JEE Main 28 January 2025 Shift 2 question paper with solutions

Given: $E_1$: Bag $B_1$ is selected $B_1$: 6W4B, $B_2$: 4W6B, $B_3$: 5W5B $E_2$: Bag $B_2$ is selected $E_3$: Bag $B_3$ is selected $A$: Drawn ball is white We have to find $P \left( \frac{E_2}{A} \right)$ $$P \left( \frac{E_2}{A} \right) = \frac{P(E_2)P \left( \frac{A}{E_2} \right)}{P(E_1)P \left( \frac{A}{E_1} \right) + P(E_2)P \left( \frac{A}{E_2} \right) + P(E_3)P \left( \frac{A}{E_3} \right)}$$ $$= \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}}$$ $$= \frac{4}{15}$$

Equation of line in the internal bisector of $OA$ and $OB$ is $(\sqrt{3} + 1)\hat{i} + (\sqrt{3} + 1)\hat{j}$. Therefore, the line will be $y = x \Rightarrow x - y = 0$. $$D = \left| \frac{a - (1-a)}{\sqrt{a^2 + (1-a)^2}} \right| = \frac{9}{\sqrt{2}}$$ $$(2a - 1)^2 = \frac{81}{2} \left( a^2 + (1-a)^2 \right)$$ $$\Rightarrow 2 \left( 4a^2 - 4a + 1 \right) = 81a^2 + 8a^2 - 162a - 81$$ $$\Rightarrow 162a^2 - 162a + 81 - 8a^2 + 8a - 2 = 0$$ $$\Rightarrow 154a^2 - 154a + 79 = 0$$ Sum of values $= -\frac{-154}{154} = 1$

Let $\vec{a}_{11} = component of \vec{a} along \vec{b}$ and $\vec{a}_1 = component of \vec{a} perpendicular to \vec{b}$. $$\vec{a}_{11} = \frac{16}{11} (3\hat{i} + \hat{j} - \hat{k})$$ $$\vec{a}_1 = \frac{1}{11} (-4\hat{i} - 5\hat{j} - 17\hat{k})$$ Therefore, $\vec{a} = \vec{a}_{11} + \vec{a}_1$. Thus, $\vec{a} = \frac{16}{11} (3\hat{i} + \hat{j} - \hat{k}) + \frac{1}{11} (-4\hat{i} - 5\hat{j} - 17\hat{k})$. $$\vec{a} = 11\hat{j} - \frac{33}{11} \hat{k}$$ $$\alpha = 4\hat{i} + \hat{j} - 3\hat{k}$$ $$\alpha^2 + \beta^2 + \gamma^2 = 16 + 1 + 9 = 26$$

Given the equation $x^2 - (3 - 2i)x - (2i - 2) = 0$. Using the quadratic formula, we have: $$x = \frac{(3 - 2i) \pm \sqrt{(3 - 2i)^2 - 4(1)(-(2i - 2))}}{2(1)}$$ Simplifying inside the square root: $$(3 - 2i) \pm \sqrt{9 - 4 - 12i + 8}$$ This becomes: $$\frac{3 - 2i \pm \sqrt{-3 - 4i}}{2}$$ Further simplifying: $$3 - 2i \pm \sqrt{(1)^2 + (2i)^2 - 2(1)(2i)}$$ This simplifies to: $$\frac{3 - 2i \pm (1 - 2i)}{2}$$ Thus, we have two solutions: $$\frac{3 - 2i + 1 - 2i}{2} or \frac{3 - 2i - 1 + 2i}{2}$$ This results in: $$2 - 2i or 1 + 0i$$ So $\alpha \gamma + \beta \delta = 2(1) + (-2)(0) = 2$$

Given the ellipse equation $E: \frac{x^2}{9} + \frac{y^2}{4} = 1$ and the tangent $T = S_1$. $$\Rightarrow \frac{\sqrt{2x}}{9} + \frac{1}{3} \left(4 \left(\frac{4}{3} y\right) - 1 \right) = \frac{2}{9} + \frac{16}{9(4)} - 1$$ $$\frac{\sqrt{2x}}{9} + \frac{y}{3} = \frac{2}{9} + \frac{4}{9}$$ $$\frac{\sqrt{2x}}{9} + \frac{y}{3} = \frac{2}{3} \Rightarrow \sqrt{2x} + 3y = 6$$ Now the point of intersection of the chord and ellipse is $$\frac{(6 - 3y)^2}{18} + \frac{y^2}{4} = 1$$ $$(2 - y)^2 + \frac{y^2}{2} = 1$$ $$2 \left(4 + y^2 - 4y\right) + y^2 = 4$$ $$\Rightarrow 3y^2 - 8y + 4 = 0$$ $$\Rightarrow y = 2, \frac{2}{3}$$ So, points are $(0, 2)$ are $\left(2\sqrt{2}, \frac{2}{3}\right)$. Length of chord $= \sqrt{(2\sqrt{2})^2 + \left(\frac{2}{3} - 2\right)^2}$ $$= \sqrt{8 + \frac{16}{9}}$$ $$= \frac{\sqrt{88}}{3} = \frac{2\sqrt{22}}{3}$$ On comparing $\alpha = 22$

Probability (P) = $\frac{favourable case}{Total case}$ (when A $\&$ E are in order) Total case = 6! Favourable case = $\binom{6}{2}$ $\cdot$ 4! $$P = \frac{(15)4!}{(30)4!}$$ Probability when not in order = 1 - $\frac{1}{2}$ = $\frac{1}{2}$

Given $g(x) = x^2 + x^{7/3}$. The derivative is $g'(x) = 2x + \frac{7}{3} x^{4/3}$. Let $f(x) = \frac{g'(x)}{x}$. Then $f(x) = 2 + \frac{7}{3} x^{1/3}$. For $f(r^3)$, we have $f(r^3) = 2 + \frac{7}{3} r$. The sum is $$\sum_{r=1}^{15} \left( 2 + \frac{7}{3} r \right) = 2(15) + \frac{7}{3} \left( \frac{15(16)}{2} \right) = 310.$$

Line L is given by $\($ $\frac{x+1}{3}$ = $\frac{y+3}{5}$ = $\frac{z+5}{7}$ $\)$. The line PQ is given by $\($ $\frac{x - \frac{15}{7}}{1}$ = $\frac{y - \frac{32}{7}}{4}$ = $\frac{z - 7}{7}$ = $\lambda$ $\)$. Therefore, point Q is $\($ $\left$( $\lambda$ + $\frac{15}{7}$, 4$\lambda$ + $\frac{32}{7}$, 7$\lambda$ + 7 $\right$) $\)$. Since Q lies on line L, we have $\($ $\frac{\lambda + \frac{15}{7} + 1}{3}$ = $\frac{7\lambda + 7 + 5}{7}$ $\)$. This simplifies to $\($ 7$\lambda$ + 22 = 21$\lambda$ + 36 $\)$, giving $\($ $\lambda$ = -1 $\)$. Thus, point Q is $\($ $\left$( $\frac{8}{7}$, $\frac{4}{7}$, 0 $\right$) $\)$. The distance PQ is $\($ $\sqrt{ \left( \frac{15}{7} - \frac{8}{7} \right)^2 + \left( \frac{32}{7} - \frac{4}{7} \right)^2 + (7 - 0)^2 }$ $\)$. Therefore, $\($ PQ = $\sqrt{66}$ $\)$ and $\($ (PQ)^2 = 66 $\)$.

Given the equations $x(1 + y^2) = 1$ and $y^2 = 2x$. From equation (1) and (2), $x(1 + 2x) = 1 \Rightarrow 2x^2 + x - 1 = 0$. Solving for $x$, we get $x = \frac{1}{2}$ and $x = -1$ (Reject). Therefore, $y^2 = 2 \left( \frac{1}{2} \right)$ which implies $y = \pm 1$. The area bounded is given by $$\int_{-1}^{1} \left( \frac{1}{1 + y^2} - \frac{y^2}{2} \right) \, dy$$ which evaluates to $$\left( \tan^{-1} y - \frac{y^3}{6} \right) \bigg|_{-1}^{1}$$ resulting in $$\frac{\pi}{2} - \frac{1}{3}.$$

Given $\($ P = $\begin{bmatrix}$ $\cos$ $\theta$ & -$\sin$ $\theta$ $\\$ $\sin$ $\theta$ & $\cos$ $\theta$ $\end{bmatrix}$ $\)$. Therefore, $\($ P^T P = I $\)$. Let $\($ B = P A P^T $\)$. Pre-multiply by $\($ P^T $\)$ (Given): $\($ P^T B = P^T P A P^T = A P^T $\)$. Now post-multiply by $\($ P $\)$: $\($ P^T B P = A P^T P = A $\)$. So $\($ A^2 = P^T B P $\underline{P^T B P}$_{I} $\)$. $\($ A^2 = P^T B^2 P $\)$. Similarly, $\($ A^{10} = P^T B^{10} P = C $\)$. Given $\($ A = $\begin{bmatrix}$ $\frac{1}{\sqrt{2}}$ & -2 $\\$ 0 & 1 $\end{bmatrix}$ $\)$. $\($ $\Rightarrow$ A^2 = $\begin{bmatrix}$ $\frac{1}{2}$ & -$\sqrt{2}$ - 2 $\\$ 0 & 1 $\end{bmatrix}$ $\)$. Similarly check $\($ A^3 $\)$ and so on since $\($ C = A^{10} $\Rightarrow$ $\)$ Sum of diagonal elements of $\($ C $\)$ is $\($ $\left$( $\frac{1}{\sqrt{2}}$ $\right$)^{10} + 1 $\)$. $\($ = $\frac{1}{32}$ + 1 = $\frac{33}{32}$ = $\frac{m}{n}$ $\)$. $\($ $\gcd$(m, n) = 1 $\)$ (Given). $\($ $\Rightarrow$ m + n = 65 $\)$.

Put $x^{1/4} = t \implies dx = 4t^3 \, dt$ $$\int 4t^3 \, dt = 4 \int \left( \frac{t^2 - 1}{t + 1} + \frac{1}{t + 1} \right) dt$$ $$f(x) = 4 \left[ \frac{x^{1/2}}{2} - x^{1/4} + \ln |x^{1/4} + 1| \right] + C$$ $$f(0) = -6$$ $$\Rightarrow C = -6$$ $$\Rightarrow f(1) = 4 \left( \log_e 2 - 2 \right)$$

Given $$\int_0^2 x F'(x) \, dx = 6$$ we have $$= x F(x) \bigg|_0^2 - \int_0^2 f(x) \, dx = 6$$ $$= 2 F(2) - \int_0^2 x F(x) \, dx = 6 [\because f(2) = 2 F(2) = 2]$$ $$\int_0^2 x F(x) \, dx = -2 \ldots (1)$$ Therefore, $$\int_0^2 F(x) \, dx = -2 \ldots (2)$$ Also, $$\int_0^2 x^2 F''(x) \, dx = x^2 F'(x) \bigg|_0^2 - 2 \int_0^2 x' F'(x) \, dx = 40$$ $$= 4 F'(2) - 2 \times 6 = 40$$ $$F'(2) = 13$$ Therefore, $$F'(2) + \int_0^2 F(x) \, dx = 13 - 2 = 11$$

Given $$S_n = \sum_{k=1}^{n} \frac{4}{k^2 + 5k + 6}$$ This can be rewritten as $$= \sum_{k=1}^{n} \frac{4}{(k+2)(k+3)} = 4 \sum_{k=1}^{n} \left( \frac{1}{k+2} - \frac{1}{k+3} \right)$$ Simplifying, we have $$= 4 \left[ \frac{1}{3} - \frac{1}{4} \right]$$ $$= 4 \left[ \frac{1}{4} - \frac{1}{5} \right]$$ $$= 4 \left[ \frac{1}{n+2} - \frac{1}{n+3} \right]$$ Thus, $$S_n = 4 \left[ \frac{1}{3} - \frac{1}{n+3} \right]$$ For $n = 2025$, $$S_{2025} = 4 \left[ \frac{1}{3} - \frac{1}{2028} \right]$$ Simplifying further, $$S_{2025} = 4 \left[ \frac{675}{2028} \right]$$ Finally, $$507 S_{2025} = 675$$

As $f(x)$ is onto, hence $A$ is the range of $f(x)$. $$f'(x) = 6x^2 - 30x + 36$$ $$= 6(x - 2)(x - 3)$$ Now $$f(2) = 16 - 60 + 72 + 7 = 35$$ $$f(3) = 54 - 135 + 108 + 7 = 34$$ $$f(0) = 7$$ Hence range $\in [7, 35] = A$. Also for range of $g(x)$ $$g(x) = 1 - \frac{1}{x^{2025} + 1} \in (0, 1) = B$$ $$s = \{0, 7, 8, \ldots, 35\}$$ hence $n(s) = 30$.

Given $f(x) = \sec^{-1}(2[x] + 1)$. $\Rightarrow 2[x] + 1 \geq 1$ or $2[x] + 1 \leq -1$ $\Rightarrow 2[x] \geq 0$ or $2[x] \leq -2$ $\Rightarrow [x] \geq 0$ or $[x] \leq -1$ $\Rightarrow x \geq 0$ or $x \leq 0$ Domain of $f(x)$ is $(-\infty, \infty)$

Given the expression $$\frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13} \frac{\sin \left[ \left( \frac{\pi}{4} + r \frac{\pi}{6} \right) - \left( \frac{\pi}{4} \right) - (r-1) \frac{\pi}{6} \right]}{\sin \left( \frac{\pi}{4} + (r-1) \frac{\pi}{6} \right) \sin \left( \frac{\pi}{4} + r \frac{\pi}{6} \right)}$$ which simplifies to $$\frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13} \left( \cot \left( \frac{\pi}{4} + (r-1) \frac{\pi}{6} \right) - \cot \left( \frac{\pi}{4} + r \frac{\pi}{6} \right) \right).$$ This evaluates to $$= 2\sqrt{3} - 2 = a \sqrt{3} + b.$$ So $$a^2 + b^2 = 8.$$

Slope of the third side $=$ slope of the perpendicular bisector of given lines $h:\;\dfrac{-x+\frac{2}{\sqrt2}y-4}{\sqrt5} =\pm\dfrac{x+\frac{1}{\sqrt2}y-4}{\sqrt2}$ $h_1:\;\sqrt2(-x+2y-4)=\sqrt5(x+y-4)$ $h_2:\;\sqrt2(-x+2y-4)=-\sqrt5(x+y-4)$ $M_{L_1}=-\dfrac{\sqrt5+\sqrt2}{\sqrt5-2\sqrt2}$ $M_{L_2}=\dfrac{\sqrt5-\sqrt2}{\sqrt5+2\sqrt2}$ $M_{L_1}+M_{L_2} =-\dfrac{\sqrt5+\sqrt2}{\sqrt5-2\sqrt2} +\dfrac{\sqrt5-\sqrt2}{\sqrt5+2\sqrt2}$ $=-\dfrac{(\sqrt5+\sqrt2)(\sqrt5+2\sqrt2)+(\sqrt5-\sqrt2)(\sqrt5-2\sqrt2)}{-3}$ $=6$

Coefficient of $T_r, T_{r+1}, T_{r+2} \rightarrow GP$ $$\Rightarrow \left(^{12}C_r\right)^2 = ^{12}C_{r-1} \cdot ^{12}C_{r+1}$$ $$\Rightarrow \left(^{12}C_r\right)^2 = ^{12}C_{r-1} \cdot ^{12}C_{r+1}$$ but no three consecutive binomial coefficient are in GP $$\Rightarrow P = 0$$ Now for $\left(3^{1/4} + 4^{1/3}\right)^{12}$, $T_{r+1} = ^{12}C_r (4)^{K/3} (3)^{\frac{12-K}{4}}$ for rational terms $K = 0, 12$ sum of rational terms $$= ^{12}C_0 4^0 \cdot 3^3 + ^{12}C_{12} \cdot 4^4 \cdot 3^0$$ $$= 27 + 256 = 283 = q$$ $$\therefore p + q = 283$$

By solving $\frac{x^2}{9} - \left( \frac{8x - x^2}{4} \right) = 1$ we have: $$4x^2 - 72x + 9x^2 = 36$$ $$\Rightarrow 13x^2 - 72x - 36 = 0$$ $$\Rightarrow 13x^2 - 78x + 6x - 36 = 0$$ $$\Rightarrow 13x(x - 6) + 6(x - 6) = 0$$ $$\Rightarrow x = 6 or -\frac{13}{6} (neglected)$$ $$\Rightarrow y^2 = 8(6) - (6)^2$$ $$\Rightarrow y = \pm \sqrt{12}$$ So, points $A$ and $B$ are $(6, \sqrt{12})$, $(6, -\sqrt{12})$. Point $P \left( h, \frac{2h+4}{3} \right)$. Centroid of $\triangle PAB$ is $\left( \frac{12+h}{3}, \frac{2h+4}{9} \right)$. By options, this centroid lies on the line $6x - 9y = 20$.

As $f(x)$ is a polynomial of degree two, let it be $$f(x) = ax^2 + bx + c (a \neq 0)$$ On satisfying given conditions, we get $C = 1$ and $a = \pm 1$. Hence, $f(x) = 1 \pm x^2$. Also, range $\in (-\infty, 1]$, hence $$f(x) = 1 - x^2$$ Now $f(k) = -2k$. $$1 - k^2 = -2k \rightarrow k^2 - 2k - 1 = 0$$ Let roots of this equation be $\alpha$ and $\beta$, then $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 4 - 2(-1) = 6$$

Let the number be $$2ab, \ a + b = 13$$ Thus, $a, b \in \{0, 9\}$ This gives 6 numbers $\{(9, 4), (8, 5), \ldots, (4, 9)\}$. Similarly, for $3ab, \ a + b = 12 \Rightarrow 7$ numbers. For $4ab, \ a + b = 11 \Rightarrow$ Numbers. For $5ab, \ a + b = 10 \Rightarrow 9$ numbers. For $6ab, \ a + b = 9 \Rightarrow 10$ numbers. For $7ab, \ a + b = 8 \Rightarrow 9$ numbers. For $8ab, \ a + b = 7 \Rightarrow 8$ numbers. For $9ab, \ a + b = 6 \Rightarrow 7$ numbers. Therefore, total ways $= 64$.

Given $$f(x) = \lim_{x \to \infty} \left( \frac{\tan(x/2^{r+1}) + \tan^3(x/2^{r+1})}{1 - \tan^2(x/2^{r+1})} \right)$$ We have $$= \lim_{x \to \infty} \frac{\tan\left(\frac{x}{2^{r+1}}\right)}{\cos\left(\frac{x}{2^r}\right)}$$ $$= \lim_{x \to \infty} \frac{\sin(x/2^{r+1})}{\cos\left(\frac{x}{2^{r+1}}\right) \cos\left(\frac{x}{2^r}\right)}$$ $$= \lim_{x \to \infty} \frac{\sin\left(\frac{x}{2^r} - \frac{x}{2^{r+1}}\right)}{\cos\left(\frac{x}{2^{r+1}}\right) \cos\left(\frac{x}{2^r}\right)}$$ $$= \lim_{x \to \infty} \tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right)$$ From condition given in the question: $$\therefore \lim_{n \to \infty} \sum_{r=0}^{n} \left[ \tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right) \right] = \tan x$$ $$\therefore \lim_{x \to 0} \left( e^x - e^{\tan x} \right)$$ $$= \lim_{x \to 0} \left( x - \tan x \right)$$ $$\Rightarrow \lim_{x \to 0} e^{\tan x} \left( \frac{e^{x - \tan x} - 1}{x - \tan x} \right)$$ $$\Rightarrow \lim_{x \to 0} e^{\tan x} \lim_{x \to 0} \left( \frac{e^{x - \tan x} - 1}{x - \tan x} \right)$$ $$= 1.1 \left( \therefore \lim_{x \to 0} \frac{e^{x-1}}{x} = 1 \right)$$ $$= 1$$

Given $\frac{n}{2} [2a + (n - 1)6] = (n - 2)180$ and $an + 3n^2 - 3n = 3n(n - 2)180$ ...(i). Given $a + (n - 1)6^\circ = 219^\circ$. Therefore, $a = 225^\circ - 6n^\circ$. Putting the value of $a$ in (i), we get $ (225 - 6n^2) + 3n^2 - 3n = 180n - 360^\circ$. Therefore, $2n^2 - 42n - 360 = 0$. Therefore, $n^2 - 14n - 120 = 0$. Therefore, $(n - 20)(n + 6) = 0$. Therefore, $n = 20, -6$ (Rejected). Therefore, $n = 20$.

To find image of $P \left(t^2, 2t\right)$ $$\frac{x - t^2}{1} = \frac{y - 2t}{1} = -2\left(t^2 + 2t + 4\right) = -(t+1)^2 - 3$$ $$x = t^2 - (t+1)^2 - 3 = -2t - 4$$ $$y = 2t - (t+1)^2 - 3 = -t^2 - 4$$ $$t = \frac{-x-4}{2}$$ $$\Rightarrow y + 4 = -\left(\frac{-x-4}{2}\right)^2$$ $$\Rightarrow (y+4) = -\frac{(x+4)^2}{4}$$ $$\Rightarrow x^2 = -4y$$ $$\Rightarrow Focus (-4, -5)$$ Also, $y = -5$ intersect $$\therefore (-4)(-1) = (x+4)^2$$ $$4 = (x+4)^2$$ $$x+4 = \pm 2$$ $$x = -2, -6$$ $$\Rightarrow d = 4$$ $$a = \frac{1}{2}\begin{vmatrix} 1 & 0 & 1 \\ -2 & -5 & 1 \\ -6 & -5 & 1 \end{vmatrix}$$ $$= \frac{1}{2} [1((-5+5) + 1(10-30))]$$ $$= \frac{1}{2} (20)$$ $$a = 10$$ $$\therefore a + d = 14$$

Given $\($ $\frac{dy}{dx}$ + $\left$( $\sin$^{-1} $\frac{x}{2}$ $\right$)^2 y = $\left$( $\sin$^{-3} $\frac{x}{2}$ $\right$)^3 $\frac{\sqrt{4-x^2}}{\sqrt{4-x^2}}$ $\)$. Multiplying both sides by $\($ e^{$\left$( $\sin$^{-1} $\frac{x}{2}$ $\right$)^2} $\)$, we have $$ ye^{\left( \sin^{-1} \frac{x}{2} \right)^2} = \int \left( \sin^{-3} \frac{x}{2} \right)^3 e^{\left( \sin^{-1} \frac{x}{2} \right)^2} \frac{1}{4-x^2} \, dx $$ Thus, $$ y = \left( \sin^{-1} \frac{x}{2} \right)^2 - 2 + c \cdot e^{-\left( \sin^{-1} \frac{x}{2} \right)^2} $$ Given $\($ y(2) = $\frac{\pi^2}{4}$ - 2 $\Rightarrow$ c = 0 $\)$ Therefore, $\($ y(0) = -2 $\)$

Given $B = 0.4 \, \mathrm{T}$, $r = 20 \, \mathrm{cm}$, $\omega = 10 \pi \, \mathrm{rad/s}$. The electromotive force $E$ is given by $$E = \frac{1}{2} B \omega R^2$$ which equals $0.2512 \, \mathrm{V}$.

Energy density = $\frac{1}{2}$ $\varepsilon$_0 E^2 = $\frac{1}{2}$ $\varepsilon$_0 $\left$( $\frac{V}{d}$ $\right$)^2 = $\frac{1}{2}$ $\left$( 8.85 $\times$ 10^{-12} $\right$) $\left$( $\frac{20}{10^{-6}}$ $\right$)^2 $\,$ $\mathrm{J/m^3}$ $\simeq$ 1.8 $\times$ 10^3 $\,$ $\mathrm{J/m^3}$

Angular impulse = $\($[ML^2 T^{-1}]$\)$ Latent Heat = $\($[M^0 L^2 T^{-2}]$\)$ Electrical resistivity = $\($[ML^3 T^{-3} A^{-2}]$\)$ Electromotive force = $\($[ML^2 T^{-3} A^{-1}]$\)$

Q2. (Vapor density) = $\($ $\left$( $\frac{Molar mass}{2}$ $\right$) $\)$, i.e., $\($ vd = $\frac{M}{2}$ $\)$. $\($ v_{rms} = $\sqrt{\frac{3RT}{M}}$ $\)$ $\($ $\frac{v_2}{v_1}$ = $\sqrt{\frac{M_1}{M_2}}$ = $\sqrt{\frac{25}{4}}$ = $\frac{5}{2}$ $\)$

Kinetic energy of translation is given by $\frac{3}{2} nRT$. Given $n = \frac{50 \, \mathrm{g}}{44 \, \mathrm{g}} = \frac{25}{22} \, \mathrm{mol}$ and $T = 17^\circ \mathrm{C} = 290 \, \mathrm{K}$. Therefore, the kinetic energy of translation is $$\frac{3}{2} \left( \frac{25}{22} \right) (8.3)(290) \, \mathrm{J}$$ which equals $4102.8 \, \mathrm{J}$.

Given the refractive indices $n_1 = 1.3$ and $n_2 = 1.4$, and the object distance $u = -13 \, \mathrm{cm}$, we use the lens formula: $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ Substituting the values: $$\frac{1.4}{v} - \frac{1.3}{-13} = \frac{0.1}{R}$$ Simplifying: $$\frac{1.4}{v} = \frac{1}{R} - \frac{1}{10R}$$ $$\frac{1.4}{v} = \frac{10R}{1 - R}$$ The magnification $m$ is given by: $$m = \frac{v/n_2}{u/n_1}$$ Substituting the values: $$-2 \times \frac{-13}{1.3} = \frac{10R}{1 - R}$$ Solving for $R$: $$R = \frac{2}{3} \, \mathrm{cm}$$

Given $f = \frac{v}{2 \pi r} = \left( \frac{V_0}{n} \right) \left( \frac{1}{2 \pi r_0 n^2} \right)$. Therefore, $$f = \frac{V_0}{2 \pi r_0} = \left( \frac{1}{n^3} \right) = f_0 \left( \frac{1}{n^3} \right).$$

Compton effect refers to scattering of a photon by free electrons. This phenomenon provides an evidence for particle nature of light.

Distance travelled = displacement when direction of velocity remains constant. Therefore, distance = area. $$Distance = Area$$ $$= \frac{1}{2} (2 \, s + 4 \, s)(10 \, m/s)$$ $$= 30 \, m$$

Magnetic field at $P$, $B = \frac{\mu_0 \frac{m}{(2l)}}{4\pi r^3}$, where $m$ is the pole strength. $$\Rightarrow \left( \frac{\Delta B}{B} \times 100 \right) = \left( \frac{\Delta l}{l} \right) \times 100 + 3 \left| \frac{\Delta r}{r} \right| \times 100$$ $$= 1\% + 3\% = 4\%$$

The escape velocity $V_{escape}$ is given by the formula $$V_{escape} = \sqrt{\frac{2GM}{R}}.$$ For the planet, the escape velocity is $$\left( V_{escape} \right)_{Planet} = \sqrt{\left( \frac{M_P}{M_E} \right) \times \left( \frac{R_E}{R_P} \right)} = \frac{1}{2}.$$ Therefore, $$\left( V_{escape} \right)_{Planet} = \frac{1}{2} \left( V_{escape} \right)_{Earth} = 5.6 \, km/s.$$

If we express position $x(t) = A \sin(\omega t + \phi)$ then $x_0 = A \sin \phi$. $v_0 = A \omega \cos \phi$ implies $\tan \phi = \frac{\omega x_0}{v_0}$. $$A = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}}$$ Hence both position and linear momentum of a particle can be expressed as a function of time if we know initial momentum and position.

Given the magnification $m = -3 = -\frac{u}{v}$. We have $u = -x$ and $v = -3x$. This implies $2x = 20 \, \mathrm{cm}$, so $x = 10 \, \mathrm{cm}$. The focal length $f$ is given by $$f = \frac{uV}{u + V} = \frac{(-10)(-30)}{(-10) + (-30)} = -7.5 \, \mathrm{cm}.$$ Also, $f = -\frac{R}{2}$, so $R = -2f = 15 \, \mathrm{cm}$.

Given $$\langle p \rangle = \frac{(2\hat{i} + 3\hat{j}) \cdot (3\hat{i} + 6\hat{j})}{4} = 6$$ $$\vec{a} = \left( \frac{\vec{F}}{m} = \frac{1}{2} \hat{i} + \frac{3}{4} \hat{j} \right)$$ $$\vec{v} at t = 4 sec = \left( \frac{1}{2} \hat{i} + \frac{3}{4} \hat{j} \right) \times 4 = (2\hat{i} + 3\hat{j})$$ $$P_{ins} = (2\hat{i} + 3)(2\hat{i} + 3\hat{j}) = 13$$ $$\langle P \rangle = \frac{6}{4}$$ $$P_{ins} = 13$$

Given $V = V_0 \sin \omega t$. The input waveform is a sinusoidal wave with both positive and negative cycles. The output waveform shows only the negative cycle of the input, indicating a half-wave rectification process.

For the given circuit, the magnetic field contributions are calculated as follows: The magnetic field due to section (1) is given by: $$B_1 = \frac{\mu_0 i}{4 \pi a} \otimes$$ The magnetic field due to section (2) is given by: $$B_2 = \frac{\mu_0 i}{4 \pi a} \left( \frac{3 \pi}{2} \right) \otimes$$ The magnetic field due to section (3) is: $$B_3 = 0$$ The total magnetic field is: $$B = \frac{\mu_0 i}{4 \pi a} \left( 1 + \frac{3 \pi}{2} \right) \otimes$$

Given $\tau_{Net} = 0$, we have $ (400 \, \mathrm{g} \times 30) = (250 \, \mathrm{g} \times 10) + (m \, \mathrm{g} \times 50)$. Solving for $m$, $$\frac{12000 - 2500}{50} = \frac{9500}{50}$$ Thus, $M = 190 \, \mathrm{g}$.

Volume of cube inside water $$= \left( \frac{Density of cube}{Density of water} \right) \times Volume of cube$$ Mass of cube $$= Density of water$$ $$= 400 \, gm$$ $$= 1 \, gm/cm^3$$ $$= 400 \, cm^3$$ Volume of cube outside water $$= Volume of cube - Volume of cube inside water$$ $$= 1000 \, cm^3 - 400 \, cm^3$$ $$= 600 \, cm^3$$

Given $$\vec{B} = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) 30 \sin \left[ \omega \left( t - \frac{z}{c} \right) \right]$$ $$\vec{E} = \vec{B} \times \vec{c} and E = B_0 c$$ Here $$\vec{E} \left( \frac{\sqrt{3}}{2} (-\hat{j}) + \frac{1}{2} \hat{i} \right)$$ $$E_0 = 30c$$ $$\vec{E} = \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 30c \sin \left[ \omega \left( t - \frac{z}{c} \right) \right]$$

Given the equations: $$F - mg = ma$$ Rearranging gives: $$F = ma + mg$$ For the second scenario: $$F - (m-x)g = (m-x)3a$$ Substitute for $F$: $$Ma + mg - mg + xg = 3ma - 3xa$$ Solving for $x$: $$x = \frac{2ma}{g + 3a}$$

As the bar moves without change in orientation, the length of bar will be proportional to its distance from the vertex. i.e. $I = c(vt)$ induced emf $E = B/v$ $= cBv^2t$ $\Rightarrow n = 1$

Work done in rotating a dipole $= \Delta U$ $$W = (-PE \cos \theta_f) - (-PE \cos \theta_i)$$ or $$= 2PE (\therefore \theta_f = 180^\circ and \theta_i = 0^\circ)$$ $$= (2 \times 6 \times 10^{-6} \times 10^6) \, \mathrm{J} = 12 \, \mathrm{J}$$

Given the formula for bulk modulus $B = \frac{\Delta P}{\frac{\Delta V}{V}}$. Calculate $\Delta V$ using the given values: $$\Delta V = \frac{7 \times 10^6}{1.4 \times 10^{11}} \times \left(10 \times 10^{-2}\right)^3$$ Finally, $\Delta V = 50 \, \mathrm{mm}^3$.

Given $V_A = V_B$, the bridge is balanced. $$\frac{10}{R} = \frac{20}{40}$$ Solving for $R$: $$R = 20 \, \Omega$$ Calculating the current $I$: $$I = \frac{40}{20} = 2 \, \mathrm{A}$$

For a thin film interference, a fringe for transmission is formed. When $$2 \mu x = n \lambda$$ Therefore, $$\frac{dx}{dt} = \left( \frac{dn}{dt} \right) \frac{\lambda}{2 \mu}$$ $$= \left( \frac{1}{12} \right) \frac{560 \times 10^{-9}}{2 \times 1.4} = \frac{5}{3} \times 10^{-8} \, \mathrm{m/s}$$ Let $V$ be the volume of the film, then $$V = \pi R^2 x$$ Differentiating with respect to time, $$\frac{dV}{dt} = \pi R^2 \frac{dx}{dt}$$ $$= \pi (1.8 \times 10^{-2})^2 \frac{5}{3} \times 10^{-8} \, \mathrm{m^3/s}$$ $$= 54 \pi \times 10^{-13} \, \mathrm{m^3/s}$$

Since the reaction is elementary, $$Rate = k[A]^1[B]^1$$ When $V$ is reduced to $\frac{1}{3} V$, concentration will be tripled. Hence, $$rate = 9 \times (rate)_{initial}$$ $$x = 9$$

$V_2O_5$ is amphoteric oxide. $V_2O_5$ gives $VO_4^{3-}$ on reaction with alkali. Oxidation state of V in $VO_4^{3-}$ is +5.

(A) Sucrose $\rightarrow \alpha_1 - \beta_2$ Glycosidic linkage (B) Maltose $\rightarrow \alpha \ 1-4$ Glycosidic linkage (C) Lactose $\rightarrow \beta \ 1-4$ Glycosidic linkage (D) Amylopectin $\rightarrow \alpha 1 - 4$ and $\alpha 1 - 6$ Glycosidic linkage A-III, B-I, C-IV, D-II

Based on the Ksp values and salt analysis cation identification, we can say that order of Ksp value is: HgS < PbS < AgBr < $\mathrm{Ca(OH)_2}$ Ksp values HgS $\to$ 4 $\times$ 10^{-53} PbS $\to$ 8 $\times$ 10^{-28} AgBr $\to$ 5 $\times$ 10^{-13} $\mathrm{Ca(OH)_2}$ $\to$ 5.5 $\times$ 10^{-6}

Phase transfer from solid to vapour directly is known as sublimation.

Q10. (A) Starch $\xrightarrow{H^+/H_2O}$ Glucose (B) Cane sugar $\xrightarrow{H^+/H_2O}$ glucose + fructose (Sucrose) 50$\%$ 50$\%$ (C) Milk sugar $\xrightarrow{H^+/H_2O}$ glucose + galactose (Lactose) (D) Amylopectin $\xrightarrow{H^+/H_2O}$ Glucose (E) Amylose $\xrightarrow{H^+/H_2O}$ Glucose So, correct options are B, C, and E only.

As internal energy $U$ is a state function, its cyclic integral must be zero in a cyclic process. Therefore, $\Delta U$ case (I) = $\Delta U$ case (II) = $\Delta U$ case (III).

75% by mass means: Therefore, $75 \, \mathrm{gHNO_3}$ in $100 \, \mathrm{g}$ solution. Therefore, $30 \, \mathrm{gHNO_3}$ in $\frac{100}{75} \times 30 \, \mathrm{g}$ solution. $$m_{solution} = \frac{100 \times 30}{75} \, \mathrm{g}$$ Therefore, $$V_{sol} = \frac{m_{sol}}{d_{sol}} = \frac{100 \times 30}{75 \times 1.25} = 32 \, \mathrm{mL}$$

Option (A) $[\mathrm{CoF}_6]^{3-}$: $\mathrm{Co}^{3+} \rightarrow 3d^6$. The hybridization is $sp^3d^2$. Option (B) $[\mathrm{NiCl}_4]^{2-}$: $\mathrm{Ni}^{2+} \rightarrow 3d^8$. The hybridization is $sp^3$. Option (C) $[\mathrm{Co(NH}_3)_6]^{3+}$: $\mathrm{Co}^{3+} \rightarrow 3d^6$. The hybridization is $d^2sp^3$. Option (D) $[\mathrm{Ni(CN)}_4]^{2-}$: $\mathrm{Ni}^{2+} \rightarrow 3d^8$. The hybridization is $dsp^2$.

The compounds having benzylic hydrogen will give benzoic acid on treatment with hot $\mathrm{KMnO_4}$. The compounds shown are:

Law of octaves was arranged in the increasing order of their atomic weight. Lothar Meyer plotted the physical properties such as atomic volume, melting point and boiling point against atomic weight.

Because the number of bacteria initially is $N_0$ and the number of bacteria at any time $t$ is $N$. Since bacterial growth is given as $$N = N_0 e^{Kt}$$ where $K$ is the growth constant for bacterial growth.

For single electron species, energy only depends on 'n' (principal quantum number). So energy of $2s = 2p$ and energy of $3d < 4s$.

Living cell = 0.9 gm in 100 gm of solution. $\n$ %w/w = 0.9. $\n$ Solution is have equal moles of glucose and water = 0.5. $\n$ Weight of solution = 0.5 $\times$ 180 + 0.5 $\times$ 18 = 99 gm $\;$ $\%$w/w $\simeq$ 90$\%$. $\n$ Concentrated solution = Cell will shrink.

(C) Only primary amine gives carbyl amine test. (D) $\mathrm{Ph - SO_2Cl}$ is the Hinsberg reagent, benzene sulphonyl chloride. (E) Tertiary amine do not react with $\mathrm{Ph - SO_2Cl}$. So correct options are (B) and (D) only.

Statement-I is true. Both are ring chain isomers. Statement-II is true. $1^\circ$ Amine and $2^\circ$ Amine are different functional groups, hence both are functional group isomers.

As$_2$S$_3$ and As$_2$S$_5$ are yellow colour sulphides, (NH$_4$)$_2$S is colourless, PbS is black, CuS is black in colour.

Strongest oxidising power among the option is $\mathrm{Mn_2O_3}$ because of $E^\circ$ value. $$E^\circ_{\mathrm{Mn^{+3}/Mn^{+2}}} = +1.57 \, \mathrm{V}$$ $\mathrm{Mn^{+3}} \rightarrow d^4$ configuration $$\mu = \sqrt{24} \, \mathrm{BM}$$ $$= 4.89 \, \mathrm{BM}$$ $$\Rightarrow 5$$

Given $\Delta H_f [\mathrm{CO_2} (g)] = -393.5 \, \mathrm{kJ/mole}$, $\Delta H_f [\mathrm{H_2O}(\ell)] = -286.0 \, \mathrm{kJ/mole}$, $\Delta H_c [\mathrm{C_6H_6}] = -3267.0 \, \mathrm{kJ/mole}$. Find $\Delta H_f \mathrm{C_6H_6} = (?)$. The reaction is $\mathrm{C_6H_6} + \frac{15}{2} \mathrm{O_2} (g) \rightarrow 6\mathrm{CO_2} (g) + 3\mathrm{H_2O}(\ell)$. The reaction enthalpy $\Delta H_R = \Delta H_C = \Sigma \Delta H_f(\mathrm{P}) - \Sigma \Delta H_f(\mathrm{R})$. Therefore, $$-3267 = 6 \times (-393.5) + 3 \times (-286) - \Delta H_f (\mathrm{C_6H_6})$$ Solving gives $\Delta H_f (\mathrm{C_6H_6}) = 48 \, \mathrm{kJ/mole}$.

Electrolysis of NaCl is NaCl + $\mathrm{H_2O(aq)}$ $\rightarrow$ $\mathrm{NaOH(aq)}$ + $\frac{1}{2}$ $\mathrm{Cl_2(g)}$ + $\frac{1}{2}$ $\mathrm{H_2(g)}$ Since during electrolysis pH changes to 12 So $[\mathrm{OH}^-] = 10^{-2}$ and $[\mathrm{H}^+] = 10^{-12}$ So by Faraday law Gram amount of substance deposited = Amount of electricity passed $$10^{-2} \times \frac{600}{1000} \times 96500 = I \times t$$ $$10^{-2} \times \frac{600}{1000} \times 96500 = I \times 5 \times 60$$ $$I = \frac{10^{-2} \times 600 \times 96500}{1000 \times 5 \times 60}$$ $I = 1.93$ ampere So, $I = 2$ ampere (nearest integer)

Phosphorus belongs to the $15^{th}$ group and forms $d\pi - d\pi$ bond with transition metal and $\mathrm{PH_3}$ is the strongest base among the other group members except $\mathrm{NH_3}$.

Q11. $\mathrm{O_2} \rightarrow 2$ unpaired electrons according to MOT. $\mathrm{O_2^+} \rightarrow 1$ unpaired electron according to MOT. $\mathrm{O_2^-} \rightarrow 1$ unpaired electron according to MOT. NO $\rightarrow$ odd electron species. $\mathrm{NO_2} \rightarrow$ odd electron species. $\mathrm{K_2[NiCl_4]} \rightarrow \mathrm{Ni^{2+}} \Rightarrow 3\, d^8$ weak Ligand, C.N. $= 4$. $\Rightarrow$ Tetrahedral, Paramagnetic with 2 unpaired electrons.