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JEE Mains · Physics · STD 11 - Friction
  • JEE Mains 2023
  • Medium
  • MCQ

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 \(g =10\) \(\left.m / s ^2\right)\)

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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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