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 R_{0}, l_{0}, A_{0} 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 =2l_{0}

When the wire is stretched its Volume remains the same.

Therefore, V= Al = A_{0}l_{0}

A(2l_{0}) = A_{0}l_{0}

A=A_{0}/2

Resistance R=ρl/A

R/R_{0} = lA_{0}/l_{0}A

R/R_{0} = 2l_{0 }A_{0}/ (l_{0 }(A_{0}/2)) = 4

R = 4R_{0} = 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 R^{1}= 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} = R^{1}/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