# RS Aggarwal Solutions Class 9 Mathematics Solutions for Volume And Surface Area Of Solids Exercise 15D in Chapter 15 - Volume And Surface Area Of Solids

Question 21 Volume And Surface Area Of Solids Exercise 15D

A cylindrical bucket with base radius 15cm is filled with water up to a height of 20cm. A heavy iron spherical ball of radius 9cm is dropped into the bucket to submerge completely in the water. Find the increase in the level of water.

It is given that

Radius of the cylindrical bucket = 15cm

Height of the cylindrical bucket = 2cm

We know that

Volume of water in bucket = π r2h

By substituting the values

Volume of water in bucket = (22/7) × 152 × 20

So we get

Volume of water in bucket = 14142.8571 cm3

It is given that

Radius of spherical ball = 9cm

We know that

Volume of spherical ball = 4/3 πr3

By substituting the values

Volume of spherical ball = 4/3 × (22/7) × 93

So we get

Volume of spherical ball = 3054.8571 cm3 ……. (1)

Consider h cm as the increase in water level

So we get

Volume of increased water level = π r2h

By substituting the values

Volume of increased water level = (22/7) × 152 × h …… (2)

By equating both the equations

3054.8571 = (22/7) × 152 × h

On further calculation

h = 3054.8571/ ((22/7) × 152) = 4.32 cm

Therefore, the increase in the level of water is 4.32cm.

Related Questions

Lido

Courses

Teachers

Book a Demo with us

Syllabus

Maths
CBSE
Maths
ICSE
Science
CBSE

Science
ICSE
English
CBSE
English
ICSE
Coding

Terms & Policies

Selina Question Bank

Maths
Physics
Biology

Allied Question Bank

Chemistry
Connect with us on social media!