Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The radius of a sphere is measured to be (7.50±0.85) cm. Suppose the percentage error in its volume is x. The value of x, to the nearest x, is ___ .
PhysicsMathematics in PhysicsJEE MainJEE Main 2021 (18 Mar Shift 2)
Solution:
2474 Upvotes Verified Answer
The correct answer is: 34

v=43πr3

taking log & then differentiate

dVV=3drr=3×0.857.5×100%=34%

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.