5.Work, Energy, Power and Collision
medium

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

A

Top of the circle

B

Bottom of the circle

C

Half way down

D

None of the above

(AIIMS-1982)

Solution

(a) $mg = 1 \times 10 = 10N,$ $\frac{{m{v^2}}}{r} = \frac{{1 \times {{(4)}^2}}}{1} = 16$

Tension at the top of circle = $\frac{{m{v^2}}}{r} – mg = 6N$

Tension at the bottom of circle = $\frac{{m{v^2}}}{r} + mg = 26N$

Standard 11
Physics

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