A wire of resistance 160 is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

Step-by-step solution

Let Initial length =l₁ Final length =l₂ Initial area = A ₁ Final area =A₁ Volume remains same A ₁ l ₁= A ₂ l ₂ ~A ₁ l ₁= A ₂ l ₁ 4 4 ~A ₁= A ₂ Initial resistance, R ₁= l₁ A₁ =160 (given) Final resistance, R ₂= l ₂ ~A ₂ R ₂ R ₁ = l ₂ ~A ₁ ~A ₂ l ₁ = l₁ 4 ~A ₁ 4 ~A ₁ l₁ R₂= 1 16 R₁= 1 16 160=10