The time taken by a block of mass to slide down from the highest point to the lowest point on a rough inclined plane is 50% more compared

Question

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 $45 °$ with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is _________. [2026]

Smart Achievers · Accepted Answer

(3)

$a_{rough} = g (\sin θ - μ \cos θ)$

$a_{smooth inclined plane} = g \sin θ$

$For Rough incline plane, t_{1} = \sqrt{\frac{2 ℓ}{g (\sin θ - μ \cos θ)}}$

$For smooth incline plane, t_{2} = \sqrt{\frac{2 ℓ}{g \sin θ}}$

$Given, t_{1} = 1.5 t_{2}$

$\sqrt{\frac{2 ℓ}{g (\sin θ - μ \cos θ)}} = \frac{3}{2} \times \sqrt{\frac{2 ℓ}{g \sin θ}}$

$(\sin θ - μ \cos θ) = \frac{4}{9} \sin θ$

$\frac{5}{9} \sin θ = μ \cos θ \Rightarrow μ = \frac{5}{9} \tan θ$ $(∵ θ = 45 °)$

$\Rightarrow μ = \frac{5}{9} \tan 45 ° = \frac{5}{9}$

Solution

1 $\frac{3}{4}$

2 $\frac{2}{3}$

3 $\frac{5}{9}$

4 $\frac{4}{9}$

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Solution

1 34

2 23

3 59

4 49

Ans. (3) arough=g(sinθ-μcosθ) asmooth inclined plane=gsinθ For Rough incline plane, t1=2ℓg(sinθ-μcosθ) For smooth incline plane, t2=2ℓgsinθ Given, t1=1.5t2 2ℓg(sinθ-μcosθ)=32×2ℓgsinθ (sinθ-μcosθ)=49sinθ 59sinθ=μcosθ⇒μ=59tanθ (∵ θ=45°) ⇒μ=59tan45°=59

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1 $\frac{3}{4}$

2 $\frac{2}{3}$

3 $\frac{5}{9}$

4 $\frac{4}{9}$