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A force of 40N acts on a body. If the units of mass and length are tripled and the unit of time is doubled, then the force in the new system becomes

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90 N

90 new units

$\frac{160}{9}$ new units

$\frac{160}{9}$ N

answer is C.

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Detailed Solution

$F = M L T^{- 2}$

$N_{1} U V_{1} = N_{2} U_{2}$

$\frac{N_{2}}{N_{1}} = [\frac{M_{1}}{M_{2}}] [\frac{L_{1}}{L_{2}}] {[[\frac{T_{1}}{T_{2}}]}^{- 2} = [\frac{M_{1}}{3 M_{1}}] [\frac{L_{1}}{3 L_{1}}] {[[\frac{T_{1}}{2 T_{1}}]}^{- 2}$

$= \frac{N_{2}}{40} = \frac{4}{9}$

$N_{2} = \frac{160}{9} N$

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