Question

A body of mass 'm' attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is
(\(g\) = gravitational acceleration)

• A. \(6\,mg\)
• B. \(3\,mg\)
• C. \(1\,mg\)
• D. zero

Hint:

Just completing the loop means the speed at the top is sqrt(gR), which makes the speed at the bottom sqrt(5gR).

Answer 1

The Correct Option is A

Step 1: Understand the concept
A body "just completing" a vertical loop has zero tension at the top, so gravity alone provides the centripetal force there: \(mg = \dfrac{mv_t^2}{R}\), so \(v_t^2 = gR\).

Step 2: Speed at the bottom
Use energy conservation between the top and bottom (height difference \(2R\)):
\[ \frac{1}{2}mv_b^2 = \frac{1}{2}mv_t^2 + mg(2R) \Rightarrow v_b^2 = gR + 4gR = 5gR \]

Step 3: Force at the bottom
At the lowest point, \(T - mg = \dfrac{mv_b^2}{R} = 5mg\), so \(T = 6mg\).

Step 4: Apparent weight
The apparent weight is the force the body exerts on the string, equal to the tension, so it is \(6mg\), option (A). The values \(3mg\) and \(mg\) do not fit \(v_b^2 = 5gR\).

Final Answer:
The apparent weight at the lowest point is 6 mg. This is option (A). \[ \boxed{\text{(A) }6mg} \]