JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A wooden block of mass \(5 \mathrm{~kg}\) rests on soft horizontal floor. When an iron cylinder of mass \(25\) \(\mathrm{kg}\) is placed on the top of the block, the floor yields and the block and the cylinder together go down with an acceleration of \(0.1 \mathrm{~ms}^{-2}\). The action force of the system on the floor is equal to _______.
- A \(297 \mathrm{~N}\)
- B \(294 \mathrm{~N}\)
- C \(291 \mathrm{~N}\)
- D \(196 \mathrm{~N}\)
Answer & Solution
Correct Answer
(C) \(291 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
Taking \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\) \(294-N=30 \times 0.1\) \(N=291\)
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