{"id":4919,"date":"2023-02-13T11:38:42","date_gmt":"2023-02-13T06:08:42","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=4919"},"modified":"2023-03-30T10:42:43","modified_gmt":"2023-03-30T05:12:43","slug":"ap-board-9th-class-maths-solutions-chapter-10-ex-10-4","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-board-9th-class-maths-solutions-chapter-10-ex-10-4\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 10 Surface Areas and Volumes Ex 10.4 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.4<\/h2>\n

\"AP<\/p>\n

Question 1.
\nThe radius of a sphere is 3.5 cm. Find its surface area and volume.
\nSolution:
\nRadius of the sphere, r = 3.5 cm
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\nThe surface area of a sphere is 1018\\(\\frac{2}{7}\\) cm2<\/sup> . What is its volume ?
\nSolution:
\nSurface area of sphere = 4\u03c0r2<\/sup>
\n= 1018\\(\\frac{2}{7}\\) cm2<\/sup>
\n\"AP
\n= 3054.857cm3<\/sup>
\n\u2245 3054.86cm3<\/sup><\/p>\n

Question 3.
\nThe length of equator of the globe is 44 cm. Find its surface area.
\nSolution:
\nLength of the equator of the globe 2\u03c0r = 44 cm.
\n2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 r = 44
\n\u2234 r = \\(\\frac{44 \\times 7}{2 \\times 22}\\) = 7cm
\n\u2234 surface area = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7
\n= 4 \u00d7 22 \u00d7 7
\n= 616cm2<\/sup><\/p>\n

\"AP<\/p>\n

Question 4.
\nThe diameter of a spherical ball is 21 cm. How much leather is required to prepare 5 such balls?
\nSolution:
\nDiameter of the spherical ball d\u2019 = 21 cm
\nThus, its radius r = \\(\\frac{d}{2}=\\frac{21}{2}\\) = 10.5 cm
\nSurface area of one ball = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac{22}{7}\\) \u00d7 10.5 \u00d7 10.5
\n= 88 \u00d7 1.5 \u00d7 10.5 = 1386 cm2<\/sup>
\n\u2234 Leather required for 5 such balls
\n= 5 \u00d7 1386 = 6930 cm2<\/sup><\/p>\n

Question 5.
\nThe ratio of radii of two spheres is 2 : 3. Find the ratio of their surface areas and volumes.
\nSolution:
\nRatio of radii r1<\/sub> : r2<\/sub> = 2 : 3
\nRatio of surface area
\n= 4\u03c0r1<\/sub>2<\/sup> : 4\u03c0r2<\/sub>2<\/sup>
\n= 22<\/sup>: 32<\/sup> = 4 : 9
\nRatio of volumes
\n= 4\/3 \u03c0r1<\/sub>3<\/sup> : 4\/3 \u03c0r2<\/sub>3<\/sup>
\n= 23<\/sup> : 33<\/sup> = 8 : 27<\/p>\n

Question 6.
\nFind the total surface area of hemisphere of radius 10 cm. (Use \u03c0 = 3.14)
\nSolution:
\nRadius of the hemisphere = 10 cm
\nTotal surface area of the hemisphere = 3\u03c0r2<\/sup>
\n= 3 \u00d7 3.14 \u00d7 10 \u00d7 10
\n= 9.42 \u00d7 100
\n= 942 cm2<\/sup><\/p>\n

\"AP<\/p>\n

Question 7.
\nThe diameter of a spherical balloon increases from 14 cm. to 28 cm. as air is being pumped into it. Find the ratio of surface areas of the balloons in the
\ntwo cases.
\nSolution:
\nThe diameter of the balloon, d = 14 cm
\nThus, its radius, r = \\(\\frac{d}{2}=\\frac{14}{2}\\) = 7 cm
\n\u2234 Surface area = 4\u03c0r2<\/sup> = 4 \u00d7 \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7
\n= 88 \u00d7 7 = 616cm2<\/sup>
\nWhen air is pumped, the diameter = 28 cm
\nthus its radius = \\(\\frac{d}{2}=\\frac{28}{2}\\) = 14 cm
\nIts surface area = 4\u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac{22}{7}\\) \u00d7 14 \u00d7 14
\n= 88 \u00d7 28 = 2464 cm2<\/sup>
\nRatio of areas = 616 : 2464
\n= 1 : 4<\/p>\n

(OR)<\/p>\n

Original radius = \\(\\frac{14}{2}\\) = 7 cm
\nIncreased radius = \\(\\frac{28}{2}\\) = 14cm
\nRatio of areas = r1<\/sub>2<\/sup> : r2<\/sub>2<\/sup>
\n= 72<\/sup> : 142<\/sup>
\n= 7 \u00d7 7 : 14 \u00d7 14
\n= 1:4<\/p>\n

\"AP<\/p>\n

Question 8.
\nA hemispherical bowl is made of brass, 0.25 cm thickness. The inner radius of the bowl is 5 cm. Find the ratio of outer surface area to inner surface area.
\nSolution:
\nInner radius of the hemisphere ‘r’ = 5 cm
\nOuter radius of the hemisphere ‘R’
\n= inner radius + thickness
\n= (5 + 0.25) cm = 5.25 cm
\nRatio of areas = 3\u03c0R2<\/sup>: 3\u03c0r2<\/sup>
\n= R2<\/sup> : r2<\/sup>
\n= (5.25)2<\/sup>: 52<\/sup>
\n= 27.5625 : 25
\n= 1.1025:1
\n= 11025 : 10000
\n= 441 : 400
\n[Note : If we read “radius as diameter” then we get the T.B. answer]<\/p>\n

Question 9.
\nThe diameter of a lead ball is 2.1 cm. The density of the lead used is 11.34 g\/c3<\/sup>. What is the, weight of the ball ?
\nSolution:
\nThe diameter of the ball = 2.1 cm
\nThus, its radius, r = \\(\\frac{d}{2}=\\frac{2.1}{2}\\) = 1.05 cm
\nVolume of the ball V’ = \\(\\frac{4}{3}\\)\u03c0r3<\/sup>
\n= \\(\\frac{4}{3} \\times \\frac{22}{7}\\) x 1.053<\/sup> = \\(\\frac{101.87}{21}\\)
\n\u2234Weight of the ball = Volume \u00d7 density
\n= 4.851 \u00d7 1.34
\n= 55.010<\/p>\n

\"AP<\/p>\n

Question 10.
\nA metallic cylinder of diameter 5 cm 1 and height 3 \\(\\frac{1}{3}\\) cm is melted and cast into a sphere. What is its diameter ?
\nSolution:
\nDiameter of the cylinder’d’ = 5 cm
\nThus, its radius, r = \\(\\frac{d}{2}=\\frac{5}{2}\\) = 2.5 cm
\nHeight of the cylinder,
\n\"AP
\nVolume of the cylinder
\n\"AP
\nGiven that cylinder melted to form sphere
\n\u2234 Volume of the sphere = Volume of the cylinder
\n\"AP
\n(Where r is the radius of the sphere)
\nr3<\/sup> = \\(\\frac{3}{4}\\) \u00d7 2.5 \u00d7 2.5 \u00d7 \\(\\frac{10}{3}\\)
\nr3<\/sup> = 2.53<\/sup>
\n\u2234 r = 2.5 cm
\nHence its diameter, d = 2r
\n= 2 \u00d7 2.5 = 5 cm<\/p>\n

\"AP<\/p>\n

Question 11.
\nHow many litres of milk can a hemispherical bowl of diameter 10.5 cm hold ?
\nSolution:
\nDiameter of the hemispherical bowl = 10.5 cm
\nThus its radius = \\(\\frac{d}{2}=\\frac{10.5}{2}\\) = 5.25cm
\nQuantity of milk, the bowl can hold = Volume of the bowl = \\(\\frac{2}{3}\\)\u03c0r3<\/sup>
\n= \\(\\frac{2}{3} \\times \\frac{22}{7}\\) \u00d7 5.25 \u00d7 5.25 \u00d7 5.25
\n= 303.1875 cm3<\/sup>
\n= \\(\\frac{303.1875}{1000}\\) lit = 0.303 lit.<\/p>\n

Question 12.
\nA hemispherical bowl has diameter 9 cm. The liquid is poured into cylindrical bottles of diameter 3 cm and height 3 cm. If a full bowl of liquid is Riled in the bottles, find how many
\nbottles are required ?
\nSolution:
\nDiameter of the hemispherical bowl ‘d’ = 9 cm
\nIts radius, r = \\(\\frac{d}{2}=\\frac{9}{2}\\) = 4.5cm
\nVolume of its liquid = Volume of the bowl = \\(\\frac{2}{3}\\) \u03c0r3<\/sup>
\n= \\(\\frac{2}{3} \\times \\frac{22}{7}\\) \u00d7 4.5 \u00d7 45 \u00d7 4.5
\nDiameter of the cylindrical bottle, d = 3 cm
\nIts radius, r = \\(\\frac{d}{2}\\)
\n= \\(\\frac{3.0}{2}\\)
\n= 1.5cm<\/p>\n

Height of the bottle, h = 3 cm
\nLet the number of bottles required = n
\nThen total volumes of these n bottles = n \u03c0r2<\/sup>h
\nBut this is equal to volume of the bowl
\nHence n. \\(\\frac{22}{7}\\) \u00d7 1.5 \u00d7 1.5 \u00d7 3
\n= \\(\\frac{2}{3} \\times \\frac{22}{7}\\) \u00d7 4.5 \u00d7 4.5 \u00d7 4.5
\n\u2234 n = \\(\\frac{2}{3} \\times \\frac{20.25}{1.5}\\) = 9
\n\u2234 Number of bottles required = 9<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.4 Question 1. The radius of a sphere is 3.5 cm. Find its surface area and volume. Solution: Radius of the … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":36049,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2023\/02\/Chapter-10-Surface-Areas-and-Volumes-Ex-10.4.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/4919"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=4919"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/4919\/revisions"}],"predecessor-version":[{"id":34196,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/4919\/revisions\/34196"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/36049"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=4919"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=4919"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=4919"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}