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AIIMS 201925 May
Exam
AIIMS
Year
2019
Date
25 May
Subject
Physics
Chapter
Laws of Motion

AIIMS 2019 · 25 May · Physics · Laws of Motion

Block A of mass 2 ~kg is placed over block B of mass 8 ~kg . The...

Block AA of mass 2 kg2 \mathrm{~kg} is placed over block BB of mass 8 kg8 \mathrm{~kg}. The combination is placed over a rough horizontal surface. Coefficient of friction between BB and the floor is 0.5 . Coefficient of friction between AandBA and B is 0.4 . A horizontal force of 10 N10 \mathrm{~N} is applied on block BB. The force of friction between AandBA and B is (g=10 m s−2)\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right) Block A of mass 2 ~kg is placed over block B of mass 8 ~kg . The combination is placed over a rough horizontal surface.

Options

  1. A100 N100 \mathrm{~N}
  2. B40 N40 \mathrm{~N}
  3. C50 N50 \mathrm{~N}
  4. Dzero

Correct answer

D. zero

Step-by-step solution

Block A of mass 2 ~kg is placed over block B of mass 8 ~kg . The combination is placed over a rough horizontal surface.
Here, mA=2 kg,mB=8 kg,μ1=0.4,μ2=0.5m_A=2 \mathrm{~kg}, m_B=8 \mathrm{~kg}, \mu_1=0.4, \mu_2=0.5, F=10 NF=10 \mathrm{~N} The frictional force between block BB and surface is f=μ2N=μ2(mA+mB)g=0.5×(2+8)×10=50 Nf=\mu_2 N=\mu_2\left(m_A+m_B\right) g=0.5 \times(2+8) \times 10=50 \mathrm{~N} As applied force F(=10 N)<f(=50 N)F(=10 \mathrm{~N}) < f(=50 \mathrm{~N}), the system will not move. Hence, the force of friction between AA and BB is zero.
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