Q. A load suspended by a massless spring produces an extension of x cm in equilibrium. When it is cut into two unequal parts, the same load produces an extension of 7.5 cm when suspended by the larger part of length 60 cm. When it is suspended by the smaller part, the extension is 5.0 cm. Then
A) $x = 12.5$
B) $x = 3.0$
C) The length of the original spring is 90 cm
D) The length of the original spring is 80 cm
Explanation
(a) Assume original length of spring = $l$
$m$g = $kx$
$k_{1}(60) = k_{2}(l - 60) = \ kl$
$\therefore\ \ \ mg = k_{1} = (7.5)$ according to question
And mg = $k_{2} = (5.0)$
$\therefore\ \ \ k_{1} = \frac{kl}{60},\ k_{2} = \frac{kl}{l - 60}\ $
$\frac{k_{1}}{k_{2}} = \frac{5.0}{7.5} = \frac{l - 60}{60}$
$\Rightarrow \ \ \frac{2}{3} = \frac{l - 60}{60}$
$\therefore\ \ l = 100\ cm\ $
And $kx = k_{1} \times 7.5$
$kx = \left( \frac{5k}{3} \right) \times 7.5\ $
$\therefore\ \ x = 12.5\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.