Q 1 :    

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 cm3/min and the thickness of the ice-cream layer decreases at the rate of 14π cm/min. The surface area (in cm2) of the chocolate ball (without the ice-cream layer) is:          [2025]

  • 128π

     

  • 225π

     

  • 196π

     

  • 256π

     

(4)

Volume of spherical chocolate,

V=43πr3

Derivative w.r.t. t, we get

dVdt=4πr2drdt

 81=4πr2×14π

 r2=81  r=9

   Surface area of chocolate = 4π(r1)2

                                                     = 4π(91)2

                                                     = 256π cm2