A solid sphere (\(A\)) of mass \(5m\) and a spherical shell (\(B\)) of mass \(m\), both having same radius, are placed on a rough surface. When a force of same magnitude is applied tangentially at the highest points of \(A\) and \(B\), they start rolling without slipping with an acceleration of \(a_A\) and \(a_B\), respectively. The ratio of \(a_A\) and \(a_B\) is __________.

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: \(5 : 21\)