TS EAMCET · Physics · Laws of Motion
A 20 ton truck is travelling along a curved path of radius 240 m . If the center of gravity of the truck above the ground is 2 m and the distance between its wheels is 1.5 m , the maximum speed of the truck with which it can travel without toppling over is (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(43 \mathrm{~ms}^{-1}\)
- B \(40 \mathrm{~ms}^{-1}\)
- C \(38 \mathrm{~ms}^{-1}\)
- D \(30 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(30 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Let the maximum speed of the truck with which it can travel without toppling \(=V_{\max }=\) ? Here \(\frac{m V_{\max }^2 \mathrm{~h}}{\mathrm{R}}=\mathrm{mg} \frac{\mathrm{d}}{2}\) where \(\mathrm{h}=\) height of centre of gravity of truck above the ground, \(R=\) radius of…
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