Answer to Question #312977 in Mechanics | Relativity for Shark

Question

A table tennis ball has a diameter of 3.80 cm and an average density of 0.0840 g/cm³. What force is required to hold it completely submerged underwater?

Answer 1

Given

$d-$ Diameter of a tenis ball=3.80cm

$\rho-$ Density of the tenis ball$=0.0840g/cm^3$

$dw-$ Density of water$=1g/cm^3$

Volume of the ball,$w-v=43 \pi d38$

$=16\pi\times(3.8)3cm^3=28.73cm^3$

$w-$ weight of the ball$=v×\rho ×g$

when the ball is submerged the buoyant force acting on the ball wil be equal to the weight of displaced water.

So the bouyant force

$Fb=v\times dw \times g$

So the net downward force to be appliied on the ball to just submerge it will be

$F_{net}=Fb-w=vdwg-v\rho g=vg(dw-\rho )$

$=28.73 \times 980 \times (1-0.084) dyne = 25790.3 dyne$