Grade 10 math problem?

Tried for hours on this....

The solid copper block is melted, all of the copper is used to make one hollow cylindrical pipe of outer diamter 14mm, inner diameter 10mm, and length xmm.

Find x, the length of the pipe.

That is all the information it gives.

Thanks guys :)

Comments

  • copper block volume ???

    your question is WRONG

    lol

  • You need more information. Are you certain that the mass of copper isn't given in the problem? The issue is, if we don't know how much copper is being used, the length of the copper pipe cannot be computed.

    Example: Let's say that the mass of copper used is 2g

    The density of copper is 8.96 g·cm−3 which I found here:

    http://en.wikipedia.org/wiki/Copper

    Then we can compute the volume of the copper block that is being used by dividing the mass by its density:

    V = m/p, V = volume, m = mass, p = density. Using our values:

    V = 2g / 8.96 g*cm^-3

    V = 0.2232 cm^3, converting this into mm^3

    V = 1000*0.223214

    V = 223.214 mm^3, and this is the volume of copper we have to make our pipe.

    Now the volume of a cylindrical pipe is:

    V = π*r1^2*L - π*r2^2*L

    V = volume, r1 = outer radius, r2 = inner radius, L = length, doing a bit of factoring:

    V = π*L*(r1^2 - r2^2)

    The problem gives measurements in terms of diameter, we cut these in half to get the radii.

    r1 = 7mm, r2 = 5mm

    and we know our volume V = 223.214 mm^3 from our previous calculation. We can plug these into our equation:

    223.214 = π*L*(7^2 - 5^2)

    223.214 = π*L*(49 - 25)

    223.214 = π*L*(24), solving for L

    (223.214) / (π*24) = L

    So for this example:

    L = (223.214) / (π*24)

    L = 2.9605 mm

    So a cylindrical copper pipe made out of 2 grams of copper with the dimensions given in the problem would have a length of 2.9605 mm. So if you know the mass of the copper in the problem, you are able to solve it.

  • Let Volume of the block= V

    Volume of holloe pipe =π(R²−r²)×x

    Where R = Outside Radius

    r = inner radius

    and x = Length of the tube

    Equate V=π(R²−r²)×x

    Substitute values and get x

  • The volume is required.

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