A metal rod of length 1 m at `20^@C` have coefficient of linear expansion of `2.5 xx 10^(-5)//""^@C` .The temperature at which it is shortened by 1 mm is `- 10x""^@C`. Find the value of x.

Answer

`02.00`

Explanation

`Deltat=(L_2-L_1)/(L_(1)alpha)`
`:. t_(2)= (L_2-L_1)/(L_1alpha)+t_1= (-10^(-1))/(100 xx2.5 xx10^(-5))+20`
`:. t_2 = -40+20`
`:. t_2 = -20^@C`