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JEE Main 20146 April
Exam
JEE Main
Year
2014
Date
6 April
Subject
Physics
Chapter
Laws of Motion

JEE Main 2014 · 6 April · Physics · Laws of Motion

A block of mass m is placed on a surface with a vertical cross...

A block of mass mm is placed on a surface with a vertical cross section given by y=x36y = \dfrac{{x}^{3}}{6}. If the coefficient of friction is 0.50 . 5, the maximum height above the ground at which the block can be placed without slipping is

Options

  1. A16m\dfrac{1}{6}\text{m}
  2. B23m\dfrac{2}{3}\text{m}
  3. C13m\dfrac{1}{3}\text{m}
  4. D12m\dfrac{1}{2}\text{m}

Correct answer

A. 16m\dfrac{1}{6}\text{m}

Step-by-step solution

For block remain stationary.

μ=tan\theta(\theta\mu = tan\text{\theta} ( \text{\theta} is angle at that point)

∴ \text\tan\text{\theta} = \mu = \dfrac{1}{2}\text{(slope)}

y=x36y=\dfrac{{x}^{3}}{6}

∴dydx=3x26=x22⇒slope=12⇒x=1∴ \dfrac{\text{d}y}{\text{d}x} = \dfrac{3{x}^{2}}{6} = \dfrac{{x}^{2}}{2} ⇒ \text{slope} = \dfrac{1}{2} ⇒ x = 1

∴y=(1)36∴ y = \dfrac{{(1)}^{3}}{6} =16= \dfrac{1}{6} m
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