Question
Download Solution PDFThe rate of change of surface area of cube is 216 cm2/s and that of the side of the cube is 2 cm/s, find the length at that time.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
Given that the side (L) of the cube changes at the rate = \(\rm{dL\over dt}\)= 2 cm/s
Surface Area of the cube V = 6L2
Rate of change of Surface Area A =\(\rm{dA\over dt}\)= \(\rm{dA\over dL}×{dL\over dt}\)
\(\rm{dA\over dt}\) = \(\rm{d\over dL}\left[6L^2\right]× {dL\over dt}\)
\(\rm{dA\over dt}\) = \(\rm 12L × 2\) = 24L
As \(\rm{dA\over dt}\) = 216 cm2/s
216 = 24 × L
L = \(\rm{216\over 24}\) = 9 cm
Last updated on Jun 18, 2025
->UPSC has extended the UPSC NDA 2 Registration Date till 20th June 2025.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced. The written examination will be held on 14th September 2025.
-> The selection process for the NDA exam includes a Written Exam and SSB Interview.
-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100.
-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential.