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Q.

The power of a motor is 200 W. If the unit of length is halved, that of mass is doubled and that of time is also doubled, then the power of the motor in the new system is

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a

3200 new units

b

12.5 W

c

3200 W

d

12.5 new units

answer is B.

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Detailed Solution

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$N_{1} U_{1} = N_{2} U_{2} 200 [M_{1} L_{1}^{2} T_{1}^{- 3}] = N_{2} [M_{2} L_{2}^{2} T_{2}^{- 3}] N_{2} = 200 [(\frac{M_{1}}{M_{2}}) {(\frac{L_{1}}{L_{2}})}^{2} {(\frac{T_{1}}{T_{2}})}^{- 3}] = {[(\frac{M_{1}}{2 M_{1}}) {(\frac{L_{1}}{L_{1} / 2})}^{2} {(\frac{T_{1}}{2 T_{1}})}^{- 3}]}^{-} = 200 [(\frac{1}{2}) (2)^{2} (2)^{3}] = 200 (2^{4}) = 200 \times 16 = 3200 New units$

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