{"id":87743,"date":"2025-06-01T06:53:18","date_gmt":"2025-06-01T06:53:18","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87743"},"modified":"2025-06-01T06:53:18","modified_gmt":"2025-06-01T06:53:18","slug":"numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c\/","title":{"rendered":"Numerically two thermometers, one in Fahrenheit scale and another in C"},"content":{"rendered":"

Numerically two thermometers, one in Fahrenheit scale and another in Celsius scale shall read same at<\/p>\n

[amp_mcq option1=”\u2013 40\u00b0” option2=”0\u00b0” option3=”\u2013 273\u00b0” option4=”100\u00b0” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2021<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe temperature at which both the Fahrenheit and Celsius scales read the same numerical value is -40\u00b0.
\n<\/section>\n
\nTo find the point where the two scales read the same, we set the temperature in Fahrenheit ($T_F$) equal to the temperature in Celsius ($T_C$) and use the conversion formula: $T_F = T_C$. Let this common temperature be $x$. The conversion formula from Celsius to Fahrenheit is $T_F = T_C \\times \\frac{9}{5} + 32$. Substituting $x$ for both $T_F$ and $T_C$: $x = x \\times \\frac{9}{5} + 32$. Solving for $x$: $x – \\frac{9x}{5} = 32 \\implies \\frac{5x – 9x}{5} = 32 \\implies \\frac{-4x}{5} = 32 \\implies -4x = 160 \\implies x = \\frac{160}{-4} = -40$. Thus, -40\u00b0C is equal to -40\u00b0F.
\n<\/section>\n
\nWater freezes at 0\u00b0C (32\u00b0F) and boils at 100\u00b0C (212\u00b0F). These fixed points are different on the two scales, but there is one point where they intersect.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Numerically two thermometers, one in Fahrenheit scale and another in Celsius scale shall read same at [amp_mcq option1=”\u2013 40\u00b0” option2=”0\u00b0” option3=”\u2013 273\u00b0” option4=”100\u00b0” correct=”option1″] This question was previously asked in UPSC NDA-1 – 2021 Download PDFAttempt Online The temperature at which both the Fahrenheit and Celsius scales read the same numerical value is -40\u00b0. To … <\/p>\n

Detailed SolutionNumerically two thermometers, one in Fahrenheit scale and another in C<\/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":[1110,1236,1128],"class_list":["post-87743","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1110","tag-measuring-instruments-and-scales","tag-physics","no-featured-image-padding"],"yoast_head":"\nNumerically two thermometers, one in Fahrenheit scale and another in C<\/title>\n<meta name=\"description\" content=\"The temperature at which both the Fahrenheit and Celsius scales read the same numerical value is -40\u00b0. To find the point where the two scales read the same, we set the temperature in Fahrenheit ($T_F$) equal to the temperature in Celsius ($T_C$) and use the conversion formula: $T_F = T_C$. Let this common temperature be $x$. The conversion formula from Celsius to Fahrenheit is $T_F = T_C times frac{9}{5} + 32$. Substituting $x$ for both $T_F$ and $T_C$: $x = x times frac{9}{5} + 32$. Solving for $x$: $x - frac{9x}{5} = 32 implies frac{5x - 9x}{5} = 32 implies frac{-4x}{5} = 32 implies -4x = 160 implies x = frac{160}{-4} = -40$. Thus, -40\u00b0C is equal to -40\u00b0F.\" \/>\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\/numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Numerically two thermometers, one in Fahrenheit scale and another in C\" \/>\n<meta property=\"og:description\" content=\"The temperature at which both the Fahrenheit and Celsius scales read the same numerical value is -40\u00b0. To find the point where the two scales read the same, we set the temperature in Fahrenheit ($T_F$) equal to the temperature in Celsius ($T_C$) and use the conversion formula: $T_F = T_C$. Let this common temperature be $x$. The conversion formula from Celsius to Fahrenheit is $T_F = T_C times frac{9}{5} + 32$. Substituting $x$ for both $T_F$ and $T_C$: $x = x times frac{9}{5} + 32$. Solving for $x$: $x - frac{9x}{5} = 32 implies frac{5x - 9x}{5} = 32 implies frac{-4x}{5} = 32 implies -4x = 160 implies x = frac{160}{-4} = -40$. Thus, -40\u00b0C is equal to -40\u00b0F.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:53:18+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":"Numerically two thermometers, one in Fahrenheit scale and another in C","description":"The temperature at which both the Fahrenheit and Celsius scales read the same numerical value is -40\u00b0. To find the point where the two scales read the same, we set the temperature in Fahrenheit ($T_F$) equal to the temperature in Celsius ($T_C$) and use the conversion formula: $T_F = T_C$. Let this common temperature be $x$. The conversion formula from Celsius to Fahrenheit is $T_F = T_C times frac{9}{5} + 32$. Substituting $x$ for both $T_F$ and $T_C$: $x = x times frac{9}{5} + 32$. Solving for $x$: $x - frac{9x}{5} = 32 implies frac{5x - 9x}{5} = 32 implies frac{-4x}{5} = 32 implies -4x = 160 implies x = frac{160}{-4} = -40$. Thus, -40\u00b0C is equal to -40\u00b0F.","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\/numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c\/","og_locale":"en_US","og_type":"article","og_title":"Numerically two thermometers, one in Fahrenheit scale and another in C","og_description":"The temperature at which both the Fahrenheit and Celsius scales read the same numerical value is -40\u00b0. To find the point where the two scales read the same, we set the temperature in Fahrenheit ($T_F$) equal to the temperature in Celsius ($T_C$) and use the conversion formula: $T_F = T_C$. Let this common temperature be $x$. The conversion formula from Celsius to Fahrenheit is $T_F = T_C times frac{9}{5} + 32$. Substituting $x$ for both $T_F$ and $T_C$: $x = x times frac{9}{5} + 32$. Solving for $x$: $x - frac{9x}{5} = 32 implies frac{5x - 9x}{5} = 32 implies frac{-4x}{5} = 32 implies -4x = 160 implies x = frac{160}{-4} = -40$. Thus, -40\u00b0C is equal to -40\u00b0F.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:53:18+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\/numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c\/","url":"https:\/\/exam.pscnotes.com\/mcq\/numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c\/","name":"Numerically two thermometers, one in Fahrenheit scale and another in C","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:53:18+00:00","dateModified":"2025-06-01T06:53:18+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The temperature at which both the Fahrenheit and Celsius scales read the same numerical value is -40\u00b0. To find the point where the two scales read the same, we set the temperature in Fahrenheit ($T_F$) equal to the temperature in Celsius ($T_C$) and use the conversion formula: $T_F = T_C$. Let this common temperature be $x$. The conversion formula from Celsius to Fahrenheit is $T_F = T_C \\times \\frac{9}{5} + 32$. Substituting $x$ for both $T_F$ and $T_C$: $x = x \\times \\frac{9}{5} + 32$. Solving for $x$: $x - \\frac{9x}{5} = 32 \\implies \\frac{5x - 9x}{5} = 32 \\implies \\frac{-4x}{5} = 32 \\implies -4x = 160 \\implies x = \\frac{160}{-4} = -40$. Thus, -40\u00b0C is equal to -40\u00b0F.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/numerically-two-thermometers-one-in-fahrenheit-scale-and-another-in-c\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-1\/"},{"@type":"ListItem","position":3,"name":"Numerically two thermometers, one in Fahrenheit scale and another in C"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87743","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=87743"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87743\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87743"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87743"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87743"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}