SOME MECHANICAL PROPERTIES OF MATTER:--
OBJECTIVE-II
1. A Student plots a graph from his readings on the determination of Young's modulus of a metal wire but forgets to put the labels (figure 14-Q4). The quantities on X and Y-axes may be respectively
(a) weight hung and length increased
(b) stress applied and length increased
(c) stress applied and strain developed
(d) length increased and the weight hung.
Figure for Q-1
ANSWER: All
Explanation: It is clear from the graph that the quantity along one axis is directly proportional to the quantity along another axis.
We know that stress/strain = Y (constant)
→W/A = Y*l/L
→W =(YA/L)*l
For a given wire, cross-sectional area A and the length L is constant. Hence
W = K*l {where YA/L =K, a constant}
→W∝l
i.e. the weight hung is proportional to the length increased. So the option (a) is true.
If we write the expression as
W/A = (Y/L)*l
→stress ∝ length increased {where Y/L is another constant}
Hence the option (b) is true.
Since the stress is directly proportional to the strain developed and the constant of proportionality is Young's modulus of elasticity. Hence the option (c) is correct.
It is the same case of option (a) simply the axes have been changed. Which is also true. Hence the option (d) is true.
Thus all options are true.
2. The properties of a surface are different from those of the bulk liquid because of the surface molecules
(a) are smaller than other molecules
(b) acquire charge due to collision from air molecules
(c) find different type of molecules in their range of influence
(d) feel a net force in one direction.
ANSWER: (c), (d)
Explanation: The surface molecules are neither smaller than other molecules nor acquire charge due to collision from air molecules.
The surface molecules have similar molecules on one side while different molecules on the other side. Due to this, they feel a net force in one direction. Hence the options (c) and (d).
3. The rise of a liquid in a capillary tube depends on
(a) the material
(b) the length
(c) the outer radius
(d) the inner radius of the tube
ANSWER: (a), (b), (d)
Explanation: The rise of a liquid in a capillary tube depends on
(a) the material because the angle of contact depends on it.
(b) the length because if the length is smaller than required the rise will be up to the tube length.
(d) the inner radius of the tube because the rise is inversely proportional to the inner radius.
The option (c) is not true because the outer radius does not play any role.
4. The contact angle between a solid and a liquid is a property of (a) the material of the solid
(b) the material of the liquid
(c) the shape of the solid
(d) the mass of the solid
ANSWER: (a), (b)
Explanation: The contact angle between a solid and a liquid depends on the materials of the solid and the liquid. The same solid has different angle of contact with different liquids. The same liquid has different angle of contact with different solids. Hence options (a) and (b) are true.
The contact angle between a solid and a liquid does not depend on the shape or the mass of the solid.
5. A liquid is contained in a vertical tube of a semicircular cross-section (figure 14-Q5). The contact angle is zero.
The forces of surface tension on the curved part and on the flat part are in the ratio
(a) 1:1
(b) 1:2
(c) π:2
(d) 2:π
Figure for Q-5
ANSWER: (c)
Ratio of forces
Explanation: The surface tension force on the curved part F = S*πr
where r is the radius of the tube and S is the surface tension of the liquid.
The surface tension force on the flat part F' = S*2r
Hence F:F' =(S*πr):(S*2r) =π:2. Hence the option (c) is true.
6. When a capillary tube is dipped into a liquid, the liquid neither rises nor falls in the capillary.
(a) The surface tension of the liquid must be zero.
(b) The contact angle must be 90°.
(c) The surface tension may be zero.
(d) The contact angle may be 90°.
ANSWER: (c), (d)
Explanation: The rise in a capillary tube is given as,
h = 2S.cosθ/rρg
The capillary tube has a certain inner radius r. So r, ρ and g are not zero. For h to be zero either surface tension S may be zero or the contact angle θ = 90° so that cosθ = 0. But we can not conclude that S must be zero or θ must be 90°. Hence the option (c) and (d) are true.
7. A solid sphere moves at a terminal velocity of 20 m/s in the air at a place where g = 9.8 m/s². The sphere is taken in a gravity-free hall having air at the same pressure and pushed down at a speed of 20 m/s.
(a) Its initial acceleration will be 9.8 m/s² downward.
(b) Its initial acceleration will be 9.8 m/s² upward.
(c) The magnitude of acceleration will decrease as time passes.
(d) It will eventually stop.
ANSWER: (b), (c), (d)
Explanation: Since there is no gravity there will be no initial acceleration downward, hence the option (a) is wrong.
Since 20 m/s is the terminal velocity, at this velocity the viscous force is equal to the weight of the sphere (assuming the buoyancy force negligible), i.e. it can produce an upward acceleration equal to g = 9.8 m/s². Hence when it is thrown down at 20 m/s, the sphere gets an upward initial acceleration =9.8 m/s². Hence option (b) is true.
Due to upward acceleration, the speed reduces and hence the viscous force. Since the force reduces the magnitude of acceleration also reduces as time passes. Hence option (c) is true.
Since the viscous force opposes the motion of the sphere, it will eventually stop. Hence the option (d) is true.
1. A Student plots a graph from his readings on the determination of Young's modulus of a metal wire but forgets to put the labels (figure 14-Q4). The quantities on X and Y-axes may be respectively
(a) weight hung and length increased
(b) stress applied and length increased
(c) stress applied and strain developed
(d) length increased and the weight hung.
 |
| Figure for Q-1 |
ANSWER: All
Explanation: It is clear from the graph that the quantity along one axis is directly proportional to the quantity along another axis.
We know that stress/strain = Y (constant)
→W/A = Y*l/L
→W =(YA/L)*l
For a given wire, cross-sectional area A and the length L is constant. Hence
W = K*l {where YA/L =K, a constant}
→W∝l
i.e. the weight hung is proportional to the length increased. So the option (a) is true.
If we write the expression as
W/A = (Y/L)*l
→stress ∝ length increased {where Y/L is another constant}
Hence the option (b) is true.
Since the stress is directly proportional to the strain developed and the constant of proportionality is Young's modulus of elasticity. Hence the option (c) is correct.
Thus all options are true.
2. The properties of a surface are different from those of the bulk liquid because of the surface molecules
(a) are smaller than other molecules
(b) acquire charge due to collision from air molecules
(c) find different type of molecules in their range of influence
(d) feel a net force in one direction.
ANSWER: (c), (d)
Explanation: The surface molecules are neither smaller than other molecules nor acquire charge due to collision from air molecules.
The surface molecules have similar molecules on one side while different molecules on the other side. Due to this, they feel a net force in one direction. Hence the options (c) and (d).
3. The rise of a liquid in a capillary tube depends on
(a) the material
(b) the length
(c) the outer radius
(d) the inner radius of the tube
ANSWER: (a), (b), (d)
Explanation: The rise of a liquid in a capillary tube depends on
(a) the material because the angle of contact depends on it.
(b) the length because if the length is smaller than required the rise will be up to the tube length.
(d) the inner radius of the tube because the rise is inversely proportional to the inner radius.
The option (c) is not true because the outer radius does not play any role.
4. The contact angle between a solid and a liquid is a property of (a) the material of the solid
(b) the material of the liquid
(c) the shape of the solid
(d) the mass of the solid
ANSWER: (a), (b)
Explanation: The contact angle between a solid and a liquid depends on the materials of the solid and the liquid. The same solid has different angle of contact with different liquids. The same liquid has different angle of contact with different solids. Hence options (a) and (b) are true.
The contact angle between a solid and a liquid does not depend on the shape or the mass of the solid.
5. A liquid is contained in a vertical tube of a semicircular cross-section (figure 14-Q5). The contact angle is zero.
The forces of surface tension on the curved part and on the flat part are in the ratio
(a) 1:1
(b) 1:2
(c) π:2
(d) 2:π
 |
| Figure for Q-5 |
ANSWER: (c)
 |
| Ratio of forces |
Explanation: The surface tension force on the curved part F = S*πr
where r is the radius of the tube and S is the surface tension of the liquid.
The surface tension force on the flat part F' = S*2r
Hence F:F' =(S*πr):(S*2r) =π:2. Hence the option (c) is true.
6. When a capillary tube is dipped into a liquid, the liquid neither rises nor falls in the capillary.
(a) The surface tension of the liquid must be zero.
(b) The contact angle must be 90°.
(c) The surface tension may be zero.
(d) The contact angle may be 90°.
ANSWER: (c), (d)
Explanation: The rise in a capillary tube is given as,
h = 2S.cosθ/rρg
The capillary tube has a certain inner radius r. So r, ρ and g are not zero. For h to be zero either surface tension S may be zero or the contact angle θ = 90° so that cosθ = 0. But we can not conclude that S must be zero or θ must be 90°. Hence the option (c) and (d) are true.
7. A solid sphere moves at a terminal velocity of 20 m/s in the air at a place where g = 9.8 m/s². The sphere is taken in a gravity-free hall having air at the same pressure and pushed down at a speed of 20 m/s.
(a) Its initial acceleration will be 9.8 m/s² downward.
(b) Its initial acceleration will be 9.8 m/s² upward.
(c) The magnitude of acceleration will decrease as time passes.
(d) It will eventually stop.
ANSWER: (b), (c), (d)
Explanation: Since there is no gravity there will be no initial acceleration downward, hence the option (a) is wrong.
Since 20 m/s is the terminal velocity, at this velocity the viscous force is equal to the weight of the sphere (assuming the buoyancy force negligible), i.e. it can produce an upward acceleration equal to g = 9.8 m/s². Hence when it is thrown down at 20 m/s, the sphere gets an upward initial acceleration =9.8 m/s². Hence option (b) is true.
Due to upward acceleration, the speed reduces and hence the viscous force. Since the force reduces the magnitude of acceleration also reduces as time passes. Hence option (c) is true.
Since the viscous force opposes the motion of the sphere, it will eventually stop. Hence the option (d) is true.
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CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
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Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
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CHAPTER- 3 - Kinematics - Rest and Motion
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Click here for EXERCISES (Question number 31 to 40)
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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