JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 4)
One end of a massless spring of spring constant k and natural length l0 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 $$\omega$$ about an axis passing through fixed end, then the elongation of the spring will be :
$${{k - m{\omega ^2}{l_0}} \over {m{\omega ^2}}}$$
$${{m{\omega ^2}{l_0}} \over {k + m{\omega ^2}}}$$
$${{m{\omega ^2}{l_0}} \over {k - m{\omega ^2}}}$$
$${{k + m{\omega ^2}{l_0}} \over {m{\omega ^2}}}$$
Explanation
$$m{\omega ^2}({l_0} + x) = kx$$
$$ \Rightarrow m{\omega ^2}{l_0} = (k - m{\omega ^2}) \times x$$
$$ \Rightarrow x = {{m{\omega ^2}{l_0}} \over {(k - m{\omega ^2})}}$$
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