Physics 2009 JAMB Past Questions - Question 17: A Plastic sphere floats in water with 50% of its volume submerged, the density of the glycerin is

Correct Answer

C

Explanation

To find the density of the glycerin, you can use the principle of buoyancy. When an object floats in a fluid, it displaces a volume of fluid equal to its own volume. In this case, you have a plastic sphere floating in water with 50% of its volume submerged, which means that 50% of its volume is displaced by the water.

Assuming that the plastic sphere has a uniform density, you can calculate the density of the glycerin by considering the buoyant force acting on the sphere in the glycerin. The buoyant force is equal to the weight of the displaced glycerin.

Let:
- V_s be the volume of the sphere.
- V_w be the volume of water displaced by the sphere.
- V_g be the volume of glycerin displaced by the sphere.
- ρ_w be the density of water.
- ρ_g be the density of glycerin.
- g be the acceleration due to gravity.

From the information given, you know that 50% of the sphere's volume is submerged in water. So:

V_w = 0.5 * V_s

The buoyant force on the sphere in the water is given by:

F_b = ρ_w * V_w * g

Since the sphere is in equilibrium (not sinking or rising), the buoyant force must equal the weight of the sphere. The weight of the sphere is given by:

F_weight = ρ_s * V_s * g

Where ρ_s is the density of the sphere.

Now, the buoyant force in water and the weight of the sphere must be equal:

ρ_w * 0.5 * V_s * g = ρ_s * V_s * g

We can cancel the "g" terms:

ρ_w * 0.5 * V_s = ρ_s * V_s

Now, we can solve for the density of the sphere:

ρ_s = (ρ_w * 0.5)

Now, you need to know the density of water to find the density of the sphere. The density of water is typically around 1000 kg/m^3. So:

ρ_s = (1000 kg/m^3 * 0.5) = 500 kg/m^3

Therefore, the density of the glycerin must be equal to the density of the sphere, which is 500 kg/m^3 in this scenario.