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The length of an elastic string is $a$ metre when the longitudinal tension is $4 \mathrm{~N}$ and $b$ metre when the longitudinal tension is $5 \mathrm{~N}$. The length of the string in metre when longitudinal tension is $9 \mathrm{~N}$ is
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Verified Answer
The correct answer is:
$5 b-4 a$
Let $L$ is the original length of the wire and $k$ is force constant of wire. Final length = initial length $+$ elongation
$$
L^{\prime}=L+\frac{F}{k}
$$
For first condition $a=L+\frac{4}{k}$ For second condition $b=L+\frac{5}{k}$
By solving Eqs. (i) and (ii), we get
$$
L=5 a-4 b \text { and } k=\frac{1}{b-a}
$$
Now, when the longitudinal tension is $9 \mathrm{~N}$, length of the string
$$
\begin{array}{l}
=L+\frac{9}{k}=5 a-4 b+9(b-a) \\
=5 b-4 a
\end{array}
$$
$$
L^{\prime}=L+\frac{F}{k}
$$
For first condition $a=L+\frac{4}{k}$ For second condition $b=L+\frac{5}{k}$
By solving Eqs. (i) and (ii), we get
$$
L=5 a-4 b \text { and } k=\frac{1}{b-a}
$$
Now, when the longitudinal tension is $9 \mathrm{~N}$, length of the string
$$
\begin{array}{l}
=L+\frac{9}{k}=5 a-4 b+9(b-a) \\
=5 b-4 a
\end{array}
$$
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