As shown in the figure, a particle of mass m and another particle of mass 10m are connected with a string. Friction is sufficient to prevent the slipping of 10 m. Mass m
is given a velocity u in vertical direction. For complete circular motion of mass m :- 

Let say speed at top position is v.

Using energy conservation for smaller mass, 

$\begin{array}{l}{U}_i+K_i=U_f+K_f\text{}\\ \Rightarrow 0+\frac{1}{2}m{u}^2=mgL+\frac{1}{2}m{v}^2\text{}\\ \Rightarrow v^2=u^2-2gL\mathrm{............}\left(1\right)\end{array}$

At top position using FBD we can write,

$\begin{array}{l}\text{T}+mg=\frac{{\displaystyle m{v}^2}}{{\displaystyle \text{L}}}\\ \text{}\Rightarrow T=\frac{{\displaystyle m{v}^2}}{{\displaystyle \text{L}}}-mg\end{array}$

Condition 1 : Tension should not become zero at top position

$\begin{array}{l}T>0\text{}\\ \Rightarrow v^2>gL\text{}\\ \Rightarrow u^2-2gL>gL\text{}\\ \Rightarrow u>\sqrt{3gL}\end{array}$

Condition 2 : The larger mass should not break contact with ground

$\begin{array}{l}T<10mg\text{}\\ \Rightarrow \frac{m{v}^2}{\text{L}}-mg<10mg\text{}\\ \Rightarrow v^2<11gL\text{}\\ \Rightarrow u^2-2gL<11gL\text{}\\ \Rightarrow u<\sqrt{13gL}\end{array}$

Thus, $\sqrt{3\mathrm{gL}}<\mathrm{u}<\sqrt{13\mathrm{gL}}$