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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 U0 are constants). Match the potential energies in column I to the corresponding statement(s) in column II.

Column I Column II
(A) U1(x)=U02[1(xa)2]2 (P) The force acting on the particle is zero at x = a.
(B) U2(x)=U02(xa)2 (Q) The force acting on the particle is zero at x = 0.
(C) U3(x)=U02(xa)2 exp[(xa)2] (R) The force acting on the particle is zero at x = – a.
(D) U4(x)=U02[xa13(xa)3] (S) The particle experiences an attractive force
towards x = 0 in the region |x|<a.
(T) The particle with total energy U04can oscillate
about the point x = – a.

A
(A)(P,Q,T);(B)(Q,T)
(C)(P,Q.R,S)
(D)(P,R,T)
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B
(A)(P,Q,R,T);(B)(Q,S)
(C)(P,Q.R,S)
(D)(P,R,T)
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C
(A)(P,Q,R,T);(B)(Q,S)
(C)(P,Q.R,S)
(D)(P,R,T)
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D
(A)(P,Q,R,T);(B)(Q,S)
(C)(P,Q,S,T)
(D)(P,R,S)
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Solution

The correct option is C (A)(P,Q,R,T);(B)(Q,S)
(C)(P,Q.R,S)
(D)(P,R,T)

For A: F=dUdx^i=U022(1(xa)2)×[2(xa)×1a]^i=2U0[1(xa)2][xa2]^i

If x=0 F=U02[2(1)×0]=0, U=U02

If x=a F=0, and U=0

If x=a F=0, and U=0
so, AP,Q,R,T

For B: F=U02×2(xa)×1a^i=U0xa2^i

If x=0 F=0, and U=0

If x=a F=U02^i and U=U02

If x=a F=+U02^i and U=U02
so, BQ,S

For C:
U3(x)=U02(xa)2 exp[(xa)2]
,
for force, F=U0ex2a2xa[x2a21]
when, x=0; U0=0F=0; x=0,x=a,x=a
so, CP,Q,R,S

For D:
U4(x)=U02[xa13(xa)3]

on differentiating the above equation.
dUdt=F=U02a[1x2a2]
So, for U=0;x=0, x=+3a,x=3a
F=0;x=+a,x=a
so, DP,R,T

Hence,
AP,Q,R,T
BQ,S
CP,Q,R,S
DP,R,T

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