JEE Main 2025 Jan 24 Shift 1, Physics Q10: Friction

JEE Main2025Jan 24, Shift 1Physics

A car of mass ' $m$ ' moves on a banked road having radius ' $r$ ' and banking angle $θ$ . To avoid slipping from banked road, the maximum permissible speed of the car is $v_{0}$ . The coefficient of friction $μ$ between the wheels of the car and the banked road is :-

A
$μ = \frac{v _{0}^{2} + r g tan θ}{r g - v _{0}^{2} tan θ}$
B
$μ = \frac{v _{0}^{2} + r g tan θ}{r g + v _{0}^{2} tan θ}$
C
$μ = \frac{v _{0}^{2} - r g tan θ}{r g + v _{0}^{2} tan θ}$
D
$μ = \frac{v _{0}^{2} - r g tan θ}{r g - v _{0}^{2} tan θ}$

Solution

At maximum permissible speed $v_{0}$ on a banked turn, the car is on the verge of sliding outward, so friction $f = μ N$ acts down the incline.

Resolve forces along radial (horizontal toward centre) and vertical directions:

Radial: $N sin θ + μ N cos θ = \frac{m v _{0}^{2}}{r}$

Vertical: $N cos θ - μ N sin θ = m g$

Divide the two equations to eliminate $N$ and $m$ :

$\frac{sin θ + μ cos θ}{cos θ - μ sin θ} = \frac{v _{0}^{2}}{r g}$

Cross-multiply and group $μ$ terms:

$r g (sin θ + μ cos θ) = v_{0}^{2} (cos θ - μ sin θ)$

$r g tan θ + μ r g = v_{0}^{2} - μ v_{0}^{2} tan θ$ (dividing through by $cos θ$ )

$μ (r g + v_{0}^{2} tan θ) = v_{0}^{2} - r g tan θ$

$μ = \frac{v _{0}^{2} - r g tan θ}{r g + v _{0}^{2} tan θ}$

$∴$ Option (3).

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JEE Main 2025 Jan 24 Shift 1, Physics Q10: Friction

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