0
$\begingroup$

In the figure below, the dimensions of an unsharpened pencil are given. The unsharpened pencil have graphite in a cylindrical shape, surrounded by wood. The diameter of the graphite is 2mm. The density of graphite is $640\text{ kg/m}^3$. The density of wood is $420\text{ kg/m}^3$. Find the mass of 500 pencils. enter image description here

My attempt was the following: I found the volume of the wood: $$3.14\times (0.004)^2\times 0.15= 0.000007536$$ The volume of graphite: $$3.14\times (0.001)^2\times 0.15= 0.000000471$$ I found the mass of 1 pencil: $$0.000007536\times 420 + 0.000000471\times 640= 0.0034\\ 0.0034\times 500= 1.7 \text{ kg}$$

I think I messed up the solution!

All help/solutions appreciated.

Thanks

$\endgroup$
0

2 Answers 2

1
$\begingroup$

first of all, I'd like to say it would be much easier for you to calculate with, say, $4.71×10^7$ instead of $0.000000471$. The mistake you're making here is counting the graphite's volume twice. When calculating the volume of the wooden part, you need to subtract the graphite's volume:

$7.536×10^{-6}-0.471×10^{-7}=10^{-7}×(75.36-0.471)=74.889×10^{-7}=7.4889×10^{-6} m^3$

Now, you have the density of both materials. That is to say, you know how many $kg$ one $m^3$ of wood/graphite weighs. From that info, you can calculate the mass of wood/graphite in one pencil:

$m_{wood}=7.4889×10^{-8}×420=3145.338×10^{-8}=3.145338×10^{-5}kg$
$m_{graphite}=0.471×10^{-7}×640=301.44×10^{-7}=3.0144×10^{-5}kg$
$m_{pencil}=m_{wood}+m_{graphite}=10^{-5}×(3.0144+3.145338)=6.159738×10^{-5}kg$

Now getting the mass of 500 pencils is simple! The answer is

$6.159738×500×10^{-5}=30.79869×10^{-3}kg \approx 30.8 gr$

Hopefully this solves your problem :)

$\endgroup$
1
$\begingroup$

You forgot to take out the volume of the graphite when calculating the volume of the wood. In fact the volume of the wood is $$[\pi (0.004)^2(0.15)-\pi (0.001)^2(0.15)]\text{ m}^3$$

$\endgroup$
1
  • $\begingroup$ Thank you so much🙏🏻 $\endgroup$
    – Ko_17
    Jan 9, 2021 at 18:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.