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In a resonance tube experiment when the tube is filled with water up to a height of 20 \mathrm{~cm} from bottom. It resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of 40 \mathrm{~cm}. If the velocity of sound in dir is 332 \mathrm{~m} / \mathrm{s} the tuning fork frequency

Option: 1

880 \mathrm{~Hz}


Option: 2

830 \mathrm{~Hz}


Option: 3

330 \mathrm{~Hz}


Option: 4

800 \mathrm{~Hz}


Answers (1)

best_answer

$$ \begin{aligned} & 40-20=\frac{\lambda}{2} \\ & 20=\frac{\lambda}{2} \\ & \lambda=40 \mathrm{~cm} \\ & u=f \lambda \\ & \delta_{0}=\frac{u}{\lambda}=\frac{332}{40 \times 10^{-2}}=8.3 \times 10^{2} \\ & f_{0}=830 \mathrm{~Hz} \end{aligned}

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Irshad Anwar

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