The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. The area and volume of the sheet to correct significant figures are

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$8.72 m^{2} a n d 0.0855 m^{3}$

$0.087 m^{2} a n d 0.0855 m^{3}$

$8.7 m^{2} a n d 0.085 m^{3}$

$0.87 m^{2} a n d 0.855 m^{3}$

answer is A.

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Detailed Solution

Given, length (l) = 4.234 m

Breadth (b) =1.005 m

Thickness (t) =2.01cm =0.0201m

Area of sheet (A)

$= 2 (ℓ \times b + b \times t + t \times ℓ)$

$= 2 [(4.234 \times 1.005) + (1.005 \times 0.0201) + (0.020 \times 4.234 )]$

$= 2 \times 43604739$

$= 8.7209478 m^{2}$

As thickness has least number of significant figures 3, therefore rounding-off area up to three significant figures, we get

Area of sheet $(A) = 8.72 m^{2}$

Volume of sheet (V) = l $\times$ b $\times$ t

$= 4.234 \times 1.005 \times 0.0201$

= 0.0855289

Rounding-off up to three significant figures, we get Volume of the sheet

$= 0.0855 m^{3}$

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