A spring of spring constant $k$ is cut into two pieces such that their lengths are in the ratio $1 : 2,$ then the force constant of the bigger piece will be

\frac{3 k}{2}

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\frac{k}{2}

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2 k

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3 k

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Solution

The correct option is C $\frac{3 k}{2}$
Let length of spring be 3 m(Since ratio is 1:2).
$∴ l_{1} = 1 m$
$∴ l_{2} = 2 m$
From formula, $∵ l_{2} i s b i g g e r, k_{2} = \frac{k (1 + 2)}{2} = \frac{3 k}{2}$ .
Hence option A is correct.

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A spring of spring constant k is cut into two pieces such that their lengths are in the ratio 1:2, then the force constant of the bigger piece will be

The correct option is C 3k2Let length of spring be 3 m(Since ratio is 1:2). ∴l1=1m∴l2=2mFrom formula, ∵l2 is bigger,k2=k(1+2)2=3k2. Hence option A is correct.

A spring of spring constant $k$ is cut into two pieces such that their lengths are in the ratio $1 : 2,$ then the force constant of the bigger piece will be

The correct option is C $\frac{3 k}{2}$
Let length of spring be 3 m(Since ratio is 1:2).
$∴ l_{1} = 1 m$
$∴ l_{2} = 2 m$
From formula, $∵ l_{2} i s b i g g e r, k_{2} = \frac{k (1 + 2)}{2} = \frac{3 k}{2}$ .
Hence option A is correct.