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Class 11 NEET Physics Laws of Motion Questions
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© examsnet.com
Question : 21
Total: 75
A block A of mass m
1
rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m
2
is suspended. The coefficient of kinetic friction between the block and the table is
µ
k
.When the block A is sliding onthe table, the tension in the string is :-
[NEET 2015 C]
(
m
2
−
µ
k
m
1
)
g
(
m
1
+
m
2
)
m
1
m
2
(
1
+
µ
k
)
g
(
m
1
+
m
2
)
m
1
m
2
(
1
−
µ
k
)
g
(
m
1
+
m
2
)
(
m
2
+
µ
k
m
1
)
g
(
m
1
+
m
2
)
Validate
Solution:
👈: Video Solution
For the motion of both blocks
m
2
g
−
T
=
m
2
a
T
−
μ
k
m
1
g
=
m
1
a
⇒
a
=
(
m
2
−
μ
k
m
1
)
g
m
1
+
m
2
For the block of the mass
m
2
m
2
g
−
T
=
m
2
[
m
2
−
μ
k
m
1
m
1
+
m
2
]
g
T
=
m
2
g
−
[
m
2
μ
k
m
1
m
1
+
m
2
]
m
2
g
=
m
2
g
[
m
1
+
μ
k
m
1
m
1
+
m
2
]
⇒
T
=
m
1
m
2
(
1
+
μ
k
)
g
m
1
+
m
2
Hence option (2)
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