If a metallic wire of resistance R is melted and recast to half of its length, the new resistance of the wire will be?
R4
Given:
Resistance of wire before recast = R
Length of the wire after recast = half of length of wire before recast
Let:
Original length be l
Length after recasting be l′
Original area of cross-section be A
Area of cross-section after recasting be A′
Volume of the metallic wire is the product of length and area of the cross section and does not change.
Al=A′l′
AA′=l′l
∴AA′=12
Resistance is given by:
R=ρlA
Ratio of resistances are:
R′R=ρl′/A′ρl/A
=(l′l)(AA′)
=(12)(12)=14
R′=R4