Physics

For uniformly charged spherical shell,

$V=\frac{kq}{R}\left( For r\le R\right)$

$\therefore V_C=V_P$

$V_C-V_P= Zero$

In the case for maximum elongation,

Stress = Elastic limit

${\delta }_{max}=\frac{{\sigma }_{elastic }*L}{ Young\text{'}s modulus }=\frac{8*10^8*1}{2*10^{11}}=4*10^{-3}$

$=4 mm$

i.e. maximum elongation is 4 mm

According to transformer ratio,

$\frac{V_S}{V_P}=\frac{N_S}{N_P}=2:1$

Excess force  $=T*2\pi R$

$=\frac{7}{100}*2*3.14*\frac{4.5}{100}$

$=197.82*10^{-4}$

$=19.8*10^{-3} N$

$=19.8mN$

Energy difference $\Delta E=\frac{hc}{\lambda }$

$\therefore \lambda \propto \frac{1}{\Delta E}$

${(\Delta E)}_{6-2}>{(\Delta E)}_{5-2}>{(\Delta E)}_{4-2}>{(\Delta E)}_{3-2}$

${\lambda }_{6-2}<{\lambda }_{5-2}<{\lambda }_{4-2}<{\lambda }_{3-2}$
 

$Y_1=\overline{A·A}$

$=\overline{A}$

$Y_2=\overline{B+B}$

$=\overline{B}$

$Y=\overline{Y_1+Y_2}$

$=\overline{\overline{A}+\overline{B}}$

$=\overline{\overline{A}}·\overline{\overline{B}}$

$=A·B is similar to output of AND Gate$

(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  $\frac{1}{4}={\mu }^2-1$

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