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Question

A particle moving with constant speed u inside a fixed smooth spherical bowl of radius a describes a horizontal circle at a distance a2 below its centre. Then,

A
the radius of the circular motion is a32
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B
the value of u is 3ag2
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C
the normal reaction of the spherical surface on the particle is mg2
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D
the magnitude of the resultant force acting on the particle is zero, in an inertial frame.
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Solution

The correct option is A the radius of the circular motion is a32
Since the body is given just enough speed to complete the vertical circle
u2=5gR
v2=u22gR
=5gR2gR
=3gR
When the velocity is vertical, the body is acted by two acceleration aT=g tangentially and centripetal acceleration ac=v2R radially.
ac=v2R=3gRR
=3g
Total acceleration =a2T+a2c
=g2+(3g)2
=g10
Hence a is the correct answer.

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