JEE Advanced · Physics · 5. Laws of Motion
A block of base \(10 \mathrm{~cm} \times 10 \mathrm{~cm}\) and height \(15 \mathrm{~cm}\) is kept on an inclined plane. The coefficient of friction between them is \(\sqrt{3}\). The inclination \(\theta\) of this inclined plane from the horizontal plane is gradually increased from \(0^{\circ}\). Then,
- A at \(\theta=30^{\circ}\), the block will start sliding down the plane
- B the block will remain at rest on the plane up to certain \(\theta\) and then it will topple
- C at \(\theta=60^{\circ}\), the block will start sliding down the plane and continue to do so at higher angles
- D at \(\theta=60^{\circ}\), the block will start sliding down the plane and on further increasing \(\theta\), it will topple at certain \(\theta\)
Answer & Solution
Correct Answer
(B) the block will remain at rest on the plane up to certain \(\theta\) and then it will topple
Step-by-step Solution
Detailed explanation
Condition of sliding is \(m g \sin \theta>\mu m g \cos \theta\)
or \(\tan \theta>\mu\)
or \(\quad \tan \theta>\sqrt{3}\)
Condition of toppling is
Torque of \(m g \sin \theta\) about \(0>\) torque of \(m g \cos \theta\) about
\(\therefore \quad(m g \sin \theta)\left(\frac{15}{2}\right)>(m g \cos \theta)\left(\frac{10}{2}\right)\)
or \(\quad \tan \theta>\frac{2}{3}\)
With increase in value of \(\theta\), condition of sliding is satisfied first.
or \(\tan \theta>\mu\)
or \(\quad \tan \theta>\sqrt{3}\)
Condition of toppling is
Torque of \(m g \sin \theta\) about \(0>\) torque of \(m g \cos \theta\) about
\(\therefore \quad(m g \sin \theta)\left(\frac{15}{2}\right)>(m g \cos \theta)\left(\frac{10}{2}\right)\)
or \(\quad \tan \theta>\frac{2}{3}\)
With increase in value of \(\theta\), condition of sliding is satisfied first.
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