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JEE Main Physics Class 11 Laws of Motion Part 2 Questions
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© examsnet.com
Question : 11
Total: 91
A
1
kg
mass is suspended from the ceiling by a rope of length
4
m
. A horizontal force '
F
' is applied at the mid point of the rope so that the rope makes an angle of
45
∘
with respect to the vertical axis as shown in figure. The magnitude of
F
is :
[9 Apr 2024 Shift 2]
10
√
2
N
1
N
1
10
×
√
2
N
10
N
Validate
Solution:
👈: Video Solution
T
1
s
i
n
45
∘
=
F
T
1
cos
45
∘
=
T
2
=
1
×
g
∴
tan
45
∘
=
F
g
∴
F
=
10
N
© examsnet.com
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