{"id":87277,"date":"2025-06-01T06:39:10","date_gmt":"2025-06-01T06:39:10","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87277"},"modified":"2025-06-01T06:39:10","modified_gmt":"2025-06-01T06:39:10","slug":"a-person-stands-on-his-two-feet-over-a-surface-and-experiences-a-press","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-person-stands-on-his-two-feet-over-a-surface-and-experiences-a-press\/","title":{"rendered":"A person stands on his two feet over a surface and experiences a press"},"content":{"rendered":"

A person stands on his two feet over a surface and experiences a pressure P. Now the person stands on only one foot. He would experience a pressure of magnitude<\/p>\n

[amp_mcq option1=”4 P” option2=”P” option3=”$\\frac{1}{2}$ P” option4=”2 P” correct=”option4″]<\/p>\n

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\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2015<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
Pressure is defined as force per unit area (P = F\/A). When a person stands on a surface, the force exerted is their weight (F = W). When standing on two feet, the weight is distributed over the total area of two feet ($A_{two}$). So the pressure is $P = W \/ A_{two}$. When standing on one foot, the weight is distributed over the area of one foot ($A_{one}$). Assuming the area of one foot is approximately half the area of two feet ($A_{one} \\approx A_{two}\/2$), the new pressure is $P’ = W \/ A_{one} \\approx W \/ (A_{two}\/2) = 2W\/A_{two} = 2P$. Thus, the pressure is doubled.<\/section>\n
Pressure is inversely proportional to the area over which the force is applied, assuming the force remains constant.<\/section>\n
This is a simple application of the definition of pressure. By reducing the contact area by half (from two feet to one foot), the pressure exerted on the surface doubles for the same force (weight).<\/section>\n","protected":false},"excerpt":{"rendered":"

A person stands on his two feet over a surface and experiences a pressure P. Now the person stands on only one foot. He would experience a pressure of magnitude [amp_mcq option1=”4 P” option2=”P” option3=”$\\frac{1}{2}$ P” option4=”2 P” correct=”option4″] This question was previously asked in UPSC NDA-1 – 2015 Download PDFAttempt Online Pressure is defined … <\/p>\n

Detailed SolutionA person stands on his two feet over a surface and experiences a press<\/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":[1443,1129,1128],"class_list":["post-87277","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1443","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA person stands on his two feet over a surface and experiences a press<\/title>\n<meta name=\"description\" content=\"Pressure is defined as force per unit area (P = F\/A). When a person stands on a surface, the force exerted is their weight (F = W). When standing on two feet, the weight is distributed over the total area of two feet ($A_{two}$). So the pressure is $P = W \/ A_{two}$. When standing on one foot, the weight is distributed over the area of one foot ($A_{one}$). Assuming the area of one foot is approximately half the area of two feet ($A_{one} approx A_{two}\/2$), the new pressure is $P' = W \/ A_{one} approx W \/ (A_{two}\/2) = 2W\/A_{two} = 2P$. Thus, the pressure is doubled. Pressure is inversely proportional to the area over which the force is applied, assuming the force remains constant.\" \/>\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\/a-person-stands-on-his-two-feet-over-a-surface-and-experiences-a-press\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A person stands on his two feet over a surface and experiences a press\" \/>\n<meta property=\"og:description\" content=\"Pressure is defined as force per unit area (P = F\/A). When a person stands on a surface, the force exerted is their weight (F = W). When standing on two feet, the weight is distributed over the total area of two feet ($A_{two}$). So the pressure is $P = W \/ A_{two}$. When standing on one foot, the weight is distributed over the area of one foot ($A_{one}$). Assuming the area of one foot is approximately half the area of two feet ($A_{one} approx A_{two}\/2$), the new pressure is $P' = W \/ A_{one} approx W \/ (A_{two}\/2) = 2W\/A_{two} = 2P$. Thus, the pressure is doubled. Pressure is inversely proportional to the area over which the force is applied, assuming the force remains constant.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-person-stands-on-his-two-feet-over-a-surface-and-experiences-a-press\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:39:10+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":"A person stands on his two feet over a surface and experiences a press","description":"Pressure is defined as force per unit area (P = F\/A). When a person stands on a surface, the force exerted is their weight (F = W). When standing on two feet, the weight is distributed over the total area of two feet ($A_{two}$). So the pressure is $P = W \/ A_{two}$. When standing on one foot, the weight is distributed over the area of one foot ($A_{one}$). Assuming the area of one foot is approximately half the area of two feet ($A_{one} approx A_{two}\/2$), the new pressure is $P' = W \/ A_{one} approx W \/ (A_{two}\/2) = 2W\/A_{two} = 2P$. Thus, the pressure is doubled. Pressure is inversely proportional to the area over which the force is applied, assuming the force remains constant.","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\/a-person-stands-on-his-two-feet-over-a-surface-and-experiences-a-press\/","og_locale":"en_US","og_type":"article","og_title":"A person stands on his two feet over a surface and experiences a press","og_description":"Pressure is defined as force per unit area (P = F\/A). When a person stands on a surface, the force exerted is their weight (F = W). When standing on two feet, the weight is distributed over the total area of two feet ($A_{two}$). So the pressure is $P = W \/ A_{two}$. When standing on one foot, the weight is distributed over the area of one foot ($A_{one}$). Assuming the area of one foot is approximately half the area of two feet ($A_{one} approx A_{two}\/2$), the new pressure is $P' = W \/ A_{one} approx W \/ (A_{two}\/2) = 2W\/A_{two} = 2P$. Thus, the pressure is doubled. Pressure is inversely proportional to the area over which the force is applied, assuming the force remains constant.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-person-stands-on-his-two-feet-over-a-surface-and-experiences-a-press\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:39:10+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\/a-person-stands-on-his-two-feet-over-a-surface-and-experiences-a-press\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-person-stands-on-his-two-feet-over-a-surface-and-experiences-a-press\/","name":"A person stands on his two feet over a surface and experiences a press","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:39:10+00:00","dateModified":"2025-06-01T06:39:10+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"Pressure is defined as force per unit area (P = F\/A). When a person stands on a surface, the force exerted is their weight (F = W). When standing on two feet, the weight is distributed over the total area of two feet ($A_{two}$). So the pressure is $P = W \/ A_{two}$. When standing on one foot, the weight is distributed over the area of one foot ($A_{one}$). Assuming the area of one foot is approximately half the area of two feet ($A_{one} \\approx A_{two}\/2$), the new pressure is $P' = W \/ A_{one} \\approx W \/ (A_{two}\/2) = 2W\/A_{two} = 2P$. Thus, the pressure is doubled. 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