Q. A solid sphere of mass ' $m$ ' and radius ' $r$ ' is allowed to roll without slipping from the highest point of an inclined plane of length ' $L$ ' and makes an angle $30^\circ$ with the horizontal. The speed of the particle at the bottom of the plane is $v_1$. If the angle of inclination is increased to $45^\circ$ while keeping $L$ constant. Then the new speed of the sphere at the bottom of the plane is $v_2$. The ratio $v_1^2:v_2^2$ is
A) $1:\sqrt2$
B) $1:\sqrt3$
C) $1:3$
D) $1:2$
Explanation
[Solution by Dr. Abubacker] For rolling without slipping, $v^2=\frac{2gL\sin\theta}{1+k^2/r^2}\propto\sin\theta$. So $v_1^2:v_2^2=\sin30^\circ:\sin45^\circ=\frac12:\frac1{\sqrt2}=1:\sqrt2$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.