\[\begin {array} {1 1} (a)\;10 & \quad (b)\;20 \\ (c)\;30 & \quad (d)\;40 \end {array}\]

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$\theta= \large\frac{1}{2} \alpha t^2$

$2 \pi \times 10 =\large\frac{1}{2}$$ \alpha (3)^2$

$ 2 \pi n= \large\frac{1}{2} $$ \alpha (6)^2$

$n$- total number of rotation in 6 seconds

$\large\frac{n}{10}=\frac{36}{9}$

$n=40$

the number of rotations in the first 3 second is 10. therefore the number of rotations in the next 3 seconds is

$\therefore \;n_1+n_2=n=40$

$n_2= 40-10=30$

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