A petrol tank is in the form of a frustum of a cone of height 20 m with diameters of its lower and upper ends as 20 m and 50 m respectively. Find the cost of petrol which can fill the tank completely at the rate of Rs. 70 per litre. Also find the surface area of the tank.

A petrol tank is in the form of a frustum of a cone of height 20 m

Question 38 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 2
Question 38 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 3 Question 38 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 4

 

Note : This is similar to Ex 13.4, 4 of NCERT – Chapter 13 Class 10

Check the answer here https://www.teachoo.com/1915/554/Ex-13.4--4---A-container--opened-from-top-and-made-up-of/category/Ex-13.4/

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Transcript

Question 38 (OR 1st question) A petrol tank is in the form of a frustum of a cone of height 20 m with diameters of its lower and upper ends as 20 m and 50 m respectively. Find the cost of petrol which can fill the tank completely at the rate of Rs. 70 per litre. Also find the surface area of the tank. In order to find Cost of petrol, we need to find volume (in litres) Volume of container = Volume of frustum = 1/3 πœ‹β„Ž(π‘Ÿ12+π‘Ÿ22+π‘Ÿ1π‘Ÿ2) Here h = height = 20 m r1 = radius of upper end = 25 m r2 = radius of lower end = 10 m Volume of container = 1/3 πœ‹β„Ž(π‘Ÿ12+π‘Ÿ22+π‘Ÿ1π‘Ÿ2) = 1/3Γ—3.14Γ—20Γ—(252+γ€–10γ€—^2+25Γ—10) = (3.14 Γ— 20)/3 (625 + 100 + 250) = (3.14 Γ— 20)/3 Γ— 975 = 3.14 Γ— 20 Γ— 325 = 20410 m3 = 20410 Γ— 1000 litre = 2,04,10,000 litre Now , Cost of 1 litre petrol = Rs 70 Cost of 2,04,10,000 litre petrol = Rs 70 Γ— 20410000 1 m3 = 1000 litre = Rs 1428700000 Now, We need to find surface area of tank Surface Area of tank = Area of frustum = πœ‹(π‘Ÿ1+π‘Ÿ2)𝑙 Here, r1 = 20 cm , r2 = 8 cm We need to find l We know that 𝑙 = √(β„Ž2+(π‘Ÿ1βˆ’π‘Ÿ2)2) 𝑙 = √(202+(25βˆ’10)2) 𝑙 = √(202+(15)2) 𝑙 = √(400+225) 𝑙 = √625 𝑙 = √(γ€–25γ€—^2 ) 𝑙 = 25 cm Area of frustum = πœ‹(π‘Ÿ1+π‘Ÿ2)𝑙 = 3.14 Γ— (25+10)Γ—25 = 3.14 Γ— 35Γ—25 = 2747.5 cm2

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.