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Circular Motion question

2023 · 24 Jan · Shift 2 · Q63

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Circular Motion
2023 · 24 Jan · Shift 2 · Q63

Circular Motion question

2023 · 24 Jan · Shift 2 · Q63

JEE MainPhysicsCircular MotionMCQ+4 / −1

A body of mass 200g 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
2 : 3
C
2 : 5
D
1 : 1

View written solutionFree

Correct answer: B

Given data

Mass of body: $m = 200\text{ g} = 0.2\text{ kg}$
Spring constant: $k = 12.5\,\text{N/m}$
Angular speed: $\omega = 5\,\text{rad/s}$

Let

natural length of spring $= l_0$
extension in spring $= x$
actual length during rotation $= r = l_0 + x$

Force providing centripetal acceleration

The body moves in a circle of radius $r$ , so required centripetal force is

$F_c = m\omega^2 r$

The only horizontal force is spring force:

$F_s = kx$

Hence,

$kx = m\omega^2 r$

Since $r = l_0 + x$ ,

$kx = m\omega^2 (l_0 + x)$

Substitute values

First compute:

$m\omega^2 = 0.2 \times 5^2 = 0.2 \times 25 = 5$

So the equation becomes

$12.5x = 5(l_0 + x)$

Expand:

$12.5x = 5l_0 + 5x$

$12.5x - 5x = 5l_0$

$7.5x = 5l_0$

$\frac{x}{l_0} = \frac{5}{7.5} = \frac{2}{3}$

Required ratio

Ratio of extension to natural length:

$x : l_0 = 2 : 3$

Option check

A: $1:2$ ❌
B: $2:3$ ✅
C: $2:5$ ❌
D: $1:1$ ❌

Therefore, the correct answer is B.

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