Physics

(A) If $c$  is the velocity of light

so,   $E=h\nu$   (Energy of photon)

(B) Velocity of photon is equal to velocity of light i.e. c.

(C)  $\lambda =\frac{h}{p}$

$p=\frac{h}{\lambda }$

$p=\frac{hv}{c}$

(D) In photon-electron collision both total energy and total momentum are conserved.

At central point on screen, path difference is zero for all wavelength. So, central bright fringe is white and other fringes depend on wavelength as $\beta =\frac{\lambda D}{d}$ .

Therefore, other fringes will be coloured.

$V.C=MSD-VSD$     … (i)

given :  $\left(N+1\right)VSD=NMSD$

$VSD=\left(\frac{N}{N+1}\right)MSD$     … (ii)

From (1) and (2)

$V.C=\left(MSD\right)-\frac{N}{N+1}\left(MSD\right)$

$=MSD\left(1-\frac{N}{N+1}\right)=\frac{MSD}{N+1}$

$=\frac{0.01}{N+1}=\frac{1}{100\left(N+1\right)}$

$A=90^{?}$

In prism,  $r_1+c=A$

$r_1=90^{?}-c$                  …(i)

$sinc=\frac{1}{\mu }\Rightarrow cosc=\frac{\sqrt[]{{\mu }^2-1}}{\mu }$

$\Rightarrow$ Apply Snell's law, on incidence surface

$1·sin30^{?}=\mu sin\left(r_1\right)\Rightarrow 1*\frac{1}{2}=\mu *sin\left(90^{?}-c\right)$

$\frac{1}{2}=\mu *\frac{\sqrt[]{{\mu }^2-1}}{\mu }$

On squaring  $\frac{1}{4}={\mu }^2-1$

$\Rightarrow {\mu }^2=\frac{5}{4}\Rightarrow \mu =\frac{\sqrt[]{5}}{2}$

A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously.

Apply energy conservation,

$U_i+K_i=U_f+K_f$

$\Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}m{v}^2$

$\Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}*m*\frac{GM}{3R}$

$\Rightarrow K_i=-\frac{1}{6}\frac{GMm}{R}+\frac{GMm}{R}$

$K_i=\frac{5}{6}\frac{GMm}{R}$

At same temperature, curve with higher volume corresponds to lower pressure.

$V_3>V_2>V_1$

$\Rightarrow P_1>P_2>P_3$

(We draw a straight line parallel to volume axis to get this)

$T^{\prime }=2\pi \sqrt[]{\frac{{?}^{\prime }}{g}}$ where $t^{\prime }=\frac{\mathcal{l}}{2}$

$T=2\pi \sqrt[]{\frac{t}{g}}$

$T^{\prime }=\frac{x}{2}T$

$2\pi \sqrt[]{\frac{t}{2g}}=\frac{x}{2}2\pi \sqrt[]{\frac{t}{g}}$

$\frac{1}{\sqrt[]{2}}=\frac{x}{2}\Rightarrow x=\sqrt[]{2}$

Capacitive Reactance $X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2*3.14*50*10*10^{-6}}$

$=\frac{1000}{3.14}$

$V_{rms}=210 V$

$i_{rms}=\frac{V_{rms}}{X_C}=\frac{210}{X_C}$

$Peak current =\sqrt[]{2}{i}_{ms}=\sqrt[]{2}*\frac{210}{1000}*3.14=0.932$

$?0.93 A$

According to modified Ampere's law

$?B.dI={\mu }_0\left(I_C+I_D\right)$

For Loop  $L_1{I}_C\ne 0$ and $I_D=0$

For Loop  $L_2{I}_C=0$ and $I_D\ne 0$

Due to  $KCL{I}_C=I_D$