Ncert Exercises & Solutions

4.17 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle at a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of the acceleration of the stone?

NEETprep Answer:

$G i v e n,$
$L e n g t h o f s t r i n g, l = 80 c m = 0.8 m$
$N u m b e r o f r e v o l u t i o n s = 14$
$T i m e, t = 25 s$
$Frequency, v = \frac{Number of revolutions}{Time taken} = \frac{14}{25} Hz$
$Angular frequency, ω = 2 πν = 2 x \frac{22}{7} x \frac{14}{25} = \frac{88}{25} rad s^{- 1}$
$Centripetal acceleration, a_{c} = ω^{2} r = {(\frac{88}{25})}^{2} x 0.8$
$= 9.91 m / s^{2}$

The direction of centripetal acceleration is always directed along the string, toward the center, at all points.

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