MHTCET 20 Apr 2026 Shift 1 Paper

Concept:Use related rates by expressing the ice volume as a function of its thickness and differentiating with respect to time.Explanation:Let the thickness of ice be $x$ cm.The radius of the iron ball is 10 cm, so the outer radius of the coated ball is $10 + x$ cm.Volume of ice is the difference between the outer sphere and the iron ball:$$V = \frac{4}{3}\pi(10+x)^3 - \frac{4}{3}\pi(10)^3$$Ice melts at $50 \ \mathrm{cm}^3/\mathrm{min}$, so:$$\frac{dV}{dt} = -50$$Differentiate $V$ with respect to $t$:$$\frac{dV}{dt} = 4\pi(10+x)^2 \frac{dx}{dt}$$Given thickness is 5 cm, i.e., $x = 5$:$$-50 = 4\pi(10+5)^2 \frac{dx}{dt}$$$$-50 = 4\pi(15)^2 \frac{dx}{dt}$$$$-50 = 900\pi \frac{dx}{dt}$$Solving:$$\frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi}$$The negative sign indicates the thickness is decreasing.Thus, the rate of decrease of ice thickness is:$$\frac{1}{18\pi} \ \mathrm{cm/min}$$Answer:Option A: $\frac{1}{18\pi} \ \mathrm{cm/min}$