Class 12th

$i_b=\frac{5-0.7}{8.6}=0.5mA$

$\Rightarrow I_c=\beta I_b=100*0.5 mA$

By using 

$V_{CE}=V_{CC}-I_C{R}_L=18-50*10^{-3}*100=13 V$

The charge on hole is positive.

$\stackrel{?}{F}=e{E}_0\stackrel{?}{j}$

${\stackrel{?}{a}}_y=\frac{e{E}_0}{m}\stackrel{?}{j}$

$=v_0\stackrel{?}{i}+\left(\frac{e{E}_{0}t}{m}\right)\stackrel{?}{j}$

$\left|\stackrel{?}{v}\right|=\sqrt[]{v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}=v_0\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}$

$\lambda =\frac{{\lambda }_0}{\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}}$

Two successive β decays increase the charge no. by 2.

$C_1=1\mu F{V}_1=100V?U_1=\frac{1}{2}*1*{10}^{-6}*{\left(100\right)}^2=5*{10}^{-3}J$  

$C_2=1\mu F$ , common potential $V=\frac{C_1{V}_1}{C_1{C}_2}=\frac{100}{2}=50V$

Final electrostatic energy stored in both the capacitors.

$=\frac{1}{2}\left(C_1+C_2\right)V^2=\frac{1}{2}*2*{\left(50\right)}^2*{10}^{-6}$

$=2.5*{10}^{-3}J$

$Energylost=2.5*{10}^{-3}$

$\%lossofenergy=\frac{2.5*{10}^{-3}}{5*{10}^{-3}}*100$

$=50\%$

When wire is stretched uniformly

$R\propto \mathcal{R}$

$\frac{R}{R^{\text{'}}}={\left(\frac{l}{nl}\right)}^2=\frac{1}{n^2}$  

$R^{\text{'}}=n^{2}R$

When this elongated wire is cut into 5 parts, resistance of each part  

Effective resistance between A and C $=\frac{n^{2}R}{5}$  

For normal emergence

$r=A$

From Snell's Law

$\frac{\sin ?i}{\sin ?r}=\frac{\mu }{1}$

$?\sin ?i=\sqrt[]{2}\sin ?{30}^0$

$?i={45}^0$

Fraction of molecule having energy $E_a$ or greater than $E_a=e^{\frac{-E_a}{RT}}$  

In case of gas in gas, true solution will be always formed.