- 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
A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, 2/5 of the water flows out.
Find the number of lead shots dropped into the vessel.
Solution:
The volume of water flown out of the vessel will be equal to the total volume of the lead shots immersed in the vessel.
Given height of the cone, h = 11 cm
Radius of the cone, r = 2.5 cm
Volume of the cone = (1/3)π r2h
= (1/3)π x2.52×11
= (11/3)π x 6.25 cm3
When lead shots are dropped into vessel, (2/5) of water flows out.
Volume of water flown out = (2/5)π x (11/3)x6.25
= (22/15)π x 6.25
= (137.5/15) π cm3
Radius of sphere, R = 0.25 cm = ¼ cm
Volume of sphere = (4/3)π R3
= (4/3)π x (1/4)3
= π /48 cm3
Number of lead shots dropped = Volume of water flown out/ Volume of sphere
= (137.5/15)π ÷π /48
= (137.5/15)π x 48/π
= 137.5×48/15
= 440
Hence the number of lead shots dropped is 440.
Facebook
Whatsapp
Copy Link
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
