A circular ring and a solid sphere of the same radius roll from rest without slipping down the same inclined plane through the same vertical height. If the ratio of their speeds at the bottom is √(x/5), find x, rounded to the nearest integer as required for the Numerical Value section.
Physics · Rotational Motion · rolling on inclined surface
JEE Mains 2025 4 April Morning Shift

Question

A circular ring and a solid sphere of the same radius roll from rest without slipping down the same inclined plane through the same vertical height. If the ratio of their speeds at the bottom is √(x/5), find x, rounded to the nearest integer as required for the Numerical Value section.

Physics · Rotational Motion · rolling on inclined surface

JEE Mains 2025 4 April Morning Shift

G1 Physics · Accepted Answer

For rolling without slipping, mgh = ½mv² + ½Iω², with ω = v/R. Thus, v² = 2gh/[1 + I/(mR²)]. For a ring, I = mR², so v r i n g 2 v_ring² v r ​ in g 2 = 2gh/(1+1) = gh. For a solid sphere, I = (2/5)mR², so v s p h e r e 2 v_sphere² v s ​ p h er e 2 = 2gh/(1+2/5) = 2gh/(7/5) = 10gh/7. Therefore, ( v r i n g / v s p h e r e ) 2 v_ring/v_sphere)² v r ​ in g / v s ​ p h er e ) 2 = (gh)/(10gh/7) = 7/10. Given v r i n g / v s p h e r e v_ring/v_sphere v r ​ in g / v s ​ p h er e = √(x/5), we have x/5…