A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness the melts at a rate of 50 cm³/min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which of the thickness of ice decreases, is :

Solution

Maths sir · 4 steps

The iron core does not change, so the ice volume changes only through the outer radius.

1. Set up volume. Let the outer radius of ball plus ice be r. The ice volume is the big sphere minus the iron ball: V = 4/3π(r³ - 10³).
2. Differentiate in time. The 10³ is a constant, so it drops out. Since r = 10 + thickness, the thickness changes at the same rate as r.

   dV/dt = 4π r² dr/dt

3. Put in numbers. When the thickness is 5 cm, r = 15 cm, so 4π r² = 900π. The ice melts, so dV/dt = -50, giving dr/dt = -50/900π = -1/18π cm/min.
4. So. The thickness of ice decreases at 1/18π cm/min, which is option (a).

The idea to keep. For a shell, write the volume as outer sphere minus inner sphere. The fixed inner part vanishes on differentiating, and dV/dt = 4π r² dr/dt uses the outer radius, not the thickness.

Answer from the NTA final answer key. Solution by PrepWiser.