If the radius of a sphere is R=(10+/-0.1)cm what is the percentage of its volume ?

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  • TomV
    Lv 7
    1 month ago

    The radius of the sphere can be anywhere from 9.9 cm to 10.1 cm

    The volume of the sphere, V = 4πr³/3, is between limits of 4064.38 and 4315.71 cm°

    The percent difference between the minimum and maximum volumes is the ratio of the volumetric difference between the limits to the average of the limiting values, said ratio expressed as a percentage:

    %Diff = (4315.71 - 4064.38)/(4315.71 + 4064.38) = 6.0%

    The percent error of any particular sphere would be the ratio of the volume of the particular sphere to the volume of a nominal sphere minus 1, said difference expressed as a percentage.

    For a minimum material sphere, the percent error would be:

    %Err = 4064.38/4188.79 - 1 = -0.0297 = -2.97%

    For a maximum material sphere, the percent error would be

    %Err = 4315.71/4188.79 - 1 = 0.0303 = 3.03%

    Note the %Err is not symmetric about the nominal volume because the volume is a non-linear function of the radius.

  • 1 month ago

    You are probably asking the ‘%error’.

    Use Calculus or Memorize the formula:

    ➤ Formula:( ΔV / V ) = 3( Δr / r )

    error =  ΔV / V

    % error = 100% (ΔV / V )

    error =  3( Δr / r )

    % error = 100% * 3( Δr / r )

    % error = 100% * 3( 0.1 / 10 )

    % error = 3%

    ━━━━━━

    ➤ Calculus:

    V = (4/3)πr³

    dV/dr = 4πr² 

    dV = 4πr² dr

    dV/V = ( 4πr² )dr / [ (4/3)πr³ ]

    dV/V = 3 dr/r

    error = dV/V

    % error = 100% * 3( dr / r )

    % error = 100% * 3( 0.1 / 10 )

    % error = 3%

    ━━━━━━

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