Answer to Question #246132 in Physics for Faisal Yusuf

Question

The apparent weight of a body holy emerged in the water is 32N and eat sweet in a is 96N Calculate the volume of the body. (Density of water =1000kgm-3,g=10ms-2

Answer 1

The apparent weight of a body holy emerged in the water is 32N and eat sweet in a is 96N Calculate the volume of the body. (Density of water =1000kgm-3,g=10ms-2

The weight of the body in the water is the resultant of two forces: weight of the body in air $F_g = 96N$ acting down and buoyant force $F_b$ acting up:

$$W = F_g - F_b$$

The buoyant force is given as follows:

$$F_b = \rho_w V g$$

where $\rho_w = 1000kg/m^3$ is the density of water and $V$ is the volume of the body. Expressing $V$ and substituting $W = 32N$, obtain:

$$V = \dfrac{F_g - W}{\rho_w g}\\ V = \dfrac{96N - 32N}{1000kg/m^3\cdot 10N/kg} = 6.4\times 10^{-3}m^3$$

Answer. $6.4\times 10^{-3}m^3$.