Two wires A and B of same material and same mass have radius 2r and r. If resistance of wire A is $34\,\Omega $, then resistance of B will be?
A. $544\,\Omega $
B. $272\,\Omega $
C. $68\,\Omega $
D. $17\,\Omega $

Hint:First try to find the relation between electric resistance and the area of rod of the given material. Then find the ratio of area of both rods using the radius of two wires given in the question. Then apply the same ratio for the resistance of the wire as resistance is inversely proportional to area.

Formula used:
Electric resistance, $R = \rho \dfrac{L}{A}$
Where, $\rho $ is resistivity, L is length and A is the area of the rod given.

Complete step by step solution:
Given information:
Let resistance of rod 1 = ${R_1}$
Resistance of rod 2 = ${R_2}$
We know that:
Electric resistance, $R = \rho \dfrac{L}{A}$
As both the rods of the same material so $\rho $ will be constant here.

For solving for Area, A:
That means:
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{A_2}}}{{{A_1}}}$= $\dfrac{{\pi r_1^2}}{{\pi r_2^2}} \\ $
$\Rightarrow \dfrac{{{A_2}}}{{{A_1}}} = \dfrac{{\pi {r^2}}}{{\pi {{\left( {2r} \right)}^2}}} = \dfrac{{\pi {r^2}}}{{\pi 4{r^2}}} = \dfrac{1}{4} \\ $
$\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{4}$

Solving for length, L:
Now, for both wire to have same mass;
Wire B has to be 4 times long, that means length of wire B is four times the length of wire A.
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{{4 \times 4}} = \dfrac{1}{{16}}$
So, the resistance of wire B = 16 times of A
Resistance of wire B = $16 \times 34 = 544\Omega $

Hence, the correct answer is option A.

Note: Here the rods that are given in the question are of same material and same mass that’s why the other quantities other than area in the formula of electric resistance was taken constant but it is not the case all the time. When different materials are used for rods then the answer will differ.