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a piece of iron weighs 44.5gf in air, 39.5gf in water and 40.3gf in oil.find (1)the relative density of iron (2)the density of oil in SI unit.
Asked by mansi | 28 Feb, 2015, 07:48: PM
answered-by-expert Expert Answer
begin mathsize 12px style Given colon Weight space of space iron space in space air comma space straight W subscript straight a space equals space 44.5 space gf space equals space 0.445 space gN Weight space of space iron space in space water comma space straight W subscript straight w space equals space 39.5 space gf space equals space 0.395 space gN Weight space of space iron space in space paraffin space oil comma space straight W subscript straight o space equals space 40.3 space gf space equals space 0.403 space gN Let space density space of space iron space be space apostrophe straight d subscript straight i apostrophe space and space straight V subscript straight i space be space the space volume space of space the space iron. As space weight space of space iron space in space air space equals space Vdg space equals space space 0.445 space straight g rightwards double arrow space straight W subscript straight a space equals space space straight V subscript straight i straight d subscript straight i space equals space space 0.445 space space space space space space space space space space left parenthesis straight i right parenthesis Density space of space water space straight d space equals space 1000 space kg divided by straight m cubed Hence comma space straight V subscript straight i space left parenthesis 1000 right parenthesis straight g space equals straight W subscript straight a space space minus space straight W subscript straight w space space equals space space 0.445 straight g space minus space 0.395 space straight g rightwards double arrow space straight V subscript straight i space left parenthesis 1000 right parenthesis space equals space 0.05 space rightwards double arrow space straight V subscript straight i space equals space 5 space cross times space 10 to the power of negative 5 end exponent space space space space space space space space space space space left parenthesis ii right parenthesis From space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space 5 space cross times space 10 to the power of negative 5 end exponent space cross times space straight d subscript straight i space equals space 0.445 straight d subscript straight i space equals space 8900 space kg divided by straight m cubed Relative space density space of space iron space equals space fraction numerator Density space of space iron over denominator density space of space water end fraction space equals space 8900 over 1000 space equals space 8.9 space Hence comma space relative space density space of space iron space is space 8.9 To space calculate space density space of space oil comma Let space straight d subscript straight o space end subscript be space the space density space of space oil space then comma straight V subscript straight i straight d subscript straight o straight g space equals space space straight W subscript straight a space minus space straight W subscript straight o space equals space 0.445 straight g space minus space 0.403 straight g straight V subscript straight i straight d subscript straight o space equals space 0.042 straight d subscript straight o space equals space fraction numerator 0.042 over denominator 5 space cross times space 10 to the power of negative 5 end exponent end fraction space space equals space 840 space kg divided by straight m cubed Hence comma space density space of space oil space is space 840 space kg divided by straight m cubed end style
Answered by Priyanka Kumbhar | 01 Mar, 2015, 01:02: PM

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