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VITEEE 2018 Paper

Section: Physics

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Question : 20 of 125

Marks: +1, -0

The length of elastic string, obeying Hooke’s law is

l_1

metres when the tension 4N and

l_2

metres when the tension is 5N. The length in metres when the tension is 9N is -

$5l_1-4l_2$
$5l_2-4l_1$
$9l_1-8l_2$
$9l_2-8l_1$

Solution:

Let

\ell_{0}

be the unstretched length and

\ell_{3}

be the length under a tension of

9\,\text{N}

.

Then

Y = \frac{4\ell_{0}}{A(\ell_{1}-\ell_{0})} = \frac{5\ell_{0}}{A(\ell_{2}-\ell_{0})}

= \frac{9\ell_{0}}{A(\ell_{3}-\ell_{0})}

These give

\frac{4}{\ell_{1}-\ell_{0}} = \frac{5}{\ell_{2}-\ell_{0}} \Rightarrow \ell_{0}=5\ell_{1}-4\ell_{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}=5\ell_{2}-4\ell_{1}

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