A car moving with a speed of 54 km/h takes a turn of radius 20 m. A...

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