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A pen of mass ' $\mathrm{m}^{\prime}$ is lying on a piece of paper of mass $\mathrm{M}$ placed on a rough table. If the coefficient of friction between the pen and paper, and, the paper and table are $\mu_{1}$ and $\mu_{2}$, respectively, then the minimum horizontal force with which the paper has to be pulled for the pen to start slipping is given by-
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The correct answer is:
$(\mathrm{m}+\mathrm{M})\left(\mu_{1}+\mu_{2}\right) \mathrm{g}$
For pen to start slipping maximum horizontal force on it is $\mathrm{f}=\mu \mathrm{mg}$ $\therefore \mathrm{a}=\mu_{1} \mathrm{~g}$ is the maximum common acceleration for both pen and paper F.B.D. for both pen and paper
$$
\begin{array}{l}
\therefore \mathrm{F}=\mathrm{f}_{1}+\mathrm{f}_{2}+\mathrm{Ma} \\
\mathrm{F}=\mu_{1} \mathrm{mg}+\mu_{2}(\mathrm{~m}+\mathrm{M}) g+\mathrm{M}\left(\mu_{1} \mathrm{~g}\right) \\
\therefore \mathrm{F}=(\mathrm{m}+\mathrm{M})\left(\mu_{1}+\mu_{2}\right) \mathrm{g}
\end{array}
$$
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