{"id":217659,"date":"2025-05-22T09:49:51","date_gmt":"2025-05-22T09:49:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217659"},"modified":"2025-05-22T09:50:05","modified_gmt":"2025-05-22T09:50:05","slug":"if-a-certain-liquid-has-a-mass-density-of-129-msl-m3-what-are-the-values-of-its-specific-weight-specific-gravity-and-specific-volume-in-metric-gravitational-and-metric-absolute-systems-of-units","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/22\/if-a-certain-liquid-has-a-mass-density-of-129-msl-m3-what-are-the-values-of-its-specific-weight-specific-gravity-and-specific-volume-in-metric-gravitational-and-metric-absolute-systems-of-units\/","title":{"rendered":"If a certain liquid has a mass density of 129 msl\/ m3, what are the values of its specific weight, specific gravity and specific volume in metric gravitational and metric absolute systems of units."},"content":{"rendered":"\n

If a certain liquid has a mass density of 129 msl\/ m3, what are the values of its specific weight, specific gravity and specific volume in metric gravitational and metric absolute systems of units.<\/p>\n\n\n\n

If 5.27 m3 of a certain oil weighs 44 kN, calculate the specific weight, mass density and specific gravity of the oil.<\/p>\n\n\n\n

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

Let\u2019s solve each part of the question step by step.<\/p>\n\n\n\n

\n\n\n\n

Part 1: Given a mass density of 129 kg\/m\u00b3 (129 msl\/m\u00b3), find:<\/strong><\/h2>\n\n\n\n
    \n
  • Specific Weight (\u03b3)<\/li>\n\n\n\n
  • Specific Gravity (SG)<\/li>\n\n\n\n
  • Specific Volume (v)<\/li>\n<\/ul>\n\n\n\n

    System 1: Metric Gravitational System<\/strong><\/h3>\n\n\n\n
      \n
    • Mass Density (\u03c1)<\/strong> = 129 kg\/m\u00b3<\/li>\n\n\n\n
    • Acceleration due to gravity (g)<\/strong> = 9.81 m\/s\u00b2<\/li>\n<\/ul>\n\n\n\n

      1. Specific Weight (\u03b3):<\/strong><\/h3>\n\n\n\n

      This is the weight per unit volume.<\/p>\n\n\n\n

      $$
      \\gamma = \\rho \\cdot g = 129 \\, \\text{kg\/m}^3 \\cdot 9.81 \\, \\text{m\/s}^2 = 1,265.49 \\, \\text{N\/m}^3
      $$<\/p>\n\n\n\n

      2. Specific Volume (v):<\/strong><\/h3>\n\n\n\n

      This is the volume per unit mass.<\/p>\n\n\n\n

      $$
      v = \\frac{1}{\\rho} = \\frac{1}{129} = 0.00775 \\, \\text{m}^3\/\\text{kg}
      $$<\/p>\n\n\n\n

      3. Specific Gravity (SG):<\/strong><\/h3>\n\n\n\n

      This is the ratio of the density of the fluid to the density of water (1000 kg\/m\u00b3).<\/p>\n\n\n\n

      $$
      SG = \\frac{\\rho}{\\rho_{water}} = \\frac{129}{1000} = 0.129
      $$<\/p>\n\n\n\n

      \n\n\n\n

      Part 2: Given 5.27 m\u00b3 of oil weighs 44 kN, find:<\/strong><\/h2>\n\n\n\n
        \n
      • Specific Weight (\u03b3)<\/li>\n\n\n\n
      • Mass Density (\u03c1)<\/li>\n\n\n\n
      • Specific Gravity (SG)<\/li>\n<\/ul>\n\n\n\n

        1. Specific Weight (\u03b3):<\/strong><\/h3>\n\n\n\n

        $$
        \\gamma = \\frac{\\text{Weight}}{\\text{Volume}} = \\frac{44\\,000 \\, \\text{N}}{5.27 \\, \\text{m}^3} = 8,347.45 \\, \\text{N\/m}^3
        $$<\/p>\n\n\n\n

        2. Mass Density (\u03c1):<\/strong><\/h3>\n\n\n\n

        Using the relation $\\gamma = \\rho \\cdot g$:<\/p>\n\n\n\n

        $$
        \\rho = \\frac{\\gamma}{g} = \\frac{8347.45}{9.81} = 850.89 \\, \\text{kg\/m}^3
        $$<\/p>\n\n\n\n

        3. Specific Gravity (SG):<\/strong><\/h3>\n\n\n\n

        $$
        SG = \\frac{\\rho}{\\rho_{water}} = \\frac{850.89}{1000} = 0.851
        $$<\/p>\n\n\n\n

        \n\n\n\n

        Explanation (300+ words):<\/strong><\/h2>\n\n\n\n

        In fluid mechanics, several important properties describe how a fluid behaves and how it compares with water\u2014these include specific weight<\/strong>, mass density<\/strong>, specific gravity<\/strong>, and specific volume<\/strong>. Each of these has a distinct physical meaning and mathematical relation.<\/p>\n\n\n\n

        For the first scenario, we\u2019re told the mass density<\/strong> is 129 kg\/m\u00b3. Mass density (\u03c1) is defined as mass per unit volume and is a fundamental property. Using this, we compute specific weight<\/strong> (\u03b3) by multiplying the density by gravitational acceleration (g = 9.81 m\/s\u00b2). The result, 1,265.49 N\/m\u00b3, tells us how much force the liquid exerts due to gravity per unit volume.<\/p>\n\n\n\n

        Next, we calculate the specific volume<\/strong>, which is simply the reciprocal of density. This gives an idea of how much space 1 kilogram of the liquid occupies\u2014in this case, about 0.00775 m\u00b3 per kg. Finally, specific gravity<\/strong> (SG) is a unitless ratio comparing the fluid\u2019s density to water\u2019s (typically 1000 kg\/m\u00b3). A value less than 1 (here, 0.129) means the liquid is much lighter than water.<\/p>\n\n\n\n

        In the second part, we reverse-engineer the problem. Given the weight<\/strong> and volume<\/strong> of oil, we compute specific weight first. Then we divide by gravitational acceleration to find the mass density, followed by calculating the specific gravity as a ratio to water\u2019s density.<\/p>\n\n\n\n

        This method allows engineers and scientists to understand fluid properties in both the metric gravitational system<\/strong> (using force units like N\/m\u00b3) and metric absolute system<\/strong> (mass-based, like kg\/m\u00b3). These conversions are crucial for accurate fluid analysis in fields such as civil engineering, hydraulics, and petroleum engineering<\/p>\n\n\n\n

        .<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        If a certain liquid has a mass density of 129 msl\/ m3, what are the values of its specific weight, specific gravity and specific volume in metric gravitational and metric absolute systems of units. If 5.27 m3 of a certain oil weighs 44 kN, calculate the specific weight, mass density and specific gravity of the […]<\/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-217659","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217659","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=217659"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217659\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217659"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217659"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217659"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}