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Question

A spehere of mass 0.2 kg is attached to an inextensible string of length 0.5m whose upper end is fixed to the cielling . The sphere is made to describe a horizontal circle of radius 0.3m The speed of the sphere will be

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Solution

m=200g
l=130cm

130cm=1.3 metre
So the time period of a pendulum doesn't depend on the mass of the Bob. It is equal to :
2πlg
(2pi×rootl/rootg) where l is length and g is gravity. Which gives us:
t2=4π2×1.3/9.8 (squaring both sides)
π2 and 9.8 are approx the same so we can cut it out, giving us:
4×1.3=520s
Therefore t2=5.2
t=2.29s

As per the second part of the question, the force on the string is simple mg = 200×10=2000 gram m/s^2
Or 200gf
Or 0.2kg×10=2N


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