A steel block of $10 \mathrm{~kg}$ rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration $0.2 \mathrm{~m} / \mathrm{s}^{2}$. The normal reaction $R^{\prime}$ by the floor if mass of the iron cylinders are equal and of $20 \mathrm{~kg}$ each, is
N. [Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ and $\left.\mu_{\mathrm{s}}=0.2\right]$
Correct Option: , 2
Writing force equation in vertical direction
$\mathrm{Mg}-\mathrm{N}=\mathrm{Ma}$
$\Rightarrow 70 \mathrm{~g}-\mathrm{N}=70 \times 0.2$
$\Rightarrow \mathrm{N}=70[\mathrm{~g}-0.2]=70 \times 9.8$
$\therefore \mathrm{N}=686$ Newton
Note : Since there is no compressive normal from the sides, hence friction will not act.
Hence option 2 .