Answer to Question #90366 in Mechanics | Relativity for Joy godson

Question

What is the upthrust on a body which displaces
0.6m^3 of water, 0.5m^3 of methylated spirit?
(density of water = 10^3 kgm^-3, density of
methylated spirit = 0.8 x 10^3 kgm^-3 and
g = 10ms^-2)

Answer 1

Upthrust is an upward force exerted by a fluid that opposes the weight of an immersed body.

The magnitude of the force F is equal to the weight of the fluid that the body displaces. This fluid weight is equal to the density ρ of the fluid multiplied by the volume V and the acceleration due to gravity g

$$F=\rho Vg$$

First calculate upthrust for water. Substituting $\rho ={{10}^{3}}kg\cdot {{m}^{-}}^{3},\,\, V=0.6{{m}^{3}}$ and $g=10m\cdot {{s}^{-}}^{2}$ into this formula we get

$$F={{10}^{3}}kg\cdot {{m}^{-}}^{3}\cdot 0.6{{m}^{3}}\cdot 10m\cdot {{s}^{-}}^{2}$$

$$=6\cdot {{10}^{3}}kg\cdot m\cdot {{s}^{-}}^{2}=6\cdot {{10}^{3}}N=6\,kN$$

Now calculate upthrust for methylated spirit. Substitute$\rho =0.8\cdot {{10}^{3}}kg\cdot {{m}^{-}}^{3}, \,\,V=0.5{{m}^{3}}$   and $g=10m\cdot {{s}^{-}}^{2}$ into formula:

$$F=0.8\cdot {{10}^{3}}kg\cdot {{m}^{-}}^{3}\cdot 0.5\,{{m}^{3}}\cdot 10m\cdot {{s}^{-}}^{2}$$

$$=4\cdot {{10}^{3}}kg\cdot m\cdot {{s}^{-}}^{2}=4\cdot {{10}^{3}}N=4\,kN$$

So upthrust for water is $F=6\,kN$ , upthrust for methylated spirit is $F=4\,kN$