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`(10)/(pi)s``(pi)/(10)s``(pi)/(sqrt(10))`s`(1)/(sqrt(10))s`

Answer :

B::DSolution :

Acceleration `a=omega^2x` <br> Maximum acceleration `a=omega^2A=((2pi)/(T))^2=(25)/(100)` <br> When block and platform are separated <br> `a=g=10` <br> `(4pi^2)/(T^2)`.`(1)/(40)impliesT^2=(pi^2)/(100)`. `T=(pi)/(10)sec`