a
Let \(a_{1}\) and \(a_{2}\) be accelerations of a ball (upward) and rod (downward), respectively.

Clearly, from the diagram

\(2 a_{1}=a_{2} \ldots( i )\)

Now, for the ball

\(2 T-\frac{9}{5} m g-\frac{9}{5} m a-(i i)\)

and for the rod, \(m g-T=m a_{2} \ldots\) \((iii)\)

On solving equations \((i)\) and \((iii),\) we get

\(a_{1}=\frac{g}{29} m / s^{2} \uparrow\) (upward)

\(a_{2}=\frac{2 g}{29} m / s ^{2} \downarrow\) (upward)

So, acceleration of ball \(w.r.t\) rod \(=a_{1}+a_{2}=\frac{3 g}{29}\)

Now, displacement of ball \(w.r.t.\) rod when it reaches the upper end of rod is \(1\, m\).

Using equation of motion,

\(s=u t+\frac{1}{2} a t^{2}\)

\(s=0+\frac{1}{2} \times \frac{3 \times 10}{29} t^{2}\)

\(t=\sqrt{\frac{58}{30}}=1.4 s\) (approx)