Question

A paperweight, when weighed in air, has a weight of W = 6.6 N. When completely immersed in water, however, it has a weight of Win water = 4.5 N. Find the volume of the paperweight.

Answer 1

buoyant force = 6.6 - 4.5 = 2.1 N

buoyant force = V*rho*g

=> 2.1 = V*1000*9.8

=> V = 2.14*10^-4 m^3

Answer 2

Data:

Weight in air, W = 6.6 N

Weight in water, W' = 3.9 N

Density of water, ρ = 1000 kg/m^3

Let

Volume of the paper weight = V

Solution:

Loss of weight, ΔW = W - W'

                              = 6.6 - 3.9

                              = 2.7 N

Loss of weight = Upthrust force

                       = Weight of the water displaced

                       = ρ V g

So, V = ΔW / ρ g

          = 2.7 / ( 1000 * 9.8 )

          = 2.76 x 10^-4 m^3

Ans:

Volume, V = 2.76 x 10^-4 m^3

Answer 3

Weight of water displaced when the paperweight is submerged inside water = 6.6 N - 4.5 N = 2.1 N
Mass of the water displaced = 2.1/g = 2.1/9.8 = 0.214285 kg
Volume of water displaced = mass of water displaced/density of water
= 0.214285/1000 (because density of water = 1000 kg/m^3)
= 0.245 x 10^-3 m^3
= 2.14285 x 10^-4 m^3

Since the paperweight is completely submerged inside water, therefore the volume of the paperweight
= 2.14285 x 10^-4 m^3

Ans: 2.14285 x 10^-4 m^3