Question

a steel rod of 25 cm long has cross section area 0.8 cm². Force that would be required to stretch this rod by the same amount as the expansion produced by heating it through 10° C is : ( coefficient of liner expansion of steel is $10^{-5}$/ °C and young's modulus of steel is 2×$10^{10}$ N/m²

Answer 1

Check the attachment

Answer 2

Dear Student,

◆ Answer -

F = 160 N

● Explanation -

# Given -

L = 25 cm = 0.25 m

A = 0.8 cm^2 = 8×10^-5 m^2

∆T = 10 °C

α = 10^-5 /°C

Y = 2×10^10 N/m^2

# Solution -

Expansion produced by stretching is -

ls = FL/YA

Expansion produced by heating is -

lh = Lα∆T

Given that ls and lh are equal.

FL/YA = Lα∆T

F = α∆T × YA

F = 10^-5 × 10 × 2×10^10 × 8×10^-5

F = 160 N

Hence, 160 N force is required to stretch the rod.

Thank you. Hope this helps you ..