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

Question

A spherical iron ball of $ 10 \mathrm{~cm} $ radius is coated with a layer of ice of uniform thickness that melts at a rate of $ 50 \mathrm{~cm}^{3} / \mathrm{min} $. When the thickness of ice is $ 5 \mathrm{~cm} $, then the rate (in $ \mathrm{cm} / \mathrm{min} $.) at which of the thickness of ice decreases, is:, $ \frac{5}{6 \pi} $, $ \frac{1}{3 \pi} $, $ \frac{1}{36 \pi} $, $ \frac{1}{18 \pi} $

MARKS App · Accepted Answer

Let thickness = x cm Total volume V = 4 3 π 10 + x 3 d V d t = 4 π 10 + x 2 d x d t … . . ( i ) Given d V d t = 50 cm 3 / min At x = 5 cm 50 = 4 π 10 + 5 2 d x d t d x d t = 1 18 π cm / min

Search any question & find its solution

Looking for more such questions to practice?