Question

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

पर्याय

•  k%

• 3k%

• 2k%

• k/3%

Answer 1

3k%
Let x be the radius of the sphere and y be its volume.
Then,

\[\frac{∆ x}{x} \times 100 = k\]

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

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

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

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

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