 # ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.5 in Chapter 17 - Mensuration

A hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted into a cone of base diameter 8 cm.

Find the height of the cone.

Solution:

When we convert one solid shape to another, its volume remains the same, no matter how different the new shape is. So here sphere is converted into the cone.

Given internal diameter of hollow sphere = 4 cm

Internal radius, r = 4/2 = 2 cm

External diameter = 8 cm

External radius, R = 8/2 = 4 cm

Volume of the hollow sphere, V = (4/3)π (R3-r3)

V = (4/3)π ×(43-23)

V = (4/3)π ×(64-8)

V = (4/3)π ×56

Base diameter of the cone = 8 cm

Radius, r = 8/2 = 4 cm

Volume of cone = (1/3)π r2h

= (1/3)π ×42 ×h

= (16/3)π ×h

Since sphere is melted and changed into a cone, their volumes remain same.

(4/3)π ×56 = (16/3)π ×h

h = 4×56/16

h = 14 cm

Hence the height of the cone is 14 cm. 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!