Question 9 (MCQ) - Approximations (using Differentiation) - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Approximations (using Differentiation)
Question 1 (ii) Deleted for CBSE Board 2025 Exams
Question 1 (iii) Deleted for CBSE Board 2025 Exams
Question 1 (iv) Deleted for CBSE Board 2025 Exams
Question 1 (v) Important Deleted for CBSE Board 2025 Exams
Question 1 (vi) Deleted for CBSE Board 2025 Exams
Question 1 (vii) Deleted for CBSE Board 2025 Exams
Question 1 (viii) Deleted for CBSE Board 2025 Exams
Question 1 (ix) Deleted for CBSE Board 2025 Exams
Question 1 (x) Deleted for CBSE Board 2025 Exams
Question 1 (xi) Important Deleted for CBSE Board 2025 Exams
Question 1 (xii) Deleted for CBSE Board 2025 Exams
Question 1 (xiii) Deleted for CBSE Board 2025 Exams
Question 1 (xiv) Important Deleted for CBSE Board 2025 Exams
Question 1 (xv) Deleted for CBSE Board 2025 Exams
Question 2 Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 5 Important Deleted for CBSE Board 2025 Exams
Question 6 Deleted for CBSE Board 2025 Exams
Question 7 Deleted for CBSE Board 2025 Exams
Question 8 (MCQ) Important Deleted for CBSE Board 2025 Exams
Question 9 (MCQ) Deleted for CBSE Board 2025 Exams You are here
Approximations (using Differentiation)
Last updated at April 16, 2024 by Teachoo
Question 9 The approximate change in the volume of a cube of side x meters caused by increasing the side by 3% is (A) 0.06 x3 m3 (B) 0.6 x3 m3 (C) 0.09 x3 m3 (D) 0.9 x3 m3Let side of the cube = x meters Increase in side = 3% = 0.03 x Hence, ∆x = 0.03 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 (0.03x) = 0.09 x3 So part (C) is the correct answer.