{"id":28323,"date":"2021-07-05T12:30:02","date_gmt":"2021-07-05T07:00:02","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=28323"},"modified":"2021-07-05T12:30:02","modified_gmt":"2021-07-05T07:00:02","slug":"ncert-solutions-for-class-8-maths-chapter-11-ex-11-4","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-11-ex-11-4\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 11 Mensuration Ex 11.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.4<\/h2>\n

Question 1.
\nGiven a cylindrical tank, in which situation will you find surface area and in which situation volume.
\n(a) To find how much it can hold.
\n(b) Number of cement bags required to plaster it.
\n(c) To find the number of smaller tanks that can be filled with water from it.
\n\"NCERT
\nAnswer:
\n(a) Volume
\n(b) Surface area
\n(c) volume<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nDiameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
\n\"NCERT
\nAnswer:
\nVolume of cylinder B will be greater than the volume of cylinder A.
\nFor cylinder A
\nr = \\(\\frac { 7 }{ 2 }\\) cm, h = 14 cm
\nVolume of the cylinder A = \u03c0r2<\/sup> h cm units
\n= \\(\\frac{22}{7}\\) \u00d7 \\(\\frac { 7 }{ 2 }\\) \u00d7 \\(\\frac { 7 }{ 2 }\\) \u00d7 14 cm3<\/sup>
\n= 11 \u00d7 7 \u00d7 7 cm3<\/sup> = 539 cm3<\/sup>
\nSurface Area of cylinder A = 2\u03c0 r(h + r)
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 \\(\\frac { 7 }{ 2 }\\) 14 + \\(\\frac { 7 }{ 2 }\\)
\n= 22 \u00d7 \\(\\frac { 35 }{ 2 }\\) = 385 cm2<\/sup><\/p>\n

For cylinder B
\nHere r = \\(\\frac { 14 }{ 2 }\\) = 7 cm , h = 7 cm
\nVolume of the cylinder B = \u03c0 r2<\/sup> h
\n= \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7 \u00d7 7 = 1078 cm3<\/sup>
\nSurface area of cylinder = 2 \u03c0 r (h + r)
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 7(7 + 7) = 44 \u00d7 14 = 616 cm2<\/sup>
\nThus, cylinder B has greater volume than cylinder A.
\nThus, the cylinder of great capacity has greater surface area.<\/p>\n

Question 3.
\nFind the height of a cuboid whose base area is 180 cm2<\/sup> and volume is 900 cm3<\/sup>?
\nAnswer:
\nBase area of the cuboid = 180 cm2<\/sup>
\n\u2234 lb = 180
\nVolume of the cuboid = 900 cm3<\/sup>
\nBase Area \u00d7 height = 900
\nlb \u00d7 h = 900
\n180 \u00d7 h = 900
\nh= \\(\\frac{900}{180}\\) = 5
\n\u2234 Height of the cuboid = 5 cm<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nA cuboid is of dimensions 60 cm \u00d7 54 cm \u00d7 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
\nAnswer:
\nHere l = 60 cm, b = 54 cm, h = 30 cm
\nVolume of the cuboid = l \u00d7 b \u00d7 h cm3<\/sup> = 60 \u00d7 54 \u00d7 30 cm3<\/sup>
\nSide of a cube = 6 cm
\nVolume of cube = l3<\/sup> cm3<\/sup> = 6 \u00d7 6 \u00d7 6 cm3<\/sup>
\nRequired number of small cubes
\n= \\(\\frac{\\text { Volume of the cuboid }}{\\text { Volume of the small cube }}\\)
\n= \\(\\frac{60 \\times 54 \\times 30}{6 \\times 6 \\times 6}\\) = 10 \u00d7 9 \u00d7 5 = 450
\nNumber of small cubes = 450.<\/p>\n

Question 5.
\nFind the height of the cylinder whose volume is 1.54 m3<\/sup> and diameter of the base is 140 cm?
\nAnswer:
\nRadius of the cylinder = \\(\\frac { 140 }{ 2 }\\) = 70 cm
\nLet the height of the cylinder be \u201ch\u201d
\nVolume of the cylinder = 1.54 m3<\/sup>
\n= 1.54 \u00d7 100 \u00d7 100 \u00d7 100 cm3<\/sup>
\n\u03c0r2<\/sup> h = 154 \u00d7 100 \u00d7 100 cm3<\/sup>
\nh = \\(\\frac{22}{7}\\) \u00d7 70 \u00d7 70 \u00d7 h = 154 \u00d7 100 \u00d7 100
\n= 100 cm = 1 m
\n\u2234 height of the cylinder = 1m<\/p>\n

Question 6.
\nA milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?
\n\"NCERT
\nAnswer:
\nRadius of the tank = 1.5 m = \\(\\frac{15}{10}\\) m
\nLength of the tank = 7 m
\nVolume of the tank = \u03c0r2<\/sup> h cu. m
\n= \\(\\frac{22}{7}\\) \u00d7 \\(\\frac{15}{10}\\) \u00d7 \\(\\frac{15}{10}\\) \u00d7 7m3<\/sup>
\n= \\(\\frac{22 \\times 15 \\times 15}{10 \\times 10}\\) \u00d7 1000 litres
\n\\(\\frac{495}{10}\\) \u00d7 1000 litres = 49500 litres
\nQuantity of milk in the tank = 49500 litres<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nif each edge of a cube is doubled.
\n(i) how many times will its surface area increase?
\n(ii) how many times will its volume increase?
\nAnswer:
\nLet the original edge be \u2018x\u2019
\n\u2234 increased edge 2x
\n\u2234 original surface area = 6x2<\/sup>
\nincreased surface area = 6(2x)2<\/sup> = 24x2<\/sup>
\n= \\(\\frac{\\text { Increase surface area }}{\\text { Original surface area }}\\) = \\(\\frac{24 x^{2}}{6 x^{2}}\\)
\nSurface area is increased 4 times.<\/p>\n

(ii) Original volume = x3<\/sup>
\nIncreased volume = (2x)3<\/sup>
\n= \\(\\frac{\\text { Increased volume }}{\\text { Original volume }}\\) = \\(\\frac{8 x^{3}}{x^{3}}\\)
\nVolume is increased by 8 times<\/p>\n

Question 8.
\nWater is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m3<\/sup>. Find the number of hours it will take to fill the reservoir.
\nAnswer:
\nVolume of the reservoir = 108 m3<\/sup>
\n= 108 \u00d7 1000 litres = 108000 litres
\nAmount of water poured in one minute = 60 litres
\nTime taken to fill the reservoir = \\(\\frac{108000}{60}\\)
\n= 1800 minutes \\(\\frac{1800}{60}\\) hours = 30 hours
\nThe required number of hours = 30.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.4 Question 1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume. (a) To …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-11-ex-11-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4 Questions and Answers are prepared by our highly skilled subject experts. 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