Question

If there is 2% error in measuring the radius of the sphere, then __________will be the percentage error in the surface area.

• A. $3\%$
• B. $1\%$
• C. $4\%$
• D. $2\%$

Answer 1

The correct option is C

$4\%$

Explanation for the correct option:

Option(C):

Step 1:Given data:

Given that there is $2\%$ error in measuring the radius of the sphere,

Hence,

$$\begin{gathered} \frac{\Delta r}{r}=2\%=\frac{2}{100} \\ dr=\Delta r=\frac{2r}{100} \end{gathered}$$

Step 2: Calculation of the change of the surface area of the sphere:

Now, the Area of the sphere is given by

$A=4\pi r^2$

Differentiating both sides with respect to r;

$$\begin{gathered} \frac{dA}{dr}=8\pi r \\ dA=8\pi r\left(dr\right) \\ dA=8\pi r\times \frac{2r}{100} \\ \Delta A=dA=\frac{16\pi r^2}{100} \end{gathered}$$

Step 3: Calculation of the percentage error in the surface area

Now, the percentage error in the surface area of the sphere can be calculated as:

$\%\left(error\right)=\frac{\Delta A}{A}\times 100$

$$\begin{gathered} \Rightarrow \frac{16\pi r^2}{100\times 4\pi r^2}\times 100 \\ \Rightarrow 4\% \end{gathered}$$

Hence, the correct option is (C).