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A steel rod of length `25cm` has a cross-sectional area of `0.8cm^(2)` . The force required to stretch this rod by the same amount as the expansion produced by heating it through `10^(@)C` is `(alpha_(steel)=10^(-5)//^(@)C` and `Y_(steel)=2xx10^(10)N//m^(2))`A. 40NB. 80NC. 120ND. 160N |
Answer» Correct Answer - D `F=yA prop Delta T` `=2xx10^(10) xx 0.8 xx 10^(-4) xx 10^(-5) xx 10=160N`. |
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