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

Options

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

Explanation

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
The cone has a fixed base radius \( R \) and a height \( h \). The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi R^2 h \]

### Step 2: Analyze the Forces Acting on the Sand
When the sand is piled up, it creates an angle \( \theta \) with the horizontal. The frictional force must balance the component of the weight of the sand acting down the slope of the cone. The maximum static frictional force can be expressed as:
\[ f_{\text{max}} = \mu N \]
where \( N \) is the normal force.

### Step 3: Relate the Angle to the Height and Radius
The angle \( \theta \) can be related to the height \( h \) and the radius \( R \) of the base of the cone using the tangent function:
\[ \tan \theta = \frac{h}{R} \]

### Step 4: Set Up the Condition for No Slipping
For the sand to not slip, the maximum static friction must be equal to the component of the weight acting down the slope. This gives us the condition:
\[ \mu N = mg \sin \theta \]
where \( m \) is the mass of the sand and \( g \) is the acceleration due to gravity.

### Step 5: Express Normal Force
The normal force \( N \) can be expressed as:
\[ N = mg \cos \theta \]

### Step 6: Substitute and Rearrange
Substituting \( N \) into the friction equation gives:
\[ \mu (mg \cos \theta) = mg \sin \theta \]
Dividing both sides by \( mg \) (assuming \( m \neq 0 \)):
\[ \mu \cos \theta = \sin \theta \]
This can be rearranged to:
\[ \tan \theta = \mu \]

### Step 7: Relate \( \theta \) to Height
From the relationship \( \tan \theta = \frac{h}{R} \), we can substitute:
\[ \frac{h}{R} = \mu \]
Thus, we can express the height \( h \) in terms of \( R \) and \( \mu \):
\[ h = \mu R \]

### Step 8: Calculate the Maximum Volume
Now substituting \( h \) back into the volume formula:
\[ V = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi R^2 (\mu R) = \frac{1}{3} \pi \mu R^3 \]

### Final Answer
The maximum volume of sand that can be piled up without slipping is:
\[ V = \frac{1}{3} \pi \mu R^3 \]

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