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Simple Harmonic Motion question

2005 · Shift 0 · Q156
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  5. /2005 · Shift 0 · Q156

Simple Harmonic Motion question

2005 · Shift 0 · Q156

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would
  1. A
    first decrease and then increase to the original value
  2. B
    first increase and then decrease to the original value
  3. C
    increase towards a saturation value
  4. D
    remain unchanged
View written solutionFree

Correct answer: B

  1. Time period of a simple pendulum

For small oscillations, T=2πℓg,T = 2\pi \sqrt{\frac{\ell}{g}},T=2πgℓ​​, where ℓ\ellℓ is the effective length of the pendulum, i.e. the distance between the point of suspension and the centre of mass of the bob.

So, to know how TTT changes, we must track how the centre of mass of the hollow spherical bob + water inside it shifts as water leaks out.


  1. What happens to the centre of mass when water starts leaking?

Initially, the ball is full of water.

  • Because the spherical shell is symmetric and the water completely fills the sphere, the centre of mass of the bob-water system is at the geometrical centre of the sphere.

Now water begins to come out from a hole near the bottom.

As the water level falls:

  • more mass remains in the lower portion at first,
  • so the centre of mass of the remaining water shifts downward below the sphere’s centre.

Hence the centre of mass of the whole bob-water system also shifts downward.

Therefore, the effective length ℓ\ellℓ increases, so T=2πℓgT = 2\pi\sqrt{\frac{\ell}{g}}T=2πgℓ​​ increases.


  1. What happens after more water has leaked out?

When a large amount of water has escaped, only a small quantity remains near the bottom. Then:

  • the mass of the remaining water becomes small compared to the shell,
  • the centre of mass of the whole system starts moving back upward toward the centre of the hollow spherical shell.

Finally, when all water has drained out, only the hollow spherical shell remains. Since the shell is symmetric, its centre of mass is again at the geometrical centre.

So the effective length returns to its original value, and hence the time period returns to its original value.

Thus during leakage, the time period:

  • first increases,
  • then decreases back to the original value.

  1. Option check
  • A: first decrease and then increase to the original value — incorrect
  • B: first increase and then decrease to the original value — correct
  • C: increase towards a saturation value — incorrect, because it eventually returns to the original value
  • D: remain unchanged — incorrect

  1. Final answer

The correct option is B\boxed{\text{B}}B​

The time period first increases and then decreases to the original value.

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