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SOME MECHANICAL PROPERTIES OF MATTER:--
QUESTIONS FOR SHORT ANSWER
1. The ratio stress/strain remains constant for small deformation of a metal wire. When the deformation is made larger, will this ratio increase or decrease?
ANSWER: For large deformations beyond the elastic limit, a small increase in the stress makes the strain larger. So the ratio Stress/Strain decreases.
2. When a block of mass M is suspended by a long wire of length L, the elastic potential energy stored in the wire is ½*stress*strain*volume. Show that it is equal to ½Mgl, where l is the extension. The loss in gravitational potential energy of the mass earth system is Mgl. Where does the remaining ½Mgl energy go?
1. The ratio stress/strain remains constant for small deformation of a metal wire. When the deformation is made larger, will this ratio increase or decrease?
ANSWER: For large deformations beyond the elastic limit, a small increase in the stress makes the strain larger. So the ratio Stress/Strain decreases.
2. When a block of mass M is suspended by a long wire of length L, the elastic potential energy stored in the wire is ½*stress*strain*volume. Show that it is equal to ½Mgl, where l is the extension. The loss in gravitational potential energy of the mass earth system is Mgl. Where does the remaining ½Mgl energy go?
ANSWER: For large deformations beyond the elastic limit, a small increase in the stress makes the strain larger. So the ratio Stress/Strain decreases.
2. When a block of mass M is suspended by a long wire of length L, the elastic potential energy stored in the wire is ½*stress*strain*volume. Show that it is equal to ½Mgl, where l is the extension. The loss in gravitational potential energy of the mass earth system is Mgl. Where does the remaining ½Mgl energy go?
ANSWER: Stress = Mg/A, {Where A is the cross-sectional area of the wire}
Strain = l/L.
Since the elastic potential energy
U = ½(Stress)*(Strain)*(Volume)
→U = ½*(Mg/A)*(l/L)*(AL) =½Mgl
The loss in gravitational potential energy will be Mgl only when the load is suddenly released from the initial position i.e. when elongation is zero. The elastic potential energy stored in the wire =½Mgl when the elongation is l, the rest of the gravitational potential energy Mgl-½Mgl =½Mgl is converted to the kinetic energy. At the time elongation is l, the mass M will have a certain speed say v. So ½Mv² =½Mgl
→v² = gl
→v =√(gl).
This point as a mean position the mass will be in a simple harmonic motion vertically. Slowly it will come to rest at the mean position by dissipating this energy in the form of heat.
3. When the skeleton of an elephant and the skeleton of a mouse are prepared in the same size, the bones of the elephant is shown thicker than those of the mouse. Explain why the bones of an elephant are thicker than proportionate. The bones are expected to withstand the stress due to the weight of the animal.
ANSWER: Stress = Mg/A, {Where A is the cross-sectional area of the wire}
Strain = l/L.
Since the elastic potential energy
U = ½(Stress)*(Strain)*(Volume)
→U = ½*(Mg/A)*(l/L)*(AL) =½Mgl
The loss in gravitational potential energy will be Mgl only when the load is suddenly released from the initial position i.e. when elongation is zero. The elastic potential energy stored in the wire =½Mgl when the elongation is l, the rest of the gravitational potential energy Mgl-½Mgl =½Mgl is converted to the kinetic energy. At the time elongation is l, the mass M will have a certain speed say v. So ½Mv² =½Mgl
→v² = gl
→v =√(gl).
This point as a mean position the mass will be in a simple harmonic motion vertically. Slowly it will come to rest at the mean position by dissipating this energy in the form of heat.
3. When the skeleton of an elephant and the skeleton of a mouse are prepared in the same size, the bones of the elephant is shown thicker than those of the mouse. Explain why the bones of an elephant are thicker than proportionate. The bones are expected to withstand the stress due to the weight of the animal.
Strain = l/L.
Since the elastic potential energy
U = ½(Stress)*(Strain)*(Volume)
→U = ½*(Mg/A)*(l/L)*(AL) =½Mgl
The loss in gravitational potential energy will be Mgl only when the load is suddenly released from the initial position i.e. when elongation is zero. The elastic potential energy stored in the wire =½Mgl when the elongation is l, the rest of the gravitational potential energy Mgl-½Mgl =½Mgl is converted to the kinetic energy. At the time elongation is l, the mass M will have a certain speed say v. So ½Mv² =½Mgl
→v² = gl
→v =√(gl).
This point as a mean position the mass will be in a simple harmonic motion vertically. Slowly it will come to rest at the mean position by dissipating this energy in the form of heat.
3. When the skeleton of an elephant and the skeleton of a mouse are prepared in the same size, the bones of the elephant is shown thicker than those of the mouse. Explain why the bones of an elephant are thicker than proportionate. The bones are expected to withstand the stress due to the weight of the animal.
ANSWER: The weight of an elephant is so much that, to bring the stress on the bones under permissible limit the cross-sectional areas of the bones are increased by nature because of stress =Force/area. So the bones of an elephant are thicker than proportionate.
4. The yield point of a typical solid is about 1%. Suppose you are lying horizontally and two persons are pulling your hands and two persons are pulling your legs along your own length. How much will be the increase in your length if the strain is 1%? Do you think your yield point is 1% or, much less than that?
ANSWER: The weight of an elephant is so much that, to bring the stress on the bones under permissible limit the cross-sectional areas of the bones are increased by nature because of stress =Force/area. So the bones of an elephant are thicker than proportionate.
4. The yield point of a typical solid is about 1%. Suppose you are lying horizontally and two persons are pulling your hands and two persons are pulling your legs along your own length. How much will be the increase in your length if the strain is 1%? Do you think your yield point is 1% or, much less than that?
ANSWER: If the height is H and the increase in height/length is h due to pulling then the strain =h/H. If it is 1% then
h/H = 1/100
→h = H/100
If the height H = 180 cm (say)
then the increase in length h = 1.80 cm.
Since our body has joints which are weaker than the bones so this increase in length will be located to the weakest joint. Any joint having 1.80 cm separation will be much damaging. So the yield point (or the bearable pain) due to pulling will be much less than 1%.
5. When rubber sheets are used in a shock absorber, what happens to the energy of vibration?
ANSWER: If the height is H and the increase in height/length is h due to pulling then the strain =h/H. If it is 1% then
h/H = 1/100
→h = H/100
If the height H = 180 cm (say)
then the increase in length h = 1.80 cm.
Since our body has joints which are weaker than the bones so this increase in length will be located to the weakest joint. Any joint having 1.80 cm separation will be much damaging. So the yield point (or the bearable pain) due to pulling will be much less than 1%.
5. When rubber sheets are used in a shock absorber, what happens to the energy of vibration?
h/H = 1/100
→h = H/100
If the height H = 180 cm (say)
then the increase in length h = 1.80 cm.
Since our body has joints which are weaker than the bones so this increase in length will be located to the weakest joint. Any joint having 1.80 cm separation will be much damaging. So the yield point (or the bearable pain) due to pulling will be much less than 1%.
5. When rubber sheets are used in a shock absorber, what happens to the energy of vibration?
ANSWER: The energy of shock due to vibration absorbed by the rubber sheets is converted to heat. It is also due to a property of rubber which is called "elastic hysteresis".
6. If a compressed spring is dissolved in acid what happens to the elastic potential energy of the spring?
ANSWER: The energy of shock due to vibration absorbed by the rubber sheets is converted to heat. It is also due to a property of rubber which is called "elastic hysteresis".
6. If a compressed spring is dissolved in acid what happens to the elastic potential energy of the spring?
6. If a compressed spring is dissolved in acid what happens to the elastic potential energy of the spring?
ANSWER: In a compressed spring the molecules are under stress due to the elastic potential energy stored in it. When the acid dissolves the spring molecules the P.E. stored in it is converted to K.E. and the new molecule it forms with acid ion has this extra K.E. Due to this the temperature of the acid increases slightly. It means that the elastic potential energy is converted to heat energy.
7. A steel blade placed gently on the surface of water floats on it. If the same blade is kept well inside the water, it sinks. Explain.
ANSWER: In a compressed spring the molecules are under stress due to the elastic potential energy stored in it. When the acid dissolves the spring molecules the P.E. stored in it is converted to K.E. and the new molecule it forms with acid ion has this extra K.E. Due to this the temperature of the acid increases slightly. It means that the elastic potential energy is converted to heat energy.
7. A steel blade placed gently on the surface of water floats on it. If the same blade is kept well inside the water, it sinks. Explain.
7. A steel blade placed gently on the surface of water floats on it. If the same blade is kept well inside the water, it sinks. Explain.
ANSWER: Due to the surface tension the surface of the water acts like a stretched membrane. The surface tension along the edges of the blade is more than the weight of the blade thus it floats.
Now the surface tension is a surface phenomenon, so when the same blade is kept well inside the water there is no surface tension to counter its weight. There is only the weight of the blade and buoyancy force on it. Since the density of the blade is more than the water, the weight is greater than the buoyancy force, thus it sinks.
8. When some wax is rubbed on a cloth, it becomes waterproof. Explain.
ANSWER: Due to the surface tension the surface of the water acts like a stretched membrane. The surface tension along the edges of the blade is more than the weight of the blade thus it floats.
Now the surface tension is a surface phenomenon, so when the same blade is kept well inside the water there is no surface tension to counter its weight. There is only the weight of the blade and buoyancy force on it. Since the density of the blade is more than the water, the weight is greater than the buoyancy force, thus it sinks.
8. When some wax is rubbed on a cloth, it becomes waterproof. Explain.
Now the surface tension is a surface phenomenon, so when the same blade is kept well inside the water there is no surface tension to counter its weight. There is only the weight of the blade and buoyancy force on it. Since the density of the blade is more than the water, the weight is greater than the buoyancy force, thus it sinks.
8. When some wax is rubbed on a cloth, it becomes waterproof. Explain.
ANSWER: The cloth is made of fine fibers and the spaces between the fine fibers act as capillary tubes. We know that when the angle of contact between the water and the side of the capillary tube is less than 90°, the water level in the tube rises. On the other hand, if the angle is more than 90°, the level of water in the tube goes down. In the case of water and the cloth, the angle of contact is nearly zero, so the water goes readily inside to wet it. The angle of contact between the water and the wax is about 107°. So when some wax is rubbed on the cloth, the water first comes in contact with the wax and the angle of contact being more than 90° it does not enter the cloth. So the cloth becomes waterproof.
9. The contact angle between pure water and pure silver is 90°. If a capillary tube made of silver is dipped at one end in pure water, will the water rise in the capillary?
ANSWER: The cloth is made of fine fibers and the spaces between the fine fibers act as capillary tubes. We know that when the angle of contact between the water and the side of the capillary tube is less than 90°, the water level in the tube rises. On the other hand, if the angle is more than 90°, the level of water in the tube goes down. In the case of water and the cloth, the angle of contact is nearly zero, so the water goes readily inside to wet it. The angle of contact between the water and the wax is about 107°. So when some wax is rubbed on the cloth, the water first comes in contact with the wax and the angle of contact being more than 90° it does not enter the cloth. So the cloth becomes waterproof.
9. The contact angle between pure water and pure silver is 90°. If a capillary tube made of silver is dipped at one end in pure water, will the water rise in the capillary?
9. The contact angle between pure water and pure silver is 90°. If a capillary tube made of silver is dipped at one end in pure water, will the water rise in the capillary?
ANSWER: No. The rise of liquid in a capillary tube is given as,
h = 2S*cosθ/(rρg)
Where S = Surface tension, θ = the angle of contact between the liquid and the wall of the tube, r = radius of the radius of the capillary tube, ρ = density of the liquid and g = acceleration due to gravity.
Here given that θ = 90°, hence cosθ = 0.
∴ h = 0.
So the water will not rise in the pure silver capillary tube.
10. It is said that a liquid rises or is depressed in a capillary due to the surface tension. If a liquid neither rises nor depresses in a capillary, can we conclude that the surface tension of the liquid is zero?
ANSWER: No. The rise of liquid in a capillary tube is given as,
h = 2S*cosθ/(rρg)
Where S = Surface tension, θ = the angle of contact between the liquid and the wall of the tube, r = radius of the radius of the capillary tube, ρ = density of the liquid and g = acceleration due to gravity.
Here given that θ = 90°, hence cosθ = 0.
∴ h = 0.
So the water will not rise in the pure silver capillary tube.
10. It is said that a liquid rises or is depressed in a capillary due to the surface tension. If a liquid neither rises nor depresses in a capillary, can we conclude that the surface tension of the liquid is zero?
h = 2S*cosθ/(rρg)
Where S = Surface tension, θ = the angle of contact between the liquid and the wall of the tube, r = radius of the radius of the capillary tube, ρ = density of the liquid and g = acceleration due to gravity.
Here given that θ = 90°, hence cosθ = 0.
∴ h = 0.
So the water will not rise in the pure silver capillary tube.
10. It is said that a liquid rises or is depressed in a capillary due to the surface tension. If a liquid neither rises nor depresses in a capillary, can we conclude that the surface tension of the liquid is zero?
ANSWER: No.
Because the rise of a liquid in a capillary tube depends on other factors also as we can see in the expression for height h,
h = 2S*cosθ/(rρg)
The rise h depends on S, θ, r and ρ. Since it is a capillary tube, r is small and it can not make h zero. Even the density of a liquid ρ will not be so high to make h zero. But the cosθ can be zero for θ =90°, thus making the rise h =0.
So if a liquid neither rises nor depresses in a capillary we can not conclude that the surface tension of the liquid is zero. The angle of contact between the liquid and the wall of the tube may be about 90°.
11. The contact angle between water and glass is 0°. When water is poured in a glass to the maximum of the capacity, the water surface is convex upward. The angle of contact on such a situation is more than 90°. Explain.
ANSWER: No.
Because the rise of a liquid in a capillary tube depends on other factors also as we can see in the expression for height h,
h = 2S*cosθ/(rρg)
The rise h depends on S, θ, r and ρ. Since it is a capillary tube, r is small and it can not make h zero. Even the density of a liquid ρ will not be so high to make h zero. But the cosθ can be zero for θ =90°, thus making the rise h =0.
So if a liquid neither rises nor depresses in a capillary we can not conclude that the surface tension of the liquid is zero. The angle of contact between the liquid and the wall of the tube may be about 90°.
11. The contact angle between water and glass is 0°. When water is poured in a glass to the maximum of the capacity, the water surface is convex upward. The angle of contact on such a situation is more than 90°. Explain.
Because the rise of a liquid in a capillary tube depends on other factors also as we can see in the expression for height h,
h = 2S*cosθ/(rρg)
The rise h depends on S, θ, r and ρ. Since it is a capillary tube, r is small and it can not make h zero. Even the density of a liquid ρ will not be so high to make h zero. But the cosθ can be zero for θ =90°, thus making the rise h =0.
So if a liquid neither rises nor depresses in a capillary we can not conclude that the surface tension of the liquid is zero. The angle of contact between the liquid and the wall of the tube may be about 90°.
11. The contact angle between water and glass is 0°. When water is poured in a glass to the maximum of the capacity, the water surface is convex upward. The angle of contact on such a situation is more than 90°. Explain.
ANSWER: Due to the surface tension, the surface of the liquid acts as a stretched membrane. So even if some more water than the rim level of the glass is poured into it, the stretched water surface which holds the glass rim due to adhesion becomes convex upward and the contact angle between the water and the glass becomes more than 90° in this situation.
The zero degrees angle of contact for water and the glass is for the situation when the water level is below the rim level of the glass. In some special situations, the angle of contact may change.
12. A uniform vertical tube of circular cross-section contains a liquid. The contact angle is 90°. Consider a diameter of the tube lying in the surface of the liquid. The surface to the right of this diameter pulls the surface on the left of it. What keeps the surface on the left in equilibrium?
ANSWER: Due to the surface tension, the surface of the liquid acts as a stretched membrane. So even if some more water than the rim level of the glass is poured into it, the stretched water surface which holds the glass rim due to adhesion becomes convex upward and the contact angle between the water and the glass becomes more than 90° in this situation.
The zero degrees angle of contact for water and the glass is for the situation when the water level is below the rim level of the glass. In some special situations, the angle of contact may change.
12. A uniform vertical tube of circular cross-section contains a liquid. The contact angle is 90°. Consider a diameter of the tube lying in the surface of the liquid. The surface to the right of this diameter pulls the surface on the left of it. What keeps the surface on the left in equilibrium?
The zero degrees angle of contact for water and the glass is for the situation when the water level is below the rim level of the glass. In some special situations, the angle of contact may change.
12. A uniform vertical tube of circular cross-section contains a liquid. The contact angle is 90°. Consider a diameter of the tube lying in the surface of the liquid. The surface to the right of this diameter pulls the surface on the left of it. What keeps the surface on the left in equilibrium?
ANSWER:
The figure for Q - 12
Let the diameter in consideration be AB as in the above figure. The surface to the right of AB pulls the surface on the left of AB (say by a total force = F) due to surface tension. The semicircular walls of the tube in contact with the surface to the left of AB also pulls it towards left with a net force F. Thus the semicircular surface ABD is pulled by two equal and opposite forces, making the net force on it zero. So it remains in equilibrium.
13. When a glass capillary tube is dipped at one end in water, water rise in the tube. The gravitational potential energy is thus increased. Is it a violation of conservation of energy?
ANSWER:
Let the diameter in consideration be AB as in the above figure. The surface to the right of AB pulls the surface on the left of AB (say by a total force = F) due to surface tension. The semicircular walls of the tube in contact with the surface to the left of AB also pulls it towards left with a net force F. Thus the semicircular surface ABD is pulled by two equal and opposite forces, making the net force on it zero. So it remains in equilibrium.
13. When a glass capillary tube is dipped at one end in water, water rise in the tube. The gravitational potential energy is thus increased. Is it a violation of conservation of energy?
 |
| The figure for Q - 12 |
13. When a glass capillary tube is dipped at one end in water, water rise in the tube. The gravitational potential energy is thus increased. Is it a violation of conservation of energy?
ANSWER: No. It is not a violation of the conservation of energy. It can be explained due to two reasons:-
1. The gravitational potential energy of the part of the water that rises in the tube increases but simultaneously the level of the rest of the water goes down depending on the open surface area of the water. So some of the increased P.E. is compensated by the decrease in this water level.
2. The work done by the vertical component of the net force due to surface tension around the meniscus is stored as the net increase in the gravitational potential energy of the water in the tube.
14. If a mosquito is dipped in water and released, it is not able to fly till it dry again. Explain.
ANSWER: No. It is not a violation of the conservation of energy. It can be explained due to two reasons:-
1. The gravitational potential energy of the part of the water that rises in the tube increases but simultaneously the level of the rest of the water goes down depending on the open surface area of the water. So some of the increased P.E. is compensated by the decrease in this water level.
2. The work done by the vertical component of the net force due to surface tension around the meniscus is stored as the net increase in the gravitational potential energy of the water in the tube.
14. If a mosquito is dipped in water and released, it is not able to fly till it dry again. Explain.
1. The gravitational potential energy of the part of the water that rises in the tube increases but simultaneously the level of the rest of the water goes down depending on the open surface area of the water. So some of the increased P.E. is compensated by the decrease in this water level.
2. The work done by the vertical component of the net force due to surface tension around the meniscus is stored as the net increase in the gravitational potential energy of the water in the tube.
14. If a mosquito is dipped in water and released, it is not able to fly till it dry again. Explain.
ANSWER: The wings are wet and stick together. To free the wings the mosquito needs to overcome the surface tension of the water which it finds unable to do. Without free wings, it is unable to fly. So it waits till the water evaporates and the wings are dry again.
15. The force of surface tension acts tangentially to the surface whereas the force due to air pressure acts perpendicularly on the surface. How is then the force due to excess pressure inside a bubble balanced by the force due to the surface tension?
ANSWER: The wings are wet and stick together. To free the wings the mosquito needs to overcome the surface tension of the water which it finds unable to do. Without free wings, it is unable to fly. So it waits till the water evaporates and the wings are dry again.
15. The force of surface tension acts tangentially to the surface whereas the force due to air pressure acts perpendicularly on the surface. How is then the force due to excess pressure inside a bubble balanced by the force due to the surface tension?
15. The force of surface tension acts tangentially to the surface whereas the force due to air pressure acts perpendicularly on the surface. How is then the force due to excess pressure inside a bubble balanced by the force due to the surface tension?
ANSWER: The surface of the bubble is curved. The forces perpendicular to the surface has different directions from point to point. The effect of the pressure inside (which is perpendicular to the surface) over an area is to produce a net force tangential to the surface that tries to increase the surface area of the bubble. While the force of the surface tension (acting also tangentially) tries to decrease the area of the bubble. Thus the pressure and the surface tension forces balance each other.
16. When the size of a soap bubble is increased by pushing more air in it, the surface area increases. Does it mean that the average separation between the surface molecules is increased?
ANSWER: The surface of the bubble is curved. The forces perpendicular to the surface has different directions from point to point. The effect of the pressure inside (which is perpendicular to the surface) over an area is to produce a net force tangential to the surface that tries to increase the surface area of the bubble. While the force of the surface tension (acting also tangentially) tries to decrease the area of the bubble. Thus the pressure and the surface tension forces balance each other.
16. When the size of a soap bubble is increased by pushing more air in it, the surface area increases. Does it mean that the average separation between the surface molecules is increased?
16. When the size of a soap bubble is increased by pushing more air in it, the surface area increases. Does it mean that the average separation between the surface molecules is increased?
ANSWER: By pushing more air in the bubble the work is done to bring more molecules from the liquid contained between the outer and inner surface of the bubble to these two surfaces. So the more surface area is due to the inclusion of more molecules not due to an increase in average separation between the surface molecules.
ANSWER: By pushing more air in the bubble the work is done to bring more molecules from the liquid contained between the outer and inner surface of the bubble to these two surfaces. So the more surface area is due to the inclusion of more molecules not due to an increase in average separation between the surface molecules.
17. The frictional force between solids operates even when they do not move with respect to each other. Do we have a viscous force acting between two layers even if there is no relative motion?
ANSWER: No. Though both frictional force and the viscous force oppose the relative motions, it is the difference between them that viscous force only comes into play when there is a relative motion between two layers.
18. Water near the bed of a deep river is quiet while that near the surface flows. Give reasons.
ANSWER: It is due to the property of a liquid which is called viscosity. The viscosity of a liquid opposes the relative motion between layers of the liquid. The layer of the water in contact with the river bed is still because the river bed has no movement. The layer above it moves with some speed due to the resistance of the lower layer. And each layer moves with some relative speed than its lower layer. Thus a velocity gradient is formed with zero at the bottom and maximum near the surface.
19. If the water in one flask and castor oil in other are violently shaken and kept on a table, which will come to rest earlier?
17. The frictional force between solids operates even when they do not move with respect to each other. Do we have a viscous force acting between two layers even if there is no relative motion?
ANSWER: No. Though both frictional force and the viscous force oppose the relative motions, it is the difference between them that viscous force only comes into play when there is a relative motion between two layers.
18. Water near the bed of a deep river is quiet while that near the surface flows. Give reasons.
ANSWER: It is due to the property of a liquid which is called viscosity. The viscosity of a liquid opposes the relative motion between layers of the liquid. The layer of the water in contact with the river bed is still because the river bed has no movement. The layer above it moves with some speed due to the resistance of the lower layer. And each layer moves with some relative speed than its lower layer. Thus a velocity gradient is formed with zero at the bottom and maximum near the surface.
19. If the water in one flask and castor oil in other are violently shaken and kept on a table, which will come to rest earlier?
ANSWER: No. Though both frictional force and the viscous force oppose the relative motions, it is the difference between them that viscous force only comes into play when there is a relative motion between two layers.
18. Water near the bed of a deep river is quiet while that near the surface flows. Give reasons.
ANSWER: It is due to the property of a liquid which is called viscosity. The viscosity of a liquid opposes the relative motion between layers of the liquid. The layer of the water in contact with the river bed is still because the river bed has no movement. The layer above it moves with some speed due to the resistance of the lower layer. And each layer moves with some relative speed than its lower layer. Thus a velocity gradient is formed with zero at the bottom and maximum near the surface.
19. If the water in one flask and castor oil in other are violently shaken and kept on a table, which will come to rest earlier?
ANSWER: The castor oil has more viscosity than the water. It means the layers of castor oil will offer more resistance in the relative movement of the layers. So the castor oil will come to rest earlier.
ANSWER: The castor oil has more viscosity than the water. It means the layers of castor oil will offer more resistance in the relative movement of the layers. So the castor oil will come to rest earlier.
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CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 13 - Fluid Mechanics
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EXERCISES Q-1 TO Q-10
EXERCISES- Q11 TO Q20
EXERCISES Q-21 TO Q30
EXERCISES Q-31 TO Q35
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
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EXERCISES Q-1 TO Q-10
EXERCISES- Q11 TO Q20
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OBJECTIVE-I
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EXERCISES Q-1 TO Q-10
EXERCISES- Q11 TO Q20
EXERCISES Q-21 TO Q30
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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
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
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HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY
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Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
HC Verma's Concepts of Physics, Chapter-7, Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
HC Verma's Concepts of Physics, Chapter-6, Friction
Click here for → Questions for Short Answer
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)
For more practice on problems on friction solve these "New Questions on Friction" .
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .
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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion
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 → 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,Exercises(Q.No. 13 to 27)
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HC Verma's Concepts of Physics, Chapter-4, 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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