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JEE Mains 03-Sep-2020 Shift 1 Solved Paper
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© examsnet.com
Question : 16
Total: 75
A block of mass
m
=
1
k
g
slides with velocity
v
=
6
m
∕
s
on a frictionless horizontal surfaceand collides with a uniform vertical rod andsticks to it as shown. The rod is pivoted about
O
and swings as a result of the collision making angle
θ
before momentarily coming to rest. If the rod has mass
M
=
2
k
g
,
and length
l
=
1
m
,
the value of
theta
is approximately :(Take
g
=
10
m
∕
s
2
)
69
°
63
°
55
°
49
°
Validate
Solution:
Angular momentum conservation
m
v
l
=
M
l
2
3
ω
+
m
l
2
ω
⇒
ω
=
1
×
6
×
1
2
3
+
1
=
18
5
Now using energy consevation
1
2
(
M
l
2
3
)
ω
2
+
1
2
(
m
l
2
)
ω
2
=
(
m
+
M
)
r
c
m
(
1
−
cos
θ
)
=
(
m
+
M
)
(
m
l
+
M
l
2
m
+
M
)
g
(
1
−
cos
θ
)
5
6
×
(
18
5
)
2
=
20
(
1
−
cos
θ
)
⇒
1
−
cos
θ
=
18
5
×
3
20
cos
θ
=
1
−
27
50
cos
θ
=
23
50
⇒
θ
≃
63
°
© examsnet.com
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