Class 12th

(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.

$A=90^{?}$

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  $\frac{1}{4}={\mu }^2-1$

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

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 $t^{\prime }=\frac{\mathcal{l}}{2}$

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

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

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

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

(1) 4 mol of $=4{}_A$ atoms

(2) 4 u of $=\frac{4u}{4u}=1$ atom

(3) 4 g of Helium $=\frac{4}{4}$ mole $=1$ mole $=N_A$ He atom

(4) $2.2710982$ of He at STP $=\frac{2.271}{22.710982}$ mole

$=0.1$

$=0.1{}_A$

The magnitude of magnetic field due to circular coil of $N$  turns is given by

$B_C=\frac{{\mu }_{0}iN}{2R}$

$=\frac{4\pi *10^{-7}*7*100}{2*0.1}$

$=4.4*10^{-3} T$

$=4.4mT$

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