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Test Index
JEE Main 28 Jan 2025 Shift 1 Paper
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Section:
Physics
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© examsnet.com
Question : 41 of 75
Marks:
+1
,
-0
A bead of mass '
m
m
m
' slides without friction on the wall of a vertical circular hoop of radius '
R
R
R
' as shown in figure. The bead moves under the combined action of gravity and a massless spring (
k
k
k
) attached to the bottom of the hoop. The equilibrium length of the spring is '
R
R
R
'. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes '
R
R
R
', would be (spring constant is '
k
k
k
',
g
g
g
is acceleration due to gravity)
[28 Jan 2025 Shift 1]
2
R
g
+
4
k
R
2
m
\sqrt{2 R g + \; \frac{4 k R^2}{m}}
2
R
g
+
m
4
k
R
2
2
R
g
+
k
R
2
m
\sqrt{2 R g + \; \frac{k R^2}{m}}
2
R
g
+
m
k
R
2
3
R
g
+
k
R
2
m
\sqrt{3 R g + \; \frac{k R^2}{m}}
3
R
g
+
m
k
R
2
2
g
R
+
k
R
2
m
2 \sqrt{g R + \; \frac{k R^2}{m}}
2
g
R
+
m
k
R
2
Validate
Solution:
👈: Video Solution
Work done by gravity
=
m
g
(
2
R
−
R
cos
6
0
∘
)
= m g (2R - R \cos 60^{\circ})
=
m
g
(
2
R
−
R
cos
6
0
∘
)
=
3
m
g
R
2
= \; \frac{3 m g R}{2}
=
2
3
m
g
R
Work done by spring
=
−
1
2
k
(
0
2
−
R
2
)
= - \; \frac{1}{2} k (0^2 - R^2)
=
−
2
1
k
(
0
2
−
R
2
)
=
1
2
k
R
2
= \; \frac{1}{2} k R^2
=
2
1
k
R
2
Net work = change in kinetic energy
i.e.
3
m
g
R
2
+
k
R
2
2
=
1
2
m
v
2
\; \frac{3 m g R}{2} + \; \frac{k R^2}{2} = \; \frac{1}{2} m v^2
2
3
m
g
R
+
2
k
R
2
=
2
1
m
v
2
or
v
2
=
3
g
R
+
k
R
2
m
\; \; v^2 = 3 g R + \; \frac{k R^2}{m}
v
2
=
3
g
R
+
m
k
R
2
or
v
=
3
g
R
+
k
R
2
m
\; \; v = \sqrt{3 g R + \; \frac{k R^2}{m}}
v
=
3
g
R
+
m
k
R
2
© examsnet.com
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