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

Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboid pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?

Answer:

**Solution:**

Here the volume of water in pipe is equal to the volume of water transferred to the pond which results in increase of water level in the pond.

Given speed of water flow = 15 km/h

Diameter of pipe = 14 cm

So radius of pipe, r = 14/2 = 7 cm = 0.07 m

Dimensions of cuboidal pond = 50 m × 44 m

Height of water in pond = 21 cm = 0.21 m

So volume of water in pond = 50×44 ×0.21

= 462 m^{3}

Volume of water in pipe = π r^{2}h

= π x 0.07^{2}×h

= 0.0049h

Volume of water in pond = Volume of water in pipe

462 = 0.0049h

h = 462/0.0049

= 462×7/0.0049×22

= 30000 m

= 30 km, (1 km = 1000 m)

Speed = distance/ time

Time = Distance/speed = 30/15 = 2 hr

**Hence the time taken is 2 hours.**

Related Questions

Was This helpful?

Exercises

Chapters

Lido

Courses

Quick Links

Terms & Policies

Terms & Policies

2022 © Quality Tutorials Pvt Ltd All rights reserved