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  • A load suspended by a massless spring produces an extension of  in equilibrium. when

A load suspended by a massless spring produces an extension of x \mathrm{~cm} in equilibrium. when it is cut into two unequal parts, the same load produces an extensions of 7 .5 \mathrm{~cm}. when suspended by the larger part of length 60 \mathrm{~cm}. When it is suspended by the smaller part, the extension is 5.0 \mathrm{~cm} then,

Option: 1

x=12.5 \mathrm{~cm}


Option: 2

x=3.0 \mathrm{~cm}


Option: 3

The length of the original spring is 90 \mathrm{~cm}


Option: 4

The length of the original spring is 80 \mathrm{~cm}


Answers (1)

best_answer


\Delta l=\frac{F l}{A Y} (can also be applied for a spring)
\begin{aligned} \therefore \quad & \Delta l \alpha l \\ \frac{7.5}{5.0} & =\frac{60}{l_2} \\ l_2 & =40 \mathrm{~cm} . \end{aligned}

\therefore length of original spring is

 \\=(60+40) \mathrm{cm} \\=100 \mathrm{~cm}
Now,

 \\\frac{x}{7.5}=\frac{100}{60}\\ x=12.5 \mathrm{~cm}

Posted by

Shailly goel

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