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Application of Derivatives question

2019 · 10 Apr · Shift 2 · Q24
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  5. /2019 · 10 Apr · Shift 2 · Q24

Application of Derivatives question

2019 · 10 Apr · Shift 2 · Q24

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3 /min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice decreases, is :
  1. A
    56π{5 \over {6\pi }}6π5​
  2. B
    19π{1 \over {9\pi }}9π1​
  3. C
    136π{1 \over {36\pi }}36π1​
  4. D
    118π{1 \over {18\pi }}18π1​
View written solutionFree

Correct answer: D

  1. Set up the geometry

Let the thickness of the ice at time ttt be xxx cm.

  • Radius of iron ball =10= 10=10 cm
  • Outer radius of ball + ice =10+x= 10 + x=10+x

So, volume of ice is V=43π((10+x)3−103).V=\frac{4}{3}\pi\big((10+x)^3-10^3\big).V=34​π((10+x)3−103).

  1. Differentiate with respect to time

Given that the ice is melting at the rate 50 cm3/min50\text{ cm}^3/\text{min}50 cm3/min, so the volume of ice is decreasing: dVdt=−50.\frac{dV}{dt}=-50.dtdV​=−50.

Differentiate: dVdt=43π⋅3(10+x)2dxdt\frac{dV}{dt}=\frac{4}{3}\pi\cdot 3(10+x)^2\frac{dx}{dt}dtdV​=34​π⋅3(10+x)2dtdx​ ⇒dVdt=4π(10+x)2dxdt.\Rightarrow \frac{dV}{dt}=4\pi(10+x)^2\frac{dx}{dt}.⇒dtdV​=4π(10+x)2dtdx​.

Thus, −50=4π(10+x)2dxdt.-50=4\pi(10+x)^2\frac{dx}{dt}.−50=4π(10+x)2dtdx​.

  1. Substitute x=5x=5x=5 cm

When the thickness is 555 cm, 10+x=15.10+x=15.10+x=15.

So, −50=4π(15)2dxdt-50=4\pi(15)^2\frac{dx}{dt}−50=4π(15)2dtdx​ −50=900πdxdt-50=900\pi\frac{dx}{dt}−50=900πdtdx​ dxdt=−50900π=−118π.\frac{dx}{dt}=-\frac{50}{900\pi}=-\frac{1}{18\pi}. dtdx​=−900π50​=−18π1​.

  1. Interpret the result

Since the question asks for the rate at which the thickness decreases, we take the positive magnitude: ∣dxdt∣=118π cm/min.\left|\frac{dx}{dt}\right|=\frac{1}{18\pi}\text{ cm/min}. ​dtdx​​=18π1​ cm/min.

  1. Match with options

This corresponds to: D 118π.\boxed{\text{D }\frac{1}{18\pi}}.D 18π1​​.

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