Sand is piled up on a horizontal ground in the form of a regular cone of a fixed base radius R. The coefficient of static friction between sand layers is `mu` . The maximum volume of sand that can be pilled up, without the sand slipping on the surface is

Options

A. `(muR^(3))/(3pi)`
B. `(muR^(3))/3`
C. `(piR^(3))/(3mu)`
D. `(mupi R^(3))/3`

Answer

To solve the problem of finding the maximum volume of sand that can be piled up in the form of a regular cone without slipping, we can follow these steps:

### Step 1: Understand the Geometry of the Cone
We have a cone with:
- Base radius \( R \)
- Height \( h \)
- Slant height \( l \)

The relationship between the height \( h \), radius \( R \), and the slant height \( l \) is given by the right triangle formed by these dimensions. The angle \( \theta \) is defined as the angle between the height and the slant height.

### Step 2: Relate Height and Radius Using the Angle
Using the definition of tangent in the context of the cone:
\[ \tan(\theta) = \frac{h}{R} \]
This means that:
\[ h = R \tan(\theta) \]

### Step 3: Apply the Condition for Static Friction
For the sand not to slip, the condition is:
\[ \tan(\theta) < \mu \]
Substituting the expression for \( h \):
\[ \frac{h}{R} < \mu \]
This implies:
\[ h < \mu R \]

### Step 4: Determine the Maximum Height
The maximum height \( h_{\text{max}} \) of the sand cone can be expressed as:
\[ h_{\text{max}} = \mu R \]

### Step 5: Calculate the Volume of the Cone
The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi R^2 h \]
Substituting \( h_{\text{max}} \) into the volume formula:
\[ V_{\text{max}} = \frac{1}{3} \pi R^2 h_{\text{max}} = \frac{1}{3} \pi R^2 (\mu R) \]
This simplifies to:
\[ V_{\text{max}} = \frac{1}{3} \pi \mu R^3 \]

### Final Answer
Thus, the maximum volume of sand that can be piled up without slipping is:
\[ V_{\text{max}} = \frac{\pi \mu R^3}{3} \]

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