JEE Main 26 February 2021 Shift 2 question paper with solutions

Given $\frac{x}{1} = -\frac{1}{y} = \frac{1}{z} = \lambda$ (let). Unit vector parallel to $x \hat{i} + y \hat{j} + z \hat{k}$ is $\pm \frac{\left( \lambda \hat{i} - \frac{1}{\lambda} \hat{j} + \frac{1}{\lambda} \hat{k} \right)}{\sqrt{\lambda^2 + \frac{2}{\lambda^2}}}$. For $\lambda = 1$, it is $\pm \frac{(\hat{i} - \hat{j} + \hat{k})}{\sqrt{3}}$.

Given $g(f(x)) = f(x)$, it implies $g(x) = x$, when $x$ is even. 5 elements in $A$ can be mapped to any 10. So, $10^5 \times 1 = 10^5$.

If $f$ is continuous at $x = -1$, then $f(-1^-) = f(-1)$. $$2 = |a - 1 + b|$$ $$|a + b - 1| = 2 \ldots (i)$$ Similarly $$f(1^-) = f(1)$$ $$|a + b + 1| = 0$$ $$a + b = -1$$

Given $f(e) + f\left(\frac{1}{e}\right) = \int_1^e \frac{\ln t}{1+t} \, dt + \int_1^{1/e} \frac{\ln t}{1+t} \, dt = I_1 + I_2$. $I_2 = \int_1^{1/e} \frac{\ln t}{1+t} \, dt$. Put $t = \frac{1}{z}$, $dt = -\frac{dz}{z^2}$. $$= \int_1^e \frac{-\ln z}{1+\frac{1}{z}} \times \left(-\frac{dz}{z^2}\right) = \int_1^e \frac{\ln z}{z(z+1)} \, dz$$ $f(e) + f\left(\frac{1}{e}\right) = \int_1^e \frac{\ln t}{1+t} \, dt + \int_1^e \frac{\ln t}{t(t+1)} \, dt = \int_1^e \frac{\ln t}{1+t} \, dt + \frac{\ln t}{t(t+1)} \, dt$. $$= \int_1^e \frac{\ln t}{t} \, dt \left\{ \ln t = u, \frac{1}{t} \, dt \right\}$$ $= du = \int_0^1 u \, du = \frac{u^2}{2} \bigg|_0^1 = \frac{1}{2}$

Given $y + z = 5$ ...(1) $$\frac{1}{y} + \frac{1}{z} = \frac{5}{6}$$ $$\Rightarrow \frac{y+z}{yz} = \frac{5}{6}$$ $$\Rightarrow \frac{5}{yz} = \frac{5}{6}$$ $$\Rightarrow yz = 6$$ Also $(y-z)^2 = (y+z)^2 - 4yz$ $$\Rightarrow (y-z)^2 = (y+z)^2 - 4yz$$ $$\Rightarrow (y-z)^2 = 25 - 4(6) = 1$$ $$\Rightarrow y-z = 1$$ From (1) and (2), $y = 3$ and $z = 2$. For calculating odd divisor of $p = 2^x \cdot 3^y \cdot 5^z$ $x$ must be zero $$P = 2^0 \cdot 3^3 \cdot 5^2$$ Therefore, total odd divisors must be $(3+1)(2+1) = 12$

Given $$g(2) = \lim_{x \to 2} \frac{(x-2)(x+1)}{(2x+3)(x-2)} = \frac{3}{7}$$ For domain of fog $(x)$ $$\left| \frac{x^2 - x - 2}{2x^2 - x - 6} \right| \leq 1$$ This implies $$(3x + 4)(x + 2) \geq 0$$ $$x \in (-\infty, -2] \cup \left( -\frac{4}{3}, \infty \right]$$

Triangle of maximum area that can be inscribed in a circle is an equilateral triangle. Let $\triangle ABC$ be inscribed in the circle. Now, in $\triangle OBD$, $OD = r \cos 60^\circ = \frac{r}{2}$. Height $= AD = \frac{3r}{2}$. Again in $\triangle ABD$, now $\sin 60^\circ = \frac{\frac{3r}{2}}{AB}$. $$\Rightarrow AB = \sqrt{3}r$$

Dr's of line $3\hat{i}-2\hat{j}+\hat{k}$ Direction ratios : $(3,-2,1)$ Point on the line : $(-2,4,0)$ Equation of the line $\frac{x+2}{3}=\frac{y-4}{-2}=\frac{z}{1}=\lambda$ Dr's of $PQ$ $(3\lambda-5,\,-2\lambda+2,\,\lambda-1)$ Dr's of the line are $(3,-2,1)$. Since $PQ\perp$ line, $3(3\lambda-5)-2(-2\lambda+2)+(\lambda-1)=0$ $\Rightarrow 9\lambda-15+4\lambda-4+\lambda-1=0$ $\Rightarrow 14\lambda-20=0$ $\Rightarrow \lambda=\frac{10}{7}$ $P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right)$ $21(\alpha+\beta+\gamma)$ $=21\left(\frac{34}{7}\right)$ $=102$

Truth table for $F_1$ $F_1$ not shows tautology and $F_2$ shows tautology

Given $\($ $\frac{dy}{dx}$ = $\frac{xy^2 + y}{x}$ $\)$. $\($ $\Rightarrow$ xdy - ydx = y^2 $\,$ dx $\)$ $\($ $\Rightarrow$ -d $\left$( $\frac{y}{x}$ $\right$) = d $\left$( $\frac{x^2}{2}$ $\right$) $\)$ $\($ $\Rightarrow$ -$\frac{y}{x}$ = $\frac{x^2}{2}$ + C $\)$ Curve intersects the line $\($ x + 2y = 4 $\)$ at $\($ x = -2 $\)$. So, $\($-2 + 2y = 4 $\Rightarrow$ y = 3 $\)$. So the curve passes through $\($(-2, 3) = $\frac{2}{3}$ = 2 + C $\)$ $\($ $\Rightarrow$ C = -$\frac{4}{3}$ $\)$ Therefore, the curve is $\($ -$\frac{y}{x}$ = $\frac{x^2}{2}$ - $\frac{4}{3}$ $\)$ It also passes through $\($(3, y) $\Rightarrow$ -$\frac{3}{y}$ = $\frac{9}{2}$ - $\frac{4}{3}$ $\)$ $\($ $\Rightarrow$ -$\frac{3}{y}$ = $\frac{19}{6}$ $\)$ $\($ $\Rightarrow$ y = -$\frac{18}{19}$ $\)$

$\therefore\ P=(2h-3,\;2k-2)\ \rightarrow\ \text{on circle}$ $(h-\tfrac{3}{2})^2+(k-1)^2=4$ $\Rightarrow\ \text{radius}=\tfrac{1}{2}\times 2=1$

Given $$D = \begin{vmatrix} 1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7 \end{vmatrix}$$ $$= 20 - 2(25) - 3(-10)$$ $$= 20 - 50 + 30 = 0$$ $$D_1 = \begin{vmatrix} a & 2 & -3 \\ b & 6 & -11 \\ c & -2 & 7 \end{vmatrix}$$ $$= 20a - 2(7b + 11c) - 3(-2b - 6c)$$ $$= 20a - 14b - 22c + 6b + 18c$$ $$= 20a - 8b - 4c$$ $$= 4(5a - 2b - c)$$ $$D_2 = \begin{vmatrix} 1 & a & -3 \\ 2 & b & -11 \\ 1 & c & 7 \end{vmatrix}$$ $$= 7b + 11c - a(25) - 3(2c - b)$$ $$= 7b + 11c - 25a - 6c + 3b$$ $$= -25a + 10b + 5c$$ $$= -5(5a - 2b - c)$$ $$D_3 = \begin{vmatrix} 1 & 2 & a \\ 2 & 6 & b \\ 1 & -2 & c \end{vmatrix}$$ $$= 6c + 2b - 2(2c - b) - 10a$$ $$= -10a + 4b + 2c$$ $$= -2(5a - 2b - c)$$ For infinite solution $$D = D_1 = D_2 = D_3 = 0$$ $$\Rightarrow 5a = 2b + c$$

Given $$\tan^{-1}\left(\frac{a+b}{1-ab}\right) = \frac{\pi}{4}$$ implies $$a+b = 1-ab$$ which implies $$(1+a)(1+b) = 2$$. Now, $$(a+b) - \left(\frac{a^2+b^2}{2}\right) + \left(\frac{a^3+b^3}{3}\right) \cdots \infty$$ $$= \left(a - \frac{a^2}{2} + \frac{a^3}{3} \cdots \right) + \left(b - \frac{b^2}{2} + \frac{b^3}{3} \cdots \right)$$ $$\log_e(1+a) + \log_e(1+b) = \log_e((1+a)(1+b)) = \log_e 2$$

Given $$\sum_{n=1}^{\infty} \frac{n^2 + 6n + 10}{(2n+1)!}$$. Put $$2n + 1 = r$$, where $$r = 3, 5, 7, \ldots$$. Thus, $$n = \frac{r-1}{2}$$. $$\frac{n^2 + 6n + 10}{(2n+1)!} = \frac{\left(\frac{r-1}{2}\right)^2 + 3r - 3 + 10}{r!} = \frac{r^2 + 10r + 29}{4r!}$$. Now, $$\sum_{r=3,5,7,\ldots} \frac{r(r-1) + 11r + 29}{4r!} = \frac{1}{4} \sum_{r=3,5,7,\ldots} \left( \frac{1}{(r-2)!} + \frac{11}{(r-1)!} + \frac{29}{r!} \right)$$. $$= \frac{1}{4} \left\{ \left( \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \ldots \right) + 11 \left( \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \ldots \right) + 29 \left( \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \ldots \right) \right\}$$ $$= \frac{1}{4} \left\{ e - \frac{1}{e} + 11 \left( \frac{e + \frac{1}{e} - 2}{2} \right) + 29 \left( \frac{e - \frac{1}{e} - 2}{2} \right) \right\}$$ $$= \frac{1}{8} \left\{ e - \frac{1}{e} + 11e + \frac{11}{e} - 22 + 29e - \frac{29}{e} - 58 \right\}$$ $$= \frac{1}{8} \left\{ 41e - \frac{19}{e} - 80 \right\}$$

By L-H rule $$L = \lim_{x \to a} \frac{f(a) - af'(x)}{1}$$ Therefore, $L = 4 - 2a$

Therefore, $PA^2 = \cos^2 \theta + (\sin \theta - 3)^2 = 10 - 6 \sin \theta$. $PB^2 = \cos^2 \theta + (\sin \theta - 6)^2 = 37 - 12 \sin \theta$. $PA^2 + PB^2 \big|_{\max} = 47 - 18 \sin \theta \big|_{\min}$. Therefore, $\theta = \frac{3\pi}{2}$. Therefore, $P$, $A$, $B$ lie on a line $x = 1$.

Image of $(1,3,5)$ in the plane $4x - 5y + 2z = 8$ is $(\alpha, \beta, \gamma)$. $$\Rightarrow \frac{\alpha - 1}{4} = \frac{\beta - 3}{-5} = \frac{\gamma - 5}{2} = \frac{-2 \left( 4(1) - 5(3) + 2(5) - 8 \right)}{4^2 + 5^2 + 2^2} = \frac{2}{5}$$ Therefore, $\alpha = 1 + 4 \left( \frac{2}{5} \right) = \frac{13}{5}$. $\beta = 3 - 5 \left( \frac{2}{5} \right) = 1 = \frac{5}{5}$. $\gamma = 5 + 2 \left( \frac{2}{5} \right) = \frac{29}{5}$. Thus, $5(\alpha + \beta + \gamma) = 5 \left( \frac{13}{5} + \frac{5}{5} + \frac{29}{5} \right) = 47$.

Given, $f(x) = \int_0^x e^t f(t) dt + e^x$ ...(1) Differentiating both sides with respect to $x$, $f'(x) = e^x \cdot f(x) + e^x$. Using Newton Leibnitz Theorem, $\Rightarrow \frac{f'(x)}{f(x) + 1} = e^x$. Integrating with respect to $x$, $$\int \frac{f'(x)}{f(x) + 1} \, dx = \int e^x \, dx$$ $$\Rightarrow \ln(f(x) + 1) = e^x + c$$ Put $x = 0$. $$2 = 1 + c (\because f(0) = 1, from equation (1))$$ $$\therefore \ln(f(x) + 1) = e^x + \ln 2 - 1$$ $$\Rightarrow f(x) + 1 = 2 \cdot e^{e^x - 1}$$ $$\Rightarrow f(x) = 2e^{e^x - 1} - 1$$

Given $A_1 + A_2 = \int_0^{\pi/2} \cos x \cdot dx = \sin x \bigg|_0^{\pi/2} = 1$. $A_1 = \int_0^{\pi/4} (\cos x - \sin x) dx = (\sin x + \cos x) \bigg|_0^{\pi/4} = \sqrt{2} - 1$. Therefore, $A_2 = 1 - (\sqrt{2} - 1) = 2 - \sqrt{2}$. Thus, $\frac{A_1}{A_2} = \frac{\sqrt{2} - 1}{\sqrt{2}(\sqrt{2} - 1)} = \frac{1}{\sqrt{2}}$.

Given $\($ n(S) = $\frac{7!}{2!3!2!}$ $\)$ and $\($ n(E) = $\frac{6!}{2!2!2!}$ $\)$. The probability $\($ P(E) = $\frac{n(E)}{n(S)}$ = $\frac{6!}{7!}$ $\times$ $\frac{2!3!2!}{2!2!2!}$ $\)$. Simplifying, $\($ $\frac{1}{7}$ $\times$ 3 = $\frac{3}{7}$ $\)$.

Given, $|z + 5| \leq 4$ $$\Rightarrow (x + 5)^2 + y^2 \leq 16 \cdots (1)$$ Also, $z(1+i) + \bar{z}(1-i) \geq -10$. $$\Rightarrow x - y \geq -5 \cdots (2)$$ From (1) and (2) Locus of $z$ is the shaded region in the diagram. $|z + 1|$ represents distance of $z'$ from $Q(-1, 0)$. Clearly $P$ is the required position of $z'$ when $|z + 1|$ is maximum. $$\therefore P \equiv (-5 - 2\sqrt{2}, -2\sqrt{2})$$ $$\therefore (PQ)^2_{max} = 32 + 16\sqrt{2}$$ $$\Rightarrow \alpha = 32$$ $$\Rightarrow \beta = 16$$ Thus, $\alpha + \beta = 48$

Let the equation of normal is $Y - y = -\frac{1}{m}(X - x)$, where, $m = \frac{dy}{dx}$. As it passes through $(a, b)$ $$b - y = -\frac{1}{m}(a - x) = -\frac{dx}{dy}(a - x)$$ $$\Rightarrow (b - y) dy = (x - a) dx$$ by $$\frac{y^2}{2} = \frac{x^2}{2} - ax + c$$ It passes through $(3, -3) \& (4, -2\sqrt{2})$ $$\therefore -3b - \frac{9}{2} = \frac{9}{2} - 3a + c$$ $$\Rightarrow -6b - 9 = 9 - 6a + 2c$$ $$\Rightarrow 6a - 6b - 2c = 18$$ $$\Rightarrow 3a - 3b - c = 9$$ Also $$-2\sqrt{2}b - 4 = 8 - 4a + c$$ $$4a - 2\sqrt{2}b - c = 12$$ Also $a - 2\sqrt{2}b = 3$ $\ldots$ (iv) (given) (ii) - (iii) $\Rightarrow -a + (2\sqrt{2} - 3)b = -3$ $\ldots$ (v) (iv) + (v) $\Rightarrow b = 0$, $a = 3$ $$\therefore a^2 + b^2 + ab = 9$$

Given, $\alpha + \beta = 1$, $\alpha \beta = -1$. Therefore, the quadratic equation with roots $\alpha$, $\beta$ is $x^2 - x + 1 = 0 \implies \alpha^2 = \alpha + 1$. Multiplying both sides by $\alpha^{n-1}$, $\alpha^{n+1} = \alpha^n + \alpha^{n-1}$ ... (1) Similarly, $\beta^{n+1} = \beta^n + \beta^{n-1}$ ... (2) Adding (1) & (2), $\alpha^{n+1} + \beta^{n+1} = (\alpha^n + \beta^n) + (\alpha^{n-1} + \beta^{n-1})$ Therefore, $P_{n+1} = P_n + P_{n-1}$ $29 = P_n + 11$ (Given, $P_{n+1} = 29$, $P_{n-1} = 11$) Therefore, $P_n = 18$ $P_n^2 = 18^2 = 324$

Given $I_{m,n} = \int_0^1 x^{m-1} \cdot (1-x)^{n-1} dx$. Put $x = \frac{1}{y+1}$, then $dx = \frac{-1}{(y+1)^2} dy$. $1-x = \frac{y}{y+1}$. Therefore, $I_{m,n} = \int_0^\infty \frac{y^{m-1}}{(y+1)^{m+n}} (-1) dy = \int_0^\infty \frac{y^{n-1}}{(y+1)^{m+n}} dy \ldots (i)$ Similarly, $I_{m,n} = \int_0^1 x^{n-1} \cdot (1-x)^{m-1} dx$ $\Rightarrow I_{m,n} = \int_0^\infty \frac{y^{n-1}}{(y+1)^{m+n}} dy \ldots (ii)$ From (i) $\&$ (ii) $2I_{m,n} = \int_0^\infty \frac{y^{m-1}+y^{n-1}}{(y+1)^{m+n}} dy$ $\Rightarrow 2I_{m,n} = \int_0^1 \frac{y^{m-1}+y^{n-1}}{(y+1)^{m+n}} dy + \int_1^\infty \frac{y^{m-1}+y^{n-1}}{(y+1)^{m+n}} dy$ Put $y = \frac{1}{z}$ in $I_2$ $dy = -\frac{1}{z^2} dz$ $\Rightarrow 2I_{m,n} = \int_0^1 \frac{y^{m-1}+y^{n-1}}{(y+1)^{m+n}} dy + \int_1^0 \frac{z^{m-1}+z^{n-1}}{(z+1)^{m+n}} (-dz)$ $\Rightarrow I_{m,n} = \int_0^1 \frac{y^{m-1}+y^{n-1}}{(y+1)^{m+n}} dy \Rightarrow \alpha = 1$

Given, $4x^2 - 9x + 5 = 0$ $\[$ $\Rightarrow$ (x - 1)(4x - 5) = 0 $\]$ $\[$ $\Rightarrow$ A. M = $\frac{5}{4}$, G. M = 1 (Q.A. M > G. M) $\]$ Again, for the series $-16, 8, -4, 2, \ldots$ $p^{th}$ term $t_p = -16 \left(-\frac{1}{2}\right)^{p-1}$ $q^{th}$ term $t_q = -16 \left(-\frac{1}{2}\right)^{q-1}$ Now, A. M $= \frac{t_p + t_q}{2} = \frac{5}{4}$ and G. M $= \sqrt{t_p t_q} = 1$ $\[$ $\Rightarrow$ 16^2 $\left$(-$\frac{1}{2}$$\right$)^{p+q-2} = 1 $\]$ $\[$ $\Rightarrow$ (-2)^8 = (-2)^{(p+q-2)} $\]$ $\[$ $\Rightarrow$ p + q = 10 $\]$

Since, required number has G.C.D with 18 as 3. It must be odd multiple of '3' but not a multiple of '9'. (i) Now, 4-digit number which are odd multiple of '3' are, 1005, 1011, 1017, $\ldots$ 9999 $\rightarrow$ 1499 (ii) 4-digit number which are odd multiple of 9 are, 1017, 1035, $\ldots$ 9999 $\rightarrow$ 499 $\therefore$ Required numbers = 1499 - 499 = 1000

Given $E: \frac{x^2}{9} + \frac{y^2}{4} = 1$ and $C: x^2 + y^2 = \frac{31}{4}$. The equation of the tangent to the ellipse is $y = mx \pm \sqrt{9m^2 + 4}$. The equation of the tangent to the circle is $$y = mx \pm \sqrt{\frac{31}{4} m^2 + \frac{31}{4}} \cdots (ii)$$ Comparing equation (i) and (ii), $9m^2 + 4 = \frac{31}{4} m^2 + \frac{31}{4}$. $$\Rightarrow 36m^2 + 16 = 31m^2 + 31$$ $$\Rightarrow 5m^2 = 15$$ $$\Rightarrow m^2 = 3$$

Let, $f(x) = 2x^5 + 5x^4 + 10x^3 + 10x^2 + 10x + 10$. Therefore, $f'(x) = 10 \left( x^4 + 2x^3 + 3x^2 + 2x + 1 \right) = 10 \left( x^2 + \frac{1}{x^2} + 2 \left( x + \frac{1}{x} \right) + 3 \right) = 10 \left( \left( x + \frac{1}{x} \right)^2 + 2 \left( x + \frac{1}{x} \right) + 1 \right) = 10 \left( \left( x + \frac{1}{x} + 1 \right)^2 \right) > 0; \forall x \in R$. Therefore, $f(x)$ is strictly increasing function. Since it is an odd degree polynomial it will have exactly one real root. Now, by observation $f(-1) = 3 > 0$. $f(-2) = -64 + 80 - 80 + 40 - 20 + 10 = -34 < 0$. Therefore, $f(x)$ has at least one root in $(-2,-1) \equiv (a, a + 1)$. Therefore, $a = -2$. Therefore, $|a| = 2$.

Given, $$\sum_{i=1}^{18} (X_i - \alpha) = 36$$ implies $$\sum x_i - 18\alpha = 36$$ implies $$\sum x_i = 18(\alpha + 2) \ldots (1)$$ Also, $$\sum_{i=1}^{18} (X_i - \beta)^2 = 90$$ implies $$\sum x_i^2 + 18\beta^2 - 2\beta \sum x_i = 90$$ implies $$\sum x_i^2 + 18\beta^2 + 2\beta \times 18(\alpha + 2) = 90$$ (using equation (1)) implies $$\sum x_i^2 = 90 - 18\beta^2 + 36\beta(\alpha + 2)$$ $$\sigma^2 = 1 \implies \frac{1}{18} \sum x_i^2 - \left( \frac{\sum x_i}{18} \right)^2 = 1 (\because \sigma = 1, given)$$ implies $$\frac{1}{18} (90 - 18\beta^2 + 36\alpha\beta + 72\beta) - \left( \frac{18(\alpha + 2)}{18} \right)^2 = 1$$ implies $$90 - 18\beta^2 + 36\alpha\beta + 72\beta - 18(\alpha + 2)^2 = 18$$ implies $$5 - \beta^2 + 2\alpha\beta + 4\beta - (\alpha + 2)^2 = 1$$ implies $$5 - \beta^2 + 2\alpha\beta + 4\beta - \alpha^2 - 4 - 4\alpha = 1$$ implies $$\alpha^2 - \beta^2 + 2\alpha\beta + 4\beta - 4\alpha = 0$$ implies $$(\alpha - \beta)(\alpha - \beta + 4) = 0$$ implies $$\alpha - \beta = -4$$ Therefore, $$|\alpha - \beta| = 4 (\alpha \neq \beta)$$

Given $$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$A^3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$A^4 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$A^{19} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^{19} & 0 \\ 3 & 0 & -1 \end{bmatrix}, A^{20} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ L.H.S = $A^{20} + \alpha A^{19} + \beta A = \begin{bmatrix} 1 + \alpha + \beta & 0 & 0 \\ 0 & 2^{20} + \alpha 2^{19} + 2 \beta & 0 \\ 3 \alpha + 3 \beta & 0 & 1 - \alpha - \beta \end{bmatrix}$ R.H.S = $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix} \Rightarrow \alpha + \beta = 0 and 2^{20} + \alpha 2^{19} + 2 \beta = 4$ $$\Rightarrow 2^{20} + \alpha (2^{19} - 2) = 4$$ $$\Rightarrow \alpha = \frac{4 - 2^{20}}{2^{19} - 2} = -2$$ $$\Rightarrow \beta = 2$$ Therefore, $\beta - \alpha = 4$

Given $v = \frac{w}{q}$ and $c = \frac{q}{v}$. Find the dimension of $\frac{c}{v}$. $$\Rightarrow \frac{q}{v^2}$$ $$\Rightarrow \frac{q}{w^2} \times q^2 \Rightarrow \frac{q^3}{w^2}$$ $$\Rightarrow \frac{w^3 \, T^3}{M^2 \, L^4 \, T^{-4}} \Rightarrow \left[ M^{-2} \, L^{-4} \, T^3 \right]$$

From Young's modulus relation $$y = \left( \frac{F}{A} \right) \left( \frac{\Delta l}{l} \right)$$ we can write for the 1st case $$\frac{T_1}{A} = \frac{y(l_1 - \ell)}{\ell}$$ we can write for the 2nd case $$\frac{T_2}{A} = \frac{y(l_2 - \ell)}{\ell}$$ $$\frac{T_1}{T_2} = \frac{l_1 - \ell}{l_2 - \ell}$$ $$T_1 l_2 - T_1 \ell = T_2 l_1 - T_2 \ell$$ $$\frac{T_2 l_1 - T_1 l_2}{T_2 - T_1} = \ell$$

The equation for the electromotive force is given by: $$\sum = B \perp v \ell$$ Using the sine of 60 degrees: $$\sin 60^\circ = \frac{B_v}{B}$$ From the diagram: $$\frac{\sqrt{3}}{2} = \frac{B_v}{B}$$ Solving for $BV$: $$BV = \frac{\sqrt{3}}{2} B$$ The electromotive force $E$ is: $$E = \frac{\sqrt{3}}{2} B \ell v$$ Substituting the given values: $$= \frac{\sqrt{3}}{2} \times 2.5 \times 10^{-4} \times 10 \times 180 \times \frac{5}{18}$$ Simplifying further: $$= \frac{\sqrt{3}}{2} \times 2.5 \times 5 \times 10^{-2} = 10.825 \times 10^{-2} = 108.25 \, \mathrm{mV}$$

Given Before Filed: 340 $\mathrm{Hz}$ $\rightarrow$ 5 beats/sec So answer should be 335 $\mathrm{Hz}$ or 345 $\mathrm{Hz}$. After Filed: 340 $\mathrm{Hz}$ $\rightarrow$ 2 beats/sec After filed beat/sec decreases only in case of 335 $\mathrm{Hz}$.

For a body performing SHM, relation between velocity and displacement $v = \omega \sqrt{A^2 - x^2}$. Now, square both sides $v^2 = \omega^2 \left(A^2 - x^2\right)$. $$\Rightarrow v^2 = \omega^2 A^2 - \omega^2 x^2$$ $$v^2 + \omega^2 x^2 = \omega^2 A^2$$ Divide the whole equation by $\omega^2 A^2$: $$\frac{v^2}{\omega^2 A^2} + \frac{\omega^2 x^2}{\omega^2 A^2} = \frac{\omega^2 x^2}{\omega^2 A^2}$$ $$\frac{v^2}{(\omega A)^2} + \frac{x^2}{(A)^2} = 1$$ The above equation is similar to the standard equation of ellipses, so the graph between velocity and displacement will be ellipses.

Given: $$y = \alpha x - \beta x^2 \cdots (1)$$ For maximum height, we should find out maximum value of $y$ from equation (1). So, for maximum value of $y$ $$\frac{dy}{dx} = 0 \Rightarrow \alpha - 2\beta x = 0$$ $$x = \frac{\alpha}{2\beta} \cdots (2)$$ Now, put value of $x$ from equation (2) in equation (1) $$y = \alpha \left( \frac{\alpha}{2\beta} \right) - \beta \left( \frac{\alpha^2}{4\beta^2} \right) \Rightarrow \frac{\alpha^2}{4\beta}$$ So, $H_{max} = \frac{\alpha^2}{4\beta} \cdots (3)$ As we know maximum height $H_{max} = \frac{u^2 \sin^2 \theta}{2g} \cdots (4)$ From (3) and (4) $$u^2 = \left( \frac{\alpha^2}{4\beta} \right) \left( \frac{2g}{\sin^2 \theta} \right)$$ And range $(R) = 2x = \frac{u^2 \times 2 \sin \theta \cos \theta}{g}$$ $$2 \left( \frac{\alpha}{2\beta} \right) = \left( \frac{\alpha^2}{4\beta} \right) \left( \frac{2g}{\sin^2 \theta} \right) \times \frac{2 \sin \theta \cos \theta}{g}$$ $$\tan \theta = \alpha \Rightarrow \theta = \tan^{-1}(\alpha)$$

Using energy conservation between A and B point: $$mgh = \frac{1}{2} m (\omega R)^2 + \frac{1}{2} I \omega^2$$ $$2mgh = \left( MR^2 + I \right) \omega^2$$ $$\omega^2 = \frac{2mgh}{I + MR^2}$$

Given $U = 3PV + 4$. $$\frac{f}{2} PV = 3PV + 4 \therefore u = \frac{f}{2} nRT$$ $$f = 6 + \frac{8}{PV} \therefore PV = nRT$$ $f > 6$. Therefore, it is a polyatomic gas.

Both obtain the same angle, since the image can be at a distance greater than 25 cm, the object can be moved closer to the eye.

Statement $-1 \rightarrow$ Correct Statement $-2 \rightarrow$ Incorrect Statement $-2 \rightarrow$ Incorrect

Momentum $(P) = \frac{\Delta E}{c} = \frac{(13.6 - 0.54) \, \mathrm{eV}}{3 \times 10^8}$. $mv = \frac{3 \times 10^8}{(13.06) \times 1.6 \times 10^{-19}}$. $v = \frac{3 \times 10^8}{3 \times 10^8 \times 1.67 \times 10^{-27}} \implies 4.17 \, \mathrm{m/s}$

Peak current in series LCR CKT $$i = \frac{v_0}{z} \Rightarrow \frac{30}{\sqrt{(X_L - X_C)^2 + R^2}}$$ $$i = \frac{30}{\sqrt{(10 - 100)^2 + (120)^2}}$$ $$i \Rightarrow \frac{30}{150} \Rightarrow \frac{1}{5} \Rightarrow 0.2 Amp$$ Therefore, $X_L = \omega \times L$ $$\Rightarrow (100) \left(100 \times 10^{-3}\right) \Rightarrow 10$$ $$X_L = \frac{1}{\omega \times C} \Rightarrow \frac{1}{100 \times 100 \times 10^{-6}}$$ $$\Rightarrow \frac{10^6}{10^4} \Rightarrow 100$$ Resonance frequency $\omega = \frac{1}{\sqrt{LC}}$ $$\omega = \frac{1}{\sqrt{100 \times 10^{-3} \times 100 \times 10^{-6}}} \Rightarrow \frac{1}{\sqrt{10^{-5}}}$$ Therefore, $\omega = 2 \pi F$ $$F = \frac{1}{2 \pi} \times \frac{1}{\sqrt{10^{-5}}}$$ $$\Rightarrow \frac{1}{2 \pi} \sqrt{10^5}$$ $$\Rightarrow \frac{100}{2 \pi} \sqrt{10}$$ $$\Rightarrow 50 Hz$$

The force is given by the equation: $$F = mg \sin \theta - (\mu N + qE \cos \theta)$$ Substituting for $N$, we have: $$F = mg \sin \theta - \mu (mg \cos \theta + qE \sin \theta) - qE \cos \theta$$ Substituting the given values: $$F = 1 \times 10 \times \sin 30 - 0.2 \left(1 \times 10 \times \cos 30 + 200 \times 5 \times 10^{-3} \sin 30 \right)$$ $$- 200 \times 5 \times 10^{-3} \cos 30$$ Calculating the force: $$F = 2.3 \, \mathrm{N}$$ The acceleration is given by: $$a = \frac{F}{m} \Rightarrow \frac{2.3}{1} \Rightarrow 2.3 \, \mathrm{m/s^2}$$ The time is calculated as: $$t = \sqrt{\frac{25}{9}} \Rightarrow \sqrt{\frac{2 \times 2}{2.3}} \Rightarrow 1.3 \, \mathrm{sec}$$

Given the forces acting on the system, we have the equation: $$F - F_s = Ma'$$ Solving for $a'$ gives: $$a' = \frac{F}{M} - 9$$ The expression for the acceleration is: $$\frac{F - ma}{M}$$

The circuit diagram shows an AND gate with inputs $A$ and $B$. The output expression is given by $y = A \cdot \overline{B}$. The truth table for the inputs is as follows: Inputs: $$ \begin{array}{|c|c|c|} \hline A & B & y = A \cdot \overline{B} \\ \hline 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \hline \end{array} $$

For activity of radioactive sample $$A = A_0 e^{-\lambda t_1} \ldots (1)$$ $$\frac{A}{5} = A_0 e^{-\lambda t_2} \ldots (2)$$ From (1)/(2) $$5 = e^{-\lambda (t_1 - t_2)}$$ $$\ln(5) = (t_2 - t_1) \lambda \Rightarrow \lambda = \frac{\ln(5)}{t_2 - t_1}$$ $$avg. life = \frac{1}{\lambda} = \frac{t_2 - t_1}{\ln(5)}$$

From given information: For 1st interval $$a_1 = \frac{v_0}{t_1}$$ $$v_0 = a_1 t_1 \ldots (1)$$ For 2nd interval $$a_2 = \frac{v_0}{t_2}$$ $$v_0 = a_2 t_2 \ldots (2)$$ From (1) and (2) $$a_1 t_1 = a_2 t_2$$ $$\frac{t_1}{t_2} = \frac{a_2}{a_1}$$

As we know time period of second's pendulum is 2 sec, so statement (1) is incorrect. Time taken between two extreme points in second's pendulum is 1 sec. Above statement is correct because time taken by particle performing SHM between two extreme position is $T/2$. Here, $T = 2$ sec. So, time $= 2/2 = 1$ sec.

For stretched or compressed wire $R \propto l^2$ $$\frac{R_1}{R_2} = \frac{l_1^2}{l_2^2}$$ $$\Rightarrow \frac{R}{R_2} = \frac{R^2}{(1.25l)^2}$$ $$\Rightarrow R_2 = 1.5625R$$ % increase $\rightarrow 56.235\%$

We see from the diagram that because of the law of reflection, the component of the unit vector $\vec{a}$ along $\vec{b}$ changes sign on reflection while the component parallel to the mirror remains unchanged. $\vec{a} = \vec{a}_{11} + \vec{a}_{\perp}$ and $\vec{a}_{\perp} = \vec{c}(\vec{a} \cdot \vec{c})$ we see that the reflected unit vector is $\vec{b} = \vec{a}_{11} - \vec{a}_{\perp} \Rightarrow \vec{a} - 2(\vec{a} \cdot \vec{c})\vec{c}$

Given: $V = kT^{2/3}$ $$V^{3/2} = (k)^{3/2}T$$ $$TV^{-3/2} = const. \ldots (1)$$ and $$TV^{\gamma - 1} = const. \ldots (2)$$ From (1) & (2) $$-\frac{3}{2} = \gamma - 1$$ $$\gamma = -\frac{1}{2}$$ Work done $(w) = \frac{nRAT}{\gamma - 1}$ $$W = \frac{1 \times R \times 90}{-\frac{1}{2} - 1}$$ $$|W| = 60R x = 60$$

Number of stations is given by the formula: $$No. of station = \frac{Band width}{2 \times Highest Band width}$$ Substituting the given values: $$\Rightarrow \frac{90}{2 \times 5}$$ Simplifying gives: $$\Rightarrow 9$$

For photoelectric effect, $E_{max} = E - \phi$. $E_1 = 2 \phi_r$, $k_1 = \phi$ $E_2 = 10 \phi_r$, $k_2 = 9 \phi$ Therefore, $V \propto \sqrt{k}$ $\left$( k = $\frac{1}{2}$ mv^2 $\right$) $$\frac{v_1}{v_2} = \sqrt{\frac{1}{9}} \implies \frac{1}{3} = \frac{x}{y}$$ $x = 1$

From similar triangles IMP and IQR, \[ \frac{QR}{25} = \frac{180}{60}. \] Therefore, \[ QR = 25 \times \frac{180}{60} = 75\,\mathrm{cm}. \] Hence, the field of view (F.O.V.) is \[ \mathrm{F.O.V.} = 2 \times 75 = 150\,\mathrm{cm}. \]

For a particle executes S.H.M $$V = \omega \sqrt{a^2 - x^2}$$ Given $V = \frac{V_{max}}{2}$ implies $$\frac{A \omega}{2}$$ $$\frac{A^2 \omega^2}{4} = \omega^2 a^2 - \omega^2 x^2$$ $$x = \frac{\sqrt{3}}{2} a$$

For self energy of sphere (conducting) $U = \frac{kq^2}{2r}$. For small drop $\rightarrow U_i = \frac{kq^2}{2r} \ldots (1)$ After combine small drops volume remains same as bigger drop. Therefore, $\frac{4}{3} \pi r^3 \times n = \frac{4}{3} \pi R^3$. $R = (n)^{\frac{1}{3}} r \ldots (2)$ For large drop $\rightarrow U_f = \frac{k(nq)^2}{2 \times 3R} \ldots (3)$ From equation (1), (2), (3) $$\frac{U_f}{U_i} = (n)^{5/3}$$ $$\Rightarrow (27)^{5/3}$$ $\Rightarrow 2$

For given $\left( \frac{5}{8} \right)$ oscillation, we can write it as $\rightarrow \left( \frac{1}{2} + \frac{1}{8} \right)$. And we know for half oscillations time $\rightarrow \frac{T}{2}$. For final point $\rightarrow \pi + \frac{\pi}{6} \implies \frac{7\pi}{6}$. Time $\rightarrow \frac{7T}{12} \rightarrow$ given $\rightarrow \frac{\alpha}{\beta} T \alpha = 7p$

Given $\($ $\frac{GM}{\left( \frac{3R}{2} \right)^2}$ = $\frac{GMr}{R^3}$ $\)$. Solving for $\($ OA $\)$, we have $\($ OA = $\frac{4R}{9}$ = r $\)$. Then, $\($ AB = R - $\frac{4R}{9}$ = $\frac{5R}{9}$ $\)$. The ratio $\($ OA : AB $\)$ is $\($ $\frac{4R}{9}$ : $\frac{5R}{9}$ $\Rightarrow$ 4 : 5 = x : y $\)$. Therefore, $\($ x = 4 $\)$.

From thermodynamics law: $$\Delta Q = \Delta U + \Delta W \ldots (1)$$ $$Q = nC_V \Delta T + \frac{Q}{5}$$ $$Q - \frac{Q}{5} = 1 \times \frac{5}{2} R \times \Delta T$$ $$Q = \frac{25}{8} R \Delta T \ldots (2) \therefore Q = nc \Delta T$$ $$c = \frac{25}{8} R given C = \frac{xR}{8}$$ $$x = 25$$

$I = \dfrac{90 - 30}{4} = 15\,\text{mA}$ $I_1 = \dfrac{30}{5\,\text{K}\Omega} = 6\,\text{mA}$ $I_2 = 15\,\text{mA} - 6\,\text{mA} = 9\,\text{mA}$

Tritium isotope of hydrogen is radioactive and emits low energy $\beta^-$ particles. It is because of high $n/p$ ratio of tritium which makes nucleus unstable.

$TlI_3$ is $\mathrm{Tl^{+}I_3^{-}}$. $CsI_3$ is $\mathrm{Cs^{+}I_3^{-}}$. Thallium shows $\mathrm{Tl^{+}}$ state due to inert pair effect.

Sucrose $\to \alpha - \mathrm{D}$ - Glucose and $\beta - \mathrm{D}$ - Fructose Lactose $\to \beta - \mathrm{D}$ - Galactose and $\beta - \mathrm{D}$ - Glucose Maltose $\to \alpha$ -D- Glucose and $\alpha$ -D- Glucose

In option (a), the lone pair on the nitrogen is localised. In options (b), (c), and (d), the lone pair on the nitrogen is delocalised.

Electron gain enthalpy of O is very low due to small size.

Seliwanoff test and Xanthaproteic test are used for identification of 'Ketoses' and proteins respectively.

The reaction involves the condensation of an aldehyde $\mathrm{R{-}CHO}$ with hydrazine $\mathrm{H_2N{-}NH{-}}$ attached to a nitrobenzene ring. The reaction proceeds with the elimination of water $\mathrm{H_2O}$ to form a hydrazone compound $\mathrm{R{-}CH{=}N{-}NH{-}}$ attached to the nitrobenzene ring.

Alcohol gives a positive test with ceric ammonium nitrate and primary amines give carbylamine test with $\mathrm{CHCl_3}$, $\mathrm{KOH}$.

$\mathrm{B_2O_3}$ and $\mathrm{SiO_2}$ both are oxides of non-metal and hence are acidic in nature.

The reaction shown is an internal aldol condensation. The starting material is a benzene ring with two propionaldehyde groups. Under basic conditions with NaOH, in the presence of ethanol and water, and with heat ($\Delta$), an intramolecular aldol condensation occurs. This results in the formation of a bicyclic compound with an aldehyde group (CHO) attached.

The reaction begins with the condensation of methoxybenzaldehyde with formaldehyde in the presence of $\mathrm{NaOH}$ to form a benzyl alcohol derivative. This is followed by the reaction with $\mathrm{CH_3CH_2Br}$ and $\mathrm{NaH}$ in $\mathrm{DMF}$ to form an ether. Finally, the ether undergoes a reaction with $\mathrm{HI}$ under heat to form the iodide derivative.

$\mathrm{Na_6(PO_3)_6}$ or $\mathrm{Na_6P_6O_{18}}$ Order of abundance of elements in earth's crust is $\mathrm{O>Si>Al>Fe>Ca>Na>Mg>K}$ So, second most abundant element in earth's crust is $\mathrm{Si}$, not $\mathrm{Ca}$.

The reactions shown are examples of Sandmeyer and Gatterman reactions. In the Sandmeyer reaction, $\mathrm{N_2^+Cl^-}$ reacts with $\mathrm{Cu_2Cl_2}$ to form chlorobenzene and $\mathrm{N_2}$. In the Gatterman reaction, $\mathrm{N_2^+Cl^-}$ reacts with $\mathrm{Cu, HCl}$ to also form chlorobenzene and $\mathrm{N_2}$. The other reactions are Wurtz and Fitting reactions. In the Wurtz reaction, $2\mathrm{CH_3 - CH_2Cl} + 2\mathrm{Na}$ in ether forms $\mathrm{C_2H_5 - C_2H_5} + 2\mathrm{NaCl}$. In the Fitting reaction, $2\mathrm{C_6H_5Cl} + 2\mathrm{Na}$ in ether forms $\mathrm{C_6H_5 - C_6H_5} + 2\mathrm{NaCl}$.

The given reaction involves the reduction of a ketone using the Clemmensen reduction method with $Zn-Hg/HCl$. This converts the ketone to an alkane. The resulting alkane is then subjected to dehydrogenation using $Cr_2O_3$, leading to aromatization and forming ethylbenzene.

As per molecular orbital theory, $\mathrm{Ne_2}$ \hspace{1cm} BO $= 0$ $\mathrm{N_2}$ \hspace{1cm} BO $= 3$ $\mathrm{F_2}$ \hspace{1cm} BO $= 1$ $\mathrm{O_2}$ \hspace{1cm} BO $= 2$

Si $- \mathrm{FeCO_3}$ Calamine $- \mathrm{ZnCO_3}$ Malachite $- \mathrm{CuCO_3 \cdot Cu(OH)_2}$ Cryolite $- \mathrm{Na_3AlF_6}$

If $\mathrm{FeCl_3}$ is added to excess of hot water, a positively charged sol of hydrated ferric oxide is formed due to adsorption of $\mathrm{Fe^{3+}}$ ions.

Given $\mathrm{Na^+} = 70 \, \mathrm{mg/mL}$. $W_{\mathrm{Nat}}$ in 50 mL solution $= 70 \times 50 \, \mathrm{mg}$ $$= 3500 \, \mathrm{mg}$$ $$= 3.5 \, \mathrm{gm}$$ Moles of $\mathrm{Na^+}$ in 50 mL solution $= \frac{3.5}{23}$. Moles of $\mathrm{NaNO_3} = moles of \mathrm{Na^+}$ $$= \frac{3.5}{23} \, \mathrm{mol}$$ Mass of $\mathrm{NaNO_3} = \frac{3.5}{23} \times 85 = 12.934$ $$\simeq 13 \, \mathrm{gm} Ans.$$

$\mathrm{Zn}(s)\ |\ \mathrm{Zn}^{2+}(0.1\,\mathrm{M})\ |\ \mathrm{Ag}^{+}(0.01\,\mathrm{M})\ |\ \mathrm{Ag}(s)$ $\mathrm{Zn}(s)+2\mathrm{Ag}^{+}\rightleftharpoons2\mathrm{Ag}(s)+\mathrm{Zn}^{2+}$ $E^\circ_{\mathrm{cell}}=0.80+0.76=1.56\,\mathrm{V}$ $Q=\frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Ag}^{+}]^2}$ $E_{\mathrm{cell}}=E^\circ_{\mathrm{cell}}-\frac{0.059}{2}\log(Q)$ $E_{\mathrm{cell}}=1.56-\frac{0.059}{2}\log\left(\frac{0.1}{(0.01)^2}\right)$ $E_{\mathrm{cell}}=1.56-\frac{0.059}{2}\log(10^3)$ $E_{\mathrm{cell}}=1.56-\frac{0.059}{2}\times3$ $E_{\mathrm{cell}}=1.4715\,\mathrm{V}$ $=147.15\times10^{-2}\,\mathrm{V}$ $x=147.15\approx147$ Answer: $147$

The reaction is given as $n \, \mathrm{PhCOOH} \rightarrow (\mathrm{PhCOOH})_n$. The van't Hoff factor $N$ is given by $N = \frac{1}{X} = i \{ As \alpha = 1 \}$. The freezing point depression is given by $$\Delta T_f = i \times k_f \times m$$. Substituting the values, we have $$0.93 = \frac{1}{n} \times 1.86 \times \frac{12.2 \times 1000}{122 \times 100}$$ Solving for $n$, we find $$n = 2$$

The reaction is given by: $$\mathrm{SF_6(g)} \longrightarrow \mathrm{S(g)} + 6 \mathrm{F(g)}$$ The enthalpy change of the reaction is: $$\Delta H^\circ_{reaction} = 6 \times E_{S--F} = \Delta H^\circ_f[\mathrm{S(g)}] + 6 \times \Delta H^\circ_f[\mathrm{F(g)}] - \Delta H^\circ_f[\mathrm{SF_6(g)}]$$ Substituting the values: $$6 \times E_{S--F} = 275 + 6 \times 80 - (-1100)$$ Simplifying: $$= 275 + 480 + 1100$$ $$6 \times E_{S--F} = 1855$$ Solving for $E_{S--F}$: $$E_{S--F} = \frac{1855}{6} = 309.1667$$ Therefore, the bond energy is approximately: $$\simeq 309 \, kJ/mol$$ Ans.

Given $m = 10g = 10^{-2} Kg$ and $v = 90 \, m/sec$. $\Delta v = v \times 5\% = 90 \times \frac{5}{100} = 4.5 \, m/sec$ Using the uncertainty principle: $$m \cdot \Delta v \cdot \Delta x \geq \frac{h}{4\pi}$$ Substituting the values: $$10^{-2} \times 4.5 \times \Delta x \geq \frac{6.63 \times 3 \times 10^{-34}}{4 \times \frac{22}{7}}$$ Simplifying: $$\Delta x \geq \frac{6.63 \times 7 \times 2 \times 10^{-34}}{9 \times 4 \times 22 \times 10^{-2}}$$ Further simplification gives: $$\Delta x \geq 1.17 \times 10^{-33} = x \times 10^{-33}$$ Thus, $x = 1.17 \approx 1$

Assuming FCC No of lattice sites = 6 face centre + 8 corner = 14 No. of octahedral voids = 13 Ratio = $\frac{13}{14}$ = 0.92857 $\approx$ 1 (Nearest integer)

Given $\mathrm{S_2O_3^{2-}} + \mathrm{MnO_4^-} \xrightarrow{Alkaline Medium} A$. $A \rightarrow \mathrm{SO_4^{2-}}$. Therefore, oxidation number of 'S' = +6 Ans.

The complex $[\mathrm{Co(ox)_2Br(NH_3)}]^2^-$ has two forms: one optically active and one optically inactive. The optically active form has a mirror image. Total stereoisomer = 2 (optically inactive) + 1 pair of enantiomers (optically active) = 3.

Given $E_a = 80.9 \, \mathrm{kJ/mol}$. Fraction of molecules able to cross energy barrier $= e^{-E_a/RT} = e^{-x}$. $$x = \frac{E_a}{RT} = \frac{80.9 \times 1000}{8.31 \times 700} = 13.91$$ Therefore, $x \approx 14$.

The reaction is $\mathrm{(NH_4)_3PO_4 \rightleftharpoons 3NH_4^+ + PO_4^{3-}}$. The concentration of hydrogen ions is given by $[\mathrm{H^+}] = K_a \times \sqrt{\frac{k_w}{k_a \times k_b}}$. The pH is calculated as follows: $$\mathrm{pH} = \mathrm{pK_a} + \frac{1}{2} \{\mathrm{pK_w} - \mathrm{pK_a} - \mathrm{pK_b}\}$$ $$\mathrm{pH} = 5.23 + \frac{1}{2} \{14 - 5.23 - 4.75\}$$ $$\mathrm{pH} = 5.23 + \frac{1}{2} (4.02) = 7.24 = 7 (Nearest integer)$$