A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed center at an angular velocity \[{\omega _0}\]​. If the length of the string and angular velocity is doubled, find the tension in the string which was initially \[{T_0}\]​.
A. \[{T_0}\]
B. \[\dfrac{{{T_0}}}{2}\]
C. \[4{T_0}\]
D. \[8{T_0}\]

Hint:In order to solve this problem we need to understand the angular velocity and tension. The angular velocity is a vector quantity that measures the speed of a rotating object. Angular velocity specifies both magnitude and direction and its unit is\[m{s^{ - 1}}\]. Tension is defined as the force that is transmitted through a rope, string, or wire when it is pulled by forces acting from opposite sides.

Formula Used:
To find the tension in the string the formula is,
\[{T_0} = m{\omega _0}^2l\]
Where, m is mass, \[{\omega _0}\] is the angular velocity and l is the length of the string.

Complete step by step solution:
Here, a mass is attached to a string and rotated in a horizontal circle. The tension developed in a string provides the necessary centripetal force. So, in the first case tension is \[{T_0}\] which is given as,
\[{T_0} = m{\omega _0}^2l\]………… (1)
If the angular velocity becomes 2 times and length becomes 2 times then,
\[{\omega _0}^1 = 2{\omega _0}\] and \[{l^1} = 2l\]

Substitute the value of \[{\omega _0}\] and l in equation (1), we obtain,
\[{T_0} = m{\left( {2{\omega _0}} \right)^2} \times 2l\]
\[\Rightarrow {T_0} = 8m{\omega _0}^2l\]
\[\therefore {T_0} = 8{T_0}\]
Therefore, the new tension in the string is \[8{T_0}\].

Hence, Option D is the correct answer.

Note:Here in the given problem it is important to remember the equation for the tension in the string and also the equation for the angular velocity. Thereby, calculating this we will get the solution.