A car moving with a speed of \(54\) km/h takes a turn of radius \(20\) m. A simple pendulum is suspended from the ceiling of the car. Determine the angle made by the string of the pendulum with the vertical during the turning. (Take \(g=10\) m/s\(^2\))

Speed of the car, \(v = 54 \times \dfrac{5}{18} = 15\) m/s.

In the frame of the car, the pendulum bob experiences a pseudo force (centrifugal force) horizontally outwards and the gravitational force downwards.

Let \(T\) be the tension in the string and \(\theta\) be the angle it makes with the vertical. For equilibrium in the car's frame, the horizontal and vertical forces balance:

\(T \sin \theta = \dfrac{mv^2}{r}\)

\(T \cos \theta = mg\)

Dividing the two equations, we get:

\(\tan \theta = \dfrac{v^2}{rg}\)

Substituting the given values \(v = 15\) m/s, \(r = 20\) m, and \(g = 10\) m/s\(^2\):

\(\tan \theta = \dfrac{(15)^2}{20 \times 10} = \dfrac{225}{200} = 1.125\)

\(\theta = \tan^{-1}(1.125)\)

Answer: \(\tan^{-1}(1.125)\)