To solve this problem, let's start by understanding how the terminal velocity of a falling drop depends on its radius. The terminal velocity
(vt) of a spherical drop falling under gravity in a viscous medium is given by:
vt∝r2 where
r is the radius of the drop. Here, we are given that 64 drops of the same radius coalesce to form a single larger drop. To find the radius of the new larger drop, we use the volume conservation principle. The volume of a sphere is given by:
V=πr3 If 64 small drops, each with radius
r, coalesce to form one larger drop, the volume of the larger drop will be 64 times the volume of one small drop. Therefore:
π(rlarge )3=64⋅πr3 From this equation, we can solve for the radius of the larger drop:
(rlarge )3=64r3rlarge =3√64⋅rrlarge =4r So, the radius of the larger drop is 4 times the radius of one of the smaller drops. Since terminal velocity is proportional to the square of the radius, we get:
vlarge ∝(rlarge )2=(4r)2=16r2 Therefore, the terminal velocity of the larger drop is 16 times the terminal velocity of one of the smaller drops. Given that the terminal velocity of the smaller drop is
0.5cm∕ s, we have:
vlarge =16⋅0.5cm∕ svlarge =8cm∕ s We need to convert this to
m∕ s as all provided options except A are in meters per second
vlarge =8cm∕ s×vlarge =0.08m∕ sTherefore, the correct answer is:
Option B
0.08ms−1