The specific weight of water at ordinary pressure and temperature is 9.81 kN\/m3. The specific gravity of mercury is 13.56. Compute the density of water and the specific weight and density of mercury.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve this problem, we will calculate the density of water and the specific weight and density of mercury. We will use the following information:<\/p>\n\n\n\n The specific weight<\/strong> (( \\gamma )) is related to the density (( \\rho )) and gravitational acceleration (( g )) by the equation:<\/p>\n\n\n\n [ We are given the specific weight of water, so we can solve for the density of water:<\/p>\n\n\n\n [ Substituting the known values:<\/p>\n\n\n\n [ [ Thus, the density of water is 1000 kg\/m\u00b3<\/strong>.<\/p>\n\n\n\n The specific gravity<\/strong> of mercury is the ratio of its density to the density of water:<\/p>\n\n\n\n [ We are given that the specific gravity of mercury is 13.56, so we can calculate the density of mercury:<\/p>\n\n\n\n [ Substituting the known values:<\/p>\n\n\n\n [ Now, we can calculate the specific weight<\/strong> of mercury using the formula:<\/p>\n\n\n\n [ Substituting the known values:<\/p>\n\n\n\n [ [ Thus, the results are as follows:<\/p>\n\n\n\n The specific weight of water at ordinary pressure and temperature is 9.81 kN\/m3. The specific gravity of mercury is 13.56. Compute the density of water and the specific weight and density of mercury. The Correct Answer and Explanation is : To solve this problem, we will calculate the density of water and the specific weight […]<\/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-187125","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187125","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=187125"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187125\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187125"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187125"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187125"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Step 1: Compute the density of water<\/h3>\n\n\n\n
\\gamma = \\rho \\times g
]<\/p>\n\n\n\n
\\rho_{\\text{water}} = \\frac{\\gamma_{\\text{water}}}{g}
]<\/p>\n\n\n\n
\\rho_{\\text{water}} = \\frac{9.81 \\, \\text{kN\/m}^3}{9.81 \\, \\text{m\/s}^2}
]<\/p>\n\n\n\n
\\rho_{\\text{water}} = 1 \\, \\text{tonne\/m}^3 = 1000 \\, \\text{kg\/m}^3
]<\/p>\n\n\n\nStep 2: Compute the specific weight and density of mercury<\/h3>\n\n\n\n
SG_{\\text{mercury}} = \\frac{\\rho_{\\text{mercury}}}{\\rho_{\\text{water}}}
]<\/p>\n\n\n\n
\\rho_{\\text{mercury}} = SG_{\\text{mercury}} \\times \\rho_{\\text{water}}
]<\/p>\n\n\n\n
\\rho_{\\text{mercury}} = 13.56 \\times 1000 \\, \\text{kg\/m}^3 = 13560 \\, \\text{kg\/m}^3
]<\/p>\n\n\n\n
\\gamma_{\\text{mercury}} = \\rho_{\\text{mercury}} \\times g
]<\/p>\n\n\n\n
\\gamma_{\\text{mercury}} = 13560 \\, \\text{kg\/m}^3 \\times 9.81 \\, \\text{m\/s}^2
]<\/p>\n\n\n\n
\\gamma_{\\text{mercury}} = 133,416.6 \\, \\text{N\/m}^3 \\approx 133.42 \\, \\text{kN\/m}^3
]<\/p>\n\n\n\n\n
Explanation:<\/h3>\n\n\n\n
\n