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BITSAT 2019 Slot 2 Solved Paper
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© examsnet.com
Question : 33
Total: 150
A cylinder of mass
2
k
g
is released from rest from the top of an inclined plane of inclination
30
°
and length
1
m
. If the cylinder rolls without slipping, then its speed when it reaches the bottom, is [Take,
g
=
10
m
∕
s
2
]
20
3
m
∕
s
√
20
3
m
∕
s
10
3
m
∕
s
√
10
3
m
∕
s
Validate
Solution:
Potential energy of cylinder at top position
U
=
m
g
(
A
B
)
=
2
×
10
[
1
.
sin
30
°
]
[
∵
sin
30
°
=
A
B
B
C
∴
A
B
=
1
sin
30
°
]
=
10
J
If v is the linear speed of cylinder when it reaches at bottom, then at bottom, total potential energy of cylinder is converted into kinetic energy,
i.e.
K
=
U
1
2
I
ω
2
+
1
2
m
v
2
=
10
1
2
.
m
r
2
2
.
v
2
r
2
+
1
2
m
v
2
=
10
[
∵
ω
=
v
r
]
⇒
3
4
m
v
2
=
10
v
=
√
40
3
m
=
√
40
3
×
2
=
√
20
3
m
∕
s
© examsnet.com
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