A current of 5A is passing through a non-linear magnesium wire of cross – section 0.04m². At every point the direction of current density is at an angle of 60° ;with unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is: (Resistivity of magnesium $\rho =44*10^{-8}\Omega m)$  

Divided into 10 parts

$R=\frac{\rho l}{A}$

$R^{\prime }=\frac{\rho l}{10A}=\frac{R}{10}$

$R_S=5\times \frac{R}{10} [series]$

$R_S=50$

$R_P=\frac{R}{50}\left[ parallel \right]$

$R_{eq }=R_S+R_P$

$=52\Omega$

For uniformly charged spherical shell,

$V=\frac{kq}{R}\left( For r\le R\right)$

$\therefore V_C=V_P$

$V_C-V_P= Zero$

In first condition R₁ = 36 $\Omega$  

In second condition R₂ = 18  $\Omega$

$P_1=\frac{V^2}{R_1}=\frac{{\left(240\right)}^2}{36}$               

$P_2=\frac{V^2}{R_2}+\frac{V^2}{R_2}=\frac{{\left(240\right)}^2}{18}+\frac{{\left(240\right)}^2}{18}$               

$P_2=\frac{{\left(240\right)}^2}{9}$               

So   $\frac{P_1}{P_2}=\frac{{\left(240\right)}^2/36}{{\left(240\right)}^2/9}=\frac{1}{4}$

x = 4

$Current in circuit i=\frac{10}{4+1}=2 A$

$Terminal voltage =E-iR$

$=10-2\times 1=8 V$

$l_1=\frac{50}{2000}=25mA,andl=\frac{V_i-V_Z}{1000}=\frac{100-50}{1000}=50mA$

$\Rightarrow l_z=l-l_1=25mA$