A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity ω0. If the length of the string and angular velocity are doubled, the tension in the string which was initially T0, is now :-T0T0/24T08T0
Question
A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity ω0. If the length of the string and angular velocity are doubled, the tension in the string which was initially T0, is now :-T0T0/24T08T0
Solution
The tension in a string of a rotating body is given by the formula T = mω²r, where m is the mass of the body, ω is the angular velocity, and r is the radius (or in this case, the length of the string).
Initially, the tension T0 = mω0²r0.
When the length of the string and the angular velocity are doubled, the new tension T is given by T = m(2ω0)²(2r0) = 8mω0²r0 = 8T0.
So, the tension in the string becomes 8 times the initial tension. Therefore, the correct answer is 8T0.
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