A cylindrical tank (height h,radius r)is full to the brim of water and its top is open to the outside air.What expression describes the speed of fluid flowing out of a hole that is opened up at height h' above the bottom of the tank?
A) - 2 g h'
B) $\sqrt { 2 g \left( h + h ^ { \prime } \right) }$
C) $\sqrt \frac { h + h ^ { l } } { 2 g }$
D) $\sqrt { 2 g \left( h - h ^ { \prime } \right) }$

Question

Quizplus · Accepted Answer

The speed of fluid flowing out of a hole in a cylindrical tank can be determined using Torricelli's law, which states that the speed of a liquid flowing out of a small hole in a container is equal to the speed of an object falling freely under gravity from a height equal to the depth of the hole below the surface of the liquid.

Therefore, the speed of fluid flowing out of a hole at height h' can be given by:

v = sqrt(2gh)

where g is the acceleration due to gravity, and h is the depth of the hole below the water surface.

Since the height of the water surface from the bottom of the tank is h, the depth of the hole from the bottom of the tank is (h - h'). Therefore, the speed of fluid flowing out of a hole at height h' can be given by:

v = sqrt(2g(h - h'))

which is option D.

A Cylindrical Tank (Height H,radius R)is Full to the Brim 2g(h+h′)\sqrt { 2 g \left( h + h ^ { \prime } \right) }2g(h+h′)​

A Cylindrical Tank (Height H,radius R)is Full to the Brim $\sqrt { 2 g \left( h + h ^ { \prime } \right) }$