
Ex 6.4
Ex 6.4, 1 (ii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (iii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (iv) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (v) Important Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (vi) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (vii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (viii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (ix) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (x) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xi) Important Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xiii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xiv) Important Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xv) Deleted for CBSE Board 2022 Exams
Ex 6.4,2 Deleted for CBSE Board 2022 Exams
Ex 6.4,3 Important Deleted for CBSE Board 2022 Exams
Ex 6.4,4 Deleted for CBSE Board 2022 Exams You are here
Ex 6.4,5 Important Deleted for CBSE Board 2022 Exams
Ex 6.4,6 Deleted for CBSE Board 2022 Exams
Ex 6.4,7 Deleted for CBSE Board 2022 Exams
Ex 6.4,8 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 6.4,9 (MCQ) Deleted for CBSE Board 2022 Exams
Last updated at April 15, 2021 by Teachoo
Ex 6.4, 4 Find the approximate change in the volume V of a cube of side x meters caused by increasing the side by 1%.Let side of the cube = x metres Increase in side = 1% = 0.01 x Hence, ∆x = 0.01 x Volume of the cube = V = x3 m3 We need to find approximate change in volume v of the cube i.e. ∆V Now, ∆V = 𝑑𝑣/𝑑𝑥 ∆x = (𝑑(𝑥^3))/𝑑𝑥 ∆x = 3x2 (0.01x) = 0.03 x3 Hence, the approximate change is 0.03 x3 m3.