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The radius of a thin wire is `0.16mm`. The area of cross section taking significant figures into consideration in square millimeter isA. 0.08B. `0.080`C. `0.0804`D. `0.080384` |
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Answer» Correct Answer - B `R`=0.16mm Hence, `A=piR^(2)=(22)/(7)xx(0.16)^(2)=0.080384` Since radius has two significant figure so answer also will have two significant figures. |
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