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SOME MECHANICAL PROPERTIES OF MATTER:--
OBJECTIVE-I
1. A rope 1 cm in diameter breaks if the tension in it exceeds 500 N.The maximum tension that may be given to a similar rope of diameter 2 cm is
(a) 500 N
(b) 250 N
(c) 1000 N
(d) 2000 N.
ANSWER: (d)
Explanation: The wire breaks when the maximum permissible stress in the cross-section of the wire is attained. Here 500 N gives maximum permissible stress to a 1 cm diameter wire. Since the area of the cross-section of a 2 cm diameter wire will be four times that of the 1 cm diameter wire, its load carrying capacity will also be four times, i.e. 4*500 N =2000 N.
1. A rope 1 cm in diameter breaks if the tension in it exceeds 500 N.The maximum tension that may be given to a similar rope of diameter 2 cm is
(a) 500 N
(b) 250 N
(c) 1000 N
(d) 2000 N.
ANSWER: (d)
Explanation: The wire breaks when the maximum permissible stress in the cross-section of the wire is attained. Here 500 N gives maximum permissible stress to a 1 cm diameter wire. Since the area of the cross-section of a 2 cm diameter wire will be four times that of the 1 cm diameter wire, its load carrying capacity will also be four times, i.e. 4*500 N =2000 N.
2. The breaking stress of a wire depends on
(a) material of the wire
(b) length of the wire
(c) radius of the wire
(d) shape of the cross-section
ANSWER: (a)
Explanation: The stress is force per unit area. The breaking stress of a wire is a property of its material.
3. A wire can sustain the weight of 20 kg before breaking. If the wire is cut into two equal parts, each part can sustain a weight of
(a) 10 kg
(b) 20 kg
(c) 40 kg
(d) 80 kg
2. The breaking stress of a wire depends on
(a) material of the wire
(b) length of the wire
(c) radius of the wire
(d) shape of the cross-section
ANSWER: (a)
Explanation: The stress is force per unit area. The breaking stress of a wire is a property of its material.
3. A wire can sustain the weight of 20 kg before breaking. If the wire is cut into two equal parts, each part can sustain a weight of
(a) 10 kg
(b) 20 kg
(c) 40 kg
(d) 80 kg
ANSWER: (b)
Explanation: The breaking stress in each part of the wire will be the same as the original because of the same material. Since the area remains the same, each part can sustain a weight of 20 kg.
ANSWER: (b)
Explanation: The breaking stress in each part of the wire will be the same as the original because of the same material. Since the area remains the same, each part can sustain a weight of 20 kg.
4. Two wires A and B are made of same material. The wire A has a length l and diameter r while the wire B has a length 2l and diameter r/2. If the two wires are stretched by the same force, the elongation in A divided by the elongation in B is
(a) 1/8
(b) 1/4
(c) 4
(d) 8.
ANSWER: (a)
Explanation: For wires of the same material the elongation is directly proportional to the length and inversely proportional to the cross-sectional area.
Δl/l = F/AY
→Δl =Fl/AY {Here F and Y are constant}
The length of A is half of B, for it, elongation of A will be ½ of B and the area of A is four times of B (Since the area is proportional to the square of the radius), for it, the elongation of A will be ¼th of B. For the combined effect of both length and the area, the elongation of A will be ½*¼ =1/8 of B.
5. A wire elongates by 1.0 mm when a load W is hanged from it. If this wire goes over a pulley and two weights, W each are hung at the two ends, the elongation of wire will be
(a) 0.5 m
(b) 1.0 mm
(c) 2.0 mm
(d) 4.0 mm
4. Two wires A and B are made of same material. The wire A has a length l and diameter r while the wire B has a length 2l and diameter r/2. If the two wires are stretched by the same force, the elongation in A divided by the elongation in B is
(a) 1/8
(b) 1/4
(c) 4
(d) 8.
ANSWER: (a)
Explanation: For wires of the same material the elongation is directly proportional to the length and inversely proportional to the cross-sectional area.
Δl/l = F/AY
→Δl =Fl/AY {Here F and Y are constant}
The length of A is half of B, for it, elongation of A will be ½ of B and the area of A is four times of B (Since the area is proportional to the square of the radius), for it, the elongation of A will be ¼th of B. For the combined effect of both length and the area, the elongation of A will be ½*¼ =1/8 of B.
5. A wire elongates by 1.0 mm when a load W is hanged from it. If this wire goes over a pulley and two weights, W each are hung at the two ends, the elongation of wire will be
(a) 0.5 m
(b) 1.0 mm
(c) 2.0 mm
(d) 4.0 mm
ANSWER: (b)
Explanation: The elongation will be the same because in both cases it is pulled by the same force at its both ends. In the case of hanging wire, the lower end is pulled by the weight W downwards and the upper end is pulled by the support upwards by the same magnitude of the force.
6. A heavy uniform rod is hanging vertically from a fixed support. It is stretched by its own weight. The diameter of the rod is
(a) smallest at the top and gradually increases down the rod
(b) largest at the top and gradually decreases down the rod
(c) uniform everywhere
(d) maximum in the middle.
ANSWER: (b)
Explanation: The elongation will be the same because in both cases it is pulled by the same force at its both ends. In the case of hanging wire, the lower end is pulled by the weight W downwards and the upper end is pulled by the support upwards by the same magnitude of the force.
6. A heavy uniform rod is hanging vertically from a fixed support. It is stretched by its own weight. The diameter of the rod is
(a) smallest at the top and gradually increases down the rod
(b) largest at the top and gradually decreases down the rod
(c) uniform everywhere
(d) maximum in the middle.
ANSWER: (a)
Explanation: For a hanged heavy uniform rod, at any cross-section, the load is only of the part of the rod which is below it. At its lowest point, the load is zero which gradually increases as we go up. Since the cross-section is the same, the stress increases as we go up. The maximum stress will be at the top and so will be the effect of elongation. Since the transverse strain is proportional to longitudinal strain, hence the diameter of the rod will be smallest at the top and gradually increase down the rod.
7. When a metal wire is stretched by a load, the fractional change in its volume ΔV/V is proportional to
(a) Δl/l
(b) (Δl/l)²
(c) √(Δl/l)
(d) none of these.
ANSWER: (a)
Explanation: For a hanged heavy uniform rod, at any cross-section, the load is only of the part of the rod which is below it. At its lowest point, the load is zero which gradually increases as we go up. Since the cross-section is the same, the stress increases as we go up. The maximum stress will be at the top and so will be the effect of elongation. Since the transverse strain is proportional to longitudinal strain, hence the diameter of the rod will be smallest at the top and gradually increase down the rod.
7. When a metal wire is stretched by a load, the fractional change in its volume ΔV/V is proportional to
(a) Δl/l
(b) (Δl/l)²
(c) √(Δl/l)
(d) none of these.
ANSWER: (a)
Explanation: When a metal wire is stretched by a load the fractional change in the transverse length is proportional to the fractional change in the longitudinal length. i.e.,
Δd/d = σ*Δl/l, where σ is the Poisson's ratio.
Let the original cross-sectional area =A and final area a.
Change in volume ΔV = a*Δl and the original volume V =Al
Hence the fractional change in the volume
ΔV/V = a*Δl/Al = (a/A)*(Δl/l) =(d'²/d²)(Δl/l)
={(d-Δd)²/d²}(Δl/l) ={(d²-2d*Δd)/d²}(Δl/l)
[taking Δd² negligible]
So, ΔV/V = {1-2*Δd/d}(Δl/l)
={1-2* σ*(Δl/l)}*(Δl/l)
=Δl/l - 2σΔl²/l²
→ΔV/V = Δl/l {Taking Δl² negligible}
Hence the option (a).
8. The length of a metal wire is l₁ when the tension in it is T₁ and is l₂ when the tension is T₂. The natural length of the wire is
(a) (l₁+l₂)/2
(b) √(l₁.l₂)
(c) (l₁T₂-l₂T₁)/(T₂-T₁)
(d) (l₁T₂+l₂T₁)/(T₂+T₁)
ANSWER: (a)
Explanation: When a metal wire is stretched by a load the fractional change in the transverse length is proportional to the fractional change in the longitudinal length. i.e.,
Δd/d = σ*Δl/l, where σ is the Poisson's ratio.
Let the original cross-sectional area =A and final area a.
Change in volume ΔV = a*Δl and the original volume V =Al
Hence the fractional change in the volume
ΔV/V = a*Δl/Al = (a/A)*(Δl/l) =(d'²/d²)(Δl/l)
={(d-Δd)²/d²}(Δl/l) ={(d²-2d*Δd)/d²}(Δl/l)
[taking Δd² negligible]
So, ΔV/V = {1-2*Δd/d}(Δl/l)
={1-2* σ*(Δl/l)}*(Δl/l)
=Δl/l - 2σΔl²/l²
→ΔV/V = Δl/l {Taking Δl² negligible}
Hence the option (a).
8. The length of a metal wire is l₁ when the tension in it is T₁ and is l₂ when the tension is T₂. The natural length of the wire is
(a) (l₁+l₂)/2
(b) √(l₁.l₂)
(c) (l₁T₂-l₂T₁)/(T₂-T₁)
(d) (l₁T₂+l₂T₁)/(T₂+T₁)
ANSWER: (c)
Explanation: If the original length of the wire be l and the area of the cross-section a, then
l₁-l = T₁*l/Ya
And l₂-l = T₂*l/Ya
Dividing we get,
(l₁ -l)*T₂ = (l₂-l)*T₁
→l₁T₂ -lT₂= l₂ T₁-T₁l
→l(T₂-T₁) = l₁T₂-l₂T₁
→l = (l₁T₂-l₂T₁)/(T₂-T₁)
Hence the option (c).
9. A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break
ANSWER: (c)
Explanation: If the original length of the wire be l and the area of the cross-section a, then
l₁-l = T₁*l/Ya
And l₂-l = T₂*l/Ya
Dividing we get,
(l₁ -l)*T₂ = (l₂-l)*T₁
→l₁T₂ -lT₂= l₂ T₁-T₁l
→l(T₂-T₁) = l₁T₂-l₂T₁
→l = (l₁T₂-l₂T₁)/(T₂-T₁)
Hence the option (c).
9. A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break
9. A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break
(a) when the mass is at the highest point.
(b) when the mass is at the lowest point
(c) when the wire is horizontal
(d) at an angle of cos⁻¹(1/3) from the upward vertical.
ANSWER: (b)
Figure for Q-9
Explanation: When the wire is whirled in a vertical circle, the weight is always downward but the centrifugal force is always outward from the center i.e. changing its direction. The resultant of the weight and the centrifugal force will be maximum when both are in the same line and in the same direction. This situation is when the rotating mass is at the lowest point.
10. When a metal wire elongates by hanging a load on it, the gravitational potential energy is decreased.
(a) this energy completely appears as the increased kinetic energy of the block.
(b) this energy completely appears as the increased elastic potential energy of the block.
(c) this energy completely appears as heat.
(d) none of these.
(a) when the mass is at the highest point.
(b) when the mass is at the lowest point
(c) when the wire is horizontal
(d) at an angle of cos⁻¹(1/3) from the upward vertical.
ANSWER: (b)
 |
| Figure for Q-9 |
Explanation: When the wire is whirled in a vertical circle, the weight is always downward but the centrifugal force is always outward from the center i.e. changing its direction. The resultant of the weight and the centrifugal force will be maximum when both are in the same line and in the same direction. This situation is when the rotating mass is at the lowest point.
10. When a metal wire elongates by hanging a load on it, the gravitational potential energy is decreased.
(a) this energy completely appears as the increased kinetic energy of the block.
(b) this energy completely appears as the increased elastic potential energy of the block.
(c) this energy completely appears as heat.
(d) none of these.
ANSWER: (d)
Explanation: When a mass is hanged by a wire the gravitational potential energy decreased is mgl (where l is the elongation) of the wire. But the elastic potential energy increased in the wire = ½mgl. Hence the reduced gravitational potential energy will never completely appear as increased kinetic energy or increased elastic potential energy or as heat.
Hence option (d).
11. By a surface of a liquid, we mean
(a) a geometrical plane like x = 0.
(b) all molecules exposed to the atmosphere.
(c) a layer of thickness of the order of 10⁻⁸ m.
(d) a layer of thickness of the order of 10⁻⁴ m.
ANSWER: (c)
Explanation: The surface tension in a liquid is effective to a 10 to 15 molecular depth from the open surface. Hence in the context of surface tension, the surface of a liquid is meant by this much thick layer which comes to the order of 10⁻⁸ m.
ANSWER: (d)
Explanation: When a mass is hanged by a wire the gravitational potential energy decreased is mgl (where l is the elongation) of the wire. But the elastic potential energy increased in the wire = ½mgl. Hence the reduced gravitational potential energy will never completely appear as increased kinetic energy or increased elastic potential energy or as heat.
Hence option (d).
11. By a surface of a liquid, we mean
(a) a geometrical plane like x = 0.
(b) all molecules exposed to the atmosphere.
(c) a layer of thickness of the order of 10⁻⁸ m.
(d) a layer of thickness of the order of 10⁻⁴ m.
ANSWER: (c)
Explanation: The surface tension in a liquid is effective to a 10 to 15 molecular depth from the open surface. Hence in the context of surface tension, the surface of a liquid is meant by this much thick layer which comes to the order of 10⁻⁸ m.
Explanation: The surface tension in a liquid is effective to a 10 to 15 molecular depth from the open surface. Hence in the context of surface tension, the surface of a liquid is meant by this much thick layer which comes to the order of 10⁻⁸ m.
12. An ice cube is suspended in vacuum in a gravity-free hall. As the ice melts it
(a) will it will retain its cubical shape.
(b) will change its shape to spherical.
(c) will fall down on the floor of the hall
(d) will fly up.
12. An ice cube is suspended in vacuum in a gravity-free hall. As the ice melts it
(a) will it will retain its cubical shape.
(b) will change its shape to spherical.
(c) will fall down on the floor of the hall
(d) will fly up.
ANSWER: (b)
Explanation: When the ice cube melts, it changes its state from solid to liquid. This melted liquid has open surfaces where the surface tension forces come into play. These surface tension forces have a tendency to minimize the surface area. Since the melted ice cube is in a gravity-free hall, it has open surface all around. The minimum surface area is achieved by getting a spherical shape.
13. When water droplets merge to form a bigger drop
(a) energy is liberated.
(b) energy is absorbed.
(c) energy is neither liberated nor absorbed.
(d) energy may either be liberated or absorbed depending on the nature of the liquid.
ANSWER: (b)
Explanation: When the ice cube melts, it changes its state from solid to liquid. This melted liquid has open surfaces where the surface tension forces come into play. These surface tension forces have a tendency to minimize the surface area. Since the melted ice cube is in a gravity-free hall, it has open surface all around. The minimum surface area is achieved by getting a spherical shape.
13. When water droplets merge to form a bigger drop
(a) energy is liberated.
(b) energy is absorbed.
(c) energy is neither liberated nor absorbed.
(d) energy may either be liberated or absorbed depending on the nature of the liquid.
ANSWER: (a)
Explanation: When water droplets merge to form a bigger drop the total surface area decreases. Since the molecules in the surface have greater potential energy, the potential energy of surface molecules in the bigger drop decreases from before the merger. In other terms, the surface energy per unit area is equal to the surface tension. Since the surface tension remains the same, the surface energy will be less in the merged bigger drop. Hence the energy is liberated in the process.
14. The dimension ML⁻¹T⁻² can correspond to
(a) moment of a force
(b) surface tension
(c) modulus of elasticity
(d) coefficient of viscosity.
ANSWER: (a)
Explanation: When water droplets merge to form a bigger drop the total surface area decreases. Since the molecules in the surface have greater potential energy, the potential energy of surface molecules in the bigger drop decreases from before the merger. In other terms, the surface energy per unit area is equal to the surface tension. Since the surface tension remains the same, the surface energy will be less in the merged bigger drop. Hence the energy is liberated in the process.
14. The dimension ML⁻¹T⁻² can correspond to
(a) moment of a force
(b) surface tension
(c) modulus of elasticity
(d) coefficient of viscosity.
ANSWER: (c)
Explanation: The dimension of force is MLT⁻² and of length is L. So the dimension of the moment of a force will be ML²T⁻². So the option (a) is not true.
The unit of surface tension is force per unit length. So the dimension of surface tension is ML⁰T⁻². So the option (b) is not true.
The unit of the coefficient of viscosity is Pascal-second. Which is Force*second/area. The dimension will be MLT⁻²*T/L² =ML⁻¹T⁻¹. So the option (d) is not true.
The modulus of elasticity is stress/strain. the strain is dimensionless. So the unit of elasticity will be the same as the unit of stress. It is force/area. The dimension will be MLT⁻²/L² =ML⁻¹T⁻². Hence the option (c).
15. Air is pushed into a soap bubble of radius r to double its radius. If the surface tension of the soap solution is S, the work done in the process is
(a) 8πr²S
(b) 12πr²S
(c) 16πr²S
(d) 24πr²S.
ANSWER: (c)
Explanation: The dimension of force is MLT⁻² and of length is L. So the dimension of the moment of a force will be ML²T⁻². So the option (a) is not true.
The unit of surface tension is force per unit length. So the dimension of surface tension is ML⁰T⁻². So the option (b) is not true.
The unit of the coefficient of viscosity is Pascal-second. Which is Force*second/area. The dimension will be MLT⁻²*T/L² =ML⁻¹T⁻¹. So the option (d) is not true.
The modulus of elasticity is stress/strain. the strain is dimensionless. So the unit of elasticity will be the same as the unit of stress. It is force/area. The dimension will be MLT⁻²/L² =ML⁻¹T⁻². Hence the option (c).
15. Air is pushed into a soap bubble of radius r to double its radius. If the surface tension of the soap solution is S, the work done in the process is
(a) 8πr²S
(b) 12πr²S
(c) 16πr²S
(d) 24πr²S.
ANSWER: (d)
Explanation: The surface tension S is equal to surface energy per unit area, S = U/A
→U = SA
The total area of the soap bubble =2*4πr² =8πr²
The total surface energy of the original bubble =8πr²S
The total surface area of the expanded bubble = 2*4π(2r)² =32πr²
The total surface energy of the expanded bubble =32πr²S
The work done in the process = increase in the surface energy of the bubble =32πr²S - 8πr²S =24πr²S
16. If more air is pushed in a soap bubble, the pressure in it
(a) decreases
(b) increases
(c) remains the same
(d) becomes zero.
ANSWER: (d)
Explanation: The surface tension S is equal to surface energy per unit area, S = U/A
→U = SA
The total area of the soap bubble =2*4πr² =8πr²
The total surface energy of the original bubble =8πr²S
The total surface area of the expanded bubble = 2*4π(2r)² =32πr²
The total surface energy of the expanded bubble =32πr²S
The work done in the process = increase in the surface energy of the bubble =32πr²S - 8πr²S =24πr²S
16. If more air is pushed in a soap bubble, the pressure in it
(a) decreases
(b) increases
(c) remains the same
(d) becomes zero.
ANSWER: (a)
Explanation: The excess pressure (w.r.t. atmospheric pressure) inside a soap bubble is given as
P'-P = 4S/R, {where P'-pressure inside the bubble, P- atmospheric pressure}
It is clear that the excess pressure inside is inversely proportional to the radius. As more air is pushed in a soap bubble its radius increases hence the excess pressure inside decreases.
17. If two soap bubbles of different radii are connected by a tube,
(a) air flows from bigger bubbles to smaller bubble till the sizes become equal
(b) air flows from bigger bubble to smaller bubble till the sizes are interchange
(c) air flows from smaller bubble to bigger.
(d) there is no flow of air
ANSWER: (a)
Explanation: The excess pressure (w.r.t. atmospheric pressure) inside a soap bubble is given as
P'-P = 4S/R, {where P'-pressure inside the bubble, P- atmospheric pressure}
It is clear that the excess pressure inside is inversely proportional to the radius. As more air is pushed in a soap bubble its radius increases hence the excess pressure inside decreases.
17. If two soap bubbles of different radii are connected by a tube,
(a) air flows from bigger bubbles to smaller bubble till the sizes become equal
(b) air flows from bigger bubble to smaller bubble till the sizes are interchange
(c) air flows from smaller bubble to bigger.
(d) there is no flow of air
ANSWER: (c)
Explanation: Since the bigger soap bubbles have lesser inside pressure in comparison to smaller bubbles, when connected by a tube the air will flow from smaller bubble to bigger.
18. Figure (14-Q1) shows a capillary tube of radius r dipped into water. If the atmospheric pressure is P₀, the pressure at point A is
(a) P₀
(b) P₀ + 2S/r
(c) P₀ - 2S/r
(d) P₀ - 4S/r.
Figure for Q-18
ANSWER: (c)
Explanation: Since the bigger soap bubbles have lesser inside pressure in comparison to smaller bubbles, when connected by a tube the air will flow from smaller bubble to bigger.
18. Figure (14-Q1) shows a capillary tube of radius r dipped into water. If the atmospheric pressure is P₀, the pressure at point A is
(a) P₀
(b) P₀ + 2S/r
(c) P₀ - 2S/r
(d) P₀ - 4S/r.
|
| Figure for Q-18 |
ANSWER: (c)
Explanation: Let pressure at A = P. Net pressure =P₀-P. Net resultant pressure on the hemispherical depression =(P₀-P)*πr² along the axis of the tube. The surface tension along the circular edge of the depression =2πrS, along the axis of the tube. Equating we get,
(P₀-P)*πr² = 2πrS
→ (P₀-P) = 2S/r
→P = P₀ - 2S/r
19. The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is
(a) 4
(b) 2
(c) 1
(d) 0.125
ANSWER: (c)
Explanation: Let pressure at A = P. Net pressure =P₀-P. Net resultant pressure on the hemispherical depression =(P₀-P)*πr² along the axis of the tube. The surface tension along the circular edge of the depression =2πrS, along the axis of the tube. Equating we get,
(P₀-P)*πr² = 2πrS
→ (P₀-P) = 2S/r
→P = P₀ - 2S/r
19. The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is
(a) 4
(b) 2
(c) 1
(d) 0.125
Explanation: Let pressure at A = P. Net pressure =P₀-P. Net resultant pressure on the hemispherical depression =(P₀-P)*πr² along the axis of the tube. The surface tension along the circular edge of the depression =2πrS, along the axis of the tube. Equating we get,
(P₀-P)*πr² = 2πrS
→ (P₀-P) = 2S/r
→P = P₀ - 2S/r
19. The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is
(a) 4
(b) 2
(c) 1
(d) 0.125
ANSWER: (d)
Explanation: Let the radius of the first soap bubble = r,
Excess pressure inside P = 4S/r
Let the radius of the second soap bubble = r'
Excess pressure inside P' = 4S/r'
Since P = 2P'
→4S/r =2* 4S/r'
→1/r = 2/r'
→r/r' = ½
The volume of first bubble = n* the volume of second bubble
→4πr³/3 = n*4πr'³/3
→r³/r'³ = n
→n = (r/r')³ = (½)³ =1/8 =0.125
20. Which of the following graphs may represent the relationship between the capillary rise h and the radius r of the capillary?
Figure for Q-20
ANSWER: (c)
Explanation: The capillary rise h is inversely proportional to the radius of the capillary r. It means when r→0, h→∞ and when r→∞, h→0. This condition is best represented by the graph (c).
ANSWER: (d)
Explanation: Let the radius of the first soap bubble = r,
Excess pressure inside P = 4S/r
Let the radius of the second soap bubble = r'
Excess pressure inside P' = 4S/r'
Since P = 2P'
→4S/r =2* 4S/r'
→1/r = 2/r'
→r/r' = ½
The volume of first bubble = n* the volume of second bubble
→4πr³/3 = n*4πr'³/3
→r³/r'³ = n
→n = (r/r')³ = (½)³ =1/8 =0.125
20. Which of the following graphs may represent the relationship between the capillary rise h and the radius r of the capillary?
 |
| Figure for Q-20 |
ANSWER: (c)
Explanation: The capillary rise h is inversely proportional to the radius of the capillary r. It means when r→0, h→∞ and when r→∞, h→0. This condition is best represented by the graph (c).
21. Water rises in a vertical capillary tube up to a length of 10 cm. If the tube is inclined at 45°, the length of water risen in the tube will be
(a) 10 cm
(b) 10√2 cm
(c) 10/√2 cm
(d) none of these
21. Water rises in a vertical capillary tube up to a length of 10 cm. If the tube is inclined at 45°, the length of water risen in the tube will be
(a) 10 cm
(b) 10√2 cm
(c) 10/√2 cm
(d) none of these
(a) 10 cm
(b) 10√2 cm
(c) 10/√2 cm
(d) none of these
ANSWER: (b)
Explanation: When the tube is inclined, the vertical rise remains the same. So the water in the inclined tube rises to 10*sec45° =10√2 cm.
22. A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in a freely falling elevator, the length of the water column in the capillary tube will be
(a) 8 cm
(b) 6 cm
(c) 10 cm
(d) 20 cm.
ANSWER: (b)
Explanation: When the tube is inclined, the vertical rise remains the same. So the water in the inclined tube rises to 10*sec45° =10√2 cm.
22. A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in a freely falling elevator, the length of the water column in the capillary tube will be
(a) 8 cm
(b) 6 cm
(c) 10 cm
(d) 20 cm.
ANSWER: (d)
Explanation: The rise in a capillary tube is given as
h = 2S.cosθ/rρg
When the entire arrangement is in a freely falling elevator, the effective g = 0. So the water will rise in the capillary up to its full length where at the top the meniscus will be convex upward. The condition is somewhat similar to a case when the length of the tube is smaller than the rise.
23. Viscosity is a property of
(a) liquids only
(b) solids only
(c) solids and liquids only
(d) liquids and gases only.
ANSWER: (d)
Explanation: The rise in a capillary tube is given as
h = 2S.cosθ/rρg
When the entire arrangement is in a freely falling elevator, the effective g = 0. So the water will rise in the capillary up to its full length where at the top the meniscus will be convex upward. The condition is somewhat similar to a case when the length of the tube is smaller than the rise.
23. Viscosity is a property of
(a) liquids only
(b) solids only
(c) solids and liquids only
(d) liquids and gases only.
ANSWER: (d)
Explanation: Viscosity comes into play when something starts to flow. Since only liquids and gasses can flow it is their property.
24. The force of viscosity is
(a) electromagnetic
(b) gravitational
(c) nuclear
(d) weak
ANSWER: (d)
Explanation: Viscosity comes into play when something starts to flow. Since only liquids and gasses can flow it is their property.
24. The force of viscosity is
(a) electromagnetic
(b) gravitational
(c) nuclear
(d) weak
ANSWER: (a)
Explanation: The force of viscosity is due to the contact of the molecules like in friction. So it is an electromagnetic force.
25. The viscous force acting between two layers of a liquid is given by F/A = η.dv/dz. This F/A may be called
(a) pressure
(b) longitudinal stress
(c) tangential stress
(d) volume stress
ANSWER: (a)
Explanation: The force of viscosity is due to the contact of the molecules like in friction. So it is an electromagnetic force.
25. The viscous force acting between two layers of a liquid is given by F/A = η.dv/dz. This F/A may be called
(a) pressure
(b) longitudinal stress
(c) tangential stress
(d) volume stress
ANSWER: (c)
Explanation: The force of viscosity acts between layers of the liquid and it acts along the surface. Thus F/A may be called tangential stress.
26. A raindrop falls near the surface of the earth with almost uniform velocity because
(a) its weight is negligible
(b) the force of surface tension balances its weight
(c) the force of viscosity of air balances its weight
(d) the drops are charged and atmospheric electric field balances its weight.
ANSWER: (c)
Explanation: When the raindrop starts falling the force of viscosity of air comes into action. This force is proportional to the velocity. Initially, this force is less than its weight. So there is a net downward force. Thus it accelerates. But a stage comes when due to increased velocity the force of viscosity is equal to its weight. So net downward force now becomes zero. Thus the raindrop keeps falling with a uniform velocity which is called the terminal velocity.
ANSWER: (c)
Explanation: The force of viscosity acts between layers of the liquid and it acts along the surface. Thus F/A may be called tangential stress.
26. A raindrop falls near the surface of the earth with almost uniform velocity because
(a) its weight is negligible
(b) the force of surface tension balances its weight
(c) the force of viscosity of air balances its weight
(d) the drops are charged and atmospheric electric field balances its weight.
ANSWER: (c)
Explanation: When the raindrop starts falling the force of viscosity of air comes into action. This force is proportional to the velocity. Initially, this force is less than its weight. So there is a net downward force. Thus it accelerates. But a stage comes when due to increased velocity the force of viscosity is equal to its weight. So net downward force now becomes zero. Thus the raindrop keeps falling with a uniform velocity which is called the terminal velocity.
27. A piece of wood is taken deep inside a long column of water and released. It will move up
(a) with a constant upward acceleration
(b) with a decreasing upward acceleration
(c) with a deceleration
(d) with a uniform velocity.
27. A piece of wood is taken deep inside a long column of water and released. It will move up
(a) with a constant upward acceleration
(b) with a decreasing upward acceleration
(c) with a deceleration
(d) with a uniform velocity.
ANSWER: (b)
Explanation: Just when the wood is released, the net force on it is B-W, {Buoyancy minus weight}upward. If the mass of the wood be m, then it has an initial acceleration a =(B-W)/m. But as the velocity increases the viscous force V acts downward. So the net force becomes B-W-V. Though B and W are constant, V increases with the velocity. Thus as the wood moves up, the net force and hence the acceleration decreases.
28. A solid sphere falls with a terminal velocity of 20 m/s in air. If it is allowed to fall in vacuum,
(a) terminal velocity will be 20 m/s
(b) terminal velocity will be less than 20 m/s
(c) terminal velocity will be more than 20 m/s
(d) there will be no terminal velocity.
ANSWER: (b)
Explanation: Just when the wood is released, the net force on it is B-W, {Buoyancy minus weight}upward. If the mass of the wood be m, then it has an initial acceleration a =(B-W)/m. But as the velocity increases the viscous force V acts downward. So the net force becomes B-W-V. Though B and W are constant, V increases with the velocity. Thus as the wood moves up, the net force and hence the acceleration decreases.
28. A solid sphere falls with a terminal velocity of 20 m/s in air. If it is allowed to fall in vacuum,
(a) terminal velocity will be 20 m/s
(b) terminal velocity will be less than 20 m/s
(c) terminal velocity will be more than 20 m/s
(d) there will be no terminal velocity.
ANSWER: (d)
Explanation: The term, terminal velocity is associated with a falling object in a fluid where the force of viscosity balances the weight and buoyancy force. Since there is no medium in the vacuum, there is no viscosity to oppose the motion. So the solid sphere will keep accelerating and there will be no terminal velocity.
29. A spherical ball is dropped in a long column of viscous liquid. the speed of the ball as a function of time may be best represented by the graph
(a) A
(b) B
(c) C
(d) D
Figure for Q-29
ANSWER: (d)
Explanation: The term, terminal velocity is associated with a falling object in a fluid where the force of viscosity balances the weight and buoyancy force. Since there is no medium in the vacuum, there is no viscosity to oppose the motion. So the solid sphere will keep accelerating and there will be no terminal velocity.
29. A spherical ball is dropped in a long column of viscous liquid. the speed of the ball as a function of time may be best represented by the graph
(a) A
(b) B
(c) C
(d) D
 |
| Figure for Q-29 |
ANSWER: (b)
Explanation: When the ball falls in the long column of viscous liquid, it starts with zero speed and with a certain acceleration. As time passes its acceleration decreases with time.
In the speed time graph, the speed is zero at t = 0 only in graph A and B. The slope of the speed-time graph gives the acceleration. The slope of A is constant, that means the acceleration is constant which is not the case here.
The slope of the graph B decreases with time i.e. the acceleration decreases with time. Thus the motion of the ball is best represented by graph B.
ANSWER: (b)
Explanation: When the ball falls in the long column of viscous liquid, it starts with zero speed and with a certain acceleration. As time passes its acceleration decreases with time.
In the speed time graph, the speed is zero at t = 0 only in graph A and B. The slope of the speed-time graph gives the acceleration. The slope of A is constant, that means the acceleration is constant which is not the case here.
The slope of the graph B decreases with time i.e. the acceleration decreases with time. Thus the motion of the ball is best represented by graph B.
Explanation: When the ball falls in the long column of viscous liquid, it starts with zero speed and with a certain acceleration. As time passes its acceleration decreases with time.
In the speed time graph, the speed is zero at t = 0 only in graph A and B. The slope of the speed-time graph gives the acceleration. The slope of A is constant, that means the acceleration is constant which is not the case here.
The slope of the graph B decreases with time i.e. the acceleration decreases with time. Thus the motion of the ball is best represented by graph B.
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
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CHAPTER- 14 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES- Q-1 TO Q-10
EXERCISES- Q-11 TO Q-20
EXERCISES- Q-21 TO Q-32
CHAPTER- 13 - Fluid Mechanics
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EXERCISES Q-1 TO Q-10
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EXERCISES Q-21 TO Q30
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EXERCISES- Q11 TO Q20
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EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
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CHAPTER- 13 - Fluid Mechanics
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Questions for Short Answers
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EXERCISES Q-1 TO Q-10
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CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
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CHAPTER- 11 - Gravitation
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CHAPTER- 7 - Circular Motion
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
The Forces-
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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