A solid sphere of gold of radius 5 cm, is being melted in a furnace such that the radius is decreasing uniformly. When the radius is 1cm, the rate at which surface area (A) is decreasing w.r.t. time (\( t \)) is equal to
(\( \displaystyle\frac{dv}{dt} \) = rate at which volume is decreasing w.r.t. \( t \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{2dv}{dt} \)
Let r be the radius at time t. The volume and the surface area of the sphere are
\( \displaystyle V=\frac{4}{3}\pi r^3 \)
\( \displaystyle A=4\pi r^2 \)
Differentiating both with respect to t:
\( \displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt} \)
\( \displaystyle \frac{dA}{dt}=8\pi r\frac{dr}{dt} \)
Dividing the second by the first:
\( \displaystyle \frac{dA/dt}{dV/dt}=\frac{8\pi r}{4\pi r^2} \)
\( \displaystyle =\frac{2}{r} \)
At r = 1, \( \frac{dA}{dt}=2\,\frac{dV}{dt} \).