A stone tied to one end of spring 80 cm long is whirled in a horizontal circle with a constant speed. If stone makes 14 revolutions in 25 s, the magnitude of acceleration of stone is  

Given, radius of the horizontal circle, r = 80 cm = 0.80 m,

n = 14 and t = 25s
Angular speed of revolution of the stone,

$\omega =\frac{2\pi n}{t}=2\pi \left(\frac{n}{t}\right)$

$\omega =2\times \frac{22}{7}\times \left(\frac{14}{25}\right)=\frac{88}{25}{\mathrm{rads}}^{-1}$

$\because$Magnitude of centripetal acceleration =  $r{\omega }^2$

$=80\times {\left(\frac{88}{25}\right)}^2=991.2\simeq 992{\mathrm{cms}}^{-2}$