Physics

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

At maximum height velocity (v) = 0

So, momentum = mv = m * 0 = 0

$\frac{sin{\theta }_C}{V_B}=\frac{sin90°}{V_A}\Rightarrow sin{\theta }_C=\frac{V_B}{V_A}=\frac{1.5*10^{10}}{2.0*10^{10}}=\frac{3}{4}$

According to question, we can write

  $\theta >\theta =si{n}^{-1}\left(\frac{3}{4}\right)$

.B.D of person w.r.t. elevator

mg – N = ma

N = mg – ma

N decreases so, person experiences weight loss

According to question, we can write

    $F_{CA}=F_{CB}=\frac{KqQ}{x^2+{\left(\frac{d}{2}\right)}^2}\Rightarrow F=2F_{CA}cos\theta =\frac{2KqQx}{{\left[x^2+{\left(\frac{d}{2}\right)}^2\right]}^{\frac{3}{2}}}$                    

For maxima of force   $\frac{dF}{dx}=0,so$

$x=\frac{d}{2\sqrt[]{2}}$

$P=\frac{\alpha }{\beta }lo{g}_e\left(\frac{kt}{\beta x}\right)$

$\frac{kt}{\beta x}$ = Dimensionless

$\beta =\frac{kt}{x}=\frac{\left[M{L}^2{T}^{-2}{k}^{-1}\right]\left[k\right]}{\left[L\right]}$

$\frac{\alpha }{\beta }=$ dimensionless

a = dimensionless of b

a = MLT^-2

This refers to the highest limit of pressure which the body can tolerate in order to regain its original shape after removing the pressure. If the pressure exceeds the elastic limit, the body will never be able to regain its original shape agian. example: rubber stretching.

According to question, we can write

$d=d_0\left(1+\alpha \Delta T\right)\Rightarrow 6.241=6.230\left(1+1.4*10^{-5}*\Delta T\right)$

$\Rightarrow T=126.18+27=153.18°C$