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A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then
  • a)
    T α 1/√m
  • b)
    T α √ρ
  • c)
    T α 1/√A
  • d)
    T α 1/α
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A rectangular block of mass m and area of cross-section A floats in a ...
Ρ. Let h be the height of the block above the liquid surface and d be the depth of the block below the liquid surface.

We can use Archimedes' principle to find the buoyant force, which is equal to the weight of the liquid displaced by the block. The weight of the liquid displaced is given by ρAd, where A is the area of the block's cross-section and d is the depth of the block below the liquid surface. Therefore, the buoyant force is ρAdg, where g is the acceleration due to gravity.

The weight of the block is mg, where m is the mass of the block. Since the block is floating, the buoyant force must be equal to the weight of the block. Therefore, we have:

ρAdg = mg

Solving for d, we get:

d = m/(ρAg)

Substituting this value of d into the equation for the buoyant force, we get:

F_b = ρAg(m/(ρAg))g = mg

Therefore, the buoyant force is equal to the weight of the block, which means the block is in equilibrium and will not sink or rise. The height of the block above the liquid surface does not affect this equilibrium.
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A rectangular block of mass m and area of cross-section A floats in a ...
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A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Thena)T α 1/√mb)T α √ρc)T α 1/√Ad)T α 1/αCorrect answer is option 'C'. Can you explain this answer?
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