Given thai, length of silver rod, L=1m Initial temperature, T1=0°C Final temperaıurc, T2=100°C Increased in length, ΔL=0.19‌cm Let α and γ be the coefficient of linear and volume expansions. Then, we know that, ΔL=LαΔT Substituting the values, we get
0.19
100
=1×α(T2−T1) ⇒
019
100
=α(100−0) ⇒α=0.19×10−4‌°C−1 We also know that, volume expansion, γ=3α=3×0.19×10−4 =5.7×10−5‌°C−1