- 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
Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.
Solution:
When we convert one solid shape to another, its volume remains the same, no matter how different the new shape is.
For same material, density will be same.
Density = mass/Volume
Mass of the smaller sphere, m1 = 1 kg
Mass of the bigger sphere, m2 = 7 kg
The spheres are melted to form a new sphere.
So the mass of new sphere, m3 = 1+7 = 8 kg
Density of smaller sphere = density of new sphere
Let V1 be volume of smaller sphere and V3 be volume of bigger sphere.
m1/V1 = m3/V3
1/V1 = 8/V3
V1/ V3 = 1/8 …(i)
Given radius of the smaller sphere, r = 3 cm
V1 = (4/3)π r3
V1 = (4/3) π 33
V1 = 36π
Let R be radius of new sphere.
V3 = (4/3)π R3
V1/ V3 = 36π ÷(4/3)π R3
V1/ V3 = 27/R3 …(ii)
From (i) and (ii)
1/8 = 27/R3
R3 = 27×8 = 216
Taking cube root on both sides,
R = 6 cm
So diameter of the new sphere = 2R = 2×6 = 12 cm
Hence diameter of the new sphere is 12 cm.
Facebook
Whatsapp
Copy Link
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
