Problem 64 - Master Resource Book in JEE Main Physics

Question

A cube has a side of length \(1.2 \times 10^{-2} \mathrm{~m} .\) Calculate its volume (a) \(1.7 \times 10^{-6} \mathrm{~m}^{3}\) (b) \(1.73 \times 10^{-6} \mathrm{~m}^{3}\) (c) \(1.70 \times 10^{-6} \mathrm{~m}^{3}\) (d) \(1.732 \times 10^{-6} \mathrm{~m}^{3}\)

Step-by-Step Solution

Answer

The cube's volume is \(1.73 \times 10^{-6} \mathrm{~m}^{3}\), option (b).

Step 1: Understanding the formula for volume of a cube

The formula to find the volume of a cube is given by \( V = a^3 \), where \( a \) is the length of a side of the cube. In this problem, the side length \( a \) is given as \( 1.2 \times 10^{-2} \) meters.

Step 2: Substitute given values into the volume formula

Substitute \( a = 1.2 \times 10^{-2} \) m into the volume formula. This results in \( V = (1.2 \times 10^{-2})^3 \).

Step 3: Compute the cube of the side length

Calculate \((1.2)^3\) and \((10^{-2})^3\) separately. We have \( (1.2)^3 = 1.728 \) and \( (10^{-2})^3 = 10^{-6} \).

Step 4: Combine the results

Multiply the results from the previous step: \( 1.728 \times 10^{-6} \).

Step 5: Round the result to select the correct option

Round \(1.728\) to three significant figures to match the given options. Thus, \( 1.728 \approx 1.73 \), resulting in \( 1.73 \times 10^{-6} \mathrm{~m}^{3} \). This matches option (b).

Key Concepts

Cube Volume Calculation

Calculating the volume of a cube is a straightforward process. The formula for finding the volume of a cube is given by \( V = a^3 \), where \( a \) is the length of one side of the cube. This means you simply take the length of the side and multiply it by itself twice more. For example, if a cube has a side length of \( 1.2 \times 10^{-2} \) meters, you would calculate the volume by finding \( (1.2 \times 10^{-2})^3 \).
It's important to note that when working with lengths given in scientific notation, it's easier to separately calculate the numeric part and the exponential part. Therefore, calculate \( (1.2)^3 \) and \( (10^{-2})^3 \) separately, and then multiply the two results.

• Calculate \( (1.2)^3 = 1.728 \)
• Calculate \( (10^{-2})^3 = 10^{-6} \)

Multiply these results to get the volume: \( 1.728 \times 10^{-6} \) cubic meters.

Significant Figures in Physics

In physics and other sciences, significant figures are crucial for indicating the precision of a measurement. When performing calculations, the number of significant figures in the result should generally reflect the number of significant figures in the least precise measurement used in the calculations. This ensures that the precision of the result is not overstated.
In our exercise, the side length of the cube is \( 1.2 \), which has two significant figures. As a rule of thumb, when you multiply or divide, your answer should have as many significant figures as the number with the least significant figures. Therefore, the computed volume should consider significant figures properly. After computing \( 1.728 \), we round it to match the precision of the original measurement, resulting in \( 1.73 \).

• Identify the least significant figures in given measurements.
• Round your final result to match this level of precision.

Scientific Notation in Calculations

Scientific notation is a method of expressing numbers that are too large or too small to be conveniently written in decimal form. It’s particularly handy in scientific calculations because it simplifies the arithmetic operations by representing numbers as powers of ten.
When calculating the cube’s volume with a side length given as \( 1.2 \times 10^{-2} \) meters, scientific notation helps manage and simplify the multiplication of small numbers. Each component of the expression, like \( 1.2 \) and \( 10^{-2} \), can be handled separately. This approach reduces errors and makes the calculation more manageable.

• Multiply the coefficients: \( (1.2)^3 = 1.728 \)
• Multiply the exponential parts, here \( (10^{-2})^3 = 10^{-6} \)

Combining these, you maintain clarity and accuracy throughout the calculation, ultimately simplifying complex multiplications.