JEE Mains · Physics · STD 11 - 4.2 friction
A car is moving with a constant speed of \(20\,m / s\) in a circular horizontal track of radius \(40\,m\). A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be : (Take \(g =10\) \(\left.m / s ^2\right)\)
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{2}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(T \cos \theta= mg\) \(T \sin \theta=\frac{ mv ^2}{ R }\) \(\tan \theta=\frac{ v ^2}{ Rg }\) \(\tan \theta=\frac{20^2}{40 \times 10}\) \(\tan \theta=1\) \(\Rightarrow \theta=\frac{\pi}{4}\)
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