Two point size bodies of same mass are knotted to a horizontal string one at the end, and the other at the midpoint of it. The string is rotated in horizontal plane with the other end as centre. If T is tension in the string between centre of circles and first body, then the tension in the string between the two bodies is

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$\frac{T}{2}$

$2 T$

$\frac{2 T}{3}$

$\frac{3 T}{2}$

answer is C.

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Detailed Solution

The tension in the string is responsible for the centripetal acceleration of the bodies.

Let the radius of the circle for the two bodies be L, 2L.

$T = {mLω}^{2} + m 2 {Lω}^{2} = 3 {mLω}^{2}$

The tension in the string that is in between the bodies is responsible for the centripetal acceleration of the body connected at the end.

$T' = m 2 {Lω}^{2} = \frac{2}{3} T$

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