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JEE Advanced Model Paper 4 with solutions for online practice
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© examsnet.com
Question : 2
Total: 54
A small sphere of mass m = 500g moving on the inner surface of a large hemispherical bowl of radius R = 5m describes a horizontal circle at a distance OC = 2.5m below the centre O of the bowl as shown in Fig. Find the force exerted by the sphere on thebowl and the time period of revolution of the sphere around the circle. Take g = 10
m
s
−
2
.
10N, 3.14sec
20N, 6.28sec
5N, 2.22sec
25N, 2sec
Validate
Solution:
Given OP = 5 m and OC = 2.5 m
Therefore cosθ =
O
C
O
P
=
1
2
⇒ θ = 60°
Radius of circle is r = CP = OPsinθ = 5 sin 60° =
5
×
√
3
2
m
The forces acting on the sphere, N is the normal reaction.
Net force towards centre C of the circle = N sin θ
⇒
m
r
ω
2
= Nsinθ (i)
Also mg = Ncosθ (ii)
From (ii) , N =
m
g
c
o
s
θ
=
0.5
×
10
c
o
s
60
°
= 10N
From (i)
m
r
ω
2
Rsinθ = Nsinθ
⇒ ω =
√
N
m
R
=
√
10
0.5
×
5
= 10
r
a
d
s
−
1
∴ Time period T =
2
π
ω
= π second = 3.14 s
© examsnet.com
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