Water running in a cylindrical pipe …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Test
Related Questions
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 1 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 0 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers
Posted by Kanika . 1 month ago
- 1 answers
Posted by Hari Anand 6 months, 1 week ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 6 years, 5 months ago
Flow rate {tex}= \frac { \text { Volumetric flow rate } } { \text { Area } }{/tex}
Area = {tex}\pi r ^ { 2 }{/tex}
{tex}= \frac { 22 } { 7 } \times \frac { 7 } { 2 } \times \frac { 7 } { 2 } = 38.5 \mathrm { cm } ^ { 2 }{/tex}
Volumetric flow rate = 192.5 l/min.
Since 1 l = 0.001 m3 = 1000 cm3
So, volumetric flow rate = 192.5 {tex}\times{/tex} 1000 cm3/min
So, flow rate per 38.5 cm2 {tex}= \frac { 192.5 \times 1000 \space cm^3 /min} { 38.5\space cm^2 } {/tex}
{tex}\Rightarrow{/tex} flow rate = 5000 cm/min
{tex}= \frac { 5000 \times 0.00001 \mathrm { km } } { \left( \frac { 1 } { 60 } \right) \mathrm { h } }{/tex}
= 3 km/hr
Thus, water is flowing at the speed of 3 km/hr in the water pipe
0Thank You