99 Percentile Qs Bank for JEE MainPhysicsOscillations
An air chamber of volume V has a neck of cross-sectional area a into which a light ball of mass m just fits and can move up and down without friction. The diameter of the ball is equal to that of the neck of the chamber. The ball is pressed down a little and released. If the bulk modulus of air is B , the time period of the oscillation of the ball is
Options
- AT = 2 π B a 2 m V
- BT = 2 π B V m a 2
- CT = 2 π m B V a 2
- DT = 2 π m V B a 2
Correct answer
D. T = 2 π m V B a 2
Step-by-step solution
The situation is as shown in the figure. Let P be pressure of air in the chamber. When the ball is pressed down a distance x , the volume of air decreases from V to say V - ∆ V . Hence, the pressure increase from P to P + ∆ P . The change in volume is, ∆ V = a x The excess pressure ∆ p is related to the bulk modulus B as ∆ P = - B ∆ V V Restoring force on ball = excess pressure × cross-sectional area or F = - B a V ∆ V or F = - B a 2 V x   ( ∵ ∆ V = a