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AIEEE 2006 Solved Paper
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© examsnet.com
Question : 57
Total: 143
A mass of
M
kg
is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of
45
∘
with the initial vertical direction is
M
g
(
√
2
+
1
)
M
g
√
2
M
g
√
2
M
g
(
√
2
−
1
)
Validate
Solution:
From work energy theorem we can say,
Work done by tension + work done by force (applied) + Work done by gravitational force
=
change in kinetic energy Here Work done by tension is zero
⇒
0
+
F
×
A
B
−
M
g
×
A
C
=
0
⇒
F
=
M
g
(
A
C
A
B
)
=
M
g
[
1
−
1
√
2
1
2
]
as
A
B
=
ℓ
s
i
n
45
∘
=
ℓ
√
2
and
A
C
=
O
C
−
O
A
=
ℓ
−
ℓ
cos
45
∘
=
ℓ
(
1
−
1
√
2
)
where ℓ= length of the string.
⇒
F
=
M
g
(
√
2
−
1
)
© examsnet.com
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