Question

A rectangular box containing water is accelerated upwards at $3m/s^2$ on an inclined plane making ${30}^o$ to the horizontal. The slope of the free liquid surface is

• A. $0.23$
• B. $\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $0.32$

Answer 1

Answer: (A)

$0.23$

So,

Net downward acceleration$=(g+\frac{3}{2})m/s^2$

Net upward acceleration$=\frac{3\sqrt{3}}{2}m/s^2$So,

A a horizontal distance l,

pressure at $L=\lambda L\frac{3\sqrt{3}}{2}=\lambda \left(\frac{2g+3}{2}\right)h$

$\Rightarrow \frac{3\sqrt{3}L}{2g+3}=h$

So,

slope of free liquid surface $=\tan \theta =\frac{\frac{3\sqrt{3}L}{2g+3}}{L}$

$\tan \theta =0.23$