NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.2
Question 2

Question. 2

A solid cylinder of radius \(r\) and height \(h\) is placed over another cylinder of same height and radius. The total surface area of the shape so formed is \(4\pi rh + 4\pi r^2\).

Answer:

false

Detailed Answer with Explanation:

Step 1: Understand the situation.

We have two identical cylinders. Each has radius \(r\) (in metres) and height \(h\) (in metres). When one is placed on top of the other, we get a single tall cylinder of:

Height = \(h + h = 2h\) (in metres).

Radius = \(r\) (remains the same, in metres).

Step 2: Recall the formula for total surface area (TSA) of a cylinder:

\[ \text{TSA} = 2\pi r h + 2\pi r^2 \]

Here:

  • \(2\pi r h\) is the curved surface area (CSA), measured in square metres (m²).
  • \(2\pi r^2\) is the area of the two circular ends (top and bottom), also in m².

Step 3: Apply the formula to the new cylinder.

For height = \(2h\), radius = \(r\):

\[ \text{TSA} = 2\pi r(2h) + 2\pi r^2 \]

Simplify:

\[ \text{TSA} = 4\pi r h + 2\pi r^2 \]

Step 4: Compare with the given statement.

The question says TSA = \(4\pi rh + 4\pi r^2\).

But we found TSA = \(4\pi rh + 2\pi r^2\).

Final Answer: The given statement is incorrect. So the answer is False.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 12: Surface Areas & Volumes – Exercise 12.2 | Detailed Answers