A solid disc of radius 'a' and mass 'm' rolls down without slipping on an inclined plane making an angle θ with the horizontal. The acceleration of the disc will be 2/b g sin θ where b is ____________. (Round off to the Nearest Integer) (g = acceleration due to gravity, θ = angle as shown in figure) Numerical answer type. Work it out, then open the answer.

Answer: 3. Rolling acceleration: a = (gsin θ)/(1 + k²/R²). Solid disc: k²/R² = 1/2 ⇒ a = 2/3gsin θ. Comparing with 2/bgsin θ: b = 3.

A solid disc of radius 'a' and mass 'm' rolls down without slipping… | JEE Main 2021 (16 Mar, Shift 2)

Q53JEE Main 2021 · 16 MAR · SHIFT 2Easy

Rolling Motion

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 $\frac{2}{b}$ g sin $\theta$ where b is ____________. (Round off to the Nearest Integer) (g = acceleration due to gravity, $\theta$ = angle as shown in figure)

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Rolling acceleration: $a = \dfrac{g\sin\theta}{1 + k^2/R^2}$ .
Solid disc: $k^2/R^2 = \frac{1}{2} \Rightarrow a = \frac{2}{3}g\sin\theta$ .
Comparing with $\frac{2}{b}g\sin\theta$ : $b = 3$ .

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Rolling Motion question from JEE Main 2021 (16 Mar, Shift 2)

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