Question

The force between two identical solid spheres each of radius \( r \) kept in contact is \( F \). If the distance of their centres is made \( 4r \), then the force between them is

• A. \( \dfrac{F}{2} \)
• B. \( \dfrac{F}{4} \)
• C. \( \dfrac{F}{8} \)
• D. \( \dfrac{F}{16} \)
• E. \( \dfrac{F}{24} \)

Hint:

Whenever force follows inverse square law, use: \[ F \propto \frac{1}{d^2} \] If the distance becomes \( n \) times, the force becomes \( \dfrac{1}{n^2} \) times.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question is based on Newton's Law of Universal Gravitation, which describes the force of attraction between two masses. The law states that the force is inversely proportional to the square of the distance between their centers.
Step 2: Key Formula or Approach:
Newton's Law of Universal Gravitation is given by:
\[ F = G \frac{m_1 m_2}{d^2} \] where \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the two objects, and \(d\) is the distance between their centers.
Since the force is proportional to \(1/d^2\), we can write \(F \propto \frac{1}{d^2}\).
Step 3: Detailed Explanation:
Let the mass of each identical sphere be \(m\).
Case 1: Initial Situation
The two spheres are kept in contact. The radius of each sphere is \(r\).
The distance between their centers, \(d_1\), is the sum of their radii: \(d_1 = r + r = 2r\).
The gravitational force between them is given as \(F\). Using the formula:
\[ F = G \frac{m \cdot m}{(d_1)^2} = G \frac{m^2}{(2r)^2} = G \frac{m^2}{4r^2} \quad \cdots (1) \] Case 2: Final Situation
The distance between the centers of the spheres is changed to \(d_2 = 4r\).
Let the new force between them be \(F'\). Using the formula again:
\[ F' = G \frac{m \cdot m}{(d_2)^2} = G \frac{m^2}{(4r)^2} = G \frac{m^2}{16r^2} \quad \cdots (2) \] To find the relation between \(F'\) and \(F\), we can take the ratio of equation (2) to equation (1):
\[ \frac{F'}{F} = \frac{G \frac{m^2}{16r^2}}{G \frac{m^2}{4r^2}} = \frac{m^2}{16r^2} \times \frac{4r^2}{m^2} \] \[ \frac{F'}{F} = \frac{4}{16} = \frac{1}{4} \] Therefore, the new force is \(F' = \frac{F}{4}\).
Step 4: Final Answer:
When the distance between the centers is doubled (from 2r to 4r), the force becomes \(1/2^2 = 1/4\) of the original force. The new force is \(\frac{F}{4}\). This corresponds to option (B).