My Channel on YouTube → SimplePhysics with KK
For links to
other chapters - See bottom of the page
Or click here → kktutor.blogspot.com
SIMPLE HARMONIC MOTION:--
EXERCISES Q-41 TO Q-50
51. A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum. How does it differ from the time period calculated using the formula for a simple pendulum?
ANSWER: (a) The time period of a physical pendulum is given by T =2π√(I/mgl).
Let us calculate I. If the radius of the sphere is r, then M.I. about the support, I = 2mr²/3 + ml²
(r = 2 cm =0.02 m, l = 0.18+0.02 =0.20 m.)
→I = m{2(0.02)²/3+(0.20²)} = 0.04027m kg-m²
So, T = 2π√(0.04027m/mgl)
=2π√(0.04027/9.8*0.20) = 0.90 s
From the formula of simple pendulum T' =2π√(l/g)
=2π√(0.20/9.8) =0.897 s
So the result is {(0.90-0.897)/0.897}*100 =0.3% higher than the calculated value.
52. A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude 2° and time period 2 s. Find the radius of the circular wire, (b) the speed of the particle farthest away from the point of suspension as it goes through its mean position. (c) the acceleration of this particle as it goes through its mean position and (d) the acceleration of this particle when it is at an extreme position. Take g = π² m/s².
ANSWER: (a) It is a physical pendulum. If r is the radius of the wire, then the distance between the support and the CoM of the pendulum, l =r.
M.I of the wire about the point of suspension, I = mr²+mr² =2mr²
Time period =2π√(I/mgl) =2π√(2mr²/mgr) =2π√(2r/g)
{It is given 2 s} so,
2π√(2r/g) =2
→√(2r/g) = 1/π
→2r =g/π² = 1
→r =1/2 m =50 cm
The figure for Q-51
(b) The distance of the point farthest from the point of suspension =2r
Let the angular velocity of the wire about the point of suspension at mean position = ⍵
K.E. at this position = ½I⍵². P.E. = 0 if we take the reference level at r distance below from the point of suspension. Total energy =½I⍵².
At the highest point, the CoM is r(1-cos 2°) m higher and velocity zero. So the total energy here =mgr(1-cos 2°)
Equating, ½I⍵² = mgr(1-cos 2°)
→½*2mr²⍵² =mgr(1-cos2°)
→r⍵² =g(1-cos2°)
→⍵² =π²(1-cos2°)/0.5 =0.012
→⍵ =0.109 radians/s
Hence the speed of the particle farthest away from the point of suspension=⍵*2r =0.11*2*0.5 =0.11 m/s =11 cm/s
(c) When going through the mean position there is no torque on the wire because the CoM and the point of suspension are in the same vertical line. So no acceleration due to the SHM but since the farthest point is moving in a circle with radius 2r having its center at the point of suspension, it will have a centripetal acceleration =v²/2r
=0.11²/(2*0.5) =0.012 m/s² =1.20 cm/s² towards the point of suspension.
(d) At the extreme position v =0, so no centripetal acceleration but there is a torque on the wire so it will have an acceleration towards the mean position.
Torque = mg*r*sinθ
Angular acceleration α =Torque/M.I.
=mgr*sinθ/2mr²
=g*sin2°/2*0.50
=π²*sin2°
=0.34 rad/s²
Acceleration =α*2r =0.34*2*0.5 =0.34 m/s² =34 cm/s² towards the mean position.
53. A uniform disc of mass m and radius r is suspended through a wire attached to its center. If the time period of the torsional oscillations be T, what is the torsional constant of the wire?
ANSWER: Let the torsional constant be k. If the torsion in the wire be 𝛕 for an angular displacement of θ, then
𝛕 = kθ
The angular acceleration α = 𝛕/I
I = mr²/2
So, α = 2kθ/mr² 2kx/mr³ {θ =x/r}
→a/r = (2k/mr³)x
→a = (2k/mr²)x
Since the acceleration is proportional to the displacement, it is a SHM with ⍵² = 2k/mr²
→⍵ = √(2k/mr²)
So the time period = T =2π/⍵
→T = 2π√(mr²/2k)
→2kT² = 4π²*mr²
→k = 2π²mr²/T²
54. Two small balls each of mass m are connected by a light rigid rod of length L. The system is suspended from its center by a thin wire of torsional constant k. The rod is rotated about the wire through an angle θ₀ and released. Find the tension in the rod as the system passes through the mean position.
ANSWER: For the angle θ₀ torsion in the wire =kθ₀
Total energy =P.E. stored at this position =½kθ₀², At the mean position, this total energy will be converted to K.E. =½I⍵²
where ⍵ is the angular velocity of the ball system at the mean position. Equating, ½I⍵² =½kθ₀²
⍵² = kθ₀²/I
{Moment of inertia of the system about the center I =2m(L/2)² =mL²/2}
→⍵² =2kθ₀²/mL²
→{v/(L/2)}² = 2kθ₀²/mL²
→4v²/L² = 2kθ₀²/mL²
→v² = kθ₀²/2m
The centrifugal force F =mv²/(L/2) =2mv²/L
=kθ₀²/L
There is another force acting on the balls which is the wight of the ball W =mg, These two forces are perpendicular to each other, hence their resultant = √[F²+W²]
=√[(kθ₀²/L)²+(mg)²]
=[k²θ₀⁴/L² + m²g²]½
Note: This tension force will be in the rod considering it not stiff or rigid. If the rod is stiff the weight portion will not contribute to tension but will try to bend the rod.
55. A particle is subjected to two simple harmonic motions of the same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a) 0°, (b) 60°, (c) 90°.
ANSWER: A₁ =3.0 cm and A₂ =4.0 cm. Since the time period and hence the frequencies of both SHMs are the same, the resultant amplitude A is given as,
A =√(A₁²+2A₁A₂cosẟ+A₂²) where ẟ is the phase difference.
(a) For ẟ =0°, cosẟ = 1
→A =√(3²+2*3*4*1+4²) =√(3+4)² =3+4 =7.0 cm
(b) For ẟ = 60°, cosẟ=1/2
→A =√(9+3*4+16) =√37 ≈6.10 cm
(c) For ẟ =90°, cosẟ = 0
→A =√(9+0+16) =√25 =5.0 cm
56. Three simple harmonic motions of equal amplitude A and equal time periods in the same direction combine. The phase of the second motion is 60° ahead of the first and the phase of the third motion is 60° ahead of the second. Find the amplitude of the resultant motion.
ANSWER:
Figure for Q-56
Let us solve the problem by vector method. The three SHMs are at a phase difference of 60° at point O. Placing them at tip-toe to find out the resultant as in the right side figure, we get the resultant OB =A*cos60°+A+A*cos60° =A/2+A+A/2 =2A.
57. A particle is subjected to two simple harmonic motions given by
x₁ =2.0 sin(100π t) and
x₂ =2.0 sin(120π t+π/3)
where x is in centimeter and t in second. Find the displacement of the particle at (a) t = 0.0125, (b) t = 0.025.
ANSWER: (a) The displacement of the particle at t = 0.0125 s is
x =x₁+x₂
→x = 2.0 sin(100π t)+2.0 sin(120π t+π/3)
→x =2.0 sin(100π *0.0125)+2.0 sin(120π *0.0125+π/3)
→x =2.0 sin(10π/8) +2.0 sin(12π/8+π/3)
→x =2.0 sin(5π/4) +2.0 sin(3π/2+π/3)
→x =-2.0*1/√2+2.0 sin(11π/6)
→x = -√2+2.0 sin(2π-π/6)
→x =-√2-2.0*0.5 =-1.41-1 = -2.41 cm
(b) At t =0.025 s, the displacement of the particle is given by
x =x₁+x₂
→x = 2.0 sin(100π t)+2.0 sin(120π t+π/3)
→x =2.0 sin(100π *0.025)+2.0 sin(120π *0.025+π/3)
→x =2.0 sin(10π/4) +2.0 sin(12π/4+π/3)
→x =2.0*sin(5π/2) +2.0 sin(10π/3)
→x =2.0*1.0+2.0*sin(2π*2-2π/3) =2.0-2.0* sin2π/3
→x =2.0-2.0*√3/2 =2.0-2.0*1.73/2 =2.0-1.73
→x = 0.27 cm
58. A particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line making an angle of 45°. with the X-axis. The two motions are given by x = x₀sin⍵t and s = s₀sin⍵t. Find the amplitude of the resultant motion.
ANSWER: Since there is no phase difference, at any instant the magnitude of individual displacements along their line will be equal. The resultant displacement can be found out by vector addition and its magnitude can be found out by
X² = x² + s² +2xs*cos45°
=(x₀sin⍵t)² + (s₀sin⍵t)² +2(x₀sin⍵t)(s₀sin⍵t)(1/√2)
=x₀²sin²⍵t + s₀²sin²⍵t +√2 x₀s₀sin²⍵t
=[x₀² + s₀² +√2 x₀s₀]sin²⍵t
→X = [x₀² + s₀² +√2 x₀s₀]½ sin⍵t
Figure for Q-58
It is the equation of SHM in the form of x =A sin⍵t. From comparison the amplitude of the resultant motion,
A = [x₀² + s₀² +√2 x₀s₀]½
Extra Questions
59. A U-tube has a liquid of density ⍴ at rest (See the figure below). The liquid in one arm is sucked up to a height x and released. Assuming that there is no loss of energy due to the adhesion between the liquid and the wall of the tube, show that the motion of the liquid in the tube is "Simple Harmonic".
Figure for Q-59
ANSWER: When the liquid in one arm is up by x distance, the liquid in other arm goes down to x distance. So the total difference between the levels in both arms =2x. If the area of the cross-section of the tube arm is A, the net force on the liquid due to level difference, F =⍴gA*2x
→F =(2A⍴g)x
Since 2A⍴g = constant,
Hence F ∝ x. Since the force is proportional to the displacement, the motion is Simple Harmonic.
60. A weightless horizontal stiff rod of length L is hinged at one end A and attached to a weightless spring having spring constant k at the mid-point. A ball of mass M is attached to the other end of the rod (See the figure below). The spring is balanced with the ball when the rod is horizontal and at rest. The ball is pulled down to a small distance d and released. Show that the motion of the ball is simple harmonic. Find its time period.
Figure for Q-60
ANSWER: When the system is in a balanced position with the system at rest, let the force between the spring and the rod at the mid-point be F. If we consider the free body diagram of the rod, the spring pulls the rod by a force F upward and the ball pulls down the rod by a force Mg. The torque of these two opposite forces about the hinge A should be equal.
F*L/2 =Mg*L
→F = 2Mg
It is as if a mass of 2M is hanged to the same spring in its balanced position. When the ball is pulled down to a distance d, the end of the spring is stretched to a distance d/2. The force on the spring F'=kd/2.
So, F' ∝ d
Hence it is a simple harmonic motion. The time period of such spring in SHM is given by
T =2π√(m/k)
Here m = 2M, so
T =2π√(2M/k)
So the midpoint of the rod and hence the ball is also in SHM with a time period 2π√(2M/k), only the amplitude of the ball will be double that of the spring end.
51. A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum. How does it differ from the time period calculated using the formula for a simple pendulum?
ANSWER: (a) The time period of a physical pendulum is given by T =2π√(I/mgl).
Let us calculate I. If the radius of the sphere is r, then M.I. about the support, I = 2mr²/3 + ml²
(r = 2 cm =0.02 m, l = 0.18+0.02 =0.20 m.)
→I = m{2(0.02)²/3+(0.20²)} = 0.04027m kg-m²
So, T = 2π√(0.04027m/mgl)
=2π√(0.04027/9.8*0.20) = 0.90 s
From the formula of simple pendulum T' =2π√(l/g)
=2π√(0.20/9.8) =0.897 s
So the result is {(0.90-0.897)/0.897}*100 =0.3% higher than the calculated value.
52. A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude 2° and time period 2 s. Find the radius of the circular wire, (b) the speed of the particle farthest away from the point of suspension as it goes through its mean position. (c) the acceleration of this particle as it goes through its mean position and (d) the acceleration of this particle when it is at an extreme position. Take g = π² m/s².
ANSWER: (a) It is a physical pendulum. If r is the radius of the wire, then the distance between the support and the CoM of the pendulum, l =r.
M.I of the wire about the point of suspension, I = mr²+mr² =2mr²
Time period =2π√(I/mgl) =2π√(2mr²/mgr) =2π√(2r/g)
{It is given 2 s} so,
2π√(2r/g) =2
→√(2r/g) = 1/π
→2r =g/π² = 1
→r =1/2 m =50 cm |
| The figure for Q-51 |
(b) The distance of the point farthest from the point of suspension =2r
Let the angular velocity of the wire about the point of suspension at mean position = ⍵
K.E. at this position = ½I⍵². P.E. = 0 if we take the reference level at r distance below from the point of suspension. Total energy =½I⍵².
At the highest point, the CoM is r(1-cos 2°) m higher and velocity zero. So the total energy here =mgr(1-cos 2°)
Equating, ½I⍵² = mgr(1-cos 2°)
→½*2mr²⍵² =mgr(1-cos2°)
→r⍵² =g(1-cos2°)
→⍵² =π²(1-cos2°)/0.5 =0.012
→⍵ =0.109 radians/s
Hence the speed of the particle farthest away from the point of suspension=⍵*2r =0.11*2*0.5 =0.11 m/s =11 cm/s(c) When going through the mean position there is no torque on the wire because the CoM and the point of suspension are in the same vertical line. So no acceleration due to the SHM but since the farthest point is moving in a circle with radius 2r having its center at the point of suspension, it will have a centripetal acceleration =v²/2r
=0.11²/(2*0.5) =0.012 m/s² =1.20 cm/s² towards the point of suspension.
(d) At the extreme position v =0, so no centripetal acceleration but there is a torque on the wire so it will have an acceleration towards the mean position.
Torque = mg*r*sinθ
Angular acceleration α =Torque/M.I.
=mgr*sinθ/2mr²
=g*sin2°/2*0.50
=π²*sin2°
=0.34 rad/s²
Acceleration =α*2r =0.34*2*0.5 =0.34 m/s² =34 cm/s² towards the mean position.
53. A uniform disc of mass m and radius r is suspended through a wire attached to its center. If the time period of the torsional oscillations be T, what is the torsional constant of the wire?
ANSWER: Let the torsional constant be k. If the torsion in the wire be 𝛕 for an angular displacement of θ, then
𝛕 = kθ
The angular acceleration α = 𝛕/I
I = mr²/2
So, α = 2kθ/mr² 2kx/mr³ {θ =x/r}
→a/r = (2k/mr³)x
→a = (2k/mr²)x
Since the acceleration is proportional to the displacement, it is a SHM with ⍵² = 2k/mr²
→⍵ = √(2k/mr²)
So the time period = T =2π/⍵
→T = 2π√(mr²/2k)
→2kT² = 4π²*mr²
→k = 2π²mr²/T²
54. Two small balls each of mass m are connected by a light rigid rod of length L. The system is suspended from its center by a thin wire of torsional constant k. The rod is rotated about the wire through an angle θ₀ and released. Find the tension in the rod as the system passes through the mean position.
ANSWER: For the angle θ₀ torsion in the wire =kθ₀
Total energy =P.E. stored at this position =½kθ₀², At the mean position, this total energy will be converted to K.E. =½I⍵²
where ⍵ is the angular velocity of the ball system at the mean position. Equating, ½I⍵² =½kθ₀²
⍵² = kθ₀²/I
{Moment of inertia of the system about the center I =2m(L/2)² =mL²/2}
→⍵² =2kθ₀²/mL²
→{v/(L/2)}² = 2kθ₀²/mL²
→4v²/L² = 2kθ₀²/mL²
→v² = kθ₀²/2m
The centrifugal force F =mv²/(L/2) =2mv²/L
=kθ₀²/L
There is another force acting on the balls which is the wight of the ball W =mg, These two forces are perpendicular to each other, hence their resultant = √[F²+W²]
=√[(kθ₀²/L)²+(mg)²]
=[k²θ₀⁴/L² + m²g²]½
Note: This tension force will be in the rod considering it not stiff or rigid. If the rod is stiff the weight portion will not contribute to tension but will try to bend the rod.
55. A particle is subjected to two simple harmonic motions of the same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a) 0°, (b) 60°, (c) 90°.
ANSWER: A₁ =3.0 cm and A₂ =4.0 cm. Since the time period and hence the frequencies of both SHMs are the same, the resultant amplitude A is given as,
A =√(A₁²+2A₁A₂cosẟ+A₂²) where ẟ is the phase difference.
(a) For ẟ =0°, cosẟ = 1
→A =√(3²+2*3*4*1+4²) =√(3+4)² =3+4 =7.0 cm
(b) For ẟ = 60°, cosẟ=1/2
→A =√(9+3*4+16) =√37 ≈6.10 cm
(c) For ẟ =90°, cosẟ = 0
→A =√(9+0+16) =√25 =5.0 cm
56. Three simple harmonic motions of equal amplitude A and equal time periods in the same direction combine. The phase of the second motion is 60° ahead of the first and the phase of the third motion is 60° ahead of the second. Find the amplitude of the resultant motion.
ANSWER:
Let us solve the problem by vector method. The three SHMs are at a phase difference of 60° at point O. Placing them at tip-toe to find out the resultant as in the right side figure, we get the resultant OB =A*cos60°+A+A*cos60° =A/2+A+A/2 =2A.
 |
| Figure for Q-56 |
57. A particle is subjected to two simple harmonic motions given by
x₁ =2.0 sin(100π t) and
x₂ =2.0 sin(120π t+π/3)
where x is in centimeter and t in second. Find the displacement of the particle at (a) t = 0.0125, (b) t = 0.025.
ANSWER: (a) The displacement of the particle at t = 0.0125 s is
x =x₁+x₂
→x = 2.0 sin(100π t)+2.0 sin(120π t+π/3)
→x =2.0 sin(100π *0.0125)+2.0 sin(120π *0.0125+π/3)
→x =2.0 sin(10π/8) +2.0 sin(12π/8+π/3)
→x =2.0 sin(5π/4) +2.0 sin(3π/2+π/3)
→x =-2.0*1/√2+2.0 sin(11π/6)
→x = -√2+2.0 sin(2π-π/6)
→x =-√2-2.0*0.5 =-1.41-1 = -2.41 cm
(b) At t =0.025 s, the displacement of the particle is given by
x =x₁+x₂
→x = 2.0 sin(100π t)+2.0 sin(120π t+π/3)
→x =2.0 sin(100π *0.025)+2.0 sin(120π *0.025+π/3)
→x =2.0 sin(10π/4) +2.0 sin(12π/4+π/3)
→x =2.0*sin(5π/2) +2.0 sin(10π/3)
→x =2.0*1.0+2.0*sin(2π*2-2π/3) =2.0-2.0* sin2π/3
→x =2.0-2.0*√3/2 =2.0-2.0*1.73/2 =2.0-1.73
→x = 0.27 cm
58. A particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line making an angle of 45°. with the X-axis. The two motions are given by x = x₀sin⍵t and s = s₀sin⍵t. Find the amplitude of the resultant motion.
ANSWER: Since there is no phase difference, at any instant the magnitude of individual displacements along their line will be equal. The resultant displacement can be found out by vector addition and its magnitude can be found out by
X² = x² + s² +2xs*cos45°
=(x₀sin⍵t)² + (s₀sin⍵t)² +2(x₀sin⍵t)(s₀sin⍵t)(1/√2)
=x₀²sin²⍵t + s₀²sin²⍵t +√2 x₀s₀sin²⍵t
=[x₀² + s₀² +√2 x₀s₀]sin²⍵t
→X = [x₀² + s₀² +√2 x₀s₀]½ sin⍵t
 |
| Figure for Q-58 |
It is the equation of SHM in the form of x =A sin⍵t. From comparison the amplitude of the resultant motion,
A = [x₀² + s₀² +√2 x₀s₀]½Extra Questions
59. A U-tube has a liquid of density ⍴ at rest (See the figure below). The liquid in one arm is sucked up to a height x and released. Assuming that there is no loss of energy due to the adhesion between the liquid and the wall of the tube, show that the motion of the liquid in the tube is "Simple Harmonic". |
| Figure for Q-59 |
→F =(2A⍴g)x
Since 2A⍴g = constant,
Hence F ∝ x. Since the force is proportional to the displacement, the motion is Simple Harmonic.
60. A weightless horizontal stiff rod of length L is hinged at one end A and attached to a weightless spring having spring constant k at the mid-point. A ball of mass M is attached to the other end of the rod (See the figure below). The spring is balanced with the ball when the rod is horizontal and at rest. The ball is pulled down to a small distance d and released. Show that the motion of the ball is simple harmonic. Find its time period.
 |
| Figure for Q-60 |
ANSWER: When the system is in a balanced position with the system at rest, let the force between the spring and the rod at the mid-point be F. If we consider the free body diagram of the rod, the spring pulls the rod by a force F upward and the ball pulls down the rod by a force Mg. The torque of these two opposite forces about the hinge A should be equal.
F*L/2 =Mg*L
→F = 2Mg
It is as if a mass of 2M is hanged to the same spring in its balanced position. When the ball is pulled down to a distance d, the end of the spring is stretched to a distance d/2. The force on the spring F'=kd/2.
So, F' ∝ d
Hence it is a simple harmonic motion. The time period of such spring in SHM is given by
T =2π√(m/k)
Here m = 2M, so
T =2π√(2M/k)
So the midpoint of the rod and hence the ball is also in SHM with a time period 2π√(2M/k), only the amplitude of the ball will be double that of the spring end.
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
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
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
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 → 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 (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 → 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" .
---------------------------------------------------------------------------------
---------------------------------------------------------------------------------
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)
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
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