Question

A small sculpture made of gold ( rho gold =19320 kg/m^ 3 ) is believed to have a secret central cavity. The weight of the sculpture in air is 11.29 N. When it is submerged in water, the weight is 9.190 N. What is the volume of the secret cavity?​

Answer 1

1.1 x 10⁻⁵m³

Step-by-step explanation:

Given parameters:

Weight in air = 11.29N

Weight in water = 9.190N

Density of gold = 19320kg/m³

Unknown:

Volume of the secret cavity = ?

Solution:

To solve this problem, we need to find the mass of the secret cavity.

Weight of the cavity = Weight of sculpture in air - Weight of sculpture in water

Weight of cavity = 11.29N - 9.190N = 2.1N

Weight = mass x acceleration due to gravity

Mass =
`(2.1)/(9.8)` = 0.2kg

So;

Volume of substance can be determined from the density.

Density =
`(mass)/(volume)`

Volume =
`(mass)/(density)`

Now insert the parameters and solve;

Volume =
`(0.2)/(19320)` = 1.1 x 10⁻⁵m³