Physics

According to question, we can write

  $d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2*125=500m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

Steam point (T₁) = 100°C = 273 + 100 = 373 K

Ice point (T₂) = 0°C = 273 K

$\eta =\left(1-\frac{T_2}{T_1}\right)*100=\left(1-\frac{273}{373}\right)*100=26.81\mathrm{\%}$  

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_P=\frac{{\mu }_{0}I}{2\pi r}*2$

$\Rightarrow 300*10^{-6}=\frac{4\pi *10^{-7}}{2\pi *4*10^{-2}}*2$                                                                         

=> I = 30 A

$g\propto r0\le r\le R$

$g\propto \frac{1}{r^2}r>R$

The minus sign in the formula indicates that:

• A gravitational force exists.
• The satellite cannot leave the orbit by itself.
• A minimum threshold of energy will be required to move the satellite from its existing orbit.

By conservation of Angular momentum

Li = Lf 

$M{R}^2\omega =\left(m{R}^2+2m{R}^2\right)\omega \text{'}$

$\frac{2M}{M+2m}=\omega \text{'}$

$\Rightarrow$ $\omega \text{'}=\frac{2M}{M+2m}$

Process-AB $\Rightarrow$ Isobaric,                                                           

Process-AC $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃

NLM at point Q

N – mg sin a =  $\frac{m{v}^2}{R}$  

$N=\frac{m{v}^2}{R}+mgsin\alpha$           - (1)

COE : between P & Q

$\frac{m{v}^2}{R}=2mgsin\alpha$               - (2)

Centripetal force = 2mg sin a

 N (Normal Reaction) = 2mg sin a + mg sin a

  = 3mg sin a

$\frac{Centripetalforce}{NormalReaction}=\frac{2mgsin\alpha }{3mgsin\alpha }=\frac{2}{3}$             

According to Law of discharging of capacitor, we can write

$Q=Q_0{e}^{-\frac{t}{\tau }}\Rightarrow \frac{Q_2}{8}=Q_0{e}^{-\frac{t_2}{\tau }}\Rightarrow t_{2}3\tau \mathcal{l}n\left(2\right),and$

^{$U=\frac{Q^2}{2C}=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow \frac{1}{2}\left(\frac{Q_0^2}{2C}\right)=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow t_1=\frac{\tau }{2}\mathcal{l}n\left(2\right)$}

^{$\Rightarrow \frac{t_1}{t_2}=\frac{1}{6}$}

The light wave contains two lights of different frequencies, so

  $E_1=h{\upsilon }_1=\frac{4.14*10^{-15}*6*10^{15}}{2\pi }=3.96eV,and$

  $E_2=h{\upsilon }_2=\frac{4.14*10^{-15}*9*10^{15}}{2\pi }=5.92eV$

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV