Waves

JEE Main level, numerical answer, medium

88 of 142

Question 88 of 142

In a closed organ pipe, the frequency of fundamental note is 30Hz30Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110Hz110Hz. If the organ pipe has a cross-sectional area of 2cm22{cm}^{2}, the amount of water poured in the organ tube is ________gg. (Take speed of sound in air is 330ms−1330m{s}^{-1} )

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Answer: 400

Fundamental frequency for closed organ pipe in the first part:
V4l1=30⇒l1=114m\dfrac{V}{4{l}_{1}}=30\Rightarrow {l}_{1}=\dfrac{11}{4}m
Similarly, for second part:
V4l2=110⇒l2=34m\dfrac{V}{4{l}_{2}}=110\Rightarrow {l}_{2}=\dfrac{3}{4}m
Therefore, change in length Δl=2m\Delta l=2m
Change in volume =AΔl=400cm3=A\Delta l=400{cm}^{3}
M=400g;(∵ρ=1gcm−3)M=400g;\left(∵\rho =1g{cm}^{-3}\right)

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