A ring of mass `m_(1)` and radius R is fixed in space at some location. An external agent brings a point mass `m_(2)` from infinity to centre of the ring. Work done by the external agent will be

Options

A. `-(Gm_(1)m_(2))/(R)`
B. `(Gm_(1)m_(2))/(R)`
C. `(Gsqrt(m_(1)^(2)+m_(2)^(2)))/(R)`
D. `(Gm_(1)m_(2))/(R(m_(1)^(2)+m_(2)^(2)))`

Explanation

To find the work done by the external agent in bringing a point mass \( m_2 \) from infinity to the center of a ring of mass \( m_1 \) and radius \( R \), we can follow these steps:

### Step 1: Understand the potential energy at infinity
When the point mass \( m_2 \) is at infinity, the gravitational potential energy \( U \) is defined to be zero. This is our initial potential energy:
\[ U_{\text{initial}} = 0 \]

### Step 2: Calculate the gravitational potential energy at the center of the ring
The gravitational potential \( V \) at a distance \( R \) from a ring of mass \( m_1 \) is given by the formula:
\[ V = -\frac{G m_1}{R} \]
where \( G \) is the universal gravitational constant.

The potential energy \( U \) of the point mass \( m_2 \) when it is at the center of the ring can be calculated using:
\[ U_{\text{final}} = m_2 V = m_2 \left(-\frac{G m_1}{R}\right) = -\frac{G m_1 m_2}{R} \]

### Step 3: Calculate the work done by the external agent
The work done \( W \) by the external agent in bringing the mass \( m_2 \) from infinity to the center of the ring is equal to the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we found:
\[ W = -\frac{G m_1 m_2}{R} - 0 = -\frac{G m_1 m_2}{R} \]

Thus, the work done by the external agent is:
\[ W = -\frac{G m_1 m_2}{R} \]

### Final Answer:
The work done by the external agent is:
\[ \boxed{-\frac{G m_1 m_2}{R}} \]

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