Question

A solid of area of cross-section 0.004 m² and length 0.60 m is completely immersed in water of density 1000 kgm³. Calculate:

1. Wt of solid in SI system
2. Upthrust acting on the solid in SI system.
3. Apparent weight of solid in water.
4. Apparent weight of solid in brine solution of density 1050 kgm³.
   [Take g = 10 N/kg; Density of solid = 7200 kgm³]

Answer 1

Area of cross-section of solid = A = 0.004 m²
Length of the solid = l = 0.60 m
Density of water = p’ = 1000 kgm³
Acceleration due to gravity = g = 10 ms^-2
Density of solid = p = 7200 kgm³

(1) Volume of solid = V = A x l
V = 0.004 x 0.60 = 0.0024 m³
Mass of solid = m = V x p
m = 0.0024 x 7200 = 17.28 kg
Weight of the solid = mg = 17.28 x 10 = 172.8 N

(2) Volume of water displaced = Volume of solid
= V = 0.0024 m³
Mass of water displaced = m’ = V x p’
m’ = 0.0024 x 1000 = 2.4 kg
Upthrust = Weight of water displaced
= m’g = 2.4 x 10 = 24 N

(3) Apparent weight of solid=Actual weight of solid – upthrust
= 172.8-24= 148.8 N
(4) Density of brine solution =ρ_b= 1050 kgm³
Volume of brine solution displaced = Volume of solid = V
V = 0.0024 m³
Mass of brine solution displaced
= m_b = V x ρ_b = 0.0024 x 1050
m_b = 2.52 kg
Upthrust acting on solid in brine solution = Weight of brine solution displaced -m_bg
= 2.52 x 10 = 25.2 N
Apparent weight of solid in brine solution
= Actual weight – Upthurst
= 172.8-25.2= 147.6 N