Question

The masses of the three wires of copper are in the ratio 5 : 3 : 1 and their lengths are in the ratio 1 : 3 : 5. The ratio of their electrical resistances is

• A. 5:03:01
• B. $\sqrt{125}$ : 15 : 1
• C. 0.053530093
• D. 1:03:05

Answer 1

Answer: (C)

0.053530093

Answer 2

Given, $l_1 = l_k, l_2 = 3k, l_3 = 5k$
and $\, \, \, \, \, \, \, \, \, m_1 = 5m, m_2 = 3m, m_3 = 1m$
Resistance $R=\frac{\rho l}{A}$
where $\rho$ is resistivity of the material of conductor.
$so, \, \, \, \, \, \, \, \, \, \, R_1 : R_2 : R_3= \frac{l_1}{A_1} : \frac{l_2}{A_2} : \frac{l_3}{A_3}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{l^2_1}{V_1} : \frac{l^2_2}{V_2} : \frac{l^2_3}{V_3}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{l^2_1}{m_1} : \frac{l^2_2}{m_1} : \frac{l^2_3}{m_3}=\frac{1}{5}:\frac{9}{3}:\frac{25}{1}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 1: 15: 125$