Question

A spherical shell made of material of conductivity $\frac{{10}^9}{\pi }(\Omega -\text{m}{)}^{-1}$ has a thickness $t=2\text{mm}$ and radius of shell is $R=10\text{cm}$. In an arrangement, it's inside surface is kept at a lower potential with respect to its outer surface. The resistance offered by the shell is equal to:

• A. $5\pi \times {10}^{-12}\Omega$
• B. $2.5\times {10}^{-11}\Omega$
• C. $5\times {10}^{-12}\Omega$
• D. $5\times {10}^{-11}\Omega$

Answer 1

The correct option is D $5\times {10}^{-11}\Omega$

According to question the inner shell is kept at a lower potential w.r.t outer shell.
It can be represented with the circuit diagram shown in figure.


Using formula of resistance

$R=\frac{\rho l}{A}......\left(i\right)$

The current will enter the outer surface and will come out from the inner surface.

i.e. $l=t=2\times {10}^{-3}\text{m}$

Now considering the thin shell, the area of cross-section for flow of current is,

$A=4\pi R^2=4\pi (10\times {10}^{-2}{)}^2{\text{m}}^2$

also, we know that

$\rho =\frac{1}{\sigma }$

$\Rightarrow \rho =\frac{1}{\left(\frac{{10}^9}{\pi }\right)}=\frac{\pi }{{10}^9}(\Omega -\text{m})$

Thus, from eq. $\left(i\right)$

$R=\frac{\pi }{{10}^9}\times \frac{2\times {10}^{-3}}{(4\pi \times 100\times {10}^{-4})}$

$\Rightarrow R=0.5\times {10}^{-10}$

$\therefore R=5\times {10}^{-11}\Omega$

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: In such problems observe the direction}\\ \text{of current flow and identify the length}\\ \text{of conductor (l) \& cross-sectional area for current flow.)}\end{array}$