The length of a wire becomes and when 100 N and 120 N tensions are applied respectively. If , the natural length of the wire will be . Here the value of is __________ . [2023]

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Solution

Q.
The length of a wire becomes $l_{1}$ and $l_{2}$ when 100 N and 120 N tensions are applied respectively. If $10 l_{2} = 11 l_{1}$ , the natural length of the wire will be $\frac{1}{x} l_{1}$ . Here the value of $x$ is __________ . [2023]

Ans.
(2)

$Let the original length be$ $' l_{0}'$

$When T_{1} = 100 N, Extension = l_{1} - l_{0}$

$When T_{2} = 120 N, Extension = l_{2} - l_{0}$

Then $100 = K (l_{1} - l_{0})$ ...(1)

and $120 = K (l_{2} - l_{0})$ ...(2)

$\Rightarrow \frac{5}{6} = \frac{l_{1} - l_{0}}{l_{2} - l_{0}}$

$5 l_{2} - 5 l_{0} = 6 l_{1} - 6 l_{0}$

$l_{0} = 6 l_{1} - 5 l_{2}$

$\Rightarrow l_{0} = 6 l_{1} - 5 (\frac{11 l_{1}}{10})$

$l_{0} = \frac{l_{1}}{2}$ $∴$ $x = 2$

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