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When the tens...

Question

When the tension on a wire is $4 N$ its length is $l_{1}$ . When the tension on the wire is $5 N$ its length is $l_{2}$ . Find its natural length.

5 l_{1} - 4 l_{2}

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4 l_{1} - 5 l_{2}

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10 l_{1} - 8 l_{2}

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8 l_{1} - l 0 l_{2}

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Solution

The correct option is A $5 l_{1} - 4 l_{2}$
$\begin{matrix} T = 4 N & length = l_{1} T = 5 N & length = l_{2} \end{matrix}$
$Let natural length = x$

$\begin{matrix} using, \frac{F}{A} = Y \frac{Δ l}{l} (1) \frac{4}{A} = Y \frac{ℓ_{1} - x}{x} \begin{matrix} (2) \frac{5}{A} = Y \frac{l_{2} - x}{x} (1) \div (2) \end{matrix} \frac{4}{5} = \frac{l_{1} - x}{ℓ_{2} - x} \Rightarrow 4 l_{2} - 4 x = 5 l_{1} - 5 x \Rightarrow x = 5 l_{1} - 4 l_{2} \end{matrix}$

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