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
Volume of the metallic wire = Length×Area of crosssection
If l and A are the original length and area of cross -section and l' and A' are their corresponding values on recasting.
Al=A′l′orl′l=AA′∵l′l=12(Given)∴AA′=12
New resistance, R′=ρl′A′
As R=ρlA
∴R′R=ρl′A′ρl′A
=(l′l)(AA′)
=(12)(12)=14
or R′=R4