Verified answer

Xanzibar is spot on, accepting his density for mercury, which may differ from what you have been given. I use 13.5.

Regardless of the sample volume or mass of that same iron, 5/9ths will be submerged in the mercury, so the volume for this problem is determined from the calculated density and the given mass.

Well the density is 5/9 that of mercury which is 14.1*5/9=7.8 gm/cm^3

and to get the volume of the iron you just do mass/density=274 g/7.8 g/cm^3=35 cm^3

You have :

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Since the piece of iron floats on the mercury:

W sub Fe = F sub BHg = ( V sub I ) ( rho sub I ) ( g / g sub C )

F sub BHg = W sub Hg Displaced = ( 5 / 9 ) ( V sub I ) ( rho sub Hg ) ( g / g sub C )

Equating F sub BHg gives:

( V sub I )( rho sub Fe ) ( g / g sub C ) = ( 5 / 9 ) ( V sub I )( rho sub Hg ) ( g / g sub C )

rho sub Fe = ( 5 / 9 ) ( rho sub Hg )

rho sub Fe = ( 5 / 9 ) ( 13,550 ) = 7528 kg. per cu. m. <----------------------------------------

rho sub Fe = ( m sub Fe ) / ( V sub Fe )

V sub Fe = ( m sub Fe ) / ( rho sub Fe )

V sub Fe = ( 0.274 ) / ( 7528 ) = 36.4 x 10^-6 cu. m. = 36.4 cu. cm. <---------------------------

Note:

g sub C = Newton's Second Law Conversion Factor

g sub C = 1.000 kg. - m. per N. - sec.^2