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NCERT Exemplar · Q38

Q.One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45° each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

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Yes -- the mercury column executes SHM, because the restoring force from the pressure imbalance between the two inclined arms is proportional to displacement along the tube. Working through the geometry carefully gives ω2=g2L\omega^2=\dfrac{g\sqrt2}{L}, so the time period is T=2πL2 g=π22 Lg≈1.69L/gT=2\pi\sqrt{\dfrac{L}{\sqrt2\,g}}=\pi\sqrt{\dfrac{2\sqrt2\,L}{g}}\approx1.69\sqrt{L/g}, where LL is the total length of the mercury column.

Why this is SHM

When the suction pump is removed, mercury in the two 45-degree-inclined arms is momentarily unbalanced. Gravity drives it back toward the level, equal-height equilibrium. The question is whether the restoring force is proportional to how far the mercury has been displaced -- the hallmark of SHM.

Setting up the geometry

Let xx be the displacement of mercury along the tube from equilibrium: if one arm's mercury surface rises by xx along the tube, the other arm's surface falls by xx along the tube (mercury is incompressible, uniform cross-section AA).

Each arm is inclined at 45∘45^\circ to the horizontal, so a displacement xx along the tube corresponds to a vertical rise or fall of xsin⁡45∘=x2x\sin45^\circ = \dfrac{x}{\sqrt2}. Since one side rises and the other falls, the total vertical height difference is:

h=2×x2=x2h = 2\times\frac{x}{\sqrt2} = x\sqrt2

Restoring force and equation of motion

This height difference creates a hydrostatic pressure imbalance:

ΔP=ρgh=ρgx2\Delta P = \rho g h = \rho g x\sqrt2

Acting over the cross-section AA, this gives a net restoring force on the whole mercury column:

F=−AΔP=−ρgAx2F = -A\Delta P = -\rho g A x\sqrt2

The total mass of mercury in the tube (length LL) is m=ρALm=\rho AL, so:

a=Fm=−ρgAx2ρAL=−g2Lxa = \frac Fm = \frac{-\rho g Ax\sqrt2}{\rho AL} = -\frac{g\sqrt2}{L}x …

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