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- Q1_to_Q10
1. Two spherical balls of mass 10 kg each are placed 10 cm apart. Find the gravitational force of attraction between them.
ANSWER: Mass of each ball, m = 10 kg
The distance between them, r = 10 cm = 0.10 m
The universal constant of gravitation, G = 6.67x10⁻¹¹ N-m²/kg²
From the universal law of gravitation the gravitational force of attraction between the balls, F = Gm.m/r²
= 6.67x10⁻¹¹ *10*10/(0.10)² N
= 6.67x10⁻¹¹ *10⁴ N
= 6.67x10⁻⁷ N
2. Four particles having masses m, 2m, 3m and 4m are placed at four corners of a square edge a. Find the gravitational force acting on a particle of mass m placed at the center.
ANSWER: The distance of the particle at the center from each corner is r = a/√2. Let the gravitational force on the central particle by the corner mass m = F then force by 2m particle = 2F, by 3m particle = 3F and by the 4m particle = 4F. (Because gravitational force is proportional to the mass). See the diagram below.
F and 3F are collinear but opposite in direction, hence net effect of these two = 3F-F = 2F towards the 3m particle (Point C). Similarly, 2F and 4F are just opposite in direction hence their net effect = 4F-2F = 2F towards the 4m particle (Point D).
So we can replace these four forces with two mutually perpendicular forces of magnitude 2F each as in the figure. Their resultant R is given by, R² = (2F)²+(2F)² = 8F²
→R =2√2F = 2√2*G*m.m/r² (Since F is the force between m and m)
→R = 2√2Gm²/(a/√2)² = 4√2Gm²/a²
And its direction will be along the angle bisector of the line joining 3m and 4m particles to the center.
(One point to be noted that we have placed the particles with increasing mass in cyclic order which is not mentioned in the problem. So there may be other configurations of the particles. For example 3m at B and 2m at C. Then the magnitude will change with R being resultant of F and F ie √2F. Similarly for other configurations or anticlockwise settings magnitudes and directions of the resultant forces on the particle at the center will change.)
3. Three equal masses m are placed at the three corners of an equilateral triangle of side a. Find the force exerted by this system on another particle of mass m placed at (a) the mid-point of a side, (b) at the center of the triangle.
ANSWER: See the figure (a). Forces on the particle by the masses at the ends of the side are equal and opposite each P, so they cancel out each other. So the force by the system on the particle is equal to the force by the particle at vertex A. Distance between them AD = r =a.sin60° =√3a/2.
So the force F = Gmm/r² =Gm²/(√3a/2)² =4Gm²/3a²
(b)
The center of an equilateral triangle is equidistant from each of the vertices and the angle between any two lines joining the center to the vertices are equal. Hence the three forces on the particle at the center will have equal magnitudes say F and the angle between any of the two = 120°. These three will cancel out each other and their resultant force will be zero.
(Resolve forces along any force Rx = F-2F.cos60° = 0
Resolve along perpendicular direction Ry = F.sin60°-F.sin60° = 0.
Hence resultant force R = 0)
4. Three uniform spheres each having a mass M and radius a are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any of the spheres due to the other two.
ANSWER: Consider the diagram below. We will calculate the force on sphere A due to B and C.
Due to similarity the magnitude of the force by B or C on the sphere A will be the same, say F.
Distance between any two spheres, r = 2a, hence
F = G.M.M/(2a)²
=GM²/4a²
So each of the sphere B and C exert a force of magnitude F on A and the angle between these forces = 60° (Because the lines joining the centers of the spheres make an equilateral triangle).
The resultant force of these two forces = √(F²+F²+2.F.F.cos60°)
=√(3F²) =√3F =√3GM²/4a²
(Direction of the resultant along the angle bisector of the forces F and F)
5. Four particles of equal masses M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.
ANSWER: Let the four masses A, B, C and D (Each of mass M) move along a circle of radius R and center at O. Assuming the system symmetrical, the masses are placed at corners of a square having its edge = √2R. See the diagram below.
Due to symmetry the speed of each mass will be the same. Let us consider the mass A. The gravitational force on A due to each of B and D = F = GM²/(√2R) =GM²/2R² {Along AB and AD each}
The gravitational force on A due to C = F' =GM²/4R² {Along AC}
The magnitude of the Resultant force of F and F = √2F
=√2GM²/2R² {Along AC}
So the resultant of gravitational forces on A due to B, C and D are,
R =√2F+F' {Along AC}
=√2GM²/2R²+GM²/4R²
={(2√2+1)/4}(GM²/R²)
The direction of this net force on A is towards the center of the circle which is also the center of the mass of the system. Let the speed of A be v, then it will have a centrifugal force P =Mv²/R
For the stability of the system P must be equal to the gravitational pull on A, hence,
Mv²/R = {(2√2+1)/4}(GM²/R²)
→v² = {(2√2+1)/4}(GM/R)
→v = √[{(2√2+1)/4}(GM/R)]
6. Find the acceleration due to the gravity of the moon at a point 1000 km above the moon's surface. The mass of the moon is 7.4x10²² kg and its radius is 1740 km.
ANSWER: The point in consideration is outside the moon's surface. The mass of the moon can be considered to be concentrated at the center. The distance of the point from the center,
h = 1000+1740 =2740 km =2.74x10⁶ m
Let a mass m be placed at the point and the acceleration due to gravity there be g. Then weight of the mass m, W=mg
But the gravitational force on the mass due to the moon,
F = GMm/h²
Bothe the forces will be the same,
W = F
mg = GMm/h²
g = GM/h²
{G=6.67x10⁻¹¹ N-m²/kg², M=7.4x10²² kg, h=2.74x10⁶ m}
→g = 6.67x10⁻¹¹*7.4x10²²/(2.74)²x10¹² m/s²
→g = 6.5x10²²⁻¹¹⁻¹² =6.5x10⁻¹ m/s²
→g = 0.65 m/s²
7. Two small bodies of masses 10 kg and 20 kg are kept at a distance 1.0 m apart and released. Assuming that only mutual gravitational forces are acting, find the speeds of the particles when the separation decreases to 0.5 m.
ANSWER: m = 10 kg, m' = 20 kg =2m
Let the final speed of m = v and that of m' = v'.
Taking the two body as a system, the gravitational force on each other is the internal force. Initially the system is at rest, so its linear momentum is zero. There is no external force hence the linear momentum will be conserved. Therefore the final linear momentum will also be zero. Due to the final velocities,
mv+m'v' = 0
→v = -m'v'/m = -20.v'/10 =-2v'
The negative sign shows that direction of the speeds are opposite.
Now the sum of P.E. and K.E. (Total enrgy) of the system will also be conserved. Initial Total energy = P.E.+K.E. = P.E.+ 0
= -Gmm'/r (r = 1.0 m)
= -Gm*2m
=-2Gm²
Final Total energy = P.E.+K.E. (Now r = 0.5 m)
=-Gmm'/0.5 +½mv² +½m'v'²
=-2Gm*2m+½m(2v')²+½*2m*v'² =-4Gm²+3mv'²
Equating Initial and final total energy
-4Gm²+3mv'² =-2Gm²
→3mv'² =2Gm²
→v'² =2Gm/3 =2*6.67x10⁻¹¹*10/3 =44.47x10⁻¹¹=4.447x10⁻¹⁰
→v' = 2.10x10⁻⁵ m/s
And v = 2v' = 4.20x10⁻⁵ m/s
8. A semi-circular wire has a length L and mass M. A particle of mass m is placed at the center of the circle. Find the gravitational attraction on the particle due to the wire.
ANSWER: Let the radius of the semi-circular wire be r. Assuming its center at the origin and its ends touching the X-axis as shown in the diagram, consider a radius at angle θ from the X-axis. For an increament in the angle by dθ we have a length of rdθ on the wire. The inceamental mass of this element dM = Mr.dθ/L
But L = πr
→r =L/π
So, dM = MLdθ/πL = Mdθ/π
Now the force by dM on a particle of mass m at the center,
dF = Gm.dM/r² =Gm(Mdθ/π)/(L²/π²) = πGMmdθ/L²
This force is directed towards the element dM.
If i and j be the unit vectors along X and Y axes respectively, we can write this force as,
dF = πGMm/L²(cosθ.i + sinθ.j)dθ
Integrating it from θ = 0 to θ = π, we can find the total gravitational force on the particle with mass m at the center.
F = ∫dF
= ∫πGMm/L²(cosθ.i + sinθ.j)dθ
=πGMm/L²*[sinθ.i -cosθ.j] {putting the limits 0 and π},
=πGMm/L²*[sinπ.i- cosπ.j-sin0.i+cos0.j]
=πGMm/L²*[0.i-(-1)j-0.i+1.j]
=πGMm/L²*[2.j]
=2πGMm/L²*j
(It has no component in X-direction)
So the gravitational force at mass m at the center of the semi-circular wire is 2πGMm/L² in magnitude and directed towards the midpoint of the ring.
9. Derive an expression for the gravitational field due to a uniform rod of length L and mass M at a point on its perpendicular bisector at a distance d from the center.
ANSWER: Mass per unit length of the rod = M/L
Let the rod be placed on the X-axis with its mid point on the origin. Cosider an infinitesimal length dx of the rod at distance x from the origin (Mid-point). It’s mass, dM = Mdx/L.
10. Two concentric spherical shells have masses M₁ and M₂ and radii R₁, R₂ (R₁<R₂). What is the force exerted by this system on a particle of mass m if it is placed at a distance (R₁+R₂)/2 from the center?
ANSWER: Since the particle is placed at (R₁+R₂)/2 distance from the center, it is outside the inner shell but inside the outer shell. The force on the particle by the outer shell = 0.
The force on the particle by the inner shell
= GM₁m/{(R₁+R₂)²/4}
= 4GM₁m/(R₁+R₂)²
And it is directed towards the center.
ANSWER: Mass of each ball, m = 10 kg
The distance between them, r = 10 cm = 0.10 m
The universal constant of gravitation, G = 6.67x10⁻¹¹ N-m²/kg²
From the universal law of gravitation the gravitational force of attraction between the balls, F = Gm.m/r²
= 6.67x10⁻¹¹ *10*10/(0.10)² N
= 6.67x10⁻¹¹ *10⁴ N
= 6.67x10⁻⁷ N
2. Four particles having masses m, 2m, 3m and 4m are placed at four corners of a square edge a. Find the gravitational force acting on a particle of mass m placed at the center.
ANSWER: The distance of the particle at the center from each corner is r = a/√2. Let the gravitational force on the central particle by the corner mass m = F then force by 2m particle = 2F, by 3m particle = 3F and by the 4m particle = 4F. (Because gravitational force is proportional to the mass). See the diagram below.
 |
| Diagram for the Q-2 |
F and 3F are collinear but opposite in direction, hence net effect of these two = 3F-F = 2F towards the 3m particle (Point C). Similarly, 2F and 4F are just opposite in direction hence their net effect = 4F-2F = 2F towards the 4m particle (Point D).
So we can replace these four forces with two mutually perpendicular forces of magnitude 2F each as in the figure. Their resultant R is given by, R² = (2F)²+(2F)² = 8F²
→R =2√2F = 2√2*G*m.m/r² (Since F is the force between m and m)
→R = 2√2Gm²/(a/√2)² = 4√2Gm²/a²
And its direction will be along the angle bisector of the line joining 3m and 4m particles to the center.
(One point to be noted that we have placed the particles with increasing mass in cyclic order which is not mentioned in the problem. So there may be other configurations of the particles. For example 3m at B and 2m at C. Then the magnitude will change with R being resultant of F and F ie √2F. Similarly for other configurations or anticlockwise settings magnitudes and directions of the resultant forces on the particle at the center will change.)
3. Three equal masses m are placed at the three corners of an equilateral triangle of side a. Find the force exerted by this system on another particle of mass m placed at (a) the mid-point of a side, (b) at the center of the triangle.
ANSWER: See the figure (a). Forces on the particle by the masses at the ends of the side are equal and opposite each P, so they cancel out each other. So the force by the system on the particle is equal to the force by the particle at vertex A. Distance between them AD = r =a.sin60° =√3a/2.
So the force F = Gmm/r² =Gm²/(√3a/2)² =4Gm²/3a²
 |
| Figure for Q-3 |
(b)
The center of an equilateral triangle is equidistant from each of the vertices and the angle between any two lines joining the center to the vertices are equal. Hence the three forces on the particle at the center will have equal magnitudes say F and the angle between any of the two = 120°. These three will cancel out each other and their resultant force will be zero.
(Resolve forces along any force Rx = F-2F.cos60° = 0
Resolve along perpendicular direction Ry = F.sin60°-F.sin60° = 0.
Hence resultant force R = 0)
4. Three uniform spheres each having a mass M and radius a are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any of the spheres due to the other two.
ANSWER: Consider the diagram below. We will calculate the force on sphere A due to B and C.
 |
| Figure for Q-4 |
Due to similarity the magnitude of the force by B or C on the sphere A will be the same, say F.
Distance between any two spheres, r = 2a, hence
F = G.M.M/(2a)²
=GM²/4a²
So each of the sphere B and C exert a force of magnitude F on A and the angle between these forces = 60° (Because the lines joining the centers of the spheres make an equilateral triangle).
The resultant force of these two forces = √(F²+F²+2.F.F.cos60°)
=√(3F²) =√3F =√3GM²/4a²
(Direction of the resultant along the angle bisector of the forces F and F)
5. Four particles of equal masses M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.
ANSWER: Let the four masses A, B, C and D (Each of mass M) move along a circle of radius R and center at O. Assuming the system symmetrical, the masses are placed at corners of a square having its edge = √2R. See the diagram below.
 |
| Diagram for the Q-5 |
The gravitational force on A due to C = F' =GM²/4R² {Along AC}
The magnitude of the Resultant force of F and F = √2F
=√2GM²/2R² {Along AC}
So the resultant of gravitational forces on A due to B, C and D are,
R =√2F+F' {Along AC}
=√2GM²/2R²+GM²/4R²
={(2√2+1)/4}(GM²/R²)
The direction of this net force on A is towards the center of the circle which is also the center of the mass of the system. Let the speed of A be v, then it will have a centrifugal force P =Mv²/R
For the stability of the system P must be equal to the gravitational pull on A, hence,
Mv²/R = {(2√2+1)/4}(GM²/R²)
→v² = {(2√2+1)/4}(GM/R)
→v = √[{(2√2+1)/4}(GM/R)]
6. Find the acceleration due to the gravity of the moon at a point 1000 km above the moon's surface. The mass of the moon is 7.4x10²² kg and its radius is 1740 km.
ANSWER: The point in consideration is outside the moon's surface. The mass of the moon can be considered to be concentrated at the center. The distance of the point from the center,
h = 1000+1740 =2740 km =2.74x10⁶ m
Let a mass m be placed at the point and the acceleration due to gravity there be g. Then weight of the mass m, W=mg
But the gravitational force on the mass due to the moon,
F = GMm/h²
Bothe the forces will be the same,
W = F
mg = GMm/h²
g = GM/h²
{G=6.67x10⁻¹¹ N-m²/kg², M=7.4x10²² kg, h=2.74x10⁶ m}
→g = 6.67x10⁻¹¹*7.4x10²²/(2.74)²x10¹² m/s²
→g = 6.5x10²²⁻¹¹⁻¹² =6.5x10⁻¹ m/s²
→g = 0.65 m/s²
7. Two small bodies of masses 10 kg and 20 kg are kept at a distance 1.0 m apart and released. Assuming that only mutual gravitational forces are acting, find the speeds of the particles when the separation decreases to 0.5 m.
ANSWER: m = 10 kg, m' = 20 kg =2m
Let the final speed of m = v and that of m' = v'.
Taking the two body as a system, the gravitational force on each other is the internal force. Initially the system is at rest, so its linear momentum is zero. There is no external force hence the linear momentum will be conserved. Therefore the final linear momentum will also be zero. Due to the final velocities,
mv+m'v' = 0
→v = -m'v'/m = -20.v'/10 =-2v'
The negative sign shows that direction of the speeds are opposite.
Now the sum of P.E. and K.E. (Total enrgy) of the system will also be conserved. Initial Total energy = P.E.+K.E. = P.E.+ 0
= -Gmm'/r (r = 1.0 m)
= -Gm*2m
=-2Gm²
Final Total energy = P.E.+K.E. (Now r = 0.5 m)
=-Gmm'/0.5 +½mv² +½m'v'²
=-2Gm*2m+½m(2v')²+½*2m*v'² =-4Gm²+3mv'²
Equating Initial and final total energy
-4Gm²+3mv'² =-2Gm²
→3mv'² =2Gm²
→v'² =2Gm/3 =2*6.67x10⁻¹¹*10/3 =44.47x10⁻¹¹=4.447x10⁻¹⁰
→v' = 2.10x10⁻⁵ m/s
And v = 2v' = 4.20x10⁻⁵ m/s
8. A semi-circular wire has a length L and mass M. A particle of mass m is placed at the center of the circle. Find the gravitational attraction on the particle due to the wire.
ANSWER: Let the radius of the semi-circular wire be r. Assuming its center at the origin and its ends touching the X-axis as shown in the diagram, consider a radius at angle θ from the X-axis. For an increament in the angle by dθ we have a length of rdθ on the wire. The inceamental mass of this element dM = Mr.dθ/L
 |
| Diagram for Q-8 |
But L = πr
→r =L/π
So, dM = MLdθ/πL = Mdθ/π
Now the force by dM on a particle of mass m at the center,
dF = Gm.dM/r² =Gm(Mdθ/π)/(L²/π²) = πGMmdθ/L²
This force is directed towards the element dM.
If i and j be the unit vectors along X and Y axes respectively, we can write this force as,
dF = πGMm/L²(cosθ.i + sinθ.j)dθ
Integrating it from θ = 0 to θ = π, we can find the total gravitational force on the particle with mass m at the center.
F = ∫dF
= ∫πGMm/L²(cosθ.i + sinθ.j)dθ
=πGMm/L²*[sinθ.i -cosθ.j] {putting the limits 0 and π},
=πGMm/L²*[sinπ.i- cosπ.j-sin0.i+cos0.j]
=πGMm/L²*[0.i-(-1)j-0.i+1.j]
=πGMm/L²*[2.j]
=2πGMm/L²*j
(It has no component in X-direction)
So the gravitational force at mass m at the center of the semi-circular wire is 2πGMm/L² in magnitude and directed towards the midpoint of the ring.
9. Derive an expression for the gravitational field due to a uniform rod of length L and mass M at a point on its perpendicular bisector at a distance d from the center.
ANSWER: Mass per unit length of the rod = M/L
Let the rod be placed on the X-axis with its mid point on the origin. Cosider an infinitesimal length dx of the rod at distance x from the origin (Mid-point). It’s mass, dM = Mdx/L.
 |
| Diagram for Q-9 |
The gravitational field at P (at a distance d from the mid-point on the perpendicular bisector) due to this small mass,
dE = G.dM/l² (l is the length of line joining P and dx)
=(GM/L)*dx/(x²+d²)
This is the magnitude of the gravitational field vector at P due to dM mass and is directed towards P-dx. If i and j be the unit vectors along the X and Y axes, then dE can be written as
dE = (GM/L)*{sinϕdx/(x²+d²)}i – (GM/L)*{cosϕdx/(x²+d²)}j
(Put the value of sinϕ and cosϕ)
=(GM/L)*{x.dx/(x²+d²)3/2}i – (GM/L)*{d.dx/(x²+d²)3/2}j
The gravitational field at P can be calculated by integrating dE from the limit x=-L/2 to x = L/2
E =∫ dE
=∫ [(GM/L)*{x.dx/(x²+d²)3/2}i – (GM/L)*{d.dx/(x²+d²)3/2}j]
To integrate it we adopt the substitution method. Put, x=d.tanϕ
dx = d.sec²ϕdϕ
Now, E=(GM/L) [(d.tanϕ*d.sec²ϕ.dϕ)i/d³(1+tan²ϕ)³/2–(d.d.sec²ϕ.dϕ)j/d³(1+tan²ϕ)³/2]
=(GM/L)[tanϕ.sec²ϕ.dϕ.i/d.sec³ϕ - sec²ϕ.dϕ.j/d.sec³ϕ]
=(GM/Ld)[sinϕ.dϕ.i-cosϕ.dϕ.j]
=(GM/Ld)[-cosϕi-sinϕ.j]
=-(GM/Ld)[cosϕi+sinϕ.j]
Since we have substituted x by the function of ϕ, our limits for x=L/2 is tanϕ = L/2d = tanα
→ϕ=α
Now our limit is from ϕ=–α to ϕ=α
Putting the limits
E =-(GM/Ld)[cosα.i+sinαj-cos(-α)i-sin(-α)j]
=-(GM/Ld) [cosα.i+sinαj-cosαi+sinαj]
=-(GM/Ld)*2sinαj
=-2GM.sinα.j/Ld
=-2GM.(L/2).j/Ld*(d²+L²/4)1/2
=-2GM.j/d√(L²+4d²)
As we see that this expression contains only unit vector j and in negative direction, so the field at P is of magnitude 2GM/d√(L²+4d²) and directed towards the midpoint of the rod.
10. Two concentric spherical shells have masses M₁ and M₂ and radii R₁, R₂ (R₁<R₂). What is the force exerted by this system on a particle of mass m if it is placed at a distance (R₁+R₂)/2 from the center?
ANSWER: Since the particle is placed at (R₁+R₂)/2 distance from the center, it is outside the inner shell but inside the outer shell. The force on the particle by the outer shell = 0.
The force on the particle by the inner shell
= GM₁m/{(R₁+R₂)²/4}
= 4GM₁m/(R₁+R₂)²
And it is directed towards the center.
===<<<O>>>===
Links to the chapters -
ALL LINKS
CHAPTER- 11 - Gravitation
Questions for Short Answers
OBJECTIVE - I
OBJECTIVE - II
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-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
--------------------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------
Thanks for such bright answers
ReplyDelete