A spherical iron ball 10 cm in radius is coated with a layer of ice of unirform thichness that melts at a rate of `50cm^3`/min. when the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is

Options

A. `(1)/(54pi)` cm/min
B. `(5)/(6pi)` cm/min
C. `(1)/(36pi)` cm/min
D. `(1)/(18pi)` cm/min

Explanation

To solve the problem, we need to find the rate at which the thickness of the ice decreases when the thickness is 5 cm. We will follow these steps:

### Step 1: Define Variables
Let:
- \( r = 10 \) cm (radius of the iron ball)
- \( x \) = thickness of the ice
- Total radius of the sphere with ice = \( R = r + x = 10 + x \)

### Step 2: Volume of the Ice
The volume of the ice can be expressed as the volume of the larger sphere (iron ball + ice) minus the volume of the iron ball:
\[ V = \frac{4}{3} \pi (R^3) - \frac{4}{3} \pi (r^3) \]
Substituting \( R = 10 + x \) and \( r = 10 \):
\[ V = \frac{4}{3} \pi ((10 + x)^3 - 10^3) \]

### Step 3: Differentiate Volume with Respect to Time
To find the rate at which the thickness of the ice decreases, we differentiate the volume \( V \) with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi ((10 + x)^3 - 10^3) \right) \]
Using the chain rule:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3(10 + x)^2 \cdot \frac{dx}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = 4 \pi (10 + x)^2 \cdot \frac{dx}{dt} \]

### Step 4: Substitute Known Values
We know that the ice melts at a rate of \( \frac{dV}{dt} = -50 \) cm³/min (negative because the volume is decreasing). When the thickness of the ice \( x = 5 \) cm:
\[ R = 10 + 5 = 15 \text{ cm} \]
Substituting into the equation:
\[ -50 = 4 \pi (15)^2 \cdot \frac{dx}{dt} \]

### Step 5: Solve for \( \frac{dx}{dt} \)
Calculating \( (15)^2 = 225 \):
\[ -50 = 4 \pi \cdot 225 \cdot \frac{dx}{dt} \]
\[ -50 = 900 \pi \cdot \frac{dx}{dt} \]
Now, solving for \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = \frac{-50}{900 \pi} = \frac{-1}{18 \pi} \text{ cm/min} \]

### Final Answer
The rate at which the thickness of the ice decreases when the thickness is 5 cm is:
\[ \frac{dx}{dt} = -\frac{1}{18 \pi} \text{ cm/min} \]