A spring mass system (mass ‘m’, spring constant ‘k’ and natural length $l_{0}$ ) rest in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of disc. If 5g mass of the body attached to the spring having spring constant $20 N m^{- 1}$ . If the disc together with spring mass system, rotates about its axis with an angular velocity of $5 {rad.s}^{- 1}$ . $(k > > m ω^{2})$ , then the percentage change in the length of the spring is

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6.25

62.5

0.0625

0.625

answer is B.

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Detailed Solution

$k Δ l \leftarrow m \to m (l_{0} + Δ l) ω^{2} f r o m f r e e b o d y d i a g r a m, f o r c e s a c t i n g o n m a s s m a r e, (k = f o r c e c o n s \tan t), r e s t o r i n g f o r c e o f s p r i n g F = - k x = - k Δ l, t o w a r d s t h e c e n t r e h e r e Δ l i s e x t e n s i o n i n l e n g t h o f s p r i n g a n d c e n t r i f u g a l f o r c e F = m w^{2} (l_{0} + Δ l), a w a y f r o m t h e c e n t r e, l_{0} = u n s t r e t c h e d l e n g t h o f s p r i n g e q u a t i n g t h e a b o v e t w o e q u a t i o n s k Δ l = m w^{2} (l_{0} + Δ l) k Δ l - m w^{2} Δ l = m w^{2} l_{0} Δ l = \frac{m w^{2} l_{0}}{k - m w^{2}} \frac{Δ l}{l_{0}} = \frac{m w^{2}}{k - m w^{2}} ≃ \frac{m w^{2}}{k} 100 (\frac{Δ l}{l_{0}}) = (\frac{m w^{2}}{k}) 100 \frac{Δ l}{l_{0}} % = (\frac{5 x 10^{- 3} x 5^{2}}{20}) 100 \frac{Δ l}{l_{0}} % = 0.625$

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