Class 9 Volume of a Cone — practice questions with answers

Mensuration · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.

1. The volume of a cone is given by which formula? (r = base radius, h = height)

A $\frac{1}{3}$ pi r² h
Bpi r² h
C $\frac{2}{3}$ pi r² h
D $\frac{1}{3}$ pi r h²

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Answer: A)

\frac{1}{3}

pi r² h

A cone is one-third of a cylinder of the same base and height, so Volume =

\frac{1}{3}

pi r² h.

2. A cone has slant height 5 cm and base radius 3 cm. Find its volume. (Use pi = 3.14)

A37.68 cm³
B47.1 cm³
C12.56 cm³
D75.36 cm³

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Answer: A) 37.68 cm³

h =

l^{2} - r^{2}

25 - 9

= 4 cm. V =

\frac{1}{3}

(3.14)(9)(4) =

\frac{1}{3}

(113.04) = 37.68 cm³.

3. The radius of a cone is halved and its height is kept the same. Its volume becomes ___ of the original.

A $\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{1}{8}$
D $\frac{1}{3}$

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Answer: A)

\frac{1}{4}

V ~ r². Halving r multiplies volume by

(\frac{1}{2})^{2} = \frac{1}{4}

4. A right triangle with legs 3 cm and 4 cm is rotated about the leg of length 4 cm.
Find the volume of the cone so generated. (Use pi = 3.14)

A75.36
B33.68
C37.68
D41.68

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Answer: C) 37.68

Rotating about the 4 cm leg gives a cone with radius 3 cm (the other leg) and height 4 cm. V =

\frac{1}{3}

(3.14)(9)(4) = 37.68 cm³.

5. Water flows at 10 m/min through a pipe of radius 7 cm into a conical tank of base radius 1 m and height 2 m, initially empty.
How many minutes are needed to fill the conical tank completely? (Use pi =

\frac{22}{7}

)

AAbout 13.6 min
BAbout 6.8 min
CAbout 27.2 min
DAbout 2 min

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Answer: A) About 13.6 min

Tank volume =

\frac{1}{3}

(

\frac{22}{7}

)(1²)(2) =

\frac{44}{21}

m³ = 2.0952 m³. Pipe area = (

\frac{22}{7}

)(0.07²) = 0.0154 m²; flow rate = 0.0154 x 10 = 0.154 m³/min. Time =

\frac{2.0952}{0.154}

= 13.6 min.

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