Question

Figure shows a three arm tube in which a liquid is filled upto levels of height l. It is now rotated at an angular frequency $\omega$ about an axis passing through arm B. The angular frequency $\omega$ at which level of liquid in arm B becomes zero.

• A. $\sqrt{\frac{2g}{3l}}$
• B. $\sqrt{\frac{g}{l}}$
• C. $\sqrt{\frac{3g}{l}}$
• D. $\sqrt{\frac{3g}{2l}}$

Answer 1

The correct option is C $\sqrt{\frac{3g}{l}}$

when middle tube gets empty half water is fitted with left arm half of the water is half of the tube

height of liquid $l+\frac{l}{2}=\frac{3l}{2}$

liquid lying in the bottom of room from the centre of the tube

$F_{centripetalforce}\times massofliquid\times \left(centripetalacceleration\right)$

$pressure\times areacross=m\times {\omega }^2\times r$

$△p\times a=m\times {\omega }^2\times r$

$\frac{3l}{2}\rho \times g\times a=a\times l\times p\times {\omega }^2\times \frac{l}{2}$

$\frac{3g}{2}=\frac{{l\omega }^2}{2}$

$3g=l{\omega }^2$

$\omega =\sqrt{\frac{3g}{l}}$