Saved Bookmarks
| 1. |
A spherical ball of salt is dissolving in water in such a manner thatthe rate of decrease of volume at any instant is proportional to the surface.Prove that the radius is decreasing at a constant rate. |
|
Answer» We have, rate of decrease of the volume of spherical ball of salt at any instant is `propto` surface. Let the radius of the spherical ball of the salt be r. `therefore` Volume of the ball (V) `=4/3pir^(3)` and surface area (S) = `4pir^(2)` `therefore (dV)/(dt) propto S rArr d/(dt) (4/3pir^(3)) propto 4pir^(2)` `rArr 4/3.pi.3r^(2).(dr)/(dt) propto 4pir^(2) rArr (dr)/(dt) propto (4pir^(2))/(4pir^(2))` `(dr)/(dt) = k.1` [where, k is the proportionally constant] `rArr (dr)/(dt) = k` Hence, the radius of ball is decreasing at a constant rate. |
|