00:01
In the question, a garden hose is given in which the initial diameter d1 is given as 2 .74 centimeter.
00:09
So converted it into meter.
00:12
Then volume of the bucket which is to be filled with the help of garden hose is 25 liter and the time of 1 .5 minute takes it to fill the bucket that is 90 second.
00:26
Now in the part a of the question we have to find the speed of water which is leaving hose.
00:33
Let's assume as v1 and in the part b we have to find the speed of water which is leaving the nozzle which is v2.
00:43
So let's try to draw the diagram of the garden hose as well as the nozzle.
00:51
So let's take this is the garden hose in which the.
00:56
First end diameter is d1.
01:00
This is d1.
01:03
Then over here we can take the next diameter, but this will be d2 as it is given in the question.
01:14
Now over here a nozzle is connected of some smaller diameter.
01:20
Let's assume this is d3.
01:23
So according to the question, we can take first of all in part a that the volume rate, it means q is equal to volume divided by time for the part d1.
01:44
So we are taking it for d1.
01:47
So this is volume will be 25 multiplied by 10 to 10 to the power minus 3.
01:54
And time is 90 second.
01:58
So we get the answer for the volume rate as, hence we get q equals to 5 divided by 18, multiplied by 10 to the power minus 3 meter cube divided by second.
02:13
Now this is not direct answer asked in the question, but we need to find out further quantities with the help of q.
02:22
Now we take the volume rate equal to the product of area and the speed.
02:29
Hence we get speed v1 equal to q by a, which is just the speed of the water which is leaving the hose.
02:38
So this is 5 multiplied by 10 to the power minus 3 by 18...