Physics

$h\nu =3.4-1.51=1.89eV$

As we know that radius of circular path in magnetic field is given as

$r=\frac{\sqrt[]{2mK}}{qB}\Rightarrow K=\frac{r^2{q}^2{B}^2}{2m}$              

$\Rightarrow K=\frac{{\left(7*10^{-3}\right)}^2*\left(1.6*10^{-19}\right)*{\left(5*10^{-4}\right)}^2}{2*9.1*10^{-31}}eV=107.7*10^{-2}eV$  = 1.08 eV

$\Rightarrow ?=h\nu -K=1.89-1.08=0.81eV$              

Method-I : Using Kirchhoff's Law for loop $\overrightarrow{BDABB},$  we can write

$-140+20l+6l_1=0\Rightarrow 10l+3l_1=70.......\left(i\right)$              

Using Kirchhoff's Law for loop $\overrightarrow{ACBBA},$  

we can write

$5\left(l-l_1\right)+90-6l_1=0$              

->-5l + 11l₁ = 90 .(ii)

Adding equation (1) with twice the equation (2), we have

$25l_1=250\Rightarrow l_1=\frac{250}{25}=10A$              

Method-II : Using concept of equivalent cell, we can write

$V_{AB}=\frac{\frac{140}{20}+\frac{0}{6}+\frac{90}{5}}{\frac{1}{20}+\frac{1}{6}+\frac{1}{5}}=\frac{2100+0+5400}{15+50+60}$

$=\frac{7500}{125}=60Volt$

$\Rightarrow l_1=\frac{60}{6}=10A$              

As we know that equivalent half life of radioactive material decays by simultaneous emissions, is given as

$\frac{1}{T}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{1}{700}+\frac{1}{1400}=\frac{3}{1400}\Rightarrow T=\frac{1400}{3}years$

$\Rightarrow t=\frac{\mathcal{l}n\left(3\right)}{\lambda }=\frac{\mathcal{l}n\left(3\right)}{\mathcal{l}n\left(2\right)}T=\frac{1.1}{0.693}*\frac{1400}{3}\approx 740.74years$

Using conservation linear momentum, we can write

$Mv=\frac{h}{\lambda }=\frac{h\nu }{c}\Rightarrow v=\frac{h\nu }{Mc}$                                                           

Loss of internal energy = Gain in Kinetic energy + energy of gamma ray

= $\frac{1}{2}M{v}^2+h\nu =\frac{1}{2}M{\left(\frac{h\nu }{Mc}\right)}^2+h\nu =h\nu \left[1+\frac{h\nu }{2M{c}^2}\right]$  

As we know that root mean square speed is given as

$v=\sqrt[]{\frac{3RT}{M}},so$

$v_A>v_B>v_C\Rightarrow \frac{1}{v_A}<\frac{1}{v_B}<\frac{1}{v_C}as{m}_A<m_B<m_C$

$\overrightarrow{A}.\overrightarrow{B}=\left|\overrightarrow{A}*\overrightarrow{B}\right|\Rightarrow ABcos\theta =ABsin\theta \Rightarrow \theta =45°$             

$\left|\overrightarrow{A}-\overrightarrow{B}\right|=\sqrt[]{A^2+B^2-2ABcos\theta }=\sqrt[]{A^2+B^2-2AB*\frac{1}{\sqrt[]{2}}}=\sqrt[]{A^2+B^2-\sqrt[]{2}AB}$                

As we know that radius of circular path in magnetic field is given as $r=\frac{mv}{qB}=\frac{\sqrt[]{2mK}}{qB},so$  

  $r=\frac{mv}{qB}=\frac{\sqrt[]{2mK}}{qB},so$

$\frac{r_d}{r_{\alpha }}=\sqrt[]{\frac{m_d}{m_{\alpha }}}*\frac{q_{\alpha }}{q_d}=\frac{1}{\sqrt[]{2}}*2=\sqrt[]{2}$               

Since process is isochoric, so

$Q=n{C}_v\Delta T=n\left(\frac{f}{2}R\right)\Delta T=4\left(\frac{5}{2}*R\right)\left(50\right)=500R$

Given ${}_{}{}^{}$ $\chi =499and{\mu }_0=4\pi *10^{-7}H/m$

As we know that

$$ ${\mu }_r=1+\chi =1+499=500$  Relative permeability

Absolute permeability = $\mu ={\mu }_r{\mu }_0=500*4\pi *10^{-7}=2\pi *10^{-4}H/m$

Let the true dip is $\delta$ $$ and apparent dip is $\delta \text{'}$ $$ , so

$tan\delta \text{'}=\frac{B_V}{B_{H}cos\theta }=\frac{tan\delta }{cos\theta }\Rightarrow tan\delta =tan\delta \text{'}cos\theta =tan45°cos30°=1*\frac{\sqrt[]{3}}{2}$

$\Rightarrow \delta =ta{n}^{-1}\left(\frac{\sqrt[]{3}}{2}\right)$