Question

A rod made up of metal is 1.2m long and 0.8cm in diameter. Its resistance is $3.5\times {{10}^{-3}}\Omega $. Another disc made up of the same metal is 2cm in diameter and 1.25mm thick. What is the resistance between the round faces of the disc?
$\text{A}\text{. }1.35\times {{10}^{-8}}\Omega $
$\text{B}\text{. 2}.70\times {{10}^{-7}}\Omega $
$\text{C}\text{. 5}.82\times {{10}^{-7}}\Omega $
$\text{D}\text{. 8}.10\times {{10}^{-6}}\Omega $

Answer 1

Hint: Use the formula for resistance. i.e. $R=\dfrac{\rho l}{A}$. First find the resistivity of the material (metal) by which the rod is made. The resistivity of the disc will be the same. Calculate A and l for the disc and then calculate the resistance of the disc using $R=\dfrac{\rho l}{A}$.

Formula used:
$R=\dfrac{\rho l}{A}$

Complete step-by-step answer:
The resistance of a rod of length l and uniform cross sectional A is given as $R=\dfrac{\rho l}{A}$ …... (i).
Here, $\rho $ is the resistivity of material that the rod is made up of.
In the question the given rod has length l = 1.2m. Let us consider that it has a uniform cross section and it is circular. The diameter of the cross section is given to be 0.8cm. Therefore, its radius will be r = 0.4cm = $0.4\times {{10}^{-2}}m$.
The area of the cross section will be A = $\pi {{r}^{2}}=\pi {{\left( 0.4\times {{10}^{-2}} \right)}^{2}}=16\pi \times {{10}^{-6}}{{m}^{2}}$.
The resistance of the rod is given to be R = $3.5\times {{10}^{-3}}\Omega $.
Substitute the values of R, l and A equation (i).
$\Rightarrow 3.5\times {{10}^{-3}}\Omega =\dfrac{\rho (1.2)}{16\pi \times {{10}^{-6}}}$
$\Rightarrow \rho =\dfrac{3.5\times {{10}^{-3}}\times 16\pi \times {{10}^{-6}}}{1.2}=46.66\pi \times {{10}^{-9}}\Omega m$
This means that the resistivity of the metal from which the rod is made is $46.66\pi \times {{10}^{-9}}\Omega m$.
When a disc is made from this metal, the resistivity of the disc will also be $\rho =46.66\pi \times {{10}^{-9}}\Omega m$.
It is given that the disc is 1.25mm thick and its diameter is 2cm. This means its radius r’ = 1cm.
Consider this disc as a cylinder. The length (l’) of the cylinder is the same as the width (thickness) of the disc. Therefore, l’ = 1.25mm = $1.25\times {{10}^{-3}}m$.
 The radius of the circular surface is r’ = 1cm = 0.01m. The cross sectional area of the cylinder will be A’ = $\pi r{{'}^{2}}=\pi {{\left( 0.01 \right)}^{2}}{{m}^{2}}$
The resistance of the disc will be $R'=\dfrac{\rho l'}{A'}$ …… (ii).
Substitute the values of $\rho $, l’, A’ in equation (ii).
$R'=\dfrac{\left( 46.66\pi \times {{10}^{-9}} \right)1.25\times {{10}^{-3}}}{\pi {{\left( 0.01 \right)}^{2}}}=5.82\times {{10}^{-7}}\Omega $
Hence, the correct option is C.

Note: Note that the resistivity is a material property. Meaning it depends on the material of a given substance. Substances of different shape and sizes but made up of the same material, have the same resistivity.