JEE Main 2026 — Motion In Two Dimensions Question with Solution

Speed of the car, $v=54\times \frac{5}{18}=15$ 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:

$T\sin \theta =\frac{m{v}^2}{r}$

$T\cos \theta =mg$

Dividing the two equations, we get:

$\tan \theta =\frac{v^2}{rg}$

Substituting the given values $v=15$ m/s, $r=20$ m, and $g=10$ m/s${}^2$:

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

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

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