Verified Solution • 2025–26 Session moderate

QJEEM18Ph1b18: JEE Main 2018 Physics - Class 11 JEEM18Ph1b18 Solution: A given object takes n times more time to slide down a rough…

A given object takes n times more time to slide down a 45∘{45^ \circ } rough inclined plane as it takes to slide down a perfectly smooth 45∘{45^ \circ } incline. The coefficient of kinetic friction between the object and the incline is :

Multiple Choice Options

  • A 12−n2{1 \over {2 - {n^2}}}
  • B 1−1n21 - {1 \over {{n^2}}}
  • C 1−1n2\sqrt {1 - {1 \over {{n^2}}}}
  • D 11−n2\sqrt {{1 \over {1 - {n^2}}}}

Step-by-Step Verified Solution

Let, t1 and t2 are time taken to move on the smooth and rough surface for smooth surface,

S = 12{1 \over 2} g sin45o t12t_1^2

⇒\Rightarrow     \,\,\,\, t1 = 22Sg\sqrt {{{2\sqrt 2 S} \over g}}

For rough surface,

S = 12{1 \over 2} g (sin45o −- μ\mu k cos45o) t22t_2^2

⇒\Rightarrow     \,\,\,\, t2 = 22Sg(1−μk)\sqrt {{{2\sqrt 2 S} \over {g\left( {1 - {\mu _k}} \right)}}}

Here μk{\mu _k} = Kinetic friction.

According to question,

t2 = n t1

⇒\Rightarrow     \,\,\,\, 22 Sg(1−μk){{2\sqrt 2 \,S} \over {g\left( {1 - {\mu _k}} \right)}} = n2 ×\times 22 Sg{{2\sqrt 2 \,S} \over g}

⇒\Rightarrow     \,\,\,\, 1 −- μ\mu k = 1n2{1 \over {{n^2}}}

⇒\Rightarrow     \,\,\,\, μk{\mu _k} = 1 −- 1n2{1 \over {{n^2}}}

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