question_answer

The coefficient of static friction, \[{{\mu }_{s}}\] between block A of mass 2 kg and the table as shown in the figure, is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and mass less:\[(g=10\text{ }m/{{s}^{2}})\] [AIPMT (S) 2004]

A) 2.0 kg

B) 4.0 kg

C) 0.2 kg

D) 0.4 kg

Correct Answer: D

Solution :

Key Idea: The tension in the string is equal to static frictional force between block A and the surface. Let the mass of the block B is M.

In equilibrium,

\[T-Mg=0\]

\[\Rightarrow \]   \[T=mg\]                                               ...(i)

If blocks do not move, then

\[T={{f}_{s}}\]

where \[{{f}_{s}}=\] frictional force \[={{\mu }_{s}}R={{\mu }_{s}}\,\,mg\]

\[\therefore \]      \[T={{\mu }_{s\,\,\,}}mg\]                                 ...(ii)

Thus, from Eqs. (i) and (ii), we have

\[mg={{\mu }_{s}}\,\,\,mg\]

or         \[M={{\mu }_{s}}\,\,m\]

Given:   \[{{\mu }_{s}}\,=0.2,\,\,m=2kg\]

\[\therefore \]      \[M=0.2\times 2=0.4\,kg\]