A charged particle accelerated through a potential difference of V volts ac quires a speed u. The particle is then made to enter perpendicularly in a uniform magnetic field B. The radius of the circular path followed by the charged particle will be proportional to

The correct option is V/u.

To find how the radius of the circular path is related to the potential difference and speed, let us break down the physical processes step-by-step.

Step 1: Acceleration through a Potential Difference
When a particle of charge $q$ and mass $m$ is accelerated from rest through a potential difference of $V$ volts, the work done on the particle is converted into kinetic energy:
$$qV=\frac{1}{2}m{u}^2$$

From this relation, we can express the mass-to-charge ratio ($\frac{m}{q}$) in terms of the potential difference $V$ and speed $u$:
$$\frac{m}{q}=\frac{2V}{u^2}$$

Step 2: Motion in a Uniform Magnetic Field
When the charged particle enters perpendicularly into a uniform magnetic field $B$, the magnetic Lorentz force acts as the centripetal force keeping the particle in a circular path of radius $r$:
$$quB=\frac{m{u}^2}{r}$$

Solving for the radius $r$ gives:
$$r=\frac{mu}{qB}=\left(\frac{m}{q}\right)\frac{u}{B}$$

Step 3: Substituting the mass-to-charge ratio
Substitute the expression for $\frac{m}{q}$ obtained from Step 1 into the radius equation:
$$r=\left(\frac{2V}{u^2}\right)\frac{u}{B}$$

Simplifying this, we get:
$$r=\frac{2V}{uB}$$

Since the magnetic field $B$ is uniform and constant, the radius $r$ is directly proportional to:
$$r\propto \frac{V}{u}$$