In the arrangement shown in figure, there is a friction force between the blocks of…
Physics · JEE Advanced · NTA Exams — Laws of Motion
In the arrangement shown in figure, there is a friction force between the blocks of masses m and 2m. The mass of the suspended block is m. The block of mass m is stationary with respect to block of mass 2 m. The minimum value of coefficient of friction between m and 2m is:
- \(\frac{1}{2}\)
- \(\frac{1}{\sqrt{2}}\)
- \(\frac{1}{4}\)
- \(\frac{1}{3}\)
Answer
(C) 1 4
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