There are two wires of same material and same length while the diameter of second wire is 2 times the diameter of first wire, then ratio of extension produced in the wiresby applying same load will be
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a
1 : 1
b
2 : 1
c
1 : 2
d
4 : 1
answer is D.
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Detailed Solution
l=FLAY∴l∝1r2(F,L and Y are constant )l1l2=r2r12=(2)2=4