JEE Main 2026 Apr 8 Shift 2, Physics Q3: Constraint Relations

JEE Main2026Apr 8, Shift 2Physics

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 $^{2}$ )

A
$tan^{- 1} (0.5)$
B
$tan^{- 1} (0.75)$
C
$tan^{- 1} (1.125)$
D
$tan^{- 1} (0.25)$

Solution

In the car's rotating frame the bob experiences gravity downward and a pseudo (centrifugal) force $m v^{2} / r$ horizontally outward. At equilibrium of the string, the tension balances both.

Horizontal balance: $T sin θ = \frac{m v ^{2}}{r}$ . Vertical balance: $T cos θ = m g$ .

Dividing: $tan θ = \frac{v ^{2}}{r g}$ .

Converting speed: $v = 54 \times \frac{5}{18} = 15$ m/s.

$tan θ = \frac{1 5 ^{2}}{20 \times 10} = \frac{225}{200} = 1.125$

$∴$ $θ = tan^{- 1} (1.125)$ .

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JEE Main 2026 Apr 8 Shift 2, Physics Q3: Constraint Relations

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