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

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KollegeApply · Accepted Answer

3k% Let x be the radius of the sphere and y be its volume. Then, ∆ x/x $\times$ 100 = k \text { Also }, y = $\frac{4}{3}$$\pi$ x^3 \Rightarrow dy/dx = 4$\pi$ x^2 \Rightarrow ∆ y/y = 4$\pi$ x^2/ydx = 4$\pi$ x^2/4/3$\pi$ x^3 $\times$ kx/100 \Rightarrow ∆ y/y $\times$ 100 = 3k \text { Hence, the error in the volume is } 3k . %

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