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(D) The initial spring constant is $k$. When a spring of length $l$ is cut into two equal parts,the spring constant of each part becomes $k' = 2k$.Whe

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$\frac{x}{4}$

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(D) The initial spring constant is $k$. When a spring of length $l$ is cut into two equal parts,the spring constant of each part becomes $k' = 2k$.When these two parts are connected in parallel,the equivalent spring constant $K_{eq}$ is the sum of the individual constants:$K_{eq} = k' + k' = 2k + 2k = 4k$.According to Hooke's Law,$F = kx$. For the original spring,$W = kx$.For the parallel combination,the same weight $W$ is applied,so $W = K_{eq} 	imes x'$,where $x'$ is the new extension.Substituting the values: $kx = 4k 	imes x'$.Solving for $x'$,we get $x' = \frac{kx}{4k} = \frac{x}{4}$.

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

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