MHT CET · Physics · Laws of Motion
A railway track is banked for a speed ' \(v\) ' by elevating outer rail by a height ' \(h\) ' above the inner rail. The distance between two rails is ' \(\mathrm{d}\) ' then the radius of curvature of track is ( \(\mathrm{g}=\) gravitational acceleration)
- A \(\frac{\mathrm{v}^2 \mathrm{~d}}{\text { gh }}\)
- B \(\frac{2 \mathrm{v}^2}{\mathrm{gdh}}\)
- C \(\frac{g d}{2 v^2 h}\)
- D \(\frac{v^2}{2 g h d}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{v}^2 \mathrm{~d}}{\text { gh }}\)
Step-by-step Solution
Detailed explanation
From figure,
\( \begin{array}{ll} & \tan \theta=\frac{\mathrm{h}}{\mathrm{d}} \\ \therefore & \frac{\mathrm{v}^2}{\mathrm{rg}}=\frac{\mathrm{h}}{\mathrm{d}} \quad \ldots .\left(\because \tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}\right) \\ \therefore & \mathrm{r}=\frac{\mathrm{v}^2 \mathrm{~d}}{\mathrm{gh}} \end{array} \)
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