MHTCET 11 May 2024 Shift 1 Paper

Concept:A resonance tube closed at one end behaves like a closed organ pipe.Resonance is obtained when the air column length $L$ equals an odd multiple of $\frac{\lambda}{4}$.Explanation:Given:Frequency, $f = 340\ \text{Hz}$Speed of sound, $v = 340\ \text{m/s}$Wavelength, $\lambda = \frac{v}{f} = \frac{340}{340} = 1\ \text{m} = 100\ \text{cm}$.For a closed pipe, resonant air column lengths are:$\frac{\lambda}{4} = 25\ \text{cm}$,$\frac{3\lambda}{4} = 75\ \text{cm}$,$\frac{5\lambda}{4} = 125\ \text{cm}$, etc.The tube height is $1.5\ \text{m} = 150\ \text{cm}$.Water is poured gradually, so the air column length decreases from $150\ \text{cm}$.The first resonance encountered is at the largest possible resonant air column, which is $125\ \text{cm}$.Therefore, height of water column $= 150 - 125 = 25\ \text{cm}$.This is the minimum height of water required for resonance.Answer:$25\ \text{cm}$ (Option D).