Angle of Pendulum Suspended in Turning Car

To find the angle made by the string of the simple pendulum with the vertical, we analyze the forces acting on the pendulum bob in the non-inertial frame of reference of the turning car.

1. Conversion of Given Quantities:

• Speed of the car, $v$: $v = 54 \text{ km/h} = 54 \times \frac{5}{18} \text{ m/s} = 15 \text{ m/s}$
• Radius of the circular turn, $R = 20 \text{ m}$
• Acceleration due to gravity, $g = 10 \text{ m/s}^2$

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2. Force Analysis in the Car's Frame:

In the non-inertial frame of reference attached to the car taking a turn:

1. Vertical Force: The downward force of gravity acting on the bob of mass $m$ is $mg$.
2. Horizontal Force: The outward pseudo-force (centrifugal force) acting on the bob is given by $\frac{mv^2}{R}$.
3. Tension in the String ($T$): The string makes an angle $\theta$ with the vertical.

Resolving the tension $T$ along the vertical and horizontal axes:

• Vertical equilibrium: $T \cos\theta = mg \quad \text{--- (1)}$
• Horizontal equilibrium: $T \sin\theta = \frac{m v^2}{R} \quad \text{--- (2)}$

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3. Calculation of Angle $\theta$:

Dividing equation $(2)$ by equation $(1)$: $\tan\theta = \frac{\left(\frac{m v^2}{R}\right)}{mg} = \frac{v^2}{R g}$

Substitute the given values into the formula: $\tan\theta = \frac{(15)^2}{20 \times 10} = \frac{225}{200} = 1.125$

Taking the inverse tangent on both sides: $\theta = \tan^{-1}(1.125)$

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Conclusion:

The angle made by the string of the pendulum with the vertical during the turn is $\tan^{-1}(1.125)$, which corresponds to Option C.