JEE Main 23 January 2025 Shift 1 question paper with solutions

Put $\ln x = t \implies \frac{1}{x} \, dx = dt$. $$I = \int_2^4 e^{(t^2+1)^{-1}} \left( e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}} \right) \, dt (i)$$ $$I = \int_2^4 e^{((6-t)^2+1)^{-1}} \left( e^{((6-t)^2+1)^{-1}} + e^{(t^2+1)^{-1}} \right) \, dt (ii)$$ Using $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$. Adding (i) and (ii) gives $$2I = \int \, dt \implies I = 1$$

Given $$I(x) = \int \frac{dx}{(x-11)^{11/13}(x+15)^{15/13}}$$ Put $$\frac{x-11}{x+15} = t \implies 26(x+5)^2 \, dx = dt$$ Then $$I(x) = \frac{1}{26} \int t^{11/13} \, dt = \frac{1}{26} \cdot \frac{t^{2/13}}{2/13}$$ Thus, $$I(x) = \frac{1}{4} \left( \frac{x-11}{x+15} \right)^{2/13} + C$$ Now, $$I(37) - I(24) = \frac{1}{4} \left( \frac{26}{52} \right)^{2/13} - \frac{1}{4} \left( \frac{13}{39} \right)^{2/13}$$ This simplifies to $$= \frac{1}{4} \left( \frac{1}{2^{2/13}} - \frac{1}{3^{2/13}} \right)$$ $$= \frac{1}{4} \left( \frac{1}{4^{1/13}} - \frac{1}{9^{1/13}} \right)$$ Therefore, $b = 4$, $c = 9$ and $$3(b+c) = 39$$

$\lim_{x \to 0^+} \dfrac{2}{x}\{\sin(k_1+1)x + \sin(k_2-1)x\} = 4$ $\Rightarrow 2(k_1+1) + 2(k_2-1) = 4$ $\Rightarrow k_1 + k_2 = 2$ $\Rightarrow \lim_{x \to 0^+} \dfrac{2}{x}\ln\left(\dfrac{2+k_1 x}{2+k_2 x}\right) = 4$ $\Rightarrow \lim_{x \to 0^+} \dfrac{1}{x}\ln\left(1 + \dfrac{(k_1-k_2)\,x}{2+k_2 x}\right) = 2$ $\Rightarrow \dfrac{k_1-k_2}{2} = 2$ $\Rightarrow k_1 - k_2 = 4$ $\therefore k_1 = 3,\, k_2 = -1$ $k_1^2 + k_2^2 = 9 + 1 = 10$

The equations are given as $3x - 2y + 12 = 0$ and $4y = 3x^2$. Therefore, $2(3x + 12) = 3x^2$. This implies $x^2 - 2x - 8 = 0$. Solving for $x$, we get $x = -2, 4$. The slopes are $m_{OA} = -3/2$ and $m_{OB} = 3$. The tangent of the angle $\theta$ is given by $$\tan \theta = \left( \frac{-3/2 - 3}{1 - 9/2} \right) = \frac{9}{7}.$$ Therefore, $\theta = \tan^{-1} \left( \frac{9}{7} \right)$ (angle will be acute).

Given $\($ $\frac{dy}{dx}$ = $\frac{2(3+y) \cdot e^{2x}}{7+e^{2x}}$ $\)$. Rewriting, $\($ $\frac{dy}{dx}$ - $\frac{2y e^{2x}}{7+e^{2x}}$ = $\frac{6 \cdot e^{2x}}{7+e^{2x}}$ $\)$. The integrating factor (I.F.) is given by $$ I.F. = e^{-\int \frac{2e^{2x}}{7+e^{2x}} \, dx} $$ which simplifies to $$ e^{-\ln(7+e^{2x})} = \frac{1}{7+e^{2x}}. $$ Multiplying through by the integrating factor, we have $$ y \cdot \frac{1}{7+e^{2x}} = \int \frac{6e^{2x}}{(7+e^{2x})^2} \, dx. $$ Solving the integral, $$ \frac{y}{7+e^{2x}} = \frac{-3}{7+e^{2x}} + C. $$ Given $\($ y(0) = 5 $\)$, $$ \frac{5}{8} = \frac{-3}{8} + C $$ which gives $\($ C = 1 $\)$. Thus, $$ y = -3 + 7 + e^{2x}. $$ Simplifying, $$ y = e^{2x} + 4. $$ Therefore, $\($ k = 8 $\)$.

Given $f(g(x)) = \ln \left( \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \right)$. Since $2x^2 - 2x + 1 > 0$ for all $x \in \mathbb{R}$, $(-2)^2 - 4(2) 0 \forall x \in \mathbb{R}$$ Therefore, $g(x) > 0 \forall x \in \mathbb{R}$. Thus, $\ln f((x)), f(x) > 0 \forall x \in \mathbb{R}$. Hence, $x \in \mathbb{R}$ is the domain.

Given $\vec{c} = \alpha \vec{a} + \beta \vec{b} \ldots (1)$ $\vec{a} \cdot \vec{c} = \alpha \vec{a} \cdot \vec{a} + \beta \vec{b} \cdot \vec{a}$ $0 = \alpha + \beta \cos 15^\circ \ldots (2)$ From (1) $\Rightarrow \vec{b} \cdot \vec{c} = \alpha \vec{a} \cdot \vec{b} + \beta \vec{b} \cdot \vec{b}$ $\Rightarrow \cos 75^\circ = \alpha \cos 15^\circ + \beta \ldots (3)$ From (2) and (3) $\Rightarrow \cos 75^\circ = -\beta \cos^2 15^\circ + \beta$ $\beta = \frac{\cos 75^\circ}{\sin^2 15^\circ} = \frac{1}{\sin 15^\circ} = \frac{2\sqrt{2}}{\sqrt{3} - 1}$ From (2) $\Rightarrow \alpha = \frac{\sin 15^\circ}{\cos 15^\circ} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ Therefore, $\vec{c} = \frac{-(\sqrt{3} + 1)}{\sqrt{3} - 1} \vec{a} + \left(\frac{2\sqrt{2}}{\sqrt{3} - 1}\right) \vec{b}$ Now, $\alpha + \sqrt{2}(\sqrt{3} - 1)\beta = \frac{-(\sqrt{3} + 1)^2}{\sqrt{3} - 1} + \frac{\sqrt{2}(\sqrt{3} - 1) \cdot 2\sqrt{2}}{\sqrt{3} - 1}$ $= \frac{-(\sqrt{3} + 1)^2 + 4}{2}$ $= \frac{-3 - 1 - 2\sqrt{3} + 8}{2}$ $= 2 - \sqrt{3}$

Given $a = 3$. $$S_4 = \frac{1}{5} \left( S_8 - S_4 \right)$$ Therefore, $$5 S_4 = S_8 - S_4$$ Thus, $$6 S_4 = S_8$$ Substituting, $$6 \cdot \frac{4}{2} \left[ 2 \times 3 + (4 - 1) \times d \right]$$ Simplifies to, $$= \frac{8}{2} \left[ 2 \times 3 + (8 - 1) d \right]$$ Therefore, $$12(6 + 3 d) = 4(6 + 7 d)$$ Which gives, $$18 + 9 d = 6 + 7 d$$ Solving for $d$, $$d = -6$$ Now, $$S_{20} = \frac{20}{2} \left[ 2 \times 3 + (20 - 1)(-6) \right]$$ This simplifies to, $$= 10[6 - 114]$$ Finally, $$= -1080$$

Given the points Q(10, -3, -1), P($\alpha$, $\beta$, $\gamma$) = (10, 1, -3), and R(3, -2, 1), we have the equations: $$\frac{x - 3}{7} = \frac{y - 2}{-1} = \frac{z + 1}{-2} = \lambda$$ This implies: $$7\lambda + 3, -\lambda + 2, -2\lambda - 1$$ The direction ratios of QP are: $$7\lambda - 7, -\lambda + 5, -2\lambda$$ Now, $$(7\lambda - 7) \cdot 7 - (-\lambda + 5) \cdot (2\lambda) \cdot 2 = 0$$ Solving gives: $$54\lambda - 54 = 0 \Rightarrow \lambda = 1$$ Therefore, $P = (10, 1, -3)$. The vector $\overrightarrow{PQ} = -4\hat{j} + 2\hat{k}$ and $\overrightarrow{PR} = -7\hat{i} - 3\hat{j} + 4\hat{k}$. The area is given by: $$Area = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & -4 & 2 \\ -7 & -3 & 4 \end{array} \right| = 3\sqrt{30}$$

Let $z = x + iy \implies \bar{z} = x - iy$. $$3|\bar{z} - i| = 1|2\bar{z} + i|$$ $$= 3 \left| x - (y + 1)i \right| = |2x + i(1 - 2y)|$$ $$= 3\sqrt{x^2 + (y + 1)^2} = \sqrt{(2x)^2 + (1 - 2y)^2}$$ $$= 9\left( x^2 + y^2 + 2y + 1 \right) = 4x^2 + 4y^2 - 4y + 1$$ $$\Rightarrow 5x^2 + 5y^2 + 22y + 8 = 0$$ $$\Rightarrow Centre \equiv \left( 0, -\frac{11}{5} \right)$$ Area of $\Delta$ $$= \frac{1}{2} |\alpha| \left| -\frac{11}{5} \right| = 11$$ $$\Rightarrow |\alpha| = 10$$ $$\Rightarrow \alpha^2 = 100$$

Given $A = \{1, 2, 3, 4\}$. For the relation to be reflexive, $R = \{(1, 2), (2, 3), (3, 3)\}$. Minimum elements added will be $(1, 1), (2, 2), (4, 4), (2, 1), (3, 2), (3, 1), (1, 3)$. Therefore, the minimum number of elements $= 7$.

Total words = 8! Total words in which vowels are together = 6! $\times$ 3! words in which all vowels are not together $$= 8! - 6! \times 3!$$ $$= 6![56 - 6]$$ $$= 720 \times 50$$ $$= 36000$$

Equation of lines $QR = 5x + 2y + 2 = 0$. Equation of lines $PR = 10x - 3y - 38 = 0$. Therefore, point $R(2, -6)$. Centroid $= \left( \frac{5 - 2 + 2}{3}, \frac{4 + 4 - 6}{3} \right) = \left( \frac{5}{3}, \frac{2}{3} \right)$. $c + 2d = \frac{5}{3} + \frac{4}{3} = 3$.

Given $\frac{12}{13}\cos x + \frac{5}{13}\sin x$ Let $\tan\alpha = \frac{5}{12}$, $\alpha \in \left(0,\frac{\pi}{2}\right)$. $\Rightarrow \sin\alpha = \frac{5}{13}$, $\cos\alpha = \frac{12}{13}$. $\Rightarrow \frac{12}{13}\cos x + \frac{5}{13}\sin x = \cos\alpha\cos x + \sin\alpha\sin x$ $= \cos(x-\alpha)$. $\Rightarrow \cos^{-1}[\cos(x-\alpha)] = x-\alpha$ $= x-\tan^{-1}\left(\frac{5}{12}\right)$.

Given $\sin 70^\circ (\cot 10^\circ \cot 70^\circ - 1)$. $$= \sin 70^\circ \cot 10^\circ \cot 70^\circ - \sin 70^\circ$$ $$= \cot 10^\circ \cos 70^\circ - \sin 70^\circ$$ $$= \cos 10^\circ \cos 70^\circ - \sin 70^\circ \sin 10^\circ$$ $$= \frac{\sin 10^\circ}{\cos(10^\circ + 70^\circ)}$$ $$= \frac{\sin 10^\circ}{\cos 80^\circ}$$ $$= \frac{\cos 80^\circ}{\sin 10^\circ} = 1$$

The median is given by the formula: $$ median = \ell + \left( \frac{N}{2} - F \right) \frac{h}{f} $$ Substituting the given values: $$ = 12 + \left( \frac{N}{2} - 18 \right) \frac{6}{12} = 14 $$ This implies: $$ \left( \frac{N}{2} - 18 \right) \times 6 = 2 $$ Solving for $N$: $$ \frac{N}{2} - 18 = 4 \implies N = 44 $$

Area of $\triangle ABC = \dfrac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right| = \dfrac{1}{2}\left|5\mathbf{i} + 3\mathbf{j} + \mathbf{k}\right| = \dfrac{1}{2}\sqrt{35}$. Volume of tetrahedron $= \dfrac{1}{3} \times \text{Base area} \times h = \dfrac{\sqrt{805}}{6\sqrt{2}}$. $$\frac{1}{3} \times \frac{1}{2}\sqrt{35} \times h = \frac{\sqrt{805}}{6\sqrt{2}}$$ $$h = \sqrt{\frac{23}{2}}$$ $AE^2 = AD^2 - DE^2 = \dfrac{13}{18}$. Therefore, $AE = \sqrt{\dfrac{13}{18}}$. $$\frac{\overrightarrow{AE}}{|\overrightarrow{AE}|} = \left(\frac{\mathbf{i} - 5\mathbf{k}}{\sqrt{26}}\right)$$ $$= \frac{\sqrt{\frac{13}{18}}}{\sqrt{26}}\left(\mathbf{i} - 5\mathbf{k}\right)$$ $$= \frac{\sqrt{\frac{13}{18}}}{\sqrt{26}} \cdot \frac{\mathbf{i} - 5\mathbf{k}}{6}$$ P.V. of $E = \dfrac{\mathbf{i} - 5\mathbf{k}}{6} + \mathbf{i} + 2\mathbf{j} + \mathbf{k} = \dfrac{1}{6}\left(7\mathbf{i} + 12\mathbf{j} + \mathbf{k}\right)$

Given the expression: $$\left[ A \left( adj \left( A^{-1} \right) + adj \left( B^{-1} \right) \right) \cdot B \right]^{-1}$$ We have: $$B^{-1} \cdot \left( adj \left( A^{-1} \right) + adj \left( B^{-1} \right) \right) \cdot A^{-1}$$ This expands to: $$B^{-1} adj \left( A^{-1} \right) A^{-1} + B^{-1} \left( adj \left( B^{-1} \right) \right) \cdot A^{-1}$$ Which simplifies to: $$B^{-1} \left| A^{-1} \right| I + \left| B^{-1} \right| IA^{-1}$$ Further simplifying: $$\frac{B^{-1}}{|A|} + \frac{A^{-1}}{|B|}$$ This leads to: $$\frac{adj B}{|B||A|} + \frac{adj A}{|A||B|}$$ Finally, we have: $$= \frac{1}{|A||B|} (adj B + adj A)$$

Given the system of equations: $$(\lambda - 1)x + (\lambda - 4)y + \lambda z = 5$$ $$\lambda x + (\lambda - 1)y + (\lambda - 4)z = 7$$ $$(\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9$$ For infinitely many solutions, $$D = \begin{vmatrix} \lambda - 1 & \lambda - 4 & \lambda \\ \lambda & \lambda - 1 & \lambda - 4 \\ \lambda + 1 & \lambda + 2 & - (\lambda + 2) \end{vmatrix} = 0$$ $$(\lambda - 3)(2\lambda + 1) = 0$$ $$D_x = \begin{vmatrix} 5 & \lambda - 4 & \lambda \\ 7 & \lambda - 1 & \lambda - 4 \\ 9 & \lambda + 2 & - (\lambda + 2) \end{vmatrix} = 0$$ $$2(3 - \lambda)(23 - 2\lambda) = 0$$ $$\lambda = 3$$ Therefore, $$\lambda^2 + \lambda = 9 + 3 = 12$$

Given: $a =$ number on dice 1 $b =$ number on dice 2 $(a, b) = (1, 3), (3, 1), (2, 2), (3, 2), (1, 4), (4, 1)$ Required probability $$= \frac{2}{6} \times \frac{2}{6} + \frac{1}{6} \times \frac{1}{6} + \frac{2}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{1}{6} + \frac{2}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6}$$ $$= \frac{18}{36} = \frac{1}{2}$$

Given $x^2 + y^2 = 5$. $x^2 + (x - 1)^2 = 25 \implies x = 4$ $x^2 + (-x + 1)^2 = 5 \implies x = -3$ $A = 25\pi - \int_{-3}^{4} \sqrt{25 - x^2} \, dx + \frac{1}{2} \times 4 \times 4 + \frac{1}{2} \times 3 \times 3$ $A = 25\pi + \frac{25}{2} - \left[ x \sqrt{25 - x^2} + \frac{25}{2} \sin^{-1} \frac{x}{5} \right]_{-3}^{4}$ $A = 25\pi + \frac{25}{2} - \left[ 6 + \frac{25}{2} \sin^{-1} \frac{4}{5} + 6 + \frac{25}{2} \sin^{-1} \frac{3}{5} \right]$ $A = 25\pi + \frac{1}{2} - \frac{25}{2} \times \frac{\pi}{2}$ $A = \frac{75\pi}{4} + \frac{1}{2}$ $A = \frac{1}{4} (75\pi + 2)$ $b = 75, c = 2$ $b + c = 75 + 2 = 77$

The general term of multinomial expansion is $$\frac{6!}{\alpha! \beta! \gamma!} (1)^\alpha \left(2^{\frac{1}{3}}\right)^\beta \left(3^{\frac{1}{2}}\right)^\gamma$$ For terms to be rational, $3 \mid \beta$ and $2 \mid \gamma$. $\begin{array}{cccc}$ $\beta$ & $\gamma$ & $\alpha$ & Term $\\$ 0 & 0 & 6 & 1 $\\$ 0 & 2 & 4 & 15 $\cdot$ 3 = 45 $\\$ 0 & 4 & 2 & 15 $\cdot$ 3^2 = 135 $\\$ 0 & 6 & 0 & 1 $\cdot$ 3^3 = 27 $\\$ 3 & 0 & 3 & 20 $\cdot$ 2 = 40 $\\$ 3 & 2 & 1 & 60 $\cdot$ 2 $\cdot$ 3 = 360 $\\$ 6 & 0 & 0 & 1 $\cdot$ 4 = 4 $\\$ $\end{array}$ $\Rightarrow$ Sum of rational terms = 1 + 45 + 135 + 27 + 40 + 360 + 4 = 612

Given $r = \left| \frac{a+1}{\sqrt{2}} \right| \Rightarrow (a+1)^2 = 2r^2$. Also $\left( \frac{3a-1}{\sqrt{13}} \right)^2 + \left( \frac{2}{\sqrt{13}} \right)^2 = r^2$. Therefore, $\[$ $\left$( $\frac{3a-1}{\sqrt{13}}$ $\right$)^2 + $\frac{4}{13}$ = $\frac{(a+1)^2}{2}$ $\]$ which simplifies to $\[$ 5a^2 - 14a - 3 = 0 $\]$ Thus, $a = -\frac{1}{5}, 3$. Since $a \neq -\frac{1}{5}$, it follows that $r = 2\sqrt{2}$. One focus of $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$ is $(3,0)$. Therefore, $\alpha e = 3$ and $2\alpha = 4\sqrt{2}$. This implies $\alpha = 2\sqrt{2} \Rightarrow \alpha^2 = 8$. Then, $\[$ $\alpha$^2 $\left$[ 1 + $\frac{\beta^2}{\alpha^2}$ $\right$] = 9 $\]$ which gives $\alpha^2 + \beta^2 = 9$. Therefore, $\beta^2 = 1$. Thus, $2\alpha^2 + 3\beta^2 = 19$.

Given $5x^3 - 15x - a = 0$. We have $f(x) = 5x^3 - 15x$. Rewriting $f(x)$, we get $f(x) = 15x^2 - 15 = 15(x - 1)(x + 1)$. From the graph, $a \in (-10, 10)$. Let $\alpha = -10$ and $\beta = 10$. Then $\beta - 2\alpha = 10 + 20 = 30$.

Given $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$. Since $x = 1$ is a root, the other root is $1$. Therefore, $\alpha + \beta = -\frac{b(c-a)}{a(b-c)} = 2$. This implies $-bc + ab = 2ab - 2ac$. Thus, $2ac = ab + bc$. Rearranging gives $2ac = b(a+c)$. Therefore, $2ac = 15b \ldots (1)$. Substituting $2ac = 15 \left(\frac{36}{5}\right) = 108$. Hence, $ac = 54$. Given $a + c = 15$, we have $a^2 + c^2 + 2ac = 225$. Therefore, $a^2 + c^2 = 225 - 108 = 117$.

(A)\quad L=\mu_f\mu_0n^2Al (B)\quad L=\mu_r\mu_vn^2AlX (C)\quad \checkmark (D)\quad \checkmark (E)\quad \checkmark

Given $g \rho a^2 x = \sigma a^3 A$. $$A = \frac{\rho \, g}{\sigma \, a \, x}$$ $$T = 2\pi \sqrt{\frac{\sigma a}{\rho g}}$$ Now, $\sigma = \frac{10 \times 10^{-3}}{10^{-3}} = 10$. Therefore, $$T = 2\pi \sqrt{\frac{10 \times 0.1}{10^3 \times 10}}$$ $$= 2\pi \times 10^{-2}$$

Q3. Hot water is less viscous than cold water. (1) Surfactant reduces surface tension.

Given $\($ $\lambda$ = $\frac{h}{p}$ = $\frac{6.63 \times 10^{-34}}{10^{-30} \times 2.21 \times 10^6}$ $\)$. This simplifies to $\($ 3 $\times$ 10^{-10} $\,$ $\mathrm{m}$ $\)$. Hence particle will behave as x-ray.

Given $\mu_1 = 1$ for air and $\mu_2 = 1.5$ for glass. PO $= u = -x$ PI $= v = x$ PO $= PI$ Using the lens formula: $$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$ Substitute the values: $$\frac{1.5}{x} + \frac{1}{x} = \frac{1}{2R}$$ Simplifying: $$\frac{5}{2x} = \frac{1}{2R}$$ Therefore, $$X = 5R$$

Given $N_2 = N_0 e^{-3\lambda t}$ and $N_1 = N_0 e^{-\lambda t}$. We have $\frac{N_2}{N_1} = e^{-2\lambda t}$. The half-life of $N$ is $t = \frac{\ln 2}{\lambda}$. Therefore, $t = \frac{\ln 2}{\lambda}$. We have $\frac{N_0}{2} = N_0 e^{-\lambda t}$, so $\lambda t = \ln 2$. Thus, $t = \frac{\ln 2}{\lambda}$. Therefore, $e^{-2\lambda \frac{\ln 2}{\lambda}} = \frac{N_2}{N_1} = \frac{1}{4}$.

Capacitor behaves like closed circuit at $t = 0$ and charge is zero.

Dimension $[x(t)] = [L]$ $[A] = [L]$ $[B] = [L]$ $[C] = [LT^{-2}]$ $[D] = [L]$ $$\left[ \frac{ABC}{D} \right] = \left[ \frac{L \times L \times LT^{-2}}{L} \right] = [L^2 T^{-2}]$$

A $\rightarrow$ P $\propto$ $\frac{1}{V}$ $\Rightarrow$ PV = constant $\Rightarrow$ nRT = const. $\Rightarrow$ T = const. Hence Isothermal III. B $\rightarrow$ IV. W $\neq$ 0, $\Delta$ U $\neq$ 0, $\Delta$ Q $\neq$ 0 [only isobaric]. C $\rightarrow$ I $\Delta$ Q = 0 Adiabatic. D $\rightarrow$ II w = 0 Isochoric. III $\ $IV $\ $I $\ $II.

Given $\tau = C \theta \Rightarrow \left[ ML^2 \, T^{-2} \right] = [C][1]$. $(B) C. S = \frac{\theta}{I} = \frac{BNA}{C}$. V.S. $= \frac{BNA}{RC}$ [R also depends on 'N'] (C) V.S. $\propto \frac{NAB}{CR}$ R $\to$ NR (D) False [Theory] (E) E [False] C. S $\propto N$ $\therefore C. S = \frac{NAB}{C}$

Given $E_1 = E_2$, taking $kq = 1$. $$\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2} = \frac{2a}{(a^2 + r^2)^{3/2}}$$ $$4ar = \frac{2a}{(r^2 - a^2)^2} = (a^2 + r^2)^{3/2}$$ $$(r^2 - a^2)^2 = 2r(a^2 + r^2)^{3/2}$$ $$\left(1 - \frac{a^2}{r^2}\right)^2 = 2\left(1 + \frac{a^2}{r^2}\right)^{3/2} \left(x = \frac{a}{i}\right)$$ $$(1 - x^2)^2 = 2(1 + x^2)^{3/2}$$ $$(1 - x^2)^2 = 2$$ $$(1 + x^2)^{3/2} = 2$$ Now for $X = 3$ We get $$\frac{64}{10\sqrt{10}} \approx 2 \Rightarrow \frac{a}{r} \approx 3$$ [But for $a > r$ point charge will be between the dipole where $\vec{E} \neq 0$]

Given the equation $625 = ms\Delta T + mL$. Step (1): $$625 = m \left[ 125 \times 300 + 2.5 \times 10^4 \right]$$ Simplifying inside the brackets: $$625 = m[37500 + 25000]$$ Further simplification gives: $$625 = m[62500]$$ Solving for $m$: $$m = \frac{1}{100} kg$$ Converting to grams: $$M = 10 grams$$

Given: $$AB = h \sec r$$ $$BC = h \sec r \sin(\overline{i - r})$$ Simplifying, we have: $$BC = \frac{h \sin(i - r)}{\cos r}$$

Using WET, $W_g = k_f - k_i$. $MgL \sin \theta = k_f - k_i$. Kinetic energy in pure rolling is $\frac{1}{2} m V_{cm}^2 + \frac{1}{2} I_{cm} \omega^2$. $$= \frac{1}{2} m V^2 + \frac{1}{2} \times \frac{2}{5} m R^2 \frac{V^2}{R^2}$$ $$= \frac{7}{10} m V^2$$ $$mgL \sin \theta = \frac{7}{10} m V_f^2 - 0$$ $$V_f^2 \propto \sin \theta$$ $$\left( \frac{V_1}{V_2} \right)^2 = \frac{\sin \theta_1}{\sin \theta_2} = \frac{\sin 30^\circ}{\sin 45^\circ} = \frac{1}{\sqrt{2}}$$

The total resistance $R_{net} = 6 \, \Omega$. The voltage across $AB$ is $V_{AB} = 0.5 \times 4 = 2 \, volt$. The voltage across $CD$ is $V_{CD} = 1 \times 4 = 4 \, volt$. The current through the ammeter is $1 \, A$. A, B, and D are correct.

Given $\($ $\alpha$ $\equiv$ $\frac{\phi}{\sigma}$ $\)$ and $\($ $\beta$ = $\frac{\phi}{\lambda}$ $\)$. Therefore, $\($ $\alpha$ $\equiv$ $\frac{\lambda}{\beta}$ $\equiv$ displacement $\)$.

Given the equation for equivalent focal length: $$\frac{1}{f_{eq}} = \frac{2}{f_L} - \frac{1}{f_m}$$ where $$f_m = -\frac{|R_2|}{2}$$ We have: $$\frac{1}{f_L} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$ Substituting into the equation for $f_{eq}$: $$\frac{1}{f_{eq}} = 2 \left(\frac{\mu_2 - \mu_1}{\mu_1} \left(\frac{R_1 + R_2}{R_1 R_2}\right) + \frac{2}{R_2}\right)$$ Simplifying further: $$= 2 \left[\frac{(\mu_2 - \mu_1)(R_1 + R_2) + \mu_1 R_1}{R_2 \mu_1 R_1}\right]$$ $$= \frac{2[\mu_2 R_1 + \mu_2 R_2 - \mu_1 R_2]}{R_2 \mu_1 R_1}$$ For the same size of image, we have: $$u = 2f$$ Thus, $$u = \frac{\mu_1 R_1 R_2}{\mu_2 R_1 + \mu_2 R_2 - \mu_1 R_2}$$

Given $\vec{K} = 3\hat{i} + 4\hat{j}$. The magnitude of $\hat{K}$ is given by $$\hat{K} = \frac{3\hat{i} + 4\hat{j}}{5}.$$ The vector $\hat{E}$ is $$\hat{E} = \frac{4\hat{i} - 3\hat{j}}{5}.$$ The cross product $\hat{B} = \hat{K} \times \hat{E}$ results in $$\hat{B} = -\hat{Z}.$$ The value of $B_0$ is given by $$B_0 = \frac{E_0}{C} = \frac{57}{3 \times 10^8}.$$

Distance is the area under the graph. $$d = 300 \times 2 + 400 \times 28.5$$ $$= 600 + 11400$$ $$= 12000 \, \mathrm{m}$$

The mass of the disc is $m$. The mass of the cut part is $\frac{m}{16}$. The center of mass $X_{com}$ is calculated as follows: $$X_{com} = \frac{m \times 0 - \frac{m}{16} \times 15}{m - \frac{m}{16}}$$ This simplifies to: $$X_{com} = 1 \, cm.$$

The electric force $F_e$ is given by: $$F_e = \frac{kq_1q_2}{r^2}$$ The gravitational force $F_g$ is given by: $$F_g = \frac{Gm_1m_2}{r^2}$$ The ratio of electric force to gravitational force is: $$\frac{F_e}{F_g} = \frac{kq_1q_2}{Gm_1m_2}$$ Substituting the given values: $$= \frac{9 \times 10^9 \times 6.67 \times 10^{-19} \times 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \times 19.2 \times 10^{-27} \times 9 \times 10^{-27}}$$ Simplifying: $$= \frac{10^{-20}}{2 \times 10^{-65}}$$ $$= 2 \times 10^{-65}$$

Given $\vec{A} \cdot \vec{B} = 0$. $4 - 6n + 8p = 0$ $|\vec{A}| = |\vec{B}|$ $4 + 9n^2 + 4 = 4 + 4 + 16p^2$ $9n^2 = 16p^2$ $P = \pm \frac{3}{4} n$ $4 - 6n \pm 6n = 0$ $12n = 4$ $n = \frac{1}{3}$

Given $\gamma = \frac{3}{2}$. The equation $T V^{\gamma - 1} = C$ is used. Substituting the values, we have: $$273 \, V_0^{0.5} = T \left( \frac{V_0}{4} \right)^{0.5}$$ Solving for $T$: $$T = 273 \times 2 = 546$$ The change in temperature $\Delta T = 273$.

Given $y = 10 - x$. $$w = \int_0^4 x^2 (10 - x) \, dx + \int_0^2 y^2 \, dy$$ $$= \left. \frac{10x^3}{3} - \frac{x^4}{4} \right|_0^4 + \left. \frac{y^3}{3} \right|_0^2$$ $$= \frac{640}{3} - \frac{256}{4} + \frac{8}{3}$$ $$= 216 \times 64$$ $$= 152 \, \mathrm{J}$$

Given $\($ $\varepsilon$ - L $\frac{dI}{dt}$ - IR = 0 $\)$. Substituting the values: $\($ 12 - 3 $\times$ (-8) - I $\times$ 12 = 0 $\)$ Solving for $\($ I $\)$: $\($ I = 3 $\)$

Palladium implies $5^{th}$ period. Iridium, Osmium, Platinum implies $6^{th}$ Period.

Given $\lambda = 900 \, \mathrm{nm}$ for a hydrogen atom $(Z = 1)$. Convert to centimeters: $\lambda = 9 \times 10^{-5} \, \mathrm{cm}$. The Rydberg constant $R_H = 10^5 \, \mathrm{cm}^{-1}$. Using the Rydberg equation: $$\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ This implies: $$\frac{1}{\lambda \times R_H} = \frac{1}{n_1^2} - \frac{1}{n_2^2}$$ Substituting the values: $$\frac{1}{9 \times 10^{-5} \, \mathrm{cm} \times 10^5 \, \mathrm{cm}^{-1}} = \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ Simplifying gives: $$\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{9}$$ It is possible when $n_1 = 3$, $n_2 = \infty$. Possible series: $\infty \to 3$

$\mathrm{SO_2}$ can act as both oxidising and reducing agents because due to intermediate oxidation state, it can oxidise and reduce as well.

Given $\Delta T_f = i K_f m$. $$0.558 = i \times 1.86 \times 0.1$$ $$i = \frac{0.558}{0.186} = 3$$ Number of ions when 100$\%$ ionisation takes place $= 3$ $$\mathrm{[Cr(NH_3)_5 Cl] \ Cl_2 \xrightarrow{aqueous solution} [Cr(NH_3)_5 Cl]^{2+} + 2Cl^-}$$ Number of ions $= 3$

Given the reactions, we need to find $E_4^\circ$. Using the relation for Gibbs free energy, we have: $$\Delta G_4^\circ = \Delta G_1^\circ + \Delta G_2^\circ$$ which implies $$-n_4 F E_4^\circ = -n_1 F E_1^\circ - n_2 F E_2^\circ$$ Therefore, $$4 E_4^\circ = 3 \times 2 + (1 \times 0.8)$$ Solving this gives: $$E_4^\circ = \frac{6.8}{4} \, \mathrm{V}$$ Thus, $$E_4^\circ = 1.7 \, \mathrm{V}$$

\begin{tabular}{|c|p{4.2cm}|c|p{8.5cm}|} \hline \multicolumn{2}{|c|}{LIST-I} & \multicolumn{2}{c|}{LIST-II} \\ \multicolumn{2}{|c|}{Name reaction} & \multicolumn{2}{c|}{Product obtainable} \\ \hline A. & Swarts reaction & I. & $Et-I \xrightarrow[\mathrm{DF}]{\mathrm{KF}} Et-F$ \\ \hline B. & Sandmeyer's reaction & II. & $PhN_2^{+}Cl^{-} \xrightarrow[\mathrm{CuCN/KCN}]{} PhCN+N_2$ \\ \hline C. & Wurtz-Fittig reaction & III. & $Ph-Cl+EtCl \xrightarrow[\mathrm{ether}]{\mathrm{Na}}$ $Ph-Et+Ph-Ph+Et-Et$ \\ \hline D. & Finkelstein reaction & IV. & $Et-Cl \xrightarrow[\mathrm{acetone}]{\mathrm{NaI}} Et-I+NaCl$ \\ \hline \end{tabular}

The reaction involves the conversion of D-Fructose to an enediol intermediate. This enediol can then rearrange to form D-Glucose and D-Mannose.

Moles of removed $\mathrm{CO_2} = \frac{10^{21}}{6.02 \times 10^{23}} \, \mathrm{mol}$ $$= 1.66 \times 10^{-3} \, \mathrm{mol}$$ Mole of $\mathrm{CO_2}$ left $= 2.8 \times 10^{-3} \, \mathrm{moles}$ Total moles of $\mathrm{CO_2}$ taken initially $$= (2.8 + 1.66) \times 10^{-3} \, \mathrm{mol}$$ Mass of $\mathrm{CO_2}$ taken initially $$= 4.46 \times 10^{-3} \times 44$$ $$= 196.24 \times 10^{-3} \, \mathrm{g}$$ $$= 196.24 \, \mathrm{mg}$$

For $\mathrm{H_2O(s)} \rightarrow \mathrm{H_2O(s)}; \Delta S_1 = \int_{268 \, \mathrm{K}}^{273 \, \mathrm{K}} C_p \, m \, dT$. For $\mathrm{H_2O(s)} \rightarrow \mathrm{H_2O(l)}; \Delta S_2 = \frac{\Delta H_{m, \mathrm{fus}}}{273}$. For $\mathrm{H_2O(l)} \rightarrow \mathrm{H_2O(l)}; \Delta S_3 = \int_{273}^{373} C_p \, m \, dT$. For $\mathrm{H_2O(l)} \rightarrow \mathrm{H_2O(g)}; \Delta S_4 = \frac{\Delta H_{m, \mathrm{vap}}}{373}$. For $\mathrm{H_2O(g)} \rightarrow \mathrm{H_2O(g)}; \Delta S_5 = \int_{373}^{383} C_p \, m \, dT$. The total entropy change is $\Delta S_{\mathrm{total}} = \Delta S_1 + \Delta S_2 + \Delta S_3 + \Delta S_4 + \Delta S_5$.

Given $\mathrm{Co^{2+}}$ complex having $\mu = 3.95 \mathrm{BM}$. Hence number of unpaired electrons $= 3$. $$\mathrm{Co^{2+}} \Rightarrow 3d^7 = t_{2g}^5 e_g^2$$

Ma_3 b_3 type complexes show Facial - Meridional isomerism. (i) $[\mathrm{Co(NH_3)_3Cl_3}] \Rightarrow \mathrm{Ma_3 \, b_3}$ (ii) $[\mathrm{Co(NH_3)_4Cl_2}]^+ \Rightarrow \mathrm{Ma_4 \, b_2}$ (iii) $[\mathrm{Co(en)_3}]^{3+} \Rightarrow \mathrm{M(AA)_3}$ (iv) $[\mathrm{Co(en)_2Cl_2}]^+ \Rightarrow \mathrm{M(AA)_2 \, b_2}$ a, b = $\mathrm{NH_3, \, Cl^-}$ AA = en

This is an example of Tollen's reaction i.e. multiple cross aldol followed by cross Cannizaro reaction. $$\mathrm{CH_3CH_2CH=O} \xrightarrow{2\mathrm{HCHO} Alkali} \mathrm{CH_3}\begin{array}{c} \mathrm{CH_2OH} \\ | \\ \mathrm{C} \\ | \\ \mathrm{CHO} \\ | \\ \mathrm{CH_2OH} \end{array}$$ $$\mathrm{HCHO} \xrightarrow{Alkali} \mathrm{CH_3}\begin{array}{c} \mathrm{CH_2OH} \\ | \\ \mathrm{C} \\ | \\ \mathrm{CH_2OH} \\ | \\ \mathrm{CH_2OH} \end{array} + \mathrm{HCOO}^-$$

q is aromatic. r is non-aromatic. p is antiaromatic. $q > r > p$ (order of stability). Aromatic $>$ non-aromatic $>$ antiaromatic.

"X" is $\mathrm{CH_3-CH^*-CH_2}$ with $\mathrm{Cl}$ attached to the second carbon. When $X$ is reacted with $\mathrm{Cl_2}$ under $h\nu$, the products are: $$\mathrm{CH_3-CH^*-CHCl + CH_3-CCl-CH_2Cl + Cl-CH_2-CH-CH_2-Cl}$$

Moles of phenol $= \frac{2}{94} = 0.021$. Therefore, moles of bromine $= 0.021 \times 3 = 0.064$. Therefore, mass of bromine $= 0.064 \times 160 = 10.22 \, \mathrm{g}$.

Q4. (1) $\mathrm{V^{2+}}$ (Violet), $\mathrm{Cr^{3+}}$ (Violet), $\mathrm{Mn^{3+}}$ (Violet) (2) $\mathrm{Zn^{2+}}$ (Colourless), $\mathrm{V^{3+}}$ (Green), $\mathrm{Fe^{3+}}$ (Yellow) (3) $\mathrm{Ti^{4+}}$ (Colourless), $\mathrm{V^{4+}}$ (Blue), $\mathrm{Mn^{2+}}$ (Pink) (4) $\mathrm{Sc^{3+}}$ (Colourless), $\mathrm{Ti^{3+}}$ (Purple), $\mathrm{Cr^{2+}}$ (Blue)

Lassaigne's test is a general test for detection of halogen, nitrogen and sulphur in an organic compound. These elements covalently bonded to the organic compounds. In order to detect them, these have to converted into ionic forms. $$3\mathrm{Na_4[Fe(CN)_6]} + 2\mathrm{Fe_2(SO_4)_3} \longrightarrow \mathrm{Fe_4[Fe(CN)_6]_3}$$ Prussian Blue

Condition for precipitation $Q_{ip} > K_{sp}$. For $[\mathrm{A(OH)_2}]$ $$[\mathrm{A^{2+}}][\mathrm{OH}^-]^2 > 9 \times 10^{-10}$$ $$[\mathrm{A^{2+}}] = 1\, \mathrm{M}$$ $$\Rightarrow [\mathrm{OH}^-] > 3 \times 10^{-5}\, \mathrm{M}$$ For $[\mathrm{B(OH)_3}]$ $$[\mathrm{B^{3+}}][\mathrm{OH}^-]^3 > 27 \times 10^{-18}$$ $$[\mathrm{B^{3+}}] = 1\, \mathrm{M}$$ $$\Rightarrow [\mathrm{OH}^-] > 3 \times 10^{-6}\, \mathrm{M}$$ So, $\mathrm{B(OH)_3}$ will precipitate before $\mathrm{A(OH)_2}$.

$\mathrm{NO}$: $7$ valence electrons $\Rightarrow$ incomplete octet (odd-electron molecule) $\mathrm{NO_2}$: $7$ valence electrons $\Rightarrow$ incomplete octet (odd-electron molecule) $\mathrm{BCl_3}$: $6$ electrons around the central atom $\Rightarrow$ incomplete octet $\mathrm{AlCl_3}$: $6$ electrons around the central atom $\Rightarrow$ incomplete octet $\mathrm{H_2SO_4}$: $12$ electrons around the central atom, $\mathrm{PCl_5}$: $10$ electrons around the central atom $\Rightarrow$ molecules with expanded octet $\mathrm{CCl_4}$: $8$ electrons around the central atom, $\mathrm{CO_2}$: $8$ electrons around the central atom $\Rightarrow$ molecules obeying the octet rule

B and D are $3^\circ$ amine which does not have replaceable H on N, so does not react.

The reaction is given by: $$\mathrm{C_2H_5NH_2(aq) + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^-}$$ The concentration is $C = 10^{-3} \, \mathrm{M}$. The expression $C(1 - \alpha)$ implies: $$C = 10^{-3}$$ The concentrations at equilibrium are: $$C\alpha = 10^{-5}$$ Thus, $1 - \alpha = 1$. Given, $\mathrm{pH} = 9$, therefore $\mathrm{pOH} = 5$ which implies $[\mathrm{OH}^-] = 10^{-5} \, \mathrm{M}$. Now, the base dissociation constant $K_b$ is given by: $$K_b = \frac{[\mathrm{C_2H_5NH_3^+}][\mathrm{OH}^-]}{[\mathrm{C_2H_5NH_2}]}$$ Substituting the values: $$\Rightarrow K_b = \frac{10^{-5} \times 10^{-5}}{10^{-3}} = 10^{-7}$$

Given $m$ mole of $\mathrm{BaSO_4} = mmoles of S = \frac{466}{233}$. Mass of $S = \frac{466}{233} \times 32 \, \mathrm{mg}$ $$= 64 \, \mathrm{mg}$$ Percentage of $S = \frac{64}{160} \times 100 = 40\%$$

The reaction sequence starts with the nitration of toluene to form a nitrotoluene derivative. Bromination occurs at the ortho position relative to the methyl group, resulting in the formation of a bromo-nitrotoluene. Reduction of the nitro group to an amino group is achieved using Sn and HCl. The amino group is then converted to a diazonium salt using NaNO2 and HCl. Finally, the diazonium group is replaced by hydrogen using hypophosphorous acid (H3PO2), yielding the final product. The molar mass of the product $\mathrm{C_7H_7Br}$ (A) is 171 g mol$^{-1}$.

Given the reaction $2 \mathrm{N_2O_5} (\mathrm{g}) \rightarrow 2 \mathrm{N_2O_4} (\mathrm{g}) + \mathrm{O_2} (\mathrm{g})$. $$k = \frac{2.303}{t} \log \frac{0.9 - 0.6}{0.9 - x}$$ $$2 \times 10^{-2} \times 100 = \log \frac{0.3}{0.9 - x}$$ $$100 = \frac{0.3}{0.9 - x}$$ $$0.9 - x = \frac{0.3}{100} = 0.01$$ $$0.9 - x = 0.003$$ $$= 897 \times 10^{-3}$$

Given $\Delta H^\circ_{rxn} = 55 \, kJ/mol$, $T = 298 \, K$ and $\Delta S^\circ_{rxn} = 175 \, J/mol$. The change in Gibbs free energy is given by $$\Delta G^\circ_{rxn} = \Delta H^\circ_{rxn} - T \Delta S^\circ_{rxn}$$ Substituting the values, $$\Rightarrow \Delta G^\circ_{rxn} = 55000 \, J/mol - 298 \times 175 \, J/mol$$ Calculating further, $$\Rightarrow \Delta G^\circ_{rxn} = 55000 - 52150$$ Finally, $$\Rightarrow \Delta G^\circ_{rxn} = 2850 \, J/mol$$