CGPET 2021 Solved Paper

(a) We know that, centripital acceleration, $g_c = \frac{v^2}{a}$ For circular path, effective acceleration of pendulum, $g_{eff}^2 = g^2 + g_c^2$ ⇒ $g_{eff} = \sqrt{g^2 + g_c^2}$ ⇒ $g_{eff} = \sqrt{g^2 + \left(\frac{v^2}{a}\right)^2}$ Now, time period, $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$ When the railway carraiage is travelling, then frequency of oscillation of simple pendulum is given as $\frac{1}{n} = 2\pi \sqrt{\frac{l}{g_{eff}}} \text{ ..... (i) } \left(\because T = \frac{1}{n}\right)$ and when $v = 0$ (stationary state), then $\frac{1}{n_1} = 2\pi \sqrt{\frac{l}{g}}$ .....(ii) On dividing Eq. (i) by Eq. (ii), we get $\frac{\frac{1}{n}}{\frac{1}{n_1}} = \frac{2\pi \sqrt{\frac{l}{g_{eff}}}}{2\pi \sqrt{\frac{l}{g}}} \Rightarrow \frac{n_1}{n} = \sqrt{\frac{g}{g_{eff}}}$ ⇒ $\frac{n_1^2}{n^2} = \frac{g}{g_{eff}} \Rightarrow \frac{n_1^2}{n^2} = \frac{g}{\sqrt{g^2 + \left(\frac{v^2}{a}\right)^2}}$ Squaring both side of the above equation, we get $\left(\frac{n_1}{n}\right)^4 = \frac{g^4}{g^2 + \left(\frac{v^2}{a}\right)^2}$ ⇒ $g^2 + \left(\frac{v^2}{a}\right)^2 = \left(\frac{n}{n_1}\right)^4 g^2$ ⇒ $(v^2)^2 = a^2 g^2 \left[ \left(\frac{n}{n_1}\right)^4 - 1 \right]$ ⇒ $v^2 = a g \sqrt{ \frac{n^4}{n_1^4} - 1 }$