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Comments on Why answer matched but, unit?

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Why answer matched but, unit?

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Mass of a timber is $20 \ g$. And, density of that timber is $0.27 \ g/cc$. That timber was bind to a metallic materials and, it was released to $0.970 \ g/cc$ water. How much the wood was submerged in water?

I was trying to solve the problem following way.

$$F=Ah\rho g$$ $$=V\rho g$$ $$=V \ m^3 \cdot 970 \ kgm^{-3} \cdot 9.8 ms^{-2}$$ $$=9506V \ N$$ $$W=mg$$ $$=20_{\times 10^3} \ kg \times 9.8 \ ms^{-2}$$ $$=196000 \ N$$ $$W=F$$ $$9506V \ N=196000 \ N$$ $$V=20.6185 \ m^3$$

I think that I didn't do any mistake while solving that problem. Even, I didn't do any mistake of Units. But, I was wondering why my book wrote that $V=20.62 \ cc$. Both answer matched. But, I was thinking of Units. I am just showing a line what they did than, you will understand what they actually did.

$$19620 \ dyne = 951.57 V \ dyne$$ $$V=20.62 \ cm^3=20.62 \ cc$$

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2 comment threads

Missing units on intermediate steps (4 comments)
What's the unit of ρ? (2 comments)
What's the unit of ρ?
Canina‭ wrote over 2 years ago · edited over 2 years ago

I don't know if this is it, but in calculating $F$, you have $\rho = 970$; but what unit is intended? Specifying density in mg/cm³ seems a somewhat odd choice.

deleted user wrote over 2 years ago · edited over 2 years ago

I have edited... It was actually $kgm^{-3}$