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QJEEM15Ph069e: JEE Main 2015 Physics - Class 11 JEEM15Ph069e Solution: Given in the figure are two blocks and of weight 20 N and 100…

JEE Main 2015 (Offline) Physics - Laws of Motion Question 112 English
Given in the figure are two blocks AA and BB of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force FF as shown. If the coefficient of friction between the blocks is 0.1 and between block BB and the wall is 0.15, the frictional force applied by the wall on block BB is :

Multiple Choice Options

  • A 120120 NN
  • B 150150 NN
  • C 100100 NN
  • D 8080 NN

Step-by-Step Verified Solution

JEE Main 2015 (Offline) Physics - Laws of Motion Question 112 English Explanation 1
Assuming both the blocks are stationary

From the F.B.D of both the blocks we can find,

N=FN=F

f1 = 20 N

f2 = 100 + 20 = 120 N

JEE Main 2015 (Offline) Physics - Laws of Motion Question 112 English Explanation 2
Considering those two blocks as one system and due to equilibrium f=120Nf=120N

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