A conical pendulum of length \(1\,m\) makes an angle \(\theta \, = 45^o\) w.r.t. \(Z-\) axis and moves in a circle in the \(XY\) plane.The radius of the circle is \(0.4\, m\) and its centre is vertically below \(O\). The speed of the pendulum, in its circular path, will be ..... \(m/s\) (Take \(g\, = 10\, ms^{-2}\))

\(\begin{array}{l} Given,\theta  = {45^ \circ },r = 0.4m,g = 10m/{s^2}\\ T\,\sin \,\theta  = \frac{{m{v^2}}}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\ T\,\cos \,\theta  = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {ii} \right)\\ From\,equation\,\left( i \right) and \left( {ii} \right)\\ We\,have,\,\tan \theta  = \frac{{{v^2}}}{{rg}}\\ {V^2} = rg\,\,\,\,\,\,\,\,\,\,\,\,\theta  = {45^ \circ }\\ Hence,\,speed\,of\,the\,pendulum\,in\,its\,circular\\ path,\\ V = \sqrt {rg}  = \sqrt {0.4 \times 10}  = 2m/s \end{array}\)