A body of mass 200 g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s, then the ratio of ext

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Solution

Q.
A body of mass 200 g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s, then the ratio of extension in the spring to its natural length will be [2023]

1 1 : 1

2 2 : 3

3 2 : 5

4 1 : 2

Ans.
(2)

$Natural length = L_{0}$

$Extension = x$

$k x = m (L_{0} + x) ω^{2}$

$\Rightarrow 12.5 x = \frac{1}{5} (L_{0} + x) 25 \Rightarrow 1.5 x = L_{0}$

$\Rightarrow \frac{x}{L_{0}} = \frac{2}{3}$

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