4.17 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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8 months ago

4.17 Given, the length of the string, l = 80 cm, No. of revolution = 14, Time taken = 25 s

We know Frequency, v = No.ofrevolutiontimetaken = 1425 Hz

Angular frequency ω = 2?v = 2 * 227 *1425 rad/s = 3.52 rad/s

Centripetal acceleration ac = ω2 r = 3.522*80100 m/s2 = 9.91 m/s2

The direction of acceleration is towards the centre.

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physics ncert solutions class 11th 2023

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