JEE Main 22 July 2021 Shift 1 question paper with solutions

$P_1:\ x-y+2z=2$ $P_2:\ 2x+y-3=2$ Let line of Intersection of planes $P_1$ and $P_2$ cuts $xy$ plane in point $Q$. $\Rightarrow$ $z$-coordinate of point $Q$ is zero $\Rightarrow \begin{cases} x - y = 2 \\ 2x + y = 2 \end{cases}$ $\Rightarrow x = \frac{4}{3}, y = \frac{-2}{3}$ $\Rightarrow Q \left( \frac{4}{3}, \frac{-2}{3}, 0 \right)$ Vector parallel to the line of intersection $\vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 1 & -1 \end{vmatrix} = -\hat{i} + 5\hat{j} + 3\hat{k}$ Equation of Line of intersection $$\frac{x - \frac{4}{3}}{-1} = \frac{y + \frac{2}{3}}{5} = \frac{z - 0}{3} = \lambda (say)$$ Let coordinates of foot of perpendicular be $F \left( -\lambda + \frac{4}{3}, 5\lambda - \frac{2}{3}, 3\lambda \right)$ $\overrightarrow{PF} = \left( -\lambda + \frac{1}{3} \right) \hat{i} + \left( 5\lambda - \frac{8}{3} \right) \hat{j} + (3\lambda) \hat{k}$ $\overrightarrow{PF} \cdot \vec{a} = 0$ $\Rightarrow \lambda - \frac{1}{3} + 25\lambda - \frac{40}{3} + 9\lambda = 0$ $\Rightarrow 35\lambda = \frac{41}{3} \Rightarrow \lambda = \frac{41}{105}$ Now, $\alpha = -\lambda + \frac{4}{3}, \beta = 5\lambda - \frac{2}{3}, \gamma = 3\lambda$ $\Rightarrow \alpha + \beta + \gamma = 7\lambda + \frac{2}{3}$ $= 7 \left( \frac{41}{105} \right) + \frac{2}{3}$ $= \frac{51}{15}$ $\Rightarrow 35(\alpha + \beta + \gamma) = \frac{51}{15} \times 35 = 119$

Given $S_{10} = 530$, we have $$\frac{10}{2} \{2a + 9d\} = 530$$ which implies $$2a + 9d = 106 \ldots (1)$$ and $S_5 = 140$, so $$\frac{5}{2} \{2a + 4d\} = 140$$ which implies $$2a + 4d = 56 \ldots (2)$$ Solving equations (1) and (2), we get $$5d = 50 \Rightarrow d = 10$$ and $$a = 8$$ Now, $$S_{20} - S_6 = \frac{20}{2} \{2a + 19d\} - \frac{6}{2} \{2a + 5d\}$$ Simplifying, we get $$= 14a + 175d$$ Substituting the values, $$= (14 \times 8) + (175 \times 10)$$ $$= 1862$$

For $x > 0$, $f'(x) = -4x^2 + 4x + 3$. $f(x)$ is increasing in $\left(-\frac{1}{2}, \frac{3}{2}\right)$. For $x \leq 0$, $f'(x) = 3e^x(1 + x)$. $f'(x) > 0 \forall x \in (-1, 0)$ $\Rightarrow f(x)$ is increasing in $(-1, 0)$. So, in complete domain $f(x)$ is increasing in $\left(-1, \frac{3}{2}\right)$.

Given $\($ $\frac{dy}{dx}$ + 2 $\sin$^2 x = 1 + y $\cos$ 2x $\)$ $\($ $\Rightarrow$ $\frac{dy}{dx}$ + (- $\cos$ 2x)y = $\cos$ 2x $\)$ I.F. = $\($ e^{$\int$ - $\cos$ 2x $\,$ dx} = e^{- $\frac{\sin 2x}{2}$} $\)$ Solution of D.E. $\($ y $\left$( e^{- $\frac{\sin 2x}{2}$} $\right$) = $\int$ ($\cos$ 2x) $\left$( e^{- $\frac{\sin 2x}{2}$} $\right$) dx + c $\)$ $\($ $\Rightarrow$ y $\left$( e^{- $\frac{\sin 2x}{2}$} $\right$) = -e^{- $\frac{\sin 2x}{2}$} + c $\)$ Given $\($ y $\left$( $\frac{\pi}{4}$ $\right$) = 0 $\)$ $\($ $\Rightarrow$ 0 = -e^{-1/2} + c $\Rightarrow$ c = e^{-1/2} $\)$ $\($ $\Rightarrow$ y $\left$( e^{- $\frac{\sin 2x}{2}$} $\right$) = -e^{- $\frac{\sin 2x}{2}$} + e^{-1/2} $\)$ at $\($ x = 0 $\)$ $\($ y = -1 + e^{-1/2} $\)$ $\($ $\Rightarrow$ y(0) = -1 + e^{-1/2} $\Rightarrow$ (y(0) + 1)^2 = e^{-1} $\)$

$A=\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ $|A|=ad-bc$ Total cases $=6^4$ For a non-singular matrix, $|A|\neq0$ $\Rightarrow ad-bc\neq0$ $\Rightarrow ad\neq bc$ And $a,\;b,\;c,\;d$ are all different numbers in the set $\{1,2,3,4,5,6\}$. Now for $ad=bc$ (i) $6\times1=2\times3$ $\Rightarrow a=6,\ b=2,\ c=3,\ d=1$ or $a=1,\ b=2,\ c=3,\ d=6$ $\Rightarrow 8$ such cases (ii) $6\times2=3\times4$ $\Rightarrow a=6,\ b=3,\ c=4,\ d=2$ or $a=2,\ b=3,\ c=4,\ d=6$ $\Rightarrow 8$ such cases Favourable cases $={}^{6}C_{4}\times4!-16$ Required probability $=\dfrac{{}^{6}C_{4}\times4!-16}{6^4}$ $=\dfrac{43}{162}$

Given $\vec{a} = \lambda \vec{b} + \mu \vec{c} = \hat{i}(2\lambda + \mu) + \hat{j}(\lambda - \mu) + \hat{k}(\lambda + \mu)$. The dot product $\vec{a} \cdot \vec{d} = 0 = 3(2\lambda + \mu) + 2(\lambda - \mu) + 6(\lambda + \mu)$. This implies $14\lambda + 7\mu = 0 \Rightarrow \mu = -2\lambda$. Thus, $\vec{a} = (0)\hat{i} - 3\lambda \hat{j} + (-\lambda)\hat{k}$. The magnitude $|\vec{a}| = \sqrt{10}|\lambda| = \sqrt{10} \Rightarrow |\lambda| = 1$. Therefore, $\lambda = 1$ or $-1$. The determinant $[\vec{a} \vec{b} \vec{c}] = 0$. Also, $[\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{d}] + [\vec{a} \vec{c} \vec{d}] = [\vec{a} \vec{b} + \vec{c} \vec{d}]$. The matrix is: $$\begin{vmatrix} 0 & -3\lambda & \lambda \\ 3 & 0 & 2 \\ 3 & 2 & 6 \end{vmatrix}$$ This equals $3\lambda(12) + \lambda(6) = 42\lambda = -42$.

Given $$I = \int_0^{100\pi} \frac{\sin^2 x}{e^{x/\pi}} \, dx = 100 \int_0^{\pi} \frac{\sin^2 x}{e^{x/\pi}} \, dx$$ We have $$100 \int_0^{\pi} e^{-x/\pi} \frac{(1 - \cos 2x)}{2} \, dx$$ This simplifies to $$= 50 \left\{ \int_0^{\pi} e^{-x/\pi} \, dx - \int_0^{\pi} e^{-x/\pi} \cos 2x \, dx \right\}$$ Let $$I_1 = \int_0^{\pi} e^{-x/\pi} \, dx = \left[ -\pi e^{-x/\pi} \right]_0^{\pi} = \pi \left( 1 - e^{-1} \right)$$ And $$I_2 = \int_0^{\pi} e^{-x/\pi} \cos 2x \, dx$$ Then $$= -\pi e^{-x/\pi} \cos 2x \bigg|_0^{\pi} - \int_0^{\pi} -\pi e^{-x/\pi} (-2 \sin 2x) \, dx$$ This gives $$= \pi \left( 1 - e^{-1} \right) - 2\pi \int_0^{\pi} e^{-x/\pi} \sin 2x \, dx$$ Thus $$\pi \left( 1 - e^{-1} \right) - 2\pi \left\{ -\pi e^{-x/\pi} \sin 2x \right\}_0^{\pi} - \int_0^{\pi} -\pi e^{-x/\pi} 2 \cos 2x \, dx$$ Finally, $$= \pi \left( 1 - e^{-1} \right) - 4\pi^2 I_2$$ Therefore, $$\Rightarrow I_2 = \frac{\pi (1 - e^{-1})}{1 + 4\pi^2}$$ Thus, $$\therefore I = 50 \left\{ \pi \left( 1 - e^{-1} \right) - \frac{\pi (1 - e^{-1})}{1 + 4\pi^2} \right\}$$ This simplifies to $$= \frac{200 (1 - e^{-1}) \pi^3}{1 + 4\pi^2}$$

(1) $\vec{a} \times ((\vec{b} + \vec{c}) \times (\vec{b} - \vec{c})) = \vec{a} \times (-\vec{b} \times \vec{c} + \vec{c} \times \vec{b}) = -2(\vec{a} \times (\vec{b} \times \vec{c})) = -2(\vec{a} \times \vec{a}) = \vec{0}$ (2) Projection of $\vec{a}$ on $\vec{b} \times \vec{c}$ $$= \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|} = \frac{\vec{a} \cdot \vec{a}}{|\vec{a}|} = |\vec{a}| = 2$$ (3) $[\vec{a}\vec{b}\vec{c}] + [\vec{c}\vec{a}\vec{b}] = 2[\vec{a}\vec{b}\vec{c}] = 2\vec{a} \cdot (\vec{b} \times \vec{c})$ $$= 2 \vec{a} \cdot \vec{a} = 2 |\vec{a}|^2 = 8$$ (4) $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{b} \times \vec{c} = \vec{a}$ Therefore, $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually perpendicular vectors. Thus, $|\vec{a} \times \vec{b}| = |\vec{c}| \Rightarrow |\vec{a}||\vec{b}| = |\vec{c}| \Rightarrow |\vec{b}| = |\vec{c}|/2$ Also, $|\vec{b} \times \vec{c}| = |\vec{a}| \Rightarrow |\vec{b}||\vec{c}| = 2 \Rightarrow |\vec{c}| = 2&|\vec{b}| = 1$ $|3\vec{a} + \vec{b} - 2\vec{c}|^2 = (3\vec{a} + \vec{b} - 2\vec{c}) \cdot (3\vec{a} + \vec{b} - 2\vec{c})$ $$= 9|\vec{a}|^2 + |\vec{b}|^2 + 4|\vec{c}|^2$$ $$= (9 \times 4) + 1 + (4 \times 4)$$ $$= 36 + 1 + 16 = 53$$

Given the equations: $$x + y + z = 6 ...(i)$$ $$3x + 5y + 5z = 26 ...(ii)$$ $$x + 2y + \lambda z = \mu ...(iii)$$ Multiplying equation (i) by 5 and subtracting equation (ii) gives: $$5 \times (i) - (ii) \Rightarrow 2x = 4 \Rightarrow x = 2$$ From equations (i) and (ii), we have: $$y + z = 4 ...(iv)$$ Substituting $x = 2$ into equation (iii) gives: $$2y + \lambda z = \mu - 2 ...(v)$$ Subtracting twice equation (iv) from equation (v) gives: $$(v) - 2 \times (iv) \Rightarrow (\lambda - 2)z = \mu - 10$$ Solving for $z$ and $y$ gives: $$z = \frac{\mu - 10}{\lambda - 2} \& y = 4 - \frac{\mu - 10}{\lambda - 2}$$ For no solution, $\lambda = 2$ and $\mu \neq 10$.

Given the lines: $$L_1: \frac{(x-1)}{2} = \frac{(y-2)}{1} = \frac{(z-1)}{3}$$ $$\vec{r_1} = 2\hat{i} + \hat{j} + 3\hat{k}$$ $$L_2: \frac{(x-2)}{1} = \frac{y-\lambda}{2} = \frac{z-3}{4}$$ $$\vec{r_2} = \hat{i} + 2\hat{j} + 4\hat{k}$$ Shortest distance is the projection of $\vec{a}$ on $\vec{r_1} \times \vec{r_2}$. $$= \frac{\left| \vec{a} \cdot (\vec{r_1} \times \vec{r_2}) \right|}{|\vec{r_1} \times \vec{r_2}|}$$ $$= \left| \begin{vmatrix} 1 & \lambda - 2 & 2 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix} \right| = |14 - 5\lambda|$$ $$|\vec{r_1} \times \vec{r_2}| = \sqrt{38}$$ Therefore, $$\frac{1}{\sqrt{38}} \cdot \frac{|14 - 5\lambda|}{\sqrt{38}} = 1$$ $$\Rightarrow |14 - 5\lambda| = 1$$ $$\Rightarrow 14 - 5\lambda = 1 or 14 - 5\lambda = -1$$ $$\Rightarrow \lambda = \frac{13}{5} or 3$$ Therefore, the integral value of $\lambda$ is 3.

Given the expressions: (1) $(p \rightarrow q) \lor (\sim q \rightarrow p)$ $$= (\sim p \lor q) \lor (q \lor p)$$ $$= (\sim p \lor p) \lor q$$ $$= t \lor q = t$$ (2) $(q \rightarrow p) \lor (\sim q \rightarrow p)$ $$= (\sim q \lor p) \lor (q \lor p)$$ $$= (\sim q \lor q) \lor p$$ $$= t \lor p = t$$ (3) $(p \rightarrow \sim q) \lor (\sim q \rightarrow p)$ $$= (\sim p \lor \sim q) \lor (q \lor p)$$ $$= (\sim p \lor q) \lor (\sim q \lor q)$$ $$= t \lor t = t$$ (4) $(\sim q \rightarrow q) \lor (\sim q \rightarrow p)$ $$= (p \lor q) \lor (q \lor p)$$ $$= (p \lor p) \lor (q \lor p)$$ $$= p \lor q$$ Which is not a tautology.

Given $$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$ Let $$x = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$ Then $$AX = \begin{bmatrix} a_{11} + a_{12} + a_{13} \\ a_{21} + a_{22} + a_{23} \\ a_{31} + a_{32} + a_{33} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$ Thus, $AX = X$. Replace $X$ by $AX$, $A^2 X = AX = X$. Replace $X$ by $AX$, $A^3 X = AX = X$. Let $$A^3 = \begin{bmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ z_1 & z_2 & z_3 \end{bmatrix}$$ Then $$A^3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} x_1 + x_2 + x_3 \\ y_1 + y_2 + y_3 \\ z_1 + z_2 + z_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$ Sum of all the elements = 3.

Given $[e^x]^2 + [e^x + 1] - 3 = 0$. This implies $[e^x]^2 + [e^x] + 1 - 3 = 0$. Let $[e^x] = t$. Then $t^2 + t - 2 = 0$. Solving, we get $t = -2, 1$. For $[e^x] = -2$ (Not possible). Or $[e^x] = 1$, therefore $1 \leq e^x < 2$. This implies $\ln(1) \leq x < \ln(2)$. Thus, $0 \leq x < \ln(2)$. Therefore, $x \in [0, \ln 2)$.

Given the equation $S : 36x^2 + 36y^2 - 108x + 120y + C = 0$. This simplifies to $x^2 + y^2 - 3x + \frac{10}{3}y + \frac{C}{36} = 0$. The centre is $(-g, -f) \equiv \left( \frac{3}{2}, -\frac{10}{6} \right)$ and the radius is $r = \sqrt{\frac{9}{4} + \frac{100}{36} - \frac{C}{36}}$. Now, $r 100$. (1) Now the point of intersection of $x - 2y = 4$ and $2x - y = 5$ is $(2, -1)$, which lies inside the circle $S$. Therefore, $S(2, -1) < 0$ implies $(2)^2 + (-1)^2 - 3(2) + \frac{10}{3}(-1) + \frac{C}{36} < 0$, which simplifies to $4 + 1 - 6 - \frac{10}{3} + \frac{C}{36} < 0$. This gives $C < 156$. (2) From (1) and (2), $100 < C < 156$.

Given $z^2 + 3z = 0$. Put $z = x + iy$. This implies $$x^2 - y^2 + 2ixy + 3(x - iy) = 0$$ $$\Rightarrow (x^2 - y^2 + 3x) + i(2xy - 3y) = 0 + i0$$ Therefore, $$x^2 - y^2 + 3x = 0 \ldots (1)$$ $$2xy - 3y = 0 \ldots (2)$$ From equation (2), $x = \frac{3}{2}, y = 0$. Put $x = \frac{3}{2}$ in equation (1): $$\frac{9}{4} - y^2 + \frac{9}{2} = 0$$ $$y^2 = \frac{27}{4} \Rightarrow y = \pm \frac{3\sqrt{3}}{2}$$ Therefore, $$(x, y) = \left( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right), \left( \frac{3}{2}, -\frac{3\sqrt{3}}{2} \right)$$ Put $y = 0$ to get $$x^2 - 0 + 3x = 0$$ $$x = 0, -3$$ Therefore, $$(x, y) = (0, 0), (-3, 0)$$ Thus, the number of solutions $= n = 4$ $$\sum_{K=0}^{\infty} \left( \frac{1}{n^k} \right) = \sum_{K=0}^{\infty} \left( \frac{1}{4^k} \right)$$ $$= \frac{1}{1} + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots$$ $$= \frac{1}{1 - \frac{1}{4}} = \frac{4}{3}$$

Given $\sin^7 x \leq \sin^2 x \leq 1 \ldots (1)$ and $\cos^7 x \leq \cos^2 x \leq 1 \ldots (2)$ also $\sin^2 x + \cos^2 x = 1$. Therefore, equality must hold for $(1)$ and $(2)$, which implies $\sin^7 x = \sin^2 x$ and $\cos^7 x = \cos^2 x$. This leads to $\sin x = 0$ and $\cos x = 1$ or $\cos x = 0$ and $\sin x = 1$. Thus, $x = 0, 2\pi, 4\pi, \frac{\pi}{2}, \frac{5\pi}{2}$. Hence, there are 5 solutions.

Given $0 \leq x^2 - x + 1 \leq 1$. This implies $x^2 - x \leq 0$. Therefore, $x \in [0, 1]$. Also, $0 < \sin^{-1}\left(\frac{2x-1}{2}\right) \leq \frac{\pi}{2}$. This implies $0 < \frac{2x-1}{2} \leq 1$. Therefore, $0 < 2x - 1 \leq 2$. This gives $1 < 2x \leq 3$. Hence, $\frac{1}{2} < x \leq \frac{3}{2}$. Taking intersection $x \in \left(\frac{1}{2}, 1\right]$. Therefore, $\alpha = \frac{1}{2}, \beta = 1$.

For continuity $\lim_{x\to0} \dfrac{x^3}{4\sin^4x} \left(\ln(1+12xe^{-2x})-2\ln(1-xe^{-x})\right) =\alpha$ $\lim_{x\to0} \dfrac{1}{4x} \left[2xe^{-2x}+2xe^{-x}\right] =\alpha$ $=\dfrac14(4)$ $\therefore\ \alpha=1$

Tangent to hyperbola of Slope $m = -2$ (given) $$y = -2x \pm \sqrt{3}(3)$$ $$\left( y = mx \pm \sqrt{x^2 m^2 - b^2} \right)$$ $$\Rightarrow y + 2x = \pm 3 \Rightarrow 2x + y = 3 (k > 0)$$ For parabola $y^2 = \alpha x$ $$y = mx + \frac{\alpha}{4m}$$ $$\Rightarrow y = -2x + \frac{\alpha}{8}$$ $$\Rightarrow \frac{\alpha}{8} = 3$$ $$\Rightarrow \alpha = -24.$$

Given the equations: $$e^2 = 1 - \frac{b^2}{a^2}$$ $$e^2 = 1 - \frac{a^2}{c^2}$$ This implies: $$\frac{b^2}{a^2} = \frac{a^2}{c^2}$$ Therefore, $$c^2 = \frac{a^4}{b^2} \Rightarrow c = \frac{a^2}{b}$$ Also, $b = ce$. Thus, $$c = \frac{b}{e}$$ $$\frac{b}{c} = \frac{a^2}{b}$$ This implies: $$e = \frac{b^2}{a^2} = 1 - e^2$$ Therefore, $$e^2 + e - 1 = 0$$ Solving for $e$, $$e = \frac{-1 + \sqrt{5}}{2}$$

Given $f(1) + f(2) = 3 - f(3)$. Therefore, $f(1) + f(2) = 3 + f(3) = 3$. The only possibility is: $0 + 1 + 2 = 3$. Therefore, elements $1, 2, 3$ in the domain can be mapped with $0, 1, 2$ only. So the number of bijective functions is $3 \times 5! = 720$.

The numbers available are 0, 2, 4, 6, 8. The sequence chosen is 2, 4, 6, 8. The number of choices for each position is 4, 4, 3, 2, 1 respectively. Therefore, the total number of combinations is calculated as follows: $$4 \times 4 \times 3 \times 2 = 96.$$

Let matrix $B = \begin{bmatrix} a & b & c \\ d & e & f \\ g & n & i \end{bmatrix}$. Therefore, $AB = BA$. $$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$\begin{bmatrix} d & e & f \\ a & b & c \\ g & h & i \end{bmatrix} = \begin{bmatrix} b & a & c \\ e & d & f \\ h & g & i \end{bmatrix}$$ This implies $d = b$, $e = a$, $f = c$, $g = h$. Therefore, Matrix $B = \begin{bmatrix} a & b & c \\ b & a & c \\ g & g & i \end{bmatrix}$. Number of ways of selecting $a, b, c, g, i$ is $5 \times 5 \times 5 \times 5 \times 5 = 5^5 = 3125$. Therefore, the number of matrices $B = 3125$.

\begin{tabular}{|c|c|c|c|} \hline \text{Class} & \text{Frequency} & $X_i$ & $F_i$ $X_i$ \\ \hline 0-6 & a & 3 & 3a \\ \hline 6-12 & b & 9 & 9b \\ \hline 12-18 & 12 & 15 & 180 \\ \hline 18-24 & 9 & 21 & 189 \\ \hline 24-30 & 5 & 27 & 135 \\ \hline & N=26+a+b & & (504+3a+9b) \\ \hline \end{tabular} Mean $\displaystyle = \frac{3a+9b+180+189+135}{a+b+26}=\frac{309}{22}$ $\Rightarrow 66a+198b+11088=309a+309b+8034$ $\Rightarrow 243a+111b=3054$ $\Rightarrow 81a+37b=1018 \qquad \ldots (1)$ Now, Median $\displaystyle = 12+\frac{\frac{a+b+26}{2}-(a+b)}{12}\times 6=14$ $\Rightarrow \frac{13}{2}-\frac{a+b}{4}=2$ $\Rightarrow \frac{a+b}{4}=\frac{9}{2}$ $\Rightarrow a+b=18 \qquad \ldots (2)$ From equations (1) and (2), $a=8,\; b=10$ $\therefore (a-b)^2=(8-10)^2=4$

Given $2040 = 2^3 \times 3 \times 5 \times 17$. $n$ should not be a multiple of $2$, $3$, $5$, and $17$. Sum of all $n = (1 + 3 + 5 + \ldots + 99) - (3 + 9 + 15 + 21 + \ldots + 99) - (5 + 25 + 35 + 55 + 65 + 85 + 95) - (17)$ $$= 2500 - \frac{17}{2}(3 + 99) - 365 - 17$$ $$= 2500 - 867 - 365 - 17$$ $$= 1251$$

Required Area (shaded) $$= 2 \left[ \int_0^2 \left( \frac{4 - y^2}{4} \right) \, dy - \int_0^1 \left( \frac{1 - x^2}{2} \right) \, dx \right]$$ $$= 2 \left[ \frac{4}{3} - \frac{1}{3} \right] = (2)$$

Given $f(x-2) = \begin{cases} \frac{3x}{2} & -4 \leq x \leq -2 \\ -\frac{3x}{2} & -2 < x \leq 0 \\ 0 & x \in (-\infty, -4) \cup (0, +\infty) \end{cases}$ $$f(x+2) = \begin{cases} \frac{3x}{2} + 6 & 0 \leq x \leq 2 \\ -\frac{3x}{2} + 6 & 2 < x \leq 4 \\ 0 & x \in (-\infty, 0) \cup (4, +\infty) \end{cases}$$ $$g(x) = f(x+2) - f(x-2) = \begin{cases} \frac{3x}{2} + 6 - \frac{3x}{2} & -4 \leq x \leq -2 \\ -\frac{3x}{2} - (-\frac{3x}{2}) & -2 < x < 2 \\ \frac{3x}{2} - 6 & 2 \leq x \leq 4 \\ 0 & x \in (-\infty, -4) \cup (4, +\infty) \end{cases}$$ From the graph, $$n = 0$$ $$m = 4 \Rightarrow (n + m = 4)$

General term = $^{10}C_R (2x^2)^{10-R} x^{-2R}$. $$\Rightarrow 2^{10-R} \, ^{10}C_R = 180 \ldots \ldots \ldots (1)$$ $\&(10-R)r - 2R = 0$ $$r = \frac{2R}{10-R}$$ $$r = \frac{2(R-10)}{10-R} + \frac{20}{10-R}$$ $$\Rightarrow r = -2 + \frac{20}{10-R} \ldots \ldots (2)$$ $R = 8$ or $5$ reject equation (1) not satisfied. At $R = 8$ $$2^{10-R} \, ^{10}C_R = 180 \Rightarrow r = 8$$ $2^{10-r}\,{}^{10}C_{r}=180$ $\Rightarrow r=8$

Given $y + 1 = Y \Rightarrow dy = dY$ and $x + 2 = X \Rightarrow dx = dX$. Therefore, $$\left( X e^{\frac{Y}{X}} + Y \right) dX = X dY$$ which implies $$X dY - Y dX = X e^{Y/X} dX$$ leading to $$d \left( \frac{Y}{X} \right) e^{-\frac{Y}{X}} = \frac{dX}{X}$$. Solving, we get $$-e^{-Y/X} = \ell |X| + c$$. For $(3, 2)$, $$-e^{-2/3} = \ell |3| + c$$. Thus, $$-e^{-\frac{t}{x}} = \ln |X| - e^{-\frac{2}{3}} - \ln 3$$. Therefore, $$e^{-\frac{Y}{X}} = e^{2/3} + \ln 3 - \ln |X| > 0$$. This implies $$\ln |X| < e^{2/3} + \ln 3 \rightarrow \lambda$$. Hence, $$\ln x + 2 < e^{\lambda}$$. Therefore, $$-e^{\lambda} < x + 2 < e^{\lambda}$$ which simplifies to $$-e^{\lambda} - 2 < x < e^{\lambda} - 2$$. Let $\alpha$ and $\beta$ be such that $$\alpha + \beta = -4 \Rightarrow |\alpha + \beta| = 4$$.

Given $11^n > 10^n + 9^n$. This implies $11^n - 9^n > 10^n$. This can be rewritten as $(10 + 1)^n - (10 - 1)^n > 10^n$. Expanding, we have $2\left\{ {^nC_1} \cdot 10^{n-1} + {^nC_3} \cdot 10^{n-3} + {^nC_5} \cdot 10^{n-5} + \cdots \right\} > 10^n$. This simplifies to $2n \cdot 10^{n-1} + 2 \left\{ {^nC_3} \cdot 10^{n-3} + {^nC_5} \cdot 10^{n-5} + \cdots \right\} > 10^n$. ...... (1) For $n = 5$ $10^5 + 2 \left\{ {^5C_3} \cdot 10^2 + {^5C_5} \right\} > 10^5$ (True) For $n = 6, 7, 8, \ldots 100$ $2n \cdot 10^{n-1} > 10^n$ This implies $2n \cdot 10^{n-1} + 2 \left\{ {^nC_3} \cdot 10^{n-3} + {^nC_5} \cdot 10^{n-5} + \cdots \right\} > 10^n$ Thus, $11^n - 9^n > 10^n$ for $n = 5, 6, 7, \ldots 100$ For $n = 4$, Inequality (1) is not satisfied $\Rightarrow$ Inequality does not hold good for $n = 1, 2, 3, 4$ So, required number of elements $= 96$

In a capacitor, current leads voltage by $\frac{\pi}{2}$.

Given $$R = \frac{R_1 R_2}{R_1 + R_2} = \frac{\ell}{A} \cdot \frac{\rho_1 \rho_2}{\rho_1 + \rho_2}$$ Substituting the values: $$R = \frac{25 \times 10^{-2}}{3 \times 10^{-6}} \times \frac{1.7 \times 2.6 \times 10^{-16}}{4.3 \times 10^{-8}}$$ Calculating gives $$R = 0.858 \, \mathrm{m\Omega}$$

(A $\cos$ $\theta$) $\hat{B}$ = A $\left$( $\frac{\vec{A} \cdot \vec{B}}{AB}$ $\right$) $\hat{B}$ = $\frac{\vec{A} \cdot \vec{B}}{B}$ $\hat{B}$ = $\frac{2}{\sqrt{2}}$ $\left$( $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$ $\right$) = $\hat{i}$ + $\hat{j}$

Given $W_{Porter} + W_{mg} = \Delta K.E. = 0$. $W_{Porter} = -W_{mg} = -mgh$. $$= -80 \times 9.8 \times 0.8 = -627.2 \, \mathrm{J}$$

Given $T_0 = 2\pi \sqrt{\frac{\ell}{g}}$. New time period $T = 2\pi \sqrt{\frac{\ell/16}{g}} = \frac{2\pi}{4} \sqrt{\frac{\ell}{g}}$. Therefore, $T = \frac{T_0}{4}$.

Given $r + r' + 90^\circ = 180^\circ$, we have $r' = 90 - r = 90 - i$. $n_1 \sin i = n_2 \sin r' = n_2 \sin(90 - i)$ $n_1 \sin i = n_2 \cos i \implies \tan i = \frac{n_2}{n_1}$ Now $\sin C = \frac{n_2}{n_1} = \tan i$ $\implies C = \sin^{-1}(\tan i) = \sin^{-1}(\tan r)$

As temperature increases, domains disintegrate so ferromagnetism decreases and above Curie temperature it becomes paramagnet.

By momentum conservation $$4 \times 10^{-3}(50 - v) - 4v = 0$$ $$v = \frac{4 \times 10^{-3} \times 50}{4 + 4 \times 10^{-3}} \approx 0.05 \, \mathrm{ms^{-1}}$$ Impulse $J = mv = 4 \times 0.05 = 0.2 \, \mathrm{kgms^{-1}}$

If the apparent dip circle is at an angle $\alpha$ with the true dip circle, then for the true dip circle: $$\tan \phi = \frac{B_V}{B_H}$$ For the apparent dip circle: $$\tan \phi' = \frac{B_V}{B_H \cos \alpha}$$ Thus, $$\tan \phi' = \frac{\tan \phi}{\cos \alpha}$$ As $\cos \alpha < 1$

The acceleration is given by $$a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}}$$ where $I$ is the moment of inertia. Comparing the moments of inertia, we have $$I_{ring} > I_{solid cylinder} > I_{solid sphere}$$ This implies $$a_{ring} < a_{solid cylinder} < a_{solid sphere}$$ Therefore, the velocities are $$v_{ring} < v_{solid cylinder} < v_{solid sphere}$$

Band gap is given by $\frac{hc}{\lambda_0}$. Here, $\lambda_0$ is the threshold wavelength. Band gap is calculated as: $$\frac{1242 \, \mathrm{eV} \cdot \mathrm{nm}}{621 \, \mathrm{nm}} = 2 \, \mathrm{eV}$$

The initial equation is given by: $$\frac{1}{2} (4m) v^2 + \frac{1}{2} (180m) \left( \frac{4v}{180} \right)^2 = 5.5 \, \mathrm{MeV}$$ This simplifies to: $$\frac{1}{2} 4mv^2 \left[ 1 + 45 \left( \frac{4}{180} \right)^2 \right] = 5.5 \, \mathrm{MeV}$$ Solving for $K \cdot E_\alpha$ gives: $$K \cdot E_\alpha = \frac{5.5}{1 + 45 \cdot \left( \frac{4}{180} \right)^2} \, \mathrm{MeV}$$ Finally, we find: $$K \cdot E_\alpha = 5.38 \, \mathrm{MeV}$$

Given $\mathrm{KE} = e \Delta V$. $$\lambda_e = \frac{h}{\sqrt{2 m_e (e \Delta V)}}$$ $$\lambda_P = \frac{h}{\sqrt{2 m_p (e \Delta V)}}$$ Therefore, $$\frac{\lambda_e}{\lambda_P} = \sqrt{\frac{m_p}{m_e}}$$

(a) For $x_L > x_C$, voltage leads the current (ii) (b) For $x_L = x_C$, voltage and current are in same phase (i) (c) For $x_L < x_C$, current leads the voltage (iv) (d) For resonant frequency $x_L = x_C$, current is maximum (iii)

Radius covered $r = \sqrt{2R H_T}$ $$150 \, \mathrm{km} = \sqrt{2 \times (6.5 \times 10^6 \, \mathrm{m}) \, H_T}$$ $$(150 \, \mathrm{km} \times 10^3)^2 = 2 \times 6.5 \times 10^6 H_T$$ $$H_T = 1731 \, \mathrm{m}$$ Population covered $= \left( \pi r^2 \right) \left( \frac{2000}{\mathrm{km}^2} \right)$ $$= 3.14 \times (150)^2 \times 2000 = 1413 \times 10^5$$

As per Equi-partition law: Each degree of freedom contributes $\frac{1}{2} k_B T$ Average Energy. In monoatomic gas D.O.F. = 3. Therefore, Average energy $= 3 \times \frac{1}{2} k_B T = \frac{3}{2} k_B T$.

Given $I_{avg} = \frac{B_0^2 C}{2 \mu_0} \& \frac{1}{\mu_0} = \epsilon_0 C^2$. $I = \frac{B_0^2}{2} \epsilon_0 C^3$. $B_0 = \sqrt{\frac{2I}{\epsilon_0 C^3}}$. $B_0 = 2.77 \times 10^{-8} \, \mathrm{T}$.

Given $x = A \sin \omega t + B \cos \omega t$. The velocity $v = \frac{dx}{dt} = A \omega \cos \omega t - B \omega \sin \omega t$. At $t = 0$, $x(0) = B$ and $v(0) = A \omega$. Therefore, $x = A \sin \omega t + B \sin (\omega t + 90^\circ)$. The net amplitude $A_{net} = \sqrt{A^2 + B^2}$. The tangent of angle $\alpha$ is $\tan \alpha = \frac{B}{A} \Rightarrow \cot \alpha = \frac{A}{B}$. Thus, $x = \sqrt{A^2 + B^2} \sin (\omega t + \alpha)$. Also, $x = \sqrt{A^2 + B^2} \cos (\omega t - (90^\circ - \alpha))$. Therefore, $x = C \cos (\omega t - \phi)$. This implies $C = \sqrt{A^2 + B^2}$. The expression for $C$ is $C = \sqrt{\frac{[v(0)]^2}{\omega^2} + [x(0)]^2}$. The phase $\phi = 90^\circ - \alpha$. Since $\tan \alpha = \cos \alpha = \frac{A}{B}$, it follows that $\tan \phi = \frac{v(0)}{x(0) \cdot \omega}$. Thus, $\phi = \tan^{-1} \left( \frac{v(0)}{x(0) \cdot \omega} \right)$.

Given $r = 10 \, \mathrm{mm}$, $x = 2$, $$\left| \vec{F}_{+q} \right| = \frac{2k\lambda}{r} \cdot q$$ $$\left| \vec{F}_{-q} \right| = \frac{2k\lambda}{r+x} \cdot q$$ $$\left| \vec{F}_{net} \right| = \frac{2k\lambda q}{r} - \frac{2k\lambda q}{r+x}$$ $$\left| \vec{F}_{net} \right| = \frac{2k\lambda q \cdot x}{r(r+x)}$$ $$4 = \frac{2 \times 9 \times 10^9 \times 3 \times 10^{-6} \times q \times 2 \, \mathrm{mm}}{10 \, \mathrm{mm} \cdot 12 \, \mathrm{mm}}$$ Therefore, $$q = 4.44 \, \mu \mathrm{C}$$

Applying energy conservation from (1) to (2) $$\frac{1}{2} \, m \cdot \left( \frac{2GM}{R_e} \right) - \frac{GMm}{R_e} = \frac{1}{2} \, mv^2 - \frac{GMm}{R + r}$$ $$\Rightarrow \frac{1}{2} \, mv^2 = \frac{GMm}{R + r}$$ $$\Rightarrow v = \sqrt{\frac{2GM}{R + r}} = \frac{dr}{dt}$$ $$\Rightarrow \sqrt{2GM} \int_0^t dt = \int_{R_e}^{R_{e+h}} (\sqrt{R + r}) dr$$ $$\sqrt{2GM} \cdot t = \frac{2}{3} \left[ (R + r)^{3/2} \right]_{R_e}^{R_e + h}$$ $$t = \frac{2}{3} \sqrt{\frac{R_e^3}{2GM}} \left[ \left( 1 + \frac{h}{R_e} \right)^{3/2} - 1 \right]$$ $$\frac{GM}{R_e^2} = g$$ $$t = \frac{1}{3} \sqrt{\frac{2R_e}{g}} \left[ \left( 1 + \frac{h}{R_e} \right)^{3/2} - 1 \right]$$

$\text{Voltage across } R_L = 5\,\text{V}$ $\Rightarrow i_2 = \dfrac{5}{R_L}$ $\text{Also voltage across } R = 50 - 5 = 45\,\text{volt}$ $\text{By } v = iR \Rightarrow R = \dfrac{v}{i} = \dfrac{45}{i_1 + i_2}$ $R = \dfrac{45}{90\,\text{mA} + \dfrac{5}{R_L}}$ $\text{Current in zener diode is maximum when } R_L \to \infty$ $(i_2 \to 0 \text{ and } i_1 = i)$ $\text{So } R = \dfrac{45}{90\,\text{mA}} = 500\,\Omega$

COM of semi-circular ring is at $\frac{2R}{\pi}$. Distance from centre $\Rightarrow x = 2$.

Terminal voltage $v = iR = \frac{ER}{R+r}$. First $\rightarrow 1.25 = \frac{E(5)}{5+r}$ $\ldots$ (i) Second $\rightarrow 1 = \frac{E(2)}{2+r}$ $\ldots$ (ii) By (i) and (ii) $r = 1 \Omega E = \frac{3}{2} V = \frac{15}{10}$ volt $\Rightarrow x = 15$

Electric flux density $$\vec{\left( \mathbf{D} \right)} = \frac{charge}{Area} \times \hat{\mathbf{r}} = \frac{Q}{4 \pi r^2} \hat{\mathbf{r}} = \varepsilon_0 \left( \frac{Q}{4 \pi \varepsilon_0 r^2} \right)$$ $$\Rightarrow \vec{\mathbf{E}} = \frac{\vec{\mathbf{D}}}{\varepsilon_0} = \frac{e^{-x} \sin y \hat{\mathbf{i}} - e^{-x} \cos y \hat{\mathbf{j}} + 2z \hat{\mathbf{k}}}{\varepsilon_0}$$ Also by Gauss's law $$\frac{\rho}{\varepsilon_0} = \left( \frac{\partial}{\partial x} \hat{\mathbf{i}} + \frac{\partial}{\partial y} \hat{\mathbf{j}} + \frac{\partial}{\partial z} \hat{\mathbf{k}} \right) \cdot \vec{\mathbf{E}} = \left( \frac{\partial}{\partial x} \hat{\mathbf{i}} + \frac{\partial}{\partial y} \hat{\mathbf{j}} + \frac{\partial}{\partial z} \hat{\mathbf{k}} \right) \cdot \frac{\vec{\mathbf{D}}}{\varepsilon_0}$$ $$\Rightarrow \rho = \frac{\partial}{\partial x} \left( e^{-x} \sin y \right) + \frac{\partial}{\partial y} \left( -e^{-x} \cos y \right) + \frac{\partial}{\partial z} \left( 2z \right)$$ $$\rho = -e^{-x} \sin y + e^{-x} \sin y + 2$$ At origin $\rho = -e^{0} \sin 0 + e^{0} \sin 0 + 2$ $$\rho = 2 \mathrm{C/m^3}$$ Charge $= \rho \times volume = 2 \times 2 \times 10^{-9} = 4 \times 10^{-9} =$ $$4 \mathrm{nC}$$

Direction of $\mathbf{P} \hat{\mathbf{v}}_1 = \pm \frac{\vec{\mathbf{A}} \times \vec{\mathbf{B}}}{|\vec{\mathbf{A}} \times \vec{\mathbf{B}}|} = \pm \frac{-\hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}}{\sqrt{3}}$ Direction of $\mathbf{Q} \hat{\mathbf{v}}_2 = \pm \frac{\vec{\mathbf{A}} \times \vec{\mathbf{C}}}{|\vec{\mathbf{A}} \times \vec{\mathbf{C}}|} = \pm \frac{2 \hat{\mathbf{k}}}{2} = \pm \hat{\mathbf{k}}$ Angle between $\hat{\mathbf{v}}_1$ and $\hat{\mathbf{v}}_2$ $$\frac{\hat{\mathbf{v}}_1 \cdot \hat{\mathbf{v}}_2}{|\hat{\mathbf{v}}_1||\hat{\mathbf{v}}_2|} = \frac{\pm 1/\sqrt{3}}{(1)(1)} = \pm \frac{1}{\sqrt{3}}$$ $\Rightarrow x = 3$

For no slipping $v_0 = \omega R$. Now $v_A = v_B = \sqrt{v_0^2 + (\omega R)^2} = \sqrt{2} v_0$. Therefore, $x = 2$.

At minimum deviation $r_1 = r_2 = \frac{A}{2}$. Also given $i = 2r_1 = A$. Now $\sin i = \sqrt{3} \sin r_1$. $\sin A = \sqrt{3} \sin \frac{A}{2}$. Therefore, $2 \sin \frac{A}{2} \cos \frac{A}{2} = \sqrt{3} \sin \frac{A}{2}$. This implies $\cos \frac{A}{2} = \frac{\sqrt{3}}{2} \Rightarrow \frac{A}{2} = 30^\circ$. Hence, $A = 60^\circ$.

Elastic energy = $\frac{Y}{2} (strain)^2 \times Area \times length$ Therefore, Elastic energy per unit length = $\frac{Y}{2} (strain)^2 \times Area$ $\left( strain = \frac{\Delta \ell}{\ell} = \alpha \Delta T = 10^{-5} \times 10 = 10^{-4} \right)$ $$= \frac{10^{11}}{2} \times (10^{-4})^2 \times 10^{-2} = 5 \, J/m$$

Given $T = 2\pi \sqrt{\frac{\ell}{g}} \implies g = \frac{4\pi^2 \ell}{T^2}$. $$\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + \frac{2 \Delta T}{T}$$ $\Delta T = \frac{least count of time (\Delta T_0)}{number of oscillations (n)}$ $$\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + \frac{2 \Delta T_0}{n T}$$ As $\Delta \ell$ and $\Delta T_0$ are same for all observations so $$\frac{\Delta g}{g} is minimum for highest value of \ell, n and T$$ Therefore, minimum percentage error in $g$ is for student number-1.

By Newton's law of cooling (with approximation) $$\frac{\Delta T}{\Delta t} = -C \left( T_{avg} - T_s \right)$$ First $$\frac{-10^\circ \mathrm{C}}{5 min} = -C \left( 70^\circ \mathrm{C} - 25^\circ \mathrm{C} \right)$$ $$\Rightarrow C = \frac{2}{45} min^{-1}$$ Second $$\frac{T - 65}{5 min} = -C \left( \frac{T + 65}{2} - 25 \right) = -\left( \frac{2}{45} \right) \left( \frac{T + 15}{2} \right)$$ $$\Rightarrow 9(T - 65) = -(T + 15)$$ $$\Rightarrow 10T = 570$$ $$\Rightarrow T = 57^\circ \mathrm{C}$$ Alternate Solution: Newton's law of cooling (without approximation) $$T_P - T_S = (T_i - T_S) e^{-Ct}$$ First $$65 - 25 = (75 - 25) e^{-5C} \Rightarrow e^{-5C} = \frac{4}{5}$$ Second $$T - 25 = (65 - 25) e^{-5C} = 40 \times \frac{4}{5} = 32$$ $$T = 57^\circ \mathrm{C}$$

On heating, concentration of $\mathrm{O_2}$ in water decreases. So boiling water and water at $80^\circ \mathrm{C}$ have less $\mathrm{O_2}$ concentration. Polluted water also has less $\mathrm{O_2}$ concentration. So water at $4^\circ \mathrm{C}$ has maximum $\mathrm{O_2}$ concentration.

At the time, he proposed the periodic table but structure of atom was unknown.

Reduced pressure distillation or vacuum distillation is used for the purification of high boiling organic liquids which decomposes at or below their boiling point.

The reaction of the given compound with $\mathrm{CH_3MgBr}$ in excess and hydrolysis leads to the formation of a tertiary alcohol. In the first reaction, the product is a tertiary alcohol. In the second reaction, the phenolic $\mathrm{-OH}$ group and tertiary alcohol are present, thus two functional groups are present in the product. In the third reaction, two tertiary alcohols are formed. Since the given question is single correct choice, the best appropriate option is (A).

Q6 $\mathrm{CH_3 - CH_2 - CH = CH - CH_2 - NH_2}$ No conjugation thus resonance is not possible.

(a) 'Ba' having outer electronic configuration $6s^2$. (b) $\mathrm{CaC_2O_4}$ is water insoluble. (c ) 'Li' is soluble in organic solvents. (d) NaOH is strong monoacidic base among given.

The reaction starts with a diazonium salt, $\mathrm{N_2^+Cl^-}$, attached to a benzene ring. In step [A], $\mathrm{H_3PO_2}$ is used for reduction, resulting in the removal of $\mathrm{N_2}$ and the formation of benzene with a hydrogen atom. In step [B], ethyl chloride ($\mathrm{Et-Cl}$) and aluminum chloride ($\mathrm{AlCl_3}$) are used to perform a Friedel-Crafts alkylation, resulting in the attachment of an ethyl group to the benzene ring.

$^1_1\mathrm{H}$ and $^2_1\mathrm{H}$ are stable while $^3_1\mathrm{H}$ is radioactive.

(a) $\mathrm{SF_4}$ -- $sp^3d$ hybridisation (b) $\mathrm{IF_5}$ -- $sp^3d^2$ hybridisation (c) $\mathrm{NO_2^+}$ -- $sp$ hybridisation (d) $\mathrm{NH_4^+}$ -- $sp^3$ hybridisation

3-Ethylhex-3-ene will not show stereo isomerism. Its diagram is shown. (1) Not show geometrical isomerism (2) Not show optical isomerism

Diborane is prepared by the reaction of $\mathrm{NaBH_4}$ with $\mathrm{I_2}$. $$2\mathrm{NaBH_4} + \mathrm{I_2} \rightarrow \mathrm{B_2H_6} + 2\mathrm{NaI} + \mathrm{H_2}$$ In diborane, ' $B$ ' is $\mathrm{sp^3}$ hybrid, it is Non-planar and two $3c - 2e^-$ bonds are present.

Among 15th group hydrides, $\mathrm{BiH_3}$ is strongest reducing agent.

$\mathrm{Cu}^{2+}:[\mathrm{Ar}]\,3d^9\,4s^0$ $\mathrm{Cr}^{3+}:[\mathrm{Ar}]\,3d^3\,4s^0$ $\mathrm{Sc}^{+}:[\mathrm{Ar}]\,3d^1\,4s^1$ All are coloured.

Vitamin $B_1$ is also known as Thiamine while vitamin $B-6$ is known as Pyridoxine.

Pb and Ag commonly exist in the form of sulphide ore like $\mathrm{PbS}$ (galena) and $\mathrm{Ag_2S}$ (Argentite). 'Al' is mainly found in the form of oxide ore whereas 'Mg' is found in the form of halide ore.

The compound $\mathrm{C_6H_6O}$ undergoes a reaction with $\mathrm{FeCl_3}$ to produce a dark green color. This compound is labeled as $[A]$. Through the Reimer-Tiemann reaction with $\mathrm{CHCl_3}$ and $\mathrm{KOH}$, compound $[A]$ is converted to compound $[B]$. Compound $[B]$ is then oxidized using PCC to form compound $[C]$.

Aniline is Lewis base give acid base reaction with $\mathrm{AlCl_3}$ and form Anilinium ion. Anilinium ion has strongest deactivated ring so further Friedel craft Alkylation not occurs.

Given $T_f - T_f' = iK_f \cdot m$. For minimum $T_f'$, $i$ should be maximum. $\mathrm{Al_2(SO_4)_3}$ $i = 5$. $\mathrm{C_6H_{12}O_6}$ $i = 1$. $\mathrm{KI}$ $i = 2$. $\mathrm{K_2SO_4}$ $i = 3$.

For the complex $[\mathrm{Co(NH_3)_6}] \mathrm{Cl}_2$, $\mathrm{Co}^{2+}$: $[\mathrm{Ar}] 3d^7 4s^0 4p^0$. For this complex $\Delta_0 \mathrm{P.E.}$ So here all electrons become paired.

The reaction is given by $\mathrm{C_6H_6+CH_3Cl\rightarrow C_6H_5CH_3+HCl}$. The calculation for the yield is as follows: $\frac{A_y}{T_y}=\%\ \mathrm{yield}=\frac{9.2}{920}\times78\times100\Rightarrow78\%$

E = E^$\circ$ - $\frac{0.059}{2}$ $\log$ $\frac{[\mathrm{Cu}^{+2}]}{[\mathrm{Ag}^+]^2}$ = 2.97 - $\frac{0.059}{2}$ $\log$ $\frac{0.25}{(10^{-3})^2}$ = 2.81 $\mathrm{V}$

Given $K_C = \frac{K_P}{RT} = \frac{47.9}{0.083 \times 288} = 2$

1 mol graphite = 12 $\mathrm{gmC}$ Ans. = $\frac{248}{12}$ = 20.67 $\,$ $\mathrm{kJ/gm}$ heat evolved

Given the density equation: $$d \left( \frac{\mathrm{gm}}{\mathrm{cc}} \right) = \frac{4 \times \frac{M}{N_A}}{(a \mathrm{cm})^3}$$ Substituting the values: $$7.62 = \frac{4 \times M / 6.022 \times 10^{23}}{(0.4518 \times 10^{-7} \mathrm{cm})^3} \Rightarrow M = 105.8 \, \mathrm{g/mol}$$

The electronic configuration of vanadium (V) is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3 4s^2$. The number of electrons in p-orbitals is equal to $12.00$.

Given the equation for K: $$K = \frac{2.303}{t} \log \frac{[\mathrm{N_2O_5}]_0}{[\mathrm{N_2O_5}]_t}$$ Substitute the given values: $$= \frac{2.303}{60} \log \frac{2.4}{1.6} = 6.76 \times 10^{-3} \, \mathrm{min}^{-1} \approx 7 \times 10^{-3} \, \mathrm{min}^{-1}$$

The concentration of glucose is calculated as follows: $$[Glucose] = \frac{C(\mathrm{gm/\ell})}{M(\mathrm{gm/mol})} = \frac{0.72}{180} = 4 \times 10^{-3} \, \mathrm{M}$$