A) \[0.6\,\,{{m}^{3}}\]
B) \[0.06\,\,{{m}^{3}}\]
C) \[0.006\,\,{{m}^{3}}\]
D) \[0.0006\,\,{{m}^{3}}\]
Correct Answer: B
Solution :
Let\[x\]be the length of an edge of a cube and\[V\]be the volume of that cube. \[\therefore \] \[V={{x}^{3}}\] On differentiating w.r.t.\[x\], we get \[\frac{dV}{dx}=3{{x}^{2}}\] Let\[5V\]be error in\[V\]and corresponding error\[\delta x\]in\[x\]. \[\therefore \] \[\delta V=\frac{dV}{dx}\delta x=3{{x}^{2}}\delta x\] Given that,\[x=2\,\,\text{m}\]and\[\delta x=0.5\,\,cm=\frac{0.5}{100}m\] \[\therefore \] \[\delta V=3{{(2)}^{2}}\left( \frac{0.5}{100} \right)\] \[=\frac{12\times 0.5}{100}=\frac{6}{100}=0.06\,\,{{m}^{3}}\]
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