Life is like riding a bicycle.  To keep your balance you must keep moving.”
Albert Einstein

Tuesday, December 4, 2007

Fluid Mechanics- Multiple Choice Practice Questions on Force of Buoyancy

In the post dated 3rd December 2007, the essential formulas you have to remember in the section 'Fluid Mechanics' were discussed. As promised, I will give you a few multiple choice questions with solution. Here is a question on force of buoyancy:

In the adjoining figure, a metallic block of volume 20 cm3 is shown suspended, using a thin nylon string, from an independent support. The entire block is immersed in water in a beaker, without touching the sides of the beaker. The reading of the electronic balance (on which the beaker is placed) is 257 gram, as shown.
If the weight of the beaker is 37 gram, what is the weight of the water in the beaker?
(a) 200 g (b) 220 g (c) 230 g (d) 240 g (e) Data insufficient for calculation
Since the metallic block is completely immersed in water, the force of buoyancy on the block is equal to the weight of 20 cm3 of water displaced by the block. In other words, water exerts an upthrust equal to the weight of 20 g of water. The block exerts an equal force downwards (reaction force) which the electronic balance registers in addition to the weight of the beaker and water.
The weight of the beaker and water together is therefore 257–20 = 237 g. Since the weight of the beaker is 37 g, the weight of water alone is 237–37 = 200 g.
Now, suppose the block was kept immersed using a rigid support instead of the string. Then also, the same explanation holds good. If you dip your finger gently in the water in the beaker, the reading of the balance will increase since the water exerts a force of buoyancy on your finger and your finger exerts an equal and opposite force of reaction.
Here is another multiple choice question (MCQ):
An iceberg floating in sea water just sinks when a mass of 360 kg is placed on it. What is the mass of the iceberg? (Relative density of sea water = 1.02; relative density of ice = 0.9)
(a) 1500 kg (b) 1800 kg (c) 2400 kg (d) 2700 kg (e) 3200 kg
The density of ice is 900 kg m3 and that of sea water is 1020 kg m3, as obtained from their relative densities. When the iceberg just sinks, the mass of displaced sea water is 1200V kg. where V is the volume of the iceberg. The mass of iceberg is 900V kg. Therefore we have, in this case (since the weight of floating body = weight of displaced liquid),
900V + 360 = 1020V, from which V = 3 m3
Therefore, mass of iceberg = 900×3 = 2700 kg.
Let us consider one more question:
A solid sphere of volume V remaining submerged in a liquid of density σ is lifted through a height ‘x’ within the liquid. If the density of the material of the sphere is ρ where ρ > σ, which one of the following statements is correct?
(a) There is no change in the potential energy of the sphere
(b) The potential energy of the sphere decreases by Vρgx
(c) The potential energy of the sphere increases by Vρgx
(d) The potential energy of the sphere increases by V[(ρ+σ)/2]gx
(e) The potential energy of the sphere increases by V(ρ σ)gx
You will have to exert a force equal to the apparent weight of the sphere for lifting it. The apparent weight of the sphere is Vρg – Vσg = V– σ)g. The first term is its real weight and the second term is the up thrust (force of buoyancy) on it.
The work done in lifting the sphere through the height ‘x’ is V(ρ – σ)gx and this is the increase in potential energy of the sphere.
More questions from Fluid Mechanics will be discussed in the next post.
Meanwhile see a useful post in this section here.

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