A stone of mass 1 kg is tied to a string 2m long and it's rotated at constant speed of 40 `ms^(-1)` in a vertical circle. The ratio of the tension at the top and the bottom is
[Take g=10m`s^(2)`]

Options

A. `81/79`
B. `79/81`
C. `19/12`
D. `12/19`

Answer

B

Explanation

Free body diagram (FBD) of a stone moving in a vertical circular path, which has tension force at point A and B as represented by TA and Ta respectively as given below in the figure,

Given, mass of stone (m) = 1 kg, length of the string ( R) = 2 m and
rotating· linear speed (v) = 40m`s^(-1)`
As, we know that the tension at position A,
`T_(A)=(mv^(2))/R+mg (thereforeF_(c)=(mv^(2))/R)`
`rArrT_(A)=(1xx(40)^(2))/2+1xx10=810N`
Similarly, tension at position B,
`rArrT_(B)=(mv^(2))/R-mg=(1xx(40)^(2))/R-mg=(1xx(40)^(2))/2-1xx10=790` N
So, the ratio of `T_(B)` and `T_(A)` i.e.,
`(T_(B))/(T_(A))=790/810=79/81`
Hence: the ratio of tension at position B and tension at position A is 79 : 81 .