Q104 / 150 · MHT CET 2026 (17 April Shift 1) · Physics
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$\dfrac{AB+BC}{3}$
- Correct answerB$\dfrac{AB+AC}{3}$
- C$\dfrac{AC+BC}{2}$
- D$\dfrac{AB+BC+AC}{3}$
Solution
Let the mass of each sphere be $m$.
Let the center of sphere A be at the origin $(0,0)$.
The coordinates of the centers of the three spheres along the X-axis are:
$x_A = 0$
$x_B = AB$
$x_C = AC$
The distance of the center of mass of the system from A (the origin) is given by the formula:
$x_{cm} = \dfrac{m x_A + m x_B + m x_C}{m + m + m}$
Substituting the coordinates into the equation:
$x_{cm} = \dfrac{m(0) + m(AB) + m(AC)}{3m}$
$x_{cm} = \dfrac{AB + AC}{3}$
Answer: $\dfrac{AB+AC}{3}$
Let the center of sphere A be at the origin $(0,0)$.
The coordinates of the centers of the three spheres along the X-axis are:
$x_A = 0$
$x_B = AB$
$x_C = AC$
The distance of the center of mass of the system from A (the origin) is given by the formula:
$x_{cm} = \dfrac{m x_A + m x_B + m x_C}{m + m + m}$
Substituting the coordinates into the equation:
$x_{cm} = \dfrac{m(0) + m(AB) + m(AC)}{3m}$
$x_{cm} = \dfrac{AB + AC}{3}$
Answer: $\dfrac{AB+AC}{3}$
Hence, the correct answer is option: 'B'.