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A particle is performing U.C.M. along the cirqumference of a circle of diameter 50 cm with frequency 2 Hz. The acceleration of the particle in `m//s^(2)` isA. `2 pi^(2)`B. ` 8 pi^(2)`C. `pi^(2)`D. `4pi^(2)` |
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Answer» Correct Answer - D (d) Given, diameter of circle ,d=50cm ` " " =50xx10^(-2)m` and frequency ,f= 2Hz The acceleration of particle in a uniform circular motion can be given as ` " "a = omega ^(2)x ` where, `omega =` angular frequency = ` 2pi f ` ` " " x =` distance from centre ` =(d)/(2)` `rArr " "a= 4pi ^(2)f^(2) xx(d)/(2) " "...(i)` Substituting given values in Eq. (i) we get ` " "a= 4pi ^(2)xx 4xx(50 xx10^(-2) )/( 2) =4pi ^(2) ` |
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