Q24. At any smooth plane a body of 2 kg is rotated in a horizontal circle at the rate of 60 revolution 1 minute at the end of a 3 m long string. Its centripetal acceleration is-
(1) 118.4 m/s² (2) 1.18 m/s² (3) 2.368 m/s² (4) 23.68 m/s²

Q26. A body is rotated in a horizontal circle with 40 cm long string. If string can withstand on a tension of 6.4 N then maximum number of revolutions per minute without relaxation in string.
(1) 1.91 (2) 19.1 (3) 191 (4) 1910

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Solution

Dear Student,

$24 . a = m ω^{2} r = 2 * {(60 * \frac{2 π}{60})}^{2} * 3 = 236.87 m / s^{2} T = m ω^{2} (0.4) m a s s o f t h e b o d y i s n o t g i v e n ? / R E g a r d s$

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Dear Student, 24.a=mω2r=2*60*2π602*3=236.87 m/s2T=mω20.4mass of the body is not given ?/REgards

Dear Student,

$24 . a = m ω^{2} r = 2 * {(60 * \frac{2 π}{60})}^{2} * 3 = 236.87 m / s^{2} T = m ω^{2} (0.4) m a s s o f t h e b o d y i s n o t g i v e n ? / R E g a r d s$