A block of mass 10 kg in contact against the inner wall of a hollow cylindrical drum of radius 1m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be `(g=10 m//s^(2))`

Options

A. `10pi rad//s`
B. `sqrt(10) rad//s`
C. `(10)/(2pi) rad//s`
D. `10 rad//s`

Explanation

To find the minimum angular velocity needed for the block to remain stationary against the inner wall of the hollow cylindrical drum, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the block, \( m = 10 \, \text{kg} \)
- Radius of the cylinder, \( r = 1 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.1 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Weight of the Block:**
\[ W = mg = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N} \]

3. **Determine the Maximum Frictional Force:**
The frictional force \( F_f \) that can act on the block is given by:
\[ F_f = \mu N \]
where \( N \) is the normal force. For the block to remain stationary, the frictional force must balance the weight of the block:
\[ F_f = W = 100 \, \text{N} \]

4. **Relate Normal Force to Centripetal Force:**
The normal force \( N \) is also related to the centripetal force required to keep the block moving in a circular path:
\[ N = \frac{mv^2}{r} \]

5. **Set Up the Equation:**
Since the maximum frictional force must equal the weight of the block:
\[ \mu N = mg \]
Substituting for \( N \):
\[ \mu \left(\frac{mv^2}{r}\right) = mg \]

6. **Solve for Velocity \( v \):**
Rearranging gives:
\[ \mu \frac{mv^2}{r} = mg \]
\[ v^2 = \frac{g r}{\mu} \]
Substituting the known values:
\[ v^2 = \frac{10 \, \text{m/s}^2 \times 1 \, \text{m}}{0.1} = 100 \, \text{m}^2/\text{s}^2 \]
\[ v = \sqrt{100} = 10 \, \text{m/s} \]

7. **Calculate Angular Velocity \( \omega \):**
The angular velocity \( \omega \) is related to the linear velocity \( v \) by the formula:
\[ \omega = \frac{v}{r} \]
Substituting the values:
\[ \omega = \frac{10 \, \text{m/s}}{1 \, \text{m}} = 10 \, \text{rad/s} \]

### Final Answer:
The minimum angular velocity needed for the cylinder to keep the block stationary is \( \omega = 10 \, \text{rad/s} \).