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Question 39
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A homogeneous solid block K, of weight W, does not dissolve in either liquid. It floats in a liquid of density ρ, as shown. At constant temperature, a second liquid of density 3ρ, which mixes uniformly with the first, is slowly added. The system is then allowed to reach a new equilibrium.
Compared with the initial state, how do the submerged volume of block K and the buoyant force acting on it change at the new equilibrium?
| Option | Submerged volume | Buoyant force |
|---|---|---|
| A | Decreases | Stays the same |
| B | Increases | Decreases |
| C | Stays the same | Stays the same |
| D | Decreases | Increases |
| E | Increases | Stays the same |
While floating, Fb = W. The weight stays constant.
Adding liquid of density 3ρ makes the mixture denser.
Since V = W / (ρmix g), submerged volume decreases.
See why the block rises while the buoyant force stays the same.
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Example geometry solution: A circle centred at O has radius 3. A is 5 units from O, and AT is tangent to the circle at T. Since OT is perpendicular to AT, Pythagoras gives 3² + AT² = 5², so AT = 4 units.
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