A body of weight $$200 \text{ N}$$ is suspended from a tree branch through a chain of mass $$10 \text{ kg}$$. The branch pulls the chain by a force equal to (if $$g = 10 \text{ m/s}^2$$) :

We need to find the force with which the branch pulls the chain.

Weight of body = 200 N, mass of chain = 10 kg, $$g = 10$$ m/s$$^2$$.

$$W_{chain} = mg = 10 \times 10 = 100 \text{ N}$$

The branch must support both the chain and the body. The total downward force at the point where the chain meets the branch is:

$$F_{total} = W_{body} + W_{chain} = 200 + 100 = 300 \text{ N}$$

The chain pulls the branch downward with 300 N. By Newton's third law, the branch pulls the chain upward with an equal and opposite force of 300 N.

The correct answer is Option 3: 300 N.