Physics

$\frac{1}{0.9}=+\frac{1}{\left(10?2t\right)}+\frac{1}{2t}$  

t = 0.5 sec and t = 4.5 sec.

Here the width of principal maxima is 2.5 mm

Then its half width is =  $\frac{\beta }{2}$  

$=\frac{2.5}{2}=1.25*10^{-3}m$             

Diffraction angle q = $\frac{\left(\frac{\beta }{2}\right)}{D}=\frac{1.25*10^{-3}}{2}$       …(i)

  $\therefore a\theta =\lambda \therefore \theta =\frac{\lambda }{a}$   …(ii)

From (i) and (ii)

  $\lambda =\theta a$          

$\frac{1.25*10^{-3}}{2}*10^{-3}$            

$\therefore \lambda =6250$ = 6.25 * 10^-7 m

Optical path $=\mu x$ =μx

Path-difference $=\mu x$ =μx

Phase-difference $=\frac{2\pi }{\lambda }*$ =2πλ*  path-difference $=\frac{2\pi }{\lambda }\mu x$ =2πλμx

To reach at point A particle must cross the peak point.

Loss in KE = gain in PE

$\frac{1}{2}*\frac{100}{1000}*v^2=\left(5-0\right)$            

v² = 100

v = 10 m/s

Area of ellipse =  $\pi ab$

Magnetic dipole moment =    $\frac{\pi abi}{2}$

Torque    $\overrightarrow{\tau }=\overrightarrow{M}*\overrightarrow{B}$

  $\tau =MBsin\theta$          

$?\overrightarrow{M}\left|\right|\overrightarrow{B}$ $\boxed{\tau =0}$               

${\left(n_i\right)}^2=n_e*n_h$ where $n_e=\frac{5*{10}^{28}}{{10}^6}$

$\begin{array}{rr}\therefore \mathrm{}{n}_h& =\frac{{\left(n_i\right)}^2}{n_e}=\frac{\left(1.5*{10}^{16}\right)\left(1.5*{10}^{16}\right)*{10}^6}{5*{10}^{28}}\\ n_h& =4.5*{10}^9/{\mathrm{m}}^3\end{array}$

  $\frac{1}{2}\mu v_r^2=\frac{1}{2}k{x}^2$  

  $x={V_{rer}}^2\sqrt[]{\frac{\mu }{k}}$          

$=2*\sqrt[]{\frac{8/6}{10}}=\sqrt[]{\frac{8}{15}}$            

From conservation of momentum

2 * 4 + 4 * 2 = 2v₁ + 4v₂

8 = v₁ + 2v₂                           ….(1)

From conservation of energy

  $\frac{1}{2}*2*4^2+\frac{1}{2}*4*2^2=\frac{1}{2}*2*v_1^2+\frac{1}{2}*4*v_2^2$           ….(ii)

On solving   $v_1=\frac{4}{3}m/s,v_2=\frac{10}{3}m/s$

${}_7^{14}\mathrm{N}$ contains 7 protons and 7 neutrons.

Mass defect,  $\left({}_{}{}_{}\right)\left(\begin{array}{c}\\ \end{array}\right)$ $\mathrm{\Delta }m=\left(7m_H+7m_n\right)-m\left(\begin{array}{c}14\\ 7\end{array}\right)=(7*1.00783u)+(7*1.00867u)-14.00307u$

$=0.11243u$

Binding energy $=0.11243*931.5\mathrm{M}\mathrm{e}\mathrm{V}=104.7\mathrm{M}\mathrm{e}\mathrm{V}$

$i_i=\frac{84}{140}=0.6A$

$i_2=\frac{84}{60}=1.4A$

p.d across capacitor = 20 v

Q = CV = 5 * 20 = 100    $?C$

After S is open

Q = Q . e^-t/RC

=100 e^-4