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**Previous Years JEE Advanced Questions**

**[IIT-JEE-2010]**

**Sol.**(A)

**[IIT-JEE-2011]**

**Sol.**5

$\mathrm{F}=\operatorname{mg} \sin \theta+\mu \operatorname{mg} \cos \theta$

$\mathrm{F}+\mu \mathrm{mg} \cos \theta=\operatorname{mg} \sin \theta$

$\mathrm{F}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta$

$\operatorname{mg} \sin \theta+\mu \operatorname{mg} \cos \theta=3 \operatorname{mg} \sin \theta-3 \mu \operatorname{mg} \cos \theta$

$4 \mu \mathrm{mg} \cos \theta=2 \mathrm{mg} \sin \mu$

$4 u \cos \theta=2 \sin \theta$

$4 \mu=2 \tan \theta$

$\mathrm{N}=10 \times \frac{1}{2}$

$\mathrm{N}=5 \quad \mu=\frac{1}{2}$

[useful information: $\left.\tan \left(5.5^{\circ}\right) \approx 0.1 ; \tan \left(11.5^{\circ}\right) \approx 0.2 ; \tan \left(16.5^{\circ}\right) \approx 0.3\right]$

**[IIT-JEE-2014]**

**Sol.**(D)

The system slip down if

$\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{g} \sin \theta>\mu \mathrm{m}_{2} \mathrm{gcos} \theta$

$\tan \theta>\frac{\mu \mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}>\frac{0.3 \times 2}{3}$

$\tan \theta>0.2$

$\Rightarrow \theta>11.5^{\circ}$

For $\mathrm{P}$ and $\mathrm{Q}$ system will remain stationary hence friction $=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{g} \sin \theta$

For $\mathrm{R}$ and $\mathrm{S}$ system will move hence limiting friction acts friction $=\mu \mathrm{m}_{2} \mathrm{g} \cos \theta$