physics ncert exemplar solution class 12th chapter one

If $P_1=P$ then $P_2=P+5\%$ of $P=1.05P.$

From Boyle's law, $PV=constant$ therefore,

$\frac{V_2}{V_1}=\frac{P_1}{P_2}=\frac{P}{1.05P}=\frac{100}{105}$

Fractional change in volume is

$\frac{\Delta V}{V}=\frac{V_2-V_1}{V_1}=\frac{100-105}{105}=-\frac{5}{105}$

Therefore, percentage change in volume is

$\frac{\Delta V}{V}*100\%=-\frac{5}{105}*100\%=-4.76\%$

That is, volume decreases by $4.76\mathrm{\%}.$

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

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

m  =l0 kg

h= 5m

By work energy theorem

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

$v^2=2$ gh 

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

$v=\sqrt[]{100}$

$v=10m/s$

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

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

By using 

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

The charge on hole is positive.

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

According to Einstein's photoelectric equation maximum kinetic energy of photoelectrons, $K{E}_{ max }=E_v-\phi$

or   $2=5-\phi$ $\therefore \phi =3eV$

When $E_v=6eV$  then, $K{E}_{ max }=6-3=3eV$

or $e\left(V_{cathode}-V_{anode}\right)=3eV$

or   $V_{cathode}-V_{anode}=3V=-V_{stopping}$

  $\therefore V_{stopping}=-3V$