- Home
- Standard 11
- Physics
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
Similar Questions
hard