Question:
The radius of sphere is measured to be $ (2.1\,\pm \,0.5)\,cm. $ Calculate its surface area with error limits:
The radius of sphere is measured to be $ (2.1\,\pm \,0.5)\,cm. $ Calculate its surface area with error limits:
Updated On: Jun 20, 2022
- $ (55.4\,\pm \,26.4)\,c{{m}^{2}} $
- $ (55.4\,\pm \,0.02)\,c{{m}^{2}} $
- $ (55.4\,\pm \,2.64)\,c{{m}^{2}} $
- $ (55.4\,\pm \,0.26)\,c{{m}^{2}} $
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The Correct Option is A
Solution and Explanation
Surface area, $ S=4\pi {{r}^{2}}=4\times \frac{22}{7}\times {{(2.1)}^{2}}=55.44\,=55.4\,c{{m}^{2}} $
Further, $ \frac{\Delta S}{S}=2.\frac{\Delta r}{r} $
or $ \Delta S=2\left( \frac{\Delta r}{r} \right)(S) $
$ =\frac{2\times 0.5\times 55.4}{2.1}=26.38=26.4\,c{{m}^{2}} $
$ \therefore $ $ S=(55.4\,\pm \,26.4)c{{m}^{2}} $
Further, $ \frac{\Delta S}{S}=2.\frac{\Delta r}{r} $
or $ \Delta S=2\left( \frac{\Delta r}{r} \right)(S) $
$ =\frac{2\times 0.5\times 55.4}{2.1}=26.38=26.4\,c{{m}^{2}} $
$ \therefore $ $ S=(55.4\,\pm \,26.4)c{{m}^{2}} $
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The physical world includes the complications of the natural world around us. It is a type of analysis of the physical world around us to understand how it works. The fundamental forces that control nature are:
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