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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)
Missing units on intermediate steps
celtschk‭ wrote over 2 years ago

You've got a number of intermediate steps with numbers but without any units (and the units don't match up if we assume true unit-less numbers). It is important to always take the units with the numbers, or else your intermediate steps are essentially meaningless (and easy to slip in unit mistakes in without noticing).

BTW, the problem statement seems incomplete, too. What exactly was the question?

deleted user wrote over 2 years ago

Ohh! Actually, I didn't know that writing units in intermediate steps is required. I always write in last steps. So, Thanks for the information.

And, I have completed the question also... :)

celtschk‭ wrote over 2 years ago

However you did it a bit wrong: $V$ is a quantity, therefore it also includes the units (otherwise your result could not be $V = 20.6185\,\rm m^3$). Remember, units are not something you tug on, it is an integral part of your quantity (you don't have a length of 5, you have a length of e.g. 5m, which is the same as 5000mm, despite 5 and 5000 being vastly different numbers). Also by convention, a unit should always follow a number (unless you're doing dimensional analysis). For example, …

celtschk‭ wrote over 2 years ago

… in your next line (the last one of calculating $F$), the result would be $9506\,\mathrm{N\,m^{-3}\cdot V}$.