Ex 6.1

Ex 6.1, 1
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Ex 6.1,2 Deleted for CBSE Board 2022 Exams

Ex 6.1,3 Deleted for CBSE Board 2022 Exams

Ex 6.1,4 Important Deleted for CBSE Board 2022 Exams

Ex 6.1,5 Important Deleted for CBSE Board 2022 Exams

Ex 6.1,6 Deleted for CBSE Board 2022 Exams

Ex 6.1,7 Deleted for CBSE Board 2022 Exams

Ex 6.1,8 Deleted for CBSE Board 2022 Exams

Ex 6.1,9 Deleted for CBSE Board 2022 Exams

Ex 6.1,10 Important Deleted for CBSE Board 2022 Exams

Ex 6.1,11 Important Deleted for CBSE Board 2022 Exams

Ex 6.1,12 Deleted for CBSE Board 2022 Exams

Ex 6.1,13 Important Deleted for CBSE Board 2022 Exams You are here

Ex 6.1,14 Deleted for CBSE Board 2022 Exams

Ex 6.1,15 Important Deleted for CBSE Board 2022 Exams

Ex 6.1,16 Deleted for CBSE Board 2022 Exams

Ex 6.1,17 (MCQ) Deleted for CBSE Board 2022 Exams

Ex 6.1, 18 (MCQ) Important Deleted for CBSE Board 2022 Exams

Last updated at April 19, 2021 by Teachoo

Ex 6.1, 13 A balloon, which always remains spherical, has a variable diameter 3/2 (2𝑥 +1). Find the rate of change of its volume with respect to 𝑥.Let d be the diameter of the balloon Given that Diameter = d = 3/2 (2x + 1) Let r be the radius of the balloon r = 𝑑/2 = 𝟑/𝟒 (2x + 1) The balloon is a spherical Volume of the balloon = 4/3 𝜋𝑟^3 We need to find rate of change of volume with respect to x i.e. 𝑑𝑉/𝑑𝑥 Now, 𝑑𝑉/𝑑𝑥 = 𝑑/𝑑𝑥 (4/3 𝜋𝑟^3 ) = 4𝜋/3 × (𝑑𝑟^3)/𝑑𝑥 = 4𝜋/3 × 𝑑/𝑑𝑥 (27/64 (2𝑥+1)^3 ) = 9𝜋/16 × (𝑑(2𝑥 + 1)^3)/𝑑𝑥 = 9𝜋/16 × 3(2x + 1)2 × 2 = 𝟐𝟕𝝅/𝟖 (𝟐𝒙+𝟏)^𝟐