{"id":928,"date":"2020-09-15T05:56:24","date_gmt":"2020-09-15T00:26:24","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=928"},"modified":"2021-07-02T15:01:52","modified_gmt":"2021-07-02T09:31:52","slug":"kseeb-sslc-class-10-maths-solutions-chapter-15-ex-15-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-15-ex-15-3\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3<\/strong>\u00a0are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3<\/h2>\n

(Take \u03c0 \\(=\\frac{22}{7}\\) unless stated otherwise)<\/p>\n

Question 1.
\nA metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
\nSolution:
\nLet the height of cylinder be ‘h’ cm.
\nRadius of cylinder, r = 6 cm.
\ni) Volume of metallic sphere, V = \\(\\frac{4}{3} \\pi r^{3}\\)
\n\"KSEEB
\nii) Volume of Cylinder, V = \\(\\pi r^{2} h\\)
\n\\(\\frac{22}{7}\\) \u00d7 36 \u00d7 h = 310.46
\n\"KSEEB
\n\u2234 Height of cylinder is 2.74 cm.<\/p>\n

Question 2.
\nMetallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
\nSolution:
\nLet the radius of the resulting sphere be ‘r’ cm.
\n(i) Volume of 3 spheres having radii 8 cm and 10 cm.
\n\"KSEEB
\n(i) Volume of the resulting Sphere, V
\n\"KSEEB
\n\u2234 Volume of the resulting sphere = Volume of three spheres
\n\"KSEEB
\n\u2234 Radius of the resulting sphere is 12 cm.<\/p>\n

Question 3.
\nA 20 m deep well with a diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
\nSolution:
\nRadius of well, ‘r’ = 3.5 m.
\nHeight of well, h = 20 m.
\n\"KSEEB
\n(i) Volume of soil dug out of well, V = \\(\\pi r^{2} h\\) (\u2235 cylinder sphere)
\n= \\(\\frac{22}{7} \\times(3.5)^{2} \\times 20\\)
\n= \\(=\\frac{22}{7} \\times \\frac{35}{10} \\times \\frac{35}{10} \\times 20\\)
\n= 770m3<\/sup>
\n(ii) Length of platform, l = 22 m.
\nBreadth, b = 14 m.
\nHeight, h =?
\nVolume of Platform, V= l \u00d7 b \u00d7 h (\u2235 Cuboid)
\n= 22 \u00d7 14 \u00d7 h
\n\u2234 Volume of Platform = Volume of soil dug out of well.
\n\"KSEEB
\n\u2234 Height of platform is 2.5 m<\/p>\n

Question 4.
\nA well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
\nSolution:
\nLet the height of the embankment be ‘h’ m.
\n\"KSEEB
\nDimeter of well, d = 3 m.
\nRadius of well, r =1.5 m.
\nHeight of well, h = 14 m.
\nVolume of an embankment = Volume of soil dug from well.
\n\"KSEEB
\n\u2234 Height of embankment = 1.125 m.<\/p>\n

Question 5.
\nA container shaped like a right cylinder diameter 12height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones cones which can be filled with ice cream.
\n\"KSEEB
\nSolution:
\nDiameter of right circular cylinder is 12 cm \u279d r = 6 cm.
\nHeight: 15 cm. \u279d h = 15 cm.
\nDiameter of Cone, 6 cm. \u279d r = 3 cm.
\nHeight 12 cm. \u279d h = 12 cm.
\n(i) Volume of the ice cream which has cone-shaped and hemisphere.
\n\"KSEEB
\n(ii) Volume of cylindrical vessel, V = \\(\\pi r^{2} h\\)
\n\"KSEEB
\nLet number of cones be ‘n’,
\nn \u00d7 54\u03c0 = \u03c0 \u00d7 36 \u00d7 15
\n54n = 36 \u00d7 15
\n\\(n=\\frac{36 \\times 15}{54}=\\frac{540}{54}\\)
\n\u2234 n = 10
\nIce cream can be filled in 10 cones.<\/p>\n

Question 6.
\nHow many silver coins, 1.75 cm. in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm \u00d7 10 cm \u00d7 3.5 cm?
\nSolution:
\nDiameter of the silver coins is 1.75 cm.
\nThickness = 2 mm.
\nMeasurement of cuboid: 5.5 \u00d7 10 \u00d7 3.5 cm.
\n\u2234 Let the required silver coins be ‘n’,
\nVolume of ‘n’ coins = Volume of cuboid
\n\\(\\pi r^{2} h\\) = l \u00d7 b \u00d7 h
\n\"KSEEB
\n\u2234 n = 400
\n\u2234 400 silver coins are required.<\/p>\n

Question 7.
\nA cylindrical bucket, 32 cm high and with a radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
\nSolution:
\nHeight of cylindrical bucket, h = 32 cm.
\nRadius, r = 18 cm.
\nHeight of conical heap of sand = 24 cm.
\nRadius, r =?
\nSlant height, l =?
\nVolume of the conical heap of sand = Volume of Cylinder.
\n\"KSEEB
\nRadius of Cone, r = 36 cm.
\nSlant height of cone, l = \\(\\sqrt{\\mathrm{h}^{2}+\\mathrm{r}^{2}}\\)
\n\"KSEEB
\n\u2234 Radius of cone, r = 36 cm.
\nSlant height of cone, h = \\(12 \\sqrt{13}\\) cm<\/p>\n

Question 8.
\nWater in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km\/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
\nSolution:
\nBreadth of a canal, b = 6 m.
\nHeight, h = 1.5 m
\nSpeed of water =10 km\/hr.
\nQuantity of water = 8 cm.
\nQuantity of water flowing in 30 minutes?
\nVolume of water flowing in 60 minutes,
\n= (10 \u00d7 1000) \u00d7 6 \u00d7 1.5 cm3<\/sup>.
\n( \u2235 Speed = 10 km\/hr.; 10 \u00d7 1000 m\/hr.)
\nVolume of water flowing in 30 minutes,
\n= \\(10000 \\times 9 \\times \\frac{30}{60} \\mathrm{m}^{3}\\)
\n\u2234 Let the area of land be ‘x’ sq.m.
\n\"KSEEB
\nx= 562500 sq.m.<\/p>\n

Question 9.
\nA farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km\/h, in how much time will the tank be filled?
\nSolution:
\nInternal diameter of a pipe = 20 cm.
\nRadius of pipe =10 cm. = \\(\\frac{1}{10} \\mathrm{m}\\)
\nDiameter of cylindrical tank = 10 m; r=5 m.
\nDepth of cylindrical tank = 2 m.
\nSpeed of water is 3 km\/hr.
\nTime required to fill the water tank =?
\nSpeed of water = 3 km\/hr.
\n= \\(\\frac{3000}{60}\\) m\/minute
\n= 50 m\/min.
\nLet the time required to fill the tank be n’ minutes.
\n\u2234 Flowing water in ‘n’ minutes = Volume of the water tank
\n\"KSEEB
\n\u2234 n = 100 minutes
\n\u2234 Time required to fill the water tank = 100 minutes<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3\u00a0are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3. Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3 (Take …<\/p>\n","protected":false},"author":1,"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\/928"}],"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\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=928"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/928\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=928"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=928"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=928"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}