Question:
A body of mass 1 kg is attached to one end of a string of 1 m length. It is rotated in a vertical circle with a constant speed of \( 4 \text{ ms}^{-1} \). When the object is at the highest point of the vertical circle, tension in the string is (\( g=10 \text{ ms}^{-2} \)):
A body of mass 1 kg is attached to one end of a string of 1 m length. It is rotated in a vertical circle with a constant speed of \( 4 \text{ ms}^{-1} \). When the object is at the highest point of the vertical circle, tension in the string is (\( g=10 \text{ ms}^{-2} \)):
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At the highest point, the tension is minimum. If the speed is less than \( \sqrt{gr} \), the string will go slack.
Updated On: Jun 9, 2026
- \( 6 \text{ N} \)
- \( 8 \text{ N} \)
- \( 10 \text{ N} \)
- \( 16 \text{ N} \)
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The Correct Option is A
Solution and Explanation
Concept:
For an object moving in a vertical circle, the forces acting at the highest point are the tension \(T\) acting downwards and the gravitational force \(mg\) acting downwards. The centripetal force required for circular motion is provided by the sum of these forces.
Step 1: Apply the centripetal force equation at the highest point.
The net force toward the center is: $$ T + mg = \frac{mv^2}{r} $$
Step 2: Substitute the given values.
Given: \( m = 1 \text{ kg} \), \( r = 1 \text{ m} \), \( v = 4 \text{ ms}^{-1} \), and \( g = 10 \text{ ms}^{-2} \). $$ T + (1 \times 10) = \frac{1 \times (4)^2}{1} $$ $$ T + 10 = 16 $$
Step 3: Solve for tension T.
$$ T = 16 - 10 = 6 \text{ N} $$ $$\boxed{6 \text{ N}}$$
Step 1: Apply the centripetal force equation at the highest point.
The net force toward the center is: $$ T + mg = \frac{mv^2}{r} $$
Step 2: Substitute the given values.
Given: \( m = 1 \text{ kg} \), \( r = 1 \text{ m} \), \( v = 4 \text{ ms}^{-1} \), and \( g = 10 \text{ ms}^{-2} \). $$ T + (1 \times 10) = \frac{1 \times (4)^2}{1} $$ $$ T + 10 = 16 $$
Step 3: Solve for tension T.
$$ T = 16 - 10 = 6 \text{ N} $$ $$\boxed{6 \text{ N}}$$
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