CBSE Class 11-science Answered

the density of water at a depth where pressure is 80 atm,given that its density at the surface is 1.03 x 10 to the power 3 kg/m cube ?given compressibility of water is 45.8x 10"-11...pa""-1,1atm=1.013 x 10"5

Asked by Aditi Srivastava | 29 Dec, 2014, 10:25: PM

Expert Answer

begin mathsize 14px style Given : Pressure space at space the space depth space of space water comma space ΔP space equals space 80 space atm space equals space 80 space cross times space 1.013 space cross times space 10 to the power of 5 space space end exponent Pa Density space at space the space surface space of space the space water comma space straight rho space equals space 1.013 space cross times space 10 to the power of 3 space end exponent kg space straight m to the power of minus 3 end exponent Let space straight m space be space the space mass space of space the space water comma and space straight rho to the power of apostrophe space end exponent be space the space density space of space water space at space depth space then comma Volume space of space the space surface space of space the space water space will space be space straight V space equals space straight m over straight rho And space volume space at space given space depth space will space be comma straight V to the power of apostrophe space equals straight m over straight rho to the power of apostrophe So comma space decrease space in space volume comma space ΔV space equals space space straight V space minus space straight V to the power of apostrophe equals space space straight m over straight rho space minus space straight m over straight rho to the power of apostrophe space equals space straight m space open parentheses 1 over straight rho space minus space 1 over straight rho to the power of apostrophe close parentheses Bulb space modulus space is space given space as comma space straight K space equals fraction numerator ΔP over denominator ΔV divided by space straight V end fraction space equals fraction numerator ΔP space straight V over denominator ΔV end fraction Compressibility space of space water comma space 1 over straight K space equals fraction numerator ΔV over denominator ΔP space straight V end fraction space equals space straight m space open parentheses 1 over straight rho space minus space 1 over straight rho to the power of apostrophe close parentheses space cross times space straight rho over straight m space cross times space 1 over ΔP space equals space 1 space minus space open parentheses straight rho over straight rho to the power of apostrophe close parentheses space cross times space 1 over ΔP space therefore comma space 45.8 space cross times space 10 to the power of minus 11 end exponent space equals space 1 space minus space open parentheses fraction numerator 1.03 space cross times space 10 cubed over denominator straight rho to the power of apostrophe end fraction close parentheses space cross times space fraction numerator 1 over denominator 80 space cross times space 1.013 space cross times space 10 to the power of 5 space space end exponent end fraction Hence comma space straight rho to the power of apostrophe space equals space fraction numerator 1.03 space cross times space 10 cubed over denominator 1 space minus space 0.0037 end fraction space straight rho to the power of apostrophe space space equals space 1.034 space cross times space 10 to the power of 3 space end exponent kg space straight m to the power of minus 3 end exponent end style

Answered by Priyanka Kumbhar | 30 Dec, 2014, 09:14: AM

CBSE Class 11-science Answered

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CBSE Class 11-science Answered

Concept Videos

Mechanical Properties of Fluids

Mechanical Properties of Fluids

Mechanical Properties of Fluids

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