A) \[\pm 1.2\]
B) \[\pm 1.0\]
C) \[\pm 0.8\]
D) \[\pm 0.6\]
Correct Answer: D
Solution :
Given, error in diameter\[=\pm 0.04\] Error in radius,\[dr=\pm 0.02\] \[\therefore \]Per cent error in the volume of sphere \[=\frac{dV}{V}\times 100=\frac{d\left( \frac{4}{3}\pi {{r}^{3}} \right)}{\frac{4}{3}\pi {{r}^{3}}}\times 100\] \[=\frac{3dr}{r}\times 100\] \[=\frac{3\times (\pm 0.02)}{10}\times 100=\pm 0.6\]
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