Physics

$\stackrel{?}{F}=e{E}_0\stackrel{?}{j}$

${\stackrel{?}{a}}_y=\frac{e{E}_0}{m}\stackrel{?}{j}$

$=v_0\stackrel{?}{i}+\left(\frac{e{E}_{0}t}{m}\right)\stackrel{?}{j}$

$\left|\stackrel{?}{v}\right|=\sqrt[]{v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}=v_0\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}$

$\lambda =\frac{{\lambda }_0}{\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}}$

Two successive β decays increase the charge no. by 2.

$C_1=1\mu F{V}_1=100V?U_1=\frac{1}{2}*1*{10}^{-6}*{\left(100\right)}^2=5*{10}^{-3}J$  

$C_2=1\mu F$ , common potential $V=\frac{C_1{V}_1}{C_1{C}_2}=\frac{100}{2}=50V$

Final electrostatic energy stored in both the capacitors.

$=\frac{1}{2}\left(C_1+C_2\right)V^2=\frac{1}{2}*2*{\left(50\right)}^2*{10}^{-6}$

$=2.5*{10}^{-3}J$

$Energylost=2.5*{10}^{-3}$

$\%lossofenergy=\frac{2.5*{10}^{-3}}{5*{10}^{-3}}*100$

$=50\%$

When wire is stretched uniformly

$R\propto \mathcal{R}$

$\frac{R}{R^{\text{'}}}={\left(\frac{l}{nl}\right)}^2=\frac{1}{n^2}$  

$R^{\text{'}}=n^{2}R$

When this elongated wire is cut into 5 parts, resistance of each part  

Effective resistance between A and C $=\frac{n^{2}R}{5}$  

Given   $F\propto \frac{1}{r^n}$

$m{\omega }^{2}R=\frac{K}{R^n}$

$?\frac{m.4{\pi }^{2}R}{T^2}=\frac{K}{R^n}$

$?T^2\propto R^{n+1}$

$T\propto R^{\frac{n+1}{2}}$

Angular momentum imparted to sprinkler per

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

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

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

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

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

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

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

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

For normal emergence

$r=A$

From Snell's Law

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

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

$?i={45}^0$

Balancing torque about COM

$T_A\left(\frac{l}{2}\right)=T_B\left(\frac{l}{6}\right)$

$?\frac{T_A}{T_B}=\frac{1}{3}$