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: Identify the initial distance between the centres.
Each solid sphere has radius \( r \).
When the two spheres are kept in contact, the distance between their centres is: \[ d_1=r+r=2r \]

Step 2: Use the inverse square law of gravitational/electrostatic force.
The force between two spherically symmetric bodies behaves as if their entire mass is concentrated at their centres.
Hence, \[ F \propto \frac{1}{d^2} \] So if distance changes, force changes inversely as the square of the distance.

Step 3: Write the given initial force relation.
Initially, when the centre-to-centre distance is \[ d_1=2r \] the force is given as \[ F_1=F \]

Step 4: Write the new distance between the centres.
It is given that the new distance between the centres becomes \[ d_2=4r \] We have to find the new force \( F_2 \).

Step 5: Apply the inverse square law ratio.
Using \[ \frac{F_2}{F_1}=\frac{d_1^2}{d_2^2} \] we get \[ \frac{F_2}{F}=\frac{(2r)^2}{(4r)^2} \] \[ =\frac{4r^2}{16r^2} \] \[ =\frac{1}{4} \]

Step 6: Find the new force.
Thus, \[ F_2=\frac{F}{4} \] So when the distance between the centres is doubled from \( 2r \) to \( 4r \), the force becomes one-fourth.

Step 7: Final conclusion.
Therefore, the force between the two spheres is \[ \boxed{\frac{F}{4}} \] Hence, the correct option is \[ \boxed{(2)\ \dfrac{F}{4}} \]