JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Laws of Motion

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) $2 : 3$ ✓ Correct
(D) $2 : 5$

Answer: (C)

Solution

On the smooth surface the stretched spring provides the centripetal force for the circular motion, so

k x = m (L_{0} + x) ω^{2}

, where

L_{0}

is the natural length and

x

the extension, and the radius of the circle is

L_{0} + x

. Substituting

k = 12.5

N/m,

m = 0.2

kg and

ω = 5

rad/s gives

12.5 x = 0.2 \times 25 (L_{0} + x) = 5 (L_{0} + x)

, that is

7.5 x = 5 L_{0}

. Therefore

\frac{x}{L _{0}} = \frac{5}{7.5} = \frac{2}{3}

, a ratio of

2 : 3

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Laws of Motion

Solution

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