Question

A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1m, by means of a string at an initial speed of 10rpm. Keeping the radius constant, the tension in the string is reduced to one-quarter of its initial value. Then find the new speed.
A. 5 rpm
B. 10 rpm
C. 20 rpm
D. 14 rpm

Answer 1

Hint:Before we start addressing the problem, we need to know about the tension in the string. Tension is defined as the force which acts along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable.

Formula Used:
To find the tension in the string the formula is,
\[T = m{\omega ^2}r\]
Where, m is mass, \[\omega \] is angular frequency and r is radius.

Complete step by step solution:
If a ball of mass 0.1 kg is whirled in a horizontal circle of radius 1m, at an initial speed of 10rpm. The tension in the string is reduced to one-quarter of its initial value. Then we need to find the new speed of the ball. The tension in the string is,
\[T = m{\omega ^2}r\]
Here, \[\omega = 2\pi n\] then,
\[T = m4{\pi ^2}{n^2}{\omega ^2}r\]

Since the mass and radius are constant, we can write above equation as,
\[T \propto {n^2}\]
\[ \Rightarrow n = \sqrt T \]
\[ \Rightarrow {n_1} = \sqrt {{T_1}} \] and \[{n_2} = \sqrt {{T_2}} \]
\[\dfrac{{{n_2}}}{{{n_1}}} = \sqrt {\dfrac{{{T_2}}}{{{T_1}}}} \] ……. (1)
Initially the tension in the string is, \[{T_1} = T\]
If the tension in the string is reduced to one-quarter of its initial value then,
\[{T_2} = \dfrac{T}{4}\]

Substitute the value of \[{T_1}\] and \[{T_2}\] in equation (1) we get,
\[\dfrac{{{n_2}}}{{{n_1}}} = \sqrt {\dfrac{{\dfrac{T}{4}}}{T}} \]
\[\Rightarrow \dfrac{{{n_2}}}{{{n_1}}} = \sqrt {\dfrac{1}{4}} \]
\[\Rightarrow \dfrac{{{n_2}}}{{{n_1}}} = \dfrac{1}{2}\]
Here, the initial speed, \[{n_1} = 10\,rpm\], we have to find \[{n_2}\].
\[{n_2} = \dfrac{{{n_1}}}{2}\]
\[\Rightarrow {n_2} = \dfrac{{10}}{2}\]
\[\therefore {n_2} = 5\,rpm\]
Therefore, the new speed is 5 rpm.

Hence, Option A is the correct answer.

Note:In this problem it is important to remember the equation for the tension in the string which is carried and how the tension in the spring depends on the speed. Thereby, calculating the speed of the ball.