The Voltage time (V-t) graph of triangular wave having peak value $V_0$ is shown. The rms value of V in interval $t=0$ to $t=\large\frac{\pi}{4}$ is

$(a)\;\frac{V_o}{\sqrt 3}\\(b)\;\frac{V_o}{2}\\ (c)\;\frac{V_o}{\sqrt 2} \\ (d)\;none\;of\;them$

How do you get the wave
equation: V=(Vo)(t)/(T/4)

$V= \large\frac{V_o}{T/4}$$t=> V=\large\frac{4 V_0}{T}$$t$
$V_{rms} =\sqrt { < v ^2 > }=\large\frac{2V_0}{T} \Large\frac{\int _0^{t/4} t^2dt}{\int _0^{\pi/4} dt}$
$\qquad= \large\frac{V_0}{\sqrt 3}$
Hence a is the correct answer.