#### Question

If an error of k% is made in measuring the radius of a sphere, then percentage error in its volume is

(a) k%

(b) 3k%

(c) 2k%

(d) k/3%

#### Solution

(b) 3*k*%

Let *x* be the radius of the sphere and *y* be its volume.

Then,

\[\frac{∆ x}{x} \times 100 = k\]

\[\text { Also }, y = \frac{4}{3}\pi x^3 \]

\[ \Rightarrow \frac{dy}{dx} = 4\pi x^2 \]

\[ \Rightarrow \frac{∆ y}{y} = \frac{4\pi x^2}{y}dx = \frac{4\pi x^2}{\frac{4}{3}\pi x^3} \times \frac{kx}{100}\]

\[ \Rightarrow \frac{∆ y}{y} \times 100 = 3k\]

\[\text { Hence, the error in the volume is } 3k % .\]

Is there an error in this question or solution?

Solution If an Error of K% is Made in Measuring the Radius of a Sphere, Then Percentage Error in Its Volume is (A) K% (B) 3k% (C) 2k% (D) K/3% Concept: Approximations.