{"id":89066,"date":"2025-06-01T07:28:44","date_gmt":"2025-06-01T07:28:44","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=89066"},"modified":"2025-06-01T07:28:44","modified_gmt":"2025-06-01T07:28:44","slug":"there-is-a-ball-of-mass-320-g-it-has-625-j-potential-energy-when-rele","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/there-is-a-ball-of-mass-320-g-it-has-625-j-potential-energy-when-rele\/","title":{"rendered":"There is a ball of mass 320 g. It has 625 J potential energy when rele"},"content":{"rendered":"

There is a ball of mass 320 g. It has 625 J potential energy when released freely from a height. The speed with which it will hit the ground is<\/p>\n

[amp_mcq option1=”62\u00b75 m\/s” option2=”2\u00b70 m\/s” option3=”50 m\/s” option4=”40 m\/s” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
The correct answer is A) 62\u00b75 m\/s.<\/section>\n
When the ball is released freely from a height, its potential energy is converted into kinetic energy as it falls. By the principle of conservation of mechanical energy (assuming no air resistance), the potential energy at the initial height is equal to the kinetic energy just before hitting the ground.<\/section>\n
\nThe mass of the ball is m = 320 g = 0.320 kg.
\nThe potential energy at the initial height is PE = 625 J.
\nAssuming the ball starts from rest, its initial kinetic energy is 0. When it hits the ground, its potential energy becomes 0 (taking the ground as the reference level).
\nBy conservation of energy, Initial Total Energy = Final Total Energy.
\nPE_initial + KE_initial = PE_final + KE_final
\n625 J + 0 J = 0 J + KE_final
\nSo, KE_final = 625 J.
\nThe kinetic energy is given by KE = (1\/2) * m * v\u00b2, where v is the speed.
\n625 = (1\/2) * 0.320 * v\u00b2
\n625 = 0.160 * v\u00b2
\nv\u00b2 = 625 \/ 0.160 = 625000 \/ 160 = 62500 \/ 16
\nv = sqrt(62500 \/ 16) = sqrt(62500) \/ sqrt(16) = 250 \/ 4 = 125 \/ 2 = 62.5 m\/s.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

There is a ball of mass 320 g. It has 625 J potential energy when released freely from a height. The speed with which it will hit the ground is [amp_mcq option1=”62\u00b75 m\/s” option2=”2\u00b70 m\/s” option3=”50 m\/s” option4=”40 m\/s” correct=”option1″] This question was previously asked in UPSC NDA-2 – 2024 Download PDFAttempt Online The correct … <\/p>\n

Detailed SolutionThere is a ball of mass 320 g. It has 625 J potential energy when rele<\/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":[1103,1421,1128],"class_list":["post-89066","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1103","tag-motion-under-gravity","tag-physics","no-featured-image-padding"],"yoast_head":"\nThere is a ball of mass 320 g. It has 625 J potential energy when rele<\/title>\n<meta name=\"description\" content=\"The correct answer is A) 62\u00b75 m\/s. When the ball is released freely from a height, its potential energy is converted into kinetic energy as it falls. 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