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A wooden ball of density 900 kg/m3 is immersed in water to a depth 18 cm below the surface of water and then it is released. Upto what height will the ball jump out of the water?
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Calculation of Height the Ball will Jump out of Water:

- Step 1: Calculate the Buoyant Force
Given density of water, ρ(water) = 1000 kg/m³
Volume of the ball, V = (4/3)πr³, where r is the radius of the ball
Mass of the ball, m = ρ(ball) * V
Buoyant force, F(b) = ρ(water) * V * g, where g is the acceleration due to gravity
- Step 2: Calculate the Weight of the Ball
Weight of the ball, W = m * g
- Step 3: Calculate the Depth at which the Ball will Float
Since the ball will float when the buoyant force equals the weight of the ball, we have:
F(b) = W
ρ(water) * V * g = ρ(ball) * V * g
ρ(water) = ρ(ball)
Solving for the depth at which the ball will float will give:
h = (ρ(ball) - ρ(water)) / ρ(water) * r
- Step 4: Calculate the Height the Ball will Jump out of Water
When the ball is released, it will experience an upward force equal to its weight minus the weight of the water displaced by the immersed volume. This will cause the ball to accelerate upwards until it reaches a height where its potential energy is equal to the work done against the buoyant force.
Height the ball will jump out of the water = h
Therefore, the height the ball will jump out of the water is 18 cm.
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A wooden ball of density 900 kg/m3 is immersed in water to a depth 18 cm below the surface of water and then it is released. Upto what height will the ball jump out of the water?
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