Question

A rock is suspended by a light string. When the rock is in air, the tension in the string is 43.8 N . When the rock is totally immersed in water, the tension is 31.3 N . When the rock is totally immersed in an unknown liquid, the tension is 18.3 N .

Answer 1

Density of unknown liquid is
`2047 kg/m^3`

Step-by-step explanation:

When rock is suspended in air then the weight of the rock is counter balanced by the tension force in the string

So here we have

`T = mg = 43.8 N`

now when the rock is immersed in water then the tension in the string is and buoyancy force due to water is counter balanced by the weight of the object

so here we have

`T_1 + F_b = mg`

`31.3 + F_b = 43.8`

`F_b = 43.8 - 31.3 = 12.5 N`

now we have

`(1000)V(9.81) = 12.5`

`V = 1.27 * 10^(-3) m^3`

now when the rock is immersed into other liquid then we have

`T_2 + F_b' = mg`

`18.3 + F_b' = 43.8`

`F_b' = 25.5 N`

now we have

`\rho(1.27 * 10^(-3))(9.81) = 25.5`

`\rho = 2047 kg/m^3`