Verified Solution • 2025–26 Session moderate

QJEEM20Ma679b: JEE Main 2020 Mathematics - Class 12 JEEM20Ma679b Solution: A spherical iron ball of 10 cm radius is coated with a layer of …

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 :

Multiple Choice Options

  • A 118π{1 \over {18\pi }}
  • B 136π{1 \over {36\pi }}
  • C 154π{1 \over {54\pi }}
  • D 56π{5 \over {6\pi }}

Step-by-Step Verified Solution

Let the thickness = h cm

Volume of ice = v = 4π3((10+h)3−103){{4\pi } \over 3}\left( {{{\left( {10 + h} \right)}^3} - {{10}^3}} \right)

⇒\Rightarrow dvdt=4π3(3(10+h)2).dhdt{{dv} \over {dt}} = {{4\pi } \over 3}\left( {3{{\left( {10 + h} \right)}^2}} \right).{{dh} \over {dt}}

Given dvdt={{dv} \over {dt}} = 50 cm3/min and h = 5 cm

∴\therefore 50 = 4π3(3(10+5)2).dhdt{{4\pi } \over 3}\left( {3{{\left( {10 + 5} \right)}^2}} \right).{{dh} \over {dt}}

⇒\Rightarrow dhdt=504π×152{{dh} \over {dt}} = {{50} \over {4\pi \times {{15}^2}}} = 118π{1 \over {18\pi }} cm/min

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