# A spherical iron ball of rdius 10cm is coated with a layer of ice of uniform thickness that melts at a rate of  $50 \: cm^{3}/min$. When the thickness of the ice is 5 cm , then the rate at which the thickness ( in cm / min ) of the icedecreases, is : Option 1) $\frac{1}{18\pi}$ Option 2) $\frac{1}{36\pi}$ Option 3) $\frac{5}{6\pi}$ Option 4) $\frac{1}{9\pi}$

Volume of ice , $V=\frac{4}{3}\pi(R+x)^{3}-\frac{4}{3}\pi R^{3}$

$\frac{dV}{dt}=4\pi (R+x)^{2}\frac{dx}{dt}=50$

$\frac{dx}{dt}=\frac{50}{4\pi (R+x)^{2}}$

at x = 5,

$\frac{dx}{dt}=\frac{50}{4\pi (10+5)^{2}}=\frac{1}{18\pi}cm/min$

So, option (1) is correct.

Option 1)

$\frac{1}{18\pi}$

Option 2)

$\frac{1}{36\pi}$

Option 3)

$\frac{5}{6\pi}$

Option 4)

$\frac{1}{9\pi}$

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