JEE Main 1-Sep-2021 Shift 2 Solved Paper

Given, the Young’s modulus of the steel rod, Y = 2 ×1011 Pa
Thermal coefficient of the steel rod, α = 10−5ºC
The length of the steel rod, I = 4 m
The area of the cross-section, A = 10 cm2
The temperature difference, ΔT = 400°C
As we know that,
Thermal strain = α ΔT
Using the Hooke’s law
Young's modulus (Y)=

Thermal‌stress

Thermal‌strain

=

F A

αΔT

Thermal stress, F = YA α Δ T
Substitute the values in the above equation, we get
F=2×1011×10×10−4×10−5×(400)
=8×105N
Comparing with, F=x×105N
The value of the x = 8.