If the coefficient of friction between an insect and bowl is `mu` and the radius of the bowl is r, find the maximum height to which the insect can crawl in the bowl.

Options

A. (a)`(r )/(sqrt(1 + mu^(2)))`
B. (b)`r[1 - (1)/(sqrt(1 + mu^(2)))]`
C. (c)`r sqrt(1 + mu^(2))`
D. (d)`r sqrt(1 + (1)/(1 + mu^(2)))`

Explanation

To solve the problem of finding the maximum height to which an insect can crawl in a bowl with a given coefficient of friction (µ) and radius (r), we can follow these steps:

### Step 1: Understand the Forces Acting on the Insect
The insect experiences two main forces:
- The gravitational force (weight) acting downwards, \( W = mg \), where \( m \) is the mass of the insect and \( g \) is the acceleration due to gravity.
- The normal force, which acts perpendicular to the surface of the bowl.

### Step 2: Analyze the Geometry of the Bowl
The bowl can be considered as a circular surface. When the insect crawls up the bowl, it makes an angle \( \theta \) with the vertical. The height \( h \) that the insect climbs can be related to the radius \( r \) of the bowl and the angle \( \theta \) using trigonometry:
- The vertical height \( h \) is given by \( h = r - r \cos \theta = r(1 - \cos \theta) \).

### Step 3: Apply the Condition for Motion
For the insect to crawl up without slipping, the frictional force must be equal to or greater than the component of the gravitational force acting down the slope of the bowl. The frictional force \( F_f \) can be expressed as:
\[ F_f = \mu N \]
where \( N \) is the normal force.

### Step 4: Determine the Normal Force
The normal force \( N \) can be expressed as:
\[ N = mg \cos \theta \]
Thus, the frictional force becomes:
\[ F_f = \mu mg \cos \theta \]

### Step 5: Set Up the Equation for Maximum Height
The component of the weight acting down the slope is:
\[ W_{\text{down}} = mg \sin \theta \]
For the insect to not slip, we set the frictional force equal to the component of weight:
\[ \mu mg \cos \theta = mg \sin \theta \]

### Step 6: Simplify the Equation
Dividing both sides by \( mg \) (assuming \( m \neq 0 \)):
\[ \mu \cos \theta = \sin \theta \]
This can be rewritten as:
\[ \tan \theta = \mu \]

### Step 7: Relate Cosine to Height
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we can express \( \cos \theta \) in terms of \( \mu \):
\[ \cos \theta = \frac{1}{\sqrt{1 + \mu^2}} \]

### Step 8: Substitute Back to Find Height
Now substituting \( \cos \theta \) back into the height equation:
\[ h = r(1 - \cos \theta) = r\left(1 - \frac{1}{\sqrt{1 + \mu^2}}\right) \]

### Step 9: Final Expression for Maximum Height
Thus, the maximum height \( h \) to which the insect can crawl in the bowl is given by:
\[ h = r\left(1 - \frac{1}{\sqrt{1 + \mu^2}}\right) \]

### Conclusion
The maximum height \( h \) that the insect can crawl in the bowl is:
\[ h = r\left(1 - \frac{1}{\sqrt{1 + \mu^2}}\right) \]