If a metallic wire of resistance R is melted and recast to half of its length, the new resistance of the wire is ___.
A
R4
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B
R2
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C
R
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D
2R
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Solution
The correct option is AR4 Volume of the metallic wire = Length×Areaofcrosssection 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