JEE Main 2019 (10 April Shift 2)

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Animated Solution for Mathematics - Differentiation: 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 $c m^{3} / min$ . When the thickness of the ice is 5cm, then the rate at which the thickness (in cm/min) of the ice decreases, is :

Select Answer:

Visualized Solution

Visualizing the Iron Ball

Radius of the iron ball, $r = 10$ cm

The Ice Coating

A layer of ice coats the ball.
Let the uniform thickness of this ice be $h$ cm at any time $t$ .

Total Radius of the Sphere

The entire system forms a larger sphere.
Total radius, $R = 10 + h$

Volume of the Ice Layer

Volume of Ice ( $V$ ) = Total Volume - Iron Ball Volume
$V = \frac{4}{3} π (10 + h)^{3} - \frac{4}{3} π (10)^{3}$

Rate of Change of Volume

Differentiate $V$ with respect to time $t$ :
$\frac{d V}{d t} = \frac{4}{3} π \cdot 3 (10 + h)^{2} \cdot \frac{d h}{d t} - 0$

Simplifying the Equation

Cancel out the $3$ in the numerator and denominator.
$\frac{d V}{d t} = 4 π (10 + h)^{2} \frac{d h}{d t}$

Applying the Given Melting Rate

Ice melts at $50 cm^{3} / min$ .
Since volume is decreasing, $\frac{d V}{d t} = - 50$

Substituting Known Values

We need the rate when thickness $h = 5$ cm.
Substitute $\frac{d V}{d t} = - 50$ and $h = 5$ :
$- 50 = 4 π (10 + 5)^{2} \frac{d h}{d t}$

Calculating the Squared Term

$- 50 = 4 π (15)^{2} \frac{d h}{d t}$
$- 50 = 4 π (225) \frac{d h}{d t}$

Isolating $\frac{d h}{d t}$

$- 50 = 900 π \frac{d h}{d t}$
$\frac{d h}{d t} = - \frac{50}{900 π}$

Final Rate of Decrease

$\frac{d h}{d t} = - \frac{1}{18 π}$ cm/min
The negative sign indicates a decrease.
Rate of decrease = $\frac{1}{18 π}$ cm/min

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The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Setup

We are dealing with a spherical iron core of radius

r = 10

cm, coated in a layer of ice of thickness

h

. The total radius of the system is given by

R = 10 + h

The volume of the ice layer,

V

, is the difference between the volume of the total sphere and the volume of the iron core:

V = \frac{4}{3} π (10 + h)^{3} - \frac{4}{3} π (10)^{3}

The Master Equation

We are given that the ice melts at a rate of

50

^{3}

/min. Since the volume is decreasing, we define the rate of change as:

\frac{d V}{d t} = - 50 cm^{3} / min

To find the rate at which the thickness

h

changes, we differentiate the volume equation with respect to time

t

using the chain rule:

\frac{d V}{d t} = \frac{4}{3} π \cdot 3 (10 + h)^{2} \cdot \frac{d h}{d t}

Note that the derivative of the constant iron core volume is zero. Simplifying the expression, we obtain:

\frac{d V}{d t} = 4 π (10 + h)^{2} \frac{d h}{d t}

Final Calculation

We substitute the known values into our derived equation at the specific moment when

h = 5

cm:

- 50 = 4 π (10 + 5)^{2} \frac{d h}{d t}

- 50 = 4 π (15)^{2} \frac{d h}{d t}

- 50 = 4 π (225) \frac{d h}{d t}

- 50 = 900 π \frac{d h}{d t}

Solving for

\frac{d h}{d t}

, we find:

\frac{d h}{d t} = - \frac{50}{900 π} = - \frac{1}{18 π} cm/min

The negative sign indicates that the thickness is decreasing. Therefore, the magnitude of the rate at which the thickness is shrinking is $\frac{1}{18 π}$ cm/min.

Common Traps & FAQs

Why do we subtract the volume of the iron ball? Can't I just calculate the volume of the ice directly?

Great question! Calculating the volume of a spherical shell directly is tricky. By subtracting the inner sphere from the outer sphere, we use the principle of superposition, which is a powerful tool in physics. It simplifies the geometry into a subtraction of two standard volumes, making the differentiation much cleaner.

Why is

\frac{d V}{d t}

negative? The question asks for the rate of decrease, which is a positive value.

You've hit on a crucial point! In calculus, the sign tells us the direction of change. Since the ice is melting, the volume is decreasing, so

\frac{d V}{d t}

must be negative. When the question asks for the 'rate of decrease,' it's asking for the magnitude of that change, which is why we interpret the negative sign as 'decrease'.

Is it possible to solve this without the chain rule?

While you could technically expand the cubic term, it would make the algebra much messier and prone to errors. The chain rule is your best friend here. It allows you to differentiate the function

V (h)

with respect to

h

and then multiply by

\frac{d h}{d t}

, which is exactly what we need to find.

Animated Solution for Mathematics - Differentiation: 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 $c m^{3} / min$ . When the thickness of the ice is 5cm, then the rate at which the thickness (in cm/min) of the ice decreases, is :

Select Answer:

Visualized Solution

Visualizing the Iron Ball

The Ice Coating

Total Radius of the Sphere

Volume of the Ice Layer

Rate of Change of Volume

Simplifying the Equation

Applying the Given Melting Rate

Substituting Known Values

Calculating the Squared Term

Isolating $\frac{d h}{d t}$

Final Rate of Decrease

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at the rate of 50 cm $^{3}$ /min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which the thickness of ice decreases, is:

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Select Answer:

Visualized Solution

Visualizing the Iron Ball

The Ice Coating

Total Radius of the Sphere

Volume of the Ice Layer

Rate of Change of Volume

Simplifying the Equation

Applying the Given Melting Rate

Substituting Known Values

Calculating the Squared Term

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtdh​

Final Rate of Decrease

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A sphere of 10 cm radius has a uniform thickness of ice around it. Ice is melting at rate 50 cm3/min when thickness is 5 cm then rate of change of thickness

The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :

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Isolating $\frac{d h}{d t}$

A sphere of $10 cm$ radius has a uniform thickness of ice around it. Ice is melting at rate $50 cm^{3} / min$ when thickness is $5 cm$ then rate of change of thickness