Physics

Uniform circular motion has constant speed. But SHM requires velocity and acceleration to vary sinusoidally with displacement. SHM is the projection of uniform circular motion. But it's not the motion itself, due to differing energy and force characteristics.

A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously.

$A_{max}=A_c+A_m$

$A_{min}=A_c-A_m$

$\frac{A_{min}}{A_{max}}=\frac{A_c-A_m}{A_c+A_m}=\frac{250-150}{250+150}$

$\Rightarrow \frac{100}{400}=\frac{50}{200}$  

By work energy theorem

Work done = change in K.E.

Work done by friction work done by spring

$=0-\frac{1}{2}m{V}^2$              

As 90% of K.E. is losed by friction so that

$-\frac{90}{100}\left(\frac{1}{2}m{V}^2\right)-\frac{1}{2}K{x}^2=-\frac{1}{2}m{V}^2$                

-K -> -16 * 10⁵

K = 16 * 10⁵

$l=\frac{m{\mathcal{l}}^2}{3}$  

Energy conservation Low

$mg\mathcal{l}=\frac{1}{2}\frac{m{\mathcal{l}}^2}{3}{\omega }^2.....\left(i\right)$                

And speed V =    $\omega r=\omega \mathcal{l}$

then    $V=\sqrt[]{6g\mathcal{l}}=\sqrt[]{6*10*.6}$

->6 m/s

for smooth surface

$a=gsin30°=\frac{g}{2}$

$S_1=ut+\frac{1}{2}a{t}^2$         

$S_1=\frac{1}{2}\frac{g}{2}{t}^2=\frac{g}{4}{t}^2.......\left(i\right)$               

for rough Surface

$S=\frac{1}{2}\frac{g}{2}\left(1-\mu \sqrt[]{3}\right){\alpha }^2{t}^2........\left(ii\right)$               

By (i) and (ii)

$\mu =\frac{1}{\sqrt[]{3}}\left(\frac{{\alpha }^2-1}{{\alpha }^2}\right)x=\sqrt[]{3}$            

For A satellite T₁ = 1hour

$So{\omega }_1=2\pi red/hour$                

for B satellite T₂ = 8 hour

given R₁ = 2 * 10³ Km

Relative  $\omega =\frac{V_1-V_2}{R_2-R_1}=\frac{2\pi *10^3}{6*10^3}$  

$\Rightarrow \frac{\pi }{3}$ rad/hour

 x = 3

K.E. energy of electron = eV

Translational K.E. of N₂ =   $\frac{3}{2}KT$

So eV = $\frac{3}{2}KT$  

$1.6*10^{-19}*0.1=\frac{3}{2}*1.38*10^{-23}*T$                

T = 773 – 273 = 500°C

Since process is isochoric

So    $\Delta U=n{C}_v\Delta T$

$\Delta U=n\left(\frac{5}{2}R\right)\Delta T--(i)\left[C_V=\frac{5}{2}R\right]$      

And external work

$\Delta W=nR\Delta T--\left(ii\right)$    

$\frac{5}{2}=\frac{x}{10}\Rightarrow x=25.00$                

In first condition R₁ = 36 $\Omega$  

In second condition R₂ = 18  $\Omega$

$P_1=\frac{V^2}{R_1}=\frac{{\left(240\right)}^2}{36}$               

$P_2=\frac{V^2}{R_2}+\frac{V^2}{R_2}=\frac{{\left(240\right)}^2}{18}+\frac{{\left(240\right)}^2}{18}$               

$P_2=\frac{{\left(240\right)}^2}{9}$               

So   $\frac{P_1}{P_2}=\frac{{\left(240\right)}^2/36}{{\left(240\right)}^2/9}=\frac{1}{4}$

x = 4