JEE Advanced 3 October 2021 Paper 1 question paper with solutions

One of the vertices is the intersection of the x-axis and $x + y + 1 = 0$, which gives $A(-1, 0)$. Let vertex $B$ be $(\alpha, \alpha - 1)$. Line $AC \perp BH$ implies $\alpha = 1$, so $B(1, -2)$. Let vertex $C$ be $(\beta, 0)$. Line $AH \perp BC$. The product of slopes $m_{AH} \cdot m_{BC} = -1$. $$\frac{1}{2} \cdot \frac{2}{\beta - 1} = -1 \implies \beta = 0$$ The centroid of $\triangle ABC$ is $$\left(0, -\frac{2}{3}\right)$$ Now $G$ (centroid) divides the line joining the circumcenter $(O)$ and orthocenter $(H)$ in the ratio $1:2$. $$\begin{array}{ccc} (h, k) & (0, -\frac{2}{3}) & (1, 1) \\ O & 1 & G & 2 & H \end{array}$$ $$2h + 1 = 0 2k + 1 = -z$$ $$h = -\frac{1}{2} k = -\frac{3}{2}$$ Therefore, the circumcenter is $$\left(-\frac{1}{2}, -\frac{3}{2}\right)$$ The equation of the circumcircle is (passing through $C(0, 0)$) $$x^2 + y^2 + x + 3y = 0$$

Given $x + y - 2 = 0$. Points are $P \left( \frac{3}{2}, \frac{1}{2} \right)$, $Q(2, 0)$, $R \left( \frac{9}{4}, 0 \right)$, $S \left( \frac{9}{4}, \frac{3}{4} \right)$. The area is calculated as follows: $$Area = \frac{1}{2} \left| \begin{array}{ccc} \frac{3}{2} & \frac{1}{2} & 1 \\ 2 & 0 & 1 \\ \frac{9}{4} & 0 & 1 \\ \frac{9}{4} & \frac{3}{4} & 1 \end{array} \right|$$ $$= \frac{1}{2} \left| (0 - 1) + (0 - 0) + \left( \frac{27}{16} - 0 \right) + \left( \frac{9}{8} - \frac{9}{8} \right) \right| = \frac{11}{32}$$

The probability is given by $$P = \frac{P(S_1 \cap (E_1 = E_3))}{P(E_1 = E_3)} = \frac{P(B_{1,2})}{P(B)}$$ where $$P(B) = P(B_{1,2}) + P(B_{1,3}) + P(B_{2,3})$$ If 1,2 are chosen at the start: $$P(B_{1,2}) = \frac{1}{3} \times \frac{1 \times \binom{3}{1}}{\binom{4}{2}} \times \frac{1}{\binom{5}{2}}$$ 1 is definitely chosen from $F_2$ and 1,2 are chosen from $G_2$. If 1,3 are chosen at the start: $$P(B_{1,3}) = \frac{1}{3} \times \frac{1 \times \binom{2}{1}}{\binom{3}{2}} \times \frac{1}{\binom{5}{2}}$$ 1 is definitely chosen from $F_2$ and 1,2 are chosen from $G_2$. If 2,3 are chosen at the start: $$P(B_{2,3}) = \frac{1}{3} \times \left[ \frac{\binom{3}{2} \times 1}{\binom{4}{2}} \times \frac{1}{\binom{4}{2}} + \frac{1 \times \binom{3}{1}}{\binom{4}{2}} \times \frac{1}{\binom{5}{2}} \right]$$ If 1 is not chosen from $F_2$ and if 1 is chosen from $F_2$. Finally, $$\frac{P(B_{1,2})}{P(B)} = \frac{1}{5}$$

Given $|z_1| = |z_2| = \ldots |z_{10}| = 1$. The angle is given by $\frac{arc}{rad}$. $\theta_2 = arc(z_1 z_2) > (z_2 > z_1)$. For $P$: $|z_2 - z_1| + \ldots + |z_1 - z_{10}| \leq \theta_1 + \theta_2 + \ldots + \theta_{10}$. Therefore, $|z_2 - z_1| + \ldots + |z_1 - z_{10}| \leq 2\pi$ which means $P$ is true. $z_1^2 = e^{i2\theta_1}$, $z_k^2 = z_{k-1}^2 \cdot e^{i2\theta_k}$. Let $2\theta_k = \alpha_k$. Then $z_1^2 = e^{i\alpha_1}$, $z_k^2 = z_{k-1}^2 \cdot e^{i\alpha_k}$. $\alpha_1 + \alpha_2 + \ldots + \alpha_k = 4\pi$. In one similar sense, $|z_1^2 - z_2^2| + \ldots + |z_1^2 - z_{10}^2| \leq 4\pi$. $Q$ is also true.

Let $p_1$ be the probability that the maximum of chosen numbers is at least 81. $p_1 = 1 -$ probability that the maximum of chosen numbers is at most 80. $$p_1 = 1 - \frac{80 \times 80 \times 80}{100 \times 100 \times 100} = 1 - \frac{64}{125}$$ $$p_1 = \frac{61}{125}$$ $$\frac{625 p_1}{4} = \frac{625}{4} \times \frac{61}{125} = \frac{305}{4} = 76.25$$ the value of $\frac{625 p_1}{4}$ is 76.25

Given $p_2 =$ probability that minimum of chosen numbers is at most 40. $= 1 -$ probability that minimum of chosen numbers is at least 41. $$= 1 - \left( \frac{600}{100} \right)^3$$ $$= 1 - \frac{27}{125} = \frac{98}{125}$$ Therefore, $$\frac{125}{4} p_2 = \frac{125}{4} \times \frac{98}{125} = 24.50$$

Given the equation $7x + 8y + 9z - (\gamma - 1) = A(4x + 5y + 6z - \beta) + B(x + 2y + 3z - \alpha)$. For $x$: $7 = 4A + B$. For $y$: $8 = 5A + 2B$. Solving these, we find $A = 2$, $B = -1$. The constant term: $-(\gamma - 1) = -A\beta - \alpha B \Rightarrow -(\gamma - 1) \equiv 2\beta + \alpha$. Thus, $\alpha - 2\beta + \gamma = 1$. The matrix $\mathbf{M}$ is given by: $$\mathbf{M} = \begin{pmatrix} \alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix} = \alpha - 2\beta + \gamma = 1.$$ The plane $P$ is given by $x - 2y + z = 1$. The perpendicular distance is $\left| \frac{3}{\sqrt{6}} \right| = P \Rightarrow D = P^2 = \frac{9}{6} = 1.5$.

Given $P(x, y)$, we have $$\left| \frac{\sqrt{2x + y - 1}}{\sqrt{3}} \right| \left| \frac{\sqrt{x - y + 1}}{\sqrt{3}} \right| = \lambda^2$$ $$\left| \frac{2x^2 - (y - 1)^2}{3} \right| = \lambda^2, \ C: \left| 2x^2 - (y - 1)^2 \right| = 3\lambda^2$$ The line is $y = 2x + 1$, and $RS = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$, where $R(x_1, y_1)$ and $S(x_2, y_2)$. We have $y_1 = 2x_1 + 1$ and $y_2 = 2x_2 + 1$, which implies $(y_1 - y_2) = 2(x_1 - x_2)$. Therefore, $$RS = \sqrt{5(x_1 - x_2)^2} = \sqrt{5} |x_1 - x_2|$$ Solving the curve $C$ and the line $y = 2x + 1$, we get $$\left| 2x^2 - (2x)^2 \right| = 3\lambda^2 \implies x^2 = \frac{3\lambda^2}{2}$$ $$RS = \sqrt{5} \left| \frac{2\sqrt{3}\lambda}{\sqrt{2}} \right| = \sqrt{30} \lambda \implies 30\lambda^2 = 270 \implies \lambda^2 = 9$$

Perpendicular bisector of RS. T = $\left$( $\frac{x_1 + x_2}{2}$, $\frac{y_1 + y_2}{2}$ $\right$) Here $x_1 + x_2 = 0$ T = (0, 1) Equation of R'S': (y - 1) = -$\frac{1}{2}$(x - 0) $\Rightarrow$ x + 2y = 2 R'(a_1, b_1) S'(a_2, b_2) D = (a_1 - a_2)^2 + (b_1 - b_2)^2 = 5(b_1 - b_2)^2 Solve $x + 2y = 2$ and $\left| 2x^2 - (y - 1)^2 \right| = 3 \lambda^2$ $\left$| 8(y - 1)^2 - (y - 1)^2 $\right$| = 3 $\lambda$^2 $\Rightarrow$ (y - 1)^2 = $\left$( $\frac{\sqrt{3 \lambda}}{\sqrt{7}}$ $\right$)^2 $y - 1 = \pm \frac{\sqrt{3 \lambda}}{\sqrt{7}} \Rightarrow b_1 = 1 + \frac{\sqrt{3 \lambda}}{\sqrt{7}}, b_2 = 1 - \frac{\sqrt{3 \lambda}}{\sqrt{17}}$ D = 5 $\left$( $\frac{2 \sqrt{3 \lambda}}{\sqrt{7}}$ $\right$)^2 = $\frac{5 \times 4 \times 3 \lambda^2}{7}$ = $\frac{5 \times 4 \times 27}{7}$ = 77.14

PEP = $\begin{pmatrix}$ 1 & 0 & 0 $\\$ 0 & 0 & 1 $\\$ 0 & 1 & 0 $\end{pmatrix}$ $\begin{pmatrix}$ 1 & 2 & 3 $\\$ 8 & 13 & 18 $\\$ 2 & 3 & 4 $\end{pmatrix}$ $\begin{pmatrix}$ 1 & 0 & 0 $\\$ 0 & 0 & 1 $\\$ 0 & 1 & 0 $\end{pmatrix}$ = $\begin{pmatrix}$ 1 & 2 & 3 $\\$ 8 & 13 & 18 $\\$ 2 & 3 & 4 $\end{pmatrix}$ $\begin{pmatrix}$ 1 & 0 & 0 $\\$ 0 & 0 & 1 $\\$ 0 & 1 & 0 $\end{pmatrix}$ = $\begin{pmatrix}$ 1 & 0 & 0 $\\$ 0 & 1 & 0 $\\$ 0 & 0 & 1 $\end{pmatrix}$ P^2 = $\begin{pmatrix}$ 1 & 0 & 0 $\\$ 0 & 0 & 1 $\\$ 0 & 1 & 0 $\end{pmatrix}$ $\begin{pmatrix}$ 1 & 0 & 0 $\\$ 0 & 0 & 1 $\\$ 0 & 1 & 0 $\end{pmatrix}$ = $\begin{pmatrix}$ 1 & 0 & 0 $\\$ 0 & 1 & 0 $\\$ 0 & 0 & 1 $\end{pmatrix}$ (B) |EQ + PFQ^{-1}| = |EQ| + |PFQ^{-1}| |E| = 0 and |F| = 0 and |Q| $\neq$ 0 |EQ| = |E||Q| = 0 , PFQ^{-1} = $\frac{|P||F|}{|Q|}$ = 0 T = EQ + PFQ^{-1} TQ = EQ^2 + PF = EQ^2 + P^2EP = EQ^2 + EP = E(Q^2 + P) |TQ| = |E(Q^2 + P)| $\Rightarrow$ |T||Q| = |E||Q^2 + P| = 0 $\Rightarrow$ |T| = 0 (as |Q| $\neq$ 0) (C) (EF)^3 > |EF|^2 Here 0 > 0 (false) (D) as P^2 = I $\Rightarrow$ P^{-1} = P so P^{-1}FP = PFP = PEPP = E so E + P^{-1}FP = E + E = 2E P^{-1}EP + F $\Rightarrow$ PEP + F = 2PEP Tr(2PEP) = 2Tr(EP) = 2Tr(EPP) = 2Tr(E)

Given $\($ f(x) = $\frac{x^2 - 3x - 6}{x^2 + 2x + 4}$ $\)$. Differentiating, we have: $$ f'(x) = \frac{(x^2 + 2x + 4)(2x - 3) - (x^2 - 3x - 6)(2x + 2)}{(x^2 + 2x + 4)^2} $$ Simplifying, $$ f'(x) = \frac{5x(x + 4)}{(x^2 + 2x + 4)^2} $$ The sign of $\($ f'(x) $\)$ is analyzed as follows: $\($ f'(x) : + - + $\)$ at $\($ -4 $\)$ and $\($ 0 $\)$. Calculating specific values, $\($ f(-4) = $\frac{11}{6}$ $\)$, $\($ f(0) = -$\frac{3}{2}$ $\)$, and $\($ $\lim$_{x $\to$ $\pm$ $\infty$} f(x) = 1 $\)$. Thus, the range is $\($ $\left$[ -$\frac{3}{2}$, $\frac{11}{6}$ $\right$] $\)$, clearly $\($ f(x) $\)$ is into.

P(E) = $\frac{1}{8}$; $\;$ P(F) = $\frac{1}{6}$; $\;$ P(G) = $\frac{1}{4}$; $\;$ P(E $\cap$ F $\cap$ G) = $\frac{1}{10}$ (C) $\;$ P(E $\cup$ F $\cup$ G) = P(E) + P(F) + P(G) - P(E $\cap$ F) - P(F $\cap$ G) - P(G $\cap$ E) + P(E $\cap$ F $\cap$ G) $$= \frac{1}{8} + \frac{1}{6} + \frac{1}{4} - \sum P(E \cap F) + \frac{1}{10}$$ $$= \frac{3 + 4 + 6}{24} + \frac{1}{10} - \sum P(E \cap F) = \frac{13}{24} + \frac{1}{10} - \sum P(E \cap F)$$ $$\Rightarrow P(E \cup F \cup G) \leq \frac{13}{24} [(C) is Correct]$$ (D) $\;$ P(E^C $\cap$ F^C $\cap$ G^C) = 1 - P(E $\cup$ F $\cup$ G) $\geq$ 1 - $\frac{13}{24}$ $$\Rightarrow P(E^C \cap F^C \cap G^C) \geq \frac{11}{24} [(D) is Incorrect]$$ (A) $\;$ P(E) = $\frac{1}{8}$ $\geq$ P(E $\cap$ F $\cap$ G^C) + P(E $\cap$ F $\cap$ G) $$\Rightarrow \frac{1}{8} \geq P(E \cap F \cap G^C) + \frac{1}{10} \Rightarrow \frac{1}{8} - \frac{1}{10} \geq P(E \cap F \cap G^C)$$ $$\Rightarrow \frac{1}{40} \geq P(E \cap F \cap G^C) [(A) is Correct]$$ (B) $\;$ P(F) = $\frac{1}{6}$ $\geq$ P(E^C $\cap$ F $\cap$ G) + P(E $\cap$ F $\cap$ G) $$\Rightarrow \frac{1}{6} - \frac{1}{10} \geq P(E^C \cap F \cap G)$$ $$\Rightarrow \frac{4}{60} \geq P(E^C \cap F \cap G)$$ $$\Rightarrow \frac{1}{15} \geq P(E^C \cap F \cap G) [(B) is Correct]$$

Given $|I - EF| \neq 0$; $G = (I - EF)^{-1} \Rightarrow G^{-1} = I - EF$. Now, $G \cdot G^{-1} = I = G^{-1} G$. Therefore, $G (I - EF) = I = (I - EF) G$. This implies $G - GEF = I = G - EFG$. Hence, $GEF = EFG$ [C is Correct]. $(I - FE)(I + FGE) = I + FGE - FE - FEFGE = I + FGE - FE - F(G - I)E = I + FGE - FE - FGE + FE = I$ [(B) is Correct]. (So 'D' is Incorrect) We have $(I - FE)(I + FGE) = I$ .....(I) Now, $FE(I + FGE) = FE + FEFGE = FE + F(G - I)E = FE + FGE - FE = FGE$ Therefore, $|FE| |I + FGE| = |FGE|$. Thus, $|FE| \times \frac{1}{|I - FE|} = |FGE|$ (from (I)) Therefore, $|FE| = |I - FE| |FGE|$ (option (A) is correct).

Given $$S_n(x) = \sum_{k=1}^{n} \tan^{-1} \left( \frac{x}{1+kx(kx+x)} \right)$$ $$= \sum_{k=1}^{n} \tan^{-1} \left( \frac{(kx+x)-(kx)}{1+(kx+x)(kx)} \right)$$ $$S_n(x) = \tan^{-1}(nx+x) - \tan^{-1} x = \tan^{-1} \left( \frac{nx}{1+(n+1)x^2} \right)$$ (A) $S_{10}(x) = \tan^{-1} \frac{10x}{1+11x^2} = \frac{\pi}{2} - \tan^{-1} \left( \frac{1+11x^2}{10x} \right)$ $(x > 0)$ (B) $\lim_{n \to \infty} \cot(S_n(x)) = \lim_{n \to \infty} \frac{1}{x} \frac{n + \left(1 + \frac{1}{n}\right)x^2}{x} = x$ $(x > 0)$ (C) $S_3(x) = \tan^{-1} \frac{-3x}{1+4x^2} = \frac{\pi}{4} \Rightarrow 4x^2 - 3x + 1 = 0 \Rightarrow x \notin \mathbb{R}$ (D) $\tan(S_n(x)) = \frac{nx}{1+(n+1)x^2} ; \forall \ n \geq 1 ; x > 0$ We need to check the validity of $$\frac{nx}{1+(n+1)x^2} \leq \frac{1}{2} \forall \ n \geq 1 ; x > 0 ; n \in \mathbb{N}$$ $$\Rightarrow 2nx \leq (n+1)x^2 + 1$$ $$\Rightarrow (n+1)x^2 - 2nx + 1 \geq 0 \forall \ n \geq 1 ; x > 0 ; n \in \mathbb{N}$$ Discriminant of $y = (n+1)x^2 - 2nx + 1$ is $$D = 4n^2 - 4(n+1)$$ and $n \in \mathbb{N}$ $D 0$ $D > 0$ for $n \geq 2 \Rightarrow \exists$ some $x > 0$ for which $y 0 ; n \in \mathbb{N}$ is false.

The argument $\($ $\arg$ $\left$( $\frac{z + \alpha}{z + \beta}$ $\right$) = $\frac{\pi}{4}$ $\)$ implies $\($ z $\)$ is on arc and $\($(-$\alpha$, 0)$\)$ and $\($(-$\beta$, 0)$\)$ subtend $\($ $\frac{\pi}{4}$ $\)$ on $\($ z $\)$. And $\($ z $\)$ lies on $\($ x^2 + y^2 + 5x - 3y + 4 = 0 $\)$. So put $\($ y = 0 $\)$; $$ x^2 + 5x + 4 = 0 \implies x = -1 ; x = -4 $$ Now, $\($ $\arg$ $\left$( $\frac{z + \alpha}{z + \beta}$ $\right$) = $\frac{\pi}{4}$ $\)$ implies $\($ z + $\alpha$ = (z + $\beta$) $\cdot$ r $\cdot$ e^{i $\frac{\pi}{4}$} $\)$. So, $\($ z + $\beta$ = z + 4 $\implies$ $\beta$ = 4 $\)$ and $\($ z + $\alpha$ = z + 1 $\implies$ $\alpha$ = 1 $\)$.

Given $3x^2 + x - 1 = 4 \lvert x^2 - 1 \rvert$. If $x \in [-1, 1]$, then $3x^2 + x - 1 = -4x^2 + 4 \Rightarrow 7x^2 + x - 5 = 0$. Say $f(x) = 7x^2 + x - 5$. Then $f(1) = 3$; $f(-1) = 1$; $f(0) = -1$. [Two Roots] If $x \in (-\infty, -1] \cup [1, \infty)$, then $3x^2 + x - 1 = 4x^2 - 4 \Rightarrow x^2 - x - 3 = 0$. Say $g(x) = x^2 - x - 3$. Then $g(-1) = -1$; $g(1) = -3$. [Two Roots] So total 4 roots.

Given $c = \sqrt{23}$, $a = 3$, $b = 4$. $$\cot A = \frac{\cos A}{\sin A} = \frac{b^2 + c^2 - a^2}{2bc \sin A}$$ $$= \frac{b^2 + c^2 - a^2}{2 \cdot 2 \Delta} \left\{ \Delta = \frac{1}{2} bc \sin A \right\}$$ $$\cot A = \frac{b^2 + c^2 - a^2}{4 \Delta}$$ Similarly, $\cot B = \frac{a^2 + c^2 - b^2}{4 \Delta}$ and $\cot C = \frac{a^2 + b^2 - c^2}{4 \Delta}$. Therefore, $$\cot A + \cot C = \frac{b^2 + c^2 - a^2 + a^2 + b^2 - c^2}{4 \Delta} = \frac{2b^2}{4 \Delta} = \frac{32}{16} = 2$$

Given, $|\mathbf{u}| = 1$; $|\mathbf{v}| = 1$; $\mathbf{u} \cdot \mathbf{v} \neq 0$; $\mathbf{u} \cdot \mathbf{w} = 1$; $\mathbf{v} \cdot \mathbf{w} = 1$; $$\mathbf{w} \cdot \mathbf{w} = |\mathbf{w}|^2 = 4 \implies |\mathbf{w}| = 2; \left[ \mathbf{u} \; \mathbf{v} \; \mathbf{w} \right] = \sqrt{2}$$ and $\left[ \mathbf{u} \; \mathbf{v} \; \mathbf{w} \right]^2 = \begin{vmatrix} \mathbf{v} \cdot \mathbf{u} & \mathbf{v} \cdot \mathbf{v} & \mathbf{v} \cdot \mathbf{w} \\ \mathbf{w} \cdot \mathbf{u} & \mathbf{w} \cdot \mathbf{v} & \mathbf{w} \cdot \mathbf{w} \end{vmatrix} = 2$ $$\begin{vmatrix} 1 & \mathbf{u} \cdot \mathbf{v} & 1 \\ \mathbf{u} \cdot \mathbf{v} & 1 & 1 \\ 1 & 1 & 4 \end{vmatrix} = 2$$ $$\implies \mathbf{u} \cdot \mathbf{v} = \frac{1}{2}$$ So, $|3\mathbf{u} + 5\mathbf{v}| = \sqrt{9|\mathbf{u}|^2 + 25|\mathbf{v}|^2 + 2 \cdot 3 \cdot 5 \mathbf{u} \cdot \mathbf{v}}$ $$= \sqrt{9 + 25 + 30 \left( \frac{1}{2} \right)} = \sqrt{49} = 7$$

Given 10 VSD = 9 MSD. $$1 VSD = \frac{9}{10} MSD$$ Least count = 1 MSD - 1 VSD $$= \left(1 - \frac{9}{10}\right) MSD$$ $$= 0.1 MSD$$ $$= 0.1 \times 0.1 cm$$ $$= 0.01 cm$$ As '0' of V.S. lie before '0' of M.S. Zero error $$= -[10 - 6] L.C.$$ $$= -4 \times 0.01 cm$$ $$= -0.04 cm$$ Reading $$= 3.1 cm + 1 \times LC$$ $$= 3.4 cm + 1 \times 0.01 cm$$ $$= 3.11 cm$$ True diameter = Reading - Zero error $$= 3.11 - (-0.04) cm = 3.15 cm$$

Process 1 P = constant, Volume increases and temperature also increases ⇒ W = positive, ΔU = positive ⇒ Heat is positive and supplied to gas Process 2 V = constant, Pressure decrease ⇒ Temperature decreases $$W = \int pdV = 0$$ ΔT is negative and $$\Delta U = \frac{f}{2} nR \Delta T$$ ⇒ ΔU is negative ΔQ = ΔU + W ∴ ΔQ → Heat is negative and rejected by gas Process 3 P = constant, Volume decreases ⇒ Temperature also decreases W = PΔV = negative $$\Delta U = \frac{f}{2} nR \Delta T = negative$$ ΔQ = W + ΔU = negative Heat is negative and rejected by gas. Process 4 V = constant, Pressure increases $$W = \int pdV = 0$$ PV = nRT ⇒ Temperature increase ⇒ $$\Delta U = \frac{f}{2} nR \Delta T$$ is positive ΔQ = ΔU + W = positive Ans. (C) step 1 and step 4

Focal length of convex lens ($f_1$) $$\frac{1}{f_1} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$$ $$= (1.5 - 1) \left[ \frac{1}{20} - \left( \frac{1}{-20} \right) \right]$$ $$\frac{1}{f_1} = \frac{1}{20}$$ Therefore, $f_1 = +20 \, \mathrm{cm}$ Focal length of concave lens ($f_2$) $$\frac{1}{f_2} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$$ $$\frac{1}{f_2} = (1.5 - 1) \left[ -\frac{1}{20} - \frac{1}{20} \right] = \frac{1}{-20}$$ Therefore, $f_2 = -20 \, \mathrm{cm}$ For lens 1 $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ $$\Rightarrow v = -20 \, \mathrm{cm}$$ $$m_1 = \frac{v}{u} = \frac{-20}{-10} = 2$$ For lens 2 $$u = -30, \ f = -20, \ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ $$v = -12 \, \mathrm{cm}$$ $$m_2 = \frac{v}{u} = \frac{-12}{-30} = \frac{2}{5}$$ Net magnification $$m = m_1 m_2 = 2 \times \frac{2}{5} = \frac{4}{5} = 0.8$$

Out of 1000 nuclei of Q, 60$\%$ may go $\alpha$-decay. Therefore, 600 nuclei may have $\alpha$-decay. $$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{20}$$ $t = 1$ hour $= 60$ minutes. Using $$N = N_0 e^{-\lambda t}$$ $$= 600 \times e^{-\frac{\ln 2}{20} \times 60}$$ $$N = 75$$ Therefore, 75 nuclei are left after one hour. So, the number of nuclei decayed $$= 600 - 75 = 525$$

Range $R = \frac{2u_x u_y}{g} = \frac{2 \times 5 \times 5}{10} = 5 \, \mathrm{m}$. Time of flight $T = \frac{2u_y}{g} = \frac{2 \times 5}{10} = 1 \, \mathrm{sec}$. Therefore, time of motion of one part falling vertically downwards is $0.5 \, \mathrm{sec} = \frac{T}{2}$. Time of motion of another part, $t = \frac{T}{2} = 0.5 \, \mathrm{sec}$. From momentum conservation $\Rightarrow P_i = P_f$. $$2m \times 5 = m \times v$$ $$v = 10 \, \mathrm{m/s}$$ Displacement of other part in $0.5 \, \mathrm{sec}$ in horizontal direction $= v \frac{T}{2}$. $$= 10 \times 0.5 = 5 \, \mathrm{m} = R$$ Therefore, total distance of second part from point 'O' is, $x = \frac{3R}{2} = 3 \times \frac{5}{2}$. $$x = 7.5 \, \mathrm{m}$$ $t = 0.5 \, \mathrm{sec}$

Switch connected to position 'P'. $V_1 = i_1 \cdot 1 = 1 + 2 - 2i_1 = V_2$. $3i_1 = 1$ $i_1 = \frac{1}{3} \mathrm{A}$ $V_1 = 1 - i_1 = 1 - \frac{1}{3}$ $V_1 = \frac{2}{3} \mathrm{volt}$ Potential drop across capacitor $\Delta V = \frac{2}{3} \mathrm{volt}$ Charge on capacitor $q = C \Delta V$ $q_1 = 1 \mu \mathrm{F} \cdot \frac{2}{3} \mathrm{V}$ $q_1 = 1.33 \mu \mathrm{C}$ Switch at Position 'Q'. $V_1 = i_1 \cdot 1 = 2 - 2i_1 = V_2$ $3i_1 = 2$ $i_1 = \frac{2}{3} \mathrm{A}$ $V_1 = 2 - i_1 = 2 - \frac{2}{3}$ $V_1 = \frac{4}{3} \mathrm{volt}$ Potential difference across capacitor $\Delta V = \frac{2}{3} \mathrm{volt}$ Charge on capacitor $q = C \Delta V$ $q_2 = 1 \mu \mathrm{F} \cdot \frac{2}{3} \mathrm{V}$ $q_2 = 0.67 \mu \mathrm{C}$

Switch connected to position 'P'. $V_1 = i_1 \cdot 1 = 1 + 2 - 2i_1 = V_0$. $3i_1 = 1$ $i_1 = \frac{1}{3} \mathrm{A}$ $q_1 = i_1 \cdot 1 = \frac{1}{3} \mathrm{C}$ $V_0 - V_1 = \frac{1}{3} \mathrm{volt}$ Potential drop across capacitor $\Delta V = \frac{1}{3} \mathrm{volt}$. Charge on capacitor $q = C \Delta V$ $q_1 = 1 \cdot \frac{1}{3}$ $q_1 = 0.33 \mu \mathrm{C}$ Switch at Position 'Q'. $V_1 = i_1 \cdot 1 = 2 - 2i_1 = V_0$ $3i_1 = 2$ $i_1 = \frac{2}{3} \mathrm{A}$ $q_1 = i_1 \cdot 1 = \frac{2}{3} \mathrm{C}$ $V_0 - V_1 = \frac{2}{3} \mathrm{volt}$ Potential difference across capacitor $\Delta V = \frac{2}{3} \mathrm{volt}$. Charge on capacitor $q = C \Delta V$ $q_1 = 1 \cdot \frac{2}{3} = 0.67 \mu \mathrm{C}$

Let a point P on the circle. $$V_p = 0 = \frac{kQ}{r_1} + \frac{kQ/\sqrt{5}}{r_2}$$ $$\frac{kQ}{r_1} = \frac{kQ/\sqrt{5}}{r_2}$$ $$\frac{1}{\sqrt{x^2 + y^2}} = \frac{1}{\sqrt{5}\sqrt{(x-2)^2 + y^2}}$$ $$3(x^2 - 2x) + 3y^2 = x^2 + y^2$$ $$3(x^2 + 4 - 4x) - x^2 + 2y^2 = 0$$ $$2x^2 + 12 - 12x + 2y^2 = 0$$ $$x^2 + 6 - 6x + y^2 = 0$$ $$(x - 3)^2 + y^2 = \left(\sqrt{3}\right)^2$$ $$R = \sqrt{3} = 1.73$$ $$b = 3$$

Given the forces and motion, we have the following equations: $$F - f = ma_c$$ $$fR = I_c \alpha$$ $$a_c - \alpha R = 0$$ Substituting, we get: $$F - \frac{I_c \alpha}{R} = ma_c$$ Solving for $a_c$: $$a_c = \frac{F}{\frac{I_c}{R^2} + m}$$ For the friction force $f$: $$f = \frac{I_c \alpha}{R} = \frac{I_c}{R^2} a_c = \frac{I_c}{R^2} \frac{F}{\left[ \frac{I_c}{R^2} + m \right]}$$ Thus, $$f = \frac{F}{\left[ m + \frac{I_c}{R^2} \right]}$$ For a thin-walled hollow cylinder, $$I_c = mR^2$$ Then, $$a_c = \frac{F}{2m}$$ Using the relation $$fR = I_c \alpha = \frac{I_c a_c}{R}$$ We have $$f = \frac{I_c a_c}{R^2} \leq \mu mg$$ Thus, $$a_c \leq \frac{\mu mg R^2}{I_c}$$ For a solid cylinder, $$I_c = \frac{mR^2}{2}$$ Then, $$a_c \leq 2 \mu g$$ The maximum acceleration $$(a_c)_{\max} = 2 \mu g$$

Given $\sin \theta_0 = \frac{1}{n_1}$ (given). i.e. $\sin \theta_1 > \frac{1}{n_1}$. $n_1 \sin \theta_1 = n_1 \sin \theta_0$. $\sin \theta_1 = n \sin \theta_0$. $\sin \theta_2 = \frac{n \sin \theta_1}{n_2}$. If $n_1 = n_2$ then $\theta_2 = \theta_1$. $n_2 \sin \theta_2 = (1) \sin \theta_0$. $\sin \theta_2 = n_1 \sin \theta_0$. $\sin \theta_1 = n_1 \sin \theta_1$. $\sin \theta_1 = \frac{\sin \theta_2}{n_1} > \frac{1}{n_1}$. $\sin \theta_1 > 1$. $\theta_3 > 90^\circ$. This means ray cannot enter air. For $n_1 > n_2$: $\sin \theta_1 = \frac{n_2}{n_1} \sin \theta_2 > \frac{1}{n_1}$. $\sin \theta_1 > \frac{1}{n_2}$. For surface 2 - air interface $n_1 \sin \theta_0 = \sin \theta_1$. $\sin \theta_1 = \frac{\sin \theta_0}{n_2} > \frac{1}{n_2}$. $\theta_2 > 90^\circ$. It means ray is reflected back in medium-2. For surface 1 - surface 2 interface $n_1 \sin \theta_0 = n_1 \sin \theta_1$. $\sin \theta_c = \frac{n_1}{n_2}$. $\theta_c$: critical angle for ray to enter medium-1. $\theta_2 n_1$: $\frac{n_2}{n_1} \sin \theta_0 > \frac{n_2}{n_1}$. $\sin \theta_0 > \frac{1}{n_2}$. For surface 2 - air interface $n_2 \sin \theta_2 = \sin \theta_3$. $\sin \theta_2 = \frac{\sin \theta_0}{n_2} > \frac{1}{n_2}$. $\theta_2 > 90$. It means ray is reflected back in medium-2. $n_2 \sin \theta_2 = n_1 \sin \theta_1$. $\sin \theta_c = \frac{n_1}{n_2}$, $\theta_c$: critical angle. For ray to enter medium-1 $\theta_2 \frac{1}{n_1}$. $\sin \theta_2 > 1 = \theta_2 > 90^\circ$. Ray is reflected back in medium-

Given $\vec{r}_A = -\hat{j}$. $S = \frac{1}{2} a t^2$ $20 = \frac{1}{2} \times 10 \times t^2$ $t = 2 sec$ $\vec{\tau}_o = \vec{r} \times \vec{F}; \; \vec{r}_B = 10 \hat{i} - \hat{j}$ $\vec{F} = m \vec{a} = 0.2 \times 10 \hat{i} = 2 \hat{i}$ $\vec{\tau}_o = (10 \hat{i} - \hat{j}) \times (2 \hat{i})$ $\vec{\tau}_o = 2 \hat{k}$ $\vec{L}_o = \vec{r}_B \times \vec{p} = \vec{r}_B \times m \vec{v}$ $\vec{v} = \vec{a} t = 10 \hat{i} \times 2 = 20 \hat{i}$ $\vec{L}_o = (0.2) \left[ (10 \hat{i} - \hat{j}) \times 20 \hat{i} \right] = 4 \hat{k}$ At point $A(0, -1)$ $\vec{\tau}_o = \vec{r}_A \times \vec{F} = (-\hat{j}) \times 2 \hat{i} = 2 \hat{k}$

For A When the transition is from any level to $n = 2$, then photon emitted belong to Balmer series. Therefore, for longest wavelength, transition occurs from $n = 3$ to $n = 2$. $$\frac{hc}{\lambda_{\max}} = \mathrm{RCh} \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]$$ and for shortest wavelength transition occurs from $n = \infty$ to $n = 2$ $$\frac{hc}{\lambda_{\min}} = \mathrm{RCh} \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right]$$ Thus, $$\frac{\lambda_{longest}}{\lambda_{shortest}} = \frac{9}{5}$$ For (B) $$\lambda_{longest} of Balmer = \frac{36}{5R}$$ $$\lambda_{shortest} of Paschen = \frac{9}{R}$$ Hence these wavelength don't overlap. For (C) For Lyman series, $$\frac{1}{\lambda} = \mathrm{R} \left[ 1 - \frac{1}{m^2} \right]$$ Also $$\frac{1}{\lambda_0} = \mathrm{R}$$ Thus, $$\frac{1}{\lambda} = \frac{1}{\lambda_0} \left[ 1 - \frac{1}{m^2} \right] \implies \lambda = \frac{\lambda_0}{1 - \frac{1}{m^2}}$$ For (D) $$\lambda_{longest} of Lyman = \frac{4}{3R}, \lambda_{shortest} of Balmer = \frac{4}{R}$$ Hence that wavelength don't overlap.

EMF developed across the emf of semi-circular rod is given by $$ \int_1^4 \frac{\mu_0 i}{2 \pi r} \, drv = \frac{\mu_0 i V}{2 \pi} \ln 4 = \frac{\mu_0 i V}{\pi} \ln 2 $$ From given value, $$ E = \frac{4 \pi \times 10^{-7} \times 2 \times 3 \times 0.7}{\pi} = 24 \times 7 \times 10^{-8} $$ $$ i_{max} = \frac{E}{R} = \frac{24 \times 7 \times 10^{-8}}{1.4} = 1.2 \times 10^{-6} \, A $$ $$ Q_{max} = C_0 E = 24 \times 7 \times 10^{-8} \times 5 \times 10^{-6} = 8.4 \times 10^{-12} \, C $$

Therefore, $P_1 - P_3 = \rho \left( g - \frac{a}{\sqrt{2}} \right) d$. $P_2 - P_3 = \rho \frac{a}{\sqrt{2}} d$. Therefore, $P_1 - P_2 = \rho d \left[ g - \frac{2a}{\sqrt{2}} \right]$. Therefore, $\frac{P_1 - P_2}{\rho g d} = \left[ 1 - \sqrt{2} \frac{a}{g} \right] = \beta$. Therefore, if $\beta = 0$, $a = \frac{g}{\sqrt{2}} \ldots (A)$. $\beta = \frac{\sqrt{2} - 1}{2}$, $a = \frac{g}{2} \ldots (C)$.

The radius is given by the equation $$r = \frac{mv}{qB} = \frac{\sqrt{2mqV}}{qB}$$. The kinetic energy is expressed as $$\frac{P^2}{2m} = K.E = qV$$. The ratio of radii is calculated as $$\frac{r_s}{r_\alpha} = \sqrt{\frac{32}{1} \times \frac{2}{4}} = 4$$. Therefore, $$\frac{r_s}{r} = 4$$.

Given $$L = \frac{Ma^2}{3} \Omega + M \left( \frac{3a}{4} \right)^2 \Omega + \frac{M \left( \frac{a}{4} \right)^2 4 \Omega}{2}$$ Then $$L = \frac{49}{48} Ma^2 \Omega$$ Finally, $$n = 49$$

Given $\sigma A T^4 = -ms \frac{dT}{dt}$. $$\int_{200}^{100} \frac{dT}{T^4} = \int_{0}^{t_1} k \, dt$$ $$\left. \frac{1}{3T^3} \right|_{200}^{100} = kt_1$$ $$\frac{1}{3} \left( \frac{1}{100^3} - \frac{1}{200^3} \right) = kt_1$$ $$\left. \frac{1}{3T^3} \right|_{200}^{50} = kt_2$$ $$\frac{1}{3} \left( \frac{1}{50^3} - \frac{1}{200^3} \right) = kt_2$$ $$\frac{t_2}{t_1} = \left( \frac{200^3 - 50^3}{200^3 - 100^3} \right) \left( \frac{100^3}{50^3} \right) = 9$$

The reaction involves the conversion of a compound with multiple triple bonds to a compound with a single double bond using $\mathrm{NaNH_2}$ and $\mathrm{Na/liq.NH_3}$. (B) is answer.

The structure shown is a meso butane-2,3-diol. It is the most stable conformation with hydrogen bonding.

Packing fraction (P.F.) is given by $$\frac{\frac{4}{3} \pi r_-^3 + \frac{4}{3} \pi r_+^3}{a^3}$$ where $$\frac{r_+}{r_-} = 0.414$$ (square planar void), and $a = 2r_-$. We get, $$P.F. = \frac{\frac{4}{3} \pi \left(r_-^3 + 3r_+^3\right)}{8r_-^3}$$ $$= \left[ \frac{\pi}{6} \left(1 + 3(0.414)^3\right) \right]$$ $$= 0.63$$

[$\mathrm{Cr(NH_3)_6}$]^{3+} $\rightarrow$ $\mathrm{Cr^{3+}}$ $\Rightarrow$ [$\mathrm{Ar}$]3d^3 In presence of NH_3 ligand $\square$ $\square$ e_g $\boxed{\uparrow \uparrow \uparrow}$ t_{2g} Number of unpaired electrons = 3 $\mu$ = $\sqrt{n(n+2)}$ B.M. $\mu$ = $\sqrt{3(3+2)}$ B.M. $\mu$ = $\sqrt{15}$ B.M. $\Rightarrow$ 3.87 B.M. [$\mathrm{CuF_6}$]^{3-} $\mathrm{Cu^{3+}}$ $\Rightarrow$ [$\mathrm{Ar}$]3d^8 In presence of $\mathrm{F^-}$ Ligand $\mathrm{Cu^{3+}}$ $\Rightarrow$ $\boxed{\uparrow \uparrow}$ e_g $\boxed{\downarrow \downarrow \downarrow \downarrow}$ t_{2g} Number of unpaired electrons = 2 $\mu$ = $\sqrt{n(n+2)}$ B.M. $\mu$ = $\sqrt{2(2+2)}$ $\Rightarrow$ $\sqrt{8}$ B.M. $\Rightarrow$ 2.84 B.M.

The reaction is given by: $$\mathrm{Mg_2C_3 + 4H_2O \rightarrow 2Mg(OH)_2 + CH_3C \equiv CH}$$ The reaction of acetylene with sodium amide and methyl iodide is: $$\mathrm{CH_3C \equiv CH \xrightarrow{NaMe_2} CH_3 - C \equiv CNa^+ \xrightarrow{MeI} CH_3 - C \equiv C - CH_3 (0.075 mmol)}$$ This compound (Q) undergoes further reaction: $$\mathrm{3CH_3 - C \equiv C - CH_3 \xrightarrow{red hot iron tube, 873K}}$$ The product is: $$\mathrm{(0.075 \times 0.4) \times = 0.01 mole}$$ The value of $x$ is calculated as: $$x = 162 \times 0.01 = 1.62 gm$$

Given (P) 0.1 mole. Using Kucherov reaction with $Hg^{2+}/H^+$ at 333K, 100%, we obtain $CH_3C=CH_3$ (0.01 mole). This undergoes a reaction with $Ba(OH)_2/\Delta$ to form $CH_3C=CH-CH_3$ (0.04 mole) and further reacts to form $CH_3C=CH-OH + CHCl_3$. Calculating the moles: $$0.1 \times \frac{82}{100} \times \frac{1}{2} = 0.04 mole$$ Further calculation: $$0.04 \times \frac{80}{100} = 0.032 mole$$ The sum $60 + 32 + 8 = 100$. The value of $Y = 0.032 \times 100 = 3.2$.

Given $\Delta G^\circ = -RT \ln \left( \frac{P}{1} \right) = \Delta H^\circ - T \Delta S^\circ$. $\newline$ $\ln$ $\left$( $\frac{P}{1}$ $\right$) = -$\frac{\Delta H^\circ}{RT}$ + $\frac{\Delta S^\circ}{R}$$. \newline Slope $= -$\frac{\Delta H^\circ}{R}$ = 10^4 $\times$ $\left$( $\frac{-4}{2}$ $\right$)$. \newline \Rightarrow \Delta H^\circ = 2 \times 10^4 \times R = 166.28 \, \mathrm{kJ/mole}$

From the plot when, $\frac{10^4}{T} = 10$ implies $T = 1000 \, \mathrm{K}$. $$\ln \left( \frac{P_2}{1} \right) = -3$$ Substituting in equation: $$\ln \left( \frac{P_2}{1} \right) = -\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}$$ We get, $$-3 = -\frac{2 \times 10^4 \times R}{R \times 1000} + \frac{\Delta S^o}{R}$$ $$\Rightarrow \Delta S^o = 17R$$ $$\Rightarrow \Delta S^o = 17 \times 8.314 \, \mathrm{J/K\cdot mol}$$ $$\Rightarrow \Delta S^o = 141.34 \, \mathrm{J/K\cdot mol}$$

The dissociation of $\mathrm{AgNO_3(aq)}$ results in $\mathrm{Ag^+(aq)}$ and $\mathrm{NO_3^-(aq)}$ with concentrations of $0.1 \, \mathrm{m}$ each. The boiling point elevation $\Delta T_b$ is calculated as $0.2 \times 0.5 = 0.1^\circ \mathrm{C} = 0.1 \, \mathrm{K}$. Therefore, the boiling point of the solution is $100.1^\circ \mathrm{C}$, which is equal to $X$.

AgNO_3($\mathrm{aq}$) $\longrightarrow$ $\mathrm{Ag^+ (aq)}$ + $\mathrm{NO_3^- (aq)}$ 0.05 $\,$ $\mathrm{m}$ 0.05 $\,$ $\mathrm{m}$ 0.05 $\,$ $\mathrm{m}$ $\mathrm{BaCl_2(aq)}$ $\longrightarrow$ $\mathrm{Ba^{2+}(aq)}$ + 2$\mathrm{Cl^-(aq)}$ 0.05 $\,$ $\mathrm{m}$ 0.05 $\,$ $\mathrm{m}$ 0.1 $\,$ $\mathrm{m}$ $\mathrm{Ag^+}$ and $\mathrm{Cl^-}$ combine to form AgCl precipitate $\mathrm{Ag^+(aq)}$ + $\mathrm{Cl^-(aq)}$ $\longrightarrow$ $\mathrm{AgCl(s)}$ t = 0 0.05 $\,$ $\mathrm{m}$ 0.1 $\,$ $\mathrm{m}$ t = $\infty$ 0 0.05 $\,$ $\mathrm{m}$ In final solution total concentration of all ions: [$\mathrm{Cl^-}$] + [$\mathrm{NO_3^-}$] + [$\mathrm{Ba^{2+}}$] = 0.05 + 0.05 + 0.05 = 0.15 $\,$ $\mathrm{m}$ $\Delta$ T_b = 0.5 $\times$ 0.15 = 0.075 $\,$ ^$\circ$ $\mathrm{C}$ B.P. of solution 'B' = 100.075^$\circ$ $\mathrm{C}$ B.P. of solution 'A' = 100.1^$\circ$ $\mathrm{C}$ |y| = 100.1 - 100.075 = 0.025 = 2.5 $\times$ 10^{-2}

The enantiomer of P has a rotation of $-52.7^\circ$ as follows. The compound with $[\alpha]_D = 52.7^\circ$ is shown. The reaction with dil. $\mathrm{HNO_3}$ leads to the formation of the enantiomer with the opposite rotation.

Both options (A) and (D) give a positive Carius test.

(A) Process of precipitating colloidal solution is called coagulation. Hence false. (B) For colloidal solutions concentration is very small due to very large molar mass and hence their colligative properties are very small as compared to true solutions. Therefore, $\Delta T_f$ is lesser for colloidal solution. Hence true. (C) At CMC surfactant form micelles. Hence true. (D) Micelles and macromolecular colloids are two different types of colloids. Hence false.

From state I to II (Reversible isothermal expansion) P decreases, V increases, T constant H constant & S increases. From state II to III (Reversible adiabatic expansion) P decreases, V increases, T decreases H decreases, S constant Therefore, plots (A), (B), (D) are correct while (C) is wrong as from II to III, H is decreasing.

(A) $\mathrm{PbS} + 2\mathrm{PbO} \rightarrow 3\mathrm{Pb} + \mathrm{SO_2}$ (self reduction) (B) Silica is added to remove impurity of Fe in the form of slag $\mathrm{FeSiO_3}$ (C) $\mathrm{CuFeS_2}$ ore is partially oxidized first by roasting and then self reduction of Cu takes place to produce blister copper. (D) $4 \mathrm{Na} [\mathrm{Au} (\mathrm{CN})_2] + 2 \mathrm{Zn} \rightarrow 2 \mathrm{Na}_2[\mathrm{Zn(CN)}_4] + 4 \mathrm{Au}$

Pb(NO_3)_2 $\xrightarrow{dil. HCl}$ PbCl_2 $\downarrow$ White Ppt. Bi(NO_3)_3 $\xrightarrow{dil. HCl}$ BiCl_3 Water Soluble Hg(NO_3)_2 $\xrightarrow{dil. HCl}$ HgCl_2 Water Soluble AgNO_3 $\xrightarrow{dil. HCl}$ AgCl $\downarrow$ White Ppt. Zn(NO_3)_2 $\xrightarrow{dil. HCl}$ ZnCl_2 Water Soluble Pb(NO_3)_2 $\xrightarrow{NaOH(dil.)}$ Pb(OH)_2 $\downarrow$ White Ppt. Zn(NO_3)_2 $\xrightarrow{NaOH(dil.)}$ Zn(OH)_2 $\downarrow$ White Ppt. Bi(NO_3)_3 $\xrightarrow{NaOH(dil.)}$ Bi(OH)_3 $\downarrow$ White Ppt. AgNO_3 $\xrightarrow{NaOH(dil.)}$ Ag_2O $\downarrow$ Brown Ppt. Hg(NO_3)_2 $\xrightarrow{NaOH(dil.)}$ HgO $\downarrow$ Yellow Ppt.

The reaction begins with the bromination of the methylcyclohexane using $\mathrm{Br_2}$ and $hv$. This results in the formation of several products. 1. The first product is formed by the substitution of a hydrogen atom with a bromine atom on the methyl group, resulting in $\mathrm{CH_2Br}$. 2. The second product involves the formation of a radical on the methyl group, which then reacts with $\mathrm{Br_2}$ to form $\mathrm{CH_2Br}$ with $R/S$ configuration. 3. Another product is formed by the substitution on the ring, resulting in a bromine atom attached to the ring with $R/S$ configuration. 4. Additional products are formed by further substitution reactions on the ring and the methyl group, each resulting in different stereochemical configurations. 5. The total number of distinct products formed is 13, considering all stereochemical variations.

The reaction involves the conversion of the given compound using $\mathrm{O_3}$ (excess) followed by $\mathrm{Zn/H_2O}$ to form an intermediate. This intermediate is then reacted with $\mathrm{NH_2OH}$ to form the final product. Total 12 atoms are $sp^2$ hybridised.