Class 12th

From Lenz law, current in the square loop will be in anticlockwise sense.

Direction of  is perpendicular to equipotential surface.

For diamagnetic ; $\chi <0andl=\chi H$

At resonance V_L = V_C

_{$\Rightarrow cos?=\frac{V_R}{\sqrt[]{V_R^2+{\left(V_L-V_C\right)}^2}}=1$}

Density of nuclei is order of 10¹⁷ kg m^-3.

Photocell works on photoelectric effect.

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

$C_2=1\mu F$ , 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.