Physics

$\frac{\mathrm{N}}{{\mathrm{N}}_0}=\frac{1}{16}={\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{{\mathrm{T}}_{1/2}}}$

$\mathrm{t}=4{\mathrm{T}}_{1/2}=20*4=80\text{minutes}$

Least count = 1 mm / 100 = 0.01 mm
Diameter = main scale reading + circular scale reading
Diameter = 0 + 52 * 0.01 mm
= 0.52 mm = 0.052 cm

$?\frac{1}{\lambda }=\mathrm{R}\left(\frac{1}{{\mathrm{n}}_2^2}-\frac{1}{{\mathrm{n}}_1^2}\right)$

$\Rightarrow \frac{1}{\lambda }=\mathrm{R}\left(\frac{1}{2^2}\right)$

$\Rightarrow \lambda =\frac{4}{\mathrm{R}}$

$\frac{1}{{\lambda }^{\mathrm{\text{'}}}}=\mathrm{R}\left(\frac{1}{4^2}\right)$

$\Rightarrow {\lambda }^{\mathrm{\text{'}}}=\frac{16}{\mathrm{R}}$

$\Rightarrow {\lambda }^{\mathrm{\text{'}}}=4\lambda$

Incident energy $=2.20\mathrm{e}\mathrm{V}$

If $?<2.20\mathrm{e}\mathrm{V}$  electron will emit.

$?>2.20\mathrm{e}\mathrm{V}\mathrm{}$ No electron emission

Only caesium will emit electron

$\mathrm{e}\mathrm{V}=$ Energy of electron

for minimum wavelength, maximum loss of energy

$\mathrm{e}\mathrm{V}=\left(\frac{\mathrm{h}\mathrm{c}}{\lambda }\right)$

$\lambda \propto \left(\frac{1}{\mathrm{V}}\right)$

$\mathrm{\Delta }\theta =\frac{\lambda }{\mathrm{d}}$

$\mathrm{\Delta }\theta$ is proportional to $\lambda$ but independent of $D$

Speed of light in air is

$\mathrm{C}=\frac{\mathrm{x}}{{\mathrm{t}}_1}$

Speed of light in another denser medium.

${\mathrm{C}}_2=\frac{10\mathrm{x}}{{\mathrm{t}}_2}$

$\Rightarrow \mu =\frac{C}{C_2}=\frac{x}{t_1}*\frac{t_2}{10x}$

$\Rightarrow \mu =\frac{{\mathrm{t}}_2}{10{\mathrm{t}}_1}$

For total internal reflection

$\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{C}=\frac{1}{\mu }$

$\Rightarrow \mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{C}=\frac{10{\mathrm{t}}_1}{{\mathrm{t}}_2}$

$\mathrm{C}={\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}?\left(\frac{10{\mathrm{t}}_1}{{\mathrm{t}}_2}\right)$

$\mathrm{B}=\frac{\mathrm{E}}{\mathrm{C}}=\frac{48}{3*{10}^8}=16*{10}^{-8}$

$=1.6*{10}^{-7}\mathrm{T}$

Capacitive reactance $=\frac{1}{\omega \mathrm{C}}={\mathrm{X}}_{\mathrm{c}}$ (say) on decreasing the operating frequency $\omega$  reduces

As ${\mathrm{X}}_{\mathrm{c}}$ is inversely proportional to $\omega$  the value of ${\mathrm{X}}_{\mathrm{c}}$  increase

$\therefore {\mathrm{I}}_{\mathrm{C}}={\mathrm{I}}_{\mathrm{D}}$

$=\frac{{\mathrm{V}}_{\mathrm{o}}}{{\mathrm{X}}_{\mathrm{c}}}$

As $X_c$  increases, therefore displacement current Id decreases.

For transformer $\left(\frac{i_p}{i_s}\right)=\left(\frac{V_s}{V_P}\right)=\left(\frac{N_s}{N_P}\right)$

For bulb, $\mathrm{V}\mathrm{s}=12\mathrm{V}$

$\begin{array}{rr}& {\mathrm{i}}_{\mathrm{s}}=\left(\frac{60}{12}\right)=5\mathrm{A}\\ & \Rightarrow \frac{{\mathrm{i}}_{\mathrm{p}}}{5}=\left(\frac{12}{220}\right)\\ & \mathrm{}{\mathrm{i}}_{\mathrm{P}}=\frac{60}{220}=\left(\frac{3}{11}\right)=0.27\mathrm{A}\end{array}$