An ice sculpture in the form of a sphere melts in such a way that it maintains its spherical shape. The volume of the sphere is decreasing at a constant rate of $2 \pi$ cubic meters per hour. At what rate, in square meters per hour, is the surface area of the sphere decreasing at the moment when the radius is 5 meters? (Note: For a sphere of radius $r$, the surface area is $4 \pi r ^ { 2 }$ and the volume is $\frac { 4 } { 3 } \pi r ^ { 3 }$.)
(A) $\frac { 4 \pi } { 5 }$
(B) $40 \pi$
(C) $80 \pi ^ { 2 }$
(D) $100 \pi$

121. The volume of a sphere is increasing at a constant rate of $3$ cubic centimeters per second. At the moment when the radius of the sphere is $8$ centimeters, the surface area of the sphere increases how many square centimeters per second?
(1) $1/2$ (2) $1/25$ (3) $1/5$ (4) $1/6$

A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at a rate of $50 \mathrm {~cm} ^ { 3 } / \mathrm { min }$. When the thickness of ice is 5 cm , then the rate (in $\mathrm { cm } / \mathrm { min }$.) at which of the thickness of ice decreases, is:
(1) $\frac { 5 } { 6 \pi }$
(2) $\frac { 1 } { 54 \pi }$
(3) $\frac { 1 } { 36 \pi }$
(4) $\frac { 1 } { 18 \pi }$

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm, the ice-cream melts at the rate of $81 \mathrm {~cm} ^ { 3 } / \mathrm { min }$ and the thickness of the ice-cream layer decreases at the rate of $\frac { 1 } { 4 \pi } \mathrm {~cm} / \mathrm { min }$. The surface area (in $\mathrm { cm } ^ { 2 }$) of the chocolate ball (without the ice-cream layer) is :
(1) $196 \pi$
(2) $256 \pi$
(3) $225 \pi$
(4) $128 \pi$