Question

[
begin{aligned}
R &=rho frac{l}{A}
text { or } R &=P frac{l}{A} times frac{l}{l}=P frac{l^{2}}{V}
end{aligned}
]
( r, R=frac{l frac{l^{2}}{left(frac{m}{d}right)}}{d}=frac{P l^{2} d}{m}left[begin{array}{l}because d=frac{m}{v} theta v=frac{m}{d}end{array}right] )
for scume wire, ( p ) & ( d ) is constant ( operatorname{sos} R x frac{e^{2}}{m} )
rence
[
R_{1}: R_{2}: R_{3}=frac{5^{2}}{1}: frac{3^{2}}{3}: frac{1^{2}}{5}
]
( =25: 3: frac{1}{5} )
mulbiply by '5 ' on each term gives ( R_{2}: R_{2}: R_{3}=125: 15: 1 )

# The masses of three wires of copper are in the ratio of 1:3:5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistances is: A 1:3:5 B 5:3:1 C 1:15:125 D 125:15:1

Solution