{"id":184815,"date":"2025-01-21T16:38:27","date_gmt":"2025-01-21T16:38:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=184815"},"modified":"2025-01-21T16:38:28","modified_gmt":"2025-01-21T16:38:28","slug":"density-of-sea-water-at-the-surface-was-measured-as-1040-kg-m3-at-an-atmosphericpressure-of-1-bar","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/21\/density-of-sea-water-at-the-surface-was-measured-as-1040-kg-m3-at-an-atmosphericpressure-of-1-bar\/","title":{"rendered":"Density of sea water at the surface was measured as 1040 kg\/m3 at an atmosphericpressure of 1 bar"},"content":{"rendered":"\n

Density of sea water at the surface was measured as 1040 kg\/m3 at an atmosphericpressure of 1 bar. At certain depth in water, the density was found to be 1055 kg\/m3. Determinethe pressure at that point. The bulk modulus is 2290 \u00d7 106 N\/m2.Bulk modulus, Ev = \u2013 dp\/(dv\/v) = \u2013 (P2 \u2013 P1 )\/ [(v2 \u2013 v1)v1]As v = 1\/?, \u2013 (P2 \u2013 P1) = Ev \u00d7 [{1\/?2) \u2013 (1\/?1)}\/(1\/?1)]= Ev \u00d7 [(?1\u2013 ?2)\/?2]P2 = P1 \u2013 Ev \u00d7 [(?1\u2013 ?2)\/?2] = 1 \u00d7105 \u2013 2290 \u00d7 106 {(1040 \u2013 1055)\/1055}= 32.659 \u00d7 106 N\/m2 or about 326.59 bar.<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Let’s reevaluate the problem carefully, ensuring the calculations and explanation are accurate.<\/p>\n\n\n\n

Problem Data:<\/h3>\n\n\n\n
    \n
  1. Surface density (( \\rho_1 )): ( 1040 \\, \\text{kg\/m}^3 )<\/li>\n\n\n\n
  2. Depth density (( \\rho_2 )): ( 1055 \\, \\text{kg\/m}^3 )<\/li>\n\n\n\n
  3. Surface pressure (( P_1 )): ( 1 \\, \\text{bar} = 10^5 \\, \\text{N\/m}^2 )<\/li>\n\n\n\n
  4. Bulk modulus (( E_v )): ( 2290 \\times 10^6 \\, \\text{N\/m}^2 )<\/li>\n<\/ol>\n\n\n\n

    Bulk Modulus Formula:<\/h3>\n\n\n\n

    The formula relating bulk modulus, pressure, and density is:
    [
    P_2 = P_1 – E_v \\times \\left(\\frac{\\rho_1 – \\rho_2}{\\rho_2}\\right)
    ]<\/p>\n\n\n\n

    Step-by-Step Calculation:<\/h3>\n\n\n\n
      \n
    1. Difference in densities:
      [
      \\rho_1 – \\rho_2 = 1040 – 1055 = -15 \\, \\text{kg\/m}^3
      ]<\/li>\n\n\n\n
    2. Fraction term:
      [
      \\frac{\\rho_1 – \\rho_2}{\\rho_2} = \\frac{-15}{1055} \\approx -0.01421
      ]<\/li>\n\n\n\n
    3. Apply the formula:
      [
      P_2 = 1 \\times 10^5 – 2290 \\times 10^6 \\times (-0.01421)
      ]
      [
      P_2 = 1 \\times 10^5 + 2290 \\times 10^6 \\times 0.01421
      ]
      [
      P_2 = 1 \\times 10^5 + 32530900
      ]
      [
      P_2 \\approx 32.631 \\times 10^6 \\, \\text{N\/m}^2
      ]<\/li>\n\n\n\n
    4. Convert to bar:
      [
      P_2 = \\frac{32.631 \\times 10^6}{10^5} = 326.31 \\, \\text{bar}
      ]<\/li>\n<\/ol>\n\n\n\n

      Correct Answer:<\/h3>\n\n\n\n

      The pressure at the given depth is approximately 326.31 bar<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

      Explanation (300 Words):<\/h3>\n\n\n\n

      The density of seawater increases with depth due to the compression of water under pressure. The relationship between pressure and density in a compressible fluid is governed by the bulk modulus, ( E_v ), which quantifies the material’s resistance to compression. The formula ( E_v = -\\frac{\\Delta P}{\\Delta V \/ V} ) connects pressure change (( \\Delta P )) to the relative volume change (( \\Delta V \/ V )). Since ( V \\propto 1 \/ \\rho ), the formula can be rewritten in terms of density.<\/p>\n\n\n\n

      Using ( P_2 = P_1 – E_v \\cdot \\frac{\\rho_1 – \\rho_2}{\\rho_2} ), we relate surface pressure, bulk modulus, and densities at different depths. Substituting the given values, we calculate ( P_2 ), finding it to be ( 32.631 \\times 10^6 \\, \\text{N\/m}^2 ), which converts to ( 326.31 \\, \\text{bar} ).<\/p>\n\n\n\n

      This method assumes constant ( E_v ), which is valid for small pressure intervals. The result demonstrates the significant pressure increase at depth due to water’s slight compressibility. Understanding such relationships is crucial in oceanography and engineering, ensuring safe designs for submersibles and undersea pipelines.<\/p>\n","protected":false},"excerpt":{"rendered":"

      Density of sea water at the surface was measured as 1040 kg\/m3 at an atmosphericpressure of 1 bar. At certain depth in water, the density was found to be 1055 kg\/m3. Determinethe pressure at that point. The bulk modulus is 2290 \u00d7 106 N\/m2.Bulk modulus, Ev = \u2013 dp\/(dv\/v) = \u2013 (P2 \u2013 P1 )\/ […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-184815","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184815","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=184815"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184815\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=184815"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=184815"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=184815"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}