Two concentric shells of different masses m1 and m2 are having a sliding particle of mass m. The forces on the particle at position A, B and C are

The correct answer is: G(m₁+m₂)m/r₁², Gm₁/r₂��, 0 (noting that in the second term, the mass of the particle m is omitted in the option text).

To understand this result, we use the Shell Theorem of gravitation:
1. Outside a spherical shell: The gravitational force exerted by a uniform spherical shell on a point mass outside it is calculated by assuming the entire mass of the shell is concentrated at its center.
2. Inside a spherical shell: The gravitational force exerted by a uniform spherical shell on a point mass placed anywhere inside it is exactly zero.

Let the two concentric shells have mass $m_1$ (inner shell) and $m_2$ (outer shell). The particle has mass $m$.

Step 1: Force at Position A (outside both shells)
Let position A be at a distance $r_1$ from the common center, where $r_1$ is greater than the radii of both shells.
Since the particle is outside both shells, both shells attract it as if their masses were concentrated at the center. Thus, the total gravitational force is:

$F_A=\frac{G(m_1+m_2)m}{{r_1}^2}$

Step 2: Force at Position B (between the two shells)
Let position B be at a distance $r_2$ from the common center, lying between the inner and outer shells.
- Since the particle is inside the outer shell ($m_2$), the gravitational force exerted on it by the outer shell is zero.
- Since the particle is outside the inner shell ($m_1$), the inner shell attracts it as if its entire mass is concentrated at the center.
Therefore, the force at B is due only to the inner shell:

$F_B=\frac{G{m}_{1}m}{{r_2}^2}$

Step 3: Force at Position C (inside both shells)
Let position C be inside both shells.
Since the particle is inside both the inner and outer shells, the gravitational force exerted on it by both shells is zero:

$F_C=0$