Class 12th

$E=\frac{hc}{\lambda }=\frac{1242eV-nm}{621nm}=2eV$

minimum 2eV is required to emit.

For capacitor : current leads emf

by 90°.

CP of 378 = SP of 525

$\begin{array}{l}\frac{378}{525}=\frac{SP}{CP}\\ \frac{SP}{CP}=0.72\end{array}$

So, 28% loss

$cose{c}^{2}xdy+2dx=\left(1+ycos2x\right)cose{c}^{2}xdx.$

$\Rightarrow \frac{dy}{dx}+2sin2x=1+ycos2x.$

$I.F.=e^{-{\displaystyle \int cos2xdx}}=e^{-\frac{sin2x}{2}}$

$\therefore Solutiony{e}^{-\frac{sin2x}{2}}={\displaystyle \int e^{-\frac{sin2x}{2}}.cos2xdx.Put\frac{sin2x}{2}=t\Rightarrow cos2xdx=dt}$

$y\left(0\right)=-1+e^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\Rightarrow {\left(y\left(0\right)+1\right)}^2={\left(e^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2=e^{-1}$

$\frac{{(27+5)}^{332}}{9}=\frac{{(5)}^{332}}{9}=\frac{5^2*{(125)}^{110}}{9}=25*{(-1)}^{110}=7$

$l=\frac{1}{2}C{\epsilon }_0{E}_0^2=\frac{1}{2}C{\epsilon }_0{\left(C{B}_0\right)}^2=\frac{1}{2}{C}^3{\epsilon }_0{B}_0^2$

$B_0=\sqrt[]{\frac{2l}{{\epsilon }_0{C}^3}}$

$=\sqrt[]{\frac{2*0.092}{8.85*10^{-12}*27*10^{24}}}$

$=2.27*10^{-8}T$

Let the speed of B = X kmph.

We have X * t = 240, (X – 15) * t = 180

or X = 60, X – 15 = 45

Using conservation of momentum

=> 180 V₁ = 4V₂

=> V₂ = 45 V₁

  $Q=K{E}_{180}+K{E}_{\alpha }$              

$Q=\frac{1}{2}180V_1^2+\frac{1}{2}4V_2^2$        

= $\frac{1}{2}*180*{\left(\frac{V_2}{45}\right)}^2+\frac{1}{2}4V_2^2$

$K{E}_{\alpha }=\frac{5.5MeV}{\left(\frac{46}{45}\right)}=5.38MeV$

The work done by 12 boys in 20 days = $\frac{4}{7}$ $\frac{}{}$

So, remaining work need to be done in 6 days = $\frac{3}{7}$

$\frac{M_1*D_1}{W_1}=\frac{M_2*D_2}{W_2}$ $\frac{}{}$

$\frac{12*20}{\frac{4}{7}}=\frac{M_2*6}{\frac{3}{7}}$

We get, M2 = 30. Hence, 18 extra boys are required.