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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,
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the block will remain at rest on the plane up to certain $\theta$ and then it will topple
the block will remain at rest on the plane up to certain $\theta$ and then it will topple
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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