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JEE Advanced Model Paper 2 with solutions for online practice
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© examsnet.com
Question : 4
Total: 54
A block of mass m is initially at rest on a frictionless horizontal surface. A time -dependent force F = at -
b
t
2
acts on the body, where a and b are positive constants.
The magnitude of the force is maximum at time
t
1
given by
a
2
b
The maximum force
F
m
a
x
is given by
a
2
4
b
The maximum impulse
I
m
a
x
imparted to the block is given by
a
3
12
b
2
The magnitude of the force is maximum at time
t
1
given by
b
2
a
Validate
Solution:
The force is maximum when
d
F
d
t
= 0 and
d
2
F
d
t
2
< 0
Now ,
d
F
d
t
=
d
d
t
(at -
b
t
2
) = a - 2bt
Putting
d
F
d
t
= 0 and t =
t
1
, we get
0 = a - 2b
t
1
⇒
t
1
=
a
2
b
Also
d
2
F
d
t
2
=
d
d
t
(a - 2bt) = -2b , which is negative
F
m
a
x
=
a
t
1
-
b
t
1
2
= a ×
a
2
b
b ×
(
a
2
b
)
2
=
a
2
4
b
Maximum impulse is given by
I
m
a
x
=
t
1
∫
0
Fdt
=
t
1
∫
0
(at -
b
t
2
) dt
=
a
t
1
2
2
−
b
t
1
2
3
=
a
2
(
a
2
b
)
2
-
b
3
(
a
2
b
)
3
=
a
3
12
b
2
© examsnet.com
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