PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for A Sphere of Radius 100 Mm Shrinks to Radius 98 Mm, Then the Approximate Decrease in Its Volume is (A) 12000 π Mm3 (B) 800 π Mm3 (C) 80000 π Mm3 (D) 120 π Mm3 - PUC Karnataka Science Class 12 - Mathematics

#### Question

A sphere of radius 100 mm shrinks to radius 98 mm, then the approximate decrease in its volume is
(a) 12000 π mm3
(b) 800 π mm3
(c) 80000 π mm3
(d) 120 π mm3

#### Solution

(c) 80000 π mm3
Let x be the radius of the sphere and y be its volume.

$x = 100, x + ∆ x = 98$

$\Rightarrow ∆ x = - 2$

$y = \frac{4}{3}\pi x^3$

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

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

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

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

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Solution A Sphere of Radius 100 Mm Shrinks to Radius 98 Mm, Then the Approximate Decrease in Its Volume is (A) 12000 π Mm3 (B) 800 π Mm3 (C) 80000 π Mm3 (D) 120 π Mm3 Concept: Approximations.
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