A spherical solid ball of raduis R and density 0.5 is imersed in water with half part inside water?What is work done if we push whole ball inside water?Find answer in terms of D(density of water),R(raduis of ball)

To determine the work done in pushing a spherical solid ball of radius R and density 0.5 completely underwater when half of it is already submerged, we need to consider the buoyant force acting on the ball and the distance we need to push it down. Let's break this down step by step.

Understanding the Forces at Play

When the ball is partially submerged, it experiences a buoyant force equal to the weight of the water displaced by the submerged part of the ball. The buoyant force can be calculated using Archimedes' principle, which states that the upward buoyant force is equal to the weight of the fluid displaced by the object.

Calculating the Weight of the Ball

The volume \( V \) of a sphere is given by the formula:

• V = \( \frac{4}{3} \pi R^3 \)

Given that the density \( \rho \) of the ball is 0.5, the mass \( m \) of the ball can be calculated as:

• m = \( \rho \times V = 0.5 \times \frac{4}{3} \pi R^3 \)

Thus, the weight \( W \) of the ball is:

• W = \( m \times g = 0.5 \times \frac{4}{3} \pi R^3 \times g \)

Calculating the Buoyant Force

Now, let's find the buoyant force \( F_b \) when half of the ball is submerged. The volume of the submerged part is half of the total volume:

• V_submerged = \( \frac{1}{2} \times V = \frac{1}{2} \times \frac{4}{3} \pi R^3 = \frac{2}{3} \pi R^3 \)

The weight of the water displaced, which is equal to the buoyant force, is given by:

• F_b = \( D \times V_{submerged} \times g = D \times \frac{2}{3} \pi R^3 \times g \)

Finding the Work Done

To push the ball completely underwater, we need to overcome the buoyant force. The distance we need to push the ball down is equal to the radius \( R \) of the ball. The work done \( W_d \) in pushing the ball down can be calculated using the formula:

• W_d = F_b \times d

Substituting the values we have:

• W_d = \( D \times \frac{2}{3} \pi R^3 \times g \times R \)

Thus, the work done in pushing the entire ball underwater is:

• W_d = \( \frac{2}{3} \pi D g R^4 \)

Final Expression

In summary, the work done to push the entire ball underwater, in terms of the density of water \( D \) and the radius of the ball \( R \), is:

• W_d = \( \frac{2}{3} \pi D g R^4 \)

This expression gives you a clear understanding of how the work done is related to the physical properties of the ball and the water it is submerged in. If you have any further questions or need clarification on any part of this process, feel free to ask!