Exercise 1.11

Transcript

Exercise 1.11 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. Writing significant figures of length, breadth and thickness Since length and breadth are in m, we convert thickness into m Quantity | Value | Significant figures Length l | 4.234 m | 4 Breadth b | 1.005 m | 4 Thickness t | 2.01 cm | 3 Convert the thickness: 𝑡=2.01 cm = 2.01 ×10^(−2) m =𝟎.𝟎𝟐𝟎𝟏 𝐦 Note: It still has three significant figures. Area Now, Area = Length × Breadth = 4.234 × 1.005 = 4.25517 m^(2) Since both Length and Breadth have 4 significant figures Area will have 4 significant figures Rounding off 4.25517 to 4.255 Thus, Area = 4.255 m^(2) Volume Now, Area = Length × Breadth × Thickness = 4.234 × 1.005 × 0.0201 = 0.085528917 m^(3) Since Length and Breadth have 4 significant figures, and Thickness has 3 significant figures We use the lowest. Thus, Volume will have 3 significant figures Rounding off 0.085528917 to 0.0855 Thus, Volume = 0.0855 m^(3) Rounding off 4.25517 to 4.255 Thus, Area = 4.255 m^(2) Volume Now, Volume = Length × Breadth × Thickness = 4.234 × 1.005 × 0.0201 = 0.085528917 m^(3) Since Length and Breadth have 4 significant figures, and Thickness has 3 significant figures We use the lowest. Thus, Volume will have 3 significant figures Rounding off 0.085528917 to 0.0855 Thus, Volume = 0.0855 m^(3) Rounding off 0.085528917 to 0.0855 Thus, Volume = 0.0855 m^(3)