Question

A table tennis ball has an average density of $0.084gc{m^{ - 3}}$ and a diameter of $3.8cm$ How large a force can just submerge the ball in water?

Answer 1

In order to solve this question, we will first calculate the weight of the ball and then will calculate the buoyant force due to the water and then the net force by subtracting the buoyant force and weight and this net force will be needed to just submerge the ball.

Complete step by step answer:
According to the question, we have given that
${\rho _{ball}} = 0.084gc{m^{ - 3}}$ density of the ball
$d = 3.8cm$ diameter of the ball
$g = 980cm{s^{ - 2}}$ acceleration due to gravity
Let V be the volume of spherical ball then,
$V = \dfrac{4}{3}\pi \dfrac{{{d^3}}}{8}$ on putting the value of parameters we get,
$V = \dfrac{{3.14}}{6} \times {(3.8)^3}$
$V = 28.73c{m^3}$
now, the weight of the ball W in downward direction can be calculated as
$W = V \times {\rho _{ball}} \times g \to (i)$
Now, buoyant force due to water in upward direction on the ball can be calculated as
${F_B} = V \times {\rho _{water}} \times g \to (ii)$ where ${\rho _{water}} = 1gc{m^{ - 3}}$ density of water, ${F_B}$ represents the buoyant force on the tennis ball.
So, net force required to keep the tennis ball submerged will be
${F_{net}} = {F_B} - W$ so, subtracting equation (i) from (ii) we get,
${F_{net}} = Vg({\rho _{water}} - {\rho _{ball}})$ so, on putting the value of parameters we get,
${F_{net}} = 28.73 \times 980 \times (1 - 0.084)$
${F_{net}} = 28155.4 \times 0.916$
$\Rightarrow {F_{net}} = 25790.3dyne$
Hence, the amount of force required to keep the tennis ball inside the water submerged is ${F_{net}} = 25790.3dyne$

Note: It should be remembered that, all the units given in question are in Centimetre Gram second units so convert all quantities in same unit while solving such numerical problems and dyne represents the unit of force in CGS systems while the standard SI unit of force is Newton.