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 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 :
Show Hint
- $40 \,ms ^{-1}$
- $20\, ms ^{-1}$
- $400\, ms ^{-1}$
- $10 \,ms ^{-1}$
The Correct Option is B
Solution and Explanation
To determine the maximum linear velocity that the stone can have without breaking the string, we need to consider the centripetal force acting on the stone as it rotates in a horizontal plane.
The formula for centripetal force \(F_c\) is given by:
\(F_c = \frac{mv^2}{r}\)
where:
- \(m\) = mass of the stone = \(1 \, \text{kg}\)
- \(v\) = linear velocity of the stone
- \(r\) = radius of the circle, which is the length of the string = \(1 \, \text{m}\)
The maximum tension the string can withstand before breaking is \(T_{\text{max}} = 400 \, \text{N}\). This tension is equal to the centripetal force when the string is about to break:
\(\frac{mv^2}{r} = T_{\text{max}}\)
Substituting the known values into the equation:
\(\frac{1 \times v^2}{1} = 400\)
Simplifying gives:
\(v^2 = 400\)
Taking the square root of both sides, we find:
\(v = \sqrt{400} = 20 \, \text{ms}^{-1}\)
Therefore, the maximum linear velocity the stone can have without breaking the string is \(20 \, \text{ms}^{-1}\).
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