Question

The maximum and minimum tensions in the string whirling in a circle of radius \(2.5\) m are in the ratio \(5:3\) then its velocity is (\(g=10\ \text{m/s}^{2}\))

• A. \(10\ \text{m/s}\)
• B. \(100\ \text{m/s}\)
• C. \(1\ \text{m/s}\)
• D. \(5\ \text{m/s}\)

Hint:

For a constant speed the tension is greatest at the bottom and least at the top. Set their ratio to 5:3 and solve for the speed.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
In a vertical circle at constant speed \(v\), the tension is maximum at the lowest point and minimum at the highest point.

Step 2: Write the Tensions
\(T_{max}=\frac{mv^{2}}{r}+mg\)
\(T_{min}=\frac{mv^{2}}{r}-mg\)

Step 3: Use the Ratio
\(\dfrac{v^{2}/r+g}{v^{2}/r-g}=\dfrac{5}{3}\)
Cross-multiplying: \(3v^{2}/r+3g=5v^{2}/r-5g\), so \(8g=2v^{2}/r\).

Step 4: Calculation
\(v^{2}=4gr=4\times 10\times 2.5=100\), so \(v=10\ \text{m/s}\).

Final Answer:
The speed of the body is 10 m/s. \[ \boxed{\text{(A) }10\ \text{m/s}} \]