A 1 kg stone at the end of 1 m long string is whirled in a vertical circle at constant speed of 4 m/sec. The tension in the string is 6 N when the stone is at `(g =10m//sec^(2))`

Options

A. Top of the circle
B. Bottom of the circle
C. Half way down
D. None of the above

Answer

A

Explanation

`mg = 10N , (mv^(2))/r = (1 xx 4^(2))/1 = 16 N`
Tension at the top of circle `= (mv^(2))/r - mg = 6 N`
Tension at the bottom of circle `=(mv^(2))/r + mg = 26 N`