Question

A child is holding a ball with a diameter of 5.70 cm and average density of 0.0835 g/cm3 under water. Determine the force (in N) needed to hold it completely submerged.

Answer 1

0.488 N

Step-by-step explanation:

Given , The ball has diameter (d)=5.70 cm = 0.0570 m

radius(r) = d/2 = 2.85 cm = 0.0285 m

Average density
`=0.0835g/cm^3 = 83.5 kg/m^3`

There are two forces acting on the ball.. i.e. The buoyancy force and weight of the ball.

The force required to completely submerged under water= Buoyancy force - Weight of ball

buoyancy force = volume×density of water×g

`=(4)/(3) \pi r^(3)dg`

`=4.18*(0.0285)^3*1000*9.8` [ density of water = 1000 kg/m3]

=0.532 N

mass of ball = average density×Volume

`=83.5*4.18*(0.0285)^3`

= 0.004529 kg

Weight of ball = mg = 0.004529×9.81 =0.044 N

Required force = 0.532 - 0.044 = 0.488 N.

Answer 2

Step-by-step explanation:

Answer:

0.8712 N

Step-by-step explanation:

diameter = 5.7 cm

radius, r = 2.85 cm

Volume of sphere, V = 4/3 x 3.14 x 2.85 x 2.85 x 2.85 x 10^-6 = 9.7 x 10^-5 m^3

Density of ball, d = 0.0835 g/cm^3 = 83.5 kg/m^3

density of water, D = 1000 kg/m^3

True weight, W = Volume x density x g = 9.7 x 10^-5 x 83.5 x 9.8

W = 0.07938 N

Buoyant force, B = Volume x density of water x g

B = 9.7 x 10^-5 x 1000 x 9.8 = 0.9506 N

Force required to hold it

F = W - B

F = 0.9506 - 0.07938

F = 0.8712 N

Thus, the force required to hold it is 0.8712 N.