MHT CET · Physics · Laws of Motion
Radius of curved road is ' R ', width of road is ' b '. The outer edge of road is raised by ' \(h\) ' with respect to inner edge so that a car with velocity ' \(V\) ' can pass safe over it, then value of ' \(h\) ' is ( \(g=\) acceleration due to gravity)
- A \(\frac{V^2 b}{R g}\)
- B \(\frac{\mathrm{V}}{\mathrm{Rgb}}\)
- C \(\frac{V^2 R}{g}\)
- D \(\frac{V^2 b}{g}\)
Answer & Solution
Correct Answer
(A) \(\frac{V^2 b}{R g}\)
Step-by-step Solution
Detailed explanation
\(\tan \theta = \frac{V^2}{Rg}\) \(\tan \theta = \frac{h}{b}\)
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