Q. The length of an elastic spring is $a$ metres when a force of 4 N is applied, and $b$ metres when the 5 N force is applied. Then the length of the spring when the 9 N force is applied is
A) $a\ + \ b$
B) $9b\ –\ 9a$
C) $5b\ - 4a$
D) $4a\ –\ 5b$
Explanation
(c) From Hooke’s law, restoring force F is $F = kl\ $where $k$ is spring constant. When L is original length of spring, and $k$ the spring constant, then
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$L + \left( \frac{5}{k} \right) = b$
Also $L + \left( \frac{4}{k} \right) = a$
$\therefore$ $\frac{5}{k} - \frac{4}{k} = b - a$
⟹ k $= \frac{1}{b - a}$
$\therefore$ $L = b - \frac{5}{k}$
⟹ $L = b - 5(b - a) = 5a - 4b$
When tension is 9 N.
Length of spring $= L + \frac{9}{k}$
Length of spring $= (5a - 4b) + 9(b - a)$
Length of spring $= 5b - 4a$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.