Question

A glass cylinder of length 12 x 10^-2 m and area of cross­section 5 x 10^-4 m² has a density of 2500 kgm^-3. It is immersed in a liquid of density 1500 kgm^-3, such that 3/8. of its length is above the liquid. Find the apparent weight of glass cylinder in newtons.

Answer 1

Length of glass cylinder = l = 12 x 10^-2 m
Area of cross-section = A = 5 x 10^-4 m²
Volume of glass cylinder = V = A x l
V= 5 x 10^-4 x 12 x 10-²
V= 0.00006 m³
Acceleration due to gravity = g = 9.8 m/s²
Density of glass cylinder = ρ = 2500 kgm³ 

Density of liquid = ρ' = 1500 kgm^-3

∵ `3/8` length of glass cylinder is above the liquid

∴ Length of glass cylinder inside the liquid = `1-3/8 = (8-3)/8 = 5/8`

∴ Volume of liquid displaced by glass cylinder

`= 5/8 xx "Volume of glass cylinder"`

V' = `5/8 = 0.00006`

V' = `0.0003/8` = 0.0000375 m³ 

Mass of glass cylinder = m = V x ρ
m = 0.00006 x 2500 m
= 0.15 kg
Weight of glass cylinder = mg = 0.15 x 10=1.5 N
Mass of liquid displaced by glass cylinder = V’ x ρ’
m’= 0.0000375 x 1500
m’ = 0.05625 kg
Upthrust = Weight of liquid displaced by the glass cylinder
= m’g = 0.05625 x 10 = 0.5625 N
Apparent weight of glass cylinder in liquid = Actual weight of glass cylinder – Upthrust
= 1.5 – 0.5625 = 0.9375 N