Q#12
A 20 N metal block is suspended by a spring balance. A beaker containing some water is placed on a weighing machine which reads 40 N. The spring balance is now lowered so that the block gets immersed in the water. The spring balance now reads 16 N. The reading of the weighing machine will be(a) 36 N
(b) 60 N
(c) 44 N
(d) 56 N.
Answer: (c)
The water in the beaker applies upward buoyancy force on the metal block = 20 – 16 = 4 N. According to Newton's third law the block applies a downward force = 4 N on the water. So the weight of the water = 40 + 4 = 44 N shown in the weighing machine.
Q#13
A piece of wood is floating in water kept in a bottle. The bottle is connected to an air pump. Neglect the compressibility of water. When more air is pushed into the bottle from the pump, the piece of wood will float with
(a) larger part in the water
(b) lesser part in the water
(c) same part in the water
(d) it will sink.
Answer: (c)
Pushing air inside the bottle will increase the air pressure. When the pressure of air inside the bottle is increased it does not increase the weight of the wood nor the density of the water. So there will be no change in the submerged part.
Q#14
A metal cube is placed in an empty vessel. When water is filled in the vessel so that the cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube
(a) will increase
(b) will decrease
(c) will remain the same
(d) will become zero.
Answer: (c)
In this problem, we should be clear that the water is filled in the vessel just to completely submerge the cube and not that water is filled over the cube up to some height.
The most common and easy explanation is given assuming that water does nor enter beneath the cube (which is most unlikely) thus there is no buoyancy and the force will remain the same. But practically the water will creep beneath the cube because the surfaces of contact are not fully plane. The cube will be in contact with the bottom of the vessel through some molecules and the gap will be filled by the water. See the figure below.
The pressure of water under the cube = ρga, and it will push the cube upward with a force B
=ρg(a)a²= ρga³ = (ρa³)g
Now, a³ is the volume of the cube and also the volume of the water displaced, ρa³g is the weight of the water displaced and the B is the force of buoyancy. So, the apparent weight of the cube becomes
W = mg – B =mg – ρga³
this weight is transferred to the vessel's surface through contact molecules. The pressure of the water beneath the cube is in all directions and it also pushes down the area beneath the cube by a force B = ρg(a)a² = ρga³.
So, the total force downward on the area a² beneath the cube
= W + B = mg – B + B = mg
which is the weight of the cube before filling the vessel with water. So, the force on the bottom of the vessel in contact with the cube remains the same.
Q#15
A wooden object floats in water kept in a beaker. the object is near a side of the beaker (figure 13-Q4). Let P₁, P₂, P₃ be the pressures at the three points A, B and C of the bottom as shown in the figure.
(a) P₁ = P₂ = P₃
(b) P₁ < P₂ < P₃
(c) P₁ > P₂ > P₃
(d) P₂ = P₃ ≠ P₁.
Answer: (a)
The pressure along a horizontal plane in a liquid at rest is the same.
Q#16
A closed cubical box is completely filled with water and accelerated horizontally towards the right with an acceleration a. The resultant normal force by the water on the top of the box
(a) passes through the center of the top
(b) passes through a point to the right of the center
(c) passes through a point to the left of the center
(d) becomes zero.
Answer: (c)
Due to the acceleration a towards the right, a pseudo force ma acts on the water towards left. The pressure due to this force increases from right to left and the normal pressure on the top of the box by the water will vary as shown colored in the figure below. The pressure due to the gravitation on the top is zero. The total force (N) of this triangular pressure distribution will act at 1/3rd distance from the left side of the triangle. So, N will pass through a point left of the center at the top of the box.
Q#17
Consider the situation of the previous problem. Let the water push the left wall by a force F₁ and the right wall by a force F₂.
(a) F₁ = F₂
(b) F₁ > F₂
(c) F₁ < F₂
(d) The information is insufficient to know the relation between F₁ and F₂.
Answer: (b)
As the pressure depends on the depth of the liquid at rest, similarly in an accelerating liquid the pressure due to acceleration depends on the horizontal length from the leading surface.
Q#18
Water enters through end A with a speed v₁ and leaves through end B with a speed v₂ of a cylindrical tube AB. The tube is always completely filled with water. In case I the tube is horizontal, in case II it is vertical with the end A upward and in Case III it is vertical with end B upward. We have v₁ = v₂ for
(a) Case I
(b) Case II
(c) Case III
(d) Each case.
Answer: (d)
Since the water is incompressible, the volume of water entering end A and leaving through end B per unit time will remain constant in each case. Since the area of the cross-section of the cylindrical tube will be constant, v₁ = v₂ for each case.
Q#19
Bernoulli's theorem is based on the conservation of
(a) momentum
(b) mass
(c) energy
(c) angular momentum.
Answer: (c)
Bernoulli's theorem is based on the work-energy theorem which states that the total work done on a system is equal to the change in its kinetic energy. It is also the conservation of energy if no work is done.
Q#20
Water is flowing through a long horizontal tube. Let Pₐ and Pᵦ be the pressures at two different two points A and B of the tube.
(a) Pₐ must be equal to Pᵦ
(b) Pₐ must be greater than Pᵦ
(c) Pₐ must be smaller than Pᵦ
(d) Pₐ = Pᵦ only if the cross-sectional area at A and B are equal.
Answer: (d)
Pₐ = Pᵦ only if the velocities are equal at A and B (From the Bernoulli's theorem it can be deduced that in a horizontal tube flow pressure is less where velocity is more). The velocities at A and B will be equal only if the cross-sectional areas at A and B are the same.
Q#21
Water and mercury are filled in two cylindrical vessels up to the same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are v₁ and v₂ respectively.
(a) v₁ = v₂
(b) v₁ = 13.6v₂
(c) v₁ = v₂/13.6
(d) v₁ = √(13.6) v₂.
Answer: (a)
The velocity of the water v = √(2gh) which is independent of the density of the liquid.
Q#22
A large cylindrical tank has a hole of area A at its bottom. Water is poured in the tank by a tube of equal cross-sectional area A ejecting water at the speed v.
(a) The water level in the tank will keep rising.
(b) No water can be stored in the tank.
(c) The water level will rise to a height v²/2g and then stop.
(d) The water level will oscillate.
Answer: (c)
Initially the velocity of the water coming out of the tank will be negligible due to the negligible height of the water. So the quantity of the water coming out of the tank per second is less than the coming in, so the level of the water in the tank will begin to rise and the velocity of the water coming out will begin to increase. The level of the water will rise up to a height h for which the velocity of the water coming out is equal to the velocity of the water coming in (= v).
So, v = √(2gh)
h = v²/2g.
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