JEE Main 6 April 2024 Shift 1 question paper with solutions

Given $$f'(x) = 3x^2 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right)$$ Then $$f''(x) = 6x \sin\left(\frac{1}{x}\right) - 3 \cos\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) - \frac{\sin\left(\frac{1}{x}\right)}{x}$$ Evaluating at $$f''\left(\frac{2}{\pi}\right) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24 - \pi^2}{2\pi}$$

The area is given by the formula: $$Area = \frac{1}{2} |\overline{BD} \times \overline{AC}|$$ The vector $\($ $\overline{BD}$ $\)$ is given by: $$\overline{BD} = \frac{5}{3} \hat{i} - \frac{5}{3} \hat{j} - \frac{2}{3} \hat{k}$$ The vector $\($ $\overline{AC}$ $\)$ is given by: $$\overline{AC} = \hat{i} - \hat{j} - 2\hat{k}$$

Divide numerator and denominator by $\cos x$. $$\int_0^{\pi/4} \frac{\tan^2 x \sec^2 x \, dx}{(1 + \tan^3 x)^2} \, dx$$ Let $1 + \tan^3 x = t$. $$\tan^2 x \sec^2 x \, dx = \frac{dt}{3}$$ $$\frac{1}{3} \int_1^2 \frac{dt}{t^2} = \frac{1}{6}$$

Mean $\left( \bar{x} \right) = 10$ $$\Rightarrow \frac{\Sigma x_i}{20} = 10$$ $$\Sigma x_i = 10 \times 20 = 200$$ If 8 is replaced by 12, then $\Sigma x_i = 200 - 8 + 12 = 204$ Therefore, correct mean $\left( \bar{x} \right) = \frac{\Sigma x_i}{20}$ $$= \frac{204}{20} = 10.2$$ Therefore, standard deviation $= 2$ Therefore, variance $= (S.D.)^2 = 2^2 = 4$ $$\Rightarrow \frac{\Sigma x_i^2}{20} - \left( \frac{\Sigma x_i}{20} \right)^2 = 4$$ $$\Rightarrow \frac{\Sigma x_i^2}{20} - (10)^2 = 4$$ $$\Rightarrow \frac{\Sigma x_i^2}{20} = 104$$ $$\Rightarrow \Sigma x_i^2 = 2080$$ Now, replaced '8' observations by '12' Then, $\Sigma x_i^2 = 2080 - 8^2 + 12^2 = 2160$ Therefore, variance of removing observations $$\Rightarrow \frac{\Sigma x_i^2}{20} - \left( \frac{\Sigma x_i}{20} \right)^2$$ $$= \frac{2160}{20} - (10.2)^2$$ $$\Rightarrow 108 - 104.04$$ $$\Rightarrow 3.96$$ Correct standard deviation $$= \sqrt{3.96}$$

Given $f(x) = \frac{(x+5)(x-3)}{x^2 - 4x + 9}$. Let $g(x) = x^2 - 4x + 9$. $D 0$ for $x \in \mathbb{R}$. Therefore, $$f(-5) = 0$$ $$f(3) = 0$$ So, $f(x)$ is many-one. Again, $$yx^2 - 4xy + 9y = x^2 + 2x - 15$$ $$x^2(y - 1) - 2x(2y + 1) + (9y + 15) = 0$$ for $\forall x \in \mathbb{R} \Rightarrow D \geq 0$ $$D = 4(2y + 1)^2 - 4(y - 1)(9y + 15) \geq 0$$ $$5y^2 + 2y + 16 \leq 0$$ $$(5y - 8)(y + 2) \leq 0$$ The range of $y$ is $y \in \left[-2, \frac{8}{5}\right]$.

Given $n(3) \Rightarrow$ multiple of $3$. The sequence is $102, 105, 108, \ldots, 699$. $$T_n = 699 = 102 + (n-1)(3)$$ Solving for $n$, we get $n = 200$. Thus, $n(3) = 200$. $n(4) \Rightarrow$ multiple of $4$. The sequence is $100, 104, 108, \ldots, 700$. $$T_n = 700 = 100 + (n-1)(4)$$ Solving for $n$, we get $n = 151$. Thus, $n(4) = 151$. For $n(3 \cap 4) \Rightarrow$ multiple of both $3$ and $4$. The sequence is $108, 120, 132, \ldots, 696$. $$T_n = 696 = 108 + (n-1)(12)$$ Solving for $n$, we get $n = 50$. Thus, $n(3 \cap 4) = 50$. Now, $n(3 \cup 4) = n(3) + n(4) - n(3 \cap 4)$ $$= 200 + 151 - 50$$ $$= 301$$ Therefore, $n(\overline{3 \cup 4}) = \text{Total} - n(3 \cup 4) =$ neither a multiple of $3$ nor a multiple of $4$ $$= 601 - 301 = 300$$

Equation of circle $$x^2 + (y - (6 - r))^2 = r^2$$ touches $\sqrt{3}x - y = 0$ $$p = r$$ $$\frac{|0 - (6 - r)|}{2} = r$$ $$|r - 6| = 2r$$ $$r = 2$$ Therefore, circle $x^2 + (y - 4)^2 = 4$ $(2, 4)$ satisfies this equation

Given $$A_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & \frac{n(3n-1)}{2} \end{vmatrix}$$ We have $$2A_{10} - A_8 = \begin{vmatrix} 20 & 1 & \frac{n^2}{2} + \alpha \\ 40 & 2 & n^2 - \beta \\ 56 & 3 & \frac{n(3n-1)}{2} \end{vmatrix} - \begin{vmatrix} 8 & 1 & \frac{n^2}{2} + \alpha \\ 16 & 2 & n^2 - \beta \\ 22 & 3 & \frac{n(3n-1)}{2} \end{vmatrix}$$ This simplifies to $$\begin{vmatrix} 12 & 1 & \frac{n^2}{2} + \alpha \\ 24 & 2 & n^2 - \beta \\ 34 & 3 & \frac{n(3n-1)}{2} \end{vmatrix}$$ Further simplifying, we get $$\begin{vmatrix} 0 & 1 & \frac{n^2}{2} + \alpha \\ 0 & 2 & n^2 - \beta \\ -2 & 3 & \frac{n(3n-1)}{2} \end{vmatrix}$$ This leads to $$\Rightarrow -2 \left( (n^2 - \beta) - (n^2 + 2\alpha) \right)$$ Finally, $$\Rightarrow -2(-\beta - 2\alpha) \Rightarrow 4\alpha + 2\beta$$

Given the equations: $$\frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5} \& \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3}$$ The scalar distance (S.D) is given by: $$S.D = \frac{\left| \left( \overline{a_2 - a_1} \right) \cdot \left( \overline{b_1 - b_2} \right) \right|}{\left| \overline{b_1 \times b_2} \right|}$$ Where: $$a_1 = 3, -15, 9$$ $$b_1 = 2, -7, 5$$ $$a_2 = -1, 1, 9$$ $$b_2 = 2, 1, -3$$ Calculate $a_2 - a_1$: $$a_2 - a_1 = -4, 16, 0$$ Calculate $\overline{b_1 \times b_2}$: $$\overline{b_1 \times b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(16) - \hat{j}(-16) + \hat{k}(16)$$ Thus: $$16(\hat{i} + \hat{j} + \hat{k})$$ Magnitude of $\overline{b_1 \times b_2}$: $$\left| \overline{b_1 \times b_2} \right| = 16\sqrt{3}$$ Therefore: $$\left( \overline{a_2 - a_1} \right) \cdot \left( \overline{b_1 - b_2} \right) = 16[-4 + 16] = (16)(12)$$ Finally, the scalar distance is: $$S.D. = \frac{(16)(12)}{16\sqrt{3}} = 4\sqrt{3}$$

\begin{tabular}{|c|c|c|} \hline & A & B \\ \hline Manufactured & 60\% & 40\% \\ \hline Standard quality & 80\% & 90\% \\ \hline \end{tabular} $P(\text{Manufactured at B } \mid \text{ found standard quality})=?$ $A:$ Found S.Q $B:$ Manufacture B $C:$ Manufacture A $P(E_1)=\dfrac{40}{100}$ $P(E_2)=\dfrac{60}{100}$ $P(A\mid E_1)=\dfrac{90}{100}$ $P(A\mid E_2)=\dfrac{80}{100}$ $\because\ P(E_1\mid A)$ $=\dfrac{P(A\mid E_1)P(E_1)}{P(A\mid E_1)P(E_1)+P(A\mid E_2)P(E_2)}$ $=\dfrac{\frac{9}{10}\times\frac{4}{10}}{\frac{9}{10}\times\frac{4}{10}+\frac{8}{10}\times\frac{6}{10}}$ $=\frac{3}{7}$ $\therefore\ 126P=54$

By Newton's theorem $$a_{n+2} - \left(t^2 - 5t + 6\right) a_{n+1} + a_n = 0$$ Therefore, $$a_{2025} + a_{2023} = \left(t^2 - 5t + 6\right) a_{2024}$$ Thus, $$\frac{a_{2025} + a_{2023}}{a_{2024}} = t^2 - 5t + 6$$ Therefore, $$t^2 - 5t + 6 = \left(t - \frac{5}{2}\right)^2 - \frac{1}{4}$$ Thus, the minimum value is $$-\frac{1}{4}$$

Given $x = \{1, 2, 3, \ldots, 20\}$. $R_1 = \{(x, y) : 2x - 3y = 2\}$ $R_2 = \{(x, y) : -5x + 4y = 0\}$ $R_1 = \{(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)\}$ $R_2 = \{(4, 5), (8, 10), (12, 15), (16, 20)\}$ In $R_1$ 6 elements are needed. In $R_2$ 4 elements are needed. So, total $6 + 4 = 10$ elements.

The equation of the line is $$y + 9 = m(x - 4)$$ Therefore, $$A = \left( \frac{9 + 4m}{m}, 0 \right)$$ $$B = (0, -9 - 4m)$$ Thus, $$OA + OB = \frac{9 + 4m}{m} + 9 + 4m$$ Since $m > 0$, $$= 13 + \frac{9}{m} + 4m$$ Therefore, $$\frac{4m + \frac{9}{m}}{2} \geq \sqrt{36} \Rightarrow 4m + \frac{9}{m} \geq 12$$ Thus, $$OA + OB \geq 25$$

Given $f(x) = x^x; x > 0$. Let $\ln n = x \ln x$. Then, $\($ $\frac{1}{y}$ $\frac{dy}{dx}$ = $\frac{x}{x}$ + $\ln$ n $\)$. Therefore, $\($ $\frac{dy}{dx}$ = x^x (1 + $\ln$ x) $\)$. For strictly increasing, $\($ $\frac{dy}{dx}$ $\geq$ 0 $\Rightarrow$ x^x (1 + $\ln$ x) $\geq$ 0 $\)$. This implies $\($ $\ln$ x $\geq$ -1 $\)$. Thus, $\($ x $\geq$ e^{-1} $\)$ or $\($ x $\geq$ $\frac{1}{e}$ $\)$. Therefore, $\($ x $\in$ $\left$[ $\frac{1}{e}$, $\infty$ $\right$) $\)$.

Given $r = \frac{\Delta}{s} = \frac{\sqrt{3}a^2}{4 \cdot \frac{3a}{2}} = \frac{a}{2\sqrt{3}} = \frac{12}{2\sqrt{3}} = 2$. Therefore, $A = r\sqrt{2} = 2\sqrt{6}$. The area is $m = A^2 = 24$. The perimeter is $n = 4A = 8\sqrt{6}$. Thus, $m + n^2 = 24 + 384 = 408$.

The number of triangles having no side common with an $n$-sided polygon is given by $$\frac{{^nC_1 \cdot ^{n-4}C_2}}{3}$$ For $n = 8$, this becomes $$\frac{{^8C_1 \cdot ^4C_2}}{3} = 16$$

$\dfrac{dy}{dx}+\dfrac{y}{1+x^2}=\dfrac{e^{\tan^{-1}x}}{1+x^2}$ I.F. $=e^{\int \frac{1}{1+x^2}\,dx}=e^{\tan^{-1}x}$ $y\cdot e^{\tan^{-1}x}=\int\left(\dfrac{e^{\tan^{-1}x}}{1+x^2}\right)e^{\tan^{-1}x}\,dx$ Let $\tan^{-1}x=z$ $\therefore\ \dfrac{dx}{1+x^2}=dz$ $\therefore\ y\cdot e^z=\int e^{2z}\,dz$ $=\dfrac{e^{2z}}{2}+C$ $\therefore\ y\cdot e^{\tan^{-1}x}=\dfrac{e^{2\tan^{-1}x}}{2}+C$ $\Rightarrow\ y=\dfrac{e^{\tan^{-1}x}}{2}+\dfrac{C}{e^{\tan^{-1}x}}$ $\therefore\ y(1)=0$ $\Rightarrow\ 0=\dfrac{e^{\pi/4}}{2}+\dfrac{C}{e^{\pi/4}}$ $\Rightarrow\ C=-\dfrac{e^{\pi/2}}{2}$ $\therefore\ y=\dfrac{e^{\tan^{-1}x}}{2}-\dfrac{e^{\pi/2}}{2e^{\tan^{-1}x}}$ $\therefore\ y(0)=\dfrac{1-e^{\pi/2}}{2}$

\[ \frac{dy}{dx}+\frac{y}{x\ln x}=\frac{3}{2x^2} \] \[ \text{I.F.}=e^{\int \frac{1}{x\ln x}\,dx} =e^{\ln(\ln x)} =\ln x \] \[ \therefore\ y\ln x=\int \frac{3\ln x}{2x^2}\,dx \] \[ =\frac{3\ln x}{2}\int x^{-2}\,dx -\int \left(\frac{3}{2x}\right)\left(-\frac{1}{x}\right)\,dx \] \[ =\frac{3\ln x}{2}\left(-\frac{1}{x}\right) -\int \frac{3}{2x^2}\,dx \] \[ y\ln x=-\frac{3\ln x}{2x}+\frac{3}{2x}+C \] \[ y(e^{-1})=0 \] \[ \therefore\ 0(-1)=\frac{3e}{2}-\frac{3e}{2}+C \Rightarrow C=0 \] \[ \therefore\ y=\frac{-3\ln x}{2x\ln x} +\frac{3}{2x\ln x} \] \[ \therefore\ y(e)=\frac{-3}{2e} +\frac{3}{2e} =\frac{-3}{e} \]

Given $y = 3x$, $2y = 27 - 3x$, and $y = 3x - x\sqrt{x}$. The area $A$ is given by: $$A = \int_0^3 3x - (3x - x\sqrt{x}) \, dx + \int_3^9 \left( \frac{27 - 3x}{2} - (3x - x\sqrt{x}) \right) \, dx$$ Simplifying, we have: $$A = \int_0^3 x^{3/2} \, dx + \int_3^9 \frac{27}{2} - \frac{9x}{2} + x^{3/2} \, dx$$ Evaluating the integrals: $$A = \left[ \frac{2x^{5/2}}{5} \right]_0^3 + \frac{27}{2} [x]_3^9 - \frac{9}{2} \left[ \frac{x^2}{2} \right]_3^9 + \left[ \frac{2x^{5/2}}{5} \right]_3^9$$ Calculating each term: $$A = \frac{2}{5} (3^{5/2}) + \frac{27}{2} (6) - \frac{9}{4} (72) + \frac{2}{5} (9^{5/2} - 3^{5/2})$$ Simplifying further: $$A = \frac{2}{5} (3^{5/2}) + 81 - 162 + \frac{2}{5} \times 3^5 - \frac{2}{5} \times 3^{5/2}$$ Finally: $$A = \frac{486}{5} - 81 = \frac{81}{5}$$ Multiplying by 10 gives $10A = 162$. Ans. = (2)

Given the function $f : (-\infty, \infty) - \{0\} \to \mathbb{R}$. We need to find $f'(1)$. First, we have: $$f'(1) = \lim_{a \to \infty} a^2 f\left(\frac{1}{a}\right)$$ Calculating the limit: $$\lim_{a \to \infty} \frac{a(a+1)}{2} \tan^{-1}\left(\frac{1}{a}\right) + a^2 - 2 \ln(a)$$ Simplifying further: $$\lim_{a \to \infty} a^2 \left( \frac{\left(1 + \frac{1}{a}\right)}{2} \tan^{-1}\left(\frac{1}{a}\right) + 1 - \frac{2}{a^2} \ln(a) \right)$$ The function is given by: $$f(x) = \frac{1}{2}(1+x) \tan^{-1}(x) + 1 - 2x^2 \ln(x)$$ Differentiating with respect to $x$: $$f'(x) = \frac{1}{2} \left( \frac{1+x}{1+x^2} + \tan^{-1}(x) + 4x \ln(x) \right) + 2x$$ Evaluating at $x = 1$: $$f'(1) = \frac{1}{2} \left( 1 + \frac{\pi}{4} \right) + 2$$ Simplifying gives: $$f'(1) = \frac{5}{2} + \frac{\pi}{8}$$

Given $\alpha \beta \gamma = 45$, $\alpha \beta \gamma \in \mathbb{R}$. The equation $\mathbf{x}(\alpha, 1, 2) + \mathbf{y}(1, \beta, 2) + \mathbf{z}(2, 3, \gamma) = (0, 0, 0)$ is given with $x, y, z \in \mathbb{R}$ and $xyz \neq 0$. The system of equations is: $$\begin{align*} \alpha x + y + 2z &= 0, \\ x + \beta y + 3z &= 0, \\ 2x + 2y + \gamma z &= 0. \end{align*}$$ Since $xyz \neq 0$, the solution is non-trivial. The determinant of the matrix is: $$\begin{vmatrix} \alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma \end{vmatrix} = 0$$ Expanding the determinant gives: $$\alpha (\beta \gamma - 6) - 1(\gamma - 6) + 2(2 - 2\beta) = 0$$ Simplifying, we have: $$\alpha \beta \gamma - 6\alpha - \gamma + 6 + 4 - 4\beta = 0$$ Thus, $6\alpha + 4\beta + \gamma = 55$.

Given $P(x, y)$ and $x \geq 3$. The slope of the line at $P(x, y)$ will be $\frac{dy}{dx} = \frac{1}{2} \left( \frac{y+5}{x-3} \right)$. $$\Rightarrow 2 \frac{dy}{(y+5)} = \frac{1}{(x-3)} dx$$ $$\Rightarrow 2 \ln(y+5) = \ln(x-3) + C$$ Passes through $(4, -2)$ $$\Rightarrow 2 \ln(3) = \ln(1) + C$$ $$\Rightarrow C = 2 \ln(3)$$ $$\Rightarrow 2 \ln(y+5) = \ln(x-3) + 2 \ln(3)$$ $$\Rightarrow 2 \left( \ln \left( \frac{y+5}{3} \right) \right) = \ln(x-3)$$ $$\Rightarrow \left( \frac{y+5}{3} \right)^2 = (x-3)$$ $$\Rightarrow (y+5)^2 = 9(x-3)$$ Parabola $$4a = 9$$ $$a = \frac{9}{4}$$ $$d = \sqrt{\left( \frac{7}{4} \right)^2 + 6^2}$$ $$d = \sqrt{\frac{625}{4}}$$ $$d = \frac{25}{4}$$ $$12d = 75$$

Given $$I_K = \int 1 \cdot (1-x^7)^K \, dx$$ We have $$I_K = (1-x^7)^K x \bigg|_0^1 + 7K \int_0^1 (1-x^7)^{K-1} x^6 \cdot x \, dx$$ This simplifies to $$I_K = -7K \int_0^1 (1-x^7)^{K-1} ((1-x^7) - 1) \, dx$$ Thus, $$I_K = -7K I_K + 7K I_{K-1}$$ This implies $$\frac{I_K}{I_{K+1}} = \frac{7K + 8}{7K + 7}$$ Therefore, $$r_K = \frac{7K + 8}{7K + 7}$$ We find $$r_K - 1 = \frac{1}{7(K+1)}$$ Thus, $$7(r_K - 1) = \frac{1}{K+1}$$ Finally, $$\sum_{K=1}^{10} (K+1) = 11(6) - 1 = 65$$

Given $$4x^4 + 8x^3 - 17x^2 - 12x + 9 = 4(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$ Put $x = 2i$ and $-2i$. $$64 - 64i + 68 - 24i + 9 = (2i - x_1)(2i - x_2)(2i - x_3)(2i - x_4)$$ $$= 141 - 88i \ldots (1)$$ $$64 + 64i + 68 + 24i + 9 = 4(-2i - x_1)(-2i - x_2)(-2i - x_3)(-2i - x_4)$$ $$= 141 + 88i \ldots (2)$$ $$\frac{125}{16} = \frac{141^2 + 88^2}{16}$$ $$m = 221$$

Given $9x^2 + 12x + 4 = -18(y - 1)$. Rewriting, we have $(3x + 2)^2 = -18(y - 1)$. This can be expressed as $$\left(x + \frac{2}{3}\right)^2 = -2(y - 1).$$ The points $Q$ and $P(0,1)$ are shown on the graph. The line $y = mx + 1$ passes through $(0,1)$. Substituting, we have $$\left(x + \frac{2}{3}\right)^2 = -2(y - 1).$$ Expanding, $(3x + 2)^2 = -18mx$. This gives $9x^2 + (12 + 18m)x + 4 = 0$. Solving, $4(6 + 9m)^2 = 4(36)$. Simplifying, $6 + 9m = 6, -6$. Thus, $m = 0, -\frac{4}{3}$. The angle $\theta$ is given by $\tan \theta = -\frac{4}{3}$. Using the identity, $$2 \tan \frac{\theta}{2} = \frac{-4}{3} \frac{1}{1 - \tan^2 \frac{\theta}{2}}.$$ Solving, $$\left(\tan \frac{\theta}{2} - 2\right)\left(2 \tan \frac{\theta}{2} + 1\right) = 0.$$ Thus, $\tan \frac{\theta}{2} = 2, -\frac{1}{2}$. The slope $m_{QR} = \tan \left(90 + \frac{\theta}{2}\right) = -\cot \frac{\theta}{2}$. Calculating, $m_1 = -\frac{1}{2}$ and $m_2 = \frac{-1}{-1/2} = 2$. Finally, $$16\left(m_1^2 + m_2^2\right) = 16\left(\frac{1}{4} + 4\right) = 4 + 64 = 68.$$

Given $^nC_1 x^{n-1} y = 135$ (i) $^nC_2 x^{n-2} y^2 = 30$ (ii) $^nC_3 x^{n-3} y^3 = \frac{10}{3}$ (iii) By (i) and (ii), $$\frac{^nC_1}{^nC_2} \frac{x}{y} = \frac{9}{2} \ldots (iv)$$ By (ii) and (iii), $$\frac{^nC_2}{^nC_3} \frac{x}{y} = 9 \ldots (v)$$ By (iv) and (v), $$\frac{^nC_1 ^nC_3}{^nC_2 ^nC_2} = \frac{1}{2}$$ $$\frac{2n^2 (n-1)(n-2)}{6} = \frac{n(n-1)}{2} \frac{n(n-1)}{2}$$ $$4n - 8 = 3n - 3$$ $$\Rightarrow n = 5$$ Put in (v), $$\frac{x}{y} = 9$$ $$x = 9y$$ Put in (i), $$5C_1 x^4 \left( \frac{x}{9} \right) = 135$$ $$x^5 = 27 \times 9$$ $$\Rightarrow x = 3, y = \frac{1}{3}$$ $$6 \left( n^3 + x^2 + y \right)$$ $$= 6 \left( 125 + 9 + \frac{1}{3} \right)$$ $$= 806$$

$T_r = 3T_{r-1} + 6^r,\ r = 2, 3, 4, \ldots n$ $T_2 = 3 \cdot T_1 + 6^2$ $T_2 = 3 \cdot 6 + 6^2 \hfill \ldots(1)$ $T_3 = 3T_2 + 6^3$ $T_3 = 3(3 \cdot 6 + 6^2) + 6^3$ $T_3 = 3^2 \cdot 6 + 3 \cdot 6^2 + 6^3 \hfill \ldots(2)$ $$T_r = 3^{r-1} \cdot 6\left[1 + \frac{6}{3} + \left(\frac{6}{3}\right)^2 + \ldots + \left(\frac{6}{3}\right)^{r-1}\right]$$ $$T_r = 3^{r-1} \cdot 6\left(1 + 2 + 2^2 + \ldots + 2^{r-1}\right)$$ $$T_r = 6 \cdot 3^{r-1} \cdot \frac{(1 - 2^r)}{(-1)}$$ $$T_r = 6 \cdot 3^{r-1} \cdot (2^r - 1)$$ $$T_r = \frac{6 \cdot 3^r}{3} \cdot (2^r - 1)$$ $$T_r = 2 \cdot (6^r - 3^r)$$ $$S_n = 2\sum(6^r - 3^r)$$ $$S_n = 2\left[\frac{6(6^n - 1)}{5} - \frac{3(3^n - 1)}{2}\right]$$ $$S_n = 2\left[\frac{12(6^n - 1) - 15(3^n - 1)}{10}\right]$$ $$S_n = \frac{3}{5}\left[4 \cdot 6^n - 5 \cdot 3^n + 1\right]$$ $\therefore\ n^2 - 12n + 39 = 3$ $n^2 - 12n + 36 = 0$ $n = 6$

Given $\cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4}$. This implies $\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{n} = \frac{\pi}{4}$. Now, $\tan^{-1} \left( \frac{46}{48} \right) + \tan^{-1} \frac{1}{n} = \frac{\pi}{4}$. This simplifies to $\tan^{-1} \left( \frac{23}{24} \right) + \tan^{-1} \frac{1}{n} = \frac{\pi}{4}$. Therefore, $\tan^{-1} \frac{1}{n} = \tan^{-1} 1 - \tan^{-1} \frac{23}{24}$. So, $\tan^{-1} \frac{1}{n} = \tan^{-1} 1 - \tan^{-1} \frac{23}{24}$. This gives $\tan^{-1} \frac{1}{n} = \tan^{-1} \left( \frac{1 - \frac{23}{24}}{1 + \frac{23}{24}} \right)$. Thus, $\tan^{-1} \frac{1}{n} = \tan^{-1} \left( \frac{1}{\frac{24}{47}} \right)$. Finally, $\tan^{-1} \frac{1}{n} = \tan^{-1} \frac{1}{47}$. Therefore, $n = 47$.

Line: $\(\frac{x+6}{-8}=\frac{y-7}{12}=\frac{z+5}{-16}\)$ \[ \frac{x+6}{2}=\frac{y-7}{-3}=\frac{z+5}{4}=\lambda \] $Q\((2\lambda-6,\;7-3\lambda,\;4\lambda-5)\)$ $QR\((2\lambda-7,\;-3\lambda,\;4\lambda-11)\)$ $\(\overrightarrow{QR}\cdot \text{dr's of line}=0\)$ \[ 4\lambda-14+9\lambda+16\lambda-44=0 \] \[ 29\lambda=58 \Rightarrow \lambda=2 \] Q\((-2,1,3)\) \[ PQ=\sqrt{144+9+16}=\sqrt{169}=13 \]

Given $\vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = (1, 8, 13)$. $\vec{a} \times (\vec{a} \times \vec{b}) + \vec{a} \times (\vec{a} \times \vec{c}) + \vec{a} \times (\vec{b} \times \vec{c})$ $= \vec{a} \times (\hat{i} + 8\hat{j} + 13\hat{k})$ $(\vec{a} \cdot \vec{b}) \vec{a} - a^2 \vec{b} + (\vec{a} \cdot \vec{c}) \vec{a} - a^2 \vec{c} + (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \vec{a} \times (\hat{i} + 8\hat{j} + 13\hat{k})$ $\Rightarrow -26 \vec{a} - 29 \vec{b} + 13 \vec{a} - 29 \vec{c} + 13 \vec{b} + 26 \vec{c} = \vec{a} \times (\hat{i} + 8\hat{j} + 13\hat{k})$ $\Rightarrow -13 \vec{a} - 16 \vec{b} - 3 \vec{c} = \vec{a} \times (\hat{i} + 8\hat{j} + 13\hat{k})$ $\Rightarrow -13 \vec{a} \cdot \vec{b} - 16 \vec{b}^2 - 3 \vec{b} \cdot \vec{c} = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 8 & 13 \\ 3 & 4 & -5 \end{vmatrix}$ $\Rightarrow (-13)(-26) - 16(50) - 3 \vec{b} \cdot \vec{c} = -396$ $\Rightarrow -462 - 3 \vec{b} \cdot \vec{c} = -396$ $\Rightarrow \vec{b} \cdot \vec{c} = -22$ Hence $24 - \vec{b} \cdot \vec{c} = 46$

Given $T = 2\pi \sqrt{\frac{m}{k}}$. $T^2 \propto \frac{m}{k}$. $$\frac{2\Delta T}{T}\% = \frac{\Delta m}{m}\% - \frac{\Delta k}{k}\%$$ $$\frac{\Delta k}{k}\% = \frac{\Delta m}{m}\% - 2\frac{\Delta T}{T}\%$$ $$\frac{\Delta k}{k}\% = (-1)\% - 2(2)\% = |-5\%| = 5\%$$

Given $K_i = \dfrac{1}{2}m(100)^2$ and $K_f = \dfrac{1}{2}m(40)^2$. The percentage loss is given by: $$\%\text{ loss} = \frac{|K_f - K_i|}{K_i} \times 100$$ Substituting the values, we have: $$= \frac{\left|\dfrac{1}{2}m(40)^2 - \dfrac{1}{2}m(100)^2\right|}{\dfrac{1}{2}m(100)^2} \times 100$$ $$= \frac{|1600 - 10000|}{10000} \times 100$$ $$= 84\%$$

The formula for the wavelength is given by $$\frac{1}{\lambda} = Rz^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ For the Lyman series, $$\frac{1}{\lambda_L} = Rz^2 \left( \frac{1}{1^2} \right)$$ For the Balmer series, $$\frac{1}{\lambda_B} = Rz^2 \left( \frac{1}{2^2} \right)$$ The ratio of wavelengths is $$\frac{\lambda_B}{\lambda_L} = 4 : 1$$

Given the equation $$\frac{1}{2} mv_e^2 = \frac{GMm}{R_E}$$. We know that $$g = \frac{GM}{R_E^2}$$ and $$K = mgR_E$$.

Given $\($ $\frac{\varepsilon_m \times \mu_m}{\varepsilon_0 \times \mu_0}$ = $\frac{\frac{1}{v^2}}{\frac{1}{c^2}}$ $\)$. $\($ $\varepsilon$_r $\times$ $\mu$_r = $\frac{c^2}{v^2}$ $\)$. $\($ $\varepsilon$_r $\times$ 2 = $\frac{(3 \times 10^8)^2}{(1.5 \times 10^8)^2}$ $\)$. $\($ $\varepsilon$_r $\times$ 2 = 4 $\)$. $\($ $\varepsilon$_r = 2 $\)$.

Photoelectric effect prove particle nature of light.

\[ \begin{aligned} \mathrm{MSR} &= 1\,\mathrm{mm}, \\ \mathrm{CSR} &= 42, \\ \text{Pitch} &= 1\,\mathrm{mm} \end{aligned} \] \[ \mathrm{LC} = \frac{\text{Pitch}}{\text{No.\ of CSD}} = \frac{1}{100} = 0.01\,\mathrm{mm} \] \[ \begin{aligned} \text{Diameter} &= \mathrm{MSR} + \mathrm{LC} \times \mathrm{CSR} \\ &= 1 + (0.01 \times 42)\,\mathrm{mm} \\ &= 1.42\,\mathrm{mm} = \frac{x}{50} \end{aligned} \] \[ \therefore\ x = 71 \]

By Gauss law $$\int \vec{E} \cdot d\vec{A} = \frac{q_{in}}{\varepsilon_0}$$ $$E dA = \frac{\sigma \times dA}{\varepsilon_0}$$ $$E = \frac{\sigma}{\varepsilon_0}$$

In case of balanced Wheatstone Bridge $$\frac{V_{AB}}{V_{AD}} = \frac{V_{BC}}{V_{CD}} \Rightarrow \frac{12}{6 + x} = \frac{0.5}{0.5}$$ $$x = 6 \, \Omega$$

Given $$dQ = du + dW$$ $$CdT = C_V dT + PdV$$ Therefore, $$PV^2 = RT$$ $$P = constant$$ $$P(2VdV) = RdT$$ $$PdV = \frac{RdT}{2V}$$ Put in equation (1) $$C = C_V + \frac{R}{2V}$$

Given $[\vec{\tau}] = [\vec{r} \times \vec{F}] = [ML^2 \, T^{-2}]$. $[F] = [qVB]$. Therefore, $B = \left( \frac{F}{qV} \right) = \left[ \frac{MLT^{-2}}{ALT^{-1}} \right] = [MA^{-1} \, T^{-2}]$. $[M] = [I \times A] = [AL^2]$. $B = \frac{\mu_0}{4\pi} \frac{Idl \sin \theta}{r^2}$. Therefore, $[\mu] = \left[ \frac{Br^2}{Idl} \right] = \left[ \frac{MT^{-2} \, A^{-1} \times L^2}{AL} \right] = [MLT^{-2} \, A^{-2}]$.

Statement-I $$I_m = \frac{V_m}{\sqrt{R^2 + (X_L - X_C)^2}}$$ at resonance $X_L = X_C$ Thus, $$I_m = \frac{V_m}{R}$$ Therefore, impedance is minimum therefore $I$ is maximum at resonance. Statement-II $$I = \left( \frac{V}{R} \right)$$ in purely resistive circuit.

The circuit consists of an AND gate followed by a NOT gate and then an OR gate. The inputs to the AND gate are $A$ and $B$. The output of the AND gate is $A \cdot B$. This output is then passed through a NOT gate, resulting in $\overline{A \cdot B}$. The OR gate takes this output and the original input $A$. The final output $Y$ is given by: $$Y = A + \overline{A \cdot B}.$$

The root mean square velocity $V_{ms}$ is given by the equation $$V_{ms} = \sqrt{\frac{3RT}{M_w}}.$$ Therefore, $$\frac{V_{O_2}}{V_{He}} = \sqrt{\frac{M_{w,He}}{M_{w,O_2}}}.$$ This simplifies to $$\sqrt{\frac{4}{32}} = \frac{1}{2\sqrt{2}}.$$ Hence, $$\frac{V_{He}}{V_{O_2}} = \frac{2\sqrt{2}}{1}.$$

The acceleration is given by the equation: $$a = \left( \frac{M_2 - M_1}{M_1 + M_2} \right) g$$ Dividing by $\sqrt{2}$, we have: $$\frac{g}{\sqrt{2}} = \left( \frac{M_2 - M_1}{M_1 + M_2} \right) g$$ Simplifying, we get: $$(M_1 + M_2) = \sqrt{2}M_2 - \sqrt{2}M_1$$ Therefore, the ratio is: $$\frac{M_1}{M_2} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$$

Given $$KE = \frac{p^2}{2m}$$. Same momentum, so less mass means more KE. So $$\frac{m}{2}$$ will have max. KE.

The magnetic field increment is given by $$\overrightarrow{\mathrm{dB}} = \frac{\mu_0 I}{4\pi} \frac{(\overrightarrow{dl} \times \overrightarrow{r})}{r^3} (Tesla)$$ Substituting the given values: $$10^{-7} \times 10 \times \left( \frac{1}{2} \times \frac{1}{100} \right) (+\hat{k})$$ The denominator is $$\left( \frac{1}{2} \right)^3$$ Thus, the result is $$= 4 \times 10^{-8} T(+\hat{k})$$

Given $$mg - F_B - F_v = ma$$ $a = 0$ for constant velocity $$mg - F_B = F_v$$ $$F_v = mg - v \rho_0 g = mg - \frac{m}{\rho} \rho_0 g = mg \left(1 - \frac{\rho_0}{\rho}\right)$$

Given $eV V_s = h \nu - \phi$. $$0.5 \, \mathrm{V} = 2.48 \, \mathrm{V} - \phi$$ The work function $(\phi)$ is calculated as: $$\phi = 2.48 \, \mathrm{V} - 0.5 \, \mathrm{V} = 1.98 \, \mathrm{V}$$

By conservation of angular momentum $$I_1 \omega_1 = I_2 \omega_2$$ $$\left( \frac{2}{5} MR^2 \right) \frac{2\pi}{T_1} = \frac{2}{5} M \left( \frac{3}{4} R \right)^2 \frac{2\pi}{T_2}$$ $$\frac{1}{T_1} = \frac{9}{16} \frac{1}{T_2}$$ $$\frac{1}{T_2} = \frac{9}{16} \times T_1 = \frac{9}{16} \times 24 hr = \frac{27}{2} hr = 13 hr 30 mins.$$

The electric field at point P is given by: $$\vec{E}_p = \left( \frac{\sigma}{2\varepsilon_0} + \frac{2\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} \right)(-\hat{i})$$ Simplifying, we have: $$= -\frac{2\sigma}{\varepsilon_0} \hat{i}$$

Given 1000 drops and a big drop, we have: $$1000 \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$ Solving for $R$, we get: $$10r = R$$ $$R = 10r$$ The ratio of the surface energy of 1000 drops to the surface energy of the big drop is: $$\frac{S.E. of 1000 drops}{S.E. of Big drop} = \frac{1000 \left(4\pi r^2\right) T}{4\pi R^2 T}$$ Simplifying, we have: $$= \frac{1000 \times r^2}{(10r)^2} = 10 = \frac{10}{x}$$ Therefore, $x = 1$.

For DC voltage, $R = \frac{V}{I} = \frac{100}{5} = 20 \, \Omega$. For AC voltage, $X_L = 20\sqrt{3} \, \Omega$ and $R = 20 \, \Omega$. The impedance $Z = \sqrt{X_L^2 + R^2} = \sqrt{3 \times 400 + 400} = 40 \, \Omega$. Power is given by $i_{rms}^2 R$. \[ \left( \frac{V_{rms}}{Z} \right)^2 \times R = \left( \frac{200}{\sqrt{2} \times 40} \right)^2 \times 20 = 250 \, \mathrm{W} \]

For $\delta_{\min}$, $i = e$. $$r_1 = r_2 = \frac{A}{2}$$ $$\frac{\delta_{\min}}{A} = 1$$ $$\frac{2i - A}{A} = 1$$ $$2i = 2A$$ $$i = A$$ Snell's law $$1 \times \sin i = \mu \sin r$$ $$\sin i = \mu \sin \left( \frac{A}{2} \right)$$ $$\sin A = \mu \sin \left( \frac{A}{2} \right)$$ $$2 \sin \frac{A}{2} \cos \frac{A}{2} = \sqrt{3} \sin \left( \frac{A}{2} \right)$$ $$\cos \left( \frac{A}{2} \right) = \frac{\sqrt{3}}{2}$$ $$\therefore \frac{A}{2} = 30^\circ$$ $$\therefore A = 60^\circ$$

We know $R = \frac{\rho l}{A}$, $R \propto \frac{l}{r^2}$. As we starch the wire, its length will increase but its radius will decrease keeping the volume constant. $V_i = V_f$ $$\pi r^2 l = \pi \frac{r^2}{4} l_f$$ $$l_f = 4l$$ $$\frac{R_{new}}{R_{old}} = \left(\frac{4l}{\frac{r^2}{4}}\right) \frac{r^2}{l} = 16$$ $$R_{new} = 16R$$ Therefore, $x = 16$

We know $$r = 0.529 \frac{n^2}{Z} \Rightarrow 8.48 = 0.529 \frac{n^2}{1}$$ $$n^2 = 16 \Rightarrow n = 4$$ We know $$E \propto \frac{1}{n^2}$$ $$E_{n^{th}} = \frac{E}{16}$$ $$x = 16$$

We know $$W_{ext} = \Delta U + \Delta KE (P.E. = -\vec{M} \cdot \vec{B})$$ $$= -\vec{M} \cdot \vec{B}_f + \vec{M} \cdot \vec{B}_i + 0$$ $$= -MB \cos 90 + MB \cos 0$$ $$= MB$$ $$= NIAB$$ $$= 200 \times 100 \times 10^{-6} \times 10^{-4} \times \frac{5}{2} \times 1 = 5 \mu J$$

We know $$V_{max} = \omega A at mean position$$ $$= \frac{2\pi}{T} A = \frac{2\pi}{\pi} \times 0.06 = 0.12 \, m/sec$$ $$V_{max} = 12 \, cm/sec$$

Given $\vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} = 15\hat{i} - 21\hat{j} + 33\hat{k}$. $\vec{A} \cdot (\vec{B} \times \vec{C}) = (-x\hat{i} - 6\hat{j} - 2\hat{k}) \cdot (15\hat{i} - 21\hat{j} + 33\hat{k})$. $0 = -15x + 126 - 66$. $15x = 60$. $x = 4$.

The group present in sulphonic acids is the sulfonic acid group, which is represented as $\mathrm{SO_3H}$.

The reaction starts with phenol. When phenol is treated with concentrated $\mathrm{H_2SO_4}$, it forms benzene sulfonic acid. This compound is then treated with concentrated $\mathrm{HNO_3}$ to form picric acid, also known as 2, 4, 6-trinitrophenol.

Metamer ⇒ Isomer having same molecular formula, same functional group but different alkyl/aryl groups on either side of functional group.

The structures shown are bases of the DNA molecule. As DNA contains four bases, which are adenine, guanine, cytosine, and thymine.

sp^3 $\rightarrow$ Tetrahedral dsp^2 $\rightarrow$ Square planar sp^3 d $\rightarrow$ Trigonal Bipyramidal sp^3 d^2 $\rightarrow$ Octahedral

Statement - I is correct. Statement - II is false. Ga is used to measure high temperature.

Option (B) is incorrect because aniline is immiscible in water.

The reaction begins with phenol reacting with chloroform and aqueous NaOH to form an intermediate. This intermediate is then treated with NaOH to form the sodium salt of salicylaldehyde. Finally, the sodium salt is protonated to form salicylaldehyde.

Graphite is conductor

As ligand field increases, light of more energy is absorbed. Energy is proportional to wave number $\propto \bar{\nu}$.

Theory based question

(A) $\mathrm{Be} + e^- \rightarrow \mathrm{Be}^-$, E.A = -ive (B) $\mathrm{N} + e^- \rightarrow \mathrm{N}^-$, E.A = -ive (C) $\mathrm{O} + e^- \rightarrow \mathrm{O}^-$ $\mathrm{O}^- + e^- \rightarrow \mathrm{O}^{2-}$, E.A = -ive (D) $\mathrm{Na} + e^- \rightarrow \mathrm{Na}^-$, E.A = +ive (E) $\mathrm{Al} + e^- \rightarrow \mathrm{Al}^-$, E.A = +ive

Cm is Actinide

Molality = $\frac{1000 \times M}{1000 \times d - M \times (\mathrm{Mw})_{solute}}$ 3 = $\frac{1000 \times x}{1000 \times 1.12 - (x \times 40)}$ x = 3

has low steric hindrance at the carbonyl carbon and high partial positive charge at the carbonyl carbon.

$\Delta n_g=0$ $n_{HI}=n_{H_2}=n_{I_2}$ so $K_p=\dfrac{(n_{HI})^2}{n_{H_2}n_{I_2}} \left(\dfrac{P_T}{n_T}\right)^{\Delta n_g}$ so $K_p=1$ $1=x\times10^{-1}$ $x=10$

Iodoform $-$ Antiseptic $\mathrm{CCl_4}$ $-$ Fire extinguisher CFC $-$ Refrigerants DDT $-$ Insecticide

Solution is already infinitely dilute, hence no change in molar conductivity upon addition of water.

Given the reaction $\mathrm{HX} \rightleftharpoons \mathrm{H^+} + \mathrm{X^-}$ with $K_a = 1.2 \times 10^{-5}$. The initial concentration is $0.03 \, \mathrm{M}$. The equilibrium concentrations are $0.03 - x$, $x$, and $x$. The expression for $K_a$ is given by: $$K_a = 1.2 \times 10^{-5} = \frac{x^2}{0.03 - x}$$ Assuming $0.03 - x \approx 0.03$ (since $K_a$ is very small), we have: $$\frac{x^2}{0.03} = 1.2 \times 10^{-5}$$ Solving for $x$, we find: $$x = 6 \times 10^{-4}$$ The final solution is $0.03 - x + x + x$: $$= 0.03 + x = 0.03 + 6 \times 10^{-4}$$ The calculation for $\Pi$ is: $$\Pi = \left(0.03 + (6 \times 10^{-4})\right) \times 0.083 \times 300$$ This simplifies to: $$= 76.19 \times 10^{-2} \approx 76 \times 10^{-2}$$

Spin only magnetic moment of Mn in $\mathrm{KMnO_4} = 0$. Spin only value of manganese product formed during titration of $\mathrm{KMnO_4}$ against oxalic acid in acidic medium is $= 6$.

Given $$K = \frac{1}{t_{99.9\%}} \ln \left( \frac{100}{0.1} \right) = \frac{1}{t_{90\%}} \ln \left( \frac{100}{10} \right)$$ $$t_{99.9\%} = t_{90\%} \frac{\ln (10^3)}{\ln 10}$$ $$t_{99.9\%} = t_{90\%} \times 3$$

The compounds $\mathrm{CH_3OH}$, $\mathrm{H_2O}$, the given aromatic compound with $\mathrm{OH}$ and $\mathrm{NO_2}$ groups, $\mathrm{HF}$, and $\mathrm{NH_3}$ can show hydrogen bonding.

Reaction suggests that 1 mole of aniline give 1 mole of orange dye. So $(mol)_{aniline} = (mol)_{orange dye}$ $$\frac{9.3 \, \mathrm{g}}{93 \, \mathrm{g \, mol^{-1}}} = \frac{mass of orange dye}{199 \, \mathrm{g \, mol^{-1}}}$$ Mass of orange dye $= 19.9 \, \mathrm{g} \simeq 20 \, \mathrm{g}$

The reaction starts with $\mathrm{CH_3-C \equiv C-CH_3}$ which is treated with $\mathrm{Na/liquid\ NH_3}$ to form $\mathrm{CH_3-CH=CH-CH_3}$. This intermediate is then treated with dilute $\mathrm{KMnO_4}$ to yield the final product $\mathrm{CH_3-CH(OH)-CH(OH)-CH_3}$, known as Product P.

Given the equation for wavelength $\lambda = \frac{h}{mv}$ and $\lambda = \frac{hv}{mv^2}$, we have $\frac{mv^2}{h} = \frac{v}{\lambda} = \nu$ (frequency). Given $\frac{1}{2}mv^2 = 2.18 \times 10^{-18} \, \mathrm{J}$, $h = 6.6 \times 10^{-34}$. Then, $$\nu = \frac{4.36 \times 10^{-18}}{6.6 \times 10^{-34}} = 660.60 \times 10^{13}$$ $$\approx 661 \times 10^{13} \, \mathrm{Hz}$$

The reaction starts with cyclohexene reacting with $\mathrm{Br_2}$ to form a dibromo compound. This compound then reacts with $\mathrm{HC\equiv C-CH_2-O^\ominus Na^+}$ to form compound (A), which is $\mathrm{O-CH_2-C\equiv CH}$. In the second reaction, the dibromo compound reacts with alcoholic $\mathrm{KOH}$ (3 equivalents) to form compound (B), which is benzene.

$\mathrm{CrO}$ is a basic oxide. $\mathrm{Cr_2O_3}$ is an amphoteric oxide. In $\mathrm{CrO}$, Cr exists as $\mathrm{Cr^{2+}}$ and has $$ \mu_{\text{spin only}} = 4.90 $$ In $\mathrm{Cr_2O_3}$, Cr exists as $\mathrm{Cr^{3+}}$ and has $$ \mu_{\text{spin only}} = 3.87 $$ Sum of spin-only magnetic moment is $$ 4.90 + 3.87 = 8.77 $$ $$ \mu_{\text{spin only}} = 877 \times 10^{-2} $$ Ans. $877$

Given $\Delta U = q + w (q = 0)$. $n C_V \Delta T = -P_{ext} (V_2 - V_1)$. $V_2 = 2 V_1$. $$\frac{nRT_2}{P_2} = \frac{2nRT_1}{P_1}$$ $P_1 = 5$, $T_1 = 298$. $$P_2 = \frac{5 \ T_2}{2 \times 298}$$ $$n \frac{5}{2} R (T_2 - T_1) = -1 \left( \frac{nRT_2}{P_1} - \frac{nRT_1}{P_1} \right)$$ Put $T_1 = 298$ and $$P_2 = \frac{5 \ T_2}{2 \times 298}$$ Solve and we get $T_2 = 274.16 \, \mathrm{K}$. $T_2 \approx 274 \, \mathrm{K}$