JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

The time taken by a block of mass m to slide down from the highest point to the lowest point on a rough inclined plane is $50%$ more compared to the time taken by the same block on identical inclined smooth plane. Both inclined planes are at $4 5^{\circ}$ with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is _______.

A
$3/4$
B
$2/3$
C
$5/9$
D
$4/9$

Solution

On the smooth incline, $a_{2} = g sin θ$ and time $t_{2} = \frac{2 ℓ}{g sin θ}$ .

On the rough incline, $a_{1} = g (sin θ - μ cos θ)$ and $t_{1} = \frac{2 ℓ}{g ( sin θ - μ cos θ )}$ .

Given $t_{1} = 1.5 t_{2}$ , square both sides:

$\frac{1}{sin θ - μ cos θ} = \frac{9}{4} \cdot \frac{1}{sin θ} \Rightarrow sin θ - μ cos θ = \frac{4}{9} sin θ .$

$μ cos θ = \frac{5}{9} sin θ \Rightarrow μ = \frac{5}{9} tan θ .$

At $θ = 4 5^{\circ}$ : $μ = \frac{5}{9} .$

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