Question

Each side of a cube is measured to be $ 7.203\,m $ . What is the total surface area and volume of the cube to appropriate significant figures?

Answer 1

Hint : To solve this question, we will use the relation of the volume of a cube with its side length. In the final answer we must have as many significant digits as the side length of the cube has and the answer should also be properly rounded to the appropriate number of significant digits.

Complete step by step answer
We’ve been given that the side length of the cube is $ 7.203\,m $ . The number of significant digits in the length is 4. We can now calculate the surface area of the cube. Since the area of one of the faces of the cube is the area of a square of length $ 7.203\,m $ and a cube has 6 square faces, we can calculate
 $\Rightarrow S = 6 \times ({\text{area of one surface)}} $
 $ \Rightarrow {\text{S = 6}} \times {{\text{l}}^2} $
Placing the value of $ l = 7.203\,m $ , we get
 $ \Rightarrow S = 6 \times {\left( {7.203} \right)^2}\,{m^2} $
Which gives us
 $\Rightarrow S = 311.299254\,{m^2} $
Since we want only 4 significant digits, we can round up the surface area to
 $\Rightarrow S = 311.3\,{m^2} $
Similarly, the volume of the cube can be calculated as
 $\Rightarrow V = {l^3} $
Placing the value of $ l = 7.203\,m $ , we get
 $\Rightarrow V = 373.714754427\,{m^3} $
Again, since we want only 4 significant digits, we can round up the volume to
 $\Rightarrow V = 373.7\,{m^3} $.

Note
While calculating the answer in the least digit form, we should be careful to round up the answer correctly. Since the 5th significant digit in the surface area is 9 which is greater than 5, we must round up the 4th significant digit to the next integer. Similarly, since the 5th significant digit in the volume is 1 which is smaller than 5, we must not change the 4th significant digit.