Laws of Motion

JEE Main level, single correct, medium

82 of 210

Question 82 of 210

A wooden block of mass 5 kg5 \mathrm{~kg} rests on a soft horizontal floor. When an iron cylinder of mass 25 kg25 \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 ms−20.1 \mathrm{~ms}^{-2}. The action force of the system on the floor is equal to:

Show answer and solution

Answer: (B) 291 N291 \mathrm{~N}

Taking g=9.8 m/s2\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2

294−N=30×0.1 N=291\begin{aligned} & 294-\mathrm{N}=30 \times 0.1 \\ & \mathrm{~N}=291\end{aligned}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Laws of motion, friction and pulleys simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Laws of Motion (Equilibrium of Forces), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.