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Physics Question on Hooke's Law

The length of elastic string, obeying Hooke�s law is ℓ1\ell_{1} metres when the tension 4N4N and ℓ2\ell_{2} metres when the tension is 5N5N. The length in metres when the tension is 9N9N is -

A

5ℓ1−4ℓ25\ell_{1} - 4\ell_{2}

B

5ℓ2−4ℓ15\ell_{2} - 4\ell_{1}

C

9ℓ1−8ℓ29\ell_{1} - 8\ell_{2}

D

9ℓ2−8ℓ19\ell_{2} - 8\ell_{1}

Answer

5ℓ2−4ℓ15\ell_{2} - 4\ell_{1}

Explanation

Solution

Let ℓ0\ell_{0} be the unstretched length and ℓ3\ell_{3} be the length under a tension of 9N9N. Then
Y=4ℓ0A(ℓ1−ℓ0)=5ℓ0A(ℓ2−ℓ0)Y = \frac{4\ell_{0}}{A\left(\ell _{1}-\ell _{0}\right)} = \frac{5\ell _{0}}{A\left(\ell _{2}-\ell _{0}\right)}
=9ℓ0A(ℓ3−ℓ0)= \frac{9\ell _{0}}{A\left(\ell _{3}-\ell _{0}\right)}
These give
4ℓ1−ℓ0=5ℓ2−ℓ0⇒ℓ0=5ℓ1−4ℓ2\frac{4}{\ell _{1}-\ell _{0}} = \frac{5}{\ell _{2}-\ell _{0}} \Rightarrow \ell_{0} = 5\ell_{1} -4\ell_{2}
Further, 4ℓ1−ℓ0=9ℓ2−ℓ0\frac{4}{\ell _{1}-\ell _{0}} = \frac{9}{\ell _{2}-\ell _{0}}
Substituting the value of ℓ0\ell_{0} and solving, we get ℓ3=5ℓ2−4ℓ1\ell _{3} = 5\ell _{2} -4\ell _{1}