{"id":87148,"date":"2025-06-01T04:30:09","date_gmt":"2025-06-01T04:30:09","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87148"},"modified":"2025-06-01T04:30:09","modified_gmt":"2025-06-01T04:30:09","slug":"an-ice-bullet-of-temperature-10c-is-fired-upon-an-object-the-bulle","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/an-ice-bullet-of-temperature-10c-is-fired-upon-an-object-the-bulle\/","title":{"rendered":"An ice bullet of temperature \u2013 10\u00b0C is fired upon an object. The bulle"},"content":{"rendered":"<p>An ice bullet of temperature \u2013 10\u00b0C is fired upon an object. The bullet spends 20% of its kinetic energy in penetrating into the object and remaining kinetic energy is spent into melting the bullet at 0\u00b0C. The latent heat of fusion of ice is 3\u00b74 \u00d7 10\u2075 J kg\u207b\u00b9 and its specific heat capacity is 2000 J kg\u207b\u00b9 K\u207b\u00b9. Which one among the following is the correct initial speed of the bullet ?<\/p>\n<p>[amp_mcq option1=&#8221;300\u221a10 m s\u207b\u00b9&#8221; option2=&#8221;3600 m s\u207b\u00b9&#8221; option3=&#8221;300\u221a5 m s\u207b\u00b9&#8221; option4=&#8221;900 m s\u207b\u00b9&#8221; correct=&#8221;option1&#8243;]<\/p>\n<div class=\"psc-box-pyq-exam-year-detail\">\n<div class=\"pyq-exam\">\n<div class=\"psc-heading\">This question was previously asked in<\/div>\n<div class=\"psc-title line-ellipsis\">UPSC Geoscientist &#8211; 2024<\/div>\n<\/div>\n<div class=\"pyq-exam-psc-buttons\"><a href=\"\/pyq\/pyq-upsc-geoscientist-2024.pdf\" target=\"_blank\" class=\"psc-pdf-button\" rel=\"noopener\">Download PDF<\/a><a href=\"\/pyq-upsc-geoscientist-2024\" target=\"_blank\" class=\"psc-attempt-button\" rel=\"noopener\">Attempt Online<\/a><\/div>\n<\/div>\n<section id=\"pyq-correct-answer\">\nThe correct option is A.<br \/>\n<\/section>\n<section id=\"pyq-key-points\">\nThe initial kinetic energy of the ice bullet is converted into different forms of energy upon impact: a portion for penetration and the rest for heating the ice to its melting point (0\u00b0C) and then melting it at 0\u00b0C.<br \/>\n<\/section>\n<section id=\"pyq-additional-information\">\nLet m be the mass of the ice bullet and v be its initial speed.<br \/>\nInitial kinetic energy (KE) = (1\/2)mv\u00b2.<br \/>\n20% of KE is spent on penetration = 0.2 * (1\/2)mv\u00b2 = 0.1mv\u00b2.<br \/>\nRemaining KE is spent on heating the ice from -10\u00b0C to 0\u00b0C and melting it at 0\u00b0C.<br \/>\nRemaining KE = KE &#8211; 0.1mv\u00b2 = 0.9mv\u00b2.<br \/>\nEnergy required to heat mass m of ice from -10\u00b0C to 0\u00b0C:<br \/>\nQ_heat = m * c_ice * \u0394T = m * (2000 J kg\u207b\u00b9 K\u207b\u00b9) * (0 &#8211; (-10)) K = m * 2000 * 10 = 20000m J.<br \/>\nEnergy required to melt mass m of ice at 0\u00b0C:<br \/>\nQ_melt = m * L = m * (3.4 \u00d7 10\u2075 J kg\u207b\u00b9) = 340000m J.<br \/>\nTotal energy required for heating and melting = Q_heat + Q_melt = 20000m + 340000m = 360000m J.<br \/>\nThis energy comes from the remaining KE:<br \/>\n0.9mv\u00b2 = 360000m<br \/>\n0.9v\u00b2 = 360000<br \/>\nv\u00b2 = 360000 \/ 0.9 = 3600000 \/ 9 = 400000.<br \/>\nv = \u221a400000 = \u221a((4 \u00d7 10\u2075) * 10\/10) = \u221a(40 \u00d7 10\u2074) = 100 * \u221a40 = 100 * \u221a(4 * 10) = 100 * 2\u221a10 = 200\u221a10 m\/s.<\/p>\n<p>Wait, let&#8217;s re-read the problem carefully: &#8220;remaining kinetic energy is spent into melting the bullet at 0\u00b0C&#8221;. This phrasing usually means only the phase change energy is considered from the remaining KE. If this interpretation is correct, the 80% KE is used *only* for melting at 0\u00b0C, implying the energy for heating from -10\u00b0C to 0\u00b0C is either negligible or sourced otherwise (e.g., from the 20% penetration energy, which is unlikely). Let&#8217;s retry with this interpretation:<\/p>\n<p>Remaining KE (80% of total KE) = 0.8 * (1\/2)mv\u00b2 = 0.4mv\u00b2.<br \/>\nThis is spent on melting the bullet at 0\u00b0C:<br \/>\n0.4mv\u00b2 = Q_melt = m * L = m * 3.4 \u00d7 10\u2075<br \/>\n0.4v\u00b2 = 3.4 \u00d7 10\u2075<br \/>\nv\u00b2 = (3.4 \u00d7 10\u2075) \/ 0.4 = (3.4 \u00d7 10\u2075) \/ (4\/10) = (3.4 \u00d7 10\u2075) * (10\/4) = (34\/4) \u00d7 10\u2075 = 8.5 \u00d7 10\u2075<br \/>\nv = \u221a(8.5 \u00d7 10\u2075) = \u221a(85 \u00d7 10\u2074) = 100 * \u221a85.<br \/>\n\u221a85 is approximately 9.22. v \u2248 922 m\/s. This doesn&#8217;t match any option well.<\/p>\n<p>Let&#8217;s go back to the first interpretation, where 80% KE covers both heating and melting.<br \/>\n0.8 * (1\/2)mv\u00b2 = m * c_ice * \u0394T + m * L<br \/>\n0.4v\u00b2 = m(c_ice * \u0394T + L)<br \/>\n0.4v\u00b2 = 2000 * 10 + 3.4 \u00d7 10\u2075 = 20000 + 340000 = 360000<br \/>\nv\u00b2 = 360000 \/ 0.4 = 900000<br \/>\nv = \u221a900000 = \u221a(90 \u00d7 10\u2074) = 100 * \u221a90 = 100 * \u221a(9 \u00d7 10) = 100 * 3\u221a10 = 300\u221a10 m\/s.<br \/>\n300\u221a10 \u2248 300 * 3.162 = 948.6 m\/s. This matches option A exactly.<\/p>\n<p>The phrase &#8220;melting the bullet at 0\u00b0C&#8221; in the problem statement, combined with providing the specific heat capacity of ice and the initial temperature of -10\u00b0C, strongly suggests that the energy required to raise the temperature to 0\u00b0C must also be accounted for by the available KE. Therefore, the 80% KE is used for the entire process from -10\u00b0C to liquid water at 0\u00b0C (heating ice + melting ice).<\/p>\n<\/section>\n","protected":false},"excerpt":{"rendered":"<p>An ice bullet of temperature \u2013 10\u00b0C is fired upon an object. The bullet spends 20% of its kinetic energy in penetrating into the object and remaining kinetic energy is spent into melting the bullet at 0\u00b0C. The latent heat of fusion of ice is 3\u00b74 \u00d7 10\u2075 J kg\u207b\u00b9 and its specific heat capacity &#8230; <\/p>\n<p class=\"read-more-container\"><a title=\"An ice bullet of temperature \u2013 10\u00b0C is fired upon an object. The bulle\" class=\"read-more button\" href=\"https:\/\/exam.pscnotes.com\/mcq\/an-ice-bullet-of-temperature-10c-is-fired-upon-an-object-the-bulle\/#more-87148\">Detailed Solution<span class=\"screen-reader-text\">An ice bullet of temperature \u2013 10\u00b0C is fired upon an object. The bulle<\/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":[1091],"tags":[1103,1130,1128],"class_list":["post-87148","post","type-post","status-publish","format-standard","hentry","category-upsc-geoscientist","tag-1103","tag-heat-and-thermodynamics","tag-physics","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>An ice bullet of temperature \u2013 10\u00b0C is fired upon an object. The bulle<\/title>\n<meta name=\"description\" content=\"The correct option is A. 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