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One end of a massless spring of spring constant $k$ and natural length $l_{0}$ is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $ω$ about an axis passing through fixed end, then the elongation of the spring will be:

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$\frac{{mω}^{2} l_{0}}{k + m ω^{2}}$

$\frac{m ω^{2} l_{0}}{k - m ω^{2}}$

$\frac{k - mω {}^{2}l_{0}}{m ω^{2}}$

$\frac{k + m ω^{2} l_{0}}{m ω^{2}}$

answer is C.

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Detailed Solution

$KΔx = m (ℓ_{0} + Δx) ω^{2}$
$\begin{array}{rcl} KΔx & = & {mℓ}_{0} ω^{2} + {mω}^{2} Δx \\ Δx & = & \frac{{mℓ}_{0} ω^{2}}{k - {mω}^{2}} \end{array}$

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