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A block of mass $m$ is lying on a rough inclined plane having an inclination $\alpha=\tan ^{-1}\left(\frac{1}{5}\right)$. The inclined plane is moving horizontally with a constant acceleration of $a=2 \mathrm{~ms}^{-2}$ as shown in the figure. The minimum value of coefficient of friction, so that the block remains stationary with respect to the inclined plane is (Take, $g=10 \mathrm{~ms}^{-2}$ )
Options:
Solution:
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Verified Answer
The correct answer is:
$\frac{5}{12}$
The block-plane system is shown in the figure,
So, from the above free body diagram (FBD), downward acceleration,
Divided the Eq. (iii) by $\sin \alpha$ and putting the values,
$$
\begin{aligned}
& \mu=\frac{10+2 \frac{1}{\tan \alpha}}{10 \frac{1}{\tan \alpha}-2}=\frac{10+2 \times 5}{10 \times 5-2} \\
& \mu=\frac{20}{48}=\frac{5}{12}
\end{aligned}
$$
Hence, the correct option is (b).
So, from the above free body diagram (FBD), downward acceleration,
Divided the Eq. (iii) by $\sin \alpha$ and putting the values,
$$
\begin{aligned}
& \mu=\frac{10+2 \frac{1}{\tan \alpha}}{10 \frac{1}{\tan \alpha}-2}=\frac{10+2 \times 5}{10 \times 5-2} \\
& \mu=\frac{20}{48}=\frac{5}{12}
\end{aligned}
$$
Hence, the correct option is (b).
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