First slide

Rate Measure

Question

A spherical iron ball of 10cm radius is located with a layer of ice of uniform thickness that melts at a rate of 50 $c m^{3} / \min$ . When the thickness of ice is 5 cm, then the rate (in cm/min.) at which of the thickness of ice decreases, is:

Easy

A

$\frac{1}{18 π}$
B

$\frac{1}{36 π}$
C

$\frac{5}{6 π}$
D

$\frac{1}{54 π}$

Solution

$\begin{array}{l} Let thickness = x cm \\ Total volume V = \frac{4}{3} π (10 + x)^{3} \\ \frac{d V}{d t} = 4 π (10 + x)^{2} \frac{d x}{d t} \\ Given \frac{d V}{d t} = - 50 {cm}^{3} / \min \end{array}$
$\begin{array}{l} At x = 5 cm, we get - 50 = 4 π (10 + 5)^{2} \frac{d x}{d t} \\ \frac{d x}{d t} = - \frac{1}{18 π} cm per min \end{array}$

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