Report Issue

JEE MAIN 2021
18-03-21 S2
Question
The radius of a sphere is measured to be $(7.50 \pm 0.85) \mathrm{cm}$. Suppose the percentage error in its volume is $x$. The value of $x$, to the nearest $x$, is, $\_\_\_\_$ .
Write Your Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Solution Image
Question Tags
JEE Main
Physics
Easy
Start Preparing for JEE with Competishun