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Question

A body of mass m=2 kg, initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB= 3 m. The height h is equal to

A
2.4 m
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B
4 m
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C
3.75 m
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D
5 m
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Solution

The correct option is C 3.75 m

Given
m = 2 kg
AB, D = 3 m
h= ?

Applying law of conservation of energy at point A & P:
PEA+KEA=PEP+KEP
12mv2=mgh
vA=2gh .....(i)

Datum is at A PEA=0
Also, vP=0,KEp=0
To just complete vertical circle, speed at point A
=5g×(D/2) .....(ii)
Equating (i) & (ii),
2gh=5g×(D/2)
h=5D4=54×3
h=3.75 m

Hence option C is the correct answer

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