Question

Three identical spheres, each of mass 'm' kg, are kept as shown in the figure, touching each other, with their centers on a straight line. If their centers are marked as A, B, C respectively, the distance of center of mass of the system from A is

• A. \(\frac{AB+BC}{3}\)
• B. \(\frac{AB+AC}{3}\)
• C. \(\frac{AC+BC}{2}\)
• D. \(\frac{AB+BC+AC}{3}\)

Hint:

Take A at the origin; the centre of mass is the average of the positions of the centres.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The figure shows three identical spheres in a row, touching, with centres A, B and C on a straight line. Each has the same mass \(m\).

Step 2: Key Formula or Approach
With A at the origin, B is at distance AB and C is at distance AC along the line.
\[ x_{cm}=\frac{m(0)+m(AB)+m(AC)}{3m} \]

Step 3: Detailed Explanation
\[ x_{cm}=\frac{AB+AC}{3} \]
The spheres touch, so \(AB=2r\) and \(AC=4r\), giving \(x_{cm}=2r\), which is the centre B, as expected from symmetry.

Final Answer:
The centre of mass is at \(\frac{AB+AC}{3}\) from A, option (B). \[ \boxed{\dfrac{AB+AC}{3}\ \text{(B)}} \]