{"id":90558,"date":"2025-06-01T10:30:59","date_gmt":"2025-06-01T10:30:59","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90558"},"modified":"2025-06-01T10:30:59","modified_gmt":"2025-06-01T10:30:59","slug":"a-wire-of-length-6-m-is-stretched-such-that-its-radius-is-reduced-by-2","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-wire-of-length-6-m-is-stretched-such-that-its-radius-is-reduced-by-2\/","title":{"rendered":"A wire of length 6 m is stretched such that its radius is reduced by 2"},"content":{"rendered":"

A wire of length 6 m is stretched such that its radius is reduced by 20%. Which one of the following is the value of increase in its length?<\/p>\n

[amp_mcq option1=”50%” option2=”56.25%” option3=”62.25%” option4=”75%” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2020<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
56.25%<\/section>\n
Assuming the wire is a cylinder and its material is incompressible, the volume of the wire remains constant when it is stretched. The volume of a cylinder is given by $V = \\pi R^2 L$.<\/section>\n
Let the original length be $L_1 = 6$ m and the original radius be $R_1$. The original volume is $V_1 = \\pi R_1^2 L_1 = 6\\pi R_1^2$.
\nThe wire is stretched such that its radius is reduced by 20%. The new radius is $R_2 = R_1 – 0.20R_1 = 0.80R_1$.
\nLet the new length be $L_2$. The new volume is $V_2 = \\pi R_2^2 L_2 = \\pi (0.80R_1)^2 L_2 = \\pi (0.64R_1^2) L_2$.
\nSince the volume remains constant, $V_1 = V_2$:
\n$6\\pi R_1^2 = \\pi (0.64R_1^2) L_2$
\nAssuming $R_1 > 0$, we can cancel $\\pi R_1^2$:
\n$6 = 0.64 L_2$
\n$L_2 = \\frac{6}{0.64} = \\frac{600}{64} = \\frac{75}{8} = 9.375$ m.
\nThe increase in length is $L_2 – L_1 = 9.375 – 6 = 3.375$ m.
\nThe percentage increase in length is $\\frac{\\text{Increase in length}}{\\text{Original length}} \\times 100\\% = \\frac{3.375}{6} \\times 100\\%$.
\n$\\frac{3.375}{6} = \\frac{3375}{6000} = \\frac{675}{1200} = \\frac{135}{240} = \\frac{27}{48} = \\frac{9}{16}$.
\nPercentage increase $= \\frac{9}{16} \\times 100\\% = 0.5625 \\times 100\\% = 56.25\\%$.<\/section>\n","protected":false},"excerpt":{"rendered":"

A wire of length 6 m is stretched such that its radius is reduced by 20%. Which one of the following is the value of increase in its length? [amp_mcq option1=”50%” option2=”56.25%” option3=”62.25%” option4=”75%” correct=”option2″] This question was previously asked in UPSC CAPF – 2020 Download PDFAttempt Online 56.25% Assuming the wire is a cylinder … <\/p>\n

Detailed SolutionA wire of length 6 m is stretched such that its radius is reduced by 2<\/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":[1085],"tags":[1288,1102],"class_list":["post-90558","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1288","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nA wire of length 6 m is stretched such that its radius is reduced by 2<\/title>\n<meta name=\"description\" content=\"56.25% Assuming the wire is a cylinder and its material is incompressible, the volume of the wire remains constant when it is stretched. 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