}{K_{27^\circ\mathrm{C}}}=5$ $\ln\left(\dfrac{k_{T_2}}{k_{T_1}}\right)=\dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ $\ln(5)=\dfrac{E_a}{R}\left(\dfrac{1}{300}-\dfrac{1}{325}\right)$ $2.303\times0.7=\dfrac{E_a}{8.314}\times\dfrac{25}{300\times325}$ $E_a=51524.96\ \mathrm{J\,mol^{-1}}$ $E_a=51.524\ \mathrm{kJ\,mol^{-1}}$ $\therefore\ 52$

Cs is used in photoelectric cell due to its very low ionization potential.

The electronic configuration of $^{29}\mathrm{Cu}^{+2}$ is $[\mathrm{Ar}]^{18}3d^9$. The diagram shows the $d^9$ configuration with one unpaired electron. The number of unpaired electrons is 1. The magnetic moment $\mu$ is given by $\mu = \sqrt{n(n+2)}$. $$\mu = \sqrt{(1)(1+2)} = \sqrt{3}B \cdot M$$ The calculated magnetic moment is approximately $1.73$.

Energy required to ionize an atom of metal $A' = \frac{hc}{\lambda} = \frac{hc}{663 \, \mathrm{nm}}$. For 1 mole atoms of $A'$, total energy required $= N_A \times \frac{hc}{\lambda}$. $$= \frac{6.023 \times 10^{23} \times 6.63 \times 10^{-34} \times 3 \times 10^8}{663 \times 10^{-9}}$$ $$= 6.023 \times 3 \times 10^{23 - 34 + 8 + 7}$$ $$= 18.04 \times 10^4 \, \mathrm{J/mol}$$ $$= 180.4 \, \mathrm{KJ/mol}$$ Nearest Integer $= 180 \, \mathrm{KJ/mol}$.

Eq. of NaOH = Eq. of oxalic acid. $[NaOH] \times 1 \times 4.4 = \frac{5}{4} \times 2 \times 10$ $$[NaOH] = \frac{100}{4 \times 4.4} = \frac{25}{4.4} = 5.68$$ Nearest integer = 6M Ans.

Moles $(n) = 5$ $T = 293 \, \mathrm{K}$ Process $= \mathrm{IsoT.} \rightarrow \mathrm{Irreversible}$ $P_{\mathrm{ini}} = 2.1 \, \mathrm{MPa}$ $P_t = 1.3 \, \mathrm{MPa}$ $P_{\mathrm{ext}} = 4.3 \, \mathrm{mPa}$ Work $= -P_{\mathrm{ext}} \Delta v$ $$= -4.3 \times \left( \frac{5 \times 293 R}{1.3} - \frac{5 \times 293}{2.1} \right)$$ $$= -5 \times 293 \times 8.314 \times 43 \left( \frac{1}{13} - \frac{1}{21} \right)$$ $$= \frac{5 \times 293 \times 8.314 \times 43 \times 8}{21 \times 13}$$ $$= -15347.7049 \, \mathrm{J} = -15.34 \, \mathrm{kJ}$$ Isothermal process, so $\Delta U = 0$ $w = -Q$ $Q = 15.34 \, \mathrm{kJ/mol}$ So answer is 15

\textbf{Anode} $$\text{Cu(s)} \rightarrow \text{Cu}^{+2} + 2e^-$$ \textbf{Cathode (1)} $\dfrac{3e^- + 4H^+ + NO_3^- \rightarrow NO + 2H_2O}{8H^- + 2NO_3^- + 3Cu(s) \rightarrow 3Cu^{+2} + 2NO + 4H_2O}$ $$Q = \frac{\left[\text{Cu}^{+2}\right]^3 \times (p_{NO})^2}{\left[NO_3^-\right]^2 \left[H^+\right]^8}$$ $$\epsilon^0_{\text{cell}} = 1.3$$ $$\epsilon_{\text{cell}} = 1.3 - \frac{0.059}{6} \log \frac{\left(\text{Cu}^{+2}\right)^3 (p_{NO})^2}{\left(NO_3^-\right)^2 \times (H^+)^8} \quad \cdots(1)$$ \textbf{Anode} $\text{Cu(s)} \rightarrow \text{Cu}^{+2} + 2e$ \textbf{Cathode} $\dfrac{e^- + 2n^+ + NO_3^- \rightarrow NO_2 + H_2O}{Cu(s) + 4H^+ + 2NO_3^- \rightarrow 2NO_2 + 2H_2O + Cu^{+2}}$ $$\epsilon^0_{\text{cell}} = 1.13$$ $$Q = \frac{(\text{Cu}^{+2})(p_{NO_2})^2}{(NO_3^-)^2 (H^+)^4}$$ $$\epsilon_{\text{cell}} = 1.13 - \frac{0.059}{2} \log \frac{(\text{Cu}^{+2})(p_{NO_2})^2}{(NO_3^-)^2 (H^+)^4}$$ $$\epsilon_{\text{cell\_2}} = \epsilon_{\text{cell\_2}}$$ $$1.3 - \frac{0.059}{6} \log(Q_1) = 1.13 - \frac{0.059}{2} \log(Q_2)$$ $$0.17 = \frac{0.059}{6} \left\{ \log(Q_1) - 3\log(Q_2) \right\}$$ $$= \frac{0.059}{6} \left\{ \log \frac{(\text{Cu}^{+2})^3 \times (p_{NO})^2 \times (NO_3^-)^6 (H^+)^{12}}{(NO_3^-)^2 (H^+)^8 \times (\text{Cu}^{+2})^3 \times (p_{NO_2})^6} \right\}$$ $$= \frac{0.059}{6} \left\{ \log \frac{[NO_3^-]^4 [H^+]^4}{(P_{NO_2})^4} \right\}$$ $$0.17 = \frac{0.059}{6} \times 8\log(HNO_3)$$ $$\log(\text{HNO}_3) = 2.16$$ $$[\text{HNO}_3] = 10^{2.16} = 10^x$$ $$x = 2.16 \Rightarrow 2x = 4.32 \approx 4$$

Given $a = 3.596 \, Å$. The formula for $d$ is given by $$d = \frac{Z \times GMM}{N_A \times a^3}$$ Substituting the values, we have $$d = \frac{4 \times 63.54 \times 10^{-3}}{6.022 \times 10^{23} \times (3.596 \times 10^{-10})^3}$$ Calculating this gives $$d = 0.9076 \times 10^4 = 9076.2 \, kg/m^3$$