{"id":85808,"date":"2025-06-01T03:28:34","date_gmt":"2025-06-01T03:28:34","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=85808"},"modified":"2025-06-01T03:28:34","modified_gmt":"2025-06-01T03:28:34","slug":"two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k\/","title":{"rendered":"Two persons are holding a rope of negligible mass horizontally. A 20 k"},"content":{"rendered":"

Two persons are holding a rope of negligible mass horizontally. A 20 kg mass is attached to the rope at the midpoint; as a result the rope deviates from the horizontal direction. The tension required to completely straighten the rope is (g = 10 m\/s2<\/sup>)<\/p>\n

[amp_mcq option1=”200 N” option2=”20 N” option3=”10 N” option4=”infinitely large” correct=”option4″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CDS-2 – 2018<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nWhen a mass is attached to the midpoint of a horizontally held rope, the rope sags downwards due to the weight of the mass (W = mg). The tension in the rope acts along the direction of the rope segments on either side of the mass. For the rope to be in equilibrium, the vector sum of the two tensions and the weight must be zero. If the rope is perfectly horizontal, the tension vectors would have only horizontal components (neglecting the mass of the rope itself). However, the weight of the attached mass acts vertically downwards. To balance this downward force, there must be an equal and opposite upward force provided by the vertical components of the tension in the rope. As the rope approaches a perfectly horizontal state, the angle (\u03b8) it makes with the horizontal approaches zero. The vertical component of the tension in each half of the rope is T * sin(\u03b8), where T is the tension. The total upward vertical force is 2 * T * sin(\u03b8). To balance the weight W, 2 * T * sin(\u03b8) = W. If the rope is to be perfectly straight and horizontal (\u03b8 = 0), then sin(\u03b8) = sin(0) = 0. For 2 * T * sin(\u03b8) to equal the non-zero weight W (20 kg * 10 m\/s\u00b2 = 200 N), the tension T must tend towards infinity as sin(\u03b8) tends towards zero. Thus, an infinitely large tension is required to completely straighten the rope.
\n<\/section>\n
\n– Static equilibrium requires the net force in all directions to be zero.
\n– When a mass hangs from a rope, the weight acts vertically downwards.
\n– Tension in the rope must have a vertical component to balance the weight.
\n– For a nearly horizontal rope, the angle with the horizontal is very small.
\n– As the angle approaches zero, the sine of the angle approaches zero, requiring tension to approach infinity to provide a non-zero vertical force.
\n<\/section>\n
\nThis scenario represents a classic physics problem illustrating the vector resolution of forces and the limit as an angle approaches zero. In any real-world scenario, the rope will always sag slightly, having a non-zero angle, because no rope can withstand infinite tension.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Two persons are holding a rope of negligible mass horizontally. A 20 kg mass is attached to the rope at the midpoint; as a result the rope deviates from the horizontal direction. The tension required to completely straighten the rope is (g = 10 m\/s2) [amp_mcq option1=”200 N” option2=”20 N” option3=”10 N” option4=”infinitely large” correct=”option4″] … <\/p>\n

Detailed SolutionTwo persons are holding a rope of negligible mass horizontally. A 20 k<\/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":[1088],"tags":[1114,1129,1128],"class_list":["post-85808","post","type-post","status-publish","format-standard","hentry","category-upsc-cds-2","tag-1114","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nTwo persons are holding a rope of negligible mass horizontally. A 20 k<\/title>\n<meta name=\"description\" content=\"When a mass is attached to the midpoint of a horizontally held rope, the rope sags downwards due to the weight of the mass (W = mg). The tension in the rope acts along the direction of the rope segments on either side of the mass. For the rope to be in equilibrium, the vector sum of the two tensions and the weight must be zero. If the rope is perfectly horizontal, the tension vectors would have only horizontal components (neglecting the mass of the rope itself). However, the weight of the attached mass acts vertically downwards. To balance this downward force, there must be an equal and opposite upward force provided by the vertical components of the tension in the rope. As the rope approaches a perfectly horizontal state, the angle (\u03b8) it makes with the horizontal approaches zero. The vertical component of the tension in each half of the rope is T * sin(\u03b8), where T is the tension. The total upward vertical force is 2 * T * sin(\u03b8). To balance the weight W, 2 * T * sin(\u03b8) = W. If the rope is to be perfectly straight and horizontal (\u03b8 = 0), then sin(\u03b8) = sin(0) = 0. For 2 * T * sin(\u03b8) to equal the non-zero weight W (20 kg * 10 m\/s\u00b2 = 200 N), the tension T must tend towards infinity as sin(\u03b8) tends towards zero. Thus, an infinitely large tension is required to completely straighten the rope. - Static equilibrium requires the net force in all directions to be zero. - When a mass hangs from a rope, the weight acts vertically downwards. - Tension in the rope must have a vertical component to balance the weight. - For a nearly horizontal rope, the angle with the horizontal is very small. - As the angle approaches zero, the sine of the angle approaches zero, requiring tension to approach infinity to provide a non-zero vertical force.\" \/>\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\/two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Two persons are holding a rope of negligible mass horizontally. A 20 k\" \/>\n<meta property=\"og:description\" content=\"When a mass is attached to the midpoint of a horizontally held rope, the rope sags downwards due to the weight of the mass (W = mg). The tension in the rope acts along the direction of the rope segments on either side of the mass. For the rope to be in equilibrium, the vector sum of the two tensions and the weight must be zero. If the rope is perfectly horizontal, the tension vectors would have only horizontal components (neglecting the mass of the rope itself). However, the weight of the attached mass acts vertically downwards. To balance this downward force, there must be an equal and opposite upward force provided by the vertical components of the tension in the rope. As the rope approaches a perfectly horizontal state, the angle (\u03b8) it makes with the horizontal approaches zero. The vertical component of the tension in each half of the rope is T * sin(\u03b8), where T is the tension. The total upward vertical force is 2 * T * sin(\u03b8). To balance the weight W, 2 * T * sin(\u03b8) = W. If the rope is to be perfectly straight and horizontal (\u03b8 = 0), then sin(\u03b8) = sin(0) = 0. For 2 * T * sin(\u03b8) to equal the non-zero weight W (20 kg * 10 m\/s\u00b2 = 200 N), the tension T must tend towards infinity as sin(\u03b8) tends towards zero. Thus, an infinitely large tension is required to completely straighten the rope. - Static equilibrium requires the net force in all directions to be zero. - When a mass hangs from a rope, the weight acts vertically downwards. - Tension in the rope must have a vertical component to balance the weight. - For a nearly horizontal rope, the angle with the horizontal is very small. - As the angle approaches zero, the sine of the angle approaches zero, requiring tension to approach infinity to provide a non-zero vertical force.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T03:28:34+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=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Two persons are holding a rope of negligible mass horizontally. A 20 k","description":"When a mass is attached to the midpoint of a horizontally held rope, the rope sags downwards due to the weight of the mass (W = mg). The tension in the rope acts along the direction of the rope segments on either side of the mass. For the rope to be in equilibrium, the vector sum of the two tensions and the weight must be zero. If the rope is perfectly horizontal, the tension vectors would have only horizontal components (neglecting the mass of the rope itself). However, the weight of the attached mass acts vertically downwards. To balance this downward force, there must be an equal and opposite upward force provided by the vertical components of the tension in the rope. As the rope approaches a perfectly horizontal state, the angle (\u03b8) it makes with the horizontal approaches zero. The vertical component of the tension in each half of the rope is T * sin(\u03b8), where T is the tension. The total upward vertical force is 2 * T * sin(\u03b8). To balance the weight W, 2 * T * sin(\u03b8) = W. If the rope is to be perfectly straight and horizontal (\u03b8 = 0), then sin(\u03b8) = sin(0) = 0. For 2 * T * sin(\u03b8) to equal the non-zero weight W (20 kg * 10 m\/s\u00b2 = 200 N), the tension T must tend towards infinity as sin(\u03b8) tends towards zero. Thus, an infinitely large tension is required to completely straighten the rope. - Static equilibrium requires the net force in all directions to be zero. - When a mass hangs from a rope, the weight acts vertically downwards. - Tension in the rope must have a vertical component to balance the weight. - For a nearly horizontal rope, the angle with the horizontal is very small. - As the angle approaches zero, the sine of the angle approaches zero, requiring tension to approach infinity to provide a non-zero vertical force.","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\/two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k\/","og_locale":"en_US","og_type":"article","og_title":"Two persons are holding a rope of negligible mass horizontally. A 20 k","og_description":"When a mass is attached to the midpoint of a horizontally held rope, the rope sags downwards due to the weight of the mass (W = mg). The tension in the rope acts along the direction of the rope segments on either side of the mass. For the rope to be in equilibrium, the vector sum of the two tensions and the weight must be zero. If the rope is perfectly horizontal, the tension vectors would have only horizontal components (neglecting the mass of the rope itself). However, the weight of the attached mass acts vertically downwards. To balance this downward force, there must be an equal and opposite upward force provided by the vertical components of the tension in the rope. As the rope approaches a perfectly horizontal state, the angle (\u03b8) it makes with the horizontal approaches zero. The vertical component of the tension in each half of the rope is T * sin(\u03b8), where T is the tension. The total upward vertical force is 2 * T * sin(\u03b8). To balance the weight W, 2 * T * sin(\u03b8) = W. If the rope is to be perfectly straight and horizontal (\u03b8 = 0), then sin(\u03b8) = sin(0) = 0. For 2 * T * sin(\u03b8) to equal the non-zero weight W (20 kg * 10 m\/s\u00b2 = 200 N), the tension T must tend towards infinity as sin(\u03b8) tends towards zero. Thus, an infinitely large tension is required to completely straighten the rope. - Static equilibrium requires the net force in all directions to be zero. - When a mass hangs from a rope, the weight acts vertically downwards. - Tension in the rope must have a vertical component to balance the weight. - For a nearly horizontal rope, the angle with the horizontal is very small. - As the angle approaches zero, the sine of the angle approaches zero, requiring tension to approach infinity to provide a non-zero vertical force.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T03:28:34+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k\/","url":"https:\/\/exam.pscnotes.com\/mcq\/two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k\/","name":"Two persons are holding a rope of negligible mass horizontally. A 20 k","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T03:28:34+00:00","dateModified":"2025-06-01T03:28:34+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"When a mass is attached to the midpoint of a horizontally held rope, the rope sags downwards due to the weight of the mass (W = mg). The tension in the rope acts along the direction of the rope segments on either side of the mass. For the rope to be in equilibrium, the vector sum of the two tensions and the weight must be zero. If the rope is perfectly horizontal, the tension vectors would have only horizontal components (neglecting the mass of the rope itself). However, the weight of the attached mass acts vertically downwards. To balance this downward force, there must be an equal and opposite upward force provided by the vertical components of the tension in the rope. As the rope approaches a perfectly horizontal state, the angle (\u03b8) it makes with the horizontal approaches zero. The vertical component of the tension in each half of the rope is T * sin(\u03b8), where T is the tension. The total upward vertical force is 2 * T * sin(\u03b8). To balance the weight W, 2 * T * sin(\u03b8) = W. If the rope is to be perfectly straight and horizontal (\u03b8 = 0), then sin(\u03b8) = sin(0) = 0. For 2 * T * sin(\u03b8) to equal the non-zero weight W (20 kg * 10 m\/s\u00b2 = 200 N), the tension T must tend towards infinity as sin(\u03b8) tends towards zero. Thus, an infinitely large tension is required to completely straighten the rope. - Static equilibrium requires the net force in all directions to be zero. - When a mass hangs from a rope, the weight acts vertically downwards. - Tension in the rope must have a vertical component to balance the weight. - For a nearly horizontal rope, the angle with the horizontal is very small. - As the angle approaches zero, the sine of the angle approaches zero, requiring tension to approach infinity to provide a non-zero vertical force.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/two-persons-are-holding-a-rope-of-negligible-mass-horizontally-a-20-k\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CDS-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cds-2\/"},{"@type":"ListItem","position":3,"name":"Two persons are holding a rope of negligible mass horizontally. 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