Physics

$t=\sqrt[]{\frac{2h}{g}}=\sqrt[]{\frac{2*19.6}{9.8}}=2sec$

$\therefore x=ut=\frac{9*10^3}{3600}*2=5m$

$Dimensionof\frac{B^2}{{\mu }_0}:-$ Magnetic energy density

$u_B=\frac{B^2}{2{\mu }_0}$

Dimension $\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.\to \left[M{L}^2{T}^{-2}{L}^{-3}\right]$ $\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{${}^{}$}\right.{}^{}{}^{}{}^{}$

$=\left[M{L}^{-1}{T}^{-2}\right]$

Yes. Escape speed from a planet or celestial body is the same for all objects. It is irrespective of their mass. This is because the gravitational force and potential energy depend on the mass of the celestial body and distance. The object's own mass cancels out when deriving escape speed. 

The escape speed from the Earth's surface is approximately 11.2 km/s. This means that the object must travel at least at this speed in the upwards direction to entirely escape Earth's gravitational pull without falling back.  

Escape velocity or escape speed is the minimum velocity required for an object to move away from the gravitational pull of a celestial body or planet. This velocity for escape does not require any additional propulsion. 

$\mathrm{}|\stackrel{?}{\mathrm{P}}+\stackrel{?}{\mathrm{Q}}|=\left|\stackrel{?}{\mathrm{P}}\right|$

^{${\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{P}\mathrm{Q}\mathrm{c}\mathrm{o}\mathrm{s}?\theta ={\mathrm{P}}^2$}

$\Rightarrow \mathrm{}Q+2P\mathrm{c}\mathrm{o}\mathrm{s}?\theta =0$

$\Rightarrow \mathrm{}\mathrm{c}\mathrm{o}\mathrm{s}?\theta =-\frac{\mathrm{Q}}{2\mathrm{P}}$

$\mathrm{t}\mathrm{a}\mathrm{n}?\alpha =\frac{2\mathrm{P}\mathrm{s}\mathrm{i}\mathrm{n}?\theta }{2\mathrm{P}\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{Q}}=\mathrm{\infty }$

$?\mathrm{}[2\mathrm{P}\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{Q}=0]$

$\alpha ={90}^{?}$

Energy of photon, $E=\frac{1240}{310}=4\mathrm{e}\mathrm{V}>2\mathrm{e}\mathrm{V}\mathrm{}$ ${}_{}{}_{}{}_{}{}_{}$  (so photoelectric effect will take place)

$=4*1.6*{10}^{-19}=6.4*{10}^{-19}\text{Joule}$

Number of photons falling per second

$=\frac{6.4*{10}^{-5}*1}{6.4*{10}^{-19}}={10}^{14}$

Number of photoelectron emitted per second

$=\frac{{10}^{14}}{{10}^3}={10}^{11}$

11

Loss of energy $=\frac{1}{2}{\mathrm{C}\mathrm{V}}^2-\frac{1}{2}\left(2\mathrm{C}\right)*{\mathrm{V}}_{\text{Common}}^2$ $\frac{}{}{}^{}\frac{}{}{}_{}^{}$

$\begin{array}{rr}& =\frac{1}{2}*60*{10}^{-12}*(20{)}^2-60*{10}^{-12}*(10{)}^2\\ & =60*{10}^{-12}(200-100)\\ & =6000*{10}^{-12}\\ & =6\mathrm{n}\mathrm{J}\end{array}$

n=12

$\mathrm{K}\mathrm{E}={\mathrm{P}\mathrm{E}}_1-{\mathrm{P}\mathrm{E}}_2=mg{\mathrm{h}}_1-{\mathrm{m}\mathrm{g}\mathrm{h}}_2$

10

Common potential after connection.

${\mathrm{V}}_{\text{common}}=\frac{{\mathrm{C}}_1{\mathrm{V}}_1+{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{60*20+0}{120}=10\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{t}$

Loss of energy $=\frac{1}{2}{\mathrm{C}\mathrm{V}}^2-\frac{1}{2}\left(2\mathrm{C}\right)*{\mathrm{V}}_{\text{Common}}^2$

$\begin{array}{rr}& =\frac{1}{2}*60*{10}^{-12}*(20{)}^2-60*{10}^{-12}*(10{)}^2\\ & =60*{10}^{-12}(200-100)\\ & =6000*{10}^{-12}\\ & =6\mathrm{n}\mathrm{J}\end{array}$