Question

A stone of mass $1 \, kg$ is tied to end of a massless string of length $1 m$ If the breaking tension of the string is $400\, N$, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is :

• A. $40 \,ms ^{-1}$
• B. $20\, ms ^{-1}$
• C. $400\, ms ^{-1}$
• D. $10 \,ms ^{-1}$

Hint:

The maximum velocity occurs when the centripetal force equals the maximum tension in the string.

Answer 1

The Correct Option is B

The correct answer is (B) : \(20\ ms^{-1}\)

Answer 2

The maximum linear velocity occurs when the centripetal force is equal to the maximum tension in the string. The centripetal force is given by: \[ F_c = \frac{mv^2}{r} \] where \( m \) is the mass of the stone, \( v \) is the linear velocity, and \( r \) is the radius (which is the length of the string). Setting the centripetal force equal to the maximum tension: \[ \frac{mv^2}{r} = T \] Substitute \( m = 1 \, \text{kg} \), \( r = 1 \, \text{m} \), and \( T = 400 \, \text{N} \): \[ \frac{1 \times v^2}{1} = 400 \] \[ v^2 = 400 \] \[ v = 20 \, \text{m/s} \]