Question

A rock is suspended by a light string. When the rock is in air, the tension in the string is 38.7 N . When the rock is totally immersed in water, the tension is 32.0 N . When the rock is totally immersed in an unknown liquid, the tension is 19.0 N .
What is the density of the unknown liquid?

Answer 1

Answer:

$\rho = 2940 kg/m^3$

Explanation:

When pendulum is in air then the tension in the string is given as

$T = mg$

$T = 38.7 N$

now when it is submerged in water then tension is decreased due to buoyancy force of water

so we will have

$T_{water} = mg - F_b$

$32 = 38.7 - F_b$

$F_b = 6.7 N$

When rock is immersed in unknown liquid then tension is given by

$T = mg - F_b'$

$19 = 38.7 - F_b'$

$F_b' = 19.7 N$

from buoyancy force of water we can say

$F_b = 1000(V)(9.8) = 6.7$

$V = 6.84 \times 10^{-4} m^3$

now for other liquid we will have

$19.7 = \rho(6.84 \times 10^{-4})(9.8)$

$\rho = 2940 kg/m^3$