Q111 / 150 · MHT CET 2026 (18 April Shift 1) · Physics
A solid cylinder and a solid sphere having the same mass and radius roll down on the same smooth inclined plane. The ratio of the acceleration of the cylinder $(a_c)$ to that of the sphere $(a_s)$ is
- Correct answerA$\dfrac{14}{15}$
- B$\dfrac{15}{14}$
- C$\dfrac{13}{14}$
- D$\dfrac{11}{15}$
Solution
The acceleration of a body rolling down an inclined plane is given by $a = \dfrac{g \sin \theta}{1 + \dfrac{I}{MR^2}}$.
For a solid cylinder, $I_c = \dfrac{1}{2}MR^2$.
$a_c = \dfrac{g \sin \theta}{1 + \dfrac{1}{2}} = \dfrac{2}{3}g \sin \theta$
For a solid sphere, $I_s = \dfrac{2}{5}MR^2$.
$a_s = \dfrac{g \sin \theta}{1 + \dfrac{2}{5}} = \dfrac{5}{7}g \sin \theta$
The ratio of the acceleration of the cylinder to that of the sphere is:
$\dfrac{a_c}{a_s} = \dfrac{\dfrac{2}{3}g \sin \theta}{\dfrac{5}{7}g \sin \theta} = \dfrac{14}{15}$
Answer: $\dfrac{14}{15}$
For a solid cylinder, $I_c = \dfrac{1}{2}MR^2$.
$a_c = \dfrac{g \sin \theta}{1 + \dfrac{1}{2}} = \dfrac{2}{3}g \sin \theta$
For a solid sphere, $I_s = \dfrac{2}{5}MR^2$.
$a_s = \dfrac{g \sin \theta}{1 + \dfrac{2}{5}} = \dfrac{5}{7}g \sin \theta$
The ratio of the acceleration of the cylinder to that of the sphere is:
$\dfrac{a_c}{a_s} = \dfrac{\dfrac{2}{3}g \sin \theta}{\dfrac{5}{7}g \sin \theta} = \dfrac{14}{15}$
Answer: $\dfrac{14}{15}$
Hence, the correct answer is option: 'A'.