A sphere of 10cm radius has a uniform thickness of ice around it. Ice is melting at rate `50 cm^3//min` when thickness is 5cm then rate of change of thickness

Options

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

Answer

To solve the problem, we need to find the rate of change of the thickness of ice around a sphere as it melts. Let's break this down step by step.

### Step 1: Understand the Problem
We have a sphere with a radius of 10 cm and a uniform thickness of ice around it. The thickness of the ice is denoted as \( h \). When \( h = 5 \) cm, we need to find the rate of change of thickness \( \frac{dh}{dt} \) given that the volume of ice melting is \( \frac{dV}{dt} = -50 \) cm³/min (negative because the volume is decreasing).

### Step 2: Define the Radii
- The radius of the inner sphere (the sphere itself) is \( r = 10 \) cm.
- The radius of the outer sphere (the sphere plus the ice) is \( R = 10 + h \).

### Step 3: Write the Volume of Ice
The volume of the ice surrounding the sphere can be expressed as the difference between the volume of the outer sphere and the inner sphere:
\[ V = \frac{4}{3} \pi R^3 - \frac{4}{3} \pi r^3 \]
Substituting \( R \) and \( r \):
\[ V = \frac{4}{3} \pi (10 + h)^3 - \frac{4}{3} \pi (10)^3 \]
This simplifies to:
\[ V = \frac{4}{3} \pi \left((10 + h)^3 - 1000\right) \]

### Step 4: Differentiate with Respect to Time
To find the rate of change of volume with respect to time, we differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3(10 + h)^2 \cdot \frac{dh}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = 4 \pi (10 + h)^2 \cdot \frac{dh}{dt} \]

### Step 5: Substitute Known Values
We know \( \frac{dV}{dt} = -50 \) cm³/min and \( h = 5 \) cm. Substituting these values into the equation:
\[ -50 = 4 \pi (10 + 5)^2 \cdot \frac{dh}{dt} \]
This simplifies to:
\[ -50 = 4 \pi (15)^2 \cdot \frac{dh}{dt} \]
Calculating \( (15)^2 = 225 \):
\[ -50 = 4 \pi \cdot 225 \cdot \frac{dh}{dt} \]

### Step 6: Solve for \( \frac{dh}{dt} \)
Now, we can solve for \( \frac{dh}{dt} \):
\[ -50 = 900 \pi \cdot \frac{dh}{dt} \]
Dividing both sides by \( 900 \pi \):
\[ \frac{dh}{dt} = \frac{-50}{900 \pi} = \frac{-1}{18 \pi} \text{ cm/min} \]

### Final Answer
Thus, the rate of change of thickness of the ice when the thickness is 5 cm is:
\[ \frac{dh}{dt} = -\frac{1}{18 \pi} \text{ cm/min} \]