NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.4
Question 9

Question. 9

A solid iron cuboid \(4.4\,\text{m}\times2.6\,\text{m}\times1\,\text{m}\) is recast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

Answer:

112 m

Detailed Answer with Explanation:

Step 1: Volume of the cuboid

The cuboid is given with dimensions: length = 4.4 m, breadth = 2.6 m, height = 1 m.

Volume of cuboid = length × breadth × height

= \(4.4 \times 2.6 \times 1 = 11.44 \, \text{m}^3\)

Step 2: Understand the pipe structure

The pipe is hollow. This means it has an outer cylinder and an inner cylinder.

  • Internal radius (r) = 30 cm = 0.30 m
  • Thickness = 5 cm = 0.05 m
  • So, outer radius (R) = internal radius + thickness = 0.30 + 0.05 = 0.35 m

Step 3: Cross-sectional area of the hollow pipe

Cross-sectional area of hollow cylinder (annulus) = \(\pi (R^2 - r^2)\)

= \(\pi (0.35^2 - 0.30^2)\)

= \(\pi (0.1225 - 0.0900)\)

= \(\pi (0.0325)\)

= \(0.0325\pi \, \text{m}^2\)

Step 4: Volume of the pipe

Volume of hollow cylinder = cross-sectional area × length

= \(0.0325\pi \times L\)

Step 5: Equating volumes

The cuboid is recast into the pipe, so volumes are equal:

\(11.44 = 0.0325\pi L\)

Step 6: Solve for length

\(L = \dfrac{11.44}{0.0325\pi}\)

\(L \approx 112\, \text{m}\)

Final Answer: The length of the pipe is 112 m.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 12: Surface Areas & Volumes – Exercise 12.4 | Detailed Answers