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If a wire having initial diameter of 2 mm produced the longitudinal strain of 0.1%, then the final diameter of wire is `(sigma = 0.5)`A. 2.002 mmB. 1.998 mmC. 1.999 mmD. 2.001 mm |
Answer» Correct Answer - C `sigma=(Delta d)/(D)xx(L)/(l)` `(Delta d)/(D)=-D xx 5 xx 10^(-4) =-2 xx 10^(-3) xx 5 xx10^(-4)` `=-1xx10^(-6)` `d_(2)-d_(1)=-0.001 mm` `d_(2)=d_(1)-0.001 =2-0.001=2.0000-0.001` `d_(2)=1.999 mm`. |
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