Time taken by bus to cover total distance with speed  x  km/h $= \frac{240}{\mathrm{x}}$

Time taken by bus to cover total distance with speed  (x 10)  km/h $= \frac{240}{\mathrm{x} - 10}$

According to the given condition,

$$\begin{gathered} \frac{240}{\mathrm{x} - 10} - \frac{240}{\mathrm{x}} = 2 \\ \Rightarrow 240 \left( \frac{1}{\mathrm{x} - 10} - \frac{1}{\mathrm{x}} \right) = 2 \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle \mathrm{x} - 10}} - \frac{{\displaystyle 1}}{{\displaystyle \mathrm{x}}} =\frac{2}{240} \\ \Rightarrow \frac{\mathrm{x} - \mathrm{x} + 10}{\mathrm{x} ( \mathrm{x} - 10 )} = \frac{1}{120} \\ \Rightarrow \frac{{\displaystyle 10}}{{\displaystyle {\mathrm{x}}^2 - 10 \mathrm{x}}} = \frac{{\displaystyle 1}}{{\displaystyle 120}} \\ \Rightarrow {\mathrm{x}}^2 - 10 \mathrm{x} = 1200 \end{gathered}$$

$$\begin{gathered} \Rightarrow {\mathrm{x}}^2 - 10 \mathrm{x} - 1200 = 0 \\ \Rightarrow {\mathrm{x}}^2 - 40 \mathrm{x} + 30 \mathrm{x} - 1200 = 0 \\ \Rightarrow \mathrm{x} ( \mathrm{x} - 40 ) + 30 ( \mathrm{x} - 40 ) = 0 \\ \Rightarrow ( \mathrm{x} - 40 ) ( \mathrm{x} + 30 ) = 0 \\ \Rightarrow \mathrm{x} - 40 = 0 \mathrm{or} \mathrm{x} + 30 = 0 \\ \Rightarrow \mathrm{x} = 40 \mathrm{or} \mathrm{x} = - 30 \end{gathered}$$

Since, the speed cannot be negative, the uniform speed is  40  km / h.