Problem 72 - Master Resource Book in JEE Main Physics

Question

A \(10 \mathrm{~m}\) long iron chain of linear mass density \(0.8 \mathrm{~kg} \mathrm{~m}^{-1}\) is hanging freely from a rigid support. If \(g=10 \mathrm{~ms}^{-2}\), then the power required to lift the chain upto the point of support in \(10 \mathrm{~s}\) is (a) \(10 \mathrm{~W}\) (b) \(20 \mathrm{~W}\) (c) \(30 \mathrm{~W}\) (d) \(40 \mathrm{~W}\)

Step-by-Step Solution

Answer

The power required is 40 W, option (d).

Step 1: Calculate the Total Mass of the Chain

The chain has a linear mass density of \(0.8 \mathrm{~kg} \mathrm{~m}^{-1}\). To find the total mass, multiply the linear density by the length of the chain: \[ m = \lambda \times L = 0.8 \times 10 = 8 \mathrm{~kg} \]

Step 2: Compute the Work Done in Lifting the Chain

Work done to lift the chain is equal to the force applied (which is equal to the weight of the chain) multiplied by the height through which the chain is lifted. The force is the weight, \( F = mg \).The height is the center of mass of the entire chain when it is lifted to the top, which is half of its length: \[ h = \frac{L}{2} = \frac{10}{2} = 5 \mathrm{~m} \]Thus, the work done is: \[ W = F \times h = mg \times h = 8 \times 10 \times 5 = 400 \mathrm{~J} \]

Step 3: Determine the Power Required

Power is the rate at which work is done. Calculate power by dividing the work done by the time: \[ P = \frac{W}{t} = \frac{400}{10} = 40 \mathrm{~W} \]

Step 4: Conclusion: Select the Correct Option

The power required to lift the chain up to the point of support is \(40 \mathrm{~W}\). Therefore, the correct option is (d) \(40 \mathrm{~W}\).

Key Concepts

Linear Mass Density

Linear mass density is a measure of how much mass is distributed over a particular length of an object. In this problem, the chain has a linear mass density of \(0.8 \, \text{kg/m}\). This means for every meter of chain, there is \(0.8\, \text{kg}\) of mass. This concept is crucial for calculating the total mass of objects that have a consistent distribution of mass over a certain length.

• To find the total mass, you multiply the linear mass density by the total length of the object.
• Here's how it worked in the problem: \( m = \lambda \times L = 0.8 \,\text{kg/m} \times 10 \, \text{m} = 8 \, \text{kg} \).

This calculation gives you the overall weight of the chain necessary for further calculations, such as determining the work needed to lift it.

Center of Mass

The center of mass is a critical concept in mechanics, representing the average position of all the mass in a objects system. For uniform objects like a chain, it simplifies to finding the geometric center. In our case, the chain is uniformly dense, so:

• The center of mass is halfway along its length, at \(5 \, \text{m}\).
• Calculating the center of mass allows us to know the effective height through which the entire chain's mass acts when lifting.

Understanding the center of mass is key to determining how energy is used to move an object uniformly. It reduces complex three-dimensional problems into simpler one-dimensional problems regarding how weight is distributed over distance.

Physical Work and Energy

Physical work involves moving an object with force over a distance. Energy is what we use to do this work, and power describes how quickly energy is being used. In the exercise, lifting the chain involves both work and energy concepts:

• The work done can be calculated using the formula \( W = F \times h \), where \( F \) is the weight of the chain \( (mg) \) and \( h \) is the height moved.
• In our problem, the force \( F \) was \( 8 \, \text{kg} \times 10 \, \text{m/s}^2 = 80 \, \text{N} \), and the height was the center of mass \( h = 5 \, \text{m}\).
• Thus, the total work done was \( W = 80 \, \text{N} \times 5 \, \text{m} = 400 \, \text{J} \).
• Power is the work done over time: \( P = \frac{400 \, \text{J}}{10 \, \text{s}} = 40 \, \text{W} \).

Grasping these concepts solidifies your understanding of how mechanical systems use energy and effort to accomplish tasks.