Answer:
0.022m or 2.2cm
Expxlanation:
Step 1:
Data obtained from the question. This includes:
Mass (m) = 2.5g = 2.5/1000 = 2.5x10^-3Kg
Tension (T) = 0.029 N
Density (ρ) = 1000 kg/m3
Acceleration due to gravity (g) = 9.81 m/s2
Diameter (d) =?
Step 2:
Finding an expression to calculate the diameter of the ball. This is illustrated below:
Tension = weight displaced - weight of the ball
Weight displaced = Mass of water x acceleration due to gravity
Mass of water = Density x volume
Mass of water = ρxV
Weight displaced = ρxVxg = ρVg
Weight of the ball = Mass of the ball x acceleration due to gravity
Weight of the ball = mg
Therefore,
Tension = weight displaced - weight of the ball
T = ρVg - mg
Make V the subject of the formula
T = ρVg - mg
T + mg = ρVg
Divide both side by ρg
V = ( T + mg) /ρg. (1)
Recall that the ball is spherical in shape and the Volume of a sphere is given by
V = 4/3πr^3
Radius (r) = diameter (d) /2
V = 4/3π(d/2)^3
V = 4/3πd^3/8
V = πd^3 /6
Substituting the value of V into equation 1, we have
V = ( T + mg) /ρg
πd^3 /6 = ( T + mg) /ρg.
Making d the subject of the formula, we have:
πd^3 /6 = (T + mg) /ρg.
d^3 = 6(T + mg) /πρg.
Taking the cube root of both sides
d = [6(T + mg) /πρg]^1/3
Step 3:
Determination of the diameter of the ball. This is illustrated below:
T = 0.029 N
m = 2.5x10^-3Kg
g = 9.81 m/s2
ρ = 1000 kg/m3
d =?
d = [6(T + mg) /πρg]^1/3
d = [6(0.029 + 2.5x10^-3x9.81)/ πx1000x9.81]^1/3
d = 0.022m
Therefore, the diameter of the ball is 0.022m or 2.2cm
, with 
being the elementary charge, and 
and 
are the initial and final voltage.
