{"id":25119,"date":"2020-03-04T18:02:12","date_gmt":"2020-03-04T12:32:12","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25119"},"modified":"2021-07-02T15:21:20","modified_gmt":"2021-07-02T09:51:20","slug":"kseeb-solutions-for-class-10-maths-chapter-15-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-15-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Additional Questions"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 15 Surface Areas and Volumes Additional Questions, 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 Additional Questions<\/h2>\n

I. Multiple Choice Questions:<\/span><\/p>\n

Question 1.
\nA cylindrical pencil sharpened at one edge is the combination of
\na. cone and cylinder
\nb. Frustum of a cone and cylinder
\nc. A hemisphere and a cylinder
\nd. two cylinders
\nAnswer:
\na. cone and cylinder<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nThe curved surface area of a solid hemisphere of radius r is
\na. 4\u03c0r2<\/sup>
\nb. 2\u03c0r2<\/sup>
\nc. 4\/3\u03c0r3<\/sup>
\nd. 3\u03c0r2<\/sup>
\nAnswer:
\nb. 2\u03c0r2<\/sup><\/p>\n

Question 3.
\nA shuttlecock used for playing badminton has the shape of a combination of
\na. a cone and a sphere
\nb. frustum of a cone and a hemisphere
\nc. a cylinder and a hemisphere
\nd. a cone and a cylinder
\nAnswer:
\nb. frustum of a cone and a hemisphere<\/p>\n

Question 4.
\nThe ratio of the radii of two spheres is 4 : 5. Find the ratio of their total surface areas is.
\na. 5 : 4
\nb. 4 : 5
\nc. 16 : 25
\nd. 25 : 16
\nAnswer:
\nc. 16 : 25<\/p>\n

Question 5.
\nIf the volume of a cube is 1331 cm3<\/sup> the length of its edge is equal to
\na. 12
\nb. 11
\nc. 15
\nd. 13
\nAnswer:
\nb. 11<\/p>\n

Question 6.
\nIf the diameter of a sphere is d, then its volume is
\na. \\(\\frac{1}{3}\\)\u03c0d3<\/sup>
\nb. \\(\\frac{1}{24}\\)\u03c0d3<\/sup>
\nc. \\(\\frac{4}{3}\\)\u03c0d3<\/sup>
\nd. \\(\\frac{1}{6}\\)\u03c0d3<\/sup>
\nAnswer:
\nd. \\(\\frac{1}{6}\\)\u03c0d3<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nThe ratio of the volume of a cube to that of the sphere which will exactly fit inside the cube is
\na. 6 : \u03c0
\nb. 4 : \u03c0
\nc. 2 : \u03c0
\nd. 3 : \u03c0
\nAnswer:
\na. 6 : \u03c0<\/p>\n

Question 8.
\nA cylinder, a cone and hemisphere are of equal base and have the same height the ratio of their volumes is
\na. 3 : 2 : 1
\nb. 2 : 3 : 1
\nc. 3 : 1 : 2
\nd. 1 : 2 : 3
\nAnswer:
\nc. 3 : 1 : 2<\/p>\n

Question 9.
\nA solid sphere of radius r is melted and cast into the shape of a solid cone of height r. The radius of the base of the cone is.
\na. r
\nb. 3r
\nc. 4r
\nd. 2r
\nAnswer:
\nd. 2r<\/p>\n

Question 10.
\nIf a rectangle revolves about one of its sides and completes a full rotation the solid formed is called
\na. sphere
\nb. cylinder
\nc. hemisphere
\nd. cone
\nAnswer:
\nb. cylinder<\/p>\n

Question 11.
\nA hollow cylinder has only
\na. curved surface area
\nb. total surface area
\nc. volume
\nd. Perimeter
\nAnswer:
\na. curved surface area<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nThe formula to find the volume of the cylinder.
\na. \u03c0r2<\/sup>h
\nb. 2\u03c0r2<\/sup>h
\nc. 2\u03c0r(r + l).
\nd. \u03c0rl
\nAnswer:
\na. \u03c0r2<\/sup>h<\/p>\n

Question 13.
\nIs a circular disc is rotated about one of its diameters the solid thus generated is called
\na. cylinder
\nb. sphere
\nc. cone
\nd. frustum
\nAnswer:
\nb. sphere<\/p>\n

Question 14.
\nThe surface area of a sphere is
\na. 4\u03c0r2<\/sup>
\nb. 4\u03c0r3<\/sup>
\nc. \u03c0rl
\nd. 2\u03c0r (r + 1).
\nAnswer:
\na. 4\u03c0r2<\/sup><\/p>\n

Question 15.
\nThe CSA of sphere is
\na. 4\u03c0r2<\/sup>
\nb. 3\u03c0r2<\/sup>
\nc. 2\u03c0r2<\/sup>
\nd. \u03c0r2<\/sup>
\nAnswer:
\na. 4\u03c0r2<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nThe surface areas of two sphere are in the ratio 1 : 4 then the ratio of their volumes is
\na. 1 : 4
\nb. 1 : 8
\nc. 1 : 16
\nd. 1 : 64
\nAnswer:
\nb. 1 : 8<\/p>\n

Question 17.
\nThe base radii if two circular cones of the same height are in the ratio 3:5. Find the ratio of their volumes.
\na. 3 : 5
\nb. 5 : 3
\nc. 9 : 25
\nd. 27: 125
\nAnswer:
\nc. 9 : 25<\/p>\n

Question 18.
\nThe volume of a cone of height 4 cm is 168\u03c0 cm3<\/sup> the radius of the cone is
\na. 6 cm
\nb. 8 cm
\nc. 10 cm
\nd. 12 cm
\nAnswer:
\na. 6 cm<\/p>\n

Question 19.
\nHow many lead balls of radius 2 cm can be made from a ball of radius 4 cm
\na. 1
\nb. 2
\nc. 4
\nd. 8
\nAnswer:
\nd. 8<\/p>\n

Question 20.
\nThe volume of a frustum of a cone of height h and ends radii r1<\/sub> and r2<\/sub> is.
\n\"KSEEB
\nAnswer:
\n\"KSEEB<\/p>\n

Question 21.
\nIf the diameter of a sphere is ‘d’ then its volume is
\na. \\(\\frac{1}{3}\\)\u03c0d3<\/sup>
\nb. \\(\\frac{1}{2}\\)\u03c0d3<\/sup>
\nc. \\(\\frac{4}{3}\\)\u03c0d3<\/sup>
\nd. \\(\\frac{1}{6}\\)\u03c0d3<\/sup>
\nAnswer:
\nd. \\(\\frac{1}{6}\\)\u03c0d3<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 22.
\nThe internal and external radii of a hemispherical bowl are r1<\/sub> and r2<\/sub> respectively the surface area of the bowl is
\n\"KSEEB
\nAnswer:
\n\"KSEEB<\/p>\n

Question 23.
\nThe formula to find the slant height of the cone.
\n\"KSEEB
\nAnswer:
\nb. l = \\(\\sqrt{h^{2}+r^{2}}\\)<\/p>\n

II. Short Answer Questions:<\/span><\/p>\n

Question 1.
\nTwo identical solid hemispheres of equal base radius r cm are struck together along their bases. What will be the total surface area of the combination.
\nAnswer:
\nThe resultant solid will be a sphere of radius r whose total surface area is 4\u03c0r2<\/sup>.<\/p>\n

Question 2.
\nA solid ball is exactly fitted inside the cubical box of side a. What is the volume of the ball?
\nAnswer:
\nDiameter of ball = edge of cube = a
\n\u2234 r = \\(\\frac{a}{2}\\)
\nVolume of a ball
\n\"KSEEB<\/p>\n

Question 3.
\nThe radii of the ends of a frustum of a cone 40 cm high are 20 cm and 11 cm. Find its slant height.
\nAnswer:
\n\"KSEEB<\/p>\n

Question 4.
\nA solid piece of iron in the form of a cuboid of dimension 49 cm \u00d7 33 cm \u00d7 24 cm is melted to form a solid sphere. Find the radius of sphere.
\nAnswer:
\n49 \u00d7 33 \u00d7 24 = 4\/3 \u03c0r3<\/sup>
\n\"KSEEB
\nr = 21 cm<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nWrite the formula to find lateral surface area of frustum of cone.
\nAnswer:
\nLSA = \u03c0l(R + r).<\/p>\n

Question 6.
\nWrite the formula to find volume of spherical shell.
\nAnswer:
\nVolume of spherical shell = 4\/3 \u03c0(R3<\/sup> – r3<\/sup>).<\/p>\n

Question 7.
\nIf two cubes of edge 5 cm each are joined end to end, find the surface area of the resulting cuboid.
\nAnswer:
\ntotal length = (l) = 5 + 5 = 10 cm
\nBreadth = b = 5 cm and
\nheight = h = 5 cm
\nSurface area = 2 (lb + bh + lh)
\n= 2 (10 \u00d7 5 + 5 \u00d7 5 + 5 \u00d7 10)
\n= 2 \u00d7 125
\n= 250 cm2<\/sup><\/p>\n

Question 8.
\nWhat is the ratio of the volume of a cube to that of a sphere which will fit inside it.
\nAnswer:
\nLet edge of the cube be ‘a’
\ndiameter of the sphere will fit inside the given cube = a
\n\u2234 volume of the cube : volume of the sphere
\n\"KSEEB<\/p>\n

Question 9.
\nWrite the formula to find slant height of an frustum cone.
\nAnswer:
\n\"KSEEB<\/p>\n

III. Long Answer Questions:<\/span><\/p>\n

Question 1.
\nA solid hemisphere of wax of radius 12 cm is melted and made into a cone of base radius 6 cm. Calculate the height of the cone.
\nAnswer:
\nRadius of the hemisphere, r1<\/sub> = 12 cm
\nRadius of the base of the cone, r2<\/sub> = 6 cm.
\nHeight of the cone = h = ?
\nvolume of wax cone = volume of wax hemisphere.
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\n21 spheres of equal radii are melted to form a cylinder of radius 14 cm and height 49 cm. Find the radius of sphere.
\nAnswer:
\nradius of cylinder = r1<\/sub> = 14 cm
\nheight h = 49 cm and radius of sphere = r2<\/sub>
\nTotal volume of 21 spheres = volume of cylinder.
\n\"KSEEB
\n\u2234 radius of the sphere is 7 cm.<\/p>\n

Question 3.
\nA hemispherical bowl of internal radius 9 cm is full of with a liquid, this liquid is to be filled into cylindrical shaped bottles of diameter 3 cm and height 4 cm. How many bottles are necessary to empty the bowl?
\nhemisphere r1<\/sub> = 9 cm
\nCylinder r2<\/sub> = \\(\\frac{d}{2}=\\frac{3}{2}\\) cm and h = 4cm
\nNumber of bottles
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nA bucket is in the shape of a frustum with the top and bottom circles of radii 15 cm and 10 cm. Its depth is 12 cm. Find its curved surface area and total surface area. (Express the answer in terms of \u03c0).
\nAnswer:
\nr1<\/sub> = 15 cm, r2<\/sub> = 10 cm and h = 12 cm
\n\"KSEEB
\nl = 13cm
\nCSA of a bucket = \u03c0(r1<\/sub> + r2<\/sub>)l
\n= \u03c0(15 + 10) 13
\n= \u03c0 \u00d7 25 \u00d7 13
\nTSA of a bucket
\n\"KSEEB
\n= \u03c0{25 \u00d7 13 + 225 + 100}
\n= \u03c0{325 + 325}
\n= 650\u03c0 sq cm<\/p>\n

Question 5.
\nFrom the top of a cone of base radius 24 cm and height 45 cm, a cone of slant height 45 cm, a cone of slant height 17 cm is cut off. What is the volume of the remaining frustum of cone.
\nAnswer:
\n\"KSEEB
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nA vessel is in the form of a frustum of a cone. Its radius at one end is 8 cm and the height is 14 cm. If its volume is
\n\\(\\frac{5676}{3}\\)cm3<\/sup>, find the radius of the other end.
\nAnswer:
\n\"KSEEB
\n\"KSEEB
\n\u2234 Radius of the other end is 5 cm<\/p>\n

Question 7.
\nA solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones each of diameter 4\\(\\frac{2}{3}\\)cm and height 3 cm. Find the number of cones so formed.
\nAnswer:
\nDiameter of sphere = 28 cm,
\nradius = r1<\/sub> = \\(\\frac{28}{2}\\) = 14cm
\n\"KSEEB
\n\"KSEEB
\n= 672 cones.<\/p>\n

Question 8.
\nA petrol tank is in the shape of a cylinder with hemispheres of same radius attached to both ends. If the total length of the tank is 6m and the . radius is 1m, what is the capacity of the tank in litres.
\nAnswer:
\n\"KSEEB
\nRadius of hemisphere = r = 1 m
\nRadius of cylinder = r = 1 m
\nheight of the cylinder = 4 m
\nCapacity of the tank = volume of 2 hemispheres + volume of cylinder
\n\"KSEEB
\n= 16.7619 m3<\/sup>. [1 m = 100 cm]
\n= 16761900 cm3<\/sup> [1000 cm3<\/sup> = 1lit]
\n= \\(\\frac{16761900}{1000}\\)
\n= 16761.9 litres.<\/p>\n

Question 9.
\nA rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other end. All of them are the same radius of 1.5 m. The total length of the rocket is 7m and height of the cone is 2m. Calculate the volume of the rocket.
\nAnswer:
\n\"KSEEB
\nRadius of the hemisphere and cylinder r = 1.5 m
\nheight of cylinder = h = 7 – (2 + 1.5)
\nh = 3.5m
\nvolume of Rocket = volume of hemisphere + volume of cone + volume of cylinder
\n\"KSEEB<\/p>\n

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

Students can Download Class 10 Maths Chapter 15 Surface Areas and Volumes Additional Questions, 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 Additional Questions I. Multiple Choice …<\/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\/25119"}],"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=25119"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25119\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25119"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25119"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25119"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}