Explanation:
It is given that,
Weight of the rock in air, W = 110 N
Since, W = mg
[tex]m=\dfrac{W}{g}[/tex]
[tex]m=\dfrac{110\ N}{9.8\ m/s^2}[/tex]
m = 11.22 kg
We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.
[tex]M=d\times V[/tex]
d is the density of water, [tex]d=1000\ kg/m^3[/tex]
V is the volume of rock, [tex]V=0.00337\ m^3[/tex]
[tex]M=1000\ kg/m^3\times 0.00337\ m^3[/tex]
M = 3.37 kg
The apparent weight in water, W = m - M
[tex]W=7.85\ kg\times 9.8\ m/s^2[/tex]
W = 76.93 N
So, the apparent weight of the rock is 76.93 N. Hence, this is the required solution.
