Question

A rock is suspended by a light string. When the rock is in air, the tension in the string is 44.8N . When the rock is totally immersed in water, the tension is 30.0N . When the rock is totally immersed in an unknown liquid, the tension is 15.8N .

What is the density of the unknown liquid?

Answer 1

We have:

The density of water is 1000kg/m^3.

F =W_air - W_water in getting the difference between the two situations. so in the problem, we are given the following values:

T_air = 44.8 N
T_water = 30.0 N
T(?) = 15.8 N

So, we use the formula:

F = T_air - T_water
= 44.8 - 30.0
= 14.8 N

The next thing to do is divide it by 9.8m/s² or the gravitational force.

That is a displaced mass equivalent of 14.8/9.8 = 1.510 kg

Then, we should find next the volume of the rock. So,
V_rock = 1.510kg/(1000kg/m³)
= 0.001510 kg/m³

Then, we should now deduct the tension in the unknown liquid from the tension of the rock in air.

30.0 - 15.8 = 14.2 N

Again, the equivalent to a displaced mass of 14.2/9.8 = 1.448 kg

Lastly, we are now going to solve for the density of the unknown liquid:

1.448/V_rock = 1.448 kg / 0.001510 kg/m³

= 959.589 kg/m³