#### Question

The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume ?

#### Solution

\[\text { Let r be the radius of the sphere }. \]

\[ r = 10 cm\]

\[r + ∆ r = 9 . 8 cm\]

\[ \Rightarrow ∆ r = 10 . 0 - 9 . 8 = 0 . 2 cm\]

\[\text { Volume of the sphere,} V = \frac{4}{3}\pi r^3 \]

\[ \Rightarrow \frac{dV}{dr} = \frac{4}{3}\pi \times 3 r^2 = 4\pi r^2 \]

\[ \Rightarrow \left( \frac{dV}{dr} \right)_{r = 10 cm} = 4\pi \left( 10 \right)^2 = 400\pi {cm}^3 /cm\]

\[\text{ Change in the volume of the sphere,} ∆ V = \frac{dV}{dr} \times dr = 400\pi \times 0 . 2 = 80\pi \ {cm}^3\]

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Solution for question: The Radius of a Sphere Shrinks from 10 to 9.8 Cm. Find Approximately the Decrease in Its Volume ? concept: Approximations. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)