Wave Mechanics Question 588
Question: A bent tube of uniform cross-section area A has a non-viscous liquid of density p. The mass of liquid in the tube is m. The time period of oscillation of the liquid is
Options:
A) $ 2\pi \sqrt{\frac{m}{\rho gA}} $
B) $ 2\pi \sqrt{\frac{m}{2\rho gA}} $
C) $ 2\pi \sqrt{\frac{2m}{\rho gA}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The restoring force $ F=-(PA) $
$ =-[\rho g(2y\sin 30^{o})]A $
$ =\rho gA(-y) $
$ \
Therefore a=\frac{F}{m}=\frac{\rho gA}{m} $ Thus $ T=2\pi \sqrt{\frac{m}{\rho gA}} $
BETA
