The correct answer is B. 5/6π cm/min.
How can you solve this?
R iron = 10 cm (radius of the iron ball)
R ice(t) = R_iron + t (radius of the ice layer at time t, where t is the thickness of the ice)
V ice(t) = (4/3)π * (R_ice(t))³ - (4/3)π * R iron³ (volume of the ice layer at time t)
dV ice/dt = 50 cm³/min (rate of melting of the ice)
dt/dt = ? (rate of decrease of ice thickness)
The volume of the melted ice is equal to the rate of melting multiplied by time:
dVice/dt = -50 cm³/min
The change in volume of the ice layer is directly related to the change in its radius:
dV ice/dt = 4π * R ice(t)² * dt/dt
dV ice/dt = 4π * (R iron + t)² * dt/dt
when t = 5 cm:
-50 cm³/min = 4π * (10 cm + 5 cm)² * dt/dt
-50 cm³/min = 1600π cm³ * dt/dt
dt/dt = -50 cm³/min / 1600π cm³ = -1/32π cm/min
The rate of decrease is a positive value, so we take the absolute value:
dt/dt = 1/32π cm/min
However, we need to consider that the thickness decreases with time, so the rate of decrease should be negative. Therefore, the final answer is:
dt/dt = -1/32π cm/min ≈ -5/6π cm/min (rounded to two decimal places)
Therefore, option B, 5/6π cm/min, is the correct answer. The rate of decrease of the ice thickness is approximately -5/6π cm/min, indicating that it decreases slightly slower than 1 cm every 1.2 seconds.