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

Options

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

Answer

To solve the problem step-by-step, we will follow these steps:

### Step 1: Understand the Problem
We have a spherical iron ball with a radius of 10 cm, and it is coated with a layer of ice that melts at a rate of 50 cm³/min. We need to find the rate at which the thickness of the ice decreases when the thickness is 5 cm.

### Step 2: Define Variables
Let:
- \( r = 10 \) cm (radius of the iron ball)
- \( W \) = thickness of the ice layer
- The total radius of the sphere with ice = \( R = r + W = 10 + W \)

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

### Step 4: Differentiate Volume with Respect to Time
We need to find the rate of change of volume with respect to time, \( \frac{dV}{dt} \). Using the chain rule:
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi ((10 + W)^3 - 10^3) \right) \]
Since \( 10^3 \) is a constant, its derivative is zero. Thus:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3(10 + W)^2 \cdot \frac{dW}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = 4 \pi (10 + W)^2 \cdot \frac{dW}{dt} \]

### Step 5: Substitute Known Values
We know that \( \frac{dV}{dt} = -50 \) cm³/min (the volume is decreasing), and we need to find \( \frac{dW}{dt} \) when \( W = 5 \) cm.
Substituting \( W = 5 \) cm:
\[ \frac{dV}{dt} = 4 \pi (10 + 5)^2 \cdot \frac{dW}{dt} \]
This becomes:
\[ -50 = 4 \pi (15)^2 \cdot \frac{dW}{dt} \]
Calculating \( (15)^2 = 225 \):
\[ -50 = 4 \pi \cdot 225 \cdot \frac{dW}{dt} \]

### Step 6: Solve for \( \frac{dW}{dt} \)
Rearranging the equation:
\[ \frac{dW}{dt} = \frac{-50}{4 \pi \cdot 225} \]
Calculating the right-hand side:
\[ \frac{dW}{dt} = \frac{-50}{900 \pi} = \frac{-1}{18 \pi} \text{ cm/min} \]

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