AP EAPCET 24-Aug-2021 Shift 2 Paper

Given that, (

1

3

) rd part of body is above the water surface. ∴ Volume of body outside the water =

1

3

(volume of body)
V0=

1

3

V
∴ Volume of body inside,
Vi=V−V0=V−

V

3

=

2V

3

Let mass of body =M
Density of bod y=σ
Density of water, ρ=103‌kg‌m−3
According to the law of floatation, body will float into the liquid (water) when weight of body is balanced by Buoyant force.
∴W=F0
⇒Mg=Viρg
Vσg=Viρg
Vσ=Viρ
σ=

Vρ

V

Subscituting the values, we get
σ=

2V

3V

ρ
=

2

3

×103=

2000

3

‌kg‌m−3
Hence, density of liquid, σ=

2000

3

‌kg‌m−3