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Question
A sphere of radius 100 mm shrinks to radius 98 mm, then the approximate decrease in its volume is
Options
12000 π mm3
800 π mm3
80000 π mm3
120 π mm3
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Solution
80000 π mm3
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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