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JEE Advanced 2009Paper 1
Exam
JEE Advanced
Year
2009
Paper
Paper 1
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2009 · Paper 1 · Physics · Laws of Motion

A block of base 10 ~cm × 10 ~cm and height 15 ~cm is kept on an...

A block of base 10 cm×10 cm10 \mathrm{~cm} \times 10 \mathrm{~cm} and height 15 cm15 \mathrm{~cm} is kept on an inclined plane. The coefficient of friction between them is 3\sqrt{3}. The inclination θ\theta of this inclined plane from the horizontal plane is gradually increased from 0∘0^{\circ}. Then,A block of base 10 ~cm × 10 ~cm and height 15 ~cm is kept on an inclined plane. The coefficient of friction between

Options

  1. Aat θ=30∘\theta=30^{\circ}, the block will start sliding down the plane
  2. Bthe block will remain at rest on the plane up to certain θ\theta and then it will topple
  3. Cat θ=60∘\theta=60^{\circ}, the block will start sliding down the plane and continue to do so at higher angles
  4. Dat θ=60∘\theta=60^{\circ}, the block will start sliding down the plane and on further increasing θ\theta, it will topple at certain θ\theta

Correct answer

B. the block will remain at rest on the plane up to certain θ\theta and then it will topple

Step-by-step solution

Condition of sliding is mgsin⁡θ>μmgcos⁡θm g \sin \theta>\mu m g \cos \theta or tan⁡θ>μ\tan \theta>\mu or tan⁡θ>3\quad \tan \theta>\sqrt{3} Condition of toppling is A block of base 10 ~cm × 10 ~cm and height 15 ~cm is kept on an inclined plane. The coefficient of friction between
Torque of mgsin⁡θm g \sin \theta about 0>0> torque of mgcos⁡θm g \cos \theta about ∴(mgsin⁡θ)\therefore \quad(m g \sin \theta)\left(\frac{15}{2}\right)>(m g \cos \theta)(102)\left(\frac{10}{2}\right) or tan⁡θ>23\quad \tan \theta>\frac{2}{3} With increase in value of θ\theta, condition of sliding is satisfied first.
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