Question

The diameter of a sphere is 4.24 m. Its surface area with due regard to significant figures is:
\[\begin{align}
  & A.\,5.65\,{{m}^{2}} \\
 & B.\,56.5\,{{m}^{2}} \\
 & C.\,565\,{{m}^{2}} \\
 & D.\,5650\,{{m}^{2}} \\
\end{align}\]

Answer 1

Hint: The radius of the sphere equals half the value of the diameter of the sphere. The surface area of the sphere is equal to 4(pi) times the square of the radius. The significant figures in the value of the diameter is 2, thus, the total number of significant figures in the value of the area should also be 2.
Formula used:
\[S=4\pi {{r}^{2}}\]

Complete answer:
From the given information, we have the data as follows.
The diameter of a sphere is 4.24 m.

Firstly, we have to find the value of the radius using the given value of the diameter of the sphere. So, we have the formula that converts the diameter into the radius, that is,
\[\begin{align}
  & r=\dfrac{d}{2} \\
 & \Rightarrow r=\dfrac{4.24}{2} \\
 & \therefore r=2.12\,m \\
\end{align}\]
Therefore, the radius of the sphere is 2.12 m.
Here keep a note that the number of digits present after the decimal point is 2.
Now, we are supposed to find the value of the surface area of the sphere. The surface area of the sphere is given by the formula as follows.
\[S=4\pi {{r}^{2}}\]
Where S is the surface area of the sphere and r is the radius of the sphere.
Substitute the obtained value of the radius of the sphere in the above equation. So, we have,
\[\begin{align}
  & S=4\pi \times {{2.12}^{2}} \\
 & \Rightarrow S=4\times \dfrac{22}{7}\times 2.12\times 2.12 \\
 & \therefore S=56.50\,{{m}^{2}} \\
\end{align}\]
Thus, the surface area of the sphere is \[56.50\,{{m}^{2}}\].
\[\therefore \] The surface area of the sphere with the diameter 4.24 m is\[56.50\,{{m}^{2}}\]

Thus, option (B) is correct.

Note:
Even we can solve this problem using the diameter of the sphere. Instead of radius, we have to replace it with the diameter of the sphere. When we replace the radius with the diameter, we get the formula as the product of (pi) and the square of the diameter.