a wire of 15 ohm resistance is gradually stretched to double its original length. it is then cut into two equal parts. these parts are connected in parallel across a 3 V cell. find the current drawn from the cell

Let R₀, l₀, A₀ be the original resistance, length and area of the wire.

Let R, l, A be the original resistance, length and area of the stretched wire. Since the length is doubled ; l =2l₀

When the wire is stretched its Volume remains the same.

Therefore, V= Al = A₀l₀

    A(2l₀) = A₀l₀

    A=A₀/2

Resistance R=ρl/A

R/R₀ = lA₀/l₀A

R/R₀ = 2l₀ A₀/ (l₀ (A₀/2)) = 4

R = 4R₀ = 4x15 = 60Ω

 When the resistance of wire R=60Ω is cut into two equal parts, the length becomes half but area remains the same. Hence resistance of each part is R­¹= R/2 = 60/2 = 30Ω

When these two wires of resistance 30Ω are connected in parallel across a 3V battery, the equivalent resistance R_p = R¹/2 = 30/2 =15 Ω

(1/R_p = (1/ 30) +( 1/ 30) = 2/30 = 1/15

R_p =15Ω )

Therefore the current through the circuit is V/R_p = 3/15 = 0.2A