Given, potential energy of the particle at a distance
r in a conservative field,
V(r)=9r22
We know that force acting on the particle,
→F=−dVdr
⇒→F=−ddr(9r22)
⇒→F=−9r (towards centre
)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1090210/original_q4.png)
At
r=1 m
⇒→F=−9×1=−9 N
From the equilibrium of force in the circular motion, we know
centripetal force will be equal to the force
→F,
∴mv2r=9 (sign removed as sign shows direction
)
Here,
r=1 m and mass of particle,
m=1 kg
and
v is the velocity of the particle as shown in figure.
Now,
1×v21=9
∴v=3 m/s
Also, we know thwt angular momentum,
→L=→r×→p.......(1)
where,
→p is linear momentum of the particle which is,
→p=m→v.......(2)
From equaltion
(1) and
(2) we have,
→L=→r×m→v
⇒→L=m→r×→v
So, magnitude of angular momentum,
⇒L=mrvsinθ
Here, angle between
→r and
→v is
90∘
⇒L=1×1×3×sin90∘
∴L=3 kg-m/s
Thus options
(b) and
(c) are correct answers.
Why this question?
This question test your concept of conservative force relation and its realtion with potential energy. Also useful in understanding of application the angular momentum.
It was asked in JEE advanced 2018. |