Question

A point mass \(m\) is moved in a vertical circle of radius \(r\) with the help of a string. The velocity of the particle is \(\sqrt{ngr}\) when the string makes an angle of \(60^{\circ }\) with the lower vertical. The tension \(T\) in the string at an angle of \(60^{\circ }\) is \(\left(\cos 60^{\circ }=\frac{1}{2}\right)\)

• A. \(mg\left(n-\frac{1}{2}\right)\)
• B. \(\frac{nmg}{2}\)
• C. \(mg\left(n+\frac{1}{2}\right)\)
• D. \(\frac{3}{2}nmg\)

Hint:

Resolve the weight along the string and set the net force towards the centre equal to the centripetal force.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
At an angle \(\theta\) from the lowest point, the component of weight along the string is \(mg\cos\theta\), directed away from the centre. The tension acts towards the centre.

Step 2: Write the equation
\(T-mg\cos60^{\circ}=\dfrac{mv^{2}}{r}\)

Step 3: Calculation
With \(v^{2}=ngr\) we get \(\dfrac{mv^{2}}{r}=nmg\). So \(T=nmg+mg\cos60^{\circ}=nmg+\dfrac{mg}{2}\).

Step 4: Result
\(T=mg\left(n+\dfrac{1}{2}\right)\)

Final Answer:
The tension is mg times n plus one half. \[ \boxed{\text{(C) }mg\left(n+\dfrac{1}{2}\right)} \]