Class 12th

D₁ → forward biased

D₂ → Reversed biased

So, current will flow through only  D₁

$l=\frac{\left(1-0.6\right)v}{\left(60+40\right)\Omega }=\frac{0.4}{100}$                

= 4 * 10^-3 A

= 4mA

$tan\delta =\frac{B_v}{B_h}$

$Bv=B_{h}tan45$                

= 4 * 10^-3

emf induced in the rod is = $B_{v}v\mathcal{l}$  

$=4*10^{-3}*20*\frac{20}{100}$                

$\begin{array}{l}=16*10^{-3}\\ =16mv\end{array}$                 

r_d = (Radius of deuteron)

$=\frac{m_d{v}_d}{q_{d}B}$                

$=\frac{\sqrt[]{2m_{d}k.E}}{q_{d}B}$                

$=\frac{\sqrt[]{2m_P}}{\sqrt[]{m_P}}*\frac{q_P}{q_d}$ $\left[q_P=q_d=1.6*10_C^{-19}\right]$                     

$=\sqrt[]{2}:1$              

Kindly go through the solution

$C_1=\frac{{\in }_{0}A}{5b}$

$C_2=\frac{{\in }_{0}A}{3b}$

$C_3=\frac{{\in }_{0}A}{b}$

$C_{eq}=C_1+C_2+C_3$

$=\frac{{\in }_{0}A}{b}\left[\frac{1}{5}+\frac{1}{3}+1\right]$

$C_{eq}=\frac{23}{15}\frac{{\in }_{0}A}{b}$

A = 8 cm

T = 6 sec

q = 60° from A to B

During reaching the point its maximum amplitude from point (A)

$\theta =60°=\frac{{\pi }^C}{3}$

$\theta =\omega t$                

$\frac{\pi }{3}=\frac{2\pi }{6}t$                              

t = 1 sec

$w=P\Delta v=nR\Delta T=400J$  - (1)

$Q=n{C}_P\Delta T=\frac{nRy}{y-1}\Delta T$                

$Q=\frac{14}{4}*400$                

Q = 1400 J

By conservation of Energy

$mgh=\frac{1}{2}{l}_P{\omega }^2$                

3gh $=\frac{1}{2}\left(M{R}^2+M{R}^2\right){\omega }^2$  

$3gh=m{v}^2$                

$v^2=\frac{3gh}{12}$                

$v=\frac{\sqrt[]{gh}}{2}=\frac{\sqrt[]{xgh}}{2}$          

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F.B.D. of point 'P'

$Tsin\theta =30$            - (1)

$Tcos\theta =100$          - (2)

$\left(1\right)÷\left(2\right)$

$tan\theta =\frac{3}{10}\Rightarrow \theta =ta{n}^{-1}\left(3*10^{-1}\right)$