Question

If value of gravitational constant in MKS is \(6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\) then value of \(\mathrm{G}\) in \(\mathrm{CGS}=\ldots \ldots \ldots \ldots . .\left[\left(\right.\right.\) dyne \(\left.\left.-\mathrm{cm}^{2}\right) /\left(\mathrm{gm}^{2}\right)\right]\) (a) \(6.67 \times 10^{-9}\) (b) \(6.67 \times 10^{-7}\) (c) \(6.67 \times 10^{-8}\) (d) \(6.67 \times 10^{-5}\)

Short answer

The value of G in the CGS system is \(6.67 \times 10^{-10} \frac{\mathrm{dyne} \cdot \mathrm{cm}^2}{\mathrm{gm}^2}\). However, none of the given options match the calculated value.

Step-by-step solution

Determine the conversion factors between the systems

: In the given exercise, we have: 1. Length: MKS (meters) to CGS (centimeters): 1 meter = 100 centimeters 2. Mass: MKS (kilograms) to CGS (grams): 1 kilogram = 1000 grams 3. Force: MKS (Newtons) to CGS (dynes): 1 Newton = 100,000 dynes

Apply the conversion factors

: The value of G in MKS is given as \(6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\). To convert this into CGS: 1. Replace meters with centimeters: \(\mathrm{Nm}^{2} \rightarrow 100 \times 100 \mathrm{cm}^{2}\) 2. Replace kilograms with grams: \(/ \mathrm{kg}^{2} \rightarrow / (1000 \times 1000 \mathrm{gm}^{2})\) 3. Replace Newtons with dynes: \(N \rightarrow 100,000 \mathrm{dyne}\) Now apply all these conversions simultaneously: \(6.67 \times 10^{-11} \times (\frac{100,000 \cdot 100 \cdot 100}{1000 \cdot 1000}) \frac{\mathrm{dyne} \cdot \mathrm{cm}^2}{\mathrm{gm}^2}\)

Simplify and compare with the given options

: Simplify the expression: \(6.67 \times 10^{-11} \times (\frac{10,000,000}{1,000,000}) \frac{\mathrm{dyne} \cdot \mathrm{cm}^2}{\mathrm{gm}^2} = 6.67 \times 10^{-11} \times 10 \frac{\mathrm{dyne} \cdot \mathrm{cm}^2}{\mathrm{gm}^2} = 6.67 \times 10^{-10} \frac{\mathrm{dyne} \cdot \mathrm{cm}^2}{\mathrm{gm}^2}\) Compare the calculated value with the given options: (a) \(6.67 \times 10^{-9}\) (b) \(6.67 \times 10^{-7}\) (c) \(6.67 \times 10^{-8}\) (d) \(6.67 \times 10^{-5}\) None of the given options match the calculated value: \(6.67 \times 10^{-10} \, \mathrm{dyne} \cdot \mathrm{cm}^2 / \mathrm{gm}^2\).

Key concepts

MKS to CGS Conversion

When working with physical measurements in physics, there are different systems of units. Two common ones are MKS and CGS, which stand for Meter-Kilogram-Second and Centimeter-Gram-Second respectively. The core difference lies in the base units used for physical quantities, such as length, mass, and force.
Here’s a simple breakdown of what conversion involves:

• Length: In MKS, distance is measured in meters. To convert to CGS, multiply by 100, since 1 meter equals 100 centimeters.
• Mass: MKS uses kilograms. To convert to grams in CGS, multiply by 1,000.
• Force: MKS measures in newtons. In CGS, this becomes dynes, with 1 newton equating to 100,000 dynes.

Performing these conversions correctly is crucial for accurate calculations across different scientific contexts that may use either of these unit systems.

Unit Conversion in Physics

In physics, unit conversion is a fundamental process allowing scientists and students to interpret data or equations in consistent or varied unit systems. It involves adjusting a value by converting it into another unit without changing its actual amount.

The procedure is typically linear and uses multiplication or division:

• Identify the relationship between the units (e.g., 1 km = 1,000 m).
• Use conversion factors based on these relationships. Conversion factors equal 1 because they signify a ratio of equivalence (e.g., 1 = 100 cm/1 m).
• Apply the conversion factor to adjust the measurement (e.g., multiply the number of meters by 100 to get centimeters).

This concept applies widely in physics to ensure calculations have dimensionally consistent results. For example, when calculating the gravitational force or adjusting energy readings. Always employ careful attention to conversion factors to avoid errors in scientific computations.

Dimensional Analysis

Dimensional analysis is a powerful tool in physics that involves checking relationships between different physical quantities by identifying their dimensions. Dimensions are the basic types of physical quantities measured, such as length \(L\), mass \(M\), and time \(T\). This method ensures that equations make sense physically.

Here’s how it works:

• Write down the equation or physical relation.
• Express each term in the equation using its fundamental dimensions (e.g., \(F = ma\) can be written as \([MLT^{-2} = MLT^{-2}]\)).
• Check for consistency: Both sides of the equation should balance dimensionally.

Dimensional analysis is indispensable for:

• Verifying equations.
• Finding relations between physical quantities without detailed knowledge about their nature.
• Converting between units accurately.

Using dimensional analysis, you can affirm that the units agree, which can lead to uncovering errors or providing insight into equation derivation.