JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Laws of Motion

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A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in radian/s) is

(A)9
(B)18
(C)27
(D)36✓ Correct

Answer: (D)

Solution

For circular motion in a horizontal plane, the horizontal component of tension provides the centripetal force. From the geometry, if the string makes angle

θ

with vertical, the horizontal radius is

r = L sin θ

.

The horizontal component of tension:

T sin θ = m ω^{2} r = m ω^{2} L sin θ

This gives:

T = m ω^{2} L

324 = 0.5 \times ω^{2} \times 0.5

ω^{2} = \frac{324}{0.25} = 1296

ω = 36 rad/s

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option

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