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))`

Answer

16

Explanation

`Y =(F//A)/(Deltal//l)`
`:. (Deltal)/(l) = alpha Delta theta`
`:. F = Y alpha A Delta theta`
`= 2 xx 10^(10) xx (0.8 xx 10^(-4)) xx 10^(-5) xx 10`
`F = 160 N`