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

Solution

## The correct option is D 125 : 15 : 1Let 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 $$(A=v/l)$$$$A_1=\dfrac{v}{5l}$$$$A_2=\dfrac{3v}{3l}$$$$A_3=\dfrac{5v}{l}$$Resistance $$\propto \dfrac{l}{A}$$$$\Rightarrow R_1=\rho \dfrac{25l^2}{v}$$$$R_2=\rho\dfrac{3l^2}{v}$$$$R_3=\rho\dfrac{1}{5}\dfrac{l^2}{v}$$ $$Ratio=R_1:R_2:R_3\ =\ \ 25:3:\dfrac15$$multiple with 5$$\therefore$$. $$Ratio =R_1:R_2:R_3 \ = \ \ 125:15:1$$Physics

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