Question

A copper tube is of internal radius $4\,mm$ and outer radius $5\,mm$.Its resistance is ${R_1}$. The tube is filled with a suitable copper wire. The resistance of the arrangement is ${R_2}$. Then $\dfrac{{{R_2}}}{{{R_1}}}$ is
A. $\dfrac{{25}}{9}$
B. $\dfrac{1}{2}$
C. 4
D. $\dfrac{9}{{25}}$

Answer 1

We know that resistance is directly proportional to length and inversely proportional to the area of cross section. It is given by the formula
$R = \dfrac{{\rho l}}{A}$
Where $\rho$ is the resistivity of the material, l is the length and A is the area of the cross section.
Using this relation find the resistance of the tube alone and then find resistance of tube with copper wire filled in it. By dividing both these we can get the final answer.

Complete step-by-step solution:
It is given that the internal radius of Copper tube is $4\,mm$, outer radius is given as $5\,mm$.
Thus inner radius ${r_1} = 4\,mm$ and outer radius ${r_2} = 5\,mm$.
The resistance is of the tube is given as ${R_1}$
Now this tube is filled with suitable copper wire and the arrangement then has a resistance ${R_2}$ in total.
We need to find the ratio $\dfrac{{{R_2}}}{{{R_1}}}$
We know that resistance is directly proportional to length, l and inversely proportional to the area of cross section. It is given by the formula
$R = \dfrac{{\rho l}}{A}$
Where $\rho$ is the resistivity of the material, l is the length and A is the area of the cross section .
So using this resistance of copper tube can be written as
${R_1} = \dfrac{{\rho l}}{A}$
We know that the area of the hollow cylinder is given as $A = \pi \left( {{r_2}^2 - {r_1}^2} \right)$. Thus we can write ,
$\Rightarrow {R_1} = \dfrac{{\rho l}}{{\pi \left( {{r_2}^2 - {r_1}^2} \right)}}$
$\Rightarrow {R_1} = \dfrac{{\rho l}}{{\pi \left( {25 - 16} \right)}}$
$\therefore {R_1} = \dfrac{{\rho l}}{{9\pi }}$
Now let us calculate resistance after inserting copper wire .
${R_2} = \dfrac{{\rho l}}{{\pi {r_2}^2}} = \dfrac{{\rho l}}{{25\pi }}$
Let us take the ratio of these two resistance .
$\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{\dfrac{{\rho l}}{{25\pi }}}}{{\dfrac{{\rho l}}{{9\pi }}}}$
$\therefore \dfrac{{{R_2}}}{{{R_1}}} = \dfrac{9}{{25}}$
So correct answer is option D

Note:- Here since both the tube and the wire inside are made of copper the resistivity of both will be the same. Resistivity is a property of the material taken hence they cancel out. But if the tube was made of some other material then the resistivity of the tube will be different from that of copper wire inside.