{"id":10119,"date":"2024-02-26T09:16:58","date_gmt":"2024-02-26T03:46:58","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=10119"},"modified":"2024-02-27T17:45:41","modified_gmt":"2024-02-27T12:15:41","slug":"ts-10th-class-maths-solutions-chapter-10-ex-10-4","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-maths-solutions-chapter-10-ex-10-4\/","title":{"rendered":"TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.4"},"content":{"rendered":"

Students can practice\u00a010th Class Maths Study Material Telangana<\/a> Chapter 10 Mensuration Ex 10.4 to get the best methods of solving problems.<\/p>\n

TS 10th Class Maths Solutions Chapter 10 Mensuration Exercise 10.4<\/h2>\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 cylinder. (AS4)
\nSolution:
\nRadius of the sphere(r) = 4.2 cm
\nVolume of the sphere = \\(\\frac{4}{3}\\) \u03c0r3<\/sup>
\n= \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 4.2 \u00d7 4.2 \u00d7 4.2
\n= \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{42}{10}\\) \u00d7 \\(\\frac{42}{10}\\) \u00d7 \\(\\frac{42}{10}\\) cm3<\/sup>
\nRadius of the cylinder (r) = 6 cm
\nHeight of the cylinder (h) = ?
\nVolume of the cylinder = \u03c0r2<\/sup>h
\n= \\(\\frac{22}{7}\\) \u00d7 6 \u00d7 6 \u00d7 h cm3<\/sup>
\nBy problem, volume of the metallic sphere = volume of the cylinder
\n\u2234 \\(\\frac{22}{7}\\) \u00d7 6 \u00d7 6 \u00d7 h = \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{42}{10}\\) \u00d7 \\(\\frac{42}{10}\\) \u00d7 \\(\\frac{42}{10}\\)
\n\u2234 h = \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{42}{10}\\) \u00d7 \\(\\frac{42}{10}\\) \u00d7 \\(\\frac{42}{10}\\) \u00d7 \\(\\frac{7}{22}\\) \u00d7 \\(\\frac{1}{6 \\times 6}\\) = 2.74 cm
\nHeight of the cylinder = 2.74 cm.<\/p>\n

\"TS<\/p>\n

Question 2.
\nThree metallic spheres of radii 6 cm; 8 cm. and 10 cm respectively are melted to from a single solid sphere. Find the radius of resulting sphere. (AS1<\/sub>)
\nSolution:
\nGiven : Radii of the three spheres
\nr1<\/sub> = 6 cm
\nr2<\/sub> = 8 cm
\nr3<\/sub> = 10 cm
\nThese three are melted to form a single sphere. Let the radius of the resulting sphere be V Then volume of the resultant sphere = sum of the the volumes of the three small spheres.
\n\u21d2 \\(\\frac{4}{3}\\)\u03c0r3<\/sup> = \\(\\frac{4}{3}\\)\u03c0r1<\/sub>3<\/sup> + \\(\\frac{4}{3}\\)\u03c0r2<\/sub>3<\/sup> + \\(\\frac{4}{3}\\)\u03c0r3<\/sub>3<\/sup>
\n\u21d2 \\(\\frac{4}{3}\\)\u03c0r3<\/sup> = \\(\\frac{4}{3}\\)\u03c0(r1<\/sub>3<\/sup> + r2<\/sub>3<\/sup> + r3<\/sub>3<\/sup>)
\nr3<\/sup> = r1<\/sub>3<\/sup> + r2<\/sub>3<\/sup> + r3<\/sub>3<\/sup>
\n= 63<\/sup> + 83<\/sup> + 103<\/sup>
\n= 216 + 512 + 1000 = 1728.
\n\u2234 1728 = (2 \u00d7 2 \u00d7 3) \u00d7 (2 \u00d7 2 \u00d7 3) \u00d7 (2 \u00d7 2 \u00d7 3)
\nr3<\/sup> = 12 \u00d7 12 \u00d7 12
\nr3<\/sup> = 123<\/sup>
\n\u2234 r = 12
\nThus the radius of the resultant sphere = 12 cm.<\/p>\n

\"TS<\/p>\n

Question 3.
\nA 20m deep well of diameter 7m is dug and the earth got by digging is evenly spread out to form a platform 22 m of base 14 m. Find the height of the platform. (AS4<\/sub>)
\nSolution:
\nDiameter of the well (d) = 7 m
\nRadius the well (r) = \\(\\frac{7}{2}\\) m
\nDepth of the well (h) = 20 m
\nQuantity of the earth dig out = Volume of the cylindrical well
\n= \u03c0r2<\/sup>h = \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{7}{2}\\) \u00d7 \\(\\frac{7}{2}\\) \u00d7 20
\n= 770 cm3<\/sup>
\nArea of the platform on which the earth is spread out = 22 \u00d7 14 = 308 cm2<\/sup>
\nLet the height of the platform be = x cm
\n\u2234 308 \u00d7 x = 770
\n\u21d2 x = \\(\\frac{770}{308}\\) = 2.5 m
\nHence, the height of the platform = 2.5m<\/p>\n

Question 4.
\nA well of diameter 14 m. is dug 15m. deep. The earth taken out of it has been spread evenly to form circular embankment of width 7m. Find the height of the embankment. (AS4<\/sub>)
\nSolution:
\nDiameter of the well (d) = 14m
\nRadius of the well = \\(\\frac{\\mathrm{d}}{2}\\) = \\(\\frac{14}{2}\\) = 7m
\nDepth of the well (h) = 15m
\n\"TS
\nQuantity of the earth dug out = \u03c0r2<\/sup>h
\n= \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7 \u00d7 15
\nArea of circular embankment = \u03c0(R + r) (R – r)
\n= 7(14 + 7)(14 – 7)
\n= \\(\\frac{22}{7}\\) \u00d7 21 \u00d7 7 = 462 m2<\/sup>
\nLet the height of the embankment be ‘x’ m.
\n462 \u00d7 x = \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7 \u00d7 15
\n\u2234 x = \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7 \u00d7 15 \u00d7 \\(\\frac{1}{462}\\)
\nHence, the height of the embankment = 5m.<\/p>\n

\"TS<\/p>\n

Question 5.
\nA Container shaped like a right circular cylinder having diameter 12 cm. and height 15cm. 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 die number of such cones which can be filled with the ice cream. (AS4<\/sub>)
\nSolution:
\nDiameter of the container (d) = 12 km
\nRadius of the cylindrical container (r) = 6 cm
\nHeight of the cylindrical container (h) = 15 cm
\n\"TS
\nVolume of the cylinder = \u03c0r2<\/sup>h
\n= \\(\\frac{22}{7}\\) \u00d7 6 \u00d7 6 \u00d7 15 cm3<\/sup> = \\(\\frac{11880}{7}\\) cm3<\/sup>
\nDiameter of the base of the cone(d) = 6 cm
\nRadius of the base of the cone (r) = \\(\\frac{\\mathrm{d}}{2}\\) = \\(\\frac{6}{2}\\) = 3 cm
\nVolume of the conical part = \\(\\frac{1}{3}\\) \u00d7 \u03c0r2<\/sup>h
\n= \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 3 \u00d7 3 \u00d7 12
\n= \\(\\frac{792}{7}\\) cm3<\/sup>
\nRadius of the hemisphere(r) = 3 cm
\n\"TS
\nRadius of the hemisphere (r) = 3 cm
\nVolume of the hemi-sphere = \\(\\frac{2}{3}\\) \u03c0r3<\/sup>
\n= \\(\\frac{2}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 3 \u00d7 3 \u00d7 3 = \\(\\frac{396}{7}\\) cm3<\/sup>
\nTotal volume of the conical part and hemispherical part
\n= \\(\\frac{792}{7}\\) + \\(\\frac{396}{7}\\) = \\(\\frac{1188}{7}\\) cm3<\/sup>
\nNumber of cones which can be filled with ice cream = \\(\\frac{11880}{7}\\) + \\(\\frac{1188}{7}\\)
\n(Q Number of icecream cones = \\(\\frac{\\text { Volume of the cylinder }}{\\text { Total volume of ice – cream cone }} \\)
\n= \\(\\frac{11880}{7}\\) \u00d7 \\(\\frac{7}{1188}\\) = 10<\/p>\n

Question 6.
\nHow many silver coins, 1.75 cm in diameter and thickness 2mm, need to be melted to form a cuboid of dimensions 5.5 cm \u00d7 10 cm \u00d7 3.5 cm ? (AS4<\/sub>)
\nSolution:
\nLet the number of silver coins needed to melt = n
\nthen total volume of n coins = volume of the cuboid.
\nn \u00d7 \u03c0r2<\/sup>h = lbh
\n[\u2235 The shape of the coins is a cylinder then v = \u03c0r2<\/sup>h]
\nn \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\left[\\frac{1.75}{2}\\right]^2\\) \u00d7 \\(\\frac{2}{10}\\) = 5.5 \u00d7 10 \u00d7 3.5
\n[\u2235 2mm = \\(\\frac{2}{10}\\) cm, r = \\(\\frac{\\mathrm{d}}{2}\\)]
\nn \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{1.75}{2}\\) \u00d7 \\(\\frac{1.75}{2}\\) \u00d7 \\(\\frac{2}{10}\\) = 55 \u00d7 3.5
\nn = 55 \u00d7 3.5 \u00d7 \\(\\frac{7 \\times 2 \\times 2 \\times 10}{22 \\times 1.75 \\times 1.75 \\times 2}\\)
\n= \\(\\frac{55 \\times 35 \\times 7 \\times 4}{22 \\times 2 \\times 1.75 \\times 1.75}\\) = \\(\\frac{5 \\times 35 \\times 7}{1.75 \\times 1.75}\\)
\n= \\(\\frac{175 \\times 7}{1.75 \\times 1.75}\\) = \\(\\frac{100 \\times 1}{0.25}\\) = 400
\n\u2234 400 Silver coins are needed.<\/p>\n

\"TS<\/p>\n

Question 7.
\nA vessel is in the form of an inverted cone. Its height is 8 cm. and the radius of its top is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, \\(\\frac{1}{4}\\) of the water flows out. Find the number of lead shots dropped into the vessel. (AS4<\/sub>)
\nSolution:
\nLet the number of lead shots dropped = n
\nthen the total volume of n – lead shots
\n= \\(\\frac{1}{4}\\) volume of the conical vessel
\n\"TS
\nLeadshots : Radius (r) = 0.5 cm
\nVolume (V) = \\(\\frac{4}{3}\\) \u03c0r3<\/sup>
\n= \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.5 \u00d7 0.5 \u00d7 0.5
\nTotal volume of n-shots
\n= n \u00d7 \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.125
\nCone : Radius (r) = 5 cm
\nheight (h) = 8 cm
\nVolume (V) = \\(\\frac{1}{3}\\) \u03c0r2<\/sup>h
\n= \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 5 \u00d7 5 \u00d7 8
\n= \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{1}{3}\\) \u00d7 200
\n\\(\\frac{1}{4}\\)th volume = \\(\\frac{1}{4}\\) \u00d7 \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 200
\n\u2234 n \u00d7 \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 0.125 = \\(\\frac{1}{4}\\) \u00d7 \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 200
\nn = \\(\\frac{1}{4}\\) \u00d7 \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 200 \u00d7 \\(\\frac{3}{4}\\) \u00d7 \\(\\frac{7}{22}\\) \u00d7 \\(\\frac{1}{0.125}\\)
\n= \\(\\frac{200 \\times 1}{4 \\times 4 \\times 0.125}\\) = \\(\\frac{200}{2}\\) = 100
\n\u2234 Number of lead shots = 100<\/p>\n

\"TS<\/p>\n

Question 8.
\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. (AS4<\/sub>)
\nSolution:
\nDiameter of the solid metallic sphere (d) = 28 cm
\n\u2234 Radius of the sphere (r) = \\(\\frac{\\mathrm{d}}{2}\\) = \\(\\frac{2 r}{2}\\) = 14 cm
\nVolume of the sphere = \\(\\frac{4}{3}\\) pr3<\/sup>
\n= \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 14 \u00d7 14 \u00d7 14
\n= \\(\\frac{176 \\times 196}{3}\\) cm3<\/sup>
\nCone : Diameter of the cone = 4\\(\\frac{2}{3}\\) cm
\n= \\(\\frac{14}{3}\\) cm
\n\u2234 Radius of the cone (r) = \\(\\frac{14}{3}\\) \u00d7 \\(\\frac{1}{2}\\) = \\(\\frac{7}{3}\\) cm
\nHeight of the cone (h) = 3 cm
\nVolume of the cone = \\(\\frac{1}{3}\\) \u03c0r2<\/sup>h
\n= \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{7}{3}\\) \u00d7 \\(\\frac{7}{3}\\) \u00d7 3
\n= \\(\\frac{154}{9}\\) cm3<\/sup>
\nMetallic sphere is recast into smaller cones.
\nTherefore, the number of cones formed
\n= \\(\\frac{176 \\times 196}{3}\\) \u00f7 \\(\\frac{154}{9}\\)
\n= \\(\\frac{176 \\times 196}{3}\\) \u00d7 \\(\\frac{9}{154}\\) = 672
\nHence, 672 smaller cones are formed.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can practice\u00a010th Class Maths Study Material Telangana Chapter 10 Mensuration Ex 10.4 to get the best methods of solving problems. TS 10th Class Maths Solutions Chapter 10 Mensuration Exercise 10.4 Question 1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[16],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10119"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=10119"}],"version-history":[{"count":1,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10119\/revisions"}],"predecessor-version":[{"id":10256,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10119\/revisions\/10256"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=10119"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=10119"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=10119"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}