Question

A spherical snow ball is melting so that its volume is decreasing at the rate of 8 c.c./sec then the rate of change of radius when the radius is 2 cm, is :

• A. The radius is increasing at the rate of \(\frac{1}{2π} \text{cm/s}\)
• B. The radius is decreasing at the rate of \(\frac{1}{2π} \text{cm/s}\)
• C. The radius is increasing at the rate of \(\frac{1}{π} \text{cm/s}\)
• D. The radius is decreasing at the rate of \(\frac{1}{π} \text{cm/s}\)

Hint:

Differentiate V = 4/3 pi r^3 with respect to time.

Answer 1

The Correct Option is B

Step 1: Volume relation
\(V = \frac43\pi r^3\), so \(\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}\).

Step 2: Substitute
The volume is decreasing, so \(\frac{dV}{dt} = -8\). With \(r=2\): \(-8 = 4\pi(4)\frac{dr}{dt} = 16\pi\frac{dr}{dt}\).

Step 3: Solve
\(\frac{dr}{dt} = -\frac{1}{2\pi}\ \text{cm/s}\). The minus sign shows the radius decreases. Option (B).

Step 4: Why not the others
Options with increasing radius have the wrong sign, and the option with \(\frac1\pi\) has the wrong magnitude.

Final Answer:
The radius decreases at 1/(2 pi) cm per second. \[ \boxed{\text{(B)}\ \text{decreasing at }\frac{1}{2\pi}\ \text{cm/s}} \]