Jump to

- 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

Question 1 Mensuration Exercise 17.5

A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water.

Calculate the rise in water level, assuming that no water overflows.

Answer:

**Solution:**

When a solid sphere is totally submerged in water, the volume of the water that it displaces is equal to its own volume.

Given internal diameter of cylindrical can = 21 cm

Radius of the cylindrical can, R = 21/2 cm

Diameter of sphere = 10.5 cm

Radius of the sphere, r = 10.5/2 = 21/4 cm

Let the rise in water level be h.

Rise in volume of water = Volume of sphere immersed

π R^{2}h = (4/3)π r^{3}

π (21/2)^{2}h = (4/3)π ×(21/4)^{3}

(21/2) ×(21/2)×h = (4/3)× (21/4)×(21/4)× (21/4)

h = 21/12

h = 7/4

h = 1.75 cm

**Hence the rise in water level is 1.75 cm.**

Related Questions

Was This helpful?

Exercises

Chapters

Lido

Courses

Quick Links

Terms & Policies

Terms & Policies

2022 © Quality Tutorials Pvt Ltd All rights reserved