#### A block of weight 100 N is pushed by a force F on a horizontal rough plane and moves with an acceleration $1 \mathrm{~m} / \mathrm{s}^2$, when force is doubled its acceleration becomes $10 \mathrm{~m} / \mathrm{s}^2$. The coefficient of friction is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$Option: 1 0.2  Option: 2 0.4Option: 3 0.6  Option: 4  0.8

Weight, W =100 N
Acceleration, $a=1 \mathrm{~m} / \mathrm{s}^2$
Acceleration when force is double, $a=10 \mathrm{~m} / \mathrm{s}^2$

Now, Let the External Force be 'F' and the friction force be ' $\mathrm{f_s}$'.
Then,
\begin{aligned} & \mathrm{F}-\mathrm{f}_{\mathrm{s}}=\mathrm{ma} \\ & \mathrm{F}-\mu \mathrm{N}=\mathrm{ma} \\ & \mathrm{F}-\mu 100=10 \times 1 \\ & \mathrm{~F}-\mu 100=10 \ldots .(\mathrm{I}) \end{aligned}

We know, When Force is doubled, Then,
\begin{aligned} & 2 \mathrm{~F}-\mathrm{f}_{\mathrm{s}}=\mathrm{ma} \\ & 2 \mathrm{~F}-\mu \mathrm{N}=\mathrm{ma} \\ & 2 \mathrm{~F}-\mu 100=10 \times 10 \\ & 2 \mathrm{~F}-\mu 100=100 \ldots \text { (II) } \end{aligned}

Now, from equations (I) and (II)
$\mathrm{F}=90 \mathrm{~N}$
Now, put the value of F in equation (I)
\begin{aligned} & 90-\mu 100=10 \\ & -\mu 100=10-90 \\ & \mu=\frac{80}{100} \\ & \mu=0.8 \end{aligned}
Hence, the coefficient of friction is 0.8