Problem 167 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A spherical iron ball \(10 \mathrm{~cm}\) in radius is coated with a layer of ice of uniform thickness than melts at a rate of \(50 \mathrm{~cm}^{3} / \mathrm{min}\). When the thickness of ice is \(5 \mathrm{~cm}\), then the rate at which the thickness of ice decreases, is (A) \(\frac{1}{36 \pi} \mathrm{cm} / \mathrm{min}\) (B) \(\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}\) (C) \(\frac{1}{54 \pi} \mathrm{cm} / \mathrm{min}\) (D) \(\frac{5}{6 \pi} \mathrm{cm} / \mathrm{min}\)

Step-by-Step Solution

Answer

(B) \(\frac{1}{18\pi}\) cm/min

Step 1: Understand the problem

We have a spherical iron ball with a radius of 10 cm, coated with a layer of ice. The total radius of the ball plus the ice is initially 15 cm. The volume of the ice that melts per minute is 50 cm³. We need to find the rate at which the thickness of the ice decreases when this thickness is 5 cm.

Step 2: Formula for the volume of a sphere

The volume of a sphere with radius \( R \) is given by the formula \( V = \frac{4}{3} \pi R^3 \). We will use this formula to calculate the volume of the ice.

Step 3: Set up the volume expressions

Let \( R \) be the total radius (15 cm) and \( r \) be the radius of the iron ball (10 cm). The initial volume (volume of the iron ball + ice) is \( V_1 = \frac{4}{3} \pi R^3 \) and the volume of the iron ball is \( V_2 = \frac{4}{3} \pi r^3 \). The volume of the ice is \( V_\text{ice} = V_1 - V_2 = \frac{4}{3} \pi (R^3 - r^3) \).

Step 4: Implicit differentiation

The rate of change of the volume of the ice is \(-50 \) cm³/min (since it is decreasing). Using the derivative formula with respect to time \( t \), we have \( V_\text{ice} = \frac{4}{3} \pi ((r+x)^3 - r^3) \), where \( x \) is the ice thickness (5 cm initially). Differentiating the equation with respect to \( t \), we get \( \frac{dV}{dt} = 4 \pi (r+x)^2 \frac{dx}{dt} \).

Step 5: Solve for the rate of change in thickness

We plug in the values: \( r = 10 \text{ cm} \), \( x = 5 \text{ cm} \), and \( \frac{dV}{dt} = -50 \text{ cm}^3/ ext{min} \). Thus, we have the derivative equation as:\[ -50 = 4 \pi (10+5)^2 \frac{dx}{dt} \]Simplifying, \( -50 = 4 \pi (15)^2 \frac{dx}{dt} \).

Step 6: Calculate \( \frac{dx}{dt} \)

Solving for \( \frac{dx}{dt} \):\[ \frac{dx}{dt} = \frac{-50}{4 \pi \times 225} = \frac{-50}{900 \pi} = \frac{1}{18 \pi} \text{ cm/min} \]

Key Concepts

Sphere Volume Calculations

Calculating the volume of a sphere is central to understanding problems involving spherical objects, such as the iron ball with ice in this exercise. To find the volume of a sphere, use the formula:
\[ V = \frac{4}{3} \pi R^3 \]where \( R \) represents the radius of the sphere.

• For a sphere, this important formula helps determine how much space is inside it.
• The sphere's volume changes depending on its radius, so even a small change in radius can lead to a significant difference in volume.

Imagine that the iron ball is the core of the sphere, with a radius of 10 cm. The outer layer of ice increases the total radius to 15 cm. By calculating the difference in volume between these two radii, we can determine the volume of the ice. This calculation is crucial as it helps in understanding the rate at which the ice melts.

Implicit Differentiation

Implicit differentiation is a technique used in calculus to find the derivative of a variable even when it is not isolated. In the context of the melting ice problem, it's used for determining how the rate of thickness change relates to the rate of volume change.

• Implicit differentiation allows you to differentiate equations where variables are mixed together rather than easily separable.
• This technique involves taking the derivative of both sides of an equation with respect to a variable and solving for the desired rate of change.

For this ice-melting scenario, the equation for the volume of the ice depends on the total radius \((r + x)\), where \(x\) is the ice thickness. The derivative connects the rate of volume change (\(-50 \text{ cm}^3/\text{min}\)) to the rate at which the ice thickness changes, making implicit differentiation a crucial tool for solving this problem.

Melting Ice Problem

The melting ice problem is a classic example of a rate of change problem in calculus. It involves understanding how two rates are related—the rate of ice volume loss and the decrease in ice thickness.
Here, the spherical iron ball is covered by melting ice:

• The ice melts at a constant rate of 50 cm³ per minute, which is the rate of volume decrease.
• The central task is to find how fast the thickness of the ice decreases at any specific thickness, such as 5 cm.

By using calculus, particularly implicit differentiation, you can relate these rates. Calculating \( \frac{dx}{dt} \) as \( \frac{1}{18 \pi} \text{ cm/min} \) shows how slight changes in thickness occur over time. Understanding this relationship helps solve similar physical problems where layers change over time due to factors like heat or pressure.