A particle moving along the circular path with a speed v and its speed increases by ‘g’ in one second. If the radius of the circular path be r, then the net acceleration of the particle is

The correct option is: [v⁴/r² +g²]⁰.⁵

Step-by-Step Explanation:

When a particle moves along a circular path of radius r with a speed v, it experiences two components of acceleration:
1. Centripetal (Radial) Acceleration ($a_c$): This component is responsible for changing the direction of the particle's velocity. It is directed towards the center of the circular path and is given by the formula:

$a_c=\frac{v^2}{r}$

2. Tangential Acceleration ($a_t$): This component is responsible for changing the magnitude of the velocity (the speed). The question states that the particle's speed increases by 'g' in one second. Since the rate of change of speed is the tangential acceleration, we have:

$a_t=\frac{dv}{dt}=g$

Since centripetal acceleration is directed radially inward and tangential acceleration is directed along the tangent to the circle, these two acceleration vectors are perpendicular to each other (the angle between them is 90°).

The net acceleration (a) of the particle is the vector sum of these two perpendicular components:

$a=\sqrt{{a_c}^2+{a_t}^2}$

Substituting the values of $a_c$ and $a_t$ into the equation:

$a={\left[{\left(\frac{v^2}{r}\right)}^2+g^2\right]}^{0.5}$

Simplifying the expression:

$a={\left[\frac{v^4}{r^2},+,g^2\right]}^{0.5}$

Thus, the net acceleration of the particle is ${\left[v^4/r^2+g^2\right]}^{0.5}$.