A hollow pipe of length $0.8 \mathrm{~m}$ is closed at one end. At its open end a $0.5 \mathrm{~m}$ long…
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
$5 \mathrm{~g}$
$10 \mathrm{~g}$
$20 \mathrm{~g}$
$40 \mathrm{~g}$
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)