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WBJEEA hollow pipe of length \(0.8 \mathrm{~m}\) is closed at one end. At...
A hollow pipe of length \(0.8 \mathrm{~m}\) is closed at one end. At its open end a \(0.5 \mathrm{~m}\) long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is \(50 \mathrm{~N}\) and the speed of sound is \(320 \mathrm{~ms}^{-1}\), the mass of the string is
Solution:
\(2\left(\frac{v_1}{2 l_1}\right)=\frac{v_2}{4 l_2}\) \(\therefore \quad \frac{\sqrt{T / \mu}}{l_1}=\frac{320}{4 l_2}\)
( \(\mu=\) mass per unit length of wire)
or \(\frac{\sqrt{50 / \mu}}{0.5}=\frac{320}{4 \times 0.8}\)
Solving we get \(\mu=0.02 \frac{\mathrm{kg}}{\mathrm{m}}=20 \frac{\mathrm{g}}{\mathrm{m}}\)
\(\therefore\) Mass of string
\(\)
=\left(20 \frac{\mathrm{g}}{\mathrm{m}}\right)(0.5 \mathrm{~m})=10 \mathrm{~g}
\(\)
\(\therefore\) The correct option is (b).
( \(\mu=\) mass per unit length of wire)
or \(\frac{\sqrt{50 / \mu}}{0.5}=\frac{320}{4 \times 0.8}\)
Solving we get \(\mu=0.02 \frac{\mathrm{kg}}{\mathrm{m}}=20 \frac{\mathrm{g}}{\mathrm{m}}\)
\(\therefore\) Mass of string
\(\)
=\left(20 \frac{\mathrm{g}}{\mathrm{m}}\right)(0.5 \mathrm{~m})=10 \mathrm{~g}
\(\)
\(\therefore\) The correct option is (b).
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