A solid disc of radius 'a' and mass 'm' rolls down

Question:

A solid disc of radius 'a' and mass 'm' rolls down without slipping on an inclined plane making an angle $\theta$ with the horizontal. The acceleration of the disc will be $\frac{2}{\mathrm{~b}} \mathrm{~g} \sin \theta$ where $\mathrm{b}$ is (Round off to the Nearest Integer)

$(\mathrm{g}=$ acceleration due to gravity $)$

$(\theta=$ angle as shown in figure $)$

Solution:

(3)

$\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{mR}^{2}}}=\frac{\mathrm{g} \sin \theta}{1+\frac{1}{2}}=\frac{2}{3} \mathrm{~g} \sin \theta$

$\mathrm{b}=3$

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