Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Now velocity of water coming out of the hole $=v=\sqrt {2gh}$

$\qquad= {2 \times 10 \times 0.6}$

$\qquad= \sqrt {12}\;m/s$

If a is area of cross section of the hole.

$a \times v$= volume of water flowing per second.

$\therefore\; Thrust =\large\frac{Volume}{Second} $$ \times \rho \times velocity$

$Thrust = \rho av^2$

$\qquad= 10^3 \times \large\frac{\pi}{4}$$ (10^{-2})^2 (\sqrt {12})^2$

$\qquad= \large\frac{3.14 \times 10^{-1} \times 12}{4}$

$\qquad= 0.942\;N$

Hence b is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...