Ashima has three cylindrical containers .
(i) P having radius 2r and height h .
(ii) Q having radius 3r and height 2h.
(iii) R having radius 3r and height `h/3` .
Arrange the cureved surface area of containers in decreasing order .

Options

A. Q,P,R
B. P,R,Q
C. P,Q,R
D. Q,R,P

Answer

To solve the problem of arranging the curved surface areas of the three cylindrical containers in decreasing order, we will follow these steps:

### Step 1: Calculate the Curved Surface Area of Container P
The formula for the curved surface area (CSA) of a cylinder is given by:
\[ \text{CSA} = 2\pi r h \]
For container P:
- Radius \( r = 2r \)
- Height \( h = h \)

Substituting the values:
\[ \text{CSA of P} = 2\pi (2r)(h) = 4\pi rh \]

### Step 2: Calculate the Curved Surface Area of Container Q
Using the same formula for container Q:
- Radius \( r = 3r \)
- Height \( h = 2h \)

Substituting the values:
\[ \text{CSA of Q} = 2\pi (3r)(2h) = 12\pi rh \]

### Step 3: Calculate the Curved Surface Area of Container R
For container R:
- Radius \( r = 3r \)
- Height \( h = \frac{h}{3} \)

Substituting the values:
\[ \text{CSA of R} = 2\pi (3r)\left(\frac{h}{3}\right) = 2\pi (3r) \cdot \frac{h}{3} = 2\pi rh \]

### Step 4: Compare the Curved Surface Areas
Now we have the curved surface areas for all three containers:
- CSA of P = \( 4\pi rh \)
- CSA of Q = \( 12\pi rh \)
- CSA of R = \( 2\pi rh \)

### Step 5: Arrange in Decreasing Order
To arrange the curved surface areas in decreasing order:
- \( CSA \, of \, Q = 12\pi rh \) (largest)
- \( CSA \, of \, P = 4\pi rh \) (middle)
- \( CSA \, of \, R = 2\pi rh \) (smallest)

Thus, the order is:
\[ Q > P > R \]
So, the final arrangement in decreasing order is:
\[ \text{Q, P, R} \]