Question

A stone tied to a string of `80 cm` long is whireled in a horizontal circle with a constant speed. If the stone makes `25` revolutions in `14 s` then, magnitude of acceleration of the same will be:
A. `990 cm//s^(2)`
B. `680 cm//s^(2)`
C. `750 cm//s^(2)`
D. `650 cm//s^(2)`

Answer 1

Correct Answer - A
When stone is tied to a `80 cm` long string and Whirled in a circle the centripetal force is providing by the tension in the string created by drawing the string inwards.
If angular velocity is `omega` and radius of circular path is `r` then
accelaration `alpha= omega^(2)r`
where `omega=(2pi)/(T)`
`T=("number of revolutions")/("time taken")= (25)/(14)= 1.78 s`
and `omega=(2pi)/(1.78)= 3.52 rad//s`
Hence, `alpha= (3.52)^(2)xx80= 991.23~~990 cm//s^(2)`