Exercise Solutions...HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics F = 204 N (b) Vertical forces (F’) includes the weight of the water and the force due to atmospheric - [PDF Document] (2024)

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

Exercise Solutions

Question 1: The surface of water in a water tank on the top of a house is 4 m above the tap level. Find the pressure of water at the tap when the tap is closed, is it necessary to specify that the tap is closed? Take g = 10 m s-2. Solution: Pressure = P = ρgh Here ρ = 1000 kg/m3, g = 10 m s-2 and h = 4 m => P = 40000 N/m2 Yes, it necessary to state that the tap is closed. If tap is open the level of water will decrease and hence the pressure will decrease, the pressure at the tap is atmospheric. Question 2: The heights of mercury surfaces in the two arms of the manometer shown in figure (below) are 2 cm and 8 cm. Atmospheric pressure = 101 × 10 N m-2. Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.

Solution:

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

(a) Pressure at the bottom of the tubes should be same when considered for both limbs. From figure, Pa + ρHg h1 g = Pg + ρHg h2 g => Pg = Pa + ρHg g(h1 - h2) (b) Pressure of mercury at the bottom of tube P = Pa + ρHg h1 g Question 3: The area of cross section of the wider tube shown in figure (below) 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.

Solution: Pressure = Force/Area F = mg = 45 x 9.8 = 441 N and A = 900 cm2 or 0.09 m2 Therefore, Pressure = 441/ 0.09 = 4900 N ...(1) Also, pressure can be represented as, P = ρg Δh Where, Δh is the height difference and ρ = density Density of water = ρ = 1000 kg/m3

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

=> P = 1000 x 9.8 x Δh ...(2) from (1) and (2) 4900 = 1000 x 9.8 x Δh or Δh = 50 cm Question 4: A glass full of water has a bottom of area 20 cm2, top of area 20 cm2, height 20 cm and volume half a liter. (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 = 10 × 105 N m-2. Density of water = 1000 kg m-3 and g = 10 m s-2. Take all numbers to be exact.

Solution: (a) Pressure at bottom of the container = pressure due to water + atmospheric pressure. P = ρg h + Patm = (1000 x 10 x 0.2) + 1 x 105 = 1.02 x 105 N/m2 Now, Force exerted on the bottom of the container = F = Px Area = (1.02 x 105) x 0.002

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

F = 204 N (b) Vertical forces (F’) includes the weight of the water and the force due to atmospheric pressure. F' = mg + [P_atm x area] = (0.5 x 9.8) + (0.002 x 1 x 10^5) F' = 205 N Therefore, total upward force = F' - F = 1 N Question 5: Suppose the glass of the previous problem is covered by a jar and the air inside the jar is completely pumped out. (a) What will be the answers to the problem? (b) Show that the answers do not change if a glass of different shape is used provided the height, the bottom area and the volume are unchanged. Solution: If the glass is covered by a jar and the air inside the jar is completely pumped out. Pressure at the bottom of the glass = P = ρ gh = 1000 x 10 x 0.2 = 2000 N/m2 The downward force is F = P x area = 2000 x 0.002 = 4N (b) Vertical force is equal to the weight of the water = mg = 0.5 x 10 = 5N The horizontal forces again cancel out. So the total upward force = 1N It glass of different shape is used provided the volume, height and area remain same, no change in answer will occur. Question 6: If water be used to construct a barometer, what would be the height of water column at standard atmospheric pressure (76 cm of mercury)? Solution:

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

Standard atmospheric pressure is always pressure exerted by 76 cm Hg column. Patm|Hg = ρHg gh = 13.6 x 103 x g x 0.76 ...(1) For water, Patm|water = ρwater gh = 1000 x g x 0.76 ...(2) Atmospheric pressure: P_atm|Hg = P_atm|water equating both the equations, we get h = 1033.6 cm, is the required height of the water. Question 7: 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? Solution: Pressure = P = ρgh = 1000 x 10 x 500 = 5 x 106 N/m2 and force = F = Pressure x Area = (5 x 106) x 2 = 107 N Question 8: Water is filled in a rectangular tank of size 3 m × 2 m × 1 m. (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 δx meter in this side, situated at a depth of x meter 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 s-2. Solution: Dimensions of rectangular tank : 3 m × 2 m × 1 m. (a) Pressure at the bottom of the tank = P = ρgh = 1000 x 10 x 1 = 104 N/m2

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

Area of the bottom of the tank = A = lb = 3x2 = 6m2 Now, force = F = Pressure x Area = 104 x 6 = 60000 N (b) P = ρgh = ρgx Area of strip = A = 2 dx So, the force on the strip = F = PA = ρgx x 2 dx = (20000xdx) N (c) Let us first find the perpendicular distance(d) of the strip from the bottom edge. d = h-x = (1-x) m Now, Torque = Force x perpendicular distance = Fd = (20000x(1-x)dx) N-m (d) Total force

(e) Total Torque

Question 9: An ornament 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 of copper is 8.9. Solution: The density of copper and gold are 8.9 and 19.3 respectively. V = x/8.9 + (36-x)/19.3 ...(1)

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

When ornament placed in water, it displaces water equal to its weight. The density of water = ρwater = 1 g/cm3 Weight of the water displaced = w = mg = v ρwater g = vg ....(2) ornament weighs 34 g in water.(Given) The buoyant force = Fb = mg = (36-34) = 2g ...(3) Buoyant force is equal to weight of the water displaced. vg = 2g => [x/8.9 + (36-x)/19.3] g = 2g or x = 23.14/10.4 = 2.22 grams (approx) The weight of copper in the ornament is 2.22 grams. Question 10: 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. Solution: Total volume of the Ornament(y)= Volume of gold + volume of cavity Total volume (y)= 36/19.3 +v Using, Principle of Flotation, The volume of the water displaced is also y cm3 The mass of the water displaced =y grams The weight of the water displaced = yg yg = 2g

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

or y = 2 Therefore, 36/19.3 +v = 2 or v = 0.13 cm3 Question 11: A metal piece of mass 160 g lies in equilibrium inside a glass of water (figure below). The piece touches the bottom of the glass at a small number of points. If the density of the metal is 8000 kg m4, find the normal force exerted by the bottom of the glass on the metal piece.

Solution: Volume of metal immersed, = V = m/ρ = 0.16/8000 = 0.00002 m3 Weight of the water displaced = Fb = V x ρwater x g = 0.00002 x 1000 x 10 = 0.2 Find the normal force exerted by the bottom of the glass on the metal piece: Vertical forces consist of the normal reaction(N) from the surface (upwards), the weight of the metal(downwards) and the buoyant force(upwards) to maintain the balance of vertical forces. N + F_b = mg => N = (0.16x10)-0.2 = 1.4 N The normal reaction is 1/4 N in total.

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

Question 12: A ferry boat has internal volume 1 m3 and weight 50 kg. (a) Neglecting the thickness of the wood, find the fraction of the volume of the boat immersed in water. (b) If a leak develops in the bottom and water starts coming in, what fraction of the boat’s volume will be filled with water before water starts coming in from the sides? Solution: (a) Buoyant force should be equal to the weight of the boat. vw g = mg v x 1000 = 50 or v = 1/20 m3 (b) Let V_b be the volume of boat filled with water before water starts coming in form the sides. here Vb 1 m3 Therefore, buoyant force = Fb = Vb ρw g = 1000 gN The volume of water = Vw = mw/ρw = 950/1000 = 0.95 m3 [Here mw + 50 = 1000 or mw = 950 kg] Now, fraction of water filled in the boat = 0.95 = 19/20 Question 13: 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. Solution: Let the edge of the cube be “x cm”. Volume of the cube = V = x3/106 m3 ...(1) And buoyant force = Fb = mmg + V ρi g = 0.5 + 900 v ...(2)

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

Also, buoyant force is equal to the weight of the water displaced => V x 1000 g = 0.5 + 900 => 100V = 0.5 Using (1), x3/106 = 0.5/100 or x = 17 cm (approx) Question 14: A cube of ice floats partly in water and partly in K.oil (figure below). Find the ratio of the volume of ice immersed in water to that in K.oil. Specific gravity of K.oil is 0.8 and that of ice is 0.9.

Solution: Density of water = ρw = 1000 kg/m3 Density of ice = ρi = 0.9 x 1000 = 900 kg/m3 Density of K.oil = ρo = 1000 kg/m3 The weight of the cube of ice = wi = volume x density x g = (a+b)900g ...(1) Where a = volume of cube immersed in water and b = volume immersed in K.oil The total buoyant force = sum of forces exerted by the liquids. => Fb = a ρw + bρo= a(1000) + b(800) ....(2)

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

From (1) and (2) a(1000) + b(800) = (a+b)900 => a/b = 1 The ratio of the volumes immersed is 1:1. Question 15: 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 m-3 and density of water = 1000 kg m-3. Solution: Density of iron = ρi = 8000 kg/m3 and density of water = 1000 kg m-3. Let “x cm” be the edge of the iron sheet. Surface area = 6x2 cm2 Volume of the iron sheets = v = 6x2 x 0.1 = (0.6/106) m3 Therefore, mass of the iron box= mb = ρi x v = 4.8 x 10-3 x2 kg ...(1) volume of the cube = V' = x3 cm3 = (x3/106) m3 buoyant force = Fb = V' ρw g = (x3/106) x 1000 x g = (x3/103) gN .....(2) Here, buoyant force is equal to the weight of the box. From (1) and (2) 4.8 x 10-3 x2 = (x3/103)g => x = 4.8 cm

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

Question 16: 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’S and that of lead is 11.3. Solution : Total mass of the system = mass of wood (mw) + mass of lead(mPb) = 200 + mPb ...(1) Density of wood = ρw = 0.8 g/cm3 Volume of the wooden block = vw = mw/ρw = 250 cm3 Density of lead = ρPb = 11.3 g/cm3 Volume of lead piece = vPb = mPb/ρPb = (mPb/11.3) cm3 Therefore, the volume of water displaced would be equal to the total volume. V = vw + vPb = 250 + (mPb/11.3) cm3 Fb = v ρw g = (250 + mPb/11.3) x 1 x g ....(2) Now, (1) = (2), we get 200 + mPb = (250 + mPb/11.3) x 1 x g => mPb = 54.8 g Question 17: Solve the previous problem if the lead piece is fastened on the top surface of the block and the block is to float with its upper surface just dipping into water. Solution: volume immersed is only due to the wooden block. The buoyant force = Fb = 250g

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

Now, Total mass = Fb => 200 + mPb = 250 g or mPb = 50 g The mass of the lead piece is 50 g. Question 18: A cubical metal block of edge 12 cm floats in mercury with one fifth of the height inside 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. Solution: Volume immersed in mercury= vm = (123/5) cm3 Buoyant force on the block is equal to the weight of the mercury displaced = ρm x vm = 13.6 x (123/5) g ....(1) Volume of block in water = vw = 123 h cm3 Volume of block in mercury = v'm = 122(12-h) cm3 Now, buoyant force = Fb = [vm ρw + vm' ρm]g = 122 h + 122 (12-h)13.6 g .....(2) The buoyant force is equal to the weight of the cube. (1)=(2) => 13.6 x (123/5) g = 122 h + 122 (12-h)13.6 g Solving above equation, we have

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

h = 10.4 cm Therefore, water needs to be poured to a height of 10.4 cm. Question 19: 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. Solution: A hollow spherical body of inner and outer radii 6 cm and 8 cm respectively floats half-submerged in water. Find density of the material of the sphere: Here, Mass of the water displaced = Mass of the material, ρw x Vi = vb x ρ => 1000 x (2/3) π 83 = (4/3)π (83 - 63)ρ Solving above equation, we have ρ = 865 kg/m3 Question 20: A solid sphere of radius 5 cm floats in water. If a maximum load of 0.1 kg can be put on it without wetting the load, find the specific gravity of the material of the sphere. Solution: A solid sphere of radius 5 cm floats in water. If a maximum load of 0.1 kg can be put on it without wetting the load Volume of sphere = v = 4πr3/3 = (5π/3) x 10-4 m3 Weight of water displaced = ρw vg = (5π/3) N Now, Weight of sphere + Load = weight of displaced water

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

ρs vg + 0.1 x 10 = ρw vg or ρs = [ρw vg - 1]/vg Therefore, specific gravity = ρs/ρw = 1 - 1/[5π/3] = 0.81 Which is the specific gravity of sphere. Question 21: Find the ratio of the weights, as measured by a spring balance, of a 1 kg block of iron and a 1 kg block of wood. Density of iron = 7800 kg m-3, density of wood = 800 kg m-3 and density of air = 1293 kg m-3. Solution: The weight of iron block = wi = mi g - vi ρa g = mi [1 - (1/ρi) ρa]g and weight of wooden block = ww = mw g - vw ρa g = mw [1 - (1/ρw) ρa]g Now, ratio of wi and ww:

Question 22: A cylindrical object of outer diameter 20 cm and mass 2 kg floats in water with its axis vertical. If it is slightly depressed and then released, find the time period of the resulting simple harmonic motion of the object. Solution: Buoyant force = Fb = v ρw g = (π r2 x)ρw g Where v = volume of the immersed object

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

and ma = (π r2 x)ρw g or a = (π r2 x ρw g)/2000 Now, time period = T = 2π √(displacement/acceleration) => T = 0.5 sec (approx) Question 23: A cylindrical object of outer diameter 10 cm, height 20 cm and density 8000 kg m-3 is supported by a vertical spring and is half dipped in water as shown in figure(below). (a) Find the elongation of the spring in equilibrium condition. (b) 1f the object is slightly depressed and released, find the time period of resulting oscillations of the object. The spring constant =500 N m-1.

Solution: (a)In the equilibrium condition, the weight of the cylinder, is supported by the spring and the buoyant force. kx + v ρw g = mg where v is volume = πr2(h/2) => 500x + (π(0.05) 2 x0.1x10x1000) = π((0.05) 2 x 0.2x8000x10) Solving above equation, we get x = 23.5 cm (b) Driving force = F = kx + vρw g

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

=> ma = kx + πr2 x ρw g => a = 2π √(displacement/acceleration)/m Time period, T = 2π √(displacement/acceleration) = 2π √0.5x/[500+(π(0.05) 2 x1000x10)x] = 0.935 sec Question 24: A wooden block of mass 0.5 kg and density 800 kg m-3 is fastened to the free end of a vertical spring of spring constant 50 N m-1 fixed at the bottom. If the entire system is completely immersed in water find (a) the elongation (for compression) of the spring in equilibrium and (b) the time-period of vertical oscillations of the block when it is slightly depressed and released. Solution: (a) The weight of the block is balanced by the spring and the buoyant force. => mg = kx + v ρw g => 0.5 x 10 = 50x + (0.5/800) x 1000 x 10 Solving for x, x = 2.5 cm (b) a = kx/m w2x = kx/m Time period = T = 2π √(m/k) = (π/5) sec Question 25: A cube of ice of edge 4 cm is placed in an empty cylindrical glass of inner diameter 6 cm. Assume that the ice melts uniformly from each side so that it always retains its cubical shape. Remembering that ice is lighter than water, find the length of

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

the edge of the ice cube at the instant it just leaves contact with the bottom of the glass. Solution: The weight of the remaining ice will be balanced by the buoyant force provided from the melted water. mg = v ρw g The mass of the ice = m = x3 x ρi => x3 x 0.9 = x2 x h x 1 => h = 0.9 x Volume of water formed = 43 - x3 = πr2 h – x2 h x = 2.26 cm Question 26: A U-tube containing a liquid is accelerated horizontally with a constant acceleration αo. If the separation between the vertical limbs is l, find the difference in the heights of the liquid in the two arms. Solution: Balance all the vertical forces on the tube. Pa A + Alphao = Pa + Ahρg Where Pa is atmospheric pressure. => ao I = hg or h = a0/g Question 27: At Deoprayag (Garhwal, UP) river Alaknanda mixes with the river Bhagjrathi and becomes river Ganga. Suppose Alaknanda has a width of 12 m, Bhagirathi has a width of 8 m and Ganga has a width of 16 m. Assume that the depth of water is same in the three rivers. Let the average speed of water in Alaknanda be 20 km

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

h-1 and in Bhagirathi be 16 km h-1. Find the average speed of water in the river Ganga. Solution: The sum of the volume flow from Alaknanda and Bhagirathi is equal to the volume flow in Ganga. Let the depth of the rivers be "d" vA x 12 x d + vB x 8 x d = vG x 16 x d => vG = 23 km/h Question 28: Water flows through a horizontal tube of variable cross section (figure below). The area of cross section at A and B are 4 mm2 and 2 mm2 respectively. If 1 cc of water enters per second through A, find (a) the speed of water at A, (b) the speed of water at B and (c) the pressure difference PA – PB.

Solution: (a)The total volume is equal to area times velocity. vA x av = v Where vA is the velocity in tube. => 4 x 10-2 x vA = 1 => VA = 25 cm/s (b) Let the velocity in tube B, vB For steady flow: vA x aA = vB x aB => 25 x 4 x 10-2 = vB x 2 x 10-2

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

=> VB = 50 cm/s (c) From Bernoulli equation, PA + (1/2)ρ vA

2 = PB + (1/2) ρ VB2

=> PA – PB = (1/2)ρ [vB

2 - vA2]

= (1/2) x 1000 x (0.52 - 0.252) = 94 N/m2 Question 29: Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same. The Separation between the cross sections at A and B is 15/16 cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = 10 m s-2. Solution: VA = 25 cm/s and VB = 50 cm/s [from previous problem solution] As changing the orientation doesn’t change the volume of water flowing from the tubes. (c) From Bernoulli’s equation

Therefore, pressure difference is 0 N/m2 Question 30: Suppose the tube in the previous problem is kept vertical with B upward. Water enters through B at the rate of 1 cm3 s-1. Repeat parts (a), (b) and (c). Note that the Speed decreases as the water falls down.

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

Solution: (a)The total volume is equal to area times velocity. V = aB x vB Where, v = 1 cm3, aA = area of cross-section of tube A = 4 mm2 and aB = area of cross-section of tube B = 2 mm2 => vB = 50 cm/s (b) let vb be the velocity in tube vA x aA = vB x aB => vA = 25 cm/s (c) From Bernoulli equation

Question 31: Water flows through a tube shown in figure (below). 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-1, find (a) the speed at B and (b) the difference in pressures at A and B.

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

Solution: (a)From equation of continuity, vA x aA = vB x av 0.1 x 1 = 0.5 x vB => vB = 20cm/s (b) From Bernoulli’s equation

Question 32: Water flows through a horizontal tube as shown in figure (below). 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 flow of water across any section.

Solution: Equation of continuity: vA x aA = vB x aB => vB = 2 vA Where vA and vB speed of water at A and speed of water at B respectively.

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

aA = Area of cross-section of tube A and aB = Area of cross-section of tube B Also, given h = difference in height in the two columns = 2 cm Pressure difference between A and B: PA - PB = ρ gh = 1000 x 10 x 0.02 = 200 N/m2 By Bernoulli’s equation, PA + (1/2)ρ[vB

2 - vA2]

200 = (1/2) x 1000 x [4vA

2 - vA2]

or vA

2 = 400/(3x100) or vA = 36.51 cm/s Rate of flow = vA x aA = 36.51 x 4 = 146 cm3/s (approx) Question 33: Water flows through the tube shown in figure (below). The areas of cross section of the wide and the narrow portions of the tube are 5 cm2 and 2 cm2 respectively. The rate of flow of water through the tube is 500 cm3 s-1. Find the difference of mercury levels in the U-tube.

Solution: Equation of continuity: vA x aA = vB x aB Where vA and vB are velocities of flow at A and B vA x 5 = 500

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

=> vA = 100 cm/s or 1 m/s Similarly, vB = 250 cm/s or 2.5 m/s Now, the pressure difference between the points, PA - PB = ρHg gh = 13.6 x 10 x h From the Bernoulli’s equation:

Question 34: Water leaks out from an open tank through a hole of area 2 mm2 in the bottom. Suppose water is filled up to a height of 80 cm and the area of cross section of the tank is 0.4 m2. The pressure at the open surface and at the hole are equal to the atmospheric pressure. Neglect the small velocity of the water near the open surface in the tank. (a) Find the initial speed of water coming out of the hole. (b) Find the speed of water coming out when half of water has leaked out. (c) Find the volume of water leaked out during a time interval dt after the height remained is h. Thus, find the decrease in height dh in terms of h and dt. (d) From the result of part (e) find the time required for half of the water to leak out. Solution: (a) The velocity of water exiting at "a" vA = √2gh => vA = √(2x10x9.8) = 4 m/s

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

(b) vB = √(2g.h/2) = √(2x10x0.4) = √8 m/s (c) volume flow = V = Area x height = Ah Therefore, dv/dt = A dh/dt ...(1) We know, V = av dt or V = a (√2gh) t Differentiating above equation, we have dV/dt = A x √2gh ...(2) From (1) and (2) a (√2gh) = A dh/dt => 2 x 10-6 (√2gh) = 0.4 dh/dt => dh/(√2gh) = 5x10-6 dt (d) integrating above equation ,we have

solving for t, t = 6.51 h Question 35: Water level is maintained in a cylindrical vessel up to a fixed height H. The vessel is kept on a horizontal plane. At what height above the bottom should a hole be made in the vessel so that the water stream coming out of the hole strikes the

HC Verma Solutions for Class 11 Physics Chapter 13 Fluid Mechanics

horizontal plane at the greatest distance from the vessel (figure below)?

Solution 35: Water level is maintained in a cylindrical vessel up to a fixed height H. Height of water above the hole = H-h Velocity with which water = v = √[2g[H-h] and time of flight be t t = √(2h/g) The horizontal distance travelled = x = vt = √[2g[H-h] x √(2h/g) => x = √[4[Hh-h2] To maximize this function: dx/dh = 0 d/dh √[4[Hh-h2] = 0 h = H/2