Energy of photon, $E=$ Rhc $(1-\frac{1}{n^2})$

$\Rightarrow \frac{\mathrm{hc}}{\lambda }=\mathrm{Rhc}(1-\frac{1}{{\mathrm{n}}^2})$

$\Rightarrow \frac{1}{\lambda }=\mathrm{R}(1-\frac{1}{{\mathrm{n}}^2})$

$\Rightarrow \frac{1}{\lambda R}=1-\frac{1}{n^2}$

$\Rightarrow \frac{1}{{\mathrm{n}}^2}=1-\frac{1}{\lambda R}=\frac{\lambda R-1}{\lambda R}$

$\Rightarrow \mathrm{n}=\sqrt{\frac{\lambda \mathrm{R}}{(\lambda R-1)}}$

The increase in the width of the depletion region in a $\mathrm{p}-\mathrm{n}$ junction diode is due to reverse bias only.

The increase in the width of the depletion region is due to the absence of the electrons and holes in the region. This occurs only in the case of the reverse bias only in a diode.

Given,

$B_E=0.4\mathrm{G}=4\times {10}^{-5}\mathrm{T}$

The radius ofthe earth$r=6.4\times {10}^6\mathrm{m}$

The equatorial magnetic field is,

$B_E=\frac{{\mu }_{0}m}{4\pi r^3}$

$m=\frac{4\times {10}^{-5}\times {(6.4\times {10}^6)}^3}{\frac{{\mu }_0}{4\pi }}$

As we know,

$\frac{{\mu }_0}{4\pi }={10}^{-7}$

$=\frac{4\times {10}^{-5}\times {(6.4\times {10}^6)}^3}{{10}^{-7}}$

$=4\times {10}^2\times {(6.4\times {10}^6)}^3$

$=1.05\times {10}^{23}{\mathrm{Am}}^2$

Magnifying power of telescope,

$MP=\frac{\beta (\text{ angle subtended by image at eye piece })}{\alpha (\text{ angle subtended by object on objective })}$

Also, $\mathrm{MP}=\frac{{\mathrm{f}}_{\mathrm{o}}}{{\mathrm{f}}_{\mathrm{e}}}=\frac{150}{5}=30$

$\alpha =\frac{50}{1000}=\frac{1}{20}\mathrm{rad}$

$\therefore \beta =\theta =\mathrm{MP}\times \alpha =30\times \frac{1}{20}=\frac{3}{2}=1.5\mathrm{rad}$

or, $\beta =1.5\times \frac{{180}^{\circ }}{\pi }\simeq {60}^{\circ }$

Given:

Capacitance, (C) $=1\mu F$,

Rate of change in voltage, $\frac{dV}{dt}={10}^{6}V/s$

The expression for displacement current $i_{d,}$ in the case of capacitance is given as:

$i_d=\frac{dq}{dt}$$$.....(i)

As we also know that, the charge on the capacitor is:

$q=CV$

Where q = charge on the capacitors

On substituting the value of q = CV in equation (i), we get,

$i_d=C\frac{dV}{dt}$$$.....(ii)

On substituting the given values in equation (ii), we get,

$i_d=({10}^{-6}\times {10}^6)A$

$i_d=1A$

Given:

$\varphi =2.8\mathrm{eV},\mathrm{E}=2\mathrm{eV}$

We know that Maximum kinetic energy $(\mathrm{E})=\mathrm{h}\nu -\varphi$

Put the given values in above formula.

$2=\mathrm{h}\nu -2.8$

$\Rightarrow \mathrm{h}\nu =4.8\mathrm{eV}$

New frequency $v^{\prime }=2\nu$

So, $E^{\prime }=h{\nu }^{\prime }-\varphi$

$\Rightarrow {\mathrm{E}}^{\prime }=2\mathrm{h}\nu -\varphi$

Put the given values in above formula.

$=2\times 4.8-2.8$

$=6.8\mathrm{eV}$

As given,

$y={\sin }^2\omega t$

$\Rightarrow y=\frac{1-\cos 2\omega t}{2}$

$\Rightarrow y=\frac{1}{2}-\frac{\cos 2\omega t}{2}$

It is a periodic motion but it is not SHM.

$\therefore$ Angular speed $=2\omega$

$\therefore$ Period $T=\frac{2\pi }{\text{ angular speed }}$

$\Rightarrow T=\frac{2\pi }{2\omega }$

$\Rightarrow T=\frac{\pi }{\omega }$

${}_{92}{\mathrm{U}}^{235}+{}_o{\mathrm{n}}^1\to {}_{\mathrm{Q}}{\mathrm{X}}^{\mathrm{P}}+{}_{36}{\mathrm{Kr}}^{89}+3{}_{\circ }{\mathrm{n}}^1+\mathrm{Q}($ energy $)$

$\sum$ Atomic number on LHS $=92$

$\sum$ Atomic number on $\mathrm{RHS}=\mathrm{Q}+36+3\times 0$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

$Q+36=92$

$Q=56$

also, $\sum$ atomic mass number on

$\mathrm{LHS}=235+1=236$

$\sum$ Atomic mass number on

$\mathrm{RHS}=\mathrm{P}+89+3\times 1$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

$P+92=236$

$P=144$

So, the element is ${}_{56}{\mathrm{Ba}}^{144}$

Given:

Refractive index of the material of the prism, $\mu =\sqrt{2}$

Angle of prism, $A={30}^{\circ }$

As we know

$\mathrm{A}={\mathrm{r}}_1+{\mathrm{r}}_2$

$30={\mathrm{r}}_1+0$

So, ${\mathrm{r}}_1=30$

According to the use of Snell's law,

$1\times \sin \mathrm{i}=\mu \times \sin \mathrm{r}$

$1\times \sin \mathrm{i}=\sqrt{2}\times \sin 30$

$\sin \mathrm{i}=\frac{1}{\sqrt{2}}$

$\mathrm{i}={45}^{\circ }$