{"id":88007,"date":"2025-06-01T06:59:32","date_gmt":"2025-06-01T06:59:32","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88007"},"modified":"2025-06-01T06:59:32","modified_gmt":"2025-06-01T06:59:32","slug":"sand-falls-vertically-on-a-conveyor-belt-at-a-rate-of-0%c2%b71-kg-s-in-ord","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/sand-falls-vertically-on-a-conveyor-belt-at-a-rate-of-0%c2%b71-kg-s-in-ord\/","title":{"rendered":"Sand falls vertically on a conveyor belt at a rate of 0\u00b71 kg\/s. In ord"},"content":{"rendered":"

Sand falls vertically on a conveyor belt at a rate of 0\u00b71 kg\/s. In order to keep the belt moving at a uniform speed of 2 m\/s, the force required to be applied on the belt is :<\/p>\n

[amp_mcq option1=”0 N” option2=”0\u00b72 N” option3=”1\u00b70 N” option4=”2\u00b70 N” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nTo keep the conveyor belt moving at a uniform speed, a force must be applied to counteract the change in momentum of the sand as it lands on the belt and accelerates to the belt’s speed. Sand falls vertically with zero initial horizontal momentum. As it lands on the belt, it acquires the belt’s horizontal velocity of 2 m\/s. The rate at which mass is added to the belt is dm\/dt = 0.1 kg\/s. The force required is equal to the rate of change of momentum in the horizontal direction. The momentum added per unit time is (dm\/dt) * v, where v is the velocity of the belt. Force F = (dm\/dt) * v = (0.1 kg\/s) * (2 m\/s) = 0.2 N. This force is needed to continuously accelerate the newly added sand horizontally from rest to the belt’s speed.
\n<\/section>\n
\nThe force required to keep the belt moving at a uniform speed is equal to the rate of change of momentum of the sand added to the belt, which is the product of the rate of mass flow and the belt’s velocity.
\n<\/section>\n
\nThis problem involves a variable mass system, specifically where mass is being added. The relevant physical principle is Newton’s second law in its momentum form: F_net = dP\/dt. In this case, the external force applied to the belt is responsible for increasing the horizontal momentum of the sand falling onto it.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Sand falls vertically on a conveyor belt at a rate of 0\u00b71 kg\/s. In order to keep the belt moving at a uniform speed of 2 m\/s, the force required to be applied on the belt is : [amp_mcq option1=”0 N” option2=”0\u00b72 N” option3=”1\u00b70 N” option4=”2\u00b70 N” correct=”option2″] This question was previously asked in UPSC … <\/p>\n

Detailed SolutionSand falls vertically on a conveyor belt at a rate of 0\u00b71 kg\/s. In ord<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1093],"tags":[1105,1129,1128],"class_list":["post-88007","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1105","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nSand falls vertically on a conveyor belt at a rate of 0\u00b71 kg\/s. In ord<\/title>\n<meta name=\"description\" content=\"To keep the conveyor belt moving at a uniform speed, a force must be applied to counteract the change in momentum of the sand as it lands on the belt and accelerates to the belt's speed. Sand falls vertically with zero initial horizontal momentum. As it lands on the belt, it acquires the belt's horizontal velocity of 2 m\/s. The rate at which mass is added to the belt is dm\/dt = 0.1 kg\/s. The force required is equal to the rate of change of momentum in the horizontal direction. The momentum added per unit time is (dm\/dt) * v, where v is the velocity of the belt. Force F = (dm\/dt) * v = (0.1 kg\/s) * (2 m\/s) = 0.2 N. This force is needed to continuously accelerate the newly added sand horizontally from rest to the belt's speed. The force required to keep the belt moving at a uniform speed is equal to the rate of change of momentum of the sand added to the belt, which is the product of the rate of mass flow and the belt's velocity.\" \/>\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\/sand-falls-vertically-on-a-conveyor-belt-at-a-rate-of-0\u00b71-kg-s-in-ord\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Sand falls vertically on a conveyor belt at a rate of 0\u00b71 kg\/s. In ord\" \/>\n<meta property=\"og:description\" content=\"To keep the conveyor belt moving at a uniform speed, a force must be applied to counteract the change in momentum of the sand as it lands on the belt and accelerates to the belt's speed. Sand falls vertically with zero initial horizontal momentum. As it lands on the belt, it acquires the belt's horizontal velocity of 2 m\/s. The rate at which mass is added to the belt is dm\/dt = 0.1 kg\/s. The force required is equal to the rate of change of momentum in the horizontal direction. The momentum added per unit time is (dm\/dt) * v, where v is the velocity of the belt. Force F = (dm\/dt) * v = (0.1 kg\/s) * (2 m\/s) = 0.2 N. This force is needed to continuously accelerate the newly added sand horizontally from rest to the belt's speed. The force required to keep the belt moving at a uniform speed is equal to the rate of change of momentum of the sand added to the belt, which is the product of the rate of mass flow and the belt's velocity.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/sand-falls-vertically-on-a-conveyor-belt-at-a-rate-of-0\u00b71-kg-s-in-ord\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:59:32+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Sand falls vertically on a conveyor belt at a rate of 0\u00b71 kg\/s. 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The force required to keep the belt moving at a uniform speed is equal to the rate of change of momentum of the sand added to the belt, which is the product of the rate of mass flow and the belt's velocity.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/sand-falls-vertically-on-a-conveyor-belt-at-a-rate-of-0\u00b71-kg-s-in-ord\/","og_locale":"en_US","og_type":"article","og_title":"Sand falls vertically on a conveyor belt at a rate of 0\u00b71 kg\/s. In ord","og_description":"To keep the conveyor belt moving at a uniform speed, a force must be applied to counteract the change in momentum of the sand as it lands on the belt and accelerates to the belt's speed. Sand falls vertically with zero initial horizontal momentum. As it lands on the belt, it acquires the belt's horizontal velocity of 2 m\/s. The rate at which mass is added to the belt is dm\/dt = 0.1 kg\/s. The force required is equal to the rate of change of momentum in the horizontal direction. The momentum added per unit time is (dm\/dt) * v, where v is the velocity of the belt. Force F = (dm\/dt) * v = (0.1 kg\/s) * (2 m\/s) = 0.2 N. This force is needed to continuously accelerate the newly added sand horizontally from rest to the belt's speed. The force required to keep the belt moving at a uniform speed is equal to the rate of change of momentum of the sand added to the belt, which is the product of the rate of mass flow and the belt's velocity.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/sand-falls-vertically-on-a-conveyor-belt-at-a-rate-of-0\u00b71-kg-s-in-ord\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:59:32+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/sand-falls-vertically-on-a-conveyor-belt-at-a-rate-of-0%c2%b71-kg-s-in-ord\/","url":"https:\/\/exam.pscnotes.com\/mcq\/sand-falls-vertically-on-a-conveyor-belt-at-a-rate-of-0%c2%b71-kg-s-in-ord\/","name":"Sand falls vertically on a conveyor belt at a rate of 0\u00b71 kg\/s. 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