The terminal voltage of the battery, whose emf is 10 V and internal resistance $1\Omega$ , when connected through an external resistance of $4\Omega$  as shown in the figure is:

 

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

$l=JAcos\theta \Rightarrow 5=J\times 0.04\times cos60°\Rightarrow J=\frac{5}{0.02}=250A/m^2$

$J=\frac{E}{\rho }\Rightarrow E=\rho J=44\times 10^{-8}\times 250=11\times 10^{-5}V/m$              

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

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