Rotational Motion

JEE Main level, numerical answer, medium

260 of 280

Question 260 of 280

A solid disc of radius aa and mass mm rolls down without slipping on an inclined plane making an angle θ\theta with the horizontal. The acceleration of the disc will be 2bgsin⁡θ\dfrac{2}{b}g\sin \theta, where bb is _______. (Round off to the Nearest Integer)
(g=g= acceleration due to gravity)
(θ=\theta = angle as shown in figure)

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Answer: 3

a=gsin⁡θ1+ImR2=gsin⁡θ1+12=23gsin⁡θa=\dfrac{g\sin \theta }{1+\dfrac{I}{{mR}^{2}}}=\dfrac{g\sin \theta }{1+\dfrac{1}{2}}=\dfrac{2}{3}g\sin \theta
b=3b=3

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