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Q.

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

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a

90 N

b

90 new units

c

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

d

$\frac{160}{9} N$

answer is B.

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Detailed Solution

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$\begin{array}{l} N_{1} U_{1} = N_{2} U_{2} \Rightarrow N_{2} = N_{2} (\frac{U_{1}}{U_{2}}) \\ U_{1} = M_{1} L_{1} T_{2}^{- 2}; U_{2} = M_{2} L_{2} T_{2}^{- 2} \\ N_{1} = 40 N; M_{2} = 2 M_{1} \\ L_{2} = 2 L_{1}; T_{2} = 3 T_{1} \\ ∴ N_{2} = 40 [\frac{M_{1} L_{1} T_{1}^{- 2}}{M_{2} L_{2} T_{2}^{- 2}}] \\ N_{2} = 40 \frac{M_{1} L_{1} T_{1}^{- 2}}{2 M_{2} 2 L_{2} {(3 T_{2})}^{- 2}} = 10 \times 3 \end{array}$
= 90 new unit

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