Question

A spring is stretched 8cm by a hanging weight. If the weight is doubled and the spring constant is not exceeded, how much will the spring stretch?
A. $4cm$
B. $8cm$
C. $12cm$
D. $16cm$

Answer 1

A spring is stretched due to some hanging weight. The spring constant is directly proportional to this weight and inversely proportional to the stretched length. Therefore, if the weight is doubled, then the spring will also stretch double to the original length.

Formula Used: The spring constant is given by: $k = \dfrac{m}{l}g$

Complete step by step solution: Consider a spring of length$L$. It is stretched by mass $m$which is suspended from it. As a result its length increases by $l$. That is, the total length becomes$\left( {L + l} \right)$. Then it is found that
$l \propto mg$ or $kl = mg$

where, $k$ is called force constant or spring constant and $g$ is acceleration due to gravity. Thus,

$k = \dfrac{m}{l}g$ $\to (1)$

Therefore, if the spring is stretched or compressed by a force$F$, and the length changes by $l$, then

$F = kl$

It is given that a spring is stretched 8cm by a hanging weight. Then, $l = 8cm$. Let hanging weight be of mass $m$. Then, the spring constant will be
${k_1} = \dfrac{m}{{{l_1}}}g = \dfrac{m}{8}g$ $\to (2)$
Then, the weight is doubled and the spring constant is not exceeded. Therefore,
${k_2} = \dfrac{{2m}}{{{l_2}}}g$ $\to (3)$

The spring constant is not exceeded. Therefore, equating equation (2) and equation (3).

\therefore {l_2} = 2m \times \dfrac{8}{m} = 16cm \\\ \ $$ Therefore, the spring will stretch by 16cm. **Hence, option (D) is the correct answer.** **Note:** A spring is stretched 8cm by a hanging mass$$m$$, then the spring is stretched by 8cm. Then, the mass is doubled as $$2m$$and the length is also doubled. But the spring constant remains to be the same in both the cases. For the spring constant to remain the same, the hanging mass and the length of the string should increase in the same proportion.