Question

A stone tied to the end of string 100cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolution in 22s, then the acceleration of the ston is
A. `16ms^(-2)`
B. `4ms^(-2)`
C. `12ms^(-2)`
D. `8ms^(-2)`

Answer 1

Correct Answer - a
Here, `r=100cm=1m `
Frequency , `v=14/22Hz`
`:. Omega=2piv=2xx22/7xx14/22=4 rads^(-1)`
The acceleration of the stone is
`a_(c)=omega^(2)r=(4)^(2)(1)=16ms^(-2)`