Question 160 of 210

A steel block of $10kg$ 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 $0.2m{s}^{-2}.$ The normal reaction ${R}^{'}$ by the floor if mass of the iron cylinders are equal and of $20kg$ each is $\left(\text{in}N\right)$,
[Take $g=10m{s}^{-2}$ and $\left.{\mu }_{s}=0.2\right]$

A$716$B$686$C$714$D$684$

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Answer: (B) $686$

Writing force equation in vertical direction,
$Mg-N=Ma$
$\Rightarrow 70g-N=70\times 0.2$
$\Rightarrow N=70[g-0.2]=70\times 9.8$
$∴N=686$ newton
Note: Since there is no compressive normal from the sides, hence friction will not act.
Hence option $2.$