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SIMPLE HARMONIC MOTION:--
OBJECTIVE-I
1. A student says that he had applied a force F = -k√x on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he has worked only with positive x and no other force acted on the particle.
(a) As x increases k increases
(b) As x increases k decreases
(c) As x increases k remains constant.
(d) The motion cannot be simple harmonic.
ANSWER: (a)
EXPLANATION: Since the motion is simple harmonic, the force F=-⍵²x where ⍵² is a constant. Equating the force applied by the student we get, -k√x=-⍵²x
→k=⍵²√x
It is clear from this relation that if x increases k increases.
2. The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. Thi point is
(a) the mean position
(b) an extreme position
(c) between the mean position and the positive extreme
(d) between the mean position and the negative extreme.
ANSWER: (b)
EXPLANATION: At the extreme position the particle is in the same phase at the consecutive appearance, hence (b) is true.
In all other options, the consecutive appearance of the particle is not in the same phase. All other options are not true.
3. The time period of a particle in simple harmonic motion is equal to the smallest time between the particle acquiring a particular velocity v. The value of v is
(a) vₘₐₓ
(b) 0
(c) between 0 and vₘₐₓ
(d) between 0 and -vₘₐₓ
ANSWER: (a)
EXPLANATION: Velocity is a vector, which means it has magnitude as well as direction. The maximum velocity at the mean position is acquired by the particle after a single time period in the same position. Other velocities occur twice between one extreme to another extreme position (i.e. within half a time period, See picture below). Hence only (a) true, others not true.
Same velocity at R and R' in less than a time period
4. The displacement of a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d)zero
ANSWER: (d)
EXPLANATION: The displacement is a vector and its magnitude is given by the line joining the initial position to the final position. Since after one time period the particle is at its initial position, hence it is zero. Option(d).
5. The distance moved by a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d)zero
ANSWER: (c)
EXPLANATION: The distance moved is a scalar quantity and in a simple harmonic motion the distance covered in one time period is 4A. For example, if the particle is at the left extreme at t=0 then at t=T it will again come back to that point.
•------------|-------------
→ A O A
The distance covered =2A+2A =4A.
6. The average acceleration in one time period in a simple harmonic motion is
(a) A⍵²
(b) A⍵²/2
(c) A⍵²/√2
(d) zero
ANSWER: (d)
EXPLANATION: Since the acceleration is proportional to the displacement and always directed towards the mean position, half of one time period it is directed towards the left and in rest half, it is towards the right. Hence the average acceleration in one time period is zero. Option (d).
7. The motion of a particle is given by x = A.sin⍵t+B.cos⍵t. The motion of the particle is
(a) not simple harmonic
(b) simple harmonic with amplitude A+B
(c) simple harmonic with amplitude (A+B)/2
(d) simple harmonic with amplitude √(A²+B²).
ANSWER: (d)
EXPLANATION: The given equation can be written as,
x = √(A²+B²){(A/√(A²+B²))sin⍵t+(B/√(A²+B²))cos⍵t}
Since the magnitudes of A/√(A²+B²) and B/√(A²+B²) are less than 1 and the sum of their square is also 1 we can find an angle α between 0 and 2π such that
sinα=B/√(A²+B²) and cosα=A/√(A²+B²)
So, x = √(A²+B²){sin⍵t.cosα+cos⍵t.sinα}
→x = √(A²+B²).sin(⍵t+α)
This is an equation of simple harmonic motion with amplitude √(A²+B²). Hence option (d).
8. The displacement of a particle is given by r=A(i.cos⍵t+j.sin⍵t). The motion of the particle is
(a) simple harmonic
(b) on a straight line
(c) on a circle
(d) with constant acceleration.
ANSWER: (c)
EXPLANATION: The magnitude of the displacement
r = √(A²cos²⍵t+A²sin²⍵t) =√A² =A
The direction of the displacement from the X-axis is
tanα = A.sin⍵t/A.cos⍵t =tan⍵t
→α = ⍵t
It means the position vector of the particle can take any angle but its magnitude is constant. So the motion is on a circle with center at the origin. Option (c).
9. A particle moves on the X-axis according to the equation x=A+B.sin⍵t. The motion is simple harmonic with amplitude
(a) A
(b) B
(c) A+B
(d) √(A²+B²)
ANSWER: (b)
EXPLANATION: The equation can be written as
x-A=B.sin⍵t
If x-A=y then
y=B.sin⍵t
This is the equation of a simple harmonic motion with amplitude B. Hence option(b).
10. Figure (12-Q1) represents two simple harmonic motions.
→ ←
A -----•----- ----------• -----•----- •----------
B -----•----- •---------- -----•----- ----------•
← →
Figure 12-Q1
The parameter which has different values in the two motions is
(a) amplitude
(b) frequency
(c) phase
(d) maximum velocity.
ANSWER: (c)
EXPLANATION: It is clear from the figure that in the next step when one time period is complete the particles will be back to their original position. Hence their frequency is the same. The maximum displacement from the mean position for both is the same, hence the amplitude is also the same. For the same amplitude and frequency, the maximum velocity is also the same. So it is the phase that is different. At the mean position, both particles are at the same time but in opposite direction. Hence their phase differs by an angle π. So option(c).
11. The total mechanical energy of a spring-mass system in simple harmonic motion is E = ½m⍵²A². Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will
(a) become 2E
(b) become E/2
(c) become √2E
(d) remain E.
ANSWER: (d)
EXPLANATION: Since no external work is done on the system the energy of the system will remain the same. Hence option (d).
12. The average energy in one time period in simple harmonic motion is
(a) ½m⍵²A²
(b) ¼m⍵²A²
(c) m⍵²A²
(d) zero
ANSWER: (a)
EXPLANATION: The total energy at any instant in a simple harmonic motion is constant and equal to ½m⍵²A². Hence the average energy in one time period will also be the same = ½m⍵²A². So option (a).
13. A particle executes simple harmonic motion with a frequency ν. The frequency with which the kinetic energy oscillates is
(a) ν/2
(b) ν
(c) 2ν
(d) zero
ANSWER: (c)
EXPLANATION: Between one extreme position to another i.e. in half the time period the K.E. increases from zero to maximum (at the mean position) and again to zero thus completing one cycle. Therefore the time period of oscillation of the K.E. = T/2 where T is the time period of the SHM. Since the frequency of SHM ν =1/T, the frequency of oscillation of the K.E. ν'=1/(T/2)=2*(1/T)=2ν. Hence the option (c).
14. A particle executes simple harmonic motion under the restoring force provided by a spring. The time period is T. If the spring is divided into two equal parts and one part is used to continue the simple harmonic motion, the time period will
(a) remain T
(b) become 2T
(c) become T/2
(d) become T/√2
ANSWER: (d)
EXPLANATION: The time period of a spring-mass system is given as T = 2π√(m/k).
When the spring is divided into two equal halves the spring constant of each half doubles, so k'=2k.
New time period T'=2π√{m/(2k)}=(1/√2)*2π√(m/k) T/√2. Hence the option (d).
15. Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant k₁ and k₂ respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is
(a) k₁/k₂
(b) √(k₁/k₂)
(c) k₂/k₁
(d) √(k₂/k₁)
ANSWER: (d)
EXPLANATION: The maximum velocity occurs at the mean position where the total energy is in the form of kinetic energy =½mv². Since the maximum velocities of both systems are equal, their total energy is also equal.
At the extreme position, the total energy is in the form of potential energy. These will also be equal. Let amplitude of A = A' and that of B = B'. Equating the total energy at extremes we get,
½k₁A'² = ½k₂B'²
→A'²/B'² = k₂/k₁
→A'/B' = √(k₂/k₁)
Hence the option (d).
16. A spring-mass system oscillates with a frequency ν. If it is taken in an elevator slowly accelerating upward, the frequency will
(a) increase
(b) decrease
(c) remain the same
(d) become zero.
ANSWER: (c)
EXPLANATION: The time period of a spring-mass system is given by T=2π√(m/k)
Since this expression is not dependent on the acceleration due to gravity, the time period T and hence the frequency (since ν=1/T) remains the same. The option (c) is correct.
17. A spring-mass system oscillates in a car. If the car accelerates on a horizontal road, the frequency of oscillation will
(a) increase
(b) decrease
(c) remain the same
(d) become zero.
ANSWER: (c)
EXPLANATION: The frequency of a spring-mass system depends only on the mass and the spring constant, none of which changes in the accelerating car. So the frequency remains the same. Option (c).
18. A pendulum clock that keeps the correct time on the earth is taken to the moon. It will run
(a) at the correct rate
(b) 6 times faster
(c) √6 times faster
(d) √6 times slower.
ANSWER: (d)
EXPLANATION: The time period of the pendulum of the clock on the earth T =2π√(l/g). On the moon g'=g/6. The time period on the moon T' =2π√6l/g) =√6*2π√(l/g) =√6T.
So the pendulum will take √6 times more time in one oscillation on the moon and it will run √6 times slower. The option (d) is true.
19. A wall clock uses a vertical spring-mass system to measure the time. Each time the mass reaches an extreme position, the clock advances by a second. The clock gives the correct time at the equator. If the clock is taken to the poles it will
(a) run slow
(b) run fast
(c) stop working
(d) give the correct time.
ANSWER: (d)
EXPLANATION: Same explanation as in Q-17. Option (d) is true.
20. A pendulum clock keeping correct time is taken to high altitudes,
(a) it will keep the correct time
(b) its length should be increased to keep correct time
(c) its length should be decreased to keep correct time
(d) it cannot keep the correct time even if the length is changed.
ANSWER: (c)
EXPLANATION: At high altitudes, the value of g decreases. Since the time period of the pendulum T = 2π√(l/g), it will increase at high altitudes. The clock will not give the correct time. To keep the correct time T should not change and to keep T constant the ratio l/g should not change. Since the value of g is getting decreased at the high altitudes, the length l should also be proportionately decreased to keep the ratio l/g constant. Hence the option (c) is true.
21. The free end of a simple pendulum is attached to the ceiling of a box. The box is taken to a height and the pendulum is oscillated. When the bob is at its lowest point, the box is released to fall freely. As seen from the box during this period, the bob will
(a) continue its oscillation as before
(b) stop
(c) will go in a circular path
(d) move on a straight line.
ANSWER: (c)
EXPLANATION: At the lowest point the bob is moving horizontally with the maximum speed. When we observe the bob from the box when free-falling, the box is a non-inertial frame with acceleration g downwards. If we want to apply Newton laws of motion here we will have to apply a pseudo force mg upwards. Thus the weight mg and the pseudo force -mg have a resultant zero force on the bob and hence no restoring force for the oscillation. Since the bob has a horizontal speed and it is tied to the ceiling with a string it will go in a circular path. Hence the option (c) is true.
===<<<O>>>===
Links to the chapters -
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
1. A student says that he had applied a force F = -k√x on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he has worked only with positive x and no other force acted on the particle.
(a) As x increases k increases
(b) As x increases k decreases
(c) As x increases k remains constant.
(d) The motion cannot be simple harmonic.
(a) As x increases k increases
(b) As x increases k decreases
(c) As x increases k remains constant.
(d) The motion cannot be simple harmonic.
ANSWER: (a)
→k=⍵²√x
It is clear from this relation that if x increases k increases.
2. The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. Thi point is
(a) the mean position
(b) an extreme position
(c) between the mean position and the positive extreme
(d) between the mean position and the negative extreme.
(a) the mean position
(b) an extreme position
(c) between the mean position and the positive extreme
(d) between the mean position and the negative extreme.
ANSWER: (b)
EXPLANATION: At the extreme position the particle is in the same phase at the consecutive appearance, hence (b) is true.
In all other options, the consecutive appearance of the particle is not in the same phase. All other options are not true.
3. The time period of a particle in simple harmonic motion is equal to the smallest time between the particle acquiring a particular velocity v. The value of v is
(a) vₘₐₓ
(b) 0
(c) between 0 and vₘₐₓ
(d) between 0 and -vₘₐₓ
ANSWER: (a)
EXPLANATION: Velocity is a vector, which means it has magnitude as well as direction. The maximum velocity at the mean position is acquired by the particle after a single time period in the same position. Other velocities occur twice between one extreme to another extreme position (i.e. within half a time period, See picture below). Hence only (a) true, others not true.
EXPLANATION: Velocity is a vector, which means it has magnitude as well as direction. The maximum velocity at the mean position is acquired by the particle after a single time period in the same position. Other velocities occur twice between one extreme to another extreme position (i.e. within half a time period, See picture below). Hence only (a) true, others not true.
 |
| Same velocity at R and R' in less than a time period |
4. The displacement of a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d)zero
ANSWER: (d)
EXPLANATION: The displacement is a vector and its magnitude is given by the line joining the initial position to the final position. Since after one time period the particle is at its initial position, hence it is zero. Option(d).
EXPLANATION: The displacement is a vector and its magnitude is given by the line joining the initial position to the final position. Since after one time period the particle is at its initial position, hence it is zero. Option(d).
5. The distance moved by a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d)zero
ANSWER: (c)
EXPLANATION: The distance moved is a scalar quantity and in a simple harmonic motion the distance covered in one time period is 4A. For example, if the particle is at the left extreme at t=0 then at t=T it will again come back to that point.
•------------|-------------
→ A O A
The distance covered =2A+2A =4A. 6. The average acceleration in one time period in a simple harmonic motion is
(a) A⍵²
(b) A⍵²/2
(c) A⍵²/√2
(d) zero
ANSWER: (d)
EXPLANATION: Since the acceleration is proportional to the displacement and always directed towards the mean position, half of one time period it is directed towards the left and in rest half, it is towards the right. Hence the average acceleration in one time period is zero. Option (d).
7. The motion of a particle is given by x = A.sin⍵t+B.cos⍵t. The motion of the particle is
(a) not simple harmonic
(b) simple harmonic with amplitude A+B
(c) simple harmonic with amplitude (A+B)/2
(d) simple harmonic with amplitude √(A²+B²).
ANSWER: (d)
EXPLANATION: The given equation can be written as,
x = √(A²+B²){(A/√(A²+B²))sin⍵t+(B/√(A²+B²))cos⍵t}
Since the magnitudes of A/√(A²+B²) and B/√(A²+B²) are less than 1 and the sum of their square is also 1 we can find an angle α between 0 and 2π such that
sinα=B/√(A²+B²) and cosα=A/√(A²+B²)
So, x = √(A²+B²){sin⍵t.cosα+cos⍵t.sinα}
→x = √(A²+B²).sin(⍵t+α)
This is an equation of simple harmonic motion with amplitude √(A²+B²). Hence option (d).
(a) simple harmonic
(b) on a straight line
(c) on a circle
(d) with constant acceleration.
ANSWER: (c)
EXPLANATION: The magnitude of the displacement
r = √(A²cos²⍵t+A²sin²⍵t) =√A² =A
The direction of the displacement from the X-axis is
tanα = A.sin⍵t/A.cos⍵t =tan⍵t
→α = ⍵t
It means the position vector of the particle can take any angle but its magnitude is constant. So the motion is on a circle with center at the origin. Option (c).
9. A particle moves on the X-axis according to the equation x=A+B.sin⍵t. The motion is simple harmonic with amplitude
(a) A
(b) B
(c) A+B
(d) √(A²+B²)
ANSWER: (b)
EXPLANATION: The equation can be written as
x-A=B.sin⍵t
If x-A=y then
y=B.sin⍵t
This is the equation of a simple harmonic motion with amplitude B. Hence option(b).
10. Figure (12-Q1) represents two simple harmonic motions.
→ ←
A -----•----- ----------• -----•----- •----------
B -----•----- •---------- -----•----- ----------•
← →
Figure 12-Q1
The parameter which has different values in the two motions is(a) amplitude
(b) frequency
(c) phase
(d) maximum velocity.
ANSWER: (c)
EXPLANATION: It is clear from the figure that in the next step when one time period is complete the particles will be back to their original position. Hence their frequency is the same. The maximum displacement from the mean position for both is the same, hence the amplitude is also the same. For the same amplitude and frequency, the maximum velocity is also the same. So it is the phase that is different. At the mean position, both particles are at the same time but in opposite direction. Hence their phase differs by an angle π. So option(c).
11. The total mechanical energy of a spring-mass system in simple harmonic motion is E = ½m⍵²A². Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will
(a) become 2E
(b) become E/2
(c) become √2E
(d) remain E.
ANSWER: (d)
EXPLANATION: Since no external work is done on the system the energy of the system will remain the same. Hence option (d).
12. The average energy in one time period in simple harmonic motion is
(a) ½m⍵²A²
(b) ¼m⍵²A²
(c) m⍵²A²
(d) zero
(a) ½m⍵²A²
(b) ¼m⍵²A²
(c) m⍵²A²
(d) zero
ANSWER: (a)
EXPLANATION: The total energy at any instant in a simple harmonic motion is constant and equal to ½m⍵²A². Hence the average energy in one time period will also be the same = ½m⍵²A². So option (a).
13. A particle executes simple harmonic motion with a frequency ν. The frequency with which the kinetic energy oscillates is
(a) ν/2
(b) ν
(c) 2ν
(d) zero
(a) ν/2
(b) ν
(c) 2ν
(d) zero
ANSWER: (c)
EXPLANATION: Between one extreme position to another i.e. in half the time period the K.E. increases from zero to maximum (at the mean position) and again to zero thus completing one cycle. Therefore the time period of oscillation of the K.E. = T/2 where T is the time period of the SHM. Since the frequency of SHM ν =1/T, the frequency of oscillation of the K.E. ν'=1/(T/2)=2*(1/T)=2ν. Hence the option (c).
14. A particle executes simple harmonic motion under the restoring force provided by a spring. The time period is T. If the spring is divided into two equal parts and one part is used to continue the simple harmonic motion, the time period will
(a) remain T
(b) become 2T
(c) become T/2
(d) become T/√2
ANSWER: (d)
EXPLANATION: The time period of a spring-mass system is given as T = 2π√(m/k).
When the spring is divided into two equal halves the spring constant of each half doubles, so k'=2k.
New time period T'=2π√{m/(2k)}=(1/√2)*2π√(m/k) T/√2. Hence the option (d).
EXPLANATION: The time period of a spring-mass system is given as T = 2π√(m/k).
When the spring is divided into two equal halves the spring constant of each half doubles, so k'=2k.
New time period T'=2π√{m/(2k)}=(1/√2)*2π√(m/k) T/√2. Hence the option (d).
15. Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant k₁ and k₂ respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is
(a) k₁/k₂
(b) √(k₁/k₂)
(c) k₂/k₁
(d) √(k₂/k₁)
(a) k₁/k₂
(b) √(k₁/k₂)
(c) k₂/k₁
(d) √(k₂/k₁)
ANSWER: (d)
EXPLANATION: The maximum velocity occurs at the mean position where the total energy is in the form of kinetic energy =½mv². Since the maximum velocities of both systems are equal, their total energy is also equal.
At the extreme position, the total energy is in the form of potential energy. These will also be equal. Let amplitude of A = A' and that of B = B'. Equating the total energy at extremes we get,
½k₁A'² = ½k₂B'²
→A'²/B'² = k₂/k₁
→A'/B' = √(k₂/k₁)
Hence the option (d).
EXPLANATION: The maximum velocity occurs at the mean position where the total energy is in the form of kinetic energy =½mv². Since the maximum velocities of both systems are equal, their total energy is also equal.
At the extreme position, the total energy is in the form of potential energy. These will also be equal. Let amplitude of A = A' and that of B = B'. Equating the total energy at extremes we get,
½k₁A'² = ½k₂B'²
→A'²/B'² = k₂/k₁
→A'/B' = √(k₂/k₁)
Hence the option (d).
16. A spring-mass system oscillates with a frequency ν. If it is taken in an elevator slowly accelerating upward, the frequency will
(a) increase
(b) decrease
(c) remain the same
(d) become zero.
(a) increase
(b) decrease
(c) remain the same
(d) become zero.
ANSWER: (c)
EXPLANATION: The time period of a spring-mass system is given by T=2π√(m/k)
Since this expression is not dependent on the acceleration due to gravity, the time period T and hence the frequency (since ν=1/T) remains the same. The option (c) is correct.
17. A spring-mass system oscillates in a car. If the car accelerates on a horizontal road, the frequency of oscillation will
ANSWER: (c)
(a) increase
(b) decrease
(c) remain the same
(d) become zero.
(b) decrease
(c) remain the same
(d) become zero.
EXPLANATION: The frequency of a spring-mass system depends only on the mass and the spring constant, none of which changes in the accelerating car. So the frequency remains the same. Option (c).
18. A pendulum clock that keeps the correct time on the earth is taken to the moon. It will run
(a) at the correct rate
(b) 6 times faster
(c) √6 times faster
(d) √6 times slower.
(b) 6 times faster
(c) √6 times faster
(d) √6 times slower.
ANSWER: (d)
EXPLANATION: The time period of the pendulum of the clock on the earth T =2π√(l/g). On the moon g'=g/6. The time period on the moon T' =2π√6l/g) =√6*2π√(l/g) =√6T.
So the pendulum will take √6 times more time in one oscillation on the moon and it will run √6 times slower. The option (d) is true.
So the pendulum will take √6 times more time in one oscillation on the moon and it will run √6 times slower. The option (d) is true.
19. A wall clock uses a vertical spring-mass system to measure the time. Each time the mass reaches an extreme position, the clock advances by a second. The clock gives the correct time at the equator. If the clock is taken to the poles it will
(a) run slow
(b) run fast
(c) stop working
(d) give the correct time.
(a) run slow
(b) run fast
(c) stop working
(d) give the correct time.
ANSWER: (d)
EXPLANATION: Same explanation as in Q-17. Option (d) is true.
20. A pendulum clock keeping correct time is taken to high altitudes,
(a) it will keep the correct time
(b) its length should be increased to keep correct time
(c) its length should be decreased to keep correct time
(d) it cannot keep the correct time even if the length is changed.
(a) it will keep the correct time
(b) its length should be increased to keep correct time
(c) its length should be decreased to keep correct time
(d) it cannot keep the correct time even if the length is changed.
EXPLANATION: At high altitudes, the value of g decreases. Since the time period of the pendulum T = 2π√(l/g), it will increase at high altitudes. The clock will not give the correct time. To keep the correct time T should not change and to keep T constant the ratio l/g should not change. Since the value of g is getting decreased at the high altitudes, the length l should also be proportionately decreased to keep the ratio l/g constant. Hence the option (c) is true.
21. The free end of a simple pendulum is attached to the ceiling of a box. The box is taken to a height and the pendulum is oscillated. When the bob is at its lowest point, the box is released to fall freely. As seen from the box during this period, the bob will
(a) continue its oscillation as before
(b) stop
(c) will go in a circular path
(d) move on a straight line.
(a) continue its oscillation as before
(b) stop
(c) will go in a circular path
(d) move on a straight line.
EXPLANATION: At the lowest point the bob is moving horizontally with the maximum speed. When we observe the bob from the box when free-falling, the box is a non-inertial frame with acceleration g downwards. If we want to apply Newton laws of motion here we will have to apply a pseudo force mg upwards. Thus the weight mg and the pseudo force -mg have a resultant zero force on the bob and hence no restoring force for the oscillation. Since the bob has a horizontal speed and it is tied to the ceiling with a string it will go in a circular path. Hence the option (c) is true.
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Links to the chapters -
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"
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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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Simple Harmonic Motion Objective I link is not correct. It is opening as Question for Short Answers.Please check
ReplyDeleteSir,
ReplyDeleteIn Shm objective 1 qn, it's F=k√x and not k/x.
Thanks Pritam Shaw! I have corrected it.
Delete