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MathsMediumMCQ2025 · 23 Jan Shift 2

Q13.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 1 cm/min. The surface area (in cm2 ) of the chocolate ball (without the ice- 4π cream layer) is : (1) 196π (2) 256π (3) 225π (4) 128π

What This Question Tests

This question is a related rates problem, requiring the application of the chain rule to relate the rate of change of volume and thickness of the ice-cream layer to find the radius of the chocolate ball, and then its surface area.

Concepts Tested

Volume of a sphereDerivative as a rate of changeChain ruleSurface area of a sphere

Formulas Used

V = (4/3)πR^3

A = 4πr^2

dV/dt = dV/dR * dR/dt

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