{"id":25048,"date":"2020-09-15T12:10:41","date_gmt":"2020-09-15T06:40:41","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25048"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-solutions-for-class-10-maths-chapter-15-ex-15-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-15-ex-15-5\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5<\/h2>\n

Question 1.
\nA copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3<\/sup>.
\nAnswer:
\nCylinder
\nDiameter = 10 cm,
\nradius = r = \\(\\frac{d}{2}=\\frac{10}{2}\\) = 5cm
\nLength of wire one round of cylinder = circumference of base
\n= 2\u03c0r = 2\u03c0(5) = 10\u03c0 cm
\nCylindrical wire diameter = 3mm
\n= \\(\\frac{3}{10}\\) cm
\nlength of cylinder = h = 12 cm
\n\"KSEEB
\nTotal round = 40
\nNow, length of wire required in one round = circumference of base of cylinder
\n2\u03c0r = 2\u03c0 \u00d7 \\(\\frac{10}{2}\\)
\n2\u03c0r = 10\u03c0 cm
\nHence in 40round the length of wire required
\n= 40 \u00d7 10\u03c0
\n= 400\u03c0 cm
\n= \\(\\frac{400 \\times 22}{7}\\) = 1257.14 cm
\n= 12.57 m.
\ntherefore total length of wire 12.57 m
\nvolume of wire = \u03c0r2<\/sup>h
\n= \u03c0 \u00d7 \\(\\left(\\frac{0.3}{2}\\right)^{2}\\) \u00d7 1257.14
\n88.898cm3<\/sup>
\nThe mass of wire of 1 cm3<\/sup> = 8.885
\n\u2234 the mass of wire of 88.898 cm3<\/sup>
\n= 8.88 \u00d7 88.898
\n= 798.413
\nHence mass of wire = 798.419 and length of wire = 12.57 m<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nA right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is -made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of \u03c0 as found appropriate.)
\nAnswer:
\n\"KSEEB
\nHence OB = radius of cone = 2.4 cm
\nVolume of double cone = volume of cone 1 + volume of cone 2.
\n\"KSEEB
\n\"KSEEB
\n= 30.14cm3<\/sup>
\nVolume of double cone = 30.14 cm3<\/sup>
\nSurface are of double cone
\n= CSA 1 + CSA 2
\n\"KSEEB<\/p>\n

Question 3.
\nA cistern, internally measuring 150 cm \u00d7 120 cm \u00d7 no cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm \u00d7 7.5 cm \u00d7 6.5 cm?
\nAnswer:
\nVolume of cistern =150 \u00d7 120 \u00d7 110
\n= 1980000 cm3<\/sup>
\nvolume of water in cistern
\n= 129600 cm3<\/sup>
\nvolume to filled in cistern
\n= 1980000 – 129600
\n= 1850400 cm3<\/sup>
\nvolume of one brick = 22.5 \u00d7 7.5 \u00d7 6.5
\n= 1096.875 cm3<\/sup>
\nLet ‘N’ no of bricks can be placed into the cistern
\nHence volume of ‘N’ no of bricks
\n= N \u00d7 1096.875 cm3<\/sup>
\nOne bricks absorb water \\(\\frac{1}{17}\\) of his volume hence water absorb bv 1 brick
\n\"KSEEB
\nHence total N no. of brick absorb water
\n\"KSEEB
\nVolume to be filled in tank + water absorbs by N no. of bricks = volume of N No of bricks.
\n\"KSEEB
\n\u2234 therefore No. of bricks can be put in cistern without over flowing water.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nIn one fortnight of a given month, there was a rainfall of 10 cm in a 5. river valley. If the area of the valley is 7280 km2<\/sup>, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
\nAnswer:
\nFor each of three rivers.
\nlength (l) = 1072 km = 1072 \u00d7 1000 m
\n= 1072000 m.
\nBreadth = b = 75 m and Depth = h = 3 m.
\nvolume of water in river = lbh
\n= 1072000 \u00d7 75 \u00d7 3
\n= 241200000 m3<\/sup>.
\nVolume of water in 3 rivers
\n= 3 \u00d7 241200000
\n= 723600000 m3<\/sup>.
\n\"KSEEB
\nArea of the valley = 7280 km2<\/sup>
\nRainfall in the valley
\n\"KSEEB
\nVolume of rainfall = Area of the valley \u00d7 height of rainfall
\n\"KSEEB
\nFrom (1) and (2) total rainfall is approximately equivalent<\/p>\n

Question 5.
\nAn oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig. 15.25).
\n\"KSEEB
\nAnswer:
\nr1<\/sub> = \\(\\frac{18}{2}\\) = 9cm , r2<\/sub> = \\(\\frac{8}{2}\\) = 4 cm
\nHeight h of frustum = 22 – 10 = 12 cm.
\nSlant height of the frustum cone =
\n\"KSEEB
\nl = 13cm
\nradius of cylindrical portion = \\(\\frac{8}{2}\\)
\n= 4 cm.
\nArea of tin sheet required = CSA of cylinder + CSA of frustum portion
\n= 2\u03c0rh + \u03c0 (r1<\/sub> + r2<\/sub>) l
\n= 2\u03c0(4)(10) + \u03c0(9 + 4) 13
\n= 80\u03c0 + \u03c0 \u00d7 13 \u00d7 13 = 80\u03c0 + 169\u03c0
\n= 249\u03c0
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25048"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=25048"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25048\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25048"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25048"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25048"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}