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Maths-

General

Easy

Question

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

$12000 π {m m}^{3}$
$800 π {m m}^{3}$
$80000 π {m m}^{3}$
$120 π {m m}^{3}$

The correct answer is: $80000 blank pi blank m m to the power of 3 end exponent$

Let $x$ be the radius and $V$ be the volume of the sphere. Then, $V equals fraction numerator 4 over denominator 3 end fraction pi x to the power of 3 end exponent$
$therefore fraction numerator d V over denominator d x end fraction equals 4 pi x to the power of 2 end exponent$

We have, $x equals 100$ and $x plus increment x equals 98$

$therefore increment x equals negative 2$

Now, $increment V equals fraction numerator d V over denominator d x end fraction increment x$

$rightwards double arrow increment V equals 4 pi cross times 100 to the power of 2 end exponent cross times negative 2 blank open square brackets because open parentheses fraction numerator d V over denominator d x end fraction close parentheses subscript x equals 100 end subscript equals 4 pi cross times 100 to the power of 2 end exponent close square brackets$

$rightwards double arrow increment V equals negative 80000 pi$

Hence, decrease in volume is $80000 pi blank m m to the power of 3 end exponent$

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