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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2019 · 10 April · Evening Shift · Physics · Laws of Motion

Two blocks A and B of masses m _ A = 1kg and m _ B = 3kg are kept...

Two blocks AA and BB of masses mA=1kg{m}_{A} = 1kg and mB=3kg{m}_{B} = 3kg are kept on the table as shown in figure. The coefficients of friction between AA and BB is 0.20.2 and between BB and the surface of the table is also 0.2.0.2 . The maximum force FF that can be applied on BB horizontally, so that the block AA does not slide over the block BB is : [Take g=10m/s2g = 10m / {s}^{2} ]
Two blocks A and B of masses m _ A = 1kg and m _ B = 3kg are kept on the table as shown in figure. The coefficients of

Options

  1. A16N16N
  2. B12N12N
  3. C40N40N
  4. D8N8N

Correct answer

A. 16N16N

Step-by-step solution

Maximum possible acceleration of 1kg1kg block a=µmgm=µg=2m/s2a = \dfrac{µmg}{m} = µg = 2m / {s}^{2}
Two blocks A and B of masses m _ A = 1kg and m _ B = 3kg are kept on the table as shown in figure. The coefficients of
For 1g1g block not to slide the maximum acceleration of both blocks should be
Two blocks A and B of masses m _ A = 1kg and m _ B = 3kg are kept on the table as shown in figure. The coefficients of
{F}_\max - µmg = m{a}_\max
{F}_\max - 0.2 \times 4 \times 10 = 4 \times 2
{F}_\max - 8 = 8
{F}_\max = 16N
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