00:01
Here in the question, we are given a solid sphere which has radius r and it acquires a terminal velocity v1 when falling through a viscous fluid which has a coefficient of viscosity eta.
00:15
Then this sphere is broken into 27 identical solid sphere which have radius r and same density as the original sphere and each of this sphere acquires a terminal velocity.
00:30
V2 when falling through the same fluid of same density and viscosity and we know that the formula we know that the terminal velocity of a sphere falling down a liquid is given by v t is equal to 2 by 9 r square by eta ro -nought minus ro l multiply g so here r signifies the radius of r signifies the radius of r -squire the radius of sphere, ro not is the density of sphere while ro l is the density of the liquid through which it is falling and g signifies the acceleration due to gravity for a given liquid and a given sphere of fixed density all parameters on the right hand side are constant except the radius except the radius of the sphere we can say that the terminal velocity of a given sphere of radius r is directly proportional to the square of the radius.
01:34
Therefore, for an initial solid sphere, we can write that v1 is directly proportional to the r square, while for each of the 27, while for each of the 27 identical sphere, we can write v2 is directly proportional to small r square.
01:57
R -square, dividing them we get v1 by v2 is equal to r -square by small r -square.
02:07
So we take it as equation number 1.
02:10
Now we need to find out the relation between the radia a of the original sphere and the 27 identical sphere.
02:17
The volume of the initial sphere will be equal to the sum of the volume of the 27 identical sphere...