The figure below represent the free body diagram for the lamp in the train.

Apply Newton’s law in vertical direction.
\(\begin{aligned}\sum {{F_y}} &= 0\\{F_T}\cos \theta - mg &= 0\\{F_T} &= \frac{{mg}}{{\cos \theta }}\end{aligned}\)
Here, \({F_T}\) is the tension force.
The lamp is in circular motion on the horizontal plane. Apply Newton’s law in horizontal direction.
\(\begin{aligned}\sum {{F_R}} &= m{a_R}\\{F_T}\sin \theta &= m\left( {\frac{{{v^2}}}{r}} \right)\\\left( {\frac{{mg}}{{\cos \theta }}} \right)\sin \theta &= m\left( {\frac{{{v^2}}}{r}} \right)\\v &= \sqrt {rg\tan \theta } \end{aligned}\)