#### Question

A sphere of radius 100 mm shrinks to radius 98 mm, then the approximate decrease in its volume is

12000 π mm

^{3}800 π mm

^{3}80000 π mm

^{3}120 π mm

^{3}

#### Solution

80000 π mm^{3}

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

\[x = 100, x + ∆ x = 98\]

\[ \Rightarrow ∆ x = - 2\]

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

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

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 100} = 40000\pi\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = 40000\pi \times \left( - 2 \right) = - 80000\pi\]

\[\text { Hence, the decrease in the volume of the sphere is } 80000\pi \text{mm}^ 3.\]

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Solution A Sphere of Radius 100 Mm Shrinks to Radius 98 Mm, Then the Approximate Decrease in Its Volume is Concept: Approximations.