> Q. The length of elastic string, obeying Hooke’s law is $ell_{1}$ metres when the tension $4N$ and $ell_{2}$ metres when the tension is $5N$. The length in metres when the tension is $9N$ is – – LIVE ANSWER TODAY

Q. The length of elastic string, obeying Hooke’s law is $ell_{1}$ metres when the tension $4N$ and $ell_{2}$ metres when the tension is $5N$. The length in metres when the tension is $9N$ is –

In This post you will get answer of Q.
The length of elastic string, obeying Hooke’s law is $ell_{1}$ metres when the tension $4N$ and $ell_{2}$ metres when the tension is $5N$. The length in metres when the tension is $9N$ is –

Solution:

Let $ell_{0}$ be the unstretched length and $ell_{3}$ be the length under a tension of $9N$. Then

$Y = frac{4ell_{0}}{Aleft(ell _{1}-ell _{0}right)} = frac{5ell _{0}}{Aleft(ell _{2}-ell _{0}right)}$

$ = frac{9ell _{0}}{Aleft(ell _{3}-ell _{0}right)}$

These give

$frac{4}{ell _{1}-ell _{0}} = frac{5}{ell _{2}-ell _{0}} Rightarrow ell_{0} = 5ell_{1} -4ell_{2}$

Further, $frac{4}{ell _{1}-ell _{0}} = frac{9}{ell _{2}-ell _{0}} $

Substituting the value of $ell_{0}$ and solving, we get $ell _{3} = 5ell _{2} -4ell _{1}$

Leave a Comment