A stress of `1kg//mm^(2)` is applied on a wire. If the modulus of elasticity of the wire is `10^(10) "dyne"//cm^(2)`, then the percentage increase in
A stress of `1kg//mm^(2)` is applied on a wire. If the modulus of elasticity of the wire is `10^(10) "dyne"//cm^(2)`, then the percentage increase in the length of the wire will be
A. 0.0098%
B. 0.0098
C. 0.098
D. 0.98
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Correct Answer - b
`(Delta I)/(I) = ("stress")/(Y) = (9.8 xx 10^(6))/(10^(9)) = 0.0098`
percentage increase in length of wire
`= (Delta I)/(I) xx 100 = 0.0098 xx 100 = 0.98%`
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