TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Given,
We know that density of fluid at depth d
ρ=‌

ρ0

1+αρ0gH

. . . (i)
where, α is constant and g be acceleration due to gravity.
As we know that,
B0=‌

−p

∆V∕V

=‌

−ρgH

∆V∕V

where, p is pressure, and ∆V∕V is relative change in volume.
Now from Eq. (i),
ρ(1+αρ0gH)=ρ0
⇒‌‌ρ+ραρ0gH=ρ0
⇒‌‌ρ−ρ0=ραρ0gH
⇒‌‌‌

∆ρ

ρ0

=αρgH . . . (ii)
and since ρ=m∕V
Taking relative error both sides, we get
‌

∆ρ

ρ0

=‌

∆m

m

+‌

∆V

V0

. . . (iii)
From Eqs. (ii) and (iii), we get
‌

∆m

m

+‌

∆V

V0

=αρgH
From Eqs. (ii) and (iii), we get
‌‌

∆m

m

+‌

∆V

V0

=αρgH
⇒‌‌‌

1

α

=‌

ρgH

∆V∕V

[∵‌

∆m

m

=0]
⇒‌‌‌

1

α

=−B0‌‌⇒‌‌α=−‌

1

B0