the fundamental units in the two system of units are in the ratio of 1:3. find the ratio of units of momentum and power

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Solution

Dear Student
$L e t i n o n e u n i t o f s y s t e m d i m e n s i o n a l f o r m u l a f o r m o m e n t u m i s p = [M^{1} L^{1} T^{- 1}] L e t i n a n o t h e r u n i t o f s y s t e m d i m e n s i o n a l f o r m u l a f o r m o m e n t u m i s p' = [M'^{1} L'^{1} T'^{- 1}] \frac{p}{p'} = \frac{[M^{1} L^{1} T^{- 1}]}{[M'^{1} L'^{1} T'^{- 1}]} A s g i v e n r a t i o o f u n i t s i n t w o s y s t e m i s 1 : 3 t h a t i s \frac{M}{M'} = \frac{L}{L'} = \frac{T}{T'} = \frac{1}{3} S o, \frac{p}{p'} = \frac{[(M^{1}) (L^{1}) (3 T')^{- 1}]}{[(3 M^{1}) (3 L^{1}) (T'^{- 1})]} = \frac{1}{27} L e t i n o n e u n i t o f s y s t e m d i m e n s i o n a l f o r m u l a f o r P o w e r i s P = [M^{1} L^{2} T^{- 3}] L e t i n a n o t h e r u n i t o f s y s t e m d i m e n s i o n a l f o r m u l a f o r m o m e n t u m i s p' = [M'^{1} L'^{2} T'^{- 3}] \frac{P}{P'} = \frac{[M^{1} L^{2} T^{- 3}]}{[M'^{1} L'^{2} T'^{- 3}]} A s g i v e n r a t i o o f u n i t s i n t w o s y s t e m i s 1 : 3 S o, \frac{P}{P'} = \frac{[M^{1} L^{2} (3 T')^{- 3}]}{[(3 M)^{1} (3 L)^{2} T'^{- 3}]} = \frac{1}{729}$
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