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SIMPLE HARMONIC MOTION:--
EXERCISES Q11 TO Q20
11. A block of mass 0.50 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.10 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.
ANSWER: The mass m = 0.50 kg, A = 0.10 m, T = 0.314 s. If the spring constant is k,
⍵ =2π/T =2π/0.314
=2*3.14/0.314
=20 s⁻¹
The magnitude of the maximum force =m⍵²A
=0.50*20²*0.10 N
=0.50*400*0.10 N
=20 N
or alternately,
T = 2π√(m/k)
→0.314 = 2π√(0.50/k)
→0.50/k = 0.314*0.314/4π²
→k = 0.50*4*π²/0.314*0.314 =200 N/m
The maximum force is at the extreme
F=k*A =200*0.10 N =20 N
When the spring is at rest with the block hanging (The mean position) the force by the spring on the block is mg = 0.50*10 =5 N. From this position the spring is stretched to the amplitude 0.10 m for which the extra force 20 N. So the total force by the spring on the block is 20+5 = 25 N.
12. A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4 s. If the oscillations are stopped and the body hangs in equilibrium, find the potential energy stored in the spring.
ANSWER: The mass m = 2 kg, T= 4 s.
T=2π√(m/k)
→m/k =T²/4π²
→k =4π²m/T² =4π²*2/4² =π²/2 =4.93 N/m
At the equilibrium extension of the spring x = F/k =2*10/4.93 =4.06 m
The potential energy stored in the spring at this position = ½kx²
=½*4.93*4.06²
=40.6 J ≈40 J
13. A spring stores 5 J of energy when stretched by 25 cm. It is kept vertical with the lower end fixed. A block fastened to its other end is made to undergo small oscillations. If the block makes 5 oscillations each second, what is the mass of the block?
ANSWER: The energy stored E =½kx²
→5 = ½k(25/100)²
→k =10/0.25² =160 N/m
𝜈 =1/T =5 Hz
→T =1/5 s =0.20 s
But T = 2π√(m/k) = 0.20
→√(m/160) =0.20/2π
→m =160*0.20*0.20/4π² =0.16 kg
14. A small block of mass m is kept on a bigger block of mass M which is attached to a vertical spring of spring constant k as shown in the figure. The system oscillates vertically. (a) Find the resultant force on the smaller block when it is displaced through a distance x above its equilibrium position. (b) Find the normal force on the smaller block at this position. When is this force smallest in magnitude? (c) What can be the maximum amplitude with which the two blocks may oscillate together?
Figure for Q-14
ANSWER: (a) The magnitude of the acceleration of the blocks at this position, a = ⍵²x.
The resultant force on the smaller block =m⍵²x = m(k/M+m)x =mkx/(M+m)
(b) Let the normal force on the smaller block at this position be N.
The weight of the block =mg.
The net force on the block =mg -N
The net force on the block as calculated in (a) above =mkx/(M+m)
Equating we get,
mg-N = mkx/(M+m)
→N =mg - mkx(M+m)
As we see the only variable in this expression is x, and N will be smallest in magnitude when x is maximum. The maximum value for x is its amplitude i.e. at the highest point.
(c) For the two blocks to move together, N must be greater than zero. For the limiting condition when both blocks are together, N approaches zero. i.e.
mg - mkx(M+m) = 0
→kx/(M+m) =g
→x = g(M+m)/k
It is the maximum amplitude that they can oscillate together.
15. The block of mass m₁ shown in figure (12-E2) is fastened to the spring and the block of mass m₂ is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance (2/k)(m₁+m₂)g.sinθ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of the blocks at the time of separation?
Figure for Q-15
ANSWER: (a) weight of the blocks = (m₁+m₂)g
The component of this weight along the slope, F = (m₁+m₂)g.sinθ
The spring constant = k
Hence the compression of the spring at equilibrium position =F/k
=(m₁+m₂)g.sinθ/k
(b) When the spring is released its motion will be an SHM for which
⍵² = k/(m₁+m₂)
Acceleration of the blocks =⍵²x
The net force on the second block =m₂⍵²x
If R is the force on the m₂ block by the m₁ block, then net force on the m₂ block = m₂g.sinθ-R
When the blocks separate R=0
So, m₂⍵²x = m₂g.sinθ
→x = g.sinθ/⍵² =g.sinθ/{k/(m₁+m₂)} =(m₁+m₂)g.sinθ/k
Which is same as compression in (a) above. That means when the spring reaches its natural length, the blocks separates.
(c) Let the common speed = v
K.E. of the system at this point = ½(m₁+m₂)v²
P.E. of the system at this point = ½kx²
=½k*(m₁+m₂)²g².sin²θ/k²
{Taking equilibrium position as zero P.E.}
=(1/2k)(m₁+m₂)²g².sin²θ
Total Energy of the system at this point = ½(m₁+m₂)v²+ (1/2k)(m₁+m₂)²g².sin²θ
Total energy of the system at full compression (only P.E.) = ½k*{(2/k)(m₁+m₂)g.sinθ}² =(2/K)(m₁+m₂)²g².sin²θ
Since the total energy of the system will be conserved, equating the two,
½(m₁+m₂)v²+ (1/2k)(m₁+m₂)²g².sin²θ = (2/K)(m₁+m₂)²g².sin²θ
→v² +(1/k)(m₁+m₂)g²sin²θ =(4/k)(m₁+m₂)g².sin²θ
→v² =(4/k-1/k)(m₁+m₂)g².sin²θ
→v² =(3/k)(m₁+m₂)g².sin²θ
→v =√{(3/k)(m₁+m₂)}*g.sinθ
16. In figure (12-E3) k=100 N/m, M=1 kg and F=10 N. (a) Find the compression of the spring in the equilibrium position. (b) A sharp blow by some external agent imparts a speed of 2 m/s to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at this instant. (c) Find the time period of the resulting simple harmonic motion. (d) Find the amplitude. (e) Write the potential energy of the spring when the block is at the left extreme. (f) Write the potential energy of the spring when the block is at the right extreme.
The answers of (b), (e) and (f) are different. Explain why this does not violet the principle of conservation of energy.
Figure for Q-16
ANSWER: (a) The compression of the spring at equilibrium position =F/k =10 N/100 N/m =0.10 m =10 cm
(b) P.E. of the spring = ½kx² =½*100*(0.10)² J
=0.50 J
K.E. of the block = ½mv² =½*1.0*2² =2.0 J
Hence the sum of the P.E. of the spring and the K.E. of the block
=0.50+2.0 =2.50 J
(c) Taking the initial equilibrium position as the mean position
Time period of the oscillation T = 2π√(m/k) =2π√(1/100) s
=2π/10 s =π/5 s
(d) At the mean position velocity v =2 m/s
Total energy at mean position =P.E.+K.E.
=0+½mv²
=½*1*2*2
=2 J
At displacement x= amplitude A, the total energy =P.E.+K.E.
=½m⍵²A²+0
=½*1*(2π/T)²*A²
=½(2π*5/π)²A²
=50A²
But this total energy of the system must be the same, hence
50A² = 2
→A² = 2/50 = 0.04
→A =0.20 m =20 cm
(e) The total displacement of the block from the unstretched position
=Displacement due to force F + displacement due to SHM produced by the given velocity
=10 cm + 20 cm
=30 cm
=0.30 m
The total P.E. stored at the left extreme =½kx²
=½*100*0.30²
=50*0.09 J
=4.50 J
(f) The displacement towards right from the mean position =A
=20 cm
The figure for Q-16
But the mean position is already compressed 10 cm towards left. So the spring is stretched 20 cm - 10 cm = 10 cm towards right.
The P.E. stored in the spring at the right extreme =½*100*0.10²
= 0.50 J
The energy is conserved in a system when no external work is done. Since we are considering the energy stored in the spring, if we take it as a system the force F does an external work on the system that is why the energy stored in the spring is not constant. But if we take the spring, force F and the block with velocity as a system which executes an SHM, the energy of this system is constant.
17. Find the time period of the oscillation of mass m in figure (12-E4 a, b, c). What is the equivalent spring constant of the pair of springs in each case.
Figure for Q-17
ANSWER: (a) Let the equivalent spring constant be k. If a displacement of x is given to the block, each of the springs will be displaced by x. The total energy stored in the springs =½k₁x²+½k₂x²
With the equivalent spring constant this energy stored =½kx²
Hence ½kx² = ½k₁x²+½k₂x²
→k = k₁+k₂
Hence the time period of the oscillation T =2π√(m/k)
=2π√[m/(k₁+k₂)]
(b) Let the equivalent spring constant be k. If a displacement of x is given to the block, each of the springs will be displaced by x. The total energy stored in the springs =½k₁x²+½k₂x²
With the equivalent spring constant this energy stored =½kx²
Hence ½kx² = ½k₁x²+½k₂x²
→k = k₁+k₂
Hence the time period of the oscillation T =2π√(m/k)
=2π√[m/(k₁+k₂)]
(c) Let us apply a force F along the springs.
Compression of the springs will be F/k₁ and F/k₂ respectively.
If the equivalent spring constant is k then total compression of spring =F/k
Now the energy stored,
½k(F/k)² = ½k₁(F/k₁)²+ ½k₁(F/k₂)²
→1/k =1/k₁+1/k₂ =(k₁+k₂)/k₁k₂
→k = k₁k₂/(k₁+k₂)
So the time period T = 2π√[m/k]
=2π√[m/{k₁k₂/(k₁+k₂)}]
=2π√[m(k₁+k₂)/k₁k₂]
18. The spring shown in figure (12-E5) is unstretched when a man starts pulling on the cord. The mass of the block is M. If the man exerts a constant force F, Find (a) the amplitude and the time period of the motion of the block. (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at this position.
Figure for Q-18
ANSWER: (a) Let the amplitude be A and when the spring is at the equilibrium position the stretching under the force be d.
d =F/k
The energy stored in the spring at this extreme position =½kd². This is the total energy of the SHM.
But according to the SHM the energy at the extreme position =½m⍵²A²
Equating,
½m⍵²A² = ½kd² =½*k*F²/k² =F²/2k
→m(k/m)A² =F²/k {Since ⍵²=k/m}
→kA² =F²/k
→A² = F²/k² =(F/k)²
→A = F/k
Hence,
Amplitude A = F/k.
The time period of the spring-mass system T = 2π√(M/k)
(b) Displacement of the end of the spring at the equilibrium position =d
The energy stored in the spring at this position =½kd²
=½k(F/k)²
=F²/2k
(c) The kinetic energy of the block at this position means the position shown in the figure. When the block executes SHM, this is the mean position. At the mean position, the total energy of the system is in the form of the kinetic energy. Since the total energy of the system =½kd² = F²/2k, the kinetic energy at this position = F²/2k.
19. A particle of mass m is attached to three springs A, B and C of equal force constants k as shown in the figure (12-E6), If the particle is pushed slightly against the spring C and released, find the time period of the oscillation.
Figure for Q-19
ANSWER: If we know the equivalent spring constant K of the system we can easily find out the time period T =2π√(m/K). To find out K let us take a displacement x along C. When x is small, elongation of each spring A and B = x/√2.
The energy stored in each spring A and B =½k(x/√2)² =kx²/4.
The energy stored in the spring C =½kx².
The total energy stored in three springs =2*kx²/4+kx²/2 =kx²
The total energy stored in terms of the equivalent spring constant K = ½Kx², Equating we get,
½Kx² = kx²
→K = 2k
Hence the time period of the oscillation T =2π√(m/2k)
Where m is the mass of the block.
20. Repeat the previous exercise if the angle between each pair of springs is 120° initially.
ANSWER: Similarly let us give a compression x towards C, due to this elongation of B = x.sin30° = x/2. See figure below:-
Figure for Q-20
From the similarity, also the elongation of the spring A = x/2.
The total energy stored in the three springs =½kx²+2*½k(x/2)²
=kx²/2+kx²/4
=3kx²/4
In terms of the equivalent spring constant K, total energy =½Kx².
Equating,
½Kx² = 3kx²/4
→K =3k/2
Hence the time period of the oscillation T =2π√(m/K)
=2π√{m/(3k/2)}
=2π√(2m/3k)
11. A block of mass 0.50 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.10 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.
ANSWER: The mass m = 0.50 kg, A = 0.10 m, T = 0.314 s. If the spring constant is k,
⍵ =2π/T =2π/0.314
=2*3.14/0.314
=20 s⁻¹
The magnitude of the maximum force =m⍵²A
=0.50*20²*0.10 N
=0.50*400*0.10 N
=20 N
or alternately,
T = 2π√(m/k)
→0.314 = 2π√(0.50/k)
→0.50/k = 0.314*0.314/4π²
→k = 0.50*4*π²/0.314*0.314 =200 N/m
The maximum force is at the extreme
F=k*A =200*0.10 N =20 N
When the spring is at rest with the block hanging (The mean position) the force by the spring on the block is mg = 0.50*10 =5 N. From this position the spring is stretched to the amplitude 0.10 m for which the extra force 20 N. So the total force by the spring on the block is 20+5 = 25 N.
12. A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4 s. If the oscillations are stopped and the body hangs in equilibrium, find the potential energy stored in the spring.
ANSWER: The mass m = 2 kg, T= 4 s.
T=2π√(m/k)
→m/k =T²/4π²
→k =4π²m/T² =4π²*2/4² =π²/2 =4.93 N/m
At the equilibrium extension of the spring x = F/k =2*10/4.93 =4.06 m
The potential energy stored in the spring at this position = ½kx²
=½*4.93*4.06²
=40.6 J ≈40 J
13. A spring stores 5 J of energy when stretched by 25 cm. It is kept vertical with the lower end fixed. A block fastened to its other end is made to undergo small oscillations. If the block makes 5 oscillations each second, what is the mass of the block?
ANSWER: The energy stored E =½kx²
→5 = ½k(25/100)²
→k =10/0.25² =160 N/m
𝜈 =1/T =5 Hz
→T =1/5 s =0.20 s
But T = 2π√(m/k) = 0.20
→√(m/160) =0.20/2π
→m =160*0.20*0.20/4π² =0.16 kg14. A small block of mass m is kept on a bigger block of mass M which is attached to a vertical spring of spring constant k as shown in the figure. The system oscillates vertically. (a) Find the resultant force on the smaller block when it is displaced through a distance x above its equilibrium position. (b) Find the normal force on the smaller block at this position. When is this force smallest in magnitude? (c) What can be the maximum amplitude with which the two blocks may oscillate together?
 |
| Figure for Q-14 |
ANSWER: (a) The magnitude of the acceleration of the blocks at this position, a = ⍵²x.
(b) Let the normal force on the smaller block at this position be N.
The weight of the block =mg.
The net force on the block =mg -N
The net force on the block as calculated in (a) above =mkx/(M+m)
Equating we get,
mg-N = mkx/(M+m)
→N =mg - mkx(M+m) As we see the only variable in this expression is x, and N will be smallest in magnitude when x is maximum. The maximum value for x is its amplitude i.e. at the highest point.
(c) For the two blocks to move together, N must be greater than zero. For the limiting condition when both blocks are together, N approaches zero. i.e.
mg - mkx(M+m) = 0
→kx/(M+m) =g
→x = g(M+m)/k
It is the maximum amplitude that they can oscillate together.
15. The block of mass m₁ shown in figure (12-E2) is fastened to the spring and the block of mass m₂ is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance (2/k)(m₁+m₂)g.sinθ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of the blocks at the time of separation?
 |
| Figure for Q-15 |
ANSWER: (a) weight of the blocks = (m₁+m₂)g
The component of this weight along the slope, F = (m₁+m₂)g.sinθ
The spring constant = k
Hence the compression of the spring at equilibrium position =F/k
=(m₁+m₂)g.sinθ/k
(b) When the spring is released its motion will be an SHM for which
⍵² = k/(m₁+m₂)
Acceleration of the blocks =⍵²x
The net force on the second block =m₂⍵²x
If R is the force on the m₂ block by the m₁ block, then net force on the m₂ block = m₂g.sinθ-R
When the blocks separate R=0
So, m₂⍵²x = m₂g.sinθ
→x = g.sinθ/⍵² =g.sinθ/{k/(m₁+m₂)} =(m₁+m₂)g.sinθ/k
Which is same as compression in (a) above. That means when the spring reaches its natural length, the blocks separates.
(c) Let the common speed = v
K.E. of the system at this point = ½(m₁+m₂)v²
P.E. of the system at this point = ½kx²
=½k*(m₁+m₂)²g².sin²θ/k²
{Taking equilibrium position as zero P.E.}
=(1/2k)(m₁+m₂)²g².sin²θ
Total Energy of the system at this point = ½(m₁+m₂)v²+ (1/2k)(m₁+m₂)²g².sin²θ
Total energy of the system at full compression (only P.E.) = ½k*{(2/k)(m₁+m₂)g.sinθ}² =(2/K)(m₁+m₂)²g².sin²θ
Since the total energy of the system will be conserved, equating the two,
½(m₁+m₂)v²+ (1/2k)(m₁+m₂)²g².sin²θ = (2/K)(m₁+m₂)²g².sin²θ
→v² +(1/k)(m₁+m₂)g²sin²θ =(4/k)(m₁+m₂)g².sin²θ
→v² =(4/k-1/k)(m₁+m₂)g².sin²θ
→v² =(3/k)(m₁+m₂)g².sin²θ
→v =√{(3/k)(m₁+m₂)}*g.sinθ
The answers of (b), (e) and (f) are different. Explain why this does not violet the principle of conservation of energy.
 |
| Figure for Q-16 |
ANSWER: (a) The compression of the spring at equilibrium position =F/k =10 N/100 N/m =0.10 m =10 cm
(b) P.E. of the spring = ½kx² =½*100*(0.10)² J
=0.50 J
K.E. of the block = ½mv² =½*1.0*2² =2.0 J
Hence the sum of the P.E. of the spring and the K.E. of the block
=0.50+2.0 =2.50 J
(c) Taking the initial equilibrium position as the mean position
Time period of the oscillation T = 2π√(m/k) =2π√(1/100) s
=2π/10 s =π/5 s
(d) At the mean position velocity v =2 m/s
Total energy at mean position =P.E.+K.E.
=0+½mv²
=½*1*2*2
=2 J
At displacement x= amplitude A, the total energy =P.E.+K.E.
=½m⍵²A²+0
=½*1*(2π/T)²*A²
=½(2π*5/π)²A²
=50A²
But this total energy of the system must be the same, hence
50A² = 2
→A² = 2/50 = 0.04
→A =0.20 m =20 cm
(e) The total displacement of the block from the unstretched position
=Displacement due to force F + displacement due to SHM produced by the given velocity
=10 cm + 20 cm
=30 cm
=0.30 m
The total P.E. stored at the left extreme =½kx²
=½*100*0.30²
=50*0.09 J
=4.50 J
(f) The displacement towards right from the mean position =A
=20 cm
 |
| The figure for Q-16 |
But the mean position is already compressed 10 cm towards left. So the spring is stretched 20 cm - 10 cm = 10 cm towards right.
The P.E. stored in the spring at the right extreme =½*100*0.10²
= 0.50 J
The energy is conserved in a system when no external work is done. Since we are considering the energy stored in the spring, if we take it as a system the force F does an external work on the system that is why the energy stored in the spring is not constant. But if we take the spring, force F and the block with velocity as a system which executes an SHM, the energy of this system is constant.
17. Find the time period of the oscillation of mass m in figure (12-E4 a, b, c). What is the equivalent spring constant of the pair of springs in each case.
 |
| Figure for Q-17 |
ANSWER: (a) Let the equivalent spring constant be k. If a displacement of x is given to the block, each of the springs will be displaced by x. The total energy stored in the springs =½k₁x²+½k₂x²
With the equivalent spring constant this energy stored =½kx²
Hence ½kx² = ½k₁x²+½k₂x²
→k = k₁+k₂
Hence the time period of the oscillation T =2π√(m/k)
=2π√[m/(k₁+k₂)]
(b) Let the equivalent spring constant be k. If a displacement of x is given to the block, each of the springs will be displaced by x. The total energy stored in the springs =½k₁x²+½k₂x²
With the equivalent spring constant this energy stored =½kx²
Hence ½kx² = ½k₁x²+½k₂x²
→k = k₁+k₂
Hence the time period of the oscillation T =2π√(m/k)
=2π√[m/(k₁+k₂)]
(c) Let us apply a force F along the springs.
Compression of the springs will be F/k₁ and F/k₂ respectively.
If the equivalent spring constant is k then total compression of spring =F/k
Now the energy stored,
½k(F/k)² = ½k₁(F/k₁)²+ ½k₁(F/k₂)²
→1/k =1/k₁+1/k₂ =(k₁+k₂)/k₁k₂
→k = k₁k₂/(k₁+k₂)
So the time period T = 2π√[m/k]
=2π√[m/{k₁k₂/(k₁+k₂)}]
=2π√[m(k₁+k₂)/k₁k₂] |
| Figure for Q-18 |
ANSWER: (a) Let the amplitude be A and when the spring is at the equilibrium position the stretching under the force be d.
d =F/k
The energy stored in the spring at this extreme position =½kd². This is the total energy of the SHM.
But according to the SHM the energy at the extreme position =½m⍵²A²
Equating,
½m⍵²A² = ½kd² =½*k*F²/k² =F²/2k
→m(k/m)A² =F²/k {Since ⍵²=k/m}
→kA² =F²/k
→A² = F²/k² =(F/k)²
→A = F/k
Hence,
Amplitude A = F/k.The time period of the spring-mass system T = 2π√(M/k)
(b) Displacement of the end of the spring at the equilibrium position =d
The energy stored in the spring at this position =½kd²
=½k(F/k)²
=F²/2k19. A particle of mass m is attached to three springs A, B and C of equal force constants k as shown in the figure (12-E6), If the particle is pushed slightly against the spring C and released, find the time period of the oscillation.
 |
| Figure for Q-19 |
ANSWER: If we know the equivalent spring constant K of the system we can easily find out the time period T =2π√(m/K). To find out K let us take a displacement x along C. When x is small, elongation of each spring A and B = x/√2.
The energy stored in each spring A and B =½k(x/√2)² =kx²/4.
The energy stored in the spring C =½kx².
The total energy stored in three springs =2*kx²/4+kx²/2 =kx²
The total energy stored in terms of the equivalent spring constant K = ½Kx², Equating we get,
½Kx² = kx²
→K = 2k
Hence the time period of the oscillation T =2π√(m/2k)
Where m is the mass of the block.
20. Repeat the previous exercise if the angle between each pair of springs is 120° initially.
ANSWER: Similarly let us give a compression x towards C, due to this elongation of B = x.sin30° = x/2. See figure below:-
 |
| Figure for Q-20 |
From the similarity, also the elongation of the spring A = x/2.
The total energy stored in the three springs =½kx²+2*½k(x/2)²
=kx²/2+kx²/4
=3kx²/4
In terms of the equivalent spring constant K, total energy =½Kx².
Equating,
½Kx² = 3kx²/4
→K =3k/2
Hence the time period of the oscillation T =2π√(m/K)
=2π√{m/(3k/2)}
=2π√(2m/3k)
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 12 - Simple Harmonic Motion
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
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
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Click here for → Question for Short Answers
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Click here for → OBJECTIVE-II
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Click here for → Exercises (11-20)
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Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
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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
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HC Verma's Concepts of Physics, Chapter-6, Friction
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For more practice on problems on friction solve these "New Questions on Friction" .
Click here for → Friction - OBJECTIVE-II
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Click here for → Exercises (11-20)
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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
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Click here for→ Newton's laws of motion - Objective - I
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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
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Click here for "The Forces" - Exercises
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