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A string vibrates with a frequency of $200 \mathrm{~Hz}$. When its length is doubled and tension is altered, it begins to vibrate with a frequency of $300 \mathrm{~Hz}$. The ratio of the new tension to the original tension is
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The correct answer is:
9:1
Given, $L_{2}=2 L_{1}, f_{1}=200 \mathrm{~Hz}, f_{2}=300 \mathrm{~Hz}$
We know that, frequency of vibrating string is given as
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}
$$
where, $\mu=$ mass per unit length of string
and $L=$ length of the string.
$$
\begin{array}{ll}
\Rightarrow & f \propto \frac{\sqrt{T}}{L} \\
\Rightarrow \quad & \frac{f_{2}}{f_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}, \frac{L_{1}}{L_{2}} \\
\Rightarrow \quad \sqrt{\frac{T_{2}}{T_{1}}}=\frac{f_{2} L_{2}}{f_{1} L_{1}}=\frac{300}{200} \times \frac{2 L_{1}}{L_{1}} \\
\Rightarrow \quad & \sqrt{\frac{T_{2}}{T_{1}}}=3 \\
& \frac{T_{2}}{T_{1}}=9 \\
\therefore \quad & T_{2}: T_{1}=9: 1
\end{array}
$$
We know that, frequency of vibrating string is given as
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}
$$
where, $\mu=$ mass per unit length of string
and $L=$ length of the string.
$$
\begin{array}{ll}
\Rightarrow & f \propto \frac{\sqrt{T}}{L} \\
\Rightarrow \quad & \frac{f_{2}}{f_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}, \frac{L_{1}}{L_{2}} \\
\Rightarrow \quad \sqrt{\frac{T_{2}}{T_{1}}}=\frac{f_{2} L_{2}}{f_{1} L_{1}}=\frac{300}{200} \times \frac{2 L_{1}}{L_{1}} \\
\Rightarrow \quad & \sqrt{\frac{T_{2}}{T_{1}}}=3 \\
& \frac{T_{2}}{T_{1}}=9 \\
\therefore \quad & T_{2}: T_{1}=9: 1
\end{array}
$$
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