- #1
harshaxnim
- 1
- 0
Explain why the thermal expansion of a spherical shell made of a homogeneous solid is equivalent to that of solid sphere of the same material.
I guess these equations would be of some help.
(ΔA)=A*2α*(ΔT)
α→ Coefficient of linear expansion.
A→ Area
T→ Temperature
(ΔV)=V*3α*(ΔT)
α→ Coefficient of linear expansion.
V→ Volume
T→ Temperature
I'm not sure if i understood the question right.
By "equivalent thermal expansion" i guess they mean to say the radius increases by same amount during the expansion.
So i set out relating the two radii.
Took two spheres, one hollow, the other solid, of same dimensions, i.e., same radii.
For the Shell,
(ΔA)=A*2α*(ΔT)
Rf2-Ri2=Ri2*2α*ΔT
Rf=Ri√(2α*ΔT+1)
For the solid sphere, similarly, relating the volume,
Rf=Ri∛(3α*ΔT+1)
But, failed to prove them to be the same.
So what exactly do they intend to ask? And how do i hit it?...
I guess these equations would be of some help.
(ΔA)=A*2α*(ΔT)
α→ Coefficient of linear expansion.
A→ Area
T→ Temperature
(ΔV)=V*3α*(ΔT)
α→ Coefficient of linear expansion.
V→ Volume
T→ Temperature
I'm not sure if i understood the question right.
By "equivalent thermal expansion" i guess they mean to say the radius increases by same amount during the expansion.
So i set out relating the two radii.
Took two spheres, one hollow, the other solid, of same dimensions, i.e., same radii.
For the Shell,
(ΔA)=A*2α*(ΔT)
Rf2-Ri2=Ri2*2α*ΔT
Rf=Ri√(2α*ΔT+1)
For the solid sphere, similarly, relating the volume,
Rf=Ri∛(3α*ΔT+1)
But, failed to prove them to be the same.
So what exactly do they intend to ask? And how do i hit it?...