Physics

$As{V}^2=V_R^2+C_C^2$

$?V_R=\sqrt[]{V^2-V_C^2}=\sqrt[]{{10}^2-8^2}=6V$

$\tan ?\theta =\frac{X_C}{X_R}=\frac{V_C}{V_R}=\frac{8}{6}=\frac{4}{3}$  

Now,

From first law of thermodynamics

$?W=?Q-?U$

$?Q=n{C}_p?T$

$?U=n{C}_v?T$

$?W=n(C_p-C_v)?T$

$\frac{?Q}{?W}=\frac{\gamma }{\gamma -1}\left. [\frac{C_p}{C_p}=\gamma \right]$

From F.B. Ds of

$9kg\to 90-T_1=9a\dots .\left(1\right)$

$6kg\to T_1-T_2=6a\dots ..\left(2\right)$

$5kg\to T_2-50=5a\dots ..\left(3\right)$

$\left(1\right)+\left(2\right)+\left(3\right)$

$40=20a?a=2m/s^2$

Bulk modulus

$B=\frac{-\Delta P}{\Delta V/V}$

$B=-V\left[\frac{\Delta P}{\Delta V}\right]$

$B=-V\left[\frac{\partial P}{\partial V}\right]$

^{$B=-V\left[\frac{\partial P}{\partial V}\right]$}

$B=-V\left[\frac{-P}{V}\right]$

$\Rightarrow B=P$

Here  $\Delta V_c=0.5V,\Delta i_c=0.05mA=0.05*10^{-3}A$

Output resistance is given by

$R_{out}=\frac{\Delta V_c}{\Delta i_c}=\frac{0.5}{0.5*10^{-3}}=10^4\Omega =10k\Omega .$

For minima

$d\cos ?\theta =\left(n+\frac{1}{2}\right)\lambda$

$?\cos ?\theta =\frac{2n+1}{8}$

$?n=-4,-3,-2,-1,0,1,2,3$

$?8$

minimas

m  =l0 kg

h= 5m

By work energy theorem

$mgh=\frac{1}{2}m{v}^2$

$v^2=2$ gh 

$v=\sqrt[]{2gh}$

$v=\sqrt[]{100}$

$v=10m/s$

From ampere's circuital law

$?\stackrel{?}{B}.\stackrel{?}{dl}={\mu }_0{i}_{enc}$

$={\mu }_0\left(3+1\right)=4{\mu }_0$

The relation between $?$ and $?$ is given as

$?=\frac{?}{1-?}$

${?}_1=\frac{\frac{20}{21}}{1-\frac{20}{21}}=20$

${?}_2=\frac{\frac{100}{101}}{1-\frac{100}{101}}=100$