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प्रश्न
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
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उत्तर
Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s
Frequency v = `"Number of revolution"/"Time taken" = 14/25 Hz`
Angular frequency, ω = 2πν
`2xx22/7xx14/25 = 88/25 rads^(-1)`
Centripetal acceleration, `a_c = omega^2 r`
=`(88/25)^2 xx 0.8`
= `0.80 xx 88/25 xx 88/25`
= 9.90 `ms^(-2)`
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.
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