Let

**h0**be the height of the fluid at equilibrium. Suppose that on the left side the fluid height is depressed by then since the total volume of the fluid in the tube should remain the same at any time, fluid height on the right side of the tube will rise. Let the rise in fluid height on the right side be . Since the area of cross section of the tube is**S**, the area of the horizontal cross section of the inclined part of the tube tube on the right side will be,The volume of fluid displaced on the left side of the tube is given by and the same of the right side of the tube is given by . Since the total volume of the liquid must be conserved we have,

Now let us consider the restoring forces acting on the fluid. Let

**S**be the cross section of the tube at the mid-point of the tube in the bottom part. Assuming the height of the liquid h0 is much larger than the radius of the tube and the variation of pressure across the crosssection of the tube is negligible, the force that the left side of the fluid exerts on the right side of the fluid is given by,

Similarly, the force that the right side of the fluid exerts on the left side of the fluid is given by,

Hence, from Newton's laws of motion we have,

From

**(6)**it is clear that the time period of oscillations is given by,

Great solution sir .

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