A body of mass 200,g is tied to a spring of spring constant 12.5,N / m , while other end of spring i

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4-1.Newton's Laws of Motion

hard

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 :

A$1: 2$

B$1: 1$

C$2: 3$

D$2: 5$

(JEE MAIN-2023)

Solution

$kx = m \left( L _0+ x \right) \omega^2$
$\Rightarrow 12.5 x =\frac{1}{5}\left( L _0+ x \right) 25 \Rightarrow 1.5 x = L _0$ $\Rightarrow \frac{ x }{ L _0}=\frac{2}{3}$

Standard 11

Physics

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easy

(NEET-2022)

Solution

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