Q 11-04-165JEE MainJEE Main 2021 (20 Jul, Shift 1)Easy
A steel block of 10 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 $0.2\ \text{m s}^{-2}$. The normal reaction $R'$ by the floor if mass of the iron cylinders are equal and of 20 kg each is (in N),
[Take $g = 10\ \text{m s}^{-2}$ and $\mu_s = 0.2$]
Answer: (B) 686
Total mass $= 10 + 3\times20 = 70$ kg, accelerating downward at 0.2 m/s²:
$$70g - R' = 70a \Rightarrow R' = 70(10 - 0.2) = 686\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics