AP EAMCET · PHYSICS · Laws of Motion
A truck of mass \(2000 \mathrm{~kg}\) is moving along a circular path having radius of curvature \(10 \mathrm{~m}\). if the banking angle is \(39^{\circ}\), then the maximum permissible speed of the truck is (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\), take \(\tan 39^{\circ}=\) \(0.81)\)
- A \(14 \mathrm{~ms}^{-1}\)
- B \(5 \mathrm{~ms}^{-1}\)
- C \(18 \mathrm{~ms}^{-1}\)
- D \(9 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(9 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Mass of truck, M=2000 kg Radius of curvature, \(\mathrm{R}=10 \mathrm{~m}\) Banking angle, \(a=39^{\circ}, \tan 39^{\circ}=0.81\) Maximum speed is given as:…
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