The radius of a right circular cyliner increases at a constant rate. Its altitude is a linear function of radius and increases three times as fast as radius. When the radius is 1 cm the altitude is 6 cm radius is 6cm, the volume is increasing at the rate of 1 Cu.cm/sec. When the radius is 36 cm Volume is increasing at a rate of n cu. cm/sec. The value of 'n' is divisible by

Options

A. 11
B. 22
C. 3
D. 33

Explanation

To solve the problem step by step, we will start by defining the variables and relationships given in the question.

### Step 1: Define Variables and Relationships
Let:
- \( r \) = radius of the cylinder (in cm)
- \( h \) = height (altitude) of the cylinder (in cm)
- \( V \) = volume of the cylinder (in cm³)

According to the problem:
- The radius \( r \) increases at a constant rate, denoted as \( \frac{dr}{dt} = k \).
- The height \( h \) is a linear function of the radius and increases three times as fast as the radius, thus:
\[ \frac{dh}{dt} = 3 \frac{dr}{dt} = 3k \]

### Step 2: Find the Relationship Between Height and Radius
From the information given:
When \( r = 1 \) cm, \( h = 6 \) cm. We can express the height \( h \) in terms of \( r \):
\[ h = 3r + c \]
To find \( c \), substitute \( r = 1 \) and \( h = 6 \):
\[ 6 = 3(1) + c \implies c = 3 \]
Thus, the equation for height becomes:
\[ h = 3r + 3 \]

### Step 3: Write the Volume Formula
The volume \( V \) of a cylinder is given by:
\[ V = \pi r^2 h \]
Substituting \( h \):
\[ V = \pi r^2 (3r + 3) = \pi r^2 (3r + 3) = 3\pi r^3 + 3\pi r^2 \]

### Step 4: Differentiate the Volume with Respect to Time
Differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}(3\pi r^3 + 3\pi r^2) = 3\pi \left(3r^2 \frac{dr}{dt} + 2r \frac{dr}{dt}\right) \]
Factoring out \( \frac{dr}{dt} \):
\[ \frac{dV}{dt} = 3\pi (3r^2 + 2r) \frac{dr}{dt} \]

### Step 5: Use Given Information to Find \( \frac{dr}{dt} \)
When \( r = 6 \) cm, the volume is increasing at a rate of 1 cm³/sec:
\[ 1 = 3\pi (3(6^2) + 2(6)) \frac{dr}{dt} \]
Calculating \( 3(6^2) + 2(6) \):
\[ 3(36) + 12 = 108 + 12 = 120 \]
Thus:
\[ 1 = 3\pi (120) \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{360\pi} \text{ cm/sec} \]

### Step 6: Find the Rate of Volume Change When \( r = 36 \) cm
Now, we need to find \( \frac{dV}{dt} \) when \( r = 36 \) cm:
\[ \frac{dV}{dt} = 3\pi (3(36^2) + 2(36)) \frac{dr}{dt} \]
Calculating \( 3(36^2) + 2(36) \):
\[ 3(1296) + 72 = 3888 + 72 = 3960 \]
Thus:
\[ \frac{dV}{dt} = 3\pi (3960) \frac{1}{360\pi} \]
The \( \pi \) cancels out:
\[ \frac{dV}{dt} = \frac{3 \times 3960}{360} = \frac{11880}{360} = 33 \text{ cm}^3/\text{sec} \]

### Step 7: Determine Divisibility of \( n \)
The value of \( n \) is 33. We need to check its divisibility:
- 33 is divisible by 1, 3, 11, and 33.

### Final Answer
The value of \( n \) is 33, and it is divisible by 1, 3, 11, and 33.