A solid sphere ( A ) of mass 5m and a spherical shell ( B ) of mass...

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 $\tau = F(2R)$.

The moment of inertia about the point of contact is $I = I_{cm} + MR^2$.

Using $\tau = I\alpha$ and the rolling without slipping condition $a = \alpha R$, we get:

$F(2R) = (I_{cm} + MR^2) \dfrac{a}{R} \Rightarrow a = \dfrac{2FR^2}{I_{cm} + MR^2}$

For the solid sphere $A$ of mass $5m$:

$I_{cm,A} = \dfrac{2}{5}(5m)R^2 = 2mR^2$

$a_A = \dfrac{2FR^2}{2mR^2 + 5mR^2} = \dfrac{2F}{7m}$

For the spherical shell $B$ of mass $m$:

$I_{cm,B} = \dfrac{2}{3}mR^2$

$a_B = \dfrac{2FR^2}{\dfrac{2}{3}mR^2 + mR^2} = \dfrac{2F}{\dfrac{5}{3}m} = \dfrac{6F}{5m}$

The ratio of their accelerations is:

$\dfrac{a_A}{a_B} = \dfrac{\dfrac{2F}{7m}}{\dfrac{6F}{5m}} = \dfrac{2}{7} \times \dfrac{5}{6} = \dfrac{5}{21}$

Answer: $5 : 21$