Class 12th

Maximum Distance = 3 (x + x) = 6x = 3 * 2x = 3 * 50 = 150 cm

 $l={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{co{s}^{2}x}{1+3^x}dx..........\left(i\right)}$

Using properties, ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$

$l={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{co{s}^{2}x}{1+3^{-x}}dx}={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{3^{x}co{s}^{2}x}{1+3^x}dx.......\left(ii\right)}$

Adding (i) and (ii), we get

$2l={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{\left(1+3^x\right)co{s}^{2}x}{1+3^x}}dx={\displaystyle \underset{-\pi /2}{\overset{\pi /2}{\int }}co{s}^{2}xdx=2}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}co{s}^{2}xdx}$

$\therefore l=\frac{\pi }{4}$

Case I :- $2?-?=\frac{1}{2}m{v}_1^2........\left(i\right)$ $\frac{}{}{}_{}^{}$

Case II :- $10?-?=\frac{1}{2}m{v}_2^2........\left(ii\right)$ $\frac{}{}{}_{}^{}$

$\Rightarrow \frac{1}{9}=\frac{v_1^2}{v_2^2}$

$\Rightarrow v_1:v_2=1:3$

x = 1

 $\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill a+2\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill a+3\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill a+4\hfill & \hfill 1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill a\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill a\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill a\hfill & \hfill 1\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill 4\hfill & \hfill 1\hfill \end{array}\right|,$ [using properties]

$=0+\left(a+1\right)\left(a+2\right)\left(-1\right)+2\left(a+3\right)\left(a+4\right)0\left(a+2\right)\left(a+3\right)+4\left(\left(a+2\right)\left(a+3\right)-3\left(a+3\right)\left(a+4\right)\right)$

$=2\left(a+2\right)-2\left(a+3\right)=2\left(a+2-a-3\right)=-2$

$\frac{si{n}^{-1}x}{a}=\frac{co{s}^{-1}x}{b}=\frac{ta{n}^{-1}y}{c}=\lambda$

$\therefore \frac{si{n}^{-1}x}{a}=\lambda$

$\therefore si{n}^{-1}x=a\lambda$

$x=sina\lambda .........\left(i\right)$

$x=cosb\lambda ..........\left(ii\right)$

$y=tanc\lambda ..........\left(iii\right)$

From (i) & (ii), we get, $sina\lambda =cosb\lambda =sin\left(\frac{\pi }{2}-b\lambda \right)$

$\therefore a\lambda =\frac{\pi }{2}-b\lambda \Rightarrow \left(a+b\right)\lambda =\frac{\pi }{2}\Rightarrow \lambda =\frac{\pi }{2\left(a+b\right)}...........\left(iv\right)$

From (iii) y = tan c $\lambda$

$\Rightarrow cos\left(\frac{2c\pi }{2\left(a+b\right)}\right)=\frac{1-y^2}{1+y^2}\Rightarrow cos\left(\frac{\pi c}{a+b}\right)=\frac{1-y^2}{1+y^2}$

$y=\overline{\overline{A}+B}=A.\overline{B}$

All the charge given to a conducting sphere resides on outer surface.

  $e^{\left(co{s}^{2}x+co{s}^{4}x+co{s}^{6}x+.....\infty \right)}lo{g}_{e}2$

$=e^{\frac{co{s}^{2}x}{1-co{s}^{2}x}lo{g}_{e}2}=e^{co{t}^{2}xlo{g}_{e}2}=2^{co{t}^{2}x}$            

t² – 9t + 8 = 0

(t – 8) (t – 1) = 0

$t=2^{co{t}^{2}x}=8=2^3$        

$\Rightarrow co{t}^{2}x=3=co{t}^2\frac{\pi }{6}$       

$\frac{2sinx}{sinx+\sqrt[]{3}cosx}=\frac{2*\frac{1}{2}}{\frac{1}{2}+\sqrt[]{3}*\frac{\sqrt[]{3}}{2}}=\frac{1}{2}$

$\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{5^2}\right)=13.6*\frac{24}{25}eV$

$\Rightarrow \frac{hc}{\lambda }=13.6*\frac{24}{25}eV..........\left(1\right)$

With the help of conservation of linear momentum, we can write

$\frac{h}{\lambda }=m_H{v}_H\Rightarrow \frac{hc}{\lambda }=c{m}_H{v}_H\Rightarrow v_H=\frac{\frac{hc}{\lambda }}{c{m}_H}=\frac{13.6*\frac{24}{25}*1.6*10^{-19}}{3*10^8*1.67*10^{-27}}=4.17m/s$

$R_i=\frac{\rho \mathcal{l}}{A}$

$R_f=\frac{\rho \left(1.25\mathcal{l}\right)}{\left(A/1.25\right)}={\left(1.25\right)}^2*\frac{\rho \mathcal{l}}{A}$

$\Rightarrow R_f=1.5625*R_i$

$\Rightarrow \frac{R_f-R_l}{R_l}*100=56.25\mathrm{\%}$