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A block $A$ of mass $M_{1}$ is connected to another block of mass $M_{2}$ by a massless string through a spring of force constant $k$ as shown in figure. The coefficient of friction between the block $A$ and the table is $μ$ . The pulley is frictionless. When the block $B$ is moving downward, the tension in the string is independent of:

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mass $M_{1}$

Coefficient of friction $μ$

mass $M_{2}$

Spring constant $k$

answer is C.

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Detailed Solution

$M_{2} g - T = M_{2} a$

and $T - μ M_{1} g = M_{1} a$

On solving, we get

$a = (\frac{M_{2} g - μ M_{1} g}{M_{1} + M_{2}})$ , and so on

$T$ will be free from force constant of spring $k$ .

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