Question

If masses of three wires of copper are in the ratio of 1 : 3 : 5 and there lengths are in the ratio of 5 : 3 : 1, then ratio of their electrical resistance is

• A. 1 : 3 : 5
• B. 5 : 3 : 1
• C. 1 : 15 : 125
• D. 125 : 15 : 1

Answer 1

The correct option is D 125 : 15 : 1

Let the masses be

$m_1=m$

$m_2=3m$

$m_3=5m$

$\Rightarrow$ Volume

$v_1=v$

$v_2=3v$

$v_3=5v$

Given lengths

$l_1=5l$

$l_2=3l$

$l_3=l$

Areas $(A=v/l)$

$A_1=\frac{v}{5l}$

$A_2=\frac{3v}{3l}$

$A_3=\frac{5v}{l}$

Resistance $\propto \frac{l}{A}$

$\Rightarrow R_1=\rho \frac{25l^2}{v}$

$R_2=\rho \frac{3l^2}{v}$

$R_3=\rho \frac{1}{5}\frac{l^2}{v}$

$Ratio=R_1:R_2:R_3=25:3:\frac{1}{5}$
multiple with 5

$\therefore$. $Ratio=R_1:R_2:R_3=125:15:1$