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SOME MECHANICAL PROPERTIES OF MATTER:--
EXERCISES, Q-21 To Q-32
21. A barometer is constructed with its tube having radius 1.0 mm. Assume that the surface of the mercury in the tube is spherical in shape. If the atmospheric pressure is equal to 76 cm of mercury, what will be the height raised in the barometer tube? The contact angle of mercury with glass =135° and surface tension of mercury =0.466 N/m. Density of mercury =13600 kg/m³.
ANSWER: If there had not been the effect of surface tension, the height of mercury would have been 76 cm. Since the tube diameter is small there will be a capillary effect due to surface tension. The rise h = 2S.cosθ/ρrg
→h = 2*0.466*cos135°/(13600*0.001*9.8)
→h = -0.005 m = - 0.5 cm
(negative sign shows that the level of mercury goes down due to the capillary effect).
So the height of mercury in the barometer tube = 76 cm - 0.5 cm
=75.5 cm.
22. A capillary tube of radius 0.50 mm is dipped vertically in a pot of water. Find the difference between the pressure of the water in the tube 5.0 cm below the surface and the atmospheric pressure. The surface tension of the water = 0.075 N/m.
ANSWER: The surface in the tube will be concave upward. So the atmospheric pressure will be 2S/r more than the water just below the surface. The atmospheric pressure = 2*0.075/(0.50/1000) N/m²
=0.150*1000/0.50 N/m²
=150*2 N/m²
=300 N/m²
The pressure 5.0 cm below the surface will be ρgh more than the pressure just below the water surface. Here pressure =1000*9.8*5/100 N/m²
=490 N/m².
So if we assume that the pressure of water just below the surface = P, then the atmospheric pressure just above the surface A =P+300 N/m² and the pressure 5.0 cm below the surface F = P+490 N/m².
Hence the difference F-A = (P+490)-(P+300) =190 N/m²
23. Find the surface energy of water kept in a cylindrical vessel of radius 6.0 cm. Surface tension of water = 0.075 J/m².
ANSWER: The surface area of the water, A = πr² =π*(0.06)² m²
→A = 0.0113 m²
Surface energy of water U = S*A = 0.075*0.0113 J
= 8.4x10⁻⁴ J
24. A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury = 0.465 J/m².
ANSWER: The radius of mercury drop r = 2 mm =0.002 m
The surface area of the drop = 4πr²
The volume of the drop V =4πr³/3
After division the volume of each droplet = V/8 =4πr³/(3*8) =πr³/6
If the radius of the droplet is r', then
V/8 =4πr'³/3
→πr³/6 = 4πr'³/3
→r³ = 8r'³
→r =2r'
→r' =r/2
The total area of the droplets =8*4πr'² =32πr²/4 =8πr²
The increase in area = 8πr² - 4πr² =4πr²
The increase in surface energy = increase in area*surface tension
=4πr²*S
= 4π*(0.002)²*0.465 J
= 2.34x10⁻⁵ J
= 23.4 µJ.
25. A capillary tube of radius 1 mm is kept vertical with the lower end in water. (a) Find the height of the water raised in the capillary. (b) If the length of the capillary tube is half the answer of part (a), find the angle θ made by the water surface in the capillary with the wall.
ANSWER: The rise h =2S*cosθ/rρg
= 2*0.075*cos0°/(0.001*1000*10)
=0.150/10 m =0.015 m =1.5 cm
{Taking S =0.075 N/m, θ = 0° and g = 10 m/s²}
If the length of the capillary tube = h/2 =1.5/2 cm =0.75 cm, then the water will rise only up to this height. Now the angle θ will be given as,
2S*cosθ/rρg = 0.75/100
→cosθ = 0.75*0.001*1000*10/(2*0.075*100) = ½
→θ = 60°
Alternately,
h' = h/2
But 2Scosθ/rρg = h'
and 2S*cos0°/rρg = h
Dividing,
→cosθ = h' /h =½
→θ = 60°
26. The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury. (a) Find the depression of the mercury column in the capillary. (b) If the length dipped inside is half the answer of part (a), find the angle made by the mercury surface at the end of the capillary with the vertical. The surface tension of mercury = 0.465 N/m and the contact angle of mercury with the glass = 135°.
ANSWER: (a) The rise h =2S*cosθ/rρg
→h = 2*0.465*cos135°/(0.001*13600*9.8)
→h = -0.0049 m
Hence the depression = 4.9 mm
(b) Let the angle made by the mercury surface at the end of the capillary with the vertical =α
then rise h' =2S*cosα/rρg
So, h'/h =(2S*cosα/rρg)/(2S*cosθ/rρg) =cosα/cosθ
But h'/h =½, so
cosα =½cosθ =½*cos135° = -0.707/2 = -0.3535
→α = 111°
27. The large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water = 0.075 N/m.
ANSWER: Let the rise of water = h
Diagram for Q-27
Consider the unit length of the glass plate. The surface tension force = 2*S*1 =2*0.075 N =0.150 N per unit length of glass plate.
The weight of water per unit length of glass plate = ρhdg*1
=1000*0.001*h*10 =10h N per unit length of the glass plate.
{Taking g = 10 N/m²}
Equating, 10h = 0.150
→h = 0.15/10 m = 0.15*100/10 cm =1.5 cm
28. Consider an ice cube of edge 1.0 cm kept in a gravity-free hall. Find the surface area of the water when the ice melts. Neglect the difference in densities of ice and water.
ANSWER: Since we are neglecting the densities of ice and water, the volume remains unchanged after the change in state.
The volume V = 0.01³ m³ =1x10⁻⁶ m³
Due to the surface tension, the melted water will change its shape to spherical because for a given volume a sphere has the least area. Let the radius of this sphere = r, the volume
4πr³/3 = 1x10⁻⁶
→r³ = 3x10⁻⁶/4π
→r = {3x10⁻⁶/4π}¹/³
The surface area of this sphere = 4πr²
=4π{3x10⁻⁶/4π}²/³ m²
={(4π)³/²}²/³*{3x10⁻⁶/4π}²/³ m²
={(4π)³/²*3x10⁻⁶/4π}²/³ m²
={√(4π)}²/³*{√(9x10⁻¹²)}²/³ m²
={4π}¹/³*{9x10⁻¹²}¹/³ *10⁴ cm²
={4π*9x10⁻¹²}¹/³*{10¹²}¹/³
=(36π)¹/³ cm²
29. A wire forming a loop is dipped into the soap solution and taken out so that a film of soap solution is formed. A loop of 6.28 cm long thread is gently put on the film and the film is pricked with a needle inside the loop. The thread loop takes the shape of a circle. Find the tension in the thread. Surface tension of soap solution = 0.030 N/m.
ANSWER: The thread has two layers of surfaces of soap solution In contact, up and down. The outward normal force due to surface tension on the loop
= 2S
Let the radius of loop = r
2πr = 6.28
r = 6.28/2π = 1 cm
Let the tension in the thread = T
Considering one semicircular part of the loop. The tension in two ends of the thread will balance the normal force of surface tension on the semicircular length. The component of this force along the direction of tension will be 2S*2r.
2T = 2S*2r
→T = 2Sr =2*0.03*(1/100) =2*0.0003 N =0.0006 N
→T = 6x10⁻⁴ N
30. A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1 cm/s, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. The density of glycerine = 1260 kg/m³ and its coefficient of viscosity at room temperature = 8.0 poise.
ANSWER: (a) The viscous force is given by Stokes' law
F =6πηrv
Given, η = 8 poise = 0.8 N-s/m², r = 1 mm =0.001 m, v = 1 cm/s = 0.01 m/s
F = 6π*0.8*0.001*0.01 =1.5x10⁻⁴ N
(b) The hydrostatic force exerted by the glycerine is the force of buoyancy which is equal to the weight of glycerine of the same volume that of the sphere.
B = (4πr³/3)*ρg = (4π*0.001³/3)*1260*9.8 N
≈52000 *10⁻⁹ N
=5.2x10⁴*10⁻⁹ N
=5.2x10⁻⁵ N
(c) Let the terminal velocity = V. Following are the forces on the sphere.
The viscous force at this terminal velocity F' = 6πηrV
→F' = 6π*0.8*0.001V = 0.0048πV {Upward}
The force of Buyancy B = 5.2x10⁻⁵ N {Upward}
The weight of the sphere W = 50*9.8/10⁶ N = 4.9x10⁻⁴ N {Downward}
So the force of the weight is balanced by the viscous force F' and the Buoyancy force B. F'+B = W
→0.0048πV+5.2x10⁻⁵ = 4.9x10⁻⁴
→0.0048πV = 49x10⁻⁵-5.2x10⁻⁵ =43.8x10⁻⁵
→V = 43.8x10⁻⁵/0.0048π =0.029 M/s
→V = 2.9 cm/s
31. Estimate the speed of vertically falling raindrops from the following data. Radius of the drops = 0.02 cm, viscosity of air = 1.8x10⁻⁴ poise, g = 9.9 m/s² and density of water = 1000 kg/m³.
ANSWER: Let the terminal speed of the raindrops = v
From Stokes' law the viscous force on it F =6πηrv
The weight of a Raindrop W = ρVg
Since the density of the air is much less than the water, the buoyancy force can be neglected. Thus the viscous force balances the weight. i.e. F =W
→6πηrv = ρVg
→v = ρVg/6πηr
→v = 1000*(4πr³/3)*9.9/6πηr
→v = 4000r²*3.3/6η
→v = 2000*(0.0002)²*1.1/1.8x10⁻⁵ m/s
→v = 8.8x10⁻⁵/1.8x10⁻⁵
→v = 4.9 m/s
→v ≈ 5.0 m/s
32. Water flows at a speed of 6 cm/s through a tube of radius 1 cm. The coefficient of viscosity of water at room temperature is 0.01 poise. calculate the Reynolds number. Is it a steady flow?
ANSWER: Velocity v = 6 cm/s = 0.06 m/s,
The density of water ρ = 1000 kg/m³,
Diameter of the tube D = 2 cm = 0.02 m,
Viscosity of water η = 0.01 poise = 0.001 N-s/m².
The Reynolds number N =ρvD/η
=1000*0.06*0.02/0.001
=1200
Since N<2000, hence the flow is steady.
21. A barometer is constructed with its tube having radius 1.0 mm. Assume that the surface of the mercury in the tube is spherical in shape. If the atmospheric pressure is equal to 76 cm of mercury, what will be the height raised in the barometer tube? The contact angle of mercury with glass =135° and surface tension of mercury =0.466 N/m. Density of mercury =13600 kg/m³.
ANSWER: If there had not been the effect of surface tension, the height of mercury would have been 76 cm. Since the tube diameter is small there will be a capillary effect due to surface tension. The rise h = 2S.cosθ/ρrg
→h = 2*0.466*cos135°/(13600*0.001*9.8)
→h = -0.005 m = - 0.5 cm
(negative sign shows that the level of mercury goes down due to the capillary effect).
So the height of mercury in the barometer tube = 76 cm - 0.5 cm
=75.5 cm. 22. A capillary tube of radius 0.50 mm is dipped vertically in a pot of water. Find the difference between the pressure of the water in the tube 5.0 cm below the surface and the atmospheric pressure. The surface tension of the water = 0.075 N/m.
ANSWER: The surface in the tube will be concave upward. So the atmospheric pressure will be 2S/r more than the water just below the surface. The atmospheric pressure = 2*0.075/(0.50/1000) N/m²
=0.150*1000/0.50 N/m²
=150*2 N/m²
=300 N/m²
The pressure 5.0 cm below the surface will be ρgh more than the pressure just below the water surface. Here pressure =1000*9.8*5/100 N/m²
=490 N/m².
So if we assume that the pressure of water just below the surface = P, then the atmospheric pressure just above the surface A =P+300 N/m² and the pressure 5.0 cm below the surface F = P+490 N/m².
Hence the difference F-A = (P+490)-(P+300) =190 N/m²
23. Find the surface energy of water kept in a cylindrical vessel of radius 6.0 cm. Surface tension of water = 0.075 J/m².
ANSWER: The surface area of the water, A = πr² =π*(0.06)² m²
→A = 0.0113 m²
Surface energy of water U = S*A = 0.075*0.0113 J
= 8.4x10⁻⁴ J
24. A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury = 0.465 J/m².
ANSWER: The radius of mercury drop r = 2 mm =0.002 m
The surface area of the drop = 4πr²
The volume of the drop V =4πr³/3
After division the volume of each droplet = V/8 =4πr³/(3*8) =πr³/6
If the radius of the droplet is r', then
V/8 =4πr'³/3
→πr³/6 = 4πr'³/3
→r³ = 8r'³
→r =2r'
→r' =r/2
The total area of the droplets =8*4πr'² =32πr²/4 =8πr²
The increase in area = 8πr² - 4πr² =4πr²
The increase in surface energy = increase in area*surface tension
=4πr²*S
= 4π*(0.002)²*0.465 J
= 2.34x10⁻⁵ J
= 23.4 µJ.
25. A capillary tube of radius 1 mm is kept vertical with the lower end in water. (a) Find the height of the water raised in the capillary. (b) If the length of the capillary tube is half the answer of part (a), find the angle θ made by the water surface in the capillary with the wall.
ANSWER: The rise h =2S*cosθ/rρg
= 2*0.075*cos0°/(0.001*1000*10)
=0.150/10 m =0.015 m =1.5 cm
{Taking S =0.075 N/m, θ = 0° and g = 10 m/s²}
If the length of the capillary tube = h/2 =1.5/2 cm =0.75 cm, then the water will rise only up to this height. Now the angle θ will be given as,
2S*cosθ/rρg = 0.75/100
→cosθ = 0.75*0.001*1000*10/(2*0.075*100) = ½
→θ = 60°Alternately,
h' = h/2
But 2Scosθ/rρg = h'
and 2S*cos0°/rρg = h
Dividing,
→cosθ = h' /h =½
→θ = 60°
26. The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury. (a) Find the depression of the mercury column in the capillary. (b) If the length dipped inside is half the answer of part (a), find the angle made by the mercury surface at the end of the capillary with the vertical. The surface tension of mercury = 0.465 N/m and the contact angle of mercury with the glass = 135°.
ANSWER: (a) The rise h =2S*cosθ/rρg
→h = 2*0.465*cos135°/(0.001*13600*9.8)
→h = -0.0049 m
Hence the depression = 4.9 mm
(b) Let the angle made by the mercury surface at the end of the capillary with the vertical =α
then rise h' =2S*cosα/rρg
So, h'/h =(2S*cosα/rρg)/(2S*cosθ/rρg) =cosα/cosθ
But h'/h =½, so
cosα =½cosθ =½*cos135° = -0.707/2 = -0.3535
→α = 111°
27. The large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water = 0.075 N/m.
ANSWER: Let the rise of water = h
 |
| Diagram for Q-27 |
Consider the unit length of the glass plate. The surface tension force = 2*S*1 =2*0.075 N =0.150 N per unit length of glass plate.
The weight of water per unit length of glass plate = ρhdg*1
=1000*0.001*h*10 =10h N per unit length of the glass plate.
{Taking g = 10 N/m²}
Equating, 10h = 0.150
→h = 0.15/10 m = 0.15*100/10 cm =1.5 cm28. Consider an ice cube of edge 1.0 cm kept in a gravity-free hall. Find the surface area of the water when the ice melts. Neglect the difference in densities of ice and water.
ANSWER: Since we are neglecting the densities of ice and water, the volume remains unchanged after the change in state.
The volume V = 0.01³ m³ =1x10⁻⁶ m³
Due to the surface tension, the melted water will change its shape to spherical because for a given volume a sphere has the least area. Let the radius of this sphere = r, the volume
4πr³/3 = 1x10⁻⁶
→r³ = 3x10⁻⁶/4π
→r = {3x10⁻⁶/4π}¹/³
The surface area of this sphere = 4πr²
=4π{3x10⁻⁶/4π}²/³ m²
={(4π)³/²}²/³*{3x10⁻⁶/4π}²/³ m²
={(4π)³/²*3x10⁻⁶/4π}²/³ m²
={√(4π)}²/³*{√(9x10⁻¹²)}²/³ m²
={4π}¹/³*{9x10⁻¹²}¹/³ *10⁴ cm²
={4π*9x10⁻¹²}¹/³*{10¹²}¹/³
=(36π)¹/³ cm²
29. A wire forming a loop is dipped into the soap solution and taken out so that a film of soap solution is formed. A loop of 6.28 cm long thread is gently put on the film and the film is pricked with a needle inside the loop. The thread loop takes the shape of a circle. Find the tension in the thread. Surface tension of soap solution = 0.030 N/m.
ANSWER: The thread has two layers of surfaces of soap solution In contact, up and down. The outward normal force due to surface tension on the loop
= 2S
Let the radius of loop = r
2πr = 6.28
r = 6.28/2π = 1 cm
Let the tension in the thread = T
Considering one semicircular part of the loop. The tension in two ends of the thread will balance the normal force of surface tension on the semicircular length. The component of this force along the direction of tension will be 2S*2r.
2T = 2S*2r
→T = 2Sr =2*0.03*(1/100) =2*0.0003 N =0.0006 N
→T = 6x10⁻⁴ N
30. A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1 cm/s, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. The density of glycerine = 1260 kg/m³ and its coefficient of viscosity at room temperature = 8.0 poise.
ANSWER: (a) The viscous force is given by Stokes' law
F =6πηrv
Given, η = 8 poise = 0.8 N-s/m², r = 1 mm =0.001 m, v = 1 cm/s = 0.01 m/s
F = 6π*0.8*0.001*0.01 =1.5x10⁻⁴ N
(b) The hydrostatic force exerted by the glycerine is the force of buoyancy which is equal to the weight of glycerine of the same volume that of the sphere.
B = (4πr³/3)*ρg = (4π*0.001³/3)*1260*9.8 N
≈52000 *10⁻⁹ N
=5.2x10⁴*10⁻⁹ N
=5.2x10⁻⁵ N
(c) Let the terminal velocity = V. Following are the forces on the sphere.
The viscous force at this terminal velocity F' = 6πηrV
→F' = 6π*0.8*0.001V = 0.0048πV {Upward}
The force of Buyancy B = 5.2x10⁻⁵ N {Upward}
The weight of the sphere W = 50*9.8/10⁶ N = 4.9x10⁻⁴ N {Downward}
So the force of the weight is balanced by the viscous force F' and the Buoyancy force B. F'+B = W
→0.0048πV+5.2x10⁻⁵ = 4.9x10⁻⁴
→0.0048πV = 49x10⁻⁵-5.2x10⁻⁵ =43.8x10⁻⁵
→V = 43.8x10⁻⁵/0.0048π =0.029 M/s
→V = 2.9 cm/s
31. Estimate the speed of vertically falling raindrops from the following data. Radius of the drops = 0.02 cm, viscosity of air = 1.8x10⁻⁴ poise, g = 9.9 m/s² and density of water = 1000 kg/m³.
ANSWER: Let the terminal speed of the raindrops = v
From Stokes' law the viscous force on it F =6πηrv
The weight of a Raindrop W = ρVg
Since the density of the air is much less than the water, the buoyancy force can be neglected. Thus the viscous force balances the weight. i.e. F =W
→6πηrv = ρVg
→v = ρVg/6πηr
→v = 1000*(4πr³/3)*9.9/6πηr
→v = 4000r²*3.3/6η
→v = 2000*(0.0002)²*1.1/1.8x10⁻⁵ m/s
→v = 8.8x10⁻⁵/1.8x10⁻⁵
→v = 4.9 m/s
→v ≈ 5.0 m/s
32. Water flows at a speed of 6 cm/s through a tube of radius 1 cm. The coefficient of viscosity of water at room temperature is 0.01 poise. calculate the Reynolds number. Is it a steady flow?
ANSWER: Velocity v = 6 cm/s = 0.06 m/s,
The density of water ρ = 1000 kg/m³,
Diameter of the tube D = 2 cm = 0.02 m,
Viscosity of water η = 0.01 poise = 0.001 N-s/m².
The Reynolds number N =ρvD/η
=1000*0.06*0.02/0.001
=1200
Since N<2000, hence the flow is steady.
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Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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