- GST
- Banking
- Shares and Dividends
- Quadratic Equations in One Variable
- Factorization
- Ratio and Proportion
- Matrices
- Arithmetic and Geometric Progression
- Reflection
- Section Formula
- Equation of Straight Line
- Similarity
- Locus
- Circles
- Constructions
- Mensuration
- Trigonometric Identities
- Trigonometric Tables
- Heights and Distances
ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.5 in Chapter 17 - Mensuration
Prev
Eight metallic spheres, each of radius 2 cm, are melted and cast into a single sphere. Calculate the radius of the new (single) sphere.
Solution:
The capacity of a sphere is its volume. It is the area that the sphere occupies. The sphere has a round, three-dimensional shape. Its shape is determined by three axes: the x, y, and z axes.
Given radius of each sphere, r = 2 cm
Volume of a sphere = (4/3)π r3
= (4/3)π ×23
= (4/3)π ×8
= (32/3) π cm3
Volume of 8 spheres = 8×(32/3)π
= (256/3)π cm3
Let R be radius of new sphere.
Volume of the new sphere = (4/3)π R3
Since 8 spheres are melted and casted into a single sphere, volume remains same.
(4/3)π R3 = (256/3)π
4R3 = 256
R3 = 256/4 = 64
Taking cube root
R = 4 cm
Hence the radius of the new sphere is 4 cm.
Facebook
Whatsapp
Copy Link
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
