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# The length of a spring is $l$ and its force constant is $K$. When a weight $W$ is suspended from it, its length increases by $x$. If the spring is cut into two equal parts and put in parallel and the same weight $W$ is suspended from them, then the extension will be ?A. $2x$B. $x$C. $\dfrac{x}{4}$D. $\dfrac{x}{2}$ Verified
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Hint:The spring constant of any spring is inversely proportional to its length.Use this property to find the spring constant of each part and find the extension of the springs. Elasticity is the property by virtue of which a material deformed under the influence of load but after the removal of the deforming load the object tends to recover its original dimension.

Formula used:
The relation between the spring constant and the length is given by,
$K \propto \dfrac{1}{l}$
where$K$ is the spring constant and $l$ is the length of the spring.
The equivalent spring constant of two springs connected in parallel is given by,
${K_{eq}} = {K_1} + {K_2}$
where, ${K_{eq}}$ is the equivalent spring constant, ${K_1}$ is the spring constant of one spring and ${K_2}$ is the spring constant of another spring.
Hooke’s law is given by,
$F = Kx$
where, $F$ is the force applied on a spring, $K$ is the spring constant and$x$ is the stretched length of the spring.

We have given here one spring that is cut to half and connected in parallel to suspend from a weight $W$ before and after it. The length of it increases by $x$ before cutting and we have to find the length after cutting it. To find the extended length of the spring we have to find the new spring constant of each of the cut springs.

Now, we know that spring constant is proportional to the length of the spring. It is given by, $K \propto \dfrac{1}{l}$, where$K$ is the spring constant and $l$ is the length of the spring. So if we cut a spring in half the spring constant of the spring will be doubled. Hence, spring constant of each half of the spring is, $2K$.

Now, if two springs are connected in parallel the equivalent spring constant becomes, ${K_{eq}} = {K_1} + {K_2}$
where, ${K_1}$ is the spring constant of one spring and\ [{K_2}\] is the spring constant of other spring.
So, for the two halves connected in parallel the spring constant will be,
${K_{eq}} = 2K + 2K = 4K$
Hence, when stretched by the same wait $W$ by Hooke’s law the extended length will be,
$W = 4Kx'$ Where, $x'$ is the extended length of the connected spring and $W = Kx$.
$Kx = 4Kx'$
$\Rightarrow x = 4x'$
$\therefore x' = \dfrac{x}{4}$
Hence, the extended length will be $\dfrac{x}{4}$.

Hence, option D is the correct answer.

Note: If two springs are connected in series the spring constant of the system becomes, $\dfrac{1}{{{K_{eq}}}} = \dfrac{1}{{{K_1}}} + \dfrac{1}{{{K_2}}}$. This convention is opposite to the convention of the electrical system. To solve this type of problem remember the formula of equivalent spring constant and the relation between the spring constant and the displacement of the spring.