{"id":88574,"date":"2025-06-01T07:15:07","date_gmt":"2025-06-01T07:15:07","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88574"},"modified":"2025-06-01T07:15:07","modified_gmt":"2025-06-01T07:15:07","slug":"a-rigid-body-of-mass-2-kg-is-dropped-from-a-stationary-balloon-kept-at","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-rigid-body-of-mass-2-kg-is-dropped-from-a-stationary-balloon-kept-at\/","title":{"rendered":"A rigid body of mass 2 kg is dropped from a stationary balloon kept at"},"content":{"rendered":"

A rigid body of mass 2 kg is dropped from a stationary balloon kept at a height of 50 m from the ground. The speed of the body when it just touches the ground and the total energy when it is dropped from the balloon are respectively (acceleration due to gravity = 9.8 m\/s\u00b2)<\/p>\n

[amp_mcq option1=”980 m.s\u207b\u00b9 and 980 J” option2=”\u221a980 m.s\u207b\u00b9 and \u221a980 J” option3=”980 m.s\u207b\u00b9 and \u221a980 J” option4=”\u221a980 m.s\u207b\u00b9 and 980 J” correct=”option4″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2019<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct option is D) \u221a980 m.s\u207b\u00b9 and 980 J.
\n<\/section>\n
\nThe problem involves calculating the final speed of a falling object using kinematics and determining the total mechanical energy at the beginning of the fall using the principle of conservation of energy.
\n<\/section>\n
\nGiven: mass (m) = 2 kg, initial height (h) = 50 m, initial velocity (u) = 0 m\/s, acceleration due to gravity (g) = 9.8 m\/s\u00b2.
\n1. Speed when it just touches the ground (v): Using the kinematic equation v\u00b2 = u\u00b2 + 2gh, we get v\u00b2 = 0\u00b2 + 2 * 9.8 * 50 = 980. So, v = \u221a980 m\/s.
\n2. Total energy when dropped: At the moment of dropping from a height h with zero initial velocity, the total mechanical energy is the sum of potential energy and kinetic energy. Potential Energy (PE) = mgh = 2 kg * 9.8 m\/s\u00b2 * 50 m = 980 J. Kinetic Energy (KE) = \u00bdmu\u00b2 = \u00bd * 2 kg * (0 m\/s)\u00b2 = 0 J. Total Energy = PE + KE = 980 J + 0 J = 980 J. (Note: By conservation of energy, the total energy remains 980 J just before hitting the ground, where all potential energy is converted to kinetic energy).
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A rigid body of mass 2 kg is dropped from a stationary balloon kept at a height of 50 m from the ground. The speed of the body when it just touches the ground and the total energy when it is dropped from the balloon are respectively (acceleration due to gravity = 9.8 m\/s\u00b2) [amp_mcq … <\/p>\n

Detailed SolutionA rigid body of mass 2 kg is dropped from a stationary balloon kept at<\/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":[1094],"tags":[1119,1421,1128],"class_list":["post-88574","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1119","tag-motion-under-gravity","tag-physics","no-featured-image-padding"],"yoast_head":"\nA rigid body of mass 2 kg is dropped from a stationary balloon kept at<\/title>\n<meta name=\"description\" content=\"The correct option is D) \u221a980 m.s\u207b\u00b9 and 980 J. 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