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Question

In the Column I below, four different paths of a particle are given as functions of time.
In these functions, α and β are positive constants of appropriate dimensions and αβ. In each case, the force acting on the particle is either zero or conservative.
In Column II, five physical quantities of the particle are mentioned: p is the linear momentum L is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy.
Match each path in Column I with those quantitites in Column II, which are conserved for that path.
Column~IColumn~II(A) r(t)=α t^i+β t^j1. p(B) r(t)=α cos ω t ^i+β sin ω t ^j2. L(C) r(t)=α(cos ω ^i+ sin ω t^j)3.K(D) r(t)=α t^i+β2t2^j4. U5. E

A
A1,2,3,4,5;B2,5;C2,3,4,5;D5
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B
A1,2,3,4,5;B3,5;C1,2,3,4;D4
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C
A1,2,3,4,5;B1,5;C2,3,4,5;D1
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D
A1,2,3,4,5;B4;C1,3,4,5;D3
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Solution

The correct option is A A1,2,3,4,5;B2,5;C2,3,4,5;D5
A1,2,3,4,5;B2,5;C2,3,4,5;D5
When force is conservative and F=0 potential energy U= constant
When force is conservative F0 Total energy E= constant
(A)r(t)=α t^i+β t ^j
drdt=v=α ^i+β^j=constantp=constant

|v|=α2+β2=constantK=constant

dvdt=a=0F=0U=constant

E=U+K=constant
L=m(r×v)=0
L=constant
A1,2,3,4,5

(B)r(t)=α cos ω t^i+β sin ω t^j
drdt=v=α ω sin ω t(^i)+β ω cos ω ^jconstantpconstant

|v|=ω(α sin ω t)2+(β cos ω t)2constantKconstant

a=dvdt=ω2r0
E=constant=K+U
But K constant
Uconstant
L=m(r×v)=mω α β(k)=constant
B2,5

(C)r(t)=α(cos ω t^i+sin ω t^j
drdt=v=α ω[sin ω t(^i)+cos ω t^jconstantpconstant

|v|=α ω=constantK=constant
a=dvdt=ω2r0E=constant,U=constant

L=m(r×v)=mω α2^k=constant)
C2,3,4,5

(D)r(t)=α t ^i+β2t2^j
drdt=v=α^i+β t^jconstantpconstant
a=dvdt=β^j0E=constant=K+U
But Kconstant
So, U constant
L=m(r×v)=12α β t2^kconstant
D5

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