A 2kg stone at the end of a string 1 m long is whirled in a vertical circle . At a certain positon , the speed of the stone is 4m /s . The tension in the string will be 52 N, when the stone is `[ g= 10 m//s^(2)]`
A
at the bottom of the circle
B
at the top of the circle
C
at a height of half the radius from the bottom
D
at the ends of the horizontal diameter
Text Solution
Verified by Experts
The correct Answer is:
A
when a body attached at the end of a string is whirled in a vertical circle , the tension in the string at any point is given by `T= (mv^2)/(r )+ mg cos theta ` `therefore 52= (2xx4xx4)/(1) +2xx10xx cos theta ` ` therefore 52-32=20 cos theta ` ` therefore cos theta =1 " " therefore theta =0^(@)` `therefore ` The stone is at the bottom of the circle .
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