 Let's solve a question on continuity equation. A pipe is joined with two narrower pipes as shown here. A stream of non-viscous liquid enters from the left, splits in the middle and flows out of the two narrow openings. We have some data. Cross-sectional area of the entry A1 that is 5 cm2 and this right here can be A1. Speed of the liquid at entry is 7 m per second, cross-sectional area of the top exit A2 that is 2 cm2, speed of the liquid at the top exit is 4 m per second and the speed of the liquid at the bottom exit V3 is 10 m per second. What is the cross-sectional area of the bottom exit? We need to choose one answer out of these four options. As always pause the video and first try this question on your own. Alright, hopefully you have given this a shot. Now let's try to make sense of all the data that has been given to us. So to begin with we have cross-sectional area of the entry A1 that is 5 cm2. Let's label that on the diagram. This right here, this is 5 cm2. Speed of the liquid is 7 m per second. First let's look at all the cross-sectional areas. The cross-sectional area of the top exit A2, this right here can be called A2 and this is 2 cm2. This is 2 cm2. We don't know the area of the bottom exit that is what we need to figure out. Now the key idea here is that the volume of the water entering, so let's say this much volume of water that is entering, this water some of it is going from the top exit, so some of it is going from here and some of it is going from here. If we take any small time interval, if we take delta t, if we take the time interval of delta t, we can say that in this time interval the amount of water that entered from the left and when we say amount we really can say the volume of water because this is not a two-dimensional tube right, it also has an area and length. So the volume of water that is entering, some of it is going from the top, some of it is going from the bottom. So we can say that the volume that is entering V1, this is equal to V2, the volume that is exiting from the top and the volume that is exiting from the bottom, the addition of these two. So this is V2 plus V3. Let's think about the volume of water that enters from the left in a time interval of delta t. So we already know the speed with which the water is moving and from the Godot equation we know that speed is equal to distance upon time. So we can write distance as, let's write that over here. We know that speed is equal to distance upon time. So distance really is speed into time and the velocity with which the water is entering from the left that is, we can call that V1, we have a variable, we will put in the numbers later but for now let's call it V1. And in a time interval of delta t, the length that the water travels would be V1 into delta t, this would be V1 into delta t. This is the distance, this is the distance that the water travels in a time interval of delta t if it is moving with the velocity of V1. So for this one, for the top exit, the speed of the liquid at the top exit, this is really V2 and this distance then would be V2 into delta t. Similarly for the bottom exit, the speed here of exit that is V3, so the distance would be, this would be V3 into delta t. And now if we think about the volume, if we think about the volume, we already know the area and V delta t, think about it, it has the unit of distance, right? Because d equals to st velocity of speed into time. So if you multiply area, if you multiply the cross section area of this part of the tube with this distance, we should be able to figure out the volume of water that entered in a time interval of delta t. So if you multiply the area, let's call this A1 for now. This is A1, this is A2 and this is A3, what we need to figure out. So the volume that entered, this is A1 into V1 delta t and this is equal to V2 that is A2, A2 into V2 delta t and adding A3 into V3 delta t. Now we can see that delta t really gets cancelled and we need to figure out A3. Everything else we already know, we know V1, V2, V3 that's the data is given to us and we also know A1 and A2. So let's keep A3 on one side and everything else on the other side. So when we keep A3 on the right hand side, let's take A2 V2 on the left hand side, this becomes A1 V1 minus A2 V2 divided by V3 and this is equal to A3. Now let's put in the values, A1 here is 5 centimeter square. So this is 5 centimeter square multiplied by V1 which is 7 meters per second. This is 7 meters per second and we are subtracting A2 V2 from it. So A2 is 2, 2 centimeter square and V2 that is 4 meters per second and this is divided by 10 meters per second. A3 is equal to this really. This is equal to A3. So now when we work this out, this is when we multiply 7 meters per second with 5 centimeters per square, this becomes equal to 35 centimeters square into meters per second and we are subtracting 8 centimeters square into meters per second with this and dividing this with 10 meters per second. So the numerator would be 27 centimeters square meter per second dividing that by 10 meters per second. So meters per second that unit gets cancelled off and 27 divided by 10 that is 2.7. 27, 35 minus 8, 27 divided by 10 that is 2.7 centimeters square. So the answer is this, that is A3. This area it is 2.7 centimeters square. All right, you can try more questions from this exercise in the lesson and if you're watching on YouTube, do check out the exercise link which is added in the description.