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Provide solution RD Sharma maths class 12 chapter 12 derivative as a Rate Measurer exercise case study base question, question 2 sub question 4

Answers (1)

Answer:

2.5cm

Hint:

 Here we use the value of cube’s surface area and volume

Given:

 We Know that

Volume = =60cm^{3}/s

And surface area = =24cm^{2}/\sec

  1. So, rate of increase of volume with respect to surface area\frac{60}{24}=2.5cm
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