Physics

Use formula for M.I.

$x=4sin\left(\frac{\pi }{2}-\omega t\right)$  - (i)

$y=4sin\left(\omega t\right)$   - (ii)

From (i) and (ii) cos^{2 $\omega t+si{n}^2\omega t={\left(\frac{x}{4}\right)}^2+{\left(\frac{y}{4}\right)}^2=1$}

$\Rightarrow x^2+y^2={\left(4\right)}^2$  

a = k2rt2

$\Rightarrow \frac{v^2}{r}=k^{2}r{t}^2$

$\Rightarrow v=krt$

$a_t=\frac{dv}{dt}=kr$

? tangential force, F_t = ma_t = mkr

$\therefore Powerdelivered,P=F_{t}v=\left(mkr\right)\left(krt\right)=m{k}^2{r}^{2}t$  

 Note ® Power delivered by centripetal force will be zero.

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f =  $\frac{1}{2\pi \sqrt[]{LC}}=\frac{1}{2\pi \sqrt[]{\left(0.5*10^{-3}\right)*\left(200*10^{-6}\right)}}$  

$\Rightarrow f=\frac{10^4}{2\pi \sqrt[]{10}}\approx 5*10^{2}Hz$         $\left[Taking\pi =\sqrt[]{10}\right]$  

According to Young's double slit experiment, we can write

$\beta =\frac{\lambda D}{d}\Rightarrow \Delta \beta ={\beta }_2-{\beta }_1=\frac{\lambda }{d}\Delta D$

$\Rightarrow \lambda =\frac{d\Delta \beta }{\Delta D}=\frac{1*10^{-3}*3*10^{-5}}{5*10^{-2}}=60*10^{-8}m=600nm$

According to Nuclear activity, we can write

$N_0\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{2}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{4}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{8}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{16}=\left(0.0625\right)N_0$  

Time required = 4 *  $t_{\frac{1}{2}}=20yrs$

According to question, we can write

$\mathcal{l}=2\pi r\Rightarrow r=\frac{\mathcal{l}}{2\pi }$  

$I_1=\frac{m{\mathcal{l}}^2}{3},and{I}_2=\frac{m{r}^2}{2}=\frac{m{\mathcal{l}}^2}{8{\pi }^2}$

$\Rightarrow \frac{I_1}{I_2}=\frac{m{\mathcal{l}}^2}{3}*\frac{8{\pi }^2}{m{\mathcal{l}}^2}=\frac{8{\pi }^2}{3}$

According to definition of displacement current, we can write

$I_d={\epsilon }_0\frac{d?}{dt}={\epsilon }_0\frac{d\left(ES\right)}{dt}={\epsilon }_0\frac{d\left(\frac{V}{\mathcal{l}}S\right)}{dt}=\frac{{\epsilon }_{0}S}{\mathcal{l}}\left(\frac{dV}{dt}\right)$  

$\Rightarrow \mathcal{l}=\frac{8.85*10^{-12}*40*10^{-4}*10^6}{4.425*10^{-6}}8*10^{-3}m$  

According to Work energy theorem, we can write

$K_f-K_i=W_{ElectricForce}\Rightarrow \frac{1}{2}m{v}^2-\frac{1}{2}m{v}_0^2=-eV\Rightarrow v^2=v_0^2-\frac{2eV}{m}$  

$\Rightarrow v^2={\left(6.0*10^5\right)}^2-\frac{2*1.6*10^{-19}}{9*10^{-31}}=\frac{324-128}{9}*10^{10}=\frac{196}{9}*10^{10}\Rightarrow V=\frac{14}{3}*10^{5}m/s$  

According to Kirchhoff's Law, we can write

-20 + 2000I + 600 * 5I = 0  $\Rightarrow I=\frac{20}{5000}A$  

Reading of voltmeter = 2000I = 2000 *  $\frac{20}{5000}=8volt$