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The length of elastic string, obeying Hooke's law is $\ell_{1}$ metres when the tension $4 \mathrm{~N}$ and $\ell_{2}$ metres when the tension is $5 \mathrm{~N}$. The length in metres when the tension is $9 \mathrm{~N}$ is $-$
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Verified Answer
The correct answer is:
$5 \ell_{2}-4 \ell_{1}$
Let $\ell_{0}$ be the unstretched length and $\ell_{3}$ be the length under a tension of $9 \mathrm{~N}$. Then
$$
\begin{aligned}
\mathrm{Y} &=\frac{4 \ell_{0}}{\mathrm{~A}\left(\ell_{1}-\ell_{0}\right)}=\frac{5 \ell_{0}}{\mathrm{~A}\left(\ell_{2}-\ell_{0}\right)} \\
&=\frac{9 \ell_{0}}{\mathrm{~A}\left(\ell_{3}-\ell_{0}\right)}
\end{aligned}
$$
These give
$$
\frac{4}{\ell_{1}-\ell_{0}}=\frac{5}{\ell_{2}-\ell_{0}} \Rightarrow \ell_{0}=5 \ell_{1}-4 \ell_{2}
$$
Further, $\frac{4}{\ell_{1}-\ell_{0}}=\frac{9}{\ell_{2}-\ell_{0}}$
Substituting the value of $\ell_{0}$ and solving, we get $\ell_{3}=5 \ell_{2}-4 \ell_{1}$
$$
\begin{aligned}
\mathrm{Y} &=\frac{4 \ell_{0}}{\mathrm{~A}\left(\ell_{1}-\ell_{0}\right)}=\frac{5 \ell_{0}}{\mathrm{~A}\left(\ell_{2}-\ell_{0}\right)} \\
&=\frac{9 \ell_{0}}{\mathrm{~A}\left(\ell_{3}-\ell_{0}\right)}
\end{aligned}
$$
These give
$$
\frac{4}{\ell_{1}-\ell_{0}}=\frac{5}{\ell_{2}-\ell_{0}} \Rightarrow \ell_{0}=5 \ell_{1}-4 \ell_{2}
$$
Further, $\frac{4}{\ell_{1}-\ell_{0}}=\frac{9}{\ell_{2}-\ell_{0}}$
Substituting the value of $\ell_{0}$ and solving, we get $\ell_{3}=5 \ell_{2}-4 \ell_{1}$
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