Question

Example 24. Two wires A and B of equal mass and of the same metal are taken. The diameter of the wire A is half the diameter of wire B. If the resistance of wire A is 24 ohm , calculate the resistance of wire B.​

Answer 1

The resistance of wire B is 6Ω.

Solution:

The formula for resistance is

R=\frac{\rho L}{A}

As both A and B wires are made up of same material with same mass, the resistivity will be constant. So

\rho_{A}=\rho_{B}

And

L_{A}=L_{B}

Thus the resistance ratio will be

\frac{R_{A}}{R_{B}}=\frac{\frac{\rho_{A} L_{A}}{A_{A}}}{\frac{\rho_{B} L_{B}}{A_{B}}}

As it is given in question that diameter of A is half of the diameter of B, the cross section area will be

d_{A}=\frac{d_{B}}{2}

Thus,

2 \times r_{A}=\frac{2 \times r_{B}}{2}=r_{B}

A_{A}=\pi r_{A}^{2}=\pi \times\left(\frac{r_{B}}{2}\right)^{2}

A_{B}=\pi r_{B}^{2}

\frac{R_{A}}{R_{B}}=\frac{A_{B}}{A_{A}}=\frac{\pi r_{B}^{2}}{\pi \times\left(\frac{r_{B}}{2}\right)^{2}}=4

As R_{A}=24 \Omega, \text { then } \mathrm{R}_{\mathrm{B}}\ is

R_{A}=4 R_{B}

R_{B}=\frac{R_{A}}{4}=\frac{24}{4}=6 \Omega

Thus, the resistance of wire B is 6Ω