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On applying a certain force, a body of mass 5 kg moves on a horizontal surface with an acceleration of $1 {ms}^{- 2}$ . On doubling the force, the acceleration becomes $3 {ms}^{- 2}$ . The coefficient of friction is $(g = 10 {ms}^{- 2})$

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0.6

0.4

0.2

0.1

answer is D.

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Detailed Solution

$F - μmg = {ma}_{1} . . . . . . . (i)$

$2 F - μmg = {ma}_{2} . . . . . . . (i)$

from (i) and (ii),

$μ = \frac{a_{2} - 2 a_{1}}{g} = 0.1$

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