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JEE Main 11 April 2015 Paper
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© examsnet.com
Question : 4
Total: 90
A particle is moving in a circle of radius r under the action of a force
F
=
α
r
2
which is directed towards center of the circle.Total mechanical energy (Kinetic energy + potential energy) of the particle is (take potential energy = 0 for r=0) :
α
r
3
1
2
α
r
3
4
3
α
r
3
5
6
α
r
3
Validate
Solution:
dU = F.dr
U
=
r
∫
0
α
r
2
d
r
=
α
r
2
3
m
v
2
r
=
α
r
2
m
2
v
2
=
m
α
r
3
2
m
(
K
E
)
=
1
2
α
r
3
Total E
=
α
r
3
3
+
α
r
3
2
=
5
3
α
r
3
© examsnet.com
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