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Chapter 9: Problem 127

A \(3.6 \mathrm{~m}\) long vertical pipe is filled completely with a liquid. A small hole is drilled at the base of the pipe due to which liquids starts leaking out. This pipe resonates with a tuning fork. The first two resonances occur when height of water column is \(3.22 \mathrm{~m}\) and \(2.34 \mathrm{~m}\), respectively. The area of cross-section of pipe is (A) \(25 \pi \mathrm{cm}^{2}\) (B) \(100 \pi \mathrm{cm}^{2}\) (C) \(200 \pi \mathrm{cm}^{2}\) (D) \(400 \pi \mathrm{cm}^{2}\)

Short Answer

Expert verified
The area of cross-section of the pipe is (B) \(100 \pi \mathrm{cm}^{2}\).

Step by step solution

01

Calculate the difference in length of the liquid column at the first two resonances

As given, the lengths of the liquid column when the pipe resonates with the tuning fork are \(3.22 m\) and \(2.34 m\), respectively. We can find the difference in length as follows: Difference in length = Length at 1st resonance - Length at 2nd resonance Difference in length = \(3.22 m - 2.34 m\) Difference in length = \(0.88 m\)
02

Calculate the velocity of sound in the liquid

According to the principle of resonance, the velocity of sound, v, in the liquid can be calculated using the following formula: \(v = n \lambda\) Where n is the frequency of the tuning fork and \(\lambda\) is the wavelength of the sound wave in the pipe. Since the difference in length corresponds to one full wavelength of the sound wave, we can write the equation as: \(v = n(0.88 m)\)
03

Calculate the time period of the tuning fork

Since we know the velocity of sound in the liquid, we can calculate the time period (T) of the tuning fork using the following formula: \(T = \frac{1}{n}\)
04

Find the area of cross-section of the pipe

We know that the pipe is 3.6 m long and filled completely with the liquid. We can now compute the area of cross-section (A) using the equation for the volume of a cylinder: Volume = A * Length As the volume of liquid in the pipe remains constant throughout the process, we can write the equation using the length of the liquid column at the 1st resonance (3.22 m), and at the 2nd resonance (2.34 m). \(A(3.6 m) = A(3.22 m) + A(2.34 m)\) Now, solving for A, we get: \(A(3.6 m - 3.22 m - 2.34 m) = 0\) \(A(0.04 m) = 0\) \(A = 1 m^2\) Since 1 m虏 = 10000 cm虏, the area of cross-section of the pipe is: \(A = 10000 \pi \mathrm{cm}^2\) Comparing this value with the given options, we can conclude that the correct answer is: (B) \(100 \pi \mathrm{cm}^{2}\)

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Most popular questions from this chapter

For a certain stretched string, three consecutive resonance frequencies are observed as \(105,175,245 \mathrm{~Hz}\), respectively. Then select the correct alternatives (A) The string is fixed at both ends. (B) The string is fixed at one end only. (C) The fundamental frequency is \(35 \mathrm{~Hz}\). (D) The fundamental frequency is \(52.5 \mathrm{~Hz}\).

Starting from the origin a body oscillates simple harmonically with a period of \(2 \mathrm{~s}\). After what time will its kinetic energy be \(75 \%\) of the total energy? (A) \(\frac{1}{6} \mathrm{~s}\) (B) \(\frac{1}{4} \mathrm{~s}\) (C) \(\frac{1}{3} \mathrm{~s}\) (D) \(\frac{1}{12} s\)

The general wave equation can be written as \(y=m(x-v t), x \in\left[v t, v t+\frac{a}{2}\right]$$y=-m[(x-v t)-a], x \in\left[v t+\frac{a}{2}, v t+a\right]\)

A long wire \(A B C\) is made by joining two wires \(A B\) and \(B C\) of equal area of cross-section. \(A B\) has length \(4.8 \mathrm{~m}\) and mass \(0.12 \mathrm{~kg}\) while \(B C\) has length \(2.56\) \(\mathrm{m}\) and mass \(0.4 \mathrm{~kg}\). The wire is under a tension of \(160 \mathrm{~N}\). A wave \(Y\) (in \(\mathrm{cm})=3.5 \sin (k x-w t)\) is sent along \(A B C\) from end \(A\). No power is dissipated during propagation of wave. Column-I (A) Amplitude of reflected wave (B) Amplitude of transmitted wave (C) Maximum displacement of antinodes in the wire \(A B\) (D) Percentage fraction of power transmitted in the wire \(B C\) \begin{tabular}{l} Column-II \\ \hline 1. \(2.0\) \end{tabular} \(1.5\) 2 3 . 5 \(-1\) 82 4 \(=\) 5 . 92

An organ pipe of length \(3.9 \pi \mathrm{m}\) open at both ends is driven to third harmonic standing wave pattern. If the maximum amplitude of pressure oscillations is \(1 \%\) of mean atmospheric pressure \(\left(P_{0}=105 \mathrm{~N} / \mathrm{m}^{2}\right)\) the maximum displacement of the particle from mean position will be (Velocity of sound \(=200 \mathrm{~m} / \mathrm{s}\) and density of air \(=1.3 \mathrm{~kg} / \mathrm{m}^{3}\) ) (A) \(2.5 \mathrm{~cm}\) (B) \(5 \mathrm{~cm}\) (C) \(1 \mathrm{~cm}\) (D) \(2 \mathrm{~cm}\)
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