Question

A metal rod has a length of 1m at 30 degrees centigrade. Alpha of metal is 2.5 into 10^-5. The temperature at which it will be shortened by 1mm is

Answer 1

initial length of metal rod, l = 1m

initial temperature , T = 30°C

coefficient of linear expansion of metal, α = 2.5 × 10^-5/°C

using formula, $L=l(1+\alpha\Delta T)$

or, $L-l=l\alpha\Delta T$

or, $\Delta l=l\alpha\Delta T$

now putting, ∆l = 1mm = 10^-3 m, l = 1m, α = 2.5 × 10^-5/°C and ∆T = (30 - T)°C

so, 10^-3 m = 1m × 2.5 × 10^-5/°C × (T - 30)°C

or, 100 = 2.5(30 - T)

or, 40 = 30 - T

or, T = -40 + 30 = -10°C

hence, The temperature at which it will be shortened by 1mm is -10°C.