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A shown in the figure, a liquid is at same levels in two arms of a U-tube of uniform cross-section when at rest. If the \(U\)-tube moves with an acceleration ' \(f\) ' towards right, the difference between liwuid height between two arms of the \(\mathrm{U}\)-tube will be, (acceleration due to gravity \(=\mathrm{g}\) )
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The correct answer is:
\(\frac{f}{g} a\)
Hint : 
\(\begin{aligned} & \tan \theta=\frac{\text { acceleration }}{\mathrm{g}} \\ & \frac{\mathrm{h}}{\mathrm{a}}=\frac{\text { acceleration }}{\mathrm{g}} \\ & \frac{\mathrm{h}}{\mathrm{a}}=\frac{\mathrm{f}}{\mathrm{g}} \\ & \mathrm{h}=\frac{\mathrm{fa}}{\mathrm{g}}\end{aligned}\)
\(\begin{aligned} & \tan \theta=\frac{\text { acceleration }}{\mathrm{g}} \\ & \frac{\mathrm{h}}{\mathrm{a}}=\frac{\text { acceleration }}{\mathrm{g}} \\ & \frac{\mathrm{h}}{\mathrm{a}}=\frac{\mathrm{f}}{\mathrm{g}} \\ & \mathrm{h}=\frac{\mathrm{fa}}{\mathrm{g}}\end{aligned}\)
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