Two solids A and B float in water .A floats with half of its volume immersed and B float s with 2/3 of its volume immersed.the ratio of their densities is ?
D1/D2 = (1/2)/(2/3) = 1/2 * 3/2 = 3/4.
To determine the ratio of densities between solids A and B, we can make use of Archimedes' principle, which states that the buoyant force experienced by an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
Let's assume the volume of solid A is V, and its density is ρA. Since A floats with half of its volume immersed, the volume of water displaced by A is also V/2.
According to Archimedes' principle, the buoyant force experienced by A is equal to the weight of the water displaced. This can be calculated using the formula:
Buoyant force = weight of water displaced = volume of water displaced * density of water * gravitational acceleration
Buoyant force on A = (V/2) * ρw * g ... (1)
where ρw is the density of water and g is the acceleration due to gravity.
Similarly, let's assume the volume of solid B is V and its density is ρB. Since B floats with 2/3 of its volume immersed, the volume of water displaced by B is (2/3) * V.
Then, the buoyant force experienced by B can be calculated as:
Buoyant force on B = (2/3) * V * ρw * g ... (2)
Now, since both solids A and B are floating, the buoyant force must be equal to the weight of each object. Therefore, we have:
Buoyant force on A = weight of A
Buoyant force on B = weight of B
Since weight is equal to mass multiplied by gravitational acceleration, we can write:
(V/2) * ρw * g = V * ρA * g ... (3)
(2/3) * V * ρw * g = V * ρB * g ... (4)
We can cancel out the common terms V, ρw, and g from equations (3) and (4) to make it simpler:
(1/2) * ρA = ρw
(2/3) * ρB = ρw
Dividing equation (2) by equation (1), we get:
(2/3) * ρB / (1/2) * ρA = ρw / ρw
This simplifies to:
4/3 * ρB / ρA = 1
Finally, rearranging the equation, we find the ratio of densities between solids A and B:
ρB / ρA = 3/4
So, the ratio of their densities is 3:4.