JEE Advanced 18 May 2025 Paper 1 question paper with solutions

Given $h(x) = f(x+1) - g(x+2)$. $$= a_1 + 10(x+1) + a_2(x+1)^2 + a_3(x+1)^3 + (x+1)^4 - b_1 - 3(x+2) - b_2(x+2)^2 - b_3(x+2)^3 - (x+2)^4$$ The coefficient of $x^3$ in $h(x)$ is $a_3 - b_3 - 4$. Since $f(x) - g(x) \neq 0$ for all $x \in \mathbb{R}$, we have: $$a_1 + 10x + a_2x^2 + a_3x^3 + x^4 - b_1 - 3x - b_2x^2 - b_3x^3 - x^4 \neq 0$$ This implies: $$x^3(a_3 - b_3) + x^2(a_2 - b_2) + 7x + (a_1 - b_1) \neq 0$$ The cubic equation will become zero at at least one value of $x$. Therefore, it will be quadratic, implying: $$a_3 - b_3 = 0$$

Given $P(U) = 1 - P(S_1' \cap S_2' \cap S_3') = \frac{1}{2}$. Therefore, $P(S_1' \cap S_2' \cap S_3') = \frac{1}{2}$; $P(S_1') \cdot P(S_2') \cdot P(S_3') = \frac{1}{2}$. This implies $(1 - P(S_1))(1 - P(S_2))(1 - P(S_3)) = \frac{1}{2}$ $\ldots$ (1) $P(V) = \frac{P(S_1 \cap S_1' \cap S_3')}{P(S_2' \cap S_3')} = \frac{1}{10}$. Thus, $P(S_1) \cdot P(S_2') P(S_3') = \frac{1}{10} P(S_2') P(S_3')$. Therefore, $P(S_1) = \frac{1}{10}$. $P(W) = P(S_2 \cap S_3') = \frac{1}{12}$. $P(S_2) \cdot P(S_3') = \frac{1}{12}$. $P(S_2)(1 - P(S_3)) = \frac{1}{12}$ $\ldots$ (2) From Eq. (1): $$\left(1 - \frac{1}{10}\right)(1 - P(S_2))(1 - P(S_3)) = \frac{1}{2}$$ $$(1 - P(S_2))(1 - P(S_3)) = \frac{9}{9}$$ $\ldots$ (3) From Eq. (2) and Eq. (3): $$\frac{P(S_2)}{1 - P(S_2)} = \frac{1}{12} \times \frac{9}{5}$$ $P(S_2) = \frac{3}{23}$. Substitute in Eq. (2): $$\frac{3}{23}(1 - P(S_3)) = \frac{1}{12}$$ $$1 - P(S_3) = \frac{23}{36}$$ $$P(S_3) = \frac{13}{36}$$ Therefore, $P(T) = \frac{13}{36}$.

RHD at $x = 0$: $$\lim_{h \to 0} \frac{\left(2 - 2h^2 - h^2 \sin \frac{1}{h}\right) - 2}{h} = 0$$ Similarly LHD at $x = 0$ is also equal to 0. Therefore, differentiable at $x = 0$. $$f'(x) = -4x - 2x \sin \frac{1}{x} - x^2 \left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)$$ $$f'(x) = -\left(4x + 2x \sin \frac{1}{x}\right) + \cos \frac{1}{x}$$ $$f'(x) = -\left(2x \left(4 - \sin \frac{1}{x}\right)\right) + \cos \frac{1}{x}$$ for $x \in (0, \delta)$ Therefore, we can't say $f(x)$ is decreasing on $(0, \delta)$ as $\cos \frac{1}{x}$ oscillates. For $x \in (-\delta, 0)$, for any $\delta > 0$ Therefore, $f(x)$ is not increasing on $(-\delta, 0)$ as $\cos \frac{1}{x}$ oscillates from $-1$ to $1$. $f(0) = 2$ $f(0 + h) < 2$ $f(0 - h) < 2$ Therefore, $x = 0$ is local maxima.

Given $Q^{-1} = Q^T \Rightarrow QQ^T = I$. $Q$ is an orthogonal matrix. Let $Q = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$. $PQ = QP \Rightarrow \begin{bmatrix} 2a_1 & 2b_1 & 2c_1 \\ 2a_2 & 2b_2 & 2c_2 \\ 3a_3 & 3b_3 & 3c_3 \end{bmatrix} = \begin{bmatrix} 2a_1 & 2b_1 & 3c_1 \\ 2a_2 & 2b_2 & 3c_2 \\ 2a_3 & 2b_3 & 3c_3 \end{bmatrix}$. $c_1 = 0$, $c_2 = 0$, $a_3 = 0$, $b_3 = 0$. $Q = \begin{bmatrix} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ 0 & 0 & c_3 \end{bmatrix}$. $a_1 a_2 + b_1 b_2 = 0$. $a_1^2 + b_1^2 = 1$, $a_2^2 + b_2^2 = 1$, $c_3^2 = 1$. $a_1$, $b_1$, $a_2$, $b_2$, $c_3$: 1, 0, 0, 0, 1, -1 -1, 0, 0, 0, 1, -1 0, 1, 1, -1, 0 0, -1, 1, -1, 0 Total 16 matrices

Let $L_1 : \mathbf{r} = \mathbf{a} + t \mathbf{b}$ $$\mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(1) + \hat{k}(1)$$ Direction ratios of $L_1$: $\langle 1, -1, 1 \rangle$ $L_1 : \frac{x+1}{1} = \frac{y-0}{-1} = \frac{z-6}{1}$ & $L_2 : \frac{x-2}{1} = \frac{y+1}{-1} = \frac{z-3}{1} = \lambda$ $M : 2x + y - 2z - 6 = 0$ Let point on $L_2 (\lambda + 2, -\lambda - 1, \lambda + 3)$ $2(\lambda + 2) - \lambda - 1 - 2\lambda - 6 = 0$ $2\lambda + 4 - 3\lambda - 13 = 0$ $\lambda = -9$ Therefore, $Q(-7, 8, -6)$ Line $PR : \frac{x-2}{2} = \frac{y+1}{1} = \frac{z-5}{-2} = \mu$ $R(2\mu + 2, \mu - 1, -2\mu + 3)$ Put in plane $2(2\mu + 2) + \mu - 1 - 2(2\mu + 3) - 6 = 0$ $4\mu + 4 + \mu - 1 + 4\mu - 6 - 6 = 0$ $9\mu - 9 = 0 \implies \mu = 1$ $R(4, 0, 1)$ $P(2, -1, 3)$ and $Q(-7, 8, -6)$ $PQ = \sqrt{81 + 81 + 81} = 9\sqrt{3}$ $Q(-7, 8, -6)$ and $R(4, 0, 1)$ $QR = \sqrt{121 + 64 + 49} = \sqrt{234}$ Area $(\Delta PQR)$ $$= \frac{1}{2} \left| \overrightarrow{QP} \times \overrightarrow{QR} \right|$$ $$= \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 9 & -9 & 9 \\ 11 & -8 & 7 \end{vmatrix}$$ $$= \frac{3}{2} \sqrt{234}$$ $\overrightarrow{PQ} = -9\hat{i} + 9\hat{j} - 9\hat{k} = -9(\hat{i} - \hat{j} + \hat{k})$ $\overrightarrow{PR} = 2\hat{i} + \hat{j} - 2\hat{k}$ $$\cos \theta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PR}}{PQ \cdot PR}$$ $$= \frac{9}{9\sqrt{3} \times 3} = \frac{1}{3\sqrt{3}}$$ $$\theta = \cos^{-1} \left( \frac{1}{3\sqrt{3}} \right)$$

Given $$f(n) = \begin{cases} \frac{(n+1)}{2} & if n is odd \\ \frac{(4-n)}{2} & if n is even \end{cases}$$ $$f(n) = \{(1, 1), (2, 1), (3, 2), (4, 0), (5, 3), (6, -1), \ldots \}$$ Therefore, $f(n)$ is many one and onto function. $$g(n) = \begin{cases} 3 + 2n & if n \geq 0 \\ -2n & if n < 0 \end{cases}$$ $$g(n) = \{(-3, 6), (-2, 4), (-1, 2), (0, 3), (1, 5), (2, 7), (3, 9), (4, 15), \ldots \}$$ Therefore, $g(n)$ is one-one and into function. $$f(g(n)) = 2 + n, \; n \in \mathbb{N}$$ fog is one-one and into. $$g(f(n)) = \begin{cases} 4 + n & if n is odd natural number \\ 7 - n & if n = 2, 4 \\ n - 4 & if n is even natural number and n \geq 6 \end{cases}$$ $$g(f(2)) = g(f(1)) = 5$$ Therefore, gof is many one and into.

Given $|x + iy - 1 - 2i| = 2 |x + iy - 3i|$. This implies $$(x - 1)^2 + (y - 2)^2 = 4 (x^2 + (y - 3)^2)$$ which simplifies to $$3x^2 + 3y^2 + 2x - 20y + 31 = 0$$ and further to $$x^2 + y^2 + \frac{2x}{3} - \frac{20y}{3} + \frac{31}{3} = 0$$ Therefore, $S$ is a circle with centre $$\left(-\frac{1}{3}, \frac{10}{3}\right)$$ and radius $$\sqrt{\frac{1}{9} + \frac{100}{9} - \frac{31}{3}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$$

For relation to be reflexive all the diagonal elements must be taken and out of remaining 30 elements there are 15 pairs and we need 2 pairs such that R contains exactly 10 elements and is both reflexive and symmetric. ∴ number of ways = $\binom{15}{2}$ = 105

Therefore, $\overrightarrow{SP} + 5 \overrightarrow{SQ} + 6 \overrightarrow{SP} = \overrightarrow{0}$. Let the position vector of $P$ be $\overrightarrow{p}$, $Q$ be $\overrightarrow{q}$, and $R$ be $\overrightarrow{r}$. $$\Rightarrow (\overrightarrow{p} - \overrightarrow{s}) + 5 (\overrightarrow{q} - \overrightarrow{s}) + 6 (\overrightarrow{r} - \overrightarrow{s}) = \overrightarrow{0}$$ $$\Rightarrow \overrightarrow{s} = \frac{\overrightarrow{p} + 5 \overrightarrow{q} + 6 \overrightarrow{r}}{12}$$ $$\Rightarrow \overrightarrow{EF} = \frac{\overrightarrow{q} - \overrightarrow{p}}{2}$$ $$\Rightarrow \overrightarrow{ES} = \frac{5 \overrightarrow{q} - 5 \overrightarrow{p}}{12} = \frac{5}{12} (\overrightarrow{q} - \overrightarrow{p})$$ $$\frac{|\overrightarrow{EF}|}{|\overrightarrow{ES}|} = \frac{6}{5} = 1.2$$

Let A $\rightarrow$ "0" appear exactly twice. and B $\rightarrow$ "1" appear exactly twice. Therefore, A $\cap$ B $\rightarrow$ "0" and "1" both appears exactly twice. For n(A): $$= \binom{6}{2} (1) (2)^5 = \frac{6 \times 5}{2} \times 2^5 = 480$$ For n(B): C-I: 1 at first place Number of ways = $$\binom{6}{1} (1) (2)^5 = 192$$ C-II: 2 at first place Number of ways = $$\binom{6}{2} (1) (2)^4 = \frac{6 \times 5}{2} \times 2^4 = 240$$ n(B) = 240 + 192 For n(A $\cap$ B): For n(A $\cap$ B): $$= \binom{6}{2} (1) \times \binom{5}{2} (1) \times (1) = \frac{6 \times 5}{2} \times \frac{5 \times 4}{2} = 150$$ Therefore, n(A $\cup$ B) = n(A) + n(B) - n(A $\cap$ B) $$= 480 + (192 + 240) - 150$$ $$= 762$$

Given the limit $$\lim_{x \to 0} \frac{\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} \, dt + \beta x \cos x}{x^3}$$ we have: $$= \lim_{x \to 0} \frac{\frac{\alpha}{2} \left( \frac{1}{1-x^2} \right) + \beta \cos x - \beta x \sin x}{3x^2}$$ $$= \frac{\alpha}{2} (1-x^2)^{-1} + \beta \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \ldots \right) - \beta x \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} \ldots \right)$$ $$= \frac{\alpha}{2} (1+x^2 + x^4 \ldots) + \beta \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \ldots \right) - \beta \left( x^2 - \frac{x^4}{3!} \ldots \right)$$ $$= \frac{\left( \frac{\alpha}{2} + \beta \right) + x^2 \left( \frac{\alpha}{2} - \frac{\beta}{2} - \beta \right) + x^4 (\ldots)}{3x^2} = 2 (Given)$$ Therefore, $$\frac{\alpha}{2} + \beta = 0$$ and $$\frac{\alpha - 3\beta}{6} = 2$$ This implies $$\alpha = -2\beta$$ and $$\alpha = 12 + 3\beta$$ Thus, $$\beta = -\frac{12}{5}$$ and $$\alpha = \frac{24}{5}$$ Therefore, $$\alpha + \beta = \frac{12}{5} = 2.40$$

Given $f(x+y) = f(x) \cdot f(y)$, it follows that $f(x) = k^x$ where $f(x) > 0$ for all $x \in \mathbb{R}$. Therefore, $f(a_{31}) = 64f(a_{25})$. This implies $k^{a_{31}} = 64 \cdot k^{a_{25}}$. Thus, $k^6 = 64$ and $k^d = 2$. The sum $\sum_{i=1}^{50} f(a_i) = f(a_1) + f(a_2) + \ldots + f(a_{50})$ can be expressed as $$k^a + k^{a+d} + \ldots + k^{a+49d} = \frac{k^{50d} - 1}{k^d - 1}.$$ Given $k^a (2^{50} - 1) = 3(2^{25} + 1)$, it follows that $k^a = \frac{3}{2^{25} - 1}$. Therefore, $\sum_{i=6}^{30} f(a_i) = k^{a+5d} + k^{a+6d} + \ldots + k^{a+29d}$ can be written as $$k^{a+5d} \left( \frac{k^{25d} - 1}{k^d - 1} \right) = k^a \cdot (k^d)^5 (2^{25} - 1).$$ This simplifies to $$\frac{3}{2^{25} - 1} \cdot 2^5 (2^{25} - 1) = 96.$$

The given differential equation is $$\frac{dy_1}{y_1} + \frac{dy_2}{y_2} + \frac{dy_3}{y_3} = \left( \sin^2 x + \cos^2 x + \frac{2 - x^3}{x^3} \right) dx$$ Taking the natural logarithm, we have: $$\ln(y_1 y_2 y_3) = \frac{-1}{x^2} + C$$ This implies: $$\ln(y_1(x)y_2(x)y_3(x)) = \frac{-1}{x^2} + C$$ Substituting the values: $$\ln\left(5 \cdot \frac{1}{3} \cdot \frac{3}{5e}\right) = \frac{-1}{x^2} + C$$ Therefore, $$C = 0$$ Thus, $$y_1(x)y_2(x)y_3(x) = e^{\frac{-1}{x^2}}$$ Now, consider the limit: $$\lim_{x \to \infty} \frac{e^{\frac{-1}{x^2}} + 2x}{e^{3x} \sin x}$$ This can be split into: $$\lim_{x \to \infty} \frac{1}{e^{3x + \frac{1}{x^2}} \sin x} + \lim_{x \to \infty} \frac{2x}{e^{3x} \sin x}$$ Simplifying further: $$\lim_{x \to \infty} \frac{1}{e^{3x} \cdot \frac{\sin x}{x}} \cdot \frac{e^{\frac{x^2}}}{x} + 2$$ This evaluates to: $$= 0 + 2$$ Therefore, the final result is: $$= 2$$

Given $\bar{x} = 7$ and $M = 6$. The table shows the calculations for $d_i = |x_i - \bar{x}|$ and $e_i = |x_i - M|$. The sums are $\sum f_i d_i = 48$, $\sum f_i e_i = 47$, and $\sum f_i d_i^2 = 146$. The values are $f_1 = 4$ and $f_2 = 3$. The calculations yield $\alpha = \frac{48}{19}$, $\beta = \frac{47}{19}$, and $\sigma^2 = \frac{146}{19}$.

Given $P(x) = 10x^3 - 45x^2 + 60x + 55$. The derivative $P'(x) = 30(x - 1)(x - 2)$. $P(x)$ decreases in $[1, 2]$. Therefore, the range of $P(x)$ is $[55, 60]$. The function $f(x) = \frac{P(x)}{n}$ has a minimum value of $n = 9$. For $g(x)$ to be increasing, $2n^2 - 13n - 15 \geq 0$. For $n \in \mathbb{N}$, $n = 8$. For $h(x) = (x^2 - 9)^n(x^2 + 2x + 3)$, the derivative $h'(x) = (x^2 - 9)^n(2x + 2) + (x^2 + 2x + 3)n(x^2 - 9)^{n-1} \cdot 2x$. This simplifies to $h'(x) = (x^2 - 9)^{n-1} [2(x^2 - 9)(x + 1) + 2nx(x^2 + 2x + 3)] = (x + 3)^{n-1}(x - 3)^{n-1} \cdot q(x)$. The derivative must change sign at $x = 3$. Therefore, $n - 1 = odd$. If $n = even$, then $n = 6$. The function $\cos \left| x - k + \frac{1}{2} \right|$ is differentiable everywhere. Therefore, $\sin |x - k|$ is not differentiable at $k = 0, 1, 2, 3, 4$.

Given $\mathbf{u} \times \mathbf{v} = \mathbf{w}$, $\mathbf{u} \perp \mathbf{v}$, $\mathbf{u} \perp \mathbf{w}$, $\mathbf{v} \times \mathbf{w} = \mathbf{u}$, $\mathbf{v} \perp \mathbf{w}$. Let $\mathbf{w} = \mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and $\mathbf{u} = \alpha \mathbf{i} + \beta \mathbf{j} + \gamma \mathbf{k}$. Then $\mathbf{u} \cdot \mathbf{w} = 0 = \alpha + \beta - 2\gamma$. $\left( \mathbf{v} \cdot \mathbf{w} \right) \times \mathbf{v} = \mathbf{u} \times \mathbf{v}$. $$\begin{vmatrix} -t & 1 & 1 \\ 1 & -t & 1 \\ 1 & 1 & -t \end{vmatrix} = 0$$ $t = -1$ or $2$. For $t = -1$, $\alpha + \beta + \gamma = 0$. For $\mathbf{w} \left| \mathbf{v} \right|^2 - 0 = \mathbf{w}$, $\left| \mathbf{v} \right| = 1(P)$. Also $\mathbf{u} \times \mathbf{v} = \mathbf{w}$, $\left| \mathbf{u} \right| \left| \mathbf{v} \right| = \left| \mathbf{w} \right|$. Thus $\Rightarrow \left| \mathbf{u} \right| = \left| \mathbf{w} \right| = \sqrt{6}$. Case-1: $t = 2$, $\alpha = \beta = \gamma$, $\alpha^2 + \beta^2 + \gamma^2 = 6$, $\alpha = \sqrt{2}, -\sqrt{2}$. (S) $t + 3 = 5$. Case-2: $t = -1$, $\alpha + \beta + \gamma = 0$, $\alpha + \beta - 2\gamma = 0$, $\gamma = 0$, $\alpha = -\beta$. (Q) $\alpha = \sqrt{3}$, $\gamma^2 = 0$, $\beta = -\sqrt{3}$. (R) $\alpha = \sqrt{3}$, $(\gamma + \beta)^2 = \beta^2 = 3$. (P) $\rightarrow$ (2), (Q) $\rightarrow$ (1), (R) $\rightarrow$ (4), (S) $\rightarrow$ (5)

The energy equation is given by $$E = \frac{1}{2}k(R-r)^2 \theta^2 + mg(R-r)(1-\cos \theta) + \frac{1}{2}mv^2 + \frac{1}{2} \frac{mr^2}{2} \omega^2.$$ Differentiating with respect to $t$, $$0 = \frac{1}{2}k(R-r)^2 \cdot 2 \theta \frac{d\theta}{dt} + mg(R-r) \cdot \frac{d}{dt} \left( \frac{\theta^2}{2} \right) + \frac{1}{2}m \cdot 2v \frac{dv}{dt} + \frac{mr^2}{4} \cdot 2\omega \frac{d\omega}{dt}.$$ This implies $$0 = k(R-r)^2 \theta \frac{d\theta}{dt} + mg(R-r) \frac{d\theta}{dt} + mv \frac{dv}{dt} + \frac{mr^2}{2} \omega \frac{d\omega}{dt}.$$ Also, $$\frac{d\theta}{dt} = \frac{V}{R-r} \implies \frac{d^2\theta}{dt^2} = \frac{1}{R-r} \frac{dv}{dt} = \frac{1}{R-r} a.$$ Therefore, $$\frac{V}{R-r} = -m\alpha r - \frac{mr^2}{2} \frac{v}{r} \alpha.$$ Thus, $$k(R-r)^2 \cdot \theta \frac{V}{R-r} + mg(R-r)\theta \frac{V}{R-r} = -\frac{3}{2}mr\alpha.$$ This implies $$k(R-r) + mg\theta = -\frac{3}{2}m(R-r) \frac{d^2\theta}{dt^2}.$$ Therefore, $$-\left[k(R-r) + mg\theta \right] = \frac{3}{2}m(R-r) \frac{d^2\theta}{dt^2}.$$ Thus, $$-\frac{2}{3} \left[ \frac{k}{m} + \frac{g}{R-r} \right] = \frac{d^2\theta}{dt^2}.$$ Comparing with the standard equation of SHM, $$\omega = \sqrt{\frac{2}{3} \left[ \frac{k}{m} + \frac{g}{R-r} \right]}.$$ Hence, the answer is option (A). Alternatively, For the second approach, $$\theta = \frac{x}{(R-r)}.$$ The equation is $$kx + mg\sin \theta - f = ma.$$ This implies $$kx + mg \frac{x}{(R-r)} - f = ma.$$ The frictional force is $$f = \frac{mr^2}{2} \cdot \alpha \implies f = \frac{ma}{2}.$$ Therefore, $$\left( \alpha + \frac{mg}{R-r} \right)x = \frac{3ma}{2}.$$ Thus, $$\omega = \sqrt{\frac{2}{3} \left[ \frac{k}{m} + \frac{g}{R-r} \right]}.$$

Given the equations: $$2mv_1 = 2mv_{1f} \cos \theta + 2mv_{2f} \cos \phi ...(i)$$ $$2m_{1f} \sin \theta = mv_{2f} \sin \phi ...(ii)$$ $$\frac{1}{2} (2m)v_1^2 + \frac{1}{2} m(0)^2 = \frac{1}{2} (2m)v_{1f}^2 + \frac{1}{2} mv_{2f}^2$$ $$2v_1^2 = 2v_{1f}^2 + v_{2f}^2 ...(iii)$$ From (i), (ii), (iii), $$3v_{1f}^2 - 4v_1 v_{1f} \cos \theta + v_1^2 = 0$$ $$(-4v_1 \cos \theta)^2 - 4(3)(v_1^2) \geq 0$$ $$\cos^2 \theta \geq \frac{3}{4}$$ $$\cos^2 \theta \geq \frac{\sqrt{3}}{2}$$ Therefore, $$\theta = \frac{\pi}{6}$$

Given $\phi = B_0 \cos \omega t A \sin \omega t = \frac{B_0 A \sin 2 \omega t}{2}$. $\varepsilon = -\frac{d\phi}{dt} = -B_0 A \cos 2 \omega t = \left( 0 \leq t \leq \frac{\pi}{\omega} \right)$. $\varepsilon = 0 \left( \frac{\pi}{\omega} \leq t \leq \frac{2\pi}{\omega} \right)$. Ans. Option (A)

10 MSD = 1 cm; 1 MSD = 0.1 $\mathrm{cm}$ 7 MSD = 10 VSD 1 VSD = 0.07 $\mathrm{cm}$ Reading = 2 MSD - VSD = 0.2 $\mathrm{cm}$ - 0.07 $\mathrm{cm}$ = 0.13 $\mathrm{cm}$ Ans. Option (C)

The rate of change of magnetic flux is given by $$ -\frac{d\phi}{dt} = \frac{d}{dt} (B_0 \times \ell \times y) = BV\ell $$ The force is given by $$ \mathbf{F} = B(i\hat{i})(-\hat{j}) $$ The equation of motion is $$ ma = -B_0 \left[ \frac{B_0 V \ell}{R} \right] (\ell) $$ Thus, $$ a = -\frac{B_0^2 \ell^2 V}{mR} $$ Also, $$ K = \frac{B_0^2 \ell^2 V}{RM} $$ So, $$ [a = -kv] $$ The differential equation is $$ \frac{dv}{dt} = -kv $$ Integrating, $$ \int_{v_0}^{v} \frac{dv}{dt} = \int_{0}^{t} -k \, dt $$ This gives $$ \ell n \frac{v}{v_0} = -kt $$ Thus, $$ [v = v_0 e^{-kt}] (i) $$ For $x \leq \ell$, $$ \frac{dx}{dt} = v_0 e^{-kt} $$ Integrating, $$ \int_{0}^{x} dx = \int_{0}^{t} v_0 e^{-kt} dt $$ This results in $$ = \frac{v_0}{k} (1 - e^{-kt}) $$ When $x = \ell$, $$ \ell = \frac{v_0}{k} (1 - e^{-kt_1}) $$ Option (D) $(v_0 = 3k\ell)$ $$ \ell = \frac{3k\ell}{k} (1 - e^{-kt}) $$ Simplifying, $$ \frac{1}{3} = 1 - e^{-kt} $$ Thus, $$ \frac{2}{3} = 2e^{-kt} $$ Solving for $t$, $$ -kt = 8n \left( \frac{2}{3} \right) $$ Therefore, $$ t = \frac{1}{k} \ln \left( \frac{2}{3} \right) $$ The complete loop will enter at $$ t = \frac{1}{k} \ln \left( \frac{2}{3} \right) $$ Option (B) $$ \frac{d\phi}{dt} = 0, \varepsilon = 0, i = 0, F = 0 $$ The answers are B, D.

Given $L = 10.5 \, \mathrm{cm}$ which has 3 significant digits. $b = 0.05 \, \mathrm{cm}$ has 1 significant digit. $t = 6.0 \, \mu\mathrm{m}$ has 2 significant digits. Volume, $V = Lbt$ must have only 1 significant digit. Therefore, $$V = 10.5 \times 0.05 \times 10^{-1} \times 6.0 \times 10^{-4} \, \mathrm{cm^3}$$ $$= 3 \times 10^{-5} \, \mathrm{cc}$$

Given $y_i = y_0 \cos (\omega t - kx)$ when wave going from rarer to denser, $y_r = A_r \cos (\omega t + kx + \pi)$. $y_r = a_1 y_0 \cos (\omega t + kx + \pi)$. Option (A) correct. (B) For transmitted from point P $$y_t = A_t \cos [\omega t - k_1 x]$$ $$\frac{k_1}{k} = \sqrt{\frac{\mu_1}{\mu}} = \frac{k_1}{k} = \sqrt{\frac{4\mu}{\mu}}$$ $$k_1 = 2k$$ $$y_t = a_2 y_0 \cos [\omega t - 2kx]$$ Option (B) incorrect. (C) When reflected from Q $$y_i = a_2 y_0 \cos [\omega t - 2kx]$$ $$y_r = a_3 y_0 \cos [\omega t + 2kx + \pi]$$ Option (C) incorrect. (D) When transmitted from Q $$y_t = a_4 y_0 \cos [\omega t = k_2 x]$$ $$\frac{k_2}{2k} = \sqrt{\frac{16\mu}{4\mu}} \implies k_2 = 4k$$ $$y_t = a_4 y_0 \cos [\omega t - 4kx]$$ Option (D) correct.

Given $y = 8 + 8 \sin \frac{2\pi t}{T}$. With respect to the elevator, variation in weight will be $\Delta W = m(\Delta a)_{\max}$. $\Delta W = m \times 2\omega^2 A$. Here the elevator is performing SHM. $$\Delta W = 2m \left( \frac{2\pi}{T} \right)^2 \times A \ \mathrm{N}$$ $$\Delta W = 2 \times 50 \times \left( \frac{2\pi}{40\pi} \right)^2 \times 8 \ \mathrm{N}$$ $$\Delta W = 2 \times 50 \times \frac{1}{400} \times 8 \ \mathrm{N}$$ $$\Delta W = \frac{800}{400} \ \mathrm{N} = 2 \ \mathrm{N}$$ Ans. is (B)

Total energy in cube = $35 \times 10^7 \times hf$ $$= 35 \times 10^7 \times 6 \times 10^{-34} \times 10^{15}$$ $$= 2.1 \times 10^{-10} \, \mathrm{J}$$ Total energy of EM waves = $\frac{B_0^2}{2 \mu_0} \times volume$ $$B_0^2 = \frac{2.1 \times 10^{-10} \times 8 \pi \times 10^{-7}}{1^3}$$ $$\Rightarrow B_0 = 22.98 \times 10^{-9} \, \mathrm{T}$$ Ans. 22.98

Initially: $$W_0 = \sigma \left( T_P^4 - T_Q^4 \right)$$ Finally: Putting heat currents equal in steady state: $$\sigma \left( T_P^4 - T_1^4 \right) = \sigma \left( T_1^4 - T_2^4 \right)$$ $$\sigma \left( T_1^4 - T_2^4 \right) = \sigma \left( T_2^4 - T_Q^4 \right)$$ Adding: $$T_P^4 - T_1^4 = T_2^4 - T_Q^4$$ $$\Rightarrow T_1^* + T_2^* = T_P^* + T_Q^*$$ and $$\Rightarrow T_1^4 - T_2^4 = T_P^4 - T_1^4$$ Adding: $$T_1^4 = \frac{2T_P^4 + T_Q^4}{3}$$ So $$W_S = \sigma \left( T_P^4 - T_1^4 \right)$$ $$= \sigma \left( T_P^4 - \left( \frac{2T_P^4 + T_Q^4}{3} \right) \right) = \sigma \left( \frac{T_P^4 - T_Q^4}{3} \right)$$ hence $$\frac{W_S}{W_0} = 3$$

Given $\tan \alpha = \sqrt{3}$. Therefore, $\alpha = 60^\circ$. $\sqrt{3} \sin \beta = 1 \times \sin \alpha \Rightarrow \beta = 30^\circ$. $\[$ $\frac{R}{2 \sin 30^\circ}$ = $\frac{x}{\sin 120^\circ}$ $\]$ $\[$ $\frac{R}{\sin 120^\circ}$ = $\frac{R \sqrt{3}}{2 \times \sin \theta}$ $\Rightarrow$ $\sin$ $\theta$ = $\frac{\sqrt{3}}{2}$ $\times$ $\frac{\sqrt{3}}{2}$ $\]$ $\[$ $\sin$ $\theta$ = $\frac{3}{4}$ $\]$

Solution-1 If we can consider $$ \frac{\Delta b}{b} = \frac{\Delta m}{m} + \frac{\Delta \lambda}{\lambda} + \frac{\Delta D}{D} + \frac{\Delta d}{d} $$ $$ \frac{\Delta b}{b} = 0 + 0 + \frac{1 \, \mathrm{cm}}{1 \, \mathrm{m}} + \frac{1 \, \mathrm{mm}}{5 \, \mathrm{mm}} = 0.21 $$ $$ b = \frac{m \lambda D}{d} = \frac{3 \times 600 \times 10^{-3} \times 1}{5 \times 10^{-3}} \, \mu \mathrm{m} = 360 \, \mu \mathrm{m} $$ $$ \Rightarrow \Delta b = 360 \times 0.21 \, \mu \mathrm{m} = 75.6 \, \mu \mathrm{m} $$ However, error in $d$ is too large (20$\%$) for the solution-1 to be correct. Hence, we propose solution-2. Solution-2 $$ b = \frac{m \lambda D}{d} = 360 \, \mu \mathrm{m} $$ $$ b_{max} = \frac{3 \times 600 \times 10^{-3} \times 1.01}{4 \times 10^{-3}} \, \mu \mathrm{m} = 454.5 \, \mu \mathrm{m} $$ $$ b_{min} = \frac{3 \times 600 \times 10^{-3} \times 0.99}{6 \times 10^{-3}} \, \mu \mathrm{m} = 297 \, \mu \mathrm{m} $$ Maximum value of $b$ gives error, $\Delta b_1 = 94.5 \, \mu \mathrm{m}$. Minimum value of $b$ gives error, $\Delta b_2 = 63 \, \mu \mathrm{m}$. Therefore, we always report the largest error, hence correct answer should be 94.5$\,$ $\mu$ $\mathrm{m}$.

Given $\($ $\frac{mv^2}{r}$ = $\frac{KZe^2}{r^2}$ $\)$. $\($ mv^2 r = $\frac{1}{4 \pi \varepsilon_0}$ Ze^2 $\)$ $\ldots$ (1) $\($ mvr = $\frac{nh}{2 \pi}$ $\)$ $\ldots$ (2) Dividing (1) by (2) gives $\[$ v = $\frac{Ze^2}{4 \pi \varepsilon_0 \frac{nh}{2 \pi}}$ = $\frac{Ze^2}{2 \varepsilon_0 nh}$ $\]$ $\[$ $\frac{h}{mv}$ = $\frac{h}{\sqrt{2m_n \cdot K_B T}}$ $\]$ $\[$ T = $\frac{mZ^2 e^4}{8 \varepsilon_0^2 n^2 h^2 m_n K_B}$ $\]$ For $\($ n = 3 $\Rightarrow$ T = $\frac{mZ^2 e^4}{72 \varepsilon_0^2 h^2 m_n K_B}$ $\)$ $\($ $\frac{Ze^2}{4 \pi \varepsilon_0}$ = $\frac{1}{mr}$ $\Rightarrow$ mr = $\frac{4 \pi \varepsilon_0}{2 nh^2}$ $\cdot$ $\frac{1}{4 \pi^2}$ $\)$ $\[$ r = $\frac{n^2 h^2 \varepsilon_0}{\pi Z e^2 \cdot m}$ $\Rightarrow$ a_0 = $\frac{h^2 \varepsilon_0}{\pi e^2 \cdot m}$ $\]$ $\[$ a_0^2 = $\frac{h^4 \varepsilon_0^2}{\pi^2 e^4 m^2}$ $\]$ $\[$ T a_0^2 = $\frac{mZ^2 e^4}{72 \varepsilon_0^2 h^2 m_n K_B}$ $\cdot$ $\frac{h^4 \varepsilon_0^2}{\pi^2 e^4 m^2}$ $\]$ $\[$ T = $\frac{h^2 Z^2}{72 \pi^2 a_0^2 m_n K_B}$ $\Rightarrow$ $\alpha$ = 72 $\]$

For (P), the net electric field is given by $$\mathbf{E}_{net} = \frac{-2kP \hat{\mathbf{j}}}{r^3}$$. For (Q), the electric fields cancel each other, so $$\mathbf{E}_{net} = 0$$. For (R), the net electric field is $$\frac{2P \hat{\mathbf{i}}}{4\pi \varepsilon_0 r^3} - \frac{P \hat{\mathbf{j}}}{4\pi \varepsilon_0 r^3}$$. For (S), the net electric field is $$\mathbf{E}_{net} = \frac{4kP \hat{\mathbf{i}}}{r^3}$$. The mapping is P → 2, Q → 1, R → 4, S → 5.

For P $$i = \frac{V}{R} = 10 \sin 400t \Rightarrow (3)$$ For Q $$X_L = \omega L = 400 \times 100 \times 10^{-3} = 40\, \Omega$$ Therefore, $$Z = 50\, \Omega$$ Thus, $$i = \frac{300}{50} \sin (400t - 53^\circ) [current will lag by \tan^{-1} \frac{X_L}{R} ] \Rightarrow (5)$$ For R $$X_C = \frac{10^6}{400 \times 50} \Omega = 50\, \Omega and X_L = 400 \times 25 \times 10^{-3} = 10\, \Omega$$ Therefore, $$Z = 50\, \Omega$$ Thus, $$i = \frac{300}{50} \sin (400t + 53^\circ) [Current will lead by \tan^{-1} \frac{X_C - X_L}{R} ] \Rightarrow (2)$$ For S $$X_C = 50\, \Omega and X_L = 400 \times 125 \times 10^{-3} = 50\, \Omega$$ $$R = 60\, \Omega$$ Thus, $$i = \frac{300}{60} \sin (400t) X_L = X_C \Rightarrow Resonance \Rightarrow (1)$$

Energy of H-like atom is $$E = -13.6 \frac{Z^2}{n^2} So$$ $$E \propto Z^2$$ P $\rightarrow$ (5) Energy of characteristic X-ray by Moseley's correction $$E = -13.6 (Z-1)^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] So$$ $$E \propto (Z-1)^2$$ Q $\rightarrow$ (1) Electrostatics binding energy is proportional to $Z(Z-1)$ R $\rightarrow$ (2) For stable nuclei with mass no. in range 30 to 170. Binding energy per nucleon is constant & graph is straight line S $\rightarrow$ (4) Ans. (C) is correct

The decomposition of $\mathrm{NH_4NO_2}$ at $60-70^\circ\mathrm{C}$ produces $\mathrm{N_2}$ and $2\mathrm{H_2O}$. $$\mathrm{NH_4NO_2} \xrightarrow{\Delta \ 60-70^\circ\mathrm{C}} \mathrm{N_2} + 2\mathrm{H_2O}$$ The decomposition of $\mathrm{NH_4NO_3}$ at $200-250^\circ\mathrm{C}$ produces $\mathrm{N_2O}$ and $2\mathrm{H_2O}$. $$\mathrm{NH_4NO_3} \xrightarrow{\Delta \ 200-250^\circ\mathrm{C}} \mathrm{N_2O} + 2\mathrm{H_2O}$$

Given $\Delta_O \propto \frac{1}{\lambda}$. Therefore, the absorb wavelength order is $[\mathrm{Co(CN)_6}]^{3-} < [\mathrm{Co(NH_3)_6}]^{3+} < [\mathrm{Co(NH_3)_5H_2O}]^{3+} < [\mathrm{Co(NH_3)_5Cl}]^{2+}$.

In a neutral solution, the reaction is given by: $$\mathrm{I^- + 2MnO_4^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-}$$

B is most acidic: It forms the most stable conjugate base.

(i) $\mathrm{Ne_2} \Rightarrow (\sigma 1s^2)(\sigma^* 1s^2)(\sigma 2s^2)(\sigma^* 2s^2)(\sigma 2p_z^2)(\pi 2p_x^2 = \pi 2p_y^2)(\pi^* 2p_x^2 = \pi^* 2p_y^2)(\sigma^* 2p_z^2)$ B.O. = $$\frac{6-6}{2} = 0$$ (ii) $\mathrm{F_2} \Rightarrow (\sigma 1s^2)(\sigma^* 1s^2)(\sigma 2s^2)(\sigma^* 2s^2)(\sigma 2p_z^2)(\pi 2p_x^2 = \pi 2p_y^2)(\pi^* 2p_x^1 = \pi^* 2p_y^1)$ (iii) $\mathrm{O_2^+} \Rightarrow (\sigma 1s^2)(\sigma^* 1s^2)(\sigma 2s^2)(\sigma^* 2s^2)(\sigma 2p_z^2)(\pi 2p_x^2 = \pi 2p_y^2)(\pi^* 2p_x^1 = \pi^* 2p_y)$ B.O. = $$\frac{6-1}{2} = 2.5$$ $\mathrm{O_2} \Rightarrow (\sigma 1s^2)(\sigma^* 1s^2)(\sigma 2s^2)(\sigma^* 2s^2)(\sigma 2p_z^2)(\pi 2p_x^2 = \pi 2p_y^2)(\pi^* 2p_x^1 = \pi^* 2p_y^1)$ B.O. = $$\frac{6-2}{2} = 2$$ (Bond order increases, Bond strength increases) (iv) Size of atom increases, Bond length increases Size of $\mathrm{Li} > \mathrm{B}$ So, Bond length of $\mathrm{Li_2} > \mathrm{B_2}$

La^{+3} $\rightarrow$ [54$\mathrm{Xe}$] $\,$ 4f^0 diamagnetic Yb^{+2} $\rightarrow$ [54$\mathrm{Xe}$] $\,$ 4f^{14} diamagnetic Lu^{+3} $\rightarrow$ [54$\mathrm{Xe}$] $\,$ 4f^{14} diamagnetic La^{+2} $\rightarrow$ [54$\mathrm{Xe}$] $\,$ 5d^1 paramagnetic Ce^{+4} $\rightarrow$ [54$\mathrm{Xe}$] $\,$ 4f^0 diamagnetic Ce^{+3} $\rightarrow$ [54$\mathrm{Xe}$] $\,$ 4f^1 paramagnetic Yb^{+3} $\rightarrow$ [54$\mathrm{Xe}$] $\,$ 4f^{13} paramagnetic Lu^{+2} $\rightarrow$ [54$\mathrm{Xe}$] $\,$ 4f^{14} $\,$ 5d^1 paramagnetic

The solution involves the Finkelstein reaction and Swart reaction. For compound P, the bond is C–Br, for Q it is C–I, and for R it is C–F. The bond lengths are 2–4, 2–5, and 2–2 respectively. As the period number increases, size increases, bond length increases, bond order decreases, and bond enthalpy decreases. (A) Wrong: bond length order Q > P > R (B) Correct: Bond enthalpy R > P > Q (C) Wrong: S_N2 reactivity Q > P > R (D) Wrong: pKa (HI P > Q (pKa)

For reduction of dichromate, balanced reaction is: $$\mathrm{Cr_2O_7^{2-}(aq) + 6e^- + 14H^+(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)}$$ 3 mol 1 mol Number of Farads required = 3 mol Let current = I amperes $$\Rightarrow \frac{I \times 48.25 \times 60}{96500} = 3$$ $$I = 100 \, \mathrm{A}$$

Because concentration of H⁺ from weak acid is less we need to consider self ionization of H₂O also. $$\mathrm{HX(aq) \rightleftharpoons H^+ + X^-(aq)}$$ $$10^{-3} - x x + y x$$ $$\mathrm{H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)}$$ $$x + y y$$ Approximation: $(10^{-3} - x) \simeq 10^{-3}$ $$\Rightarrow \frac{x(x+y)}{10^{-3}} = K_a = 4 \times 10^{-11} \ldots (1)$$ $$\Rightarrow y(x+y) = K_w = 10^{-14} \ldots (2)$$ Add (1) + (2) $$\Rightarrow (x+y)^2 = 5 \times 10^{-14}$$ $$\Rightarrow x+y = [H^+] = \sqrt{5} \times 10^{-7}$$ $$\Rightarrow x = \sqrt{5} = 2.236$$ Answer 2.23 or 2.24.

Given $\($ $\left$( P + $\frac{a}{V_m^2}$ $\right$) (V_m - b) = RT $\)$ $\($ PV_m - bP + $\frac{a}{V_m}$ - $\frac{ab}{V_m^2}$ - RT = 0 $\)$ $\($ $\Rightarrow$ PV_m^2 - (bP + RT) V_m^2 + aV_m - ab = 0 $\)$ Coefficient of $\($ V_m^2 = -(bP + RT) $\)$ Coefficient of $\($ V_m = a $\)$ Ratio = $\($-$\frac{(bP + RT)}{a}$ = -$\left$[ $\frac{0.06 \times 300 + 24.6}{6}$ $\right$] = -7.1. $\)$

The reaction is given by $\mathrm{H_2O(l)} \rightarrow \mathrm{H_2(g)} + \frac{1}{2} \mathrm{O_2(g)}$. The masses and moles are as follows: 144 g corresponds to 8 mol of $\mathrm{H_2O}$, producing 8 mol of $\mathrm{H_2}$ and 4 mol of $\mathrm{O_2}$. The work done $W$ is given by $W = -P \Delta V = -(\Delta n)RT$. The change in gaseous moles is 12. Therefore, $$W = -\frac{12 \times 8.3 \times 300}{1000} kJ = -29.88 kJ.$$

Hexamethylene diamine series give (+) carbylamine test. $\mathrm{C_x H_y N_z} + (2x + 1/2)\mathrm{CuO} \rightarrow x\mathrm{CO_2} + \frac{y}{2} \mathrm{H_2O} + \frac{z}{2} \mathrm{N_2}$ (Dumas method) $\mathrm{C_6H_{16}N_2}$ then $\frac{z}{2} \mathrm{N_2} = \frac{2}{2} \mathrm{N_2} = 1$ mole Given 10 mole in this reaction so $w_{\mathrm{N_2}} = 10 \times 28 = 280 \mathrm{gm}$

Given: $n = 16$ mole. The reaction sequence starts with the given compound undergoing a reaction with Na/dry ether followed by $\mathrm{H_3O^+}$ to form compound $\mathrm{(P)}$ with $\mathrm{CHO}$ groups and $M = 210$, $n = 8$. Next, compound $\mathrm{(P)}$ is treated with $\mathrm{NaOH/\Delta}$ and $\mathrm{H_3O^+}$ to form compound $\mathrm{(Q)}$ with $\mathrm{CH_2OH}$ and $\mathrm{COOH}$ groups, $n = 4$. Compound $\mathrm{(R)}$ with $\mathrm{OH}$ group, $n = 2$, is reacted with $\mathrm{NaOH/CaO}$ at $\Delta$ to form compound $\mathrm{(T)}$ with $\mathrm{Br}$, $n = 1$. Compound $\mathrm{(T)}$ undergoes a Williamson ether synthesis (WES) with $\mathrm{R}$, $\mathrm{NaH}$, and $\mathrm{(C_2H_5)_2O}$ to form the final product $\mathrm{(S)}$. The molecular weight $M$ of the product $\mathrm{S}$ is $350$. The weight of the product of $\mathrm{S}$ is calculated as $350 \times 1/2 = 175$ gm. Ans. *WES: Williamson ether synthesis.

Mn^{+2} + H_2S + NH_4OH $\rightarrow$ MnS $\downarrow$ Pink/buff ppt. Ba^{+2} + (NH_4)_2CO_3 + NH_4OH $\rightarrow$ BaCO_3 $\downarrow$ White ppt. Al^{+3} + NH_4Cl + NH_4OH $\rightarrow$ Al(OH)_3 $\downarrow$ White ppt. Cu^{+2} + H_2S + HCl (dil.) $\rightarrow$ CuS $\downarrow$ Black ppt.

1. The reaction of toluene with $\mathrm{CrO_2Cl_2/CS_2}$ followed by $\mathrm{H_2O^+}$ gives benzaldehyde, which undergoes the Cannizzaro reaction (S). 2. Benzoic acid reacts with $\mathrm{PCl_5}$ to form benzoyl chloride, which then reacts with $\mathrm{NH_3}$ to form benzamide. This undergoes the Stephen reaction with $\mathrm{P_4O_{10}/\Delta}$ to form benzonitrile (P). 3. Nitrobenzene is reduced with $\mathrm{Fe/HCl}$ to form aniline, which is then diazotized with $\mathrm{HCl/NaNO_2}$ at $0-5^\circ\mathrm{C}$ to form the diazonium salt. This undergoes the Sandmeyer reaction (Q). 4. Toluene is chlorinated with $\mathrm{Cl_2/hv}$ to form benzyl chloride, which is oxidized to benzoic acid. This reacts with Tollen's reagent to form benzaldehyde, which is then converted to benzoyl chloride using $\mathrm{SOCl_2}$. Benzoyl chloride reacts with $\mathrm{NH_3}$ to form benzamide, which undergoes the Hoffmann bromamide reaction (R). 5. Acetanilide is nitrated with $\mathrm{HNO_3}$ to form p-nitroacetanilide, which is then hydrolyzed with $\mathrm{NaOH}$ to form p-nitroaniline.

Gives $\mathrm{FeCl_3}$ test because of presence of phenolic group and ninhydrin test. Amino acid obtained gives (+) Ninhydrin test but not phthalein dye test. Reaction with phenyl diazonium salt gives yellow dye. Compound will form hydrazone derivative with aldehydic group of glucose.