NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.3
Question 3

Question. 3

A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm, respectively. Find the height of the bucket.

Answer:

15 cm

Detailed Answer with Explanation:

Step 1: Write down the known values.

  • Volume of water = \(28.490\,\text{litres}\).
  • Top radius (\(R\)) = \(28\,\text{cm}\).
  • Bottom radius (\(r\)) = \(21\,\text{cm}\).
  • Height (\(h\)) = ? (to be found)

Step 2: Convert volume into cubic centimetres (SI unit for volume).

We know \(1\,\text{litre} = 1000\,\text{cm}^3\).

So, \(28.490\,\text{litres} = 28.490 \times 1000 = 28490\,\text{cm}^3\).

Step 3: Recall the formula for the volume of a frustum of a cone.

\[ V = \dfrac{1}{3}\pi h \big(R^2 + r^2 + Rr\big) \]

Step 4: Substitute the values of \(R\) and \(r\).

\(R^2 = 28^2 = 784\)

\(r^2 = 21^2 = 441\)

\(Rr = 28 \times 21 = 588\)

So, \(R^2 + r^2 + Rr = 784 + 441 + 588 = 1813\).

Step 5: Put values in the volume formula.

\(28490 = \dfrac{1}{3}\pi h (1813)\)

Step 6: Simplify the equation.

Multiply denominator first: \(\dfrac{1}{3} \times 1813 = 604.33\).

So, equation becomes: \(28490 = 604.33 \pi h\).

Step 7: Use \(\pi \approx 3.1416\).

\(604.33 \pi \approx 604.33 \times 3.1416 = 1899.56\).

So, \(28490 = 1899.56 h\).

Step 8: Solve for \(h\).

\(h = \dfrac{28490}{1899.56} \approx 15.0\,\text{cm}\).

Final Answer: The height of the bucket is 15 cm.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 12: Surface Areas & Volumes – Exercise 12.3 | Detailed Answers