{"id":7577,"date":"2024-04-15T02:47:07","date_gmt":"2024-04-15T02:47:07","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=7577"},"modified":"2024-04-15T02:47:07","modified_gmt":"2024-04-15T02:47:07","slug":"if-the-depletion-of-oxygen-is-found-to-be-2-5-mg-litre-after-incubating-2-5-ml-of-sewage-diluted-to-250-ml-for-5-days-at-20ac-b-o-d-of-the-sewage-is-a-50-mg-l-b-100-mg-l-c-150-mg-l-d-250-m","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/if-the-depletion-of-oxygen-is-found-to-be-2-5-mg-litre-after-incubating-2-5-ml-of-sewage-diluted-to-250-ml-for-5-days-at-20ac-b-o-d-of-the-sewage-is-a-50-mg-l-b-100-mg-l-c-150-mg-l-d-250-m\/","title":{"rendered":"If the depletion of oxygen is found to be 2.5 mg\/litre after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20\u00c2\u00b0C, B.O.D. of the sewage is A. 50 mg\/l B. 100 mg\/l C. 150 mg\/l D. 250 mg\/l"},"content":{"rendered":"<p>[amp_mcq option1=&#8221;50 mg\/l&#8221; option2=&#8221;100 mg\/l&#8221; option3=&#8221;150 mg\/l&#8221; option4=&#8221;of the sewage is A. 50 mg\/l B. 100 mg\/l C. 150 mg\/l D. 250 mg\/l&#8221; correct=&#8221;option1&#8243;]<!--more--><\/p>\n<p>The correct answer is A. 50 mg\/l.<\/p>\n<p>Biological oxygen demand (BOD) is the amount of oxygen required by microorganisms to decompose organic matter in a given water sample at a specified temperature over a specified time period. It is a measure of the organic pollution in water.<\/p>\n<p>The BOD test is conducted by incubating a sample of water with a known amount of dissolved oxygen for a specified time period, usually five days at 20\u00c2\u00b0C. The amount of oxygen consumed during the incubation period is then measured.<\/p>\n<p>The BOD of a sample of water is expressed in milligrams of oxygen per liter (mg\/l). A high BOD indicates that the water is polluted with organic matter.<\/p>\n<p>In the question, the depletion of oxygen is found to be 2.5 mg\/litre after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20\u00c2\u00b0C. The BOD of the sewage is therefore 2.5 x 100 = 50 mg\/l.<\/p>\n<p>Option A is the correct answer because it is the only option that is within the range of BOD values for sewage. The BOD of sewage typically ranges from 100 to 500 mg\/l.<\/p>\n<p>Option B is incorrect because it is twice the actual BOD value.<\/p>\n<p>Option C is incorrect because it is three times the actual BOD value.<\/p>\n<p>Option D is incorrect because it is five times the actual BOD value.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>[amp_mcq option1=&#8221;50 mg\/l&#8221; option2=&#8221;100 mg\/l&#8221; option3=&#8221;150 mg\/l&#8221; option4=&#8221;of the sewage is A. 50 mg\/l B. 100 mg\/l C. 150 mg\/l D. 250 mg\/l&#8221; correct=&#8221;option1&#8243;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[650],"tags":[],"class_list":["post-7577","post","type-post","status-publish","format-standard","hentry","category-waste-water-engineering","no-featured-image-padding"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>If the depletion of oxygen is found to be 2.5 mg\/litre after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20\u00c2\u00b0C, B.O.D. of the sewage is A. 50 mg\/l B. 100 mg\/l C. 150 mg\/l D. 250 mg\/l<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/if-the-depletion-of-oxygen-is-found-to-be-2-5-mg-litre-after-incubating-2-5-ml-of-sewage-diluted-to-250-ml-for-5-days-at-20ac-b-o-d-of-the-sewage-is-a-50-mg-l-b-100-mg-l-c-150-mg-l-d-250-m\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"If the depletion of oxygen is found to be 2.5 mg\/litre after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20\u00c2\u00b0C, B.O.D. of the sewage is A. 50 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