Chat with tutor

Ask Questions, Get Answers


If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be :

$\begin {array} {1 1} (1)\;2 \% & \quad (2)\;4 \% \\ (3)\;6 \% & \quad (4)\;8 \% \end {array}$


1 Answer

(3) is correct. Solution: Given error 2% For a sphere, V = 4\3 × π r^3 Taking log both sides, Log V = 3 log r + log (4\3 π) Diff. Both sides V d\dV log V = 3 d\dV r + d\dV log (4\3 π) 1\V = 3\r dr\dV + 0 dV\V = 3×2% = 6%. ANSWER (given dr\r = 2%).
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.