Question

A ball of mass 0.25 kg attached to the ends of a string of length 1.96 m is rotating in a horizontal circle. The string will break, if tension is more than 25 N. What is the maximum velocity with which the ball can be rotated?

• A. 3 $ ms^{ - 1} $
• B. 5 $ ms^{ - 1} $
• C. 9 $ ms^{ - 1} $
• D. 14 $ ms^{ - 1} $

Answer 1

The Correct Option is D

To find the maximum velocity with which the ball can be rotated without breaking the string, we need to consider the relationship between the tension in the string, the mass of the ball, the radius of the circle, and the velocity of the ball. The tension in the string provides the necessary centripetal force to keep the ball moving in a circle.

The formula that relates these quantities is given by the equation for centripetal force:

T = \frac{mv^2}{r}

Where:

• T is the tension in the string (in newtons),
• m is the mass of the ball (in kilograms),
• v is the velocity of the ball (in meters per second),
• r is the radius of the circle (in meters).

We are given:

• The maximum tension (T) = 25 \, N,
• The mass of the ball (m) = 0.25 \, kg,
• The length of the string (which is the radius of the circle) (r) = 1.96 \, m.

Substituting these values into the centripetal force formula, we have:

25 = \frac{0.25 \times v^2}{1.96}

Solving for v^2:

25 \times 1.96 = 0.25 \times v^2

49 = 0.25 \times v^2

v^2 = \frac{49}{0.25}

v^2 = 196

Finding the square root of both sides gives us the maximum velocity:

v = \sqrt{196} = 14 \, ms^{-1}

Therefore, the maximum velocity with which the ball can be rotated without breaking the string is 14 ms^-1.