The centripetal force $F$ acting on a particle moving uniformly in a circle may depend upon mass $(m)$ , velocity $(v)$ and radius $r$ of the circle. Derive the formula for $F$ using the method of dimensions.

F = \frac{k m v^{2}}{r}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

F = \frac{k m v}{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F = \frac{k m v^{2}}{r^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F = \frac{k m^{2} v^{2}}{r^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $F = \frac{k m v^{2}}{r}$
Dimensions of Force, $F = M L T^{- 2}$
Dimensions of Velocity, $V = L T^{- 1}$
Dimensions of mass, $M$
Dimensions of radius, $r = L$
comparing with dimensions of force,
Power of $M = 1$
Power of $v = 2$
Power of $r = - 1$
Hence, Option A is correct.

Suggest Corrections

Similar questions

R

x

K = a x^{2},

a

F = \frac{- C}{r^{2}}

F = \frac{- C}{r^{2}}

When a solid sphere moves through a liquid, the liquid opposes the motion with a force F. The

magnitude of F depends on the coefficient of viscosity $η$ of the liquid, the radius r of the sphere and the speed v of the sphere.

Assuming that F is proportional to different powers of these quantities, guess a formula for F using the

method of dimensions. Use F = - $\frac{η A (v_{2} - v_{1})}{x_{2} - x_{1}}$

Join BYJU'S Learning Program