{"id":29159,"date":"2021-07-15T19:44:11","date_gmt":"2021-07-15T14:14:11","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=29159"},"modified":"2022-03-02T10:29:59","modified_gmt":"2022-03-02T04:59:59","slug":"ncert-solutions-for-class-10-maths-chapter-13-ex-13-5","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-13-ex-13-5\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 13 Surface Areas and Volumes Ex 13.5 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nA copper wire, 3 mm in diameter, is wound about in cylinder whose length is 12 cm, and diameter 10 cm, so as the cover the curve surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm\u00b3.
\nSolution:
\nOne round of wire covers 3 m = \\(\\frac { 3 }{ 10 }\\) cm in thickness of the surface of the cylinder.
\nLength of the cylinder = 12 cm
\nNumber of the rounds to cover 12 cm
\n\"NCERT
\nDiameter of the cylinder = 10 cm.
\nRadius = 5 cm
\nLength of the wire in completing one round
\n= 2 x \u03c0 x cm = 10\u03c0 cm
\nLength of the wire in covering the whole surface = Length of the wire in completely 40 rounds
\n(10\u03c0 x 40) cm = 400\u03c0 cm
\n= (400 x 3.14) cm.
\n= 1256 cm
\n= 125.6 m
\nRadius of copper wire = \\(\\frac { 3 }{ 20 }\\) cm
\nVolume of wire = \u03c0 x r\u00b2h
\n= \u03c0 x \\(\\frac { 3 }{ 20 }\\) x \\(\\frac { 3 }{ 20 }\\) x 400\u03c0
\n= 9\u03c0\u00b2 cm\u00b3.
\nSo, mass of the wire = 9\u03c0\u00b2 x 8,88 per cm\u00b3.
\n= 787.98 gm
\nLength of wire = 12.56 m and mass of the wire = 787.98 gm.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nA right triangle, whose sides 3 cm and 4 cm (other than hypotenuse) is made to resolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of \u03c0 as found appropriate),
\nSolution:
\nLet \u0394ABC be the right angle triangled at B whose sides AB and BC are 3 cm, 4 cm respectively.
\nLength of hypotenuse = \\(\\sqrt{(3)^{2}+(4)^{2}}\\) = \\(\\sqrt{25}\\) = 5 cm.
\n\"NCERT
\nBO and DO are the common base of double cone formed by revolving the right triangle about AC,
\nSlant height = 3 cm (for cone ABD)
\nSlant height = 4 cm (for cone BCD)
\n(\u2235 AOD = 90 = ADC and DAC is common)
\n\\(\\frac { AO }{ 3 }\\) = \\(\\frac { 3 }{ 4 }\\) = \\(\\frac { OD }{ 4 }\\)
\n\u21d2 AO = \\(\\frac { 9 }{ 5 }\\) = 1.8 cm
\nh = 1.8 cm
\nCO = AC – OA
\n\u21d2 CO = 5 – 18 = 3.2 cm (H = 3.2 cm)
\nSimilarly OD = \\(\\frac { 12 }{ 5 }\\) 2.4 cm = r
\nNow, volume of double cone
\n\"NCERT
\nCurved surface area of double cone
\n= \u03c0r(l1<\/sub>) + \u03c0r(l2<\/sub>)
\n= \u03c0r(3) + \u03c0r(4)
\n– 3.14 x 2.4 (7)
\n= 52.75 cm\u00b2.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nA cistern, internally measuring 150 cm x 120 cm x 110 cm, has 129600 cm\u00b3 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 being 22.5 cm x 7.5 cm x 6.5 cm
\nSolution:
\nArea of cistern = 150 cm x 120 cm x 110 cm – 1980000 cm\u00b3.
\nEmpty volume of cistern = (1980000 – 129600)cm\u00b3.
\nLet number of bricks dropped in the water be n
\nSo the volume of bricks = (22.5 7.5 x 6.5) n = 1096.875 cm\u00b3.
\nThe volume of water absorbed by each brick = \\(\\frac { 1 }{ 17 }\\) x (1096.S95) cm\u00b3.
\nVolume of total bricks = (1096.875) \\(\\frac { n }{ 17 }\\) cm\u00b3.
\nSo, 1850400 cm\u00b3 = 1096.875
\nn = (1096.875) \\(\\frac { n }{ 17 }\\)
\n\"NCERT
\n\u2234 Number of bricks = 1792.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nIn one fortnight of a given month, there was rainfall of 10 cm in a river valley. If the area of one valley is 97280 km\u00b2, 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.
\nSolution:
\nVolume of a river = 1072 km x 75 m x 3 m
\n= 1072000 m x 75 m 3 m
\n= 241200000 m\u00b3
\nVolume of 3 rivers
\n= 3 x 241200000
\n= 723600000 m\u00b3.
\nNow value of rainfall in the valley
\n= height of rainfall x Area
\n= 10 cm x 97280 km\u00b2.
\n= \\(\\frac { 10 }{ 100 }\\) m x 97280 x (100)\u00b2 m\u00b2.
\n= \\(\\frac { 10 }{ 100 }\\) x 97280000000
\n= 9728000000 m\u00b3.
\nNow 9728000000 m\u00b3 > 73360000 m\u00b3.
\nHence, the additional water i,e. of the 3 river cannot be equivalent to the rainfall.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nAn oil funnel made of tin sheet consist 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 diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (See the adjoining figure).
\n\"NCERT
\nSolution:
\nLet l be the slant height of the frustum part of the funnel.
\nThen, l = \\(\\sqrt{(9-4)^{2}+12^{2}}\\) = \\(\\sqrt{25+144}\\) cm
\n= \\(\\sqrt{169}\\) cm = 13 cm
\nNow, Tin required = Curved surface area of cylindrical portion + Curved surface area of frustum potion
\n= 2\u03c0r1<\/sub>h + \u03c0(r1<\/sub> + r2<\/sub>)h
\n= [2\u03c0 x 4 x 10 + \u03c0 (4 + 9) x 13] cm\u00b2
\n= (80\u03c0 + 169\u03c0) cm\u00b2 = 249\u03c0 cm\u00b2
\n= \\(\\left(249 \\times \\frac{22}{7}\\right)\\) cm\u00b2
\n= \\(\\frac { 5478 }{ 7 }\\)
\n= 782 \\(\\frac { 4 }{ 7 }\\) cm\u00b2.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nDerive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in section 13.5, using the symbols as explained.
\nSolution:
\nAPO’B and APQR are similar
\n\"NCERT
\nNow curved surface area of the fur stum is given by
\n= curved surface area of cone PQR (bigger)- curved surface and of cone PAB (smaller).
\n= \u03c0r1<\/sub>l1<\/sub> – \u03c0r2<\/sub>l2<\/sub> … (3)
\n\u2234 From (2) and (3)
\nCurved surface area of the frustum
\n\"NCERT
\n= \u03c0(r1<\/sub> + r2<\/sub>)l [Using (2)]
\n\u2234 Total surface area of the frustum curved surface area of the frustum + area of both circular bases.
\n= \u03c0(r1<\/sub> + r2<\/sub>) l + \u03c0r\u00b21<\/sub> + \u03c0r\u00b22<\/sub>
\n= \u03c0(r1<\/sub> + r2<\/sub>) l + r\u00b21<\/sub> + r\u00b22.<\/sub><\/p>\n

\"NCERT<\/p>\n

Question 7.
\nDerive the formula for the volume of the frustum of a cone, given to you in section 13.5, using the symbols as explained.
\nSolution:
\nSince \u0394PO’B and \u0394PQR are similar
\n\"NCERT
\nNow volume of the frustum = Volume of cone PQR (bigger) – Voolume of cone PAQ (smaller)
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5 Question 1. A copper wire, 3 mm in diameter, is wound about in cylinder whose …<\/p>\n

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