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JEE Main Physics Class 11 Work, Energy and Power Part 2 Questions
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© examsnet.com
Question : 39
Total: 100
Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane
A
B
along its surface with a velocity just sufficient to carry it to the top Point
B
at a height
10
m
. After reaching the Point
B
the block slides down on inclined plane
BC
. Time it takes to reach to the point
C
from point
A
is
t
(
√
2
+
1
)
s. The value of
t
is _______.
(use
g
=
10
m
∕
s
2
)
[27-Jul-2022-Shift-2]
Your Answer:
Validate
Solution:
From E.C.
=
1
2
m
v
0
2
=
mgh
v
0
=
10
√
2
For
A
→
B
at
B
,
v
=
0
a
=
−
g
s
i
n
45
∘
=
−
10
√
2
v
=
u
+
at
1
⇒
0
=
10
√
2
−
10
√
2
t
1
⇒
t
1
=
2
s
e
c
For
B
→
C
s
=
u
t
2
+
1
2
at
2
2
10
s
i
n
30
∘
=
1
2
(
10
s
i
n
30
∘
)
t
2
2
t
2
=
2
√
2
So total time
T
=
t
1
+
t
2
T
=
t
1
+
t
2
=
2
√
2
+
2
=
2
(
√
2
+
1
)
s
e
c
© examsnet.com
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