FLUIDS 1.
Sol.
A glass tube scaled at both ends is 1m long. It lies horizontally with the middle 10 cm containing Hg. the two ends of the tube equal in length contain air at 279C and pressure 76 cm of Hg. The temperature at one end is kept 00C and at the other end it is 1270C. Neglect the change in length of Hg column. Then the change in length on two sides is [Old-DPP_A Batch_05-06_8] (A) 12.3 cm (B) 10.311 cm (C) 9.9 cm (D*) 8.489 cm (D) Initially l = 45 cm (2l + 10 = 100 cm) P1 = P2 P (say) ...........(1)
Applying gas law at end A. 45 AP ( 45 x )AP1 300 273
.........(2)
45 AP ( 45 x )AP2 ( 45 x ) 45 x = From (1),(2) and (3) = = 8.49 cm, 300 400 273 400 A thin tube of uniform cross-section is seated at both ends. If lies horizontally. The middle 5 cm containing Hg and two equal ends contain air at the same pressure P0. When the tube is held at an angle of 600 with the vertical, the length of the air column above and below the Hg are 46 cm 44.5 cm. Calculate pressure P0 in cm of Hg. Assume temperature of the system constant. [Old-DPP_A Batch_05-06_9] (A) 55 cm of Hg (B) 65 cm of Hg (C) 70.4 cm of Hg (D) 75.4 cm of Hg (D) 2L + 5 = 46 + 5 + 44.5 = L = 45.25 cm
At end B, 2.
Sol.
In case (ii) PA = PB + 5 cos60 or PA – PB = 2.5 cm of Hg PA (44.5) = PB(46) or
46 1 PB = 2.5 44.5
or
2.5 44.5 1.5
46 2.5 44.5 46 = × = 75.4 cm of Hg 45.25 1.5 45.25 A mercury barometer roads 75 cm. If tube be inclined by 600 from the vertical, the length of mercury in the tube will be [Old-DPP_A Batch_05-06_10]
P0(45.25) = PB× 46. Thus
3.
PB =
(A)
75 3 2
(B) 100 cm
P0 = PB
(C) 37.5 cm
(D*) 150 cm
4.
A liquid can easily change its shape but a solid can not because [ HCV I (Ch.13)_Ob.I_Q.1 ] (A) the density of a liquid is smaller than that of a solid (B*) the forces between the molecules is stronger in solid than in liquids (C) the atoms combine to form bigger molecules in a solid (D) the average separation between the molecules is larger in solids.
5.
Consider the equations
6.
F P = Lim s 0 S and P1 – P2 = gz. In a elevator accelerating upward [ HCV I (Ch.13)_Ob.I_Q.2 ] (A) both the equations are valid (B*) the first is valid but not the second (C) the second is valid but not the first (D) both are invalid Three vessels shown in figure have same base area. Equal volumes of a liquid are poured in the three vessels. The force on the base will be [ HCV I (Ch.13)_Ob.I_Q.3 ]
(A) maximum in vessel A (C*) maximum in vessel C
(B) maximum in vessel B (D) equal in all the vessels.
7.
Equal mass of three liquids are kept in three identical cylindrical vessels A, B and C. The densities are A, B, C with A < B < C. The force the base will be [ HCV I (Ch.13)_Ob.I_Q.4 ] (A) maximum in vessel (B) maximum in vessel B (C) maximum in vessel (D*) equal in all the vessels
8.
The pressure in a liquid at two points in the same horizontal plane are equal. Consider an elevator accelerating upward and a car accelerating on a horizontal road. The above statement is correct in [ HCV I (Ch.13)_Ob.I_Q.7 ] (A) the car only (B*) the elevator only (C) both of them (D) neither of them
9.
Suppose the pressure at the surface of mercury in a barometer tube is P1 and the pressure at the surface of mercury in the cup is P2. [ HCV I (Ch.13)_Ob.I_Q.8 ] (A*) P1 = 0, P2 = atmospheric pressure (B) P1 = atmospheric pressure, P2 = 0 (C) P1 = P2 = atmospheric pressure (D) P1 = P2 = 0.
10.
A barometer kept in an elevator reads 76 cm when it is at rest. If the elevator goes up with increasing speed, the reading will be [ HCV I (Ch.13)_Ob.I_Q.9 ] (A) zero (B) 76 cm (C*) < 76 cm (D) > 76 cm
11.
A barometer kept in an elevator accelerating upward reads 76 cm. The air pressure in the elevator is [ HCV I (Ch.13)_Ob.I_Q.10 ] (A) 76 cm (B) < 76 cm (C*) > 76 cm (D) zero
12.
To construct a barometer, a tube of length 1 m is filled completely with mercury and is inverted in a mercury cup. The barometer reading on a particular day is 76 cm. Suppose a 1 m tube is filled with mercury up to 76 cm. It is inverted in a mercury cup. The height of mercury column in the tube over the surface in the cup will be [ HCV I (Ch.13)_Ob.I_Q.11 ] (A) zero (B) 76 cm (C) > 76 cm (D*) < 76 cm [Q.11/HCV-1/Ch-13/Obj-I]
Sol. Due to the trapped air P0 > O Hence by Patm = P0 + egh = 76 cm of Hg h < 76 cm Ans. (D) 13.
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 [ HCV I (Ch.13)_Ob.I_Q.12 ] (A) 36 N (B) 60 N (C*) 44 N (D) 56 N
14.
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 [ HCV I (Ch.13)_Ob.I_Q.13 ] (A) larger part in the water (B) lesser part in the water (C*) same part in the water (D) it will sink. 15.
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 [ HCV I (Ch.13)_Ob.I_Q.14 ] (A) will increase (B) will decrease (C*) will remain the same (D) will become zero
16.
A wooden object floats in water kept in a beaker. The object is near a side of the beaker. Let P1, P2, P3 be the pressures at the three points A, B and C of the bottom as shown in the figure. [ HCV I (Ch.13)_Ob.I_Q.15 ]
(A*) P1 = P2 = P3
(B) P1 < P2 < P3
(C) P1 > P2 > P3
(D) P2 = P3 P1
17.
A closed cubical box is completely filled with water and is accelerated horizontally towards right with an acceleration a. The resultant normal force by the water on the top of the box [ HCV I (Ch.13)_Ob.I_Q.16 ] (A) passes through the centre of the top (B) passes through a point to the right of the centre (C*) passes through a point to the left of the centre (D) becomes zero
18.
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. [ HCV I (Ch.13)_Ob.I_Q.17 ] (A) F1 = F2 (B*) F1 > F2 (C) F1 < F2 (D) The information is insufficient to know the relation between F1 and F2
19.
A solid floats in a liquid in a partially dipped position. [ HCV I (Ch.13)_Ob.II_Q.1 ] (A*) the solid exerts a force equal to its weight on the liquid (B*) the liquid exerts a force of buoyancy on the solid which is equal to the weight of the solid (C*) the weight of the displaced liquid equals the weight of the solid (D) the weight of the dipped part of the solid is equal to the weight of the displaced liquid
20.
The weight of an empty balloon on a spring balance is W 1. The weight becomes W 2 when the balloon is filled with air. Let the weight of the air itself be w. Neglect the thickness of the balloon when it is filled with air. Also neglect the difference in the densities of air inside and outside the balloon. [ HCV I (Ch.13)_Ob.II_Q.2 ] (A*) W 2 = W 1 (B) W 2 = W 1 + w (C*) W 2 < W 1 + w (D) W 2 > W 1
21.
A solid is completely immersed in a liquid. The force exerted by the liquid on the solid will [ HCV I (Ch.13)_Ob.II_Q.3 ] (A) increase if it is pushed deeper inside the liquid (B) change if its orientation is changed (C*) decrease if it is taken partially out of the liquid (D*) be in the vertically upward direction
22.
A closed vessel is half filled with water. There is a hole near the top of the vessel and air is pumped out from this hole [ HCV I (Ch.13)_Ob.II_Q.4 ] (A) the water level will rise up in the vessel (B*) the pressure at the surface of the water will decrease (C*) the force by the water on the bottom of the vessel will decrease (D) the density of the liquid will decrease
23.
The surface of water in a water tank on the top of a house is 4 m above the top level. Find the pressure of water at the tap when the tap is closed. It is necessary to specify that the tap is closed? Take g = 10 m/s2. [ HCV I (Ch.23)_Ex._Q.30 ] 3. 40000 N/m2, Yes
24.
The heights of mercury surfaces in the two arms of the manometer shown in figure are 2 cm and 8 cm. Atmospheric pressure = 1.01 × 105 N/m2. Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U-tube. [ HCV I (Ch.13)_Ex._Q.2 ]
4. 25.
26.
Sol.
(a) 1.09 × 105 N/m2
(b) 1.12 × 105 N/m2
The area of cross-section of the wider tube shown in figure is 900 cm2. If the boy standing on the piston weighs 45 kg, find the difference in the levels of water in the two tubes. [ HCV I (Ch.13)_Ex._Q.3 ]
5. 50 cm Find the force exerted by the water on a 2 m2 plane surface of a large stone placed at the bottom of a sea 500 m deep. Does the force depend on the orientation of the surface? [ HCV I (Ch.13)_Ex._Q.7 ] [Q.7/HCV-1/CH-13/Exercise] [2] P = 1000 × 10 × 500 = 5 × 106 F = PA = (5 × 106) (2) = 107 N. No the force on stone is independent of its orientation. Ans. 107 N, No
27.
An ornanment weighing 36 g in air, weighs only 34 g in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. Specific gravity of gold is 19.3 and that the copper is 8.9. [ HCV I (Ch.13)_Ex._Q.9 ] 7. 2.2 g
28.
Refer to the previous problem. Suppose, the goldsmith argues that he has not mixed copper or any other material with gold, rather some cavities might have been left inside the ornament. Calculate the volume of the cavities left that will allow the weights given in that problem. [ HCV I (Ch.13)_Ex._Q.10 ] 3 8. 0.112 cm
29.
Water flows through a tube shown in figure. The areas of cross-section at A and B are 1 cm2 and 0.5 cm2 respectively. The height difference between A and B is 5 cm. If the speed of water at A is 10 cm/s find (a) the speed at B and (b) the difference in pressures at A and B. [ HCV I (Ch.13)_Ex._Q.31 ]
9. 30.
(b) 485 N/m2
Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes is 2 cm, and the areas of cross-section at A and B are 4 cm2 and 2 cm2 respectively, find the rate of flows of water across any section. [ HCV I (Ch.13)_Ex._Q.32 ]
10. 31.
(a) 20 cm/s,
146 cc/s,
A mercury barometer roads 75 cm. If tube be inclined by 600 from the vertical, the length of mercury in the
tube will be
[ S-05-06/XI(A,B,N1)/DPP-76/Q.10 ]
75 3 (B) 100 cm (C) 37.5 cm (D*) 150 cm 2 An ice cube is floating in water contained in a beaker. When the ice melts, what will happen to the level of water in the beaker ? [ S-05-06/XI(A,B,N1)/DPP-76/Q.11 ] (A) it will fall (B) it will rise (C*) it will remain same (D) first it will fall and then rise to the same level as before. (A)
32.
33.
A steel ball is floating in a trough of mercury. if we fill the empty part of the through with water, what will happen to the steel ball ? [ S-05-06/XI(A,B,N1)/DPP-76/Q.12 ] (A) it will move down (B) it will execute vertical oscillations (C) it will continue in its position (D*) it will move up
34.
A wooden block is floating in a trough of water. If the trough falls freely, the upward thrust on the wooden block will be [ S-05-06/XI(A,B,N1)/DPP-76/Q.13 ] (A) equal to the weight of the block in air (B) same as before (C*) zero (D) more than earlier
35.
By what fraction an ice cube will lost in sea water of density 1.2 gcm–3. Density of ice = 0.9 gcm–3 [ S-05-06/XI(A,B,N1)/DPP-76/Q.14 ] (A) 1/4 (B*) 3/4 (C) 0.9 (D) 2/3
36.
A body of mass 0.5 kg is attached to a thread and it just floats in a liquid. The tension in the thread is [ S-05-06/XI(A,B,N1)/DPP-76/Q.16 ] (A) less than 0.5 kg wt (B) 0.5 kg wt (C) more than 0.5 kg wt (D*) zero
37.
A hose can deliver 0.5 m3 of water per minute from a hole of water that hits the wall drops down vertically, the force on the wall is [ S-05-06/XI(A,B,N1)/DPP-76/Q.17 ] (A) 98 N (B) 196 N (C*) 220 N (D) 284 N
38.
Two holes are made in the side of the tank such that the jets of water flowing out of them meet at the same point on a surface at the level of the bottom, If one hole is at the height h above the bottom, then what will be distance of the other hole from the top ? [ S-05-06/XI(A,B,N1)/DPP-76/Q.18 ] (A) h/4 (B) h/3 (C) h/2 (D*) h
39.
Two circular metal plates of radius 1m and 2m are placed horizontally in a liquid at rest at the same same depth. The ratio of thrusts on them is [ S-05-06/XI(A,B,N1)/DPP-76/Q.19 ] (A) 4 : 1 (B) 1 : 2 (C) 1 : 1 (D*) 1 : 4
40.
A U-tube is partially filled with water. OH which does not mix with water is poured into one side until water rises by 25 cm on the other side. If the density of oil be 0.8, the oil level will stand higher then the water level by [ S-05-06/XI(A,B,N1)/DPP-76/Q.21 ] (A) 31.25 cm (B) 20 cm (C*) 12.50 cm (D) 6.25 cm
41.
A tank is filled with water and an orifice is made in the wall so that the range x of water rushing out is maximum. If H is the height of water above the orifice in the tank, then [ S-05-06/XI(A,B,N1)/DPP-76/Q.15 ] (A) 3x = H (B) 2x = H (C) 4x = H (D*) x = 2H
42.
As a bubble comes from the bottom of a lake to the top its radius should [ S-05-06/XI(A,B,N1)/DPP-76/Q.22 ] (A) zero (B*) increase (C) decrease (D) remain constant
43.
An inverted (bell) lying at the bottom of a lake 47.6 m deep has 50 cm3 of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 gm cm–3. [ S-05-06/XI(A,B,N1)/DPP-76/Q.23 ] 3 3 (A) 22 cm (B) 250 cm (C*) 200 cm3 (D) 350 cm3
44.
A piston of cross-sectional area 100 cm2 is used in hydraulic press of exert a force 107 dynes on the water.
The cross-sectional area of the other pison which supports an object having a mass 2000 kg is [ S-05-06/XI(A,B,N1)/DPP-76/Q.25 ] 10 2 2 4 2 (A) 2×10 cm (B) 100 cm (C*) 2×10 cm (D) 109 cm2 45.
A fisherman hooks an old log of wood of weight 12 N and volume 1000 cm3. He pulls the log half way out of water. The tension in the string at this instant is [ S-05-06/XI(A,B,N1)/DPP-76/Q.26 ] (A*) 7 N (B) 8 N (C) 10 N (D) 12 N
46.
An iceberg is floating partially immersed in sea water. If the density of sea water is 1.03 g cm–3 and that of ice is 0.92 g cm–3. The fraction of the total volume of iceberg above the level of sea water is [ S-05-06/XI(A,B,N1)/DPP-76/Q.27 ] (A) 89% (B) 34% (C*) 11% (D) 8%
47.
A beaker containing water weights 100 g. It is placed on the pan of a balance and a piece of metal weighing 70 g and having volume of 10 cm3 is placed inside the water in the beaker. The weight of the beaker and the metal would be [ S-05-06/XI(A,B,N1)/DPP-76/Q.28 ] (A) 30 g (B) 100 g (C) 160 g (D*) 170 g
48.
A weightless rubber balloon has 100 g of water in it. Its weight in water will be [ S-05-06/XI(A,B,N1)/DPP-76/Q.29 ] (A*) zero (B) 50 g (C) 100 g (D) 200 g
49.
A boat 3m long and 2m wide is floating in a lake. When a man climbs over it, it sinks 1 cm further into water. The mass of the man is [ S-05-06/XI(A,B,N1)/DPP-76/Q.30 ] (A*) 60 kg (B) 64 kg (C) 70 kg (D) 72 kg
50.
A vessel contains oil (density 0.8 gcm–3) over mercury (density 13.6 gcm–3). A homogeneous sphere floats with half in oil. The density of the material of the sphere in gcm–3 is[ S-05-06/XI(A,B,N1)/DPP-76/Q.31 ] (A) 12.8 (B*) 7.2 (C) 6.4 (D) 3.3
51.
The centre of buoyancy of a floating object is [Made AKS 2006 A,B,N] (A) at the centre of gravity of the object. [M.Bank_Fluid_47] (B) at the centre of gravity of the submerged part of the object. (C) at the centre of gravity of the remaining part outside the fluid of the object. (D*) at the centre of gravity of the fluid displaced by the submerged part of the object. Figure shows the points described in the question.
Sol.
C.G. of submerged part of the object will be at the centre of buoyancy if the object is uniform, Otherwise shifts from this point. From figure (D) is correct. 52.
Water coming out of the mouth of a tap and falling vertically in streamline flow forms a tapering column, i.e. the area of cross-section of the liquid column decreases as it moves down. Which of the following is the most accurate explanation for this? MCQ_1.3_46
(A) as the water moves down, its speed increases and hence its pressure decreases. It is then compressed by the atmosphere (B) f alling water tries to reach a terminal v elocity and hence reduces the area of crosssection to balance upward and downward forces (C*) the mass of water flowing past any cross-section must remain constant. Also, water is almost incompressible. Hence, the rate of volume flow must remain constant. As this is equal to velocity area, the area decreases as velocity increases
(D) the surface tension causes the exposed surface area of the liquid to decrease continuously. 53.
A cylindrical drum, open at the top, contains 30 litres of water. It drains out through a small opening at the bottom. 10 litres of water comes out in time t1, the next 10 litres in further time t2 and the last 10 litres in further time t3. Then: MCQ_1.3_48 (A) t1 = t2 = t3 (B) t1 > t2 > t3 (C*) t1 < t2 < t3 (D) t2 > t1 = t3
54.
There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holds is h. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to: MCQ_1.3_49
(A)
h
(B*) h
(C) h3/2
(D) h2
55.
A light cylindrical tube ‘T’ of length and radius ‘r’ containing air is inverted in water (density d1). One end of the tube is open and the other is closed. A block ‘B’ of density d2 (>d1) is kept on the tube as shown in the figure. The tube stays in equilibrium in the position shown. Find the volume of ‘B’. The density of the air is negligible as compared with the density of water. What is the pressure of the air in the tube assuming that the atmospheric pressure is P0. [Made RKV 2005]
Sol.
The free body diagram of block + tube inclusive of water is as shown below Since the block + tube system shown in figure is in equilibrium Net weight of system = buoyant force 2 d2Vg + r2 d g = (V + r2) d1g 3 1 r 2 d1 r 2 d1 where V is the volume of block B (d2 – d1) V = or V= d2 d1 3 3 The pressure of the air trapped inside the tube is same as pressure at point A in the water as shown in figure.
P = P0 + h d1g 3
Ans.
Pa = P0 + d1g h 3
56.
A piano wire weighing 6.00 g and having a length of 90.0 cm emits a fundamental frequency corresponding to the “Middle C” (v = 261.63 Hz). Find the tension in the wire. HCV_15_Ex._37 Ans. 1480 N
57.
A sonometer wire having a length of 1.50 m between the bridges vibrates in its second harmonic in resonance with a tuning fork of frequency 256 Hz. What is the speed of the transverse wave on the wire ? HCV_15_Ex._38 Ans. 384 m/s
58.
The length of the wire shown in figure between the pulleys is 1.5 m and its mass is 12.0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest. HCV_15_Ex._39 Ans. 70 Hz
59.
A water tank stands on the roof of a building as shown. Then the value of h for which the distance covered by the water 'x' is greatest is M.Bank_Fluids_58
(A) 0.5 m Sol.
Vefflux =
(B) 0.67 m
(C*) 1 m
(D) none of these
2gh
time of fall t =
( 4 h)2 g
x = Veffloy t = 2 h( 4 h) the roots of x are (0,4) and the maximum of x is at h = 2. The permitted value of h is 0 to 1 clearly h = 1 will give the maximum value of x is this interval. Aliter : If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum range x. Farther the hole is from this midpoint, lower the range. Here the nearest point possible to this midpoint is the base of the container. Hence h = 1m. 60.
A bent tube is lowered into a water stream as shown in . The velocity of the stream relative to the tube is equal to v = 2.5 m/s. The closed upper end of the tube located at the height h0 = 12 cm has a small orifice. To what height h will the water jet spurt? Irodov_1.320
[Ans : h = 1/2v2g – h0 = 20 cm] 61.
A container of a large uniform cross-sectional area A resting on a horizontal surface holds two immiscible, non-viscous and incompressible liquids of densities ' d ' and ' 2 d ' each of height (1/2)H as shown. The
smaller density liquid is open to atmosphere. A homogeneous solid cylinder of length L 1 H cross-
2
sectional area (1/5) A is immersed such that it floats with its axis vertical to the liquid-liquid interface with length (1/4) L in denser liquid. If D is the density of the solid cylinder then : [ Made 2004 ] [M.Bank_Fluids_1.5]
(A) D =
3d 2
(B) D =
d 2
(C) D =
2d 3
(D*) D =
5d 4
62.
A body of density is dropped from rest from a height 'h' (from the surface of water) into a lake of density of water ( >). Neglecting all dissipative effects, the acceleration of body while it is in the lake is: [ Made 2004] [M.Bank_Fluids_1.3]
(C) g upwards
(D) g downwards
(A*) g 1 upwards
63.
In the above problem, the maximum depth the body sinks before returning is: [ Made 2004] [M.Bank_Fluids_1.4] (A*)
64.
Sol.
(B) g 1 downwards
h
(B)
h
(C) h
(D) h
A cylindrical vessel filled with water is released on a fixed inclined surface of angle as shown in figure. The friction coefficient of surface with vessel is ( < tan ). Then the constant angle made by the surface of water with the incline will be: (Neglect the viscosity of liquid) [ Made 2004] [7, Fluid] [M.Bank_Fluid_1.7]
(A*) tan1 (B) tan1 (C) + tan1 (D) cot1 figure shows forces acting on a 'particle' on the surface, with respect to vessel.
(mg sin & mg cos are pseudo forces). tan = = tan–1 . is angle between normal to the inclined surface and the resultant force. The same angle will be formed between the surface of water & the inclined surface. { free surface is to the resultant force acting on it.} 65.
The velocity of the liquid coming out of a small hole of a large vessel containing two different liquids of densities 2and as shown in figureis [M.Bank(07-08)_Fluid_1.33]
(A) 6gh Sol.
(B*) 2 gh
(D) gh
(B) Pressure at (1) : P1 = Patm + g (2h) Applying Bernoulli's theorum between points (1) and (2) [Patm + 2 g h] + g(2h) + = Patm + (2 ) g (0) +
66.
(C) 2 2gh
v = 2 gh
1 (2 ) (0)2 2
1 (2 ) v 2 2
Ans.
For a fluid which is flowing steadily, the level in the vertical tubes is best represented by [Made RA 2006] [M_Bank(07-08)_Fluids_1.36]
(A*)
(B)
(C)
(D)
Sol.
(A) From continuity equation, velocity at cross-section (1) is more than that at cross-section (2). Hence ; P1 < P2 Hence (A)
67.
A cylindrical wooden float whose base area S and the height H drifts on the water surface. Density of wood d and density of water is . What minimum work must be performed to take the float out of the water? [POM_Module_17_Ex.1] [4] [M_Bank(07-08)_Fluids_1.37] Applying Newton’s law vertical direction
Sol.
mg = F B
dSH × g = wSh × g
Now when force F is applied, for minimum work a = 0 F – mg + Sxg = 0 F = mg – Sxg W=
h=
dH
(for a = 0, F is minimum)
Fdx
(mg Sxg)dx = mg dx Sg xdx =
Sg Sgh 2 Sgh 2 Sgh 2 W = mgh – = (Sh) gh – = = 2 2 2 2
68.
A non uniform cylinder of mass m, length and radius r is having its cetnre of mass at a distance /4 from the centre and lying on the axis of the cylinder. The cylinder is kept in a liquid of uniform density The moment of inertia of the rod about the centre of mass is . The angular acceleration of point A relative to point B just after the rod is released from the position shown in figure is [M_Bank(07-08)_Fluids_1.34]
g 2r 2 g 2r 2 (B*) 4 (B) Torque about CM : Fb . = 4 (A)
Sol.
2
2 2 dH = Sgd H Ans. 2ρ
(C)
g 2r 2 2
(D)
3g 2r 2 4
a=
1 ( r2) () () (g). 4 I
r 2 2g = 4I '' will be same for all points. Hence (B). 69.
A tube in vertical plane is shown in figure. It is filled with a liquid of density and its end B is closedThen the force exerted by the fluid on the tube at end B will be : [Neglect atmospheric pressure and assume the radius of the tube to be negligible in comparison to ]
[ Made 2004]
Sol.
70.
[M.Bank_Fluids_1.12]
(A) 0 (B*) g A0 (C) 2g A0 (D) 4g A0 Pressure exerted by fluid at closed end B is P = g force exerted by fluid at closed end B is F = PA = g A0 Three containers of same base area, same height are filled with three different liquids of same mass as shown in the figure. If F 1, F 2, F 3 are the force exerted by the liquid on the base of the container in case I, II and II respectively, then we have the relation: [Made 2004] [M.Bank_Fluids_1.19]
(A) F 1 = F 2 = F 3
(B) F 1 > F 2 > F 3
(C) F 3 > F 2 > F 1
(D*) F 2 < F 3 < F 1
71.
A light cylindrical tube ‘T’ of length and radius ‘r’ containing air is inverted in water (density d1). One end of the tube is open and the other is closed. A block ‘B’ of density d2 (>d1) is kept on the tube as shown in the figure. The tube stays in equilibrium in the position shown. Find the volume of ‘B’. The density of the air is negligible as compared with the density of water. What is the pressure of the air in the tube assuming that the atmospheric pressure is P0. [Made RKV 2005] [24, Fluid] [M.Bank_Fluids_1.24]
Sol.
The free body diagram of block + tube inclusive of water is as shown below
Since the block + tube system shown in figure is in equilibrium Net weight of system = buoyant force d2Vg + r2
2 d g = (V + r2) d1g 3 1
where V is the volume of block B
(d2 – d1) V =
r 2 d1 3
or
d1 r 2 V = d d 3 2 1
The pressure of the air trapped inside the tube is same as pressure at point A in the water as shown in figure.
P = P0 + h d1g 3
Pa = P0 + d1g h = 3 In the figure shown water is filled in a symmetrical container. Four pistons of equal area A are used at the four opening to keep the water in equilibrium. Now an additional force F is applied at each piston. The increase in the pressure at the centre of the container due to this addition is [M.Bank_Fluids_1.25]
Ans. 72.
[Made 2005, RKV, BKM, AS]
(A*) Sol.
F A
(B)
2F A
(C)
4F A
[BM_Fluids_45] The four piston are initially in equilibrium. As additional force F is applied to each piston, the pressure in fluid at each point must be increased by
F so that each piston retains state of equilibrium. A
Thus the increment in pressure at each point is P = 73.
(D) 0
F (by Pascal’s law) A
A U-tube of base length “” filled with same volume of two liquids of densities and 2 is moving with an acceleration “a” on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by : [M.Bank_Fluids_1.32]
(A)
a 2g
(B*)
3a 2g
a (C) g
(D)
2a 3g
Sol.
(B) For the given situation, liquid of density 2 should be behind that of . From right limb : PA = Patm + gh PB = PA + a
= Patm + gh + a 2 2
3 = Patm + gh + a 2 2 But from left limb : PC = Patm + (2) gh From (1) and (2) :
PC = PB + (2) a
Patm + gh + 74.
3 a = Patm + 2 gh 2
.... (1) .... (2)
h =
3a 2g
Ans.
A narrow tube completely filled with a liquid is lying on a series of cylinders as shown in figure. Assuming no sliding between any surfaces, the value of acceleration of the cylinders for which liquid will not come out of the tube from anywhere is given by[Made RA_2006_GRSTX] [M.Bank_Fluids_1.43]
gH gH gH 2gH (B) (C) (D) 2L 2L L L (A) No sliding pure rolling Therefore, acceleration of the tube = 2a (since COM of cylinders are moving at 'a')
(A*) Sol.
PA = Patm + (2a) L Also ; PA = Patm + g H 75.
gH a = 2L
Ans.
A small uniform tube is bent into a circular tube of radius R and kept in the vertical plane. Equal volumes of two liquids of densities and (> ) fill half of the tube as shown. is the angle which the radius passing through the interface makes with the vertical. [Made 2004] [M.Bank_Fluids_1.23]
(A*) = tan–1 Sol.
(From horizontal limb) (From vertical limb)
(B) = tan–1
(C) = tan–1
(D) = tan–1
(A) Pressure at 'A' from both side must balance. Figure is self–explanatory.
O Rsin Rcos
(/2–
/2 h2
A h2g = h1g sin (45º + ) = R [cos – sin] [cos + sin] = [cos – sin] tan =
76.
A cube (density 0.5 gm/cc) of side 10 cm is floating in water kept in a cylindrical beaker of base area 1500 cm2. When a mass m is kept on wooden block the level of water rises in the beaker by 2mm. Find the mass m. [Made-2006, P1-P6,PKS] [4] [M.Bank_Fluids_1.29]
Sol.
Let the cube dips further by y cm and water level rises by 2 mm.
Then equating the volumes (/// volume = \\\ volume in figure) volume of water raised = volume of extra depth of wood
100 y = (1500 – 100)
y = 2.8 cm
Extra upthrust
2 2 = 1400 × = 280 10 10
water × (2.8 + 0.2) × 100 g = mg m = 300 gm. 77.
Sol.
78.
m = 300 gm.
....Ans.
A block is partially immersed in a liquid and the vessel is accelerating upwards with an acceleration “a”. The block is observed by two observers O 1 and O 2 , one at rest and the other accelerating with an acceleration “a” upward. The total buoyant force on the block is : [M.Bank_Fluids_1.35]
(A*) same for O 1 and O 2 (B) greater for O 1 than O 2 (C) greater for O 2 than O 1 (D) data is not sufficient Buoyant force = Fb = Vsub . . g where, Vsub , and g all are same w.r.t. O 1 and O 2. Hence (A) A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole of side 'L' at a depth '4y' from the top and the other is a circular hole of radius 'R' at a depth ‘y’ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, 'R' is equal to: [ Made 2004 ] [BM_Fluids_25] [M.Bank(07-08)_Fulids_1.6]
L
(A) Sol.
(B) 2L
2
(C*)
2 .L
(D)
L 2
Let v 1 and v 2 be the velocity of efflux from square and circular hole respectively. S1 and S2 be cross-section areas of square and circular holes.
v1 =
8gy
and
v2 =
2g( y )
The volume of water coming out of square and circular hole per second is Q1 = v 1S1 =
R=
8gy L2
; Q2 = v 2S2 =
2gy R2
Q1 = Q2
2 .L
Comprehension # (4 - 8) (Read the following passage and answer the questions numbered 1 to 5. They have only one correct option)Heat Transfer By Radiation [M_Bank_Fluid_3.1] The figure shows the commonly observed decrease in diameter of a water stream as it falls from a tap. The tap has internal diameter D0 and is connected to a large tank of water. The surface of the water is at a height b above the end of the tap. By considering the dynamics of a thin “cylinder” of water in the stream answer the following: (Ignore any resistance to the flow and any effects of surface tension, given w = density of water)
79.
Sol.
Equation for the flow rate, i.e. the mass of water flowing through a given point in the stream per unit time, as function of the water speed v will be (A*) v w D2 / 4 (B) v w D2 / 4 – D02 /4) 2 (C) v w D / 2 (D) v w D02 / 4 (A) As, dm = W v dt
dm = AW v dt
dm D2 = VW dt 4 where ‘D’ is the diameter of stream.
80.
Which of the following equation expresses the fact that the flow rate at the tap is the same as at the stream point with diameter D and velocity v (i.e. D in terms of D 0 and v will be) : (A) D =
D0 v 0 v
(B) D =
D 0 v 02 v
2
D0v (C) D = v 0
(D*) D = D0
v0 v
Sol.
(D) V1A1 = V2A2
81.
2 v v D2 v 0D0 = D = D0 0 . v 4 4 The equation for the water speed v as a function of the distance x below the tap will be :
(A) v =
2gb
(B*) v = [2g (b + x)] 1/2
(C) v =
2gx
(D) v = [2g (b – x)] 1/2
Sol.
v=
2gh 2g(b x ) .
82.
Equation for the stream diameter D in terms of x and D 0 will be : (A*) D = D0 (b/b + x)1/4 (B) D = D0 (b/b + x)1/2 (C) D = D0 (b/b + x) (D) D = D0 (b/b + x)2
Sol.
(A) Applying continuity equation at points with diameter D 0 & D : =
83.
Sol.
84.
.D 0 2 2gb . = 4
b D = D0 b x
D 2 2g(b x ) 4
1/ 4
A student observes after setting up this experiment that for a tap with D 0 = 1 cm at x = 0.3 m the stream diameter D = 0.9 cm. The heights b of the water above the tap in this case will be : (A) 5.7 cm (B*) 57 cm (C) 27 cm (D) 2.7 cm (B) Solving the preceding formula for the tank height h gives : h = x(D/D0)4/(1 – (D/D0)4) = x D4 / (D04 – D4) substituting the given parameter values gives h = (0.3 ) (0.0094) / (0.014 – 0.0094) = 0.57 m So the height of the water above the tap is 0.57 m or 57 cm. A portion of a tube is shown in the figure. Fluid is flowing from cross-section area A 1 to A2. The two cross-sections are at distance ' l ' from each other. The velocity of the fluid at section A 2 is
g . If the 2
pressures at A1 & A2 are same, then the angle made by the tube with the horizontal will be :
[M.Bank_Fluids_1.21]
(A) 37º
85.
(B*) sin1
3 4
(C) 53º
(D) none of these
The cubical container ABCDEFGH which is completely filled with an ideal (nonviscous and incompressible) fluid, moves in a gravity free space with a acceleration of [M.Bank_Fluids_1.42]
a = a0 (ˆi ˆj kˆ ) where a0 is a positive constant. Then the only point in the container where pressure can be zero is
[Made VSS 2006, GRSTX]
(A*) H Sol.
a = a0 ( ˆi – ˆj + kˆ )
(B) C
(C) E
(D) F
As there is no gravity; the pressure difference will be only due to the acceleration. All points other than point 'B', are acted upon by a pseudo force. Hence, at point 'H' pressure developed is zero. 86.
A fixed cylindrical tank having large cross-section area is filled with two liquids of densities and 2 and in equal volumes as shown in the figure. A small hole of area of crosssection ' a ' is made at height h/2 from the bottom. Calculate : [M.Bank_Fluids_1.16]
(i) (ii)
Sol.
[Made 2003]
2 a ] 3 A cylindrical vessel filled with water is released on a fixed inclined surface of angle as shown in figure. The friction coefficient of surface with vessel is ( < tan ). Then the constant angle made by the surface of water with the incline will be: (Neglect the viscosity of liquid) [ Made 2004] [7, Fluid] [M.Bank_Fluid_1.7]
[Ans. 87.
Velocity of efflux initially. The area of cross-section of stream of liquid just before it hits the ground. (i) v =
2gh ,
(ii)
(A*) tan1 (B) tan1 (C) + tan1 (D) cot1 figure shows forces acting on a 'particle' on the surface, with respect to vessel.
(mg sin & mg cos are pseudo forces). tan = = tan–1 . is angle between normal to the inclined surface and the resultant force. The same angle will be formed between the surface of water & the inclined surface. { free surface is to the resultant force acting on it.} 88.
A water tank stands on the roof of a building as shown. Then the value of h for which the distance covered by the water 'x' is greatest is M.Bank_Fluids_58 [S-(06-07)_XIII (GRSTU)_DPP-72_4]
(A) 0.5 m Sol.
Vefflux =
(B) 0.67 m
2gh
time of fall t =
( 4 h)2 g
(C*) 1 m
(D) none of these
x = Veffloy t = 2 h( 4 h) the roots of x are (0,4) and the maximum of x is at h = 2. The permitted value of h is 0 to 1 clearly h = 1 will give the maximum value of x is this interval. Aliter : If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum range x. Farther the hole is from this midpoint, lower the range. Here the nearest point possible to this midpoint is the base of the container. Hence h = 1m. 89.
A light cylindrical tube ‘T’ of length and radius ‘r’ containing air is inverted in water (density d1). One end of the tube is open and the other is closed. A block ‘B’ of density d2 (>d1) is kept on the tube as shown in the figure. The tube stays in equilibrium in the position shown. Find the volume of ‘B’. The density of the air is negligible as compared with the density of water. What is the pressure of the air in the tube assuming that the atmospheric pressure is P0. [Made RKV 2005] [24, Fluid] [S-(06-07)_XIII (GRSTU)_DPP-68_11]
Sol.
The free body diagram of block + tube inclusive of water is as shown below
Since the block + tube system shown in figure is in equilibrium Net weight of system = buoyant force d2Vg + r2
2 d g = (V + r2) d1g 3 1
d1 r 2 d1 r 2 or V = d d 3 3 2 1 The pressure of the air trapped inside the tube is same as pressure at point A in the water as shown in figure.
where V is the volume of block B
90.
Sol.
P = P0 + h d1g 3
(d2 – d1) V =
Ans.
Pa = P0 + d1g h 3
A block is partially immersed in a liquid and the vessel is accelerating upwards with an acceleration “a”. The block is observed by two observers O 1 and O 2 , one at rest and the other accelerating with an acceleration “a” upward. The total buoyant force on the block is : [M.Bank_Fluids_1.35]
(A*) same for O 1 and O 2 (B) greater for O 1 than O 2 (C) greater for O 2 than O 1 (D) data is not sufficient Buoyant force = Fb = Vsub . . g where, Vsub , and g all are same w.r.t. O 1 and O 2. Hence (A)
91.
Sol.
A cube (density 0.5 gm/cc) of side 10 cm is floating in water kept in a cylindrical beaker of base area 1500 cm2. When a mass m is kept on wooden block the level of water rises in the beaker by 2mm. Find the mass m. [Made-2006, P1-P6,PKS] [4] [M.Bank_Fluids_1.29] Let the cube dips further by y cm and water level rises by 2 mm.
Then equating the volumes (/// volume = \\\ volume in figure) volume of water raised = volume of extra depth of wood 2 2 = 1400 × = 280 10 10
100 y = (1500 – 100)
y = 2.8 cm Extra upthrust water × (2.8 + 0.2) × 100 g = mg
m = 300 gm.
m = 300 gm. .... Ans. Comprehension # (3 - 7) (Read the following passage and answer the questions numbered 1 to 5. They have only one correct option)Heat Transfer By Radiation [M_Bank_Fluid_3.1] The figure shows the commonly observed decrease in diameter of a water stream as it falls from a tap. The tap has internal diameter D0 and is connected to a large tank of water. The surface of the water is at a height b above the end of the tap. By considering the dynamics of a thin “cylinder” of water in the stream answer the following: (Ignore any resistance to the flow and any effects of surface tension, given w = density of water)
92.
Sol.
Equation for the flow rate, i.e. the mass of water flowing through a given point in the stream per unit time, as function of the water speed v will be (A*) v w D2 / 4 (B) v w D2 / 4 – D02 /4) 2 (C) v w D / 2 (D) v w D02 / 4 (A) As, dm = A W v dt
dm = AW v dt
dm D2 = VW dt 4 where ‘D’ is the diameter of stream.
93.
Which of the following equation expresses the fact that the flow rate at the tap is the same as at the stream point with diameter D and velocity v (i.e. D in terms of D 0 and v will be) : D0 v 0 v (D) V1A1 = V2A2
(A) D = Sol.
v 0D0 4
94.
(B) D =
2
=
v D2 4
D 0 v 02 v
2
(C) D =
D0v v0
D = D0
v0 . v
(D*) D = D0
v0 v
The equation for the water speed v as a function of the distance x below the tap will be :
(A) v =
(B*) v = [2g (b + x)] 1/2
2gb
(C) v =
2gx
Sol.
v=
95.
Equation for the stream diameter D in terms of x and D 0 will be : (A*) D = D0 (b/b + x)1/4 (B) D = D0 (b/b + x)1/2 (C) D = D0 (b/b + x) (D) D = D0 (b/b + x)2
Sol.
(A) Applying continuity equation at points with diameter D 0 & D : =
96.
Sol.
(D) v = [2g (b – x)] 1/2
2gh 2g(b x ) .
.D 0 2 2gb . = 4
b D = D0 b x
D 2 2g(b x ) 4
1/ 4
A student observes after setting up this experiment that for a tap with D 0 = 1 cm at x = 0.3 m the stream diameter D = 0.9 cm. The heights b of the water above the tap in this case will be : (A) 5.7 cm (B*) 57 cm (C) 27 cm (D) 2.7 cm (B) Solving the preceding formula for the tank height h gives : h = x(D/D0)4/(1 – (D/D0)4) = x D4 / (D04 – D4) substituting the given parameter values gives h = (0.3 ) (0.0094) / (0.014 – 0.0094) = 0.57 m So the height of the water above the tap is 0.57 m or 57 cm.
97.
Sol.
An incompressible liquid flows through a horizontal tube as shown in the figure. Then the velocity ' v ' of the fluid is : modified_Fluid_39]
(A) 3.0 m/s (B) 1.5 m/s from equation of continuity, (A × 3) = (A × 1.5) + (1.5 A × V)
(C*) 1.0 m/s
(D) 2.25 m/s
V = 1 m/s2
READ THE SOLVED EXAMPLE AND SOLVE THE UNSOLVED PROBLEM. Example : Water is filled in a flask upto a height of 20 cm. The bottom of the flask is circular with radius 10 cm. If the atmospheric pressure is 1.01 × 105 Pa, find the force exerted by the water on the bottom. Take g = 10 m/s2 and density of water = 1000 kg/m3. HCV_Ch-13_Sol.Ex._1 [Fluids] Sol.
The pressure at the surface of the water is equal to the atmospheric pressure P0. The pressure at the bottom is P = P0 + hg = 1.01 × 105 Pa + (0.20 m) (1000 kg/m3) (10 m/s2) = 1.01 × 105 Pa + 0.02 × 105 Pa = 1.03 × 105 Pa. The area of the bottom = r2 = 3.14 × (0.1 m)2 = 0.0314 m2. The force on the bottom is, therefore, F = P r2 5 = (1.03 × 10 Pa) × (0.0314 m2) = 3230 N.
98.
Water is fully filled in a closed rectangular tank of size 3 m × 2m × 1m. (a) Find the total force exerted by the water on the bottom surface of the tank. (b) Consider a vertical side of area 2 m × 1 m. Take a horizontal strip of width x metre in this side, stiutated at a depth of x metre from the surface of water. Find the force by the water on this strip. (c) Find the torque of the force calculated in part (b) about the bottom edge of this side. (d) Find the total force by the water on this side. (e) Find the total torque by the water on the side about the bottom edge. Neglect the atmospheric pressure and take g = 10 m/s2. HCV_Ch-13_Ex._8 Ans. (a) 60000 N, (b) 20000 × x N (c) 20000 × (1 – x)x N-m (d) 10000 N, (e) 10000/3N-m
READ THE SOLVED EXAMPLE AND SOLVE THE UNSOLVED PROBLEM. Example : A beaker of circular cross-section of radius 4 cm is filled with mercury upto a height of 10 cm. Find the force exerted by the mercury on the bottom of the beaker. The atmospheric pressure = 105 N/m2. Density of mercury = 13600 kg/m3 . Take g = 10 m/s2. HCV_Ch-13_WOE_1 Sol. The pressure at the surface = atmospheric pressure = 105 N/m2. The pressure at the bottom = 105 N/m2 + hg = 105 N/m2 + hg kg m = 105 N/m2 + (0.1 m) 13600 3 10 2 m s 5 2 2 = 10 N/m + 13600 N/m = 1.136 × 105 N/m2. The force exerted by the mercury on the bottom = (1.136 × 105 N/m2) × (3.14 × 0.04 m × 0.04 m) = 571 N.
99.
A glass full of water has a bottom of area 20 cm2, top of area 20 cm2, height 20 cm and volume half a litre.
HCV_Ch-13_Ex._4
(a) find the force exerted by the water on the bottom. (b) Considering the equilibrium of the water, find the resultant force exerted by the sides of the glass on the water. Atmospheric pressure = 1.0 × 105 N/m2. Density of water 1000 kg/m3 and g = 10 m/s2. Take all numbers to be exact. Ans.
(a) 204 N
(b) 1 N upward
READ THE SOLVED EXAMPLE AND SOLVE THE UNSOLVED PROBLEM. Example : The liquid shown in figure in the two arms are mercury (specific gravity = 13.6) and water. If the difference of heights of the mercury columns is 2 cm, find the height h of the water column. HCV_Ch-13_WOE_3
h 2cm A
Sol.
B
Suppose the atmospheric pressure = P0 kg Pressure at A = P0 + h 100 3 m
g.
kg Pressure at B = P0 + (0.02 m) 13600 3 m
g.
These pressures are equal as A and B are at the same horizontal level. Thus, h = (0.02 m) 13.6 = 0.27 m = 27 cm. 100.
The heights of mercury surfaces in the two arms of the manometer shown in figure are 2 cm and 8 cm. Atmospheric pressure = 1.01 × 105 N/m2. Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U-tube. (Specific gravity of mercury = 13.6)
Gas HCV_Ch-13_Ex_2
Ans.
(a) 1.09 × 105 N/m2
(b) 1.12 × 105 N/m2
READ THE SOLVED EXAMPLE AND SOLVE THE UNSOLVED PROBLEM. Example : A 700 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water ? Density of water = 1000 kg/m3. HCV_Ch-13_Sol.Ex._2 Sol. The weight of the cube is balanced by the buoyant force. The buoyant force is equal to the weight of the water displaced. If a volume V of the cube is inside the water, the weight of the displaced water = Vg, where is the density of water. Thus, Vg = (0.7 kg) g V=
0.7 kg 0.7 kg = = 7 × 10–4 m3 = 700 cm3. 1000 kg / m3
The total volume of the cube = (10 cm)3 = 1000 cm3. The volume outside the water is 1000 cm3 – 700 cm3 = 300 cm3. 101.
A cubical block of ice floating in water has to support a metal piece weighing 0.5 kg. What can be the minimum edge of the block so that it does not sink in water? Specific gravity of ice = 0.9. HCV_Ch-13_Ex._13
Ans. 102.
17 cm
A cubical box is to be constructed with iron sheets 1 mm in thickness. What can be the minimum value of the external edge so that the cube does not sink in water? Density of iron = 8000 kg/m3 and density of water = 1000 kg/m3. HCV_Ch-13_Ex_15 Ans. 4.8 cm
READ THE SOLVED EXAMPLE AND SOLVE THE UNSOLVED PROBLEM. Example : Ex. The area of cross-section of the two arms of a hydraulic press are 1 cm2 and 10 cm2 respectively (figure). A force of 5 N is applied on the water in the thinner arm. What force should be applied on the water in the thicker arms so that the water may remain in equilibrium ? HCV_Ch-13_WOE_5
Sol.
In equilibrium, the pressures at the two surfaces should be equal as they lie in the same horizontal level. If the atmospheric pressure is P and a force F is applied to maintain the equilibrium, the pressures are This givens F = 50 N.
103.
The area of cross-section of the wider tube shown in figure is 900 cm2. If the boy standing on the light piston weighs 45 kg, find the difference in the levels of water in the two tubes.
Ans. 104.
Sol.
50 cm
A beaker of circular cross-section of radius 4 cm is filled with mercury upto a height of 10 cm. Find the force exerted by the mercury on the bottom of the beaker. The atmospheric pressure = 105 N/m2. Density of mercury = 13600 kg/m3 . Take g = 10 m/s2. HCV_Ch-13_WOE_1 The pressure at the surface = atmospheric pressure = 105 N/m2. The pressure at the bottom = 105 N/m2 + hg = 105 N/m2 + hg = 105 N/m2 + (0.1 m) = 105 N/m2 + 13600 N/m2 = 1.136 × 105 N/m2. The force exerted by the mercury on the bottom = (1.136 × 105 N/m2) × (3.14 × 0.04 m × 0.04 m) = 571 N.
105.
A cubical block of wood weighing 200 g has a lead piece fastened underneath. Find the mass of the lead piece which will just allow the block to float in water. Specific gravity of wood is 0.8 and that of lead is 11.3. HCV_Ch-13_Ex._16
Ans.
54.8 g
106.
A cubical metal block of edge 12 cm floats in mercury with one fifth of the height insde the mercury. Water is poured till the surface of the block is just immersed in it. Find the height of the water column to be poured. Specific gravity of mercury = 13.6. HCV_Ch-13_Ex_18 Ans. 10.4 cm
107.
A hollow spherical body of inner and outer radii 6 cm and 8 cm respectively floats half submerged in water. Find the density of the material of the sphere. (dwater = 1000 kg/m3) HCV_Ch-13_Ex_19 Ans. 865 kg/m3
108.
A body of density is dropped from rest from a height 'h' (from the surface of water) into a lake of density of water ( >). Neglecting all dissipative effects, the acceleration of body while it is in the lake is: [ Made 2004] M.Bank_Fluids_1.3
(C) g upwards
(A*) g 1 upwards
109.
(D) g downwards
(B) g 1 downwards
A tube in vertical plane is shown in figure. It is filled with a liquid of density and its end B is closedThen the force exerted by the fluid on the tube at end B will be : [Neglect atmospheric pressure and assume the radius of the tube to be negligible in comparison to ]
[ Made 2004] [BM_Fluids_1.12]
[M.Bank_Fluids_1.12] (B*) g A0 (D) Cannot be determined
(A) 0 (C) 2g A0 Sol.
Pressure exerted by fluid at closed end B is P = g
force exerted by fluid at closed end B is F = PA = g A0
110.
A block of silver of mass 4 kg hanging from a string is immersed in a liquid of relative density 0.72. If relative density of silver is 10, then tension in the string will be:[ take g = 10 m/s2 ] [ Made 2004] [M.Bank_Fluid_11] (A*) 37.12 N (B) 42 N (C) 73 N (D) 21 N
Sol.
Let S, L be the density of silver and liquid. Also m and V be the mass and volume of silver block. Tension in string = mg – bouyant force T = SVg – L Vg = (S – L) Vg Also
m V = s
3 S s mg = (10 0.72) 10 × 4 × 10 T = s 10 10 3
= 37.12 N. 111.
Sol.
A cylindrical vessel filled with water is released on an inclined surface of angle as shown in figure. The friction coefficient of surface with vessel is ( < tan ). Then the constant angle made by the surface of water with the incline will be: [ Made 2004] [7, Fluid]
(A*) tan1 (B) tan1 (C) + tan1 (D) cot1 figure shows forces acting on a 'particle' on the surface, with respect to vessel.
(mg sin & mg cos are pseudo forces). tan = = tan–1 . is angle between normal to the inclined surface and the resultant force. The same angle will be formed between the surface of water & the inclined surface. { free surface is to the resultant force acting on it.} 112.
A light cylindrical tube ‘T’ of length and radius ‘r’ containing air is inverted in water (density d1). One end of the tube is open and the other is closed. A block ‘B’ of density d2 (>d1) is kept on the tube as shown in the figure. The tube stays in equilibrium in the position shown. Find the volume of ‘B’. The density of the air is negligible as compared with the density of water. What is the pressure of the air in the tube assuming that the atmospheric pressure is P0. [Made RKV 2005] [24, Fluid]
Sol.
The free body diagram of block + tube inclusive of water is as shown below
Since the block + tube system shown in figure is in equilibrium Net weight of system = buoyant force d2Vg + r2
2 d g = (V + r2) d1g 3 1
d1 r 2 d1 r 2 or V = d d 3 3 2 1 The pressure of the air trapped inside the tube is same as pressure at point A in the water as shown in figure.
where V is the volume of block B
P = P0 + h d1g 3
113.
(d2 – d1) V =
Ans.
Pa = P0 + d1g h 3
A large open tank has two small holes in the wall. One is a square hole of side 'L' at a depth '4y' from the top and the other is a circular hole of radius 'R' at a depth ‘y’ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, 'R' is equal to: [Q.6/DPP-42/P1-P6] [ Made 2004 ] M.Bank_Fluids_6 L
(A) Sol.
(B) 2L
2
(C*)
2 .L
(D)
L 2
Let v 1 and v 2 be the velocity of efflux from square and circular hole respectively. S1 and S2 be cross-section areas of square and circular holes. v1 =
8gy
and
v2 =
2g( y )
The volume of water coming out of square and circular hole per second is Q1 = v 1S1 = 114.
R=
8gy L2
; Q2 = v 2S2 =
2gy R2
Q1 = Q2
2 .L
The velocity of the liquid coming out of a small hole of a vessel containing two different liquids of densities 2and as shown in figureis M.Bank_Fluids_33
(A) 6gh Sol.
(B*) 2 gh
(C) 2 2gh
(D) gh
Pressure at (1) : P1 = Patm + g (2h) Applying Bernoulli's theorum between points (1) and (2) [Patm + 2 g h] + g(2h) + = Patm + (2 ) g (0) +
1 (2 ) (0)2 2
1 (2 ) v 2 2
v = 2 gh
Ans.
115.
Fountains usually seen in gardens are generated by a wide pipe with an enclosure at one end having many small holes. Consider one such fountain which is produced by a pipe of internal diameter 2 cm in which water flows at a rate 3ms1. The enclosure has 100 holes each of diameter 0.05 cm. The velocity of water coming out of the holes is (in ms1) M.Bank_Fluids_74 (A) 0.48 (B) 96 (C) 24 (D*) 48
116.
There is a small hole in the bottom of a fixed container containing a liquid upto height ‘h’. The top of the liquid as well as the hole at the bottom are exposed to atmosphere. As the liquid comes out of the hole. (Area of the hole is ‘a’ and that of the top surface is ‘A’) : M.Bank_Fluids_51 [Made 2006, RKV, GRSTX] (A) the top surface of the liquid accelerates with acceleration = g (B) the top surface of the liquid accelerates with acceleration = g (C) the top surface of the liquid retards with retardation = g
(D*) the top surface of the liquid retards with retardation =
a2 A2
a A
ga 2 A2
2gh
Sol.
The velocity of fluid at the hole is :
V2 =
1 (a 2 / A 2 )
Using continuity equation at the two cross-sections (1) and (2) : V1 A = V2 a
V1 =
a V A 2
acceleration (of top surface) = – V1 a2 dV2 a1 = – 2 V2 =– 2 A dh A a2
117.
=–
a d a V2 V2 A dh A
1
a1 =
2 h
ga2 A2
A water tank stands on the roof of a building as shown. Then the value of h for which the distance covered by the water 'x' is greatest is M.Bank_Fluids_58
(A) 0.5 m Sol.
2 gh 2 g.
dV1 dh
Vefflux =
(B) 0.67 m
2gh
time of fall t =
( 4 h)2 g
(C*) 1 m
(D) none of these
x = Veffloy t = 2 h( 4 h) the roots of x are (0,4) and the maximum of x is at h = 2. The permitted value of h is 0 to 1 clearly h = 1 will give the maximum value of x is this interval. Aliter : If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum range x. Farther the hole is from this midpoint, lower the range. Here the nearest point possible to this midpoint is the base of the container. Hence h = 1m. 118.
Fountains usually seen in gardens are generated by a wide pipe with an enclosure at one end having many small holes. Consider one such fountain which is produced by a pipe of internal diameter 2 cm in which water flows at a rate 3ms1. The enclosure has 100 holes each of diameter 0.05 cm. The velocity of water coming out of the holes is (in ms1) [ 74_Fluids ] (A) 0.48 (B) 96 (C) 24 (D*) 48
119.
Two very large open tanks A & F both contain the same liquid. A horizontal pipe BCD, having a small constriction at C, leads out of the bottom of tank A, and a vertical pipe E containing air opens into the constriction at C and dips into the liquid in tank F. Assume streamline flow and no viscosity. If the cross section area at C is one-half that at D, and if D is at distance h1 below the level of the liquid in A, to what height h2 will liquid rise in pipe E? Express your answer in terms of h 1. [Neglect changes in atmospheric pressure with elevation. In the containers there is atmosphere above the water surface and D is also open to atmosphere.] M.Bank_Fluid_70 [Q.12/P1-P6/DPP-43] [Q.4/RK_Topic/Fluid]
[4]
Sol.
Apply Bernolli’s equation b/w point P & D 1 1 Vp 2 = Patm + g(o) + v D2 2 2 [Assume zero level D]
Patm + gh1 +
Vp 0 Area of cross section is very large VD =
2gh1
Since area at C is half than area at D so according to continuity equation V CA C = V DA D Now
VC = 2VD = 2 2gh1
for point P & C according to bernolli’s equation. Pp + gh1 +
1 Vp2 2
= PC + g(0) +
1 .VC2 2
Vp 0
Patm + gh1 = PC +
Patm = PC + 3gh1
for point E
1 .(2 2gh1 )2 2
.........(i)
PE = PC + gh2 = Patm
.........(ii)
from (i) & (ii) 3gh1 = gh2 h 2 = 3h 1
Ans.
120.
A fixed container of height ' H ' with large cross-sectional area ' A ' is completely filled with water. Two small orifice of cross-sectional area ' a ' are made, one at the bottom and the other on the vertical side of the container at a distance H/2 from the top of the container. Find the time taken by the water level to reach a height of H/2 from the bottom of the container. [ Made 2003] [Q.20/RK_BM/Fluid] M.Bank_Fluids._1
Sol.
v1 =
2 g ( h H / 2)
2g h
v2 =
By continuity equation
dh = a (v + v ) 1 2 dt
A
dh = a dt
A
or
121.
Sol.
A
2 g (h H / 2) 2 g h
H/2
a 2g H
dh
t
=
h h H/2
dt
0
t=
2A 3a
2 1
H g
Ans.
A container of cross-section area ' S ' and height ' h ' is filled with mercury up to the brim. Then the container is sealed airtight and a hole of small cross section area ' S/n ' (where ‘ n ‘ is a positive constant) is punched in its bottom. Find out the time interval upto which the mercury will come out from the bottom hole. [Take the atmospheric pressure to be equal to h0 height of mercury column: h > h0 ] [ Made 2004] M.Bank_Fluids_15 Let the velocity of efflux of mercury coming out of hole be at an instant mercury level in container is y. At same instant the speed of top surface of fluids v. from equation of continuity
S v = Sv n
....(1)
S << S >> v n applying Bernolis therram between A and B
1 2 1 ev = Patm + e2 2 2 v << higher powers of v can be neglected and Patm = h0eg
Yeg +
=
2g( Y h0 )
.........(2)
Hence mercury flows out of both till y = h0. from equation (1)
v=–
dy 1 1 = = dt h h
dy
2g( Y h0 )
1 2gY h0 = h dt
or
integrating between limits h0
at
122.
l
=0
y=n
and t = T, Y = h0 ;
n
dy 2g ( Y h 0 )
=
1 n
T
dt ;
2 t = n g (h h 0 )
0
W ater flows through a frictionless duct with a cross-section varying as shown in figure. Pressure p at points along the axis is represented by: M.Bank_Fluid_67
(A*)
(B)
(C)
(D)
