 Hello and welcome to the session. Let us understand the following question which says sand is pouring from a pipe at the rate of 12 centimeter cube per second. The falling sand forms a cone on the ground in such a way that the height of the cone is always one sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 centimeter? Now let us proceed on to the solution. Here we can see a cone of height h and radius r. It is given to us that and is pouring from the pipe at the rate of 12 centimeter per second that is dv by dt is equal to 12. Also it is given to us that falling sand forms a cone on the ground in such a way that height of the cone is always one sixth of the radius of the base that is h is equal to 1 by 6 of r which implies r is equal to 6h. We know volume of cone is given by v is equal to 1 by 3 r square h that is v is equal to now substituting the value of r is equal to 6h we get 1 by 3 pi 6h the whole square multiplied by h. It implies v is equal to 1 by 3 multiplied by pi multiplied by 6h multiplied by 6h multiplied by h. Here 3 gets cancelled at 6 and we get here 2. It implies v is equal to 2 multiplied by 6 12 pi h cube. Now differentiating it with respect to t we get dv by dt is equal to 12 pi multiplied by 3h square dh by dt. It implies putting the value of dv by dt and h we get 12 is equal to 12 pi multiplied by 3 h is given to us as 4. Height is 4 centimetre so 3 multiplied by 4 square multiplied by dh by dt. This 12 and 12 gets cancelled so it implies dh by dt is equal to 1 by pi 3 multiplied by 4 12 multiplied by 4 is equal to 48. Therefore dh by dt is equal to 1 by 48 pi hence height is increasing at the rate of 1 by 48 pi centimetre per second which is the required answer. I hope you understood this question. Bye and have a nice day.