On applying a stress of $20 \times {10^8}N/{m^2}$ the length of a perfectly elastic wire is doubled. Its Young’s modulus will be
A$40 \times {10^8}N/{m^2}$
B$20 \times {10^8}N/{m^2}$
C$10 \times {10^8}N/{m^2}$
D$5 \times {10^8}N/{m^2}$
Medium
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B$20 \times {10^8}N/{m^2}$
b (b) Young’s modules =$\frac{{{\rm{stress}}}}{{{\rm{strain}}}}$
As the length of wire get doubled therefore strain $= 1$
$Y =$ strain $= 20 \times 10^8 N/ m^2$
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