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A pipe open at one end has length $0.8 \mathrm{~m}$. At the open end of the tube a string $0.5 \mathrm{~m}$ long is vibrating in its $1^{\text {st }}$ overtone and resonates with fundamental frequency of pipe. If tension in the string is $50 \mathrm{~N}$, the mass of string is (speed of sound $=320 \mathrm{~m} / \mathrm{s}$)A)25gB)15gC)20gD)10g
A pipe open at one end has length $0.8 \mathrm{~m}$. At the open end of the tube a string $0.5 \mathrm{~m}$ long is vibrating in its $1^{\text {st }}$ overtone and resonates with fundamental frequency of pipe. If tension in the string is $50 \mathrm{~N}$, the mass of string is (speed of sound $=320 \mathrm{~m} / \mathrm{s}$)A)25gB)15gC)20gD)10g
Physics
A pipe open at one end has length $0.8 \mathrm{~m}$. At the open end of the tube a string $0.5 \mathrm{~m}$ long is vibrating in its $1^{\text {st }}$ overtone and resonates with fundamental frequency of pipe. If tension in the string is $50 \mathrm{~N}$, the mass of string is (speed of sound $=320 \mathrm{~m} / \mathrm{s}$)A)25gB)15gC)20gD)10g

Correct option is 10g

Explanation:

Given, 2 nd harmonic of $\mathrm{I}=$ Fundamental of II
$
\begin{array}{l}
\therefore \quad 2\left(\frac{v_{1}}{2 l_{1}}\right)=\frac{v_{2}}{4 l_{2}} \Rightarrow \frac{T / \mu}{l_{1}}=\frac{v_{2}}{4 l_{2}} \\
\Rightarrow \quad \mu=\frac{16 T l_{2}^{2}}{v_{2}^{2} l_{1}^{2}}=\frac{16 \times 50 \times(0.8)^{2}}{(320)^{2} \times(0.5)^{2}} \\
\quad=0.02 \mathrm{~kg} / \mathrm{m}
\end{array}
$
$
\therefore \quad m_{1}=\mu l_{1}=(0.02)(0.2)
$
$
=0.01 k g=10 g
$

i

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