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JEE Advanced 2015 Paper 1
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© examsnet.com
Question : 20
Total: 60
A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is conserved.Four possible forms of the potential energy of the particle are given in column I(a and
U
0
are constants).Match the potential energies i column I to the corresponding statement(s) in column II.
Column I
Column II
(A)
U
1
(
x
)
=
U
0
2
[
1
−
(
x
a
)
2
]
2
(P)
The force acting on the particle is zero at
x
=
a
(B)
U
2
(
x
)
=
U
0
2
(
x
a
)
2
(Q)
The force acting on the particle is zero at
x
=
0
(C)
U
3
(
x
)
=
U
0
2
(
x
a
)
2
exp
[
−
(
x
a
)
2
]
(R)
The force acting on the particle is zero at
x
=
−
a
(D)
U
4
(
x
)
=
U
0
2
[
x
a
−
1
3
(
x
a
)
3
]
(S)
The particle experiences an attractive force towards
x
=
0
in the region
|
x
|
<
a
(T)
The particle with the total energy
U
0
4
can oscillate about the point
x
=
−
a
(A)
→ (P, Q, R, T );
(B)
→ (Q, R);
(C)
→ (P, Q, R, T);
(D)
→ (P, R, S)
(A)
→ (P, Q, R, T );
(B)
→ (Q, R );
(C)
→ (P, Q, R, S);
(D)
→ (P, Q, R)
(A)
→ (P, Q, R, T );
(B)
→ (Q, S );
(C)
→ (P, Q, R, S);
(D)
→ (P, R, T)
(A)
→ (P, Q, R, S );
(B)
→ (Q, S );
(C)
→ (P, Q, R, T);
(D)
→ (P, R, T)
Validate
Solution:
(A)→(P,Q,R,T);(B)→(Q,S);(C)→(P,Q,R,S);(D)→(P,R,T)
(A)
→
F
=
−
d
U
d
x
∧
i
=
−
U
0
2
2
(
1
−
(
x
a
)
2
)
×
[
−
2
(
x
a
)
×
1
a
]
∧
i
=
2
U
0
[
1
−
(
x
a
)
2
]
[
x
a
2
]
∧
i
If
x
=
0
⇒
→
F
=
U
0
2
[
2
(
1
)
×
0
]
=
→
0
,
U
=
U
0
2
If
x
=
a
⇒
→
F
=
→
0
.
&
U
=
0
If
x
=
−
a
⇒
→
F
=
→
0
.
&
U
=
0
(B)
→
F
=
−
U
0
2
×
2
(
x
a
)
×
1
a
∧
i
=
−
U
0
x
a
2
∧
i
If
x
=
0
⇒
→
F
=
0
and
U
=
0
If
x
=
a
⇒
→
F
=
−
U
0
a
∧
i
and
U
=
U
0
2
If
x
=
−
a
⇒
→
F
=
+
U
0
a
∧
i
and
U
=
U
0
2
For (C)and(D) similarly we can solve
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