Problem 44 - A Complete Resource Book in Physics for JEE Main

Question

A sphere of density \(\rho\) and radius \(a\) has a concentric cavity of radius \(b\) as shown in the Fig. \(7.13\). Gravitation force \(F\) exerted by the sphere on the particle of mass \(m\), located at a distance \(r\) from the centre of sphere as a function \(r\) when \(b

Step-by-Step Solution

Answer

(B) \(F_{r}=\frac{4}{3} \pi G \rho m\left(\frac{a^{3}-b^{3}}{r^{2}}\right)\)

Step 1: Understand the situation

The sphere of radius \(a\) with a hole of radius \(b\). So what matters here is only the solid volume. The volume of a sphere is given by \( \frac{4}{3} \pi r^{3} \). The volume of the complete sphere is \( \frac{4}{3} \pi a^{3} \) and the volume of the cavity is \( \frac{4}{3} \pi b^{3} \) thus the volume of the solid portion of the sphere is \( \frac{4}{3} \pi a^{3} - \frac{4}{3} \pi b^{3} \).

Step 2: Apply the gravitational force formula

The gravitational force is given by \( F = \frac{GMm}{r^{2}} \), where \(G\) is the gravitational constant, \(M\) is the mass of the sphere and \(m\) is the mass of the particle. The mass \(M\) is given by the density \(\rho\) times the volume, which is \( \rho\left(\frac{4}{3} \pi a^{3} - \frac{4}{3} \pi b^{3}\right) \). Substituting \(M\) into the gravitational formula yields: \( F = \frac{G\rho m \left(\frac{4}{3} \pi a^{3} - \frac{4}{3} \pi b^{3}\right)}{r^{2}} \). Simplifying the formula gives \(F = \frac{4}{3} \pi G \rho m\left(\frac{a^{3}-b^{3}}{r^{2}}\right) \).

Step 3: Compare with the options

The derived formula for the gravitational force exerted by the sphere on the particle matches with option (B) in the answer choices. Thus, the answer is (B).

Key Concepts

Gravitational Force Formula

Understanding how objects attract each other due to gravity is critical in physics. The gravitational force formula, represented by the equation \( F = G\frac{m_1m_2}{r^2} \), is fundamental in calculating the force between two masses. Here, \( F \) is the force of gravity, \( G \) is the universal gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between the centers of the two masses.

For solving problems involving the gravitational force inside or outside a spherical shell, such as the exercise given, this formula can be modified accounting for the different mass distributions. It's important to remember that within a uniform spherical shell, the gravitational forces exerted by the shell on an object inside it cancel each other out. Therefore, when calculating the gravitational force inside a spherical cavity, only the mass outside the radius of the object's location is considered.

Density in Physics

Density, denoted as \( \rho \), is a measure of mass per unit volume and is a significant concept in physics when discussing materials and their properties. The density formula is \( \rho = \frac{m}{V} \), where \( m \) is the mass and \( V \) is the volume of the material. Density plays a vital role when we calculate the mass of an object if its volume and density are known.

In the given exercise, the density \( \rho \) of the sphere is a key parameter in determining the mass involved in the gravitational force calculations. A common mistake is to overlook the density variation or apply it incorrectly, but in this exercise, we assume the density to be uniform for the solid part of the sphere which allows us to directly relate mass, density, and volume.

Volume of Sphere

The volume of a sphere is a geometric formula that calculates the space enclosed by a spherical object. Given a radius \( r \), the volume \( V \) of a sphere can be calculated using the formula \( V = \frac{4}{3}\pi r^3 \). This formula is critical in our sphere-with-cavity exercise.

Understanding how to compute the volume of a sphere allows us to solve for the mass of the sphere when combined with density. The mass of the sphere is derived by multiplying the volume of the physical sphere (excluding the cavity) by its density. In our case, subtracting the volume of the cavity from the volume of the entire sphere gives us the volume of the material that will exert gravitational force.

Cavity in Sphere

A sphere with a cavity presents an interesting problem in gravitational calculations. The cavity is a spherical volume from which material has been removed, thus creating a concentric empty space. When dealing with the gravitational force inside a sphere with a cavity, it's essential to consider the volume of the cavity in our calculations.

In the problem provided, we find the gravitational force on an object within a larger sphere that has a smaller sphere removed from its center. The key takeaway is that the material that would have occupied the cavity does not contribute to the gravitational force at the point where the external particle is located. This understanding helps us reach the correct force formula by considering only the mass of the spherical shell that resides between the cavity and the outer surface.