Question

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

• A. 1:03:05
• B. 5:03:01
• C. 0.053530093
• D. 125:15:01

Answer 1

Answer: (C)

0.053530093

Answer 2

Here $m_1 = 5m, m_2 = 3m$ and $m_3 = m$ and $l_1 = l, l_2 = 3l$ and $l_3 = 5 l$ Using density, D = $\frac{M}{V}$ we get M = VD Then $m = A_3 \times 5l \times D$ $3m = A_2 \times 3l \times D$ $5m = A_1 \times l \times D$ From $(i)$ and $(ii)$ $A_2 = \frac{A_1}{5}$ and From $(i)$ and $(iii)$, $A_3 = \frac{A_1}{25}$ Using, R = $\rho \frac{l}{A}$ we get $R_1 = \rho \frac{l}{A_1}$ $R_2 = \rho \frac{l_2}{A_2} = \frac{\rho 3 l}{(A_1/5)} = 15 R_1$ $R_3 = \rho \frac{l_3}{A_3} = \frac{\rho 5l}{(A_1 / 25)} = \frac{125\rho l}{A_1} = 125 \, R_1$ $\therefore$ $R_1 : R_2 : R_3 = 1 : 15 : 125$