Gravitation

JEE Main level, numerical answer, medium

213 of 223

Question 213 of 223

When a body slides down from rest along a smooth inclined plane making an angle of 30∘30^{\circ} with the horizontal, it takes time T.T. When the same body slides down from the rest along a rough inclined plane making the same angle and through the same distance, it takes time αT,\alpha T, where α\alpha is a constant greater than 1.1. The co-efficient of friction between the body and the rough plane is 1x(α2−1α2)\dfrac{1}{\sqrt{x}}\left(\dfrac{{\alpha }^{2}-1}{{\alpha }^{2}}\right) where x=__________.

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Answer: 3

On smooth incline,
a=gsin⁡30∘a=g\sin 30^{\circ}
by S=ut+12at2S=ut+\dfrac{1}{2}a{t}^{2}
S=12g2T2=g4T2...(i)S=\dfrac{1}{2}\dfrac{g}{2}{T}^{2}=\dfrac{g}{4}{T}^{2}...(i)
On rough incline
a=gsin⁡30∘−μgcos⁡30∘a=g\sin 30^{\circ}-\mu g\cos 30^{\circ}
by S=ut+12at2S=ut+\dfrac{1}{2}a{t}^{2}
S=14g(1−3μ)(αT)2...(ii)S=\dfrac{1}{4}g(1-\sqrt{3}\mu )(\alpha T{)}^{2}...(ii)
By (i)(i) and (ii)(ii)
14gT2=14g(1−3μ)α2T2\dfrac{1}{4}g{T}^{2}=\dfrac{1}{4}g(1-\sqrt{3}\mu ){\alpha }^{2}{T}^{2}
⇒1−3μ=1α2⇒μ=(α2−1α2)13\Rightarrow 1-\sqrt{3}\mu =\dfrac{1}{{\alpha }^{2}}\Rightarrow \mu =\left(\dfrac{{\alpha }^{2}-1}{{\alpha }^{2}}\right)\dfrac{1}{\sqrt{3}}
⇒x=3.00\Rightarrow x=3.00

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