JEE Main 2015 — Laws Of Motion Question with Solution
From: JEE Main 2015 (Offline)
Question
Given in the figure are two blocks and of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force as shown. If the coefficient of friction between the blocks is 0.1 and between block and the wall is 0.15, the frictional force applied by the wall on block is :
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
Assuming both the blocks are stationary
From the F.B.D of both the blocks we can find,
f1 = 20 N
f2 = 100 + 20 = 120 N
Considering those two blocks as one system and due to equilibrium
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This is a previous-year question from JEE Main 2015, covering the Laws Of Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.