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JEE Advanced Model Paper 4 with solutions for online practice
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© examsnet.com
Question : 11
Total: 54
A weightless rigid rod AB of length I connects two equal masses m one particle is fixed at the end B and the other at the middle of the rod as shown in the figure. The rod can rotate in the vertical plane freely around the hinge point A.
The minimum horizontal velocity required to be given to the paiticle B soas to make the rod go around in the complete vertical circle is
√
24
g
l
5
The minimum horizontal velocity required to be given to the particle B soas to make the rod go around in the complete vertical circle is
√
24
g
l
9
The ratio of compressive force in the rods AC and BC is 2: 1 when themasses are at highest point.
The ratio of compressive force in the rods AC and BC is 3: 1 when themasses are at highest point.
Validate
Solution:
If v is speed imported to B, total kinetic energy of rod is
K =
1
2
m
v
2
+
1
2
m
(
v
2
)
2
=
5
8
m
v
2
To reach the highest point and complete the round we use
5
8
m
v
2
+
m
g
l
2
= 2mgl +
3
m
g
l
2
⇒ v =
√
24
5
g
l
At the highest point rod will be rest and
if
T
A
C
and
T
B
C
are tension in rod reactions AB and BC.
We have
T
B
C
= mg and
T
A
C
= 2 gm ⇒
T
A
C
T
B
C
=
2
1
© examsnet.com
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