The bulk modulus of a spherical object is ‘B’. If it is subjected to uniform pressure ‘p’, the fractional decrease in radius is

Question

A 2 $\text{kg}$ mass is attached to a spring with spring constant $ k = 200, \text{N/m} $. If the mass is displaced by $ 0.1, \text{m} $, what is the potential energy stored in the spring?

Answer 1

To determine the fractional decrease in radius of a spherical object subjected to uniform pressure, we need to use the concept of bulk modulus. The bulk modulus $ B $ is defined as the ratio of the applied pressure change $ \Delta p $ to the relative change in volume $ \frac{\Delta V}{V} $:

B = \frac{\Delta p}{\frac{\Delta V}{V}}

Here, $ \Delta p = p $ since the pressure applied is $ p $, and $ \Delta V $ is the change in volume.

For a sphere, the volume $ V $ is given by:

V = \frac{4}{3} \pi r^3

where $ r $ is the radius of the sphere.

If there's a small change in radius $ \Delta r $, the new volume $ V' $ becomes:

V' = \frac{4}{3} \pi (r + \Delta r)^3

The change in volume $ \Delta V $ is:

\Delta V = V' - V = \frac{4}{3} \pi \left[(r + \Delta r)^3 - r^3\right]

For small changes, this can be approximated using a derivative. The derivative of the volume with respect to the radius is:

\frac{dV}{dr} = 4 \pi r^2

Thus, the change in volume $ \Delta V $ is approximately:

\Delta V \approx 4 \pi r^2 \Delta r

Substituting into the formula for bulk modulus, we have:

B = \frac{p}{\frac{\Delta V}{V}} = \frac{p}{\frac{4 \pi r^2 \Delta r}{\frac{4}{3} \pi r^3}}

Simplifying further gives:

B = \frac{p}{3 \frac{\Delta r}{r}}

Solving for the fractional change in radius, we find:

\frac{\Delta r}{r} = \frac{p}{3B}

Thus, the fractional decrease in radius is \frac{p}{3B}.

Therefore, the correct answer is \frac{p}{3B}.

Answer 2