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Manipal Entrance Test 2013 Physics Paper
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© examsnet.com
Question : 14
Total: 60
A wet open umbrella is held vertical and it whirld about the handle at a uniform rate of 21 revolutions in 44s. If the rim of the umbrella is circle of 1 m in diameter and the height of the rim above the flour is 4.9 m, the locus of the drop is a circle of radius
√
2.5
m
1 m
3 m
1.5 m
Validate
Solution:
From equation of motion
h
=
u
t
+
1
2
g
t
2
where u is initial velocity and t is time.
Since, u = 0
∴
t
=
√
2
h
g
=
√
2
×
4.9
9.8
=
1
s
The horizontal range of the drop = x, then
x
=
(
v
t
o
)
t
Also
,
ω
=
Δ
θ
Δ
t
=
21
×
2
π
44
=
3
rad
∕
s
Tangential speed
v
t
=
r
ω
=
0.5
×
3
×
1.5
m
∕
s
x
=
1.5
×
1
=
1.5
m
Locus of drop
=
√
x
2
+
r
2
=
√
(
1.5
)
2
+
(
0.5
)
2
=
√
2.5
m
© examsnet.com
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