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(a) How many revolutions per minute must the apparatus shown in figure make about a vertical axis so that the cord makes an angle of `45^(@)` with the vertical ? (b)What is the tension in the cord then? Given, `l=sqrt(2)m,a=20 cm` and `m=5.0 kg`? |
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Answer» Correct Answer - C `(a) r=a+l sin 45^(@)=(0.2)+(sqrt(2))(1/(sqrt(2)))=1.2 m` Now, `T cos 45^(@)=mg.....(i)` and `T sin 45^(@)=mr omega^(2).....(ii)` From eqn. (i) and (ii), we have `omega=2n pi =sqrt(g/r)` `:. n=1/(2pi)sqrt(g/r)=60/(pi)sqrt(9.8/1.2)rpm=27.3 r pm ` (b). From eq. (i), we have `T=sqrt(2)mg=(sqrt(2))(5.0)(9.8)` `=69.3 N` |
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