Class 12th

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

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

Final electrostatic energy stored in both the capacitors.

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

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

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

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

$=50\%$  

$\Rightarrow \frac{R}{100}=\frac{40}{60}$

$\Rightarrow \Delta R=\left(\frac{\Delta l}{l}+\frac{\Delta l}{\left(100-l\right)}\right)R$

$=\left(\frac{0.1}{40}+\frac{0.1}{60}\right)*66.7=0.27=0.3$

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 $=\frac{n^{2}R}{5}$  

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

$X_C=\frac{1}{\omega C}$

On increasing $\omega$ $$ XC decreases so Z decreases, current in circuit increases.

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

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

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

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

$\Rightarrow v=\frac{0.6*10^{-9}*3*10^8}{589*10^{-9}}$ = 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

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

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

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

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