Manipal Entrance Test 2015 Physics Paper

Let R be the radius of sphere, V its volume and ρ its density. Then,
ΔR=RαΔθ
and percentage change in radius

ΔR

R

×100=100αΔθ ...(i)
Now, ΔV=γvΔθ =3αvΔθ (∵γ=3α)
∴ Percentage increase in volume =

ΔV

V

×100=300αΔθ ...(ii)
Again, ρ′=

ρ

1+γΔθ

=

ρ

1+3αΔθ

∴Δρ=ρ−ρ′
=ρ[1−

1

1+3αΔθ

] =

(3αΔθ)ρ

1+3αΔθ

Percentage change in density =

Δρ

ρ

×100=

300αΔθ

1+3αΔθ

...(iii)
From Eqs. (i), (ii) and (iii), we see that the percentage change is maximum in volume.