Class 12th

$\frac{H_{D.C.}}{H_{A.C.}}=\frac{I^{2}R}{I_{rms}^{2}R}=1$

$B_0=\frac{E_0}{v}v=\frac{\omega }{k}$

$\Rightarrow \mu =\frac{c}{v}=\frac{3*10^8}{\frac{9}{4}*10^8}=\frac{4}{3}$

$\frac{\Delta \lambda }{\lambda }=\frac{v}{c}$

$\Rightarrow v=\frac{0.6*10^{-9}*3*10^8}{589*10^{-9}}$ = 3.06 * 10⁵ m/s

^{$e=\left|\frac{d?}{dt}\right|=4t+8$}

^{$\Rightarrow i=\frac{e}{R}=\frac{4t+8}{4}=t+2$}

^{$\Rightarrow$ Heat produced (H)= ${\displaystyle \underset{0}{\overset{2}{\int }}{i}^{2}Rdt}$}

^{$\Rightarrow H=4\lfloor \frac{8}{3}+8+8\rfloor =\frac{224}{3}J$}

Applying conservation of momentum for collision of blocks

$2mv+0=3m{v}^{\text{'}}$

$?v^{\text{'}}=\frac{2v}{3}\dots \left(1\right)$

Now, from conservation of energy

$\frac{1}{2}*3m*{v^{\text{'}}}^2=\frac{1}{2}k{x}_0^2$

$x_0=\sqrt[]{\frac{3m{v}^{\text{'}2}}{k}}=v\text{'}\sqrt[]{\frac{3m}{k}}$

$=\frac{2v}{3}\sqrt[]{\frac{3m}{k}}=v\sqrt[]{\frac{4m}{3k}}$

Angular momentum imparted to sprinkler per

$second=vr\frac{dm}{dt}=\frac{0.6}{60}*0.5*1=0.005N.m$

$?l\frac{d\omega }{dt}=0.005$

$?\frac{d\omega }{dt}=\frac{0.005}{500*{10}^{-7}}$

$=100\frac{rad}{s^2}$

For normal emergence

$r=A$

From Snell's Law

$\frac{\sin ?i}{\sin ?r}=\frac{\mu }{1}$

$?\sin ?i=\sqrt[]{2}\sin ?{30}^0$

$?i={45}^0$

$\begin{array}{l}t=\frac{T}{2}\\ \Rightarrow N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{T}}\end{array}$

$\Rightarrow N=N_0{\left(\frac{1}{2}\right)}^{\frac{1}{2}}=\frac{N_0}{\sqrt[]{2}}$

Magnify power of telescope for far point is given as

$m=\frac{{\theta }_i}{{\theta }_0}=\frac{f_0}{f_e}$

$?{\theta }_i={\theta }_0\left(\frac{f_0}{f_e}\right){=1}^0\left(\frac{100}{5}\right)={20}^0$

$01*10^4\Omega \pm 5\mathrm{\%}=10*10^3\Omega \pm 5\mathrm{\%}$