{"id":86544,"date":"2025-06-01T03:50:00","date_gmt":"2025-06-01T03:50:00","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=86544"},"modified":"2025-06-01T03:50:00","modified_gmt":"2025-06-01T03:50:00","slug":"a-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me\/","title":{"rendered":"A sound wave having frequency of 300 Hz is travelling in an unknown me"},"content":{"rendered":"

A sound wave having frequency of 300 Hz is travelling in an unknown medium. Its wavelength is not known. It travels a distance equal to 150 times its wavelength in time t. The value of t is :<\/p>\n

[amp_mcq option1=”0.5 s” option2=”1 s” option3=”1.5 s” option4=”2 s” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CDS-2 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe time taken for the sound wave to travel the given distance is 0.5 s.
\n<\/section>\n
\n– The speed of a wave (v) is given by the product of its frequency (f) and wavelength (\u03bb): v = f * \u03bb.
\n– The distance travelled (d) is given as 150 times the wavelength: d = 150\u03bb.
\n– The relationship between speed, distance, and time (t) is v = d \/ t.
\n– Combining these equations, we get: f * \u03bb = (150\u03bb) \/ t.
\n– We can cancel out \u03bb from both sides (assuming \u03bb is non-zero): f = 150 \/ t.
\n– We are given the frequency f = 300 Hz.
\n– Rearranging the equation to find t: t = 150 \/ f.
\n– Substituting the value of f: t = 150 \/ 300 = 0.5 s.
\n<\/section>\n
\nThis calculation assumes the speed of the sound wave is constant in the unknown medium. The wavelength of the sound wave in the medium is not required to find the time, as it cancels out in the calculation.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A sound wave having frequency of 300 Hz is travelling in an unknown medium. Its wavelength is not known. It travels a distance equal to 150 times its wavelength in time t. The value of t is : [amp_mcq option1=”0.5 s” option2=”1 s” option3=”1.5 s” option4=”2 s” correct=”option1″] This question was previously asked in UPSC … <\/p>\n

Detailed SolutionA sound wave having frequency of 300 Hz is travelling in an unknown me<\/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":[1088],"tags":[1103,1128,1147],"class_list":["post-86544","post","type-post","status-publish","format-standard","hentry","category-upsc-cds-2","tag-1103","tag-physics","tag-sound","no-featured-image-padding"],"yoast_head":"\nA sound wave having frequency of 300 Hz is travelling in an unknown me<\/title>\n<meta name=\"description\" content=\"The time taken for the sound wave to travel the given distance is 0.5 s. - The speed of a wave (v) is given by the product of its frequency (f) and wavelength (\u03bb): v = f * \u03bb. - The distance travelled (d) is given as 150 times the wavelength: d = 150\u03bb. - The relationship between speed, distance, and time (t) is v = d \/ t. - Combining these equations, we get: f * \u03bb = (150\u03bb) \/ t. - We can cancel out \u03bb from both sides (assuming \u03bb is non-zero): f = 150 \/ t. - We are given the frequency f = 300 Hz. - Rearranging the equation to find t: t = 150 \/ f. - Substituting the value of f: t = 150 \/ 300 = 0.5 s.\" \/>\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-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A sound wave having frequency of 300 Hz is travelling in an unknown me\" \/>\n<meta property=\"og:description\" content=\"The time taken for the sound wave to travel the given distance is 0.5 s. - The speed of a wave (v) is given by the product of its frequency (f) and wavelength (\u03bb): v = f * \u03bb. - The distance travelled (d) is given as 150 times the wavelength: d = 150\u03bb. - The relationship between speed, distance, and time (t) is v = d \/ t. - Combining these equations, we get: f * \u03bb = (150\u03bb) \/ t. - We can cancel out \u03bb from both sides (assuming \u03bb is non-zero): f = 150 \/ t. - We are given the frequency f = 300 Hz. - Rearranging the equation to find t: t = 150 \/ f. - Substituting the value of f: t = 150 \/ 300 = 0.5 s.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T03:50:00+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 sound wave having frequency of 300 Hz is travelling in an unknown me","description":"The time taken for the sound wave to travel the given distance is 0.5 s. - The speed of a wave (v) is given by the product of its frequency (f) and wavelength (\u03bb): v = f * \u03bb. - The distance travelled (d) is given as 150 times the wavelength: d = 150\u03bb. - The relationship between speed, distance, and time (t) is v = d \/ t. - Combining these equations, we get: f * \u03bb = (150\u03bb) \/ t. - We can cancel out \u03bb from both sides (assuming \u03bb is non-zero): f = 150 \/ t. - We are given the frequency f = 300 Hz. - Rearranging the equation to find t: t = 150 \/ f. - Substituting the value of f: t = 150 \/ 300 = 0.5 s.","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-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me\/","og_locale":"en_US","og_type":"article","og_title":"A sound wave having frequency of 300 Hz is travelling in an unknown me","og_description":"The time taken for the sound wave to travel the given distance is 0.5 s. - The speed of a wave (v) is given by the product of its frequency (f) and wavelength (\u03bb): v = f * \u03bb. - The distance travelled (d) is given as 150 times the wavelength: d = 150\u03bb. - The relationship between speed, distance, and time (t) is v = d \/ t. - Combining these equations, we get: f * \u03bb = (150\u03bb) \/ t. - We can cancel out \u03bb from both sides (assuming \u03bb is non-zero): f = 150 \/ t. - We are given the frequency f = 300 Hz. - Rearranging the equation to find t: t = 150 \/ f. - Substituting the value of f: t = 150 \/ 300 = 0.5 s.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T03:50:00+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-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me\/","name":"A sound wave having frequency of 300 Hz is travelling in an unknown me","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T03:50:00+00:00","dateModified":"2025-06-01T03:50:00+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The time taken for the sound wave to travel the given distance is 0.5 s. - The speed of a wave (v) is given by the product of its frequency (f) and wavelength (\u03bb): v = f * \u03bb. - The distance travelled (d) is given as 150 times the wavelength: d = 150\u03bb. - The relationship between speed, distance, and time (t) is v = d \/ t. - Combining these equations, we get: f * \u03bb = (150\u03bb) \/ t. - We can cancel out \u03bb from both sides (assuming \u03bb is non-zero): f = 150 \/ t. - We are given the frequency f = 300 Hz. - Rearranging the equation to find t: t = 150 \/ f. - Substituting the value of f: t = 150 \/ 300 = 0.5 s.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-sound-wave-having-frequency-of-300-hz-is-travelling-in-an-unknown-me\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CDS-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cds-2\/"},{"@type":"ListItem","position":3,"name":"A sound wave having frequency of 300 Hz is travelling in an unknown me"}]},{"@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\/86544","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=86544"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/86544\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=86544"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=86544"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=86544"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}