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: Use the standard results
For a loop of radius $R$ with the minimum speed, the speed at the top is $\sqrt{gR}$ and the speed at the bottom is $\sqrt{5gR}$.

Step 2: Net force at the bottom
Centripetal requirement: $\dfrac{m(5gR)}{R} = 5mg$ directed upward toward the centre.

Step 3: Add the weight
The string pulls up with $T$ and gravity pulls down with $mg$, so $T = 5mg + mg = 6mg$.

Step 4: Meaning
The body presses on the string with a force $6mg$, six times its true weight, which is the apparent weight.

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