The potential energy of a particle in a certain field has the form U=ar2−br, where a and b are postive constants, r is the distance from the centre of the field.
A
the value of r0 corresponding to the equilibrium position of the particle is ab
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Particle is in unstable equiltibrium.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
the minimum attractive force is −b227a3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
the value of r0 corresponding to the equilibrium position of the particle is 2ab
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct options are C the minimum attractive force is −b227a3 D the value of r0 corresponding to the equilibrium position of the particle is 2ab U=ar2−br
F=−dUdr
U=ar−2−br−1
−dUdr=−[a(−2)r−2−1−b(−1)r−1−1)]
=−[−2ar−3+br−2]
F=2ar3−br2
At equilibrium, F=0
Hence, 2ar30−br20=0
⇒1r20(2ar0−b)=0
⇒2ar0−b=0
⇒2ar0=b⇒r0=2ab
Hence, option (d) is correct
For option (b)
U=ar2−br=ar−2−br−1
dUdr=a(−2)r−2−1−b(−1)r−1−1
=−2ar−3+br−2
d2Udr2=−2a(−3)(r)−4+(−2)br−3
d2Udr2=6ar4−2br3
at r0=2ab
=6a(2a)4×b4−2b(2a)3×b3
=6ab416a4−2b48a3
=b4a3(38−14)
=b4a3(3−28)=18b4a3
This is positive value, so it is in stable equilibrium.Hence option (b) is incorrect.
For option (c)
We have F=2ar3−br2
for maximum or minimum value of force, dFdr=0
dFdr=2a(−3)r−4−(−2)br−3=0=−6ar4+2br3=0
⟹rm=3ab
d2Fdr=ddr(−6ar−4+2br−3)=−6(−4)ar−5+2(−3)br−4
=6r4(4ar−b)
4ar−b is positive for rm=3ab which says it is a local minima.
We have rm=3ab. So,