Even though you have been studying basic properties of fluids from lower classes, questions based on them are repeatedly found in Medical and Engineering Entrance Examinations. Here are some questions which may be interesting and useful to you:
(1) A cubical block of wood with a glass bead placed on it, is floating in water contained in a beaker. The height of water column in the beaker in this condition is ‘h’ and the extent to which the wooden block is within water is ‘d’. If the glass bead is gently transferred to the water in the beaker,
(a) ‘h’ will increase and ‘d’ will decrease (b) ‘h’ will decrease and ‘d’ will increase (c) ‘h’ and ‘d’ will be unchanged (d) ‘h’ and ‘d’ will increase (e) ‘h’ and ‘d’ will decrease
The glass bead can displace more water when it is resting on the wooden block since a floating body will displace a volume of liquid having the weight of the floating body. Inside the water, it can displace water having its own volume only. This volume is less since the density of glass is greater than that of water. Therefore, ‘h’ is decreased on transferring the glass bead to the water in the beaker.
The extent (d) to which the wooden block is within water also is decreased since the weight of the glass bead is relieved from the block. So, the correct option is (e).
(2) A tank is filled with water up to height ‘H’. A hole is made on the side of the tank at a distance ‘h’ below the level of water. What will be the horizontal range of the water jet?
(a) 2√[h(H-h)] (b) 4√[h(H+h)] (c)2√[h(H-h)]
(1) A cubical block of wood with a glass bead placed on it, is floating in water contained in a beaker. The height of water column in the beaker in this condition is ‘h’ and the extent to which the wooden block is within water is ‘d’. If the glass bead is gently transferred to the water in the beaker,
(a) ‘h’ will increase and ‘d’ will decrease (b) ‘h’ will decrease and ‘d’ will increase (c) ‘h’ and ‘d’ will be unchanged (d) ‘h’ and ‘d’ will increase (e) ‘h’ and ‘d’ will decrease
The glass bead can displace more water when it is resting on the wooden block since a floating body will displace a volume of liquid having the weight of the floating body. Inside the water, it can displace water having its own volume only. This volume is less since the density of glass is greater than that of water. Therefore, ‘h’ is decreased on transferring the glass bead to the water in the beaker.
The extent (d) to which the wooden block is within water also is decreased since the weight of the glass bead is relieved from the block. So, the correct option is (e).
(2) A tank is filled with water up to height ‘H’. A hole is made on the side of the tank at a distance ‘h’ below the level of water. What will be the horizontal range of the water jet?
(a) 2√[h(H-h)] (b) 4√[h(H+h)] (c)2√[h(H-h)]
(d) 2√[h(H+h)] (e) √[h(H+h)]
The water jet at the hole on the side of the tank will be directed horizontally with a velocity (of efflux) equal to √(2gh), as given by Torricelli’s theorem. The vertical fall of the jet is through the distance (H-h) and the time of fall (t) is √[2(H-h)/g], given by the equation of linear motion, H-h = 0 + ½ gt2.
The horizontal range R = Horizontal velocity × Time = √(2gh) ×√[2(H-h)/g] = 2√[h(H-h)].
So, the correct option is (c). [ It is interesting to note that this is independent of g so that the range on the moon will be the same as that on the earth for given values of H and h].
By putting dR/dh equal to zero, you can show that the horizontal range ‘R’ is maximum when h = H/2, which means that the hole should be at the middle of the water column.
(3) Two identical cylindrical vessels with base area ‘A’ containing a liquid of density ‘ρ’ are placed on a horizontal table. The heights of liquid column in them are h1 and h2. If they are connected by a narrow pipe to make the liquid levels in them same, the work done by gravity is
(a) ½ Aρg(h1+h2) (b) ½ Aρg(h1+h2)2 (c) ½ Aρg(h1 - h2)2
The water jet at the hole on the side of the tank will be directed horizontally with a velocity (of efflux) equal to √(2gh), as given by Torricelli’s theorem. The vertical fall of the jet is through the distance (H-h) and the time of fall (t) is √[2(H-h)/g], given by the equation of linear motion, H-h = 0 + ½ gt2.
The horizontal range R = Horizontal velocity × Time = √(2gh) ×√[2(H-h)/g] = 2√[h(H-h)].
So, the correct option is (c). [ It is interesting to note that this is independent of g so that the range on the moon will be the same as that on the earth for given values of H and h].
By putting dR/dh equal to zero, you can show that the horizontal range ‘R’ is maximum when h = H/2, which means that the hole should be at the middle of the water column.
(3) Two identical cylindrical vessels with base area ‘A’ containing a liquid of density ‘ρ’ are placed on a horizontal table. The heights of liquid column in them are h1 and h2. If they are connected by a narrow pipe to make the liquid levels in them same, the work done by gravity is
(a) ½ Aρg(h1+h2) (b) ½ Aρg(h1+h2)2 (c) ½ Aρg(h1 - h2)2
(d) ¼ Aρg(h1 + h2)2 (e) ¼ Aρg(h1 - h2)2
The work done in obtaining a cylindrical liquid column of height ‘h’ and cross section area ‘A’ is ½Aρgh2 and this is stored as potential energy in the liquid column. [ You can easily obtain this by integrating Aρgxdx between limits 0 and h. Note that (Adxρg) is the weight of liquid column (slice) of small height dx and hence (Adxρg)x is the work done against gravity to lift this liquid slice through a height x. The total work done is the integral between the limits 0 and h].
The work done in obtaining a cylindrical liquid column of height ‘h’ and cross section area ‘A’ is ½Aρgh2 and this is stored as potential energy in the liquid column. [ You can easily obtain this by integrating Aρgxdx between limits 0 and h. Note that (Adxρg) is the weight of liquid column (slice) of small height dx and hence (Adxρg)x is the work done against gravity to lift this liquid slice through a height x. The total work done is the integral between the limits 0 and h].
[Alternatively, you can get it by arguing that the weight of the entire liquid column is Ahρg and its centre of gravity has been raised through a height h/2 so that the work done against gravity is Ahρg(h/2), which is equal to ½Aρgh2].
The work done by gravity on equalizing the levels in the cylinders is the difference between the initial and final potential energies of the two liquid columns.
Therefore, work done = (½Aρgh12 + ½Aρgh22) - 2×½Aρg [(h1+ h2)/2]2. Note that finally the two liquid columns have the same height [(h1+ h2)/2].
The above expression for work done simplifies to (¼)Aρg(h1 - h2)2 given in option (e).
Let us now discuss the following MCQ which appeared in Har PMT 2000 question paper:
When a bubble rises from the bottom of a lake to the surface, its radius doubles. The atmospheric pressure is equal to that of a column of water of height H. The depth of the lake is
(a) 8H (b) 2H (c) 7H (d) H
This simple question involves the application of Boyle’s law: P1V1 = P2V2 with usual notations. The initial pressure is the total pressure due to the atmosphere and the water column of height ‘h’ in the lake. So, the initial pressure of the bubble is equivalent to that due to water column of height H+h. If the initial volume is V, the final volume is 8V since the radius of the bubble is doubled. The final pressure is the atmospheric pressure ‘H’.
Expressing the pressures in terms of height of water column itself, we have,
(H+h)V = H×8V, from which h = 7H [Option (c)].
You will find more questions (with solution) on properties of fluids here as well as here.
The work done by gravity on equalizing the levels in the cylinders is the difference between the initial and final potential energies of the two liquid columns.
Therefore, work done = (½Aρgh12 + ½Aρgh22) - 2×½Aρg [(h1+ h2)/2]2. Note that finally the two liquid columns have the same height [(h1+ h2)/2].
The above expression for work done simplifies to (¼)Aρg(h1 - h2)2 given in option (e).
Let us now discuss the following MCQ which appeared in Har PMT 2000 question paper:
When a bubble rises from the bottom of a lake to the surface, its radius doubles. The atmospheric pressure is equal to that of a column of water of height H. The depth of the lake is
(a) 8H (b) 2H (c) 7H (d) H
This simple question involves the application of Boyle’s law: P1V1 = P2V2 with usual notations. The initial pressure is the total pressure due to the atmosphere and the water column of height ‘h’ in the lake. So, the initial pressure of the bubble is equivalent to that due to water column of height H+h. If the initial volume is V, the final volume is 8V since the radius of the bubble is doubled. The final pressure is the atmospheric pressure ‘H’.
Expressing the pressures in terms of height of water column itself, we have,
(H+h)V = H×8V, from which h = 7H [Option (c)].
You will find more questions (with solution) on properties of fluids here as well as here.
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