My Channel on YouTube → SimplePhysics with KK
For links to
other chapters - See bottom of the page
Or click here → kktutor.blogspot.com
GRAVITATION:--
EXERCISES- Q11_to_Q20
11. A tunnel is dug along a diameter of the earth. Find the force on a particle of mass m placed in the tunnel at a distance x from the center.
ANSWER: Let the Mass of the earth = M. For a particle placed in a tunnel at a distance x from the center it can be assumed to be on the surface of a sphere of radius x and inside a uniform shell of thickness R-x, where R is the radius of the earth. Since the gravitational field inside a uniform shell is zero, the particle of mass m will only experience the gravitational force from the spherical part of radius x. Assuming the same density, the mass of a sphere is directly proportional to the cube of its radius (because the volume of a sphere is directly proportional to the cube of the radius), Let its mass be M', then
M' = Mx³/R³
The gravitational force on a particle in a tunnel at a distance x from the center =GM'm/x²
=G(Mx³/R³)m/x²
=GMmx³/x²R³
=GMmx/R³
12. A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the center of the earth. The wall of the tunnel may be assumed to be frictionless. Find the force exerted by the wall on a particle of mass m when it is at a distance x from the center of the tunnel.
ANSWER: Since the tunnel is frictionless, the force (N) exerted by the wall on the particle of mass m will be perpendicular to the tunnel at that point. This force will be equal and opposite to the component of the gravitational force by the earth on the particle along the perpendicular at that point. Let the distance of the particle from the center of the earth CA = l. The gravitational force on the particle will be the force exerted by the mass of the earth within the spherical region of radius l. As we have seen in the previous problem the magnitude of this force will be F=GMml/R³ and its direction will be towards the center of the earth.
See the diagram below:-
The component of F perpendicular to the tunnel at that point
= F.cosα =N
→N =F*(BC/AC)
=F*R/2l
=(GMml/R³)*(R/2l)
=GMm/2R²
13. A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown in figure (11-E1). A particle of mass m' is placed on the line joining the two centers at a distance x from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if (a) r<x<2r, (b) 2r<x<2R and (c) x>2R.
ANSWER: Until x < 2R, the particle will be within the shell and hence it will not exert a force on the particle. It is the situation in (a) and (b).
(a) When r<x<2r
The distance of the particle m' from the center of the solid sphere
s =x-r
The resultant gravitational force on m' will be the force due to the solid sphere of radius s =x-r
This force = Gmm's/r³ (As derived in the solution of Q-11)
=Gmm'(x-r)/r³
(b) When 2r<x<2R
In this case, the particle m' is between the space of the solid sphere and the shell. The resultant gravitational force on the particle will be due to the solid sphere only and its magnitude will be
= Gmm'/s²
{Here s = x-r, the distance of the particle from the center of the solid sphere}
= GMM'/(x-r)²
(c) When x>2R
Now both the shell and the solid sphere will exert the gravitational force on the particle m'. The masses of the sphere and the shell will be assumed to be concentrated at their respective centers.
Force by the solid sphere:-
The distance of the particle from the center of the solid sphere
= s =x-r
Hence force = Gmm'/s² = Gmm'/(x-r)²
Force by the shell:-
The distance of the particle from the center of the shell
s' = x-R
Hence force = GMm'/s'²
=GMm'/(x-R)²
Since both of these forces are in the same direction (towards the line joining two centers) hence the magnitude of the resultant force will be added, i.e.
The resultant gravitational force on m'
= Gmm'/(x-r)² + GMm'/(x-R)²
14. A uniform metal sphere of radius a and mass M is surrounded by a thin uniform spherical shell of equal mass and radius 4a (figure 11-E2). The center of the shell falls on the surface of the inner sphere. Find the gravitational field at the points P₁ and P₂ shown in the figure.
ANSWER: The gravitational field at P₁
This point falls inside the hollow shell so the field by the shell at this point will be zero. The distance of the point from the center of the metal sphere = 4a-a+a =4a (As is clear from the given figure)
Hence the gravitational field at this point = GM/(4a)²
=GM/16a²
The gravitational field at P₂
Since this point is outside the shell, both of the objects will have gravitational fields at this point and along the same direction i.e. towards their centers. So their magnitudes can be added. To calculate the field we shall assume the masses of the objects to be concentrated at their respective centers.
The distance of the point from the shell's center = 4a+a =5a and the distance of the point from the metal sphere = 5a+a =6a.
Hence the magnitude of the resultant field
=GM/(5a)² + GM/(6a)²
=(1/25+1/36)GM/a²
={(25+36)/900}GM/a²
=61GM/900a²
15. A thin spherical shell having uniform density is cut in two parts by a plane and kept separated as shown in figure (11-E3). The point A is the center of the plane section of the first part and B is the center of the plane section of the second part. Show that the gravitational field at A due to the first is equal in magnitude to the gravitational field at B due to the second part.
ANSWER: Assume the field at A due to the first part is F and the field at B due to the second part is F'. When both of the parts are joined at their planes to form the thin uniform spherical shell, the points A and B coincide and the field at this point due to both of the parts (The whole shell) is zero. Hence,
F + F' = 0
→F = -F'
So the magnitudes of the gravitational fields at A and B due to the first part and the second part respectively are equal (and opposite in direction).
16. Two small bodies of masses 2.00 kg and 4.00 kg are kept at rest at a separation of 2.0 m. Where should a particle of mass 0.10 kg be placed to experience no net gravitational force from these bodies? The particle is placed at this point. What is the gravitational potential energy of the system of three particles with usual reference level?
ANSWER: Obviously such point will be on the line joining the centers of the two bodies so that the force by each body would be equal and opposite. Let this point be at x meters from the 2.0 kg body. The gravitational field due to the 2.0 kg body at this point
= G*2/x² N/kg
= 2G/x² N/kg
The gravitational field due to the 4.00 kg body at the same point
= G*4/(2-x)² N/kg
=4G/(2-x)²
These two magnitudes will be equal, so
2G/x² =4G/(2-x)²
→(2-x)² = 2x²
→2-x = √2x
→(√2+1)x = 2
→(1.41+1)x = 2
→x = 2/2.41 =0.83 m
Due to zero net gravitational field at this point, the body placed at this point will experience no net gravitational force.
The mass of the third particle =0.10 kg. The gravitational potential energy of the three particle system will be calculated by calculating the P.E. of each pair and then adding them.
P.E. of 2.0 kg and 4.0 kg pair = -G*2*4/2 J
=-4G J
P.E. of 2.0 kg and 0.10 kg pair = -G*2*0.10/0.83 J
=-0.24G J
P.E. of 4.0 kg and 0.10 kg pair = -G*4*0.10/(2-0.83) J
=-0.34G J
The gravitational potential energy of the system
=-4G +(-0.24)G+ (-0.34)G = 4.58G J
=-4.58*6.67*10⁻¹¹ J
=-30.55*10⁻¹¹ J
=-3.06x10⁻¹⁰ J
17. Three particles of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.
ANSWER: First, let us find the gravitational potential energy of the three particle system initially, which is
P = -3Gm²/a
(Adding the P.E. of each pair)
The gravitational potential energy of the system when the side of the equilateral triangle is 2a will be
P' = -3Gm²/2a
Hence the work to be done on the system to increase the sides of the triangle to 2a = P' - P
= -3Gm²/2a -(-3Gm²/a)
= (3Gm²/a)*(1-½)
= 3Gm²/2a
18. A particle of mass 100 g is kept on a surface of a uniform sphere of mass 10 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle away from the sphere.
ANSWER: Mass of the sphere M = 10 kg
Mass of the particle m = 100 g = 0.10 kg
The radius of the sphere r = 10 cm = 0.10 m
Initial P.E. of the particle = -GMm/r
=-6.67x10⁻¹¹*10*0.10/0.10 J
=-6.67x10⁻¹⁰ J
When the particle is taken away, the final P.E. = 0
Hence the required work to be done = Final P.E. - Initial P.E.
= 0 - (-6.67x10⁻¹⁰) J
=6.67x10⁻¹⁰ J
19. The gravitational field in a region is given by E=(5 N/kg)i+ (12 N/kg)j. (a) Find the magnitude of the gravitational force acting on a particle of mass 2 kg placed at the origin. (b) Find the potential at the points (12m, 0) and (0, 5m) if the potential at the origin is taken to be zero. (c) Find the change in the gravitational potential energy if a particle of mass 2 kg is taken from the origin to the point (12m, 5m). (d) Find the change in potential energy if the particle is taken from (12m, 0) to (0, 5m).
ANSWER: (a) The gravitational force on the particle of mass 2 kg
= mE
= 2*[(5 N/kg)i+ (12 N/kg)j]
= (10 N/kg)i+ (24 N/kg)j
The magnitude of the gravitational force =√{(10)²+(24)²} N
= √676 N
= 26 N
(b) Since the gravitational field in this region is not dependent on r,
the potential at any point with respect to the origin can be given as
V = -E.r
So, the potential at the point (12 m, 0)
= -{(5 N/kg)i+ (12 N/kg)j}.{(12 m)i}
= -5*12 N-m/kg
= -60 J/kg
Similarly the potential at the point (0, 5 m)
= -{(5 N/kg)i+ (12 N/kg)j}.{(5 m)j}
= (-12 N/kg)*(5 m)
= -60 J/kg
(c) The gravitational potential at the point (12 m, 5m)
= -{(5 N/kg)i+ (12 N/kg)j}.{(12 m)i+(5 m)j}
= -(5*12+12*5) J/kg
= -120 J/kg
So the gravitational potential energy for 2.0 kg particle at this point
= -120 J/kg * 2.0 kg = -240 J
The gravitational potential energy at the origin = 0
The change in potential energy = -240 J -0
= -240 J
(d) Since the gravitational potential at both the points (12 m, 0) and (0, 5 m) is the same and equal to -60 J/kg. So the gravitational potential energy of the particle at these two points will also be the same = -60*2 =-120 J. So the change in the gravitational potential energy
= -120-(-120) =0 (Zero).
20. The gravitational potential in a region is given by V=(20 N/kg)(x+y). (a) Show that the equation is dimensionally correct. (b) Find the gravitational field at the point (x,y). Leave your answer in terms of the unit vectors i, j, k. (c) Calculate the magnitude of the gravitational force on a particle of mass 500 g placed at the origin.
ANSWER: (a) The Dimension of the V = {[MLT⁻²]/[M]}*[L]
=[L²T⁻²]
The unit of the gravitational potential = J/kg
The dimension =[ML²T⁻²]/[M]
=[L²T⁻²]
Hence the equation is dimenionally correct.
(b) The gravitational potential is given
V=(20 N/kg)(x+y)
So the gravitational field in the x direction is
Eₓ = -∂V/∂x =-20 N/kg
Similarly, the gravitational field in the y-direction is
Eᵧ = -∂V/∂y = -20 N/kg
Hence the gravitational field at the point (x,y)
= E = iEₓ+jEᵧ
= (-20 N/kg)i+(-20 N/kg)j
= -20(i+j) N/Kg
(c) Mass of the particle = 500 g = 0.50 kg
The gravitational field at a point (x,y) is E =-20(i+j) N/Kg which is independent of x or y. Hence the gravitational field at (0,0) is also =-20(i+j) N/Kg
So, the gravitational force on the 500 g particle = mE
= 0.50 kg*{-20(i+j) N/Kg}
= -10i -10j N
Hence its magnitude = √{(-10)² + (-10)²} =√200 =10√2 N
ANSWER: Let the Mass of the earth = M. For a particle placed in a tunnel at a distance x from the center it can be assumed to be on the surface of a sphere of radius x and inside a uniform shell of thickness R-x, where R is the radius of the earth. Since the gravitational field inside a uniform shell is zero, the particle of mass m will only experience the gravitational force from the spherical part of radius x. Assuming the same density, the mass of a sphere is directly proportional to the cube of its radius (because the volume of a sphere is directly proportional to the cube of the radius), Let its mass be M', then
M' = Mx³/R³
 |
| Diagram for Q-11 |
The gravitational force on a particle in a tunnel at a distance x from the center =GM'm/x²
=G(Mx³/R³)m/x²
=GMmx³/x²R³
=GMmx/R³
12. A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the center of the earth. The wall of the tunnel may be assumed to be frictionless. Find the force exerted by the wall on a particle of mass m when it is at a distance x from the center of the tunnel.
ANSWER: Since the tunnel is frictionless, the force (N) exerted by the wall on the particle of mass m will be perpendicular to the tunnel at that point. This force will be equal and opposite to the component of the gravitational force by the earth on the particle along the perpendicular at that point. Let the distance of the particle from the center of the earth CA = l. The gravitational force on the particle will be the force exerted by the mass of the earth within the spherical region of radius l. As we have seen in the previous problem the magnitude of this force will be F=GMml/R³ and its direction will be towards the center of the earth.
See the diagram below:-
 |
| Diagram for Q-12 |
The component of F perpendicular to the tunnel at that point
= F.cosα =N
→N =F*(BC/AC)
=F*R/2l
=(GMml/R³)*(R/2l)
=GMm/2R²
13. A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown in figure (11-E1). A particle of mass m' is placed on the line joining the two centers at a distance x from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if (a) r<x<2r, (b) 2r<x<2R and (c) x>2R.
 |
| Figure for Q-13 |
ANSWER: Until x < 2R, the particle will be within the shell and hence it will not exert a force on the particle. It is the situation in (a) and (b).
(a) When r<x<2r
The distance of the particle m' from the center of the solid sphere
s =x-r
The resultant gravitational force on m' will be the force due to the solid sphere of radius s =x-r
This force = Gmm's/r³ (As derived in the solution of Q-11)
=Gmm'(x-r)/r³
(b) When 2r<x<2R
In this case, the particle m' is between the space of the solid sphere and the shell. The resultant gravitational force on the particle will be due to the solid sphere only and its magnitude will be
= Gmm'/s²
{Here s = x-r, the distance of the particle from the center of the solid sphere}
= GMM'/(x-r)²
(c) When x>2R
Now both the shell and the solid sphere will exert the gravitational force on the particle m'. The masses of the sphere and the shell will be assumed to be concentrated at their respective centers.
Force by the solid sphere:-
The distance of the particle from the center of the solid sphere
= s =x-r
Hence force = Gmm'/s² = Gmm'/(x-r)²
Force by the shell:-
The distance of the particle from the center of the shell
s' = x-R
Hence force = GMm'/s'²
=GMm'/(x-R)²
Since both of these forces are in the same direction (towards the line joining two centers) hence the magnitude of the resultant force will be added, i.e.
The resultant gravitational force on m'
= Gmm'/(x-r)² + GMm'/(x-R)²
14. A uniform metal sphere of radius a and mass M is surrounded by a thin uniform spherical shell of equal mass and radius 4a (figure 11-E2). The center of the shell falls on the surface of the inner sphere. Find the gravitational field at the points P₁ and P₂ shown in the figure.
 |
| Figure for Q-14 |
ANSWER: The gravitational field at P₁
This point falls inside the hollow shell so the field by the shell at this point will be zero. The distance of the point from the center of the metal sphere = 4a-a+a =4a (As is clear from the given figure)
Hence the gravitational field at this point = GM/(4a)²
=GM/16a²
The gravitational field at P₂
Since this point is outside the shell, both of the objects will have gravitational fields at this point and along the same direction i.e. towards their centers. So their magnitudes can be added. To calculate the field we shall assume the masses of the objects to be concentrated at their respective centers.
The distance of the point from the shell's center = 4a+a =5a and the distance of the point from the metal sphere = 5a+a =6a.
Hence the magnitude of the resultant field
=GM/(5a)² + GM/(6a)²
=(1/25+1/36)GM/a²
={(25+36)/900}GM/a²
=61GM/900a²
15. A thin spherical shell having uniform density is cut in two parts by a plane and kept separated as shown in figure (11-E3). The point A is the center of the plane section of the first part and B is the center of the plane section of the second part. Show that the gravitational field at A due to the first is equal in magnitude to the gravitational field at B due to the second part.
 |
| Figure for Q-15 |
F + F' = 0
→F = -F'
So the magnitudes of the gravitational fields at A and B due to the first part and the second part respectively are equal (and opposite in direction).
16. Two small bodies of masses 2.00 kg and 4.00 kg are kept at rest at a separation of 2.0 m. Where should a particle of mass 0.10 kg be placed to experience no net gravitational force from these bodies? The particle is placed at this point. What is the gravitational potential energy of the system of three particles with usual reference level?
ANSWER: Obviously such point will be on the line joining the centers of the two bodies so that the force by each body would be equal and opposite. Let this point be at x meters from the 2.0 kg body. The gravitational field due to the 2.0 kg body at this point
= G*2/x² N/kg
= 2G/x² N/kg
The gravitational field due to the 4.00 kg body at the same point
= G*4/(2-x)² N/kg
=4G/(2-x)²
These two magnitudes will be equal, so
2G/x² =4G/(2-x)²
→(2-x)² = 2x²
→2-x = √2x
→(√2+1)x = 2
→(1.41+1)x = 2
→x = 2/2.41 =0.83 m
Due to zero net gravitational field at this point, the body placed at this point will experience no net gravitational force.
The mass of the third particle =0.10 kg. The gravitational potential energy of the three particle system will be calculated by calculating the P.E. of each pair and then adding them.
P.E. of 2.0 kg and 4.0 kg pair = -G*2*4/2 J
=-4G J
P.E. of 2.0 kg and 0.10 kg pair = -G*2*0.10/0.83 J
=-0.24G J
P.E. of 4.0 kg and 0.10 kg pair = -G*4*0.10/(2-0.83) J
=-0.34G J
The gravitational potential energy of the system
=-4G +(-0.24)G+ (-0.34)G = 4.58G J
=-4.58*6.67*10⁻¹¹ J
=-30.55*10⁻¹¹ J
=-3.06x10⁻¹⁰ J
17. Three particles of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.
ANSWER: First, let us find the gravitational potential energy of the three particle system initially, which is
P = -3Gm²/a
(Adding the P.E. of each pair)
The gravitational potential energy of the system when the side of the equilateral triangle is 2a will be
P' = -3Gm²/2a
Hence the work to be done on the system to increase the sides of the triangle to 2a = P' - P
= -3Gm²/2a -(-3Gm²/a)
= (3Gm²/a)*(1-½)
= 3Gm²/2a
18. A particle of mass 100 g is kept on a surface of a uniform sphere of mass 10 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle away from the sphere.
ANSWER: Mass of the sphere M = 10 kg
Mass of the particle m = 100 g = 0.10 kg
The radius of the sphere r = 10 cm = 0.10 m
Initial P.E. of the particle = -GMm/r
=-6.67x10⁻¹¹*10*0.10/0.10 J
=-6.67x10⁻¹⁰ J
When the particle is taken away, the final P.E. = 0
Hence the required work to be done = Final P.E. - Initial P.E.
= 0 - (-6.67x10⁻¹⁰) J
=6.67x10⁻¹⁰ J
19. The gravitational field in a region is given by E=(5 N/kg)i+ (12 N/kg)j. (a) Find the magnitude of the gravitational force acting on a particle of mass 2 kg placed at the origin. (b) Find the potential at the points (12m, 0) and (0, 5m) if the potential at the origin is taken to be zero. (c) Find the change in the gravitational potential energy if a particle of mass 2 kg is taken from the origin to the point (12m, 5m). (d) Find the change in potential energy if the particle is taken from (12m, 0) to (0, 5m).
ANSWER: (a) The gravitational force on the particle of mass 2 kg
= mE
= 2*[(5 N/kg)i+ (12 N/kg)j]
= (10 N/kg)i+ (24 N/kg)j
The magnitude of the gravitational force =√{(10)²+(24)²} N
= √676 N
= 26 N
(b) Since the gravitational field in this region is not dependent on r,
the potential at any point with respect to the origin can be given as
V = -E.r
So, the potential at the point (12 m, 0)
= -{(5 N/kg)i+ (12 N/kg)j}.{(12 m)i}
= -5*12 N-m/kg
= -60 J/kg
Similarly the potential at the point (0, 5 m)
= -{(5 N/kg)i+ (12 N/kg)j}.{(5 m)j}
= (-12 N/kg)*(5 m)
= -60 J/kg
(c) The gravitational potential at the point (12 m, 5m)
= -{(5 N/kg)i+ (12 N/kg)j}.{(12 m)i+(5 m)j}
= -(5*12+12*5) J/kg
= -120 J/kg
So the gravitational potential energy for 2.0 kg particle at this point
= -120 J/kg * 2.0 kg = -240 J
The gravitational potential energy at the origin = 0
The change in potential energy = -240 J -0
= -240 J
(d) Since the gravitational potential at both the points (12 m, 0) and (0, 5 m) is the same and equal to -60 J/kg. So the gravitational potential energy of the particle at these two points will also be the same = -60*2 =-120 J. So the change in the gravitational potential energy
= -120-(-120) =0 (Zero).
20. The gravitational potential in a region is given by V=(20 N/kg)(x+y). (a) Show that the equation is dimensionally correct. (b) Find the gravitational field at the point (x,y). Leave your answer in terms of the unit vectors i, j, k. (c) Calculate the magnitude of the gravitational force on a particle of mass 500 g placed at the origin.
ANSWER: (a) The Dimension of the V = {[MLT⁻²]/[M]}*[L]
=[L²T⁻²]
The unit of the gravitational potential = J/kg
The dimension =[ML²T⁻²]/[M]
=[L²T⁻²]
Hence the equation is dimenionally correct.
(b) The gravitational potential is given
V=(20 N/kg)(x+y)
So the gravitational field in the x direction is
Eₓ = -∂V/∂x =-20 N/kg
Similarly, the gravitational field in the y-direction is
Eᵧ = -∂V/∂y = -20 N/kg
Hence the gravitational field at the point (x,y)
= E = iEₓ+jEᵧ
= (-20 N/kg)i+(-20 N/kg)j
= -20(i+j) N/Kg
(c) Mass of the particle = 500 g = 0.50 kg
The gravitational field at a point (x,y) is E =-20(i+j) N/Kg which is independent of x or y. Hence the gravitational field at (0,0) is also =-20(i+j) N/Kg
So, the gravitational force on the 500 g particle = mE
= 0.50 kg*{-20(i+j) N/Kg}
= -10i -10j N
Hence its magnitude = √{(-10)² + (-10)²} =√200 =10√2 N
===<<<O>>>===
Links to the chapters -
ALL LINKS
CHAPTER- 11 - Gravitation
Questions for Short Answers
OBJECTIVE - I
OBJECTIVE - II
EXERCISES - Q_1 to Q_10
EXERCISES -Q-11 TO Q-20
CHAPTER- 10 - Rotational Mechanics
Questions for Short Answers
OBJECTIVE - I
OBJECTIVE - II
EXERCISES - Q-1 TO Q-15
EXERCISES - Q-16 TO Q-30
EXERCISES Q-31 TO Q-45
EXERCISES Q-46 TO Q-60
EXERCISES Q-61 TO Q-75
EXERCISES Q-76 TO Q-86
CHAPTER- 11 - Gravitation
Questions for Short Answers
OBJECTIVE - II
EXERCISES - Q_1 to Q_10
EXERCISES -Q-11 TO Q-20
CHAPTER- 10 - Rotational Mechanics
Questions for Short Answers
OBJECTIVE - I
OBJECTIVE - II
EXERCISES - Q-1 TO Q-15
EXERCISES - Q-16 TO Q-30
EXERCISES Q-31 TO Q-45
EXERCISES Q-61 TO Q-75
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY
Click here for → Question for Short Answers
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 → Exercises (31-42)
Click here for → Exercise(43-54)
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 → 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 → Friction OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → Friction - OBJECTIVE-II
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" .
---------------------------------------------------------------------------------
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→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
-------------------------------------------------------------------------------
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
--------------------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------
No comments:
Post a Comment