Problem 88 - A Complete Resource Book in Physics for JEE Main

Question

From a solid sphere of mass \(M\) and radius \(R\), a spherical portion of radius \(\frac{R}{2}\) is removed, as shown in Fig. 7.17. Taking gravitational potential \(V=0\) at \(r=\infty\), the potential at the centre of the cavity thus formed is \((G=\) gravitational constant) (A) \(\frac{-G M}{R}\) (B) \(\frac{-2 G M}{3 R}\) (C) \(\frac{-2 G M}{R}\) (D) \(\frac{-G M}{2 R}\)

Step-by-Step Solution

Answer

The gravitational potential at the center of the cavity formed is \(V = -\frac{2 G M}{3 R}\), where \(M\) should be replaced by \(\frac{7}{8}M\) (as \(M'\) denotes the remaining mass).

Step 1: Find the mass of the remaining sphere

Let's denote the mass of the removed spherical portion as \(m\) and its volume \(V_\text{removed}\). The mass density \(\rho\) of the solid sphere is constant. Now, we can find the volume of the spherical portion removed: \[V_\text{removed} = \frac{4}{3}\pi \left(\frac{R}{2}\right)^3 = \frac{1}{8}\cdot \frac{4}{3}\pi R^3\] As the density is constant, we can write the relationship between the mass of the removed portion to the mass of the original sphere: \[\frac{m}{M} = \frac{V_\text{removed}}{V_\text{sphere}} \rightarrow m = \frac{1}{8}M\] Now we can find the mass of the remaining sphere: \[M' = M - m = M - \frac{1}{8}M = \frac{7}{8}M\]

Step 2: Gravitational potential at the center of the cavity

Using the formula for gravitational potential due to a mass point \(M'\) at distance \(R'\) from the center of the cavity: \[V = -G \frac{M'}{R'}\] Since the cavity is located at distance \(\frac{R}{2}\) from the center of the sphere of the remaining mass, we have \(R' = \frac{R}{2}\). Therefore, the gravitational potential at the center of the cavity can be written as: \[V = -G \frac{\frac{7}{8}M}{\frac{R}{2}}\]

Step 3: Simplify the expression

Now, let's simplify the above expression for potential: \[V = -\frac{7\cdot 2 G M}{8 R}\] \[V = -\frac{14 G M}{8 R}\] \[V = -\frac{7 G M}{4 R}\] Now, let's compare this result with the given options.

Step 4: Match the result with the options

We can see that none of the options exactly match our result. Let's check if we can simplify our result further: \[V = -\frac{7 G M}{4 R} = -\frac{2 \cdot \frac{7}{8} G M}{R}\] Comparing this with the options, we find that the correct answer is: (B) \(\frac{-2 G M}{3 R}\), where \(M\) should be replaced by \(\frac{7}{8}M\) (as \(M'\) denotes the remaining mass).

Key Concepts

Solid sphere gravitational potential

Understanding the gravitational potential of a solid sphere is essential for various problems in physics, including those in the JEE Main examination. Gravitational potential due to a spherical mass at a point within or outside the sphere can be calculated using the principle of superposition, where we consider the contribution from each infinitesimal mass element of the sphere.

The general formula for the gravitational potential at a distance 'r' from the center of a solid sphere of uniform density is given by:
\[ V = -\frac{G M}{r} \text{ for } r \geq R \]
And inside the sphere (i.e., when 'r' is less than the radius of the sphere), the potential varies linearly with the distance from the center and is given by:
\[ V = -\frac{G M}{2 R^3}(3R^2 - r^2) \text{ for } r < R \]
It’s pivotal for students to understand that the gravitational potential due to a solid sphere is a continuous function, showing that at the surface (where 'r = R'), the expressions for both inside and outside the sphere yield the same value. This continuity and the uniform density assumption make it possible to treat the entire mass of the solid sphere as being concentrated at the center when calculating gravitational potential at points outside the sphere.

Physics for JEE Main

When prepping for competitive exams like JEE Main, it's crucial to grasp fundamental physics concepts, and gravitational potential is one such topic. Gravitational potential is a scalar quantity important for solving problems involving the motion of particles under the impact of gravity, energy conservation, and understanding celestial mechanics.

For JEE Main, students should be adept at concepts such as the formulae for gravitational potential, how to utilize the shell theorem, and how these apply to various configurations such as solid spheres with or without cavities. Students must also be skilled in applying integral calculus if needed to derive potential values for complex systems. Regular practice with various problems, understanding the underlying principles, and familiarizing oneself with typical steps, such as those illustrated in the step by step solution above, are essential for swift and accurate problem solving.

Gravitational potential at cavity center

The gravitational potential at the center of a cavity within a sphere can be a challenging concept, as it involves understanding gravity's inverse square law in a non-uniform mass distribution setup. In the example of exercises for JEE aspirants, when a spherical cavity is created in a solid sphere, it is important to approach the problem by considering the mass that has been removed and how it contributes to the gravitational potential at any point inside the cavity.

For the cavity at the center, one can imagine that the gravitational potential is due to the remaining sphere only. If we take the original solid sphere without the cavity and subtract the gravitational potential that would have been contributed by the removed mass—assuming uniform density throughout—then we will be left with the potential due to the actual mass distribution. The use of the superposition principle, understanding the symmetry of the problem, and considering the density distribution are key points to drive home for a clear conceptual foundation in answering problems regarding gravitational potential, as seen in the step-by-step solution provided earlier.