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. \( 6 \text{ N} \)
• B. \( 8 \text{ N} \)
• C. \( 10 \text{ N} \)
• D. \( 16 \text{ N} \)

Hint:

At the highest point, the tension is minimum. If the speed is less than \( \sqrt{gr} \), the string will go slack.

Answer 1

The Correct Option is A

Step 1: Focus on the highest point.
At the very top of a vertical circle, both the string tension $T$ and the weight $mg$ point downward, straight toward the centre of the circle.
Step 2: Write the centripetal condition.
The two downward forces together supply the centripetal force: \[ T + mg = \frac{mv^2}{r} \]
Step 3: List the values.
$m = 1$ kg, $r = 1$ m, $v = 4$ ms$^{-1}$, $g = 10$ ms$^{-2}$.
Step 4: Compute the required centripetal force.
$\dfrac{mv^2}{r} = \dfrac{1\times 4^2}{1} = 16$ N.
Step 5: Compute the weight.
$mg = 1\times 10 = 10$ N.
Step 6: Solve for tension.
$T = 16 - 10 = 6$ N.
\[ \boxed{6\ \text{N}} \]