Physics

$\frac{l_{max}}{l_{min}}=\frac{{\left(\sqrt[]{\frac{l_1}{l_2}}+1\right)}^2}{{\left(\sqrt[]{\frac{l_1}{l_2}}-1\right)}^2}\Rightarrow \frac{l_{max}-l_{min}}{l_{max}+l_{min}}=\frac{2.\sqrt[]{\frac{l_1}{l_2}}}{\frac{l_1}{l_2}+1}=\frac{2\sqrt[]{2x}}{2x+1}$

$\overrightarrow{AB}+\overrightarrow{AH}=\overrightarrow{AO}$

$\overrightarrow{AC}+\overrightarrow{AG}=2\overrightarrow{AO}$

$\overrightarrow{AD}+\overrightarrow{AF}=3\overrightarrow{AO}$

$\overrightarrow{AE}=2\overrightarrow{AO}$

Adding All,  

$\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}+\overrightarrow{AG}+\overrightarrow{AH}=8\overrightarrow{AO}=16\widehat{i}+24\widehat{j}-32\widehat{k}$

At t = 0, no current flows through inductor.

$i=\frac{E}{\frac{6*9}{6+9}}=\frac{5E}{18}$

At t = $\infty$ $$ , current flows through circuit as if inductor is shorted.

$i=\frac{E}{\frac{5}{2}+\frac{4}{5}}=\frac{10E}{33}$

Key is open.

$i_{rms}=\frac{15}{60}=\frac{1}{4}A$

$20=\frac{1}{4}*100*L\Rightarrow L=0.8H$

$10=\frac{1}{4}.\frac{1}{100C}\Rightarrow C=2.5*10^{-4}F=250\mu F$

K = qV

$p=\sqrt[]{2mk}=\sqrt[]{2mqV}$

$\lambda =\frac{h}{p}=\frac{h}{\sqrt[]{2mqV}}$

$\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt[]{\frac{m_{\alpha }}{m_p}.\frac{q_{\alpha }}{q_p}}=\sqrt[]{4*2}=2*1.4=2.8$

$F_1=\frac{GMm}{9R^2}$

$F_2=\frac{GMm}{9R^2}-\frac{G\frac{M}{8}m}{{\left(\frac{5R}{2}\right)}^2}=\frac{GMm}{R^2}\left(\frac{1}{9}-\frac{1}{50}\right)$

$\frac{F_1}{F_2}=\frac{1}{1-\frac{9}{50}}=\frac{50}{41}$

$l=\frac{1}{2}c{\epsilon }_0{E}_0^2\Rightarrow \frac{8}{4\pi {\left(10\right)}^2}*\frac{1}{2}=\frac{1}{4}*c*\frac{1}{c^2{\mu }_0}*E_0^2\Rightarrow E_0=\frac{2}{10}\sqrt[]{\frac{{\mu }_{0}c}{\pi }}\Rightarrow x=2$

After contact, charges an each sphere will be

$\frac{q_1+q_2}{2}=1nc\Rightarrow F=\frac{k{q}_1{q}_2}{r^2}=36*10^{-9}N$

If angle between them is 180° then

$\overrightarrow{P}*\overrightarrow{Q}=\overrightarrow{q}*\overrightarrow{P}=0$

Using conservation of energy :

$\frac{1}{2}m{v}_i^2-\frac{Gmm}{R}=0-\frac{Gmm}{11R}$

$v_i^2=\frac{20}{11}\frac{Gm}{R}$

Escape velocity is defined as $\sqrt[]{\frac{2GM}{R}}$

$v_i=\sqrt[]{\frac{10}{11}{v}_e}$