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TS EAMCET 20-July-2022 Shift 1 Solved Paper
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© examsnet.com
Question : 87
Total: 160
At time
t
=
0
, a force
F
=
α
t
, where
t
is time in seconds, is applied to a body of mass
1
k
g
, resting on a smooth horizontal plane. If the direction of the force makes an angle of
45
∘
with the horizontal, then the velocity of the body at the moment of its breaking off the plane is
100
α
m
∕
s
50
√
2
α
m
∕
s
50
α
√
2
m
∕
s
50
α
m
∕
s
Validate
Solution:
The given situation is shown below
Body loose contact with surface when reaction is zero, thus
F
s
i
n
45
∘
=
m
g
⇒
F
=
m
g
s
i
n
45
∘
=
1
×
10
1
√
2
Or
F
=
10
√
2
But,
F
=
α
t
Hence,
α
t
=
10
√
2
or
t
=
10
√
2
α
Now, acceleration of body,
a
=
F
m
cos
45
∘
⇒
d
v
d
t
=
α
t
m
cos
45
∘
or
d
v
=
α
√
2
×
1
×
t
×
dt
⇒
d
v
=
t
⋅
α
√
2
⋅
dt
⇒
v
∫
0
d
v
=
α
√
2
10
√
2
α
∫
0
t
.
dt
⇒
v
−
0
=
α
√
2
(
t
2
2
)
0
10
√
2
α
⇒
v
=
α
√
2
×
100
×
2
2
×
α
2
=
50
√
2
α
m
∕
s
© examsnet.com
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