A cylindrical plastic bottle of negligible ma | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The restoring force $F$ when the bottle is pushed down by a distance $x$ is given by the weight of the displaced water: $F = A \rho g x$,where $A

Vedclass · Accepted Answer

$7.9$

Vedclass · Answer

(B) The restoring force $F$ when the bottle is pushed down by a distance $x$ is given by the weight of the displaced water: $F = A ho g x$,where $A = \pi r^2$ is the cross-sectional area of the bottle.For simple harmonic motion,$F = m \omega^2 x$,where $m$ is the mass of the water inside the bottle.Equating the two: $\pi r^2 ho g x = m \omega^2 x$.Thus,$\omega = \sqrt{\frac{\pi r^2 ho g}{m}}$.Given $m = ho V$,where $V = 310\, ml = 310 	imes 10^{-6}\, m^3$ and $r = 2.5 	imes 10^{-2}\, m$.Substituting the values: $\omega = \sqrt{\frac{\pi r^2 ho g}{ho V}} = r \sqrt{\frac{\pi g}{V}}$.$\omega = (2.5 	imes 10^{-2}) \sqrt{\frac{3.14 	imes 10}{310 	imes 10^{-6}}} = (2.5 	imes 10^{-2}) \sqrt{\frac{31.4}{310 	imes 10^{-6}}} \approx (2.5 	imes 10^{-2}) \sqrt{10^5} \approx 2.5 	imes 316 	imes 10^{-2} \approx 7.9\, rad\, s^{-1}$.

Similar Questions

The time period of a simple pendulum is $T$. The time taken to complete $5/8$ oscillations starting from the mean position is $\frac{\alpha}{\beta} T$. The value of $\alpha$ is ..... .

From a fixed support,two small identical spheres are suspended by means of strings of length $1 \,m$ each. They are pulled aside as shown and then released. $B$ is the mean position. Then the two spheres collide

$A$ simple pendulum of length $L$ has mass $m$ and it oscillates freely with amplitude $A$. At the extreme position,its potential energy is (where $g$ is the acceleration due to gravity):

$A$ simple pendulum is vibrating in an evacuated chamber. It will oscillate with:

Vedclass Test Series

Exam Paper Generator

Online Exam Module