Problem 93 - Master Resource Book in JEE Main Physics

Question

The length, breadth and thickness of a metal block is given by \(l=90 \mathrm{~cm}, b=8 \mathrm{~cm}, t=2.45 \mathrm{~cm}\). The volume of the block is (a) \(2 \times 10^{2} \mathrm{~cm}^{3}\) (b) \(1.8 \times 10^{2} \mathrm{~cm}^{3}\) (c) \(1.77 \times 10^{2} \mathrm{~cm}^{3}\) (d) \(1.764 \times 10^{2} \mathrm{~cm}^{2}\)

Step-by-Step Solution

Answer

(d) \(1.764 \times 10^{2} \mathrm{~cm}^{3}\)

Step 1: Identify the Dimensions

The metal block has the following dimensions: length \( l = 90 \) cm, breadth \( b = 8 \) cm, and thickness \( t = 2.45 \) cm.

Step 2: Apply the Volume Formula

To find the volume of a rectangular block, use the formula \( V = l \times b \times t \). In this case, substitute the given values: \( V = 90 \times 8 \times 2.45 \).

Step 3: Perform the Calculation

Calculate the product: \( 90 \times 8 = 720 \), and then \( 720 \times 2.45 = 1764 \). Thus, the volume \( V = 1764 \) cm\(^3\).

Step 4: Express the Volume in Scientific Notation

Convert 1764 into scientific notation for precise comparison with the given options. \( 1764 \) cm\(^3\) is expressed as \( 1.764 \times 10^{2} \) cm\(^3\).

Step 5: Match with Given Options

Compare the calculated volume \( 1.764 \times 10^{2} \) cm\(^3\) with the options. Identify option (d) as the matching answer.

Key Concepts

Volume Calculation

To find the volume of a rectangular prism, we use the simple formula: \[ V = l \times b \times t \]where \(V\) represents volume, \(l\) is the length, \(b\) is the breadth, and \(t\) is the thickness (or height) of the prism. This formula essentially computes the amount of space the object occupies in three-dimensional space.

• Multiply the three dimensions together to get the volume.
• In this exercise, we used dimensions: length \(= 90\) cm, breadth \(= 8\) cm, and thickness \(= 2.45\) cm.

Performing the calculation: 1. Multiply length and breadth: \( 90 \times 8 = 720 \) 2. Next, multiply this result by the thickness: \( 720 \times 2.45 = 1764 \)Thus, the volume of the rectangular prism is \( 1764 \) cm\(^3\). It's a fundamental concept in geometry that applies to many practical physical and mathematical problems.

Scientific Notation

Scientific notation is a method of expressing numbers that are too large or small in a concise format. It is extremely useful for simplifying calculations and clearly defining the precision of measurements. This notation follows the structure:\[ a \times 10^{n} \]where \(a\) is a number, typically between 1 and 10, and \(n\) is an integer representing the power of 10.

• This way, we can represent numbers without writing long sequences of zeros.
• In our example, the volume calculated was \( 1764 \) cm\(^3\).

To express this in scientific notation, we move the decimal point two places to the left, turning it into \( 1.764 \times 10^{2} \) cm\(^3\). This helps in comparing the calculated volume with multiple-choice options succinctly.

Rectangular Prism Dimensions

The dimensions of a rectangular prism determine its size and shape. It's critical to correctly measure and multiply these to find the volume. The three orthogonal dimensions typically are:

• Length - the longest side of the prism.
• Breadth (or width) - the second longest side, perpendicular to the length.
• Thickness (or height) - the side perpendicular to both the length and breadth.

For instance, in our problem, the metal block's dimensions are: - Length (\(l = 90\) cm) is the longest side,- Breadth (\(b = 8\) cm) is perpendicular to the length, and- Thickness (\(t = 2.45\) cm) is the smallest dimension that shapes its height.

JEE Main Physics

Understanding how to calculate the volume of a rectangular prism is crucial for students preparing for exams like JEE Main Physics. Such problems test not only your mathematical computation skills but also your grasp of physical dimensions and measurements.

• Volume calculations are foundational concepts in mechanics, fluid dynamics, and thermodynamics.
• Accurately determining the space an object occupies is helpful for solving real-world physics problems that require precise measurement.

On exams like JEE Main, utilizing scientific notation can help streamline calculations and improve answer accuracy. Practicing problems involving volume, conversion, and scientific notation ensures a comprehensive understanding necessary to tackle more complex physics questions.