Four blocks of the same material, $m_{1}, m_{2}, m_{3}$ and $m_{4}$ have masses $2 k g, 4 k g, 2.5 k g$ and $2 k g$ respectively. All of them have equal surface area in contact with the concrete surface over which they are kept. Let the frictional force acting on them be $F_{1}, F_{2}, F_{3}$ and $F_{4}$ respectively.
The correct order of friction force acting on them when they slide is:

F_{2} > F_{3} > F_{1} = F_{4}

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F_{1} = F_{4} > F_{3} > F_{2}

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F_{1} = F_{4} = F_{3} = F_{2}

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F_{2} > F_{1} = F_{4} > F_{3}

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Solution

The correct option is A $F_{2} > F_{3} > F_{1} = F_{4}$
Greater the weight of an object, greater will be the friction between the object and surface. So the correct order will be:
$F 2 > F 3 > F 1 = F 4$

Here, the surface area in contact is the same for every block. Frictional force is independent of it anyways.

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The correct option is A F2>F3>F1=F4Greater the weight of an object, greater will be the friction between the object and surface. So the correct order will be: F2>F3>F1=F4 Here, the surface area in contact is the same for every block. Frictional force is independent of it anyways.