Question

two wires A and B of equal mass and of the same metal are taken. The diamete of the wire A is half the diameter of wire B. If the resistanceof wire A is 24 ohm, calculatethe resistance of the wire B.

Answer 1

Answer:  

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.

Answer 2

This is the right answer.