Verified Solution • 2025–26 Session moderate

Q13.18: NCERT Physics - Part 2 OSCILLATIONS 13.18 Solution: One end of a U-tube containing mercury is connected to a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the pressure difference created by a displacement $y$ as $\rho g A y$ (as if only one arm of the U‑tube contributes) instead of the correct $2\rho g A y$ from the two arms, leading to an incorrect $omega^2 = g/L$ and period $T=2\pi\sqrt{L/g}$. Remember that the height difference between the arms is $2y$, so the net hydrostatic pressure difference is $2\rho g y$, and the entire column of length $L$ provides the inertial mass.

One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Step-by-Step Verified Solution

**Answer:** T=2πL2gT = 2\pi\sqrt{\frac{L}{2g}} where L is total length of mercury column **Step-by-Step Explanation:** When mercury is displaced by y from equilibrium, the difference in levels between the two arms is 2y. Restoring force = weight of mercury column of height 2y:
F=−(A×2y×ρ×g)=−2AρgyF = -(A \times 2y \times \rho \times g) = -2A\rho g y
Acceleration: a=F/ma = F/m where m=ALρm = AL\rho (L = total length).
a=−2AρgyALρ=−2gLya = -\frac{2A\rho g y}{AL\rho} = -\frac{2g}{L}y
Since a∝−ya \propto -y, motion is SHM with:
ω2=2gL⇒T=2πL2g\omega^2 = \frac{2g}{L} \Rightarrow T = 2\pi\sqrt{\frac{L}{2g}}
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All Questions in this Chapter

Q13.1 Q13.2 Q13.3 Q13.4 Q13.5 Q13.6 Q13.7 Q13.8 Q13.9 Q13.10 Q13.11 Q13.12 Q13.13 Q13.14 Q13.15 Q13.16 Q13.17 Q13.18

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