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Question: A rod made up of metal is 1.2m long and 0.8cm in diameter. Its resistance is \(3.5\times {{10}^{-3}}...

A rod made up of metal is 1.2m long and 0.8cm in diameter. Its resistance is 3.5×10−3Ω3.5\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?
A. 1.35×10−8Ω\text{A}\text{. }1.35\times {{10}^{-8}}\Omega
B. 2.70×10−7Ω\text{B}\text{. 2}.70\times {{10}^{-7}}\Omega
C. 5.82×10−7Ω\text{C}\text{. 5}.82\times {{10}^{-7}}\Omega
D. 8.10×10−6Ω\text{D}\text{. 8}.10\times {{10}^{-6}}\Omega

Explanation

Solution

Use the formula for resistance. i.e. R=ρlAR=\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=ρlAR=\dfrac{\rho l}{A}.

Formula used:
R=ρlAR=\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=ρlAR=\dfrac{\rho l}{A} …... (i).
Here, ρ\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 = 0.4×10−2m0.4\times {{10}^{-2}}m.
The area of the cross section will be A = πr2=π(0.4×10−2)2=16π×10−6m2\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 = 3.5×10−3Ω3.5\times {{10}^{-3}}\Omega .
Substitute the values of R, l and A equation (i).
⇒3.5×10−3Ω=ρ(1.2)16π×10−6\Rightarrow 3.5\times {{10}^{-3}}\Omega =\dfrac{\rho (1.2)}{16\pi \times {{10}^{-6}}}
⇒ρ=3.5×10−3×16π×10−61.2=46.66π×10−9Ωm\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 46.66π×10−9Ωm46.66\pi \times {{10}^{-9}}\Omega m.
When a disc is made from this metal, the resistivity of the disc will also be ρ=46.66π×10−9Ωm\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 = 1.25×10−3m1.25\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’ = πr′2=π(0.01)2m2\pi r{{'}^{2}}=\pi {{\left( 0.01 \right)}^{2}}{{m}^{2}}
The resistance of the disc will be R′=ρl′A′R'=\dfrac{\rho l'}{A'} …… (ii).
Substitute the values of ρ\rho , l’, A’ in equation (ii).
R′=(46.66π×10−9)1.25×10−3π(0.01)2=5.82×10−7ΩR'=\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.