First slide

Ohm's law

Question

A material B has twice the specific resistance of A. A circular wire made of B has twice the diameter of a wire made of A. Then for the two wires to have the same resistance, the ratio of l_B / l_A of their respective lengths must be

Easy

A

1
B

1/2
C

1/4
D

2

Solution

Let the specific resistances, lengths, radii and areas of wires A and B be $(ρ_{A}, l_{A}, r_{A}, A_{A}) and (ρ_{B}, l_{B}, r_{B}, A_{B})$ respectively.
Resistance of A = $R_{A} = \frac{ρ_{A} l_{A}}{{πr}_{A}^{2}}$
Resistance of B= $R_{B} = \frac{ρ_{B} l_{B}}{{πr}_{B}^{2}}$
According to given question, $ρ_{B} = 2 ρ_{A}, r_{B} = 2 r_{A}$ and $R_{A} = R_{B}$ Therefore
$\frac{ρ_{A} l_{A}}{{πr}_{A}^{2}} = \frac{ρ_{B} l_{B}}{{πr}_{B}^{2}}$
Further, $\frac{ρ_{A} l_{A}}{{πr}_{A}^{2}} = \frac{(2 ρ_{A}) (l_{B})}{π {(2 r_{A})}^{2}}$
$∴ \frac{l_{B}}{l_{A}} = \frac{2}{1} = 2$

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