For a body of mass M and radius R rolling without slipping under a force F applied tangentially at the highest point, the torque about the point of contact with the ground is τ=F(2R).
The moment of inertia about the point of contact is I=Icm+MR2.
Using τ=Iα and the rolling without slipping condition a=αR, we get:
F(2R)=(Icm+MR2)Ra⇒a=Icm+MR22FR2
For the solid sphere A of mass 5m:
Icm,A=52(5m)R2=2mR2
aA=2mR2+5mR22FR2=7m2F
For the spherical shell B of mass m:
Icm,B=32mR2
aB=32mR2+mR22FR2=35m2F=5m6F
The ratio of their accelerations is:
aBaA=5m6F7m2F=72×65=215
Answer: 5:21