A railway track is banked for a speed ' $v$ ' by elevating outer rail by a height ' $h$ ' above the inner…
- $\frac{\mathrm{v}^2 \mathrm{~d}}{\text { gh }}$
- $\frac{2 \mathrm{v}^2}{\mathrm{gdh}}$
- $\frac{g d}{2 v^2 h}$
- $\frac{v^2}{2 g h d}$
Solution
From figure,
$\begin{array}{ll}
& \tan \theta=\frac{\mathrm{h}}{\mathrm{d}} \\
\therefore \quad & \frac{\mathrm{v}^2}{\mathrm{rg}}=\frac{\mathrm{h}}{\mathrm{d}} \quad \ldots .\left(\because \tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}\right) \\
\therefore \quad & \mathrm{r}=\frac{\mathrm{v}^2 \mathrm{~d}}{\mathrm{gh}}
\end{array}$Asked in: MHT CET 2023 (12 May Shift 1)