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Units and Measurement

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Question : 11 of 33
Marks: +1, -0
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Solution:  
Given, length, (l)=4.234 m,(l)=4.234\ \mathrm{m},
breadth (b)=1.005 m(b)=1.005\ \mathrm{m}
thickness, d=2.01 cmd=2.01\ \mathrm{cm}
=2.01×10−2 m=2.01\times10^{-2}\ \mathrm{m}
Area of sheet=2(lb+bd+dl)\text{Area of sheet}=2(lb+bd+dl)
=2(4.234×1.005+1.005×0.0201+0.0201×4.234)=2(4.234\times1.005+1.005\times0.0201+0.0201\times4.234)
=2(4.3604739)=2(4.3604739)
=8.7209478 m2=8.7209478\ \mathrm{m}^{2}
As the least number of significant figure in thickness is 3. Therefore, area has 3 significant figure, Area =8.72 m2=8.72\ \mathrm{m}^{2}
volume of metal sheet =l×bt×d=l\times b t\times d
=4.234×1.005×0.0201 m3=4.234\times1.005\times0.0201\ \mathrm{m}^{3}
=0.085528917 m3=0.085528917\ \mathrm{m}^{3}
After rounding off =0.0855 m3=0.0855\ \mathrm{m}^{3}
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