 Hello and welcome to the session. In this session we discussed the falling question which says a cylindrical tub of radius 9 cm contains water to a depth of 20 cm. A spherical island ball is dropped into the tub and thus the level of water is raised by 1.5 cm. What is the radius of the ball? We should first know that the volume of the cylinder is equal to pi r square H where this r is the radius of the base of the cylinder H is the height of the cylinder. Then the volume of sphere is equal to 4 upon 3 pi r2 and here this r is the radius of the sphere. This is the key idea for this question. Now we move on to the solution. Now the radius of the cylindrical tub say r1 is given as 9 cm. Now the depth of the water in the tub say H1 is given as 20 cm. The level of water in the tub is raised by 1.5 cm. So the volume of the cylindrical tub is given by pi 9 square into 1.5 cm2. We consider let the radius of the spherical ball be r cm. So the volume of the spherical ball would be equal to 4 upon 3 pi r cube cm2. Now as in the question is given to us that the spherical ball is dropped into the tub and this raises the level of water in the tub by 1.5 cm. So we have that the volume of the cylindrical tub is equal to the volume of the spherical ball. That is we get pi 9 square into 1.5 is equal to 4 upon 3 pi r cube. So from here we get r cube is equal to pi 9 square into 1.5 upon 4 upon 3 pi. Now here pi and pi cancels. This is further equal to 9 into 9 into 3 into 15 upon 4 into 10. Now 5 3 times is 15 and 5 2 times is 10. So this is equal to 9 into 9 into 3 into 3 upon 4 into 2. This is r cube or we can say r cube is equal to 3 into 3 into 3 into 3 into 3 into 3 upon 2 into 2 these are the triplet forms. So we get r is equal to 3 into 3 upon 2 equal to 9 upon 2. So this means we have r is equal to 4.5 cm or we can say the radius of the spherical ball is equal to 4.5 cm. So this is our required answer. This completes the session. Hope you have understood the question.