Question 21: Units , Dimensions and Error Analysis

Question

If units of mass, length and gravitational constant are chosen to fundamental units, the dimensions of time would be

Accepted Answer

To determine the dimensions of time when using mass (M), length (L), and the gravitational constant (G) as fundamental units, follow these steps:
Recognize that the gravitational constant has the dimensions:
$$[G] = \frac{L^3}{M \, T^2}$$
Rearrange this equation to solve for time (T). Start by writing:
$$T^2 = \frac{L^3}{M \, G}$$
Take the square root of both sides to isolate T:
$$T = \sqrt{\frac{L^3}{M \, G}}$$
Express the result using exponents:
$$T = L^{3/2} M^{-1/2} G^{-1/2}$$
Comparing with the provided options, this matches:
Option C:
$$\mathrm{M}^{-1/2} \mathrm{L}^{3/2} \mathrm{G}^{-1/2}$$
Thus, the correct answer is Option C.

Answer

$$
\mathrm{M}^{1 / 2} \mathrm{~L}^{1 / 2} \mathrm{G}^{-1 / 2}
$$

Answer

$$
\mathrm{M}^1 \mathrm{~L}^{1 / 2} \mathrm{G}^2
$$

Answer

$$
\mathrm{M}^{-1 / 2} \mathrm{~L}^{3 / 2} \mathrm{G}^{-1 / 2}
$$

Answer

$$
\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{G}^{-1}
$$

Units , Dimensions and Error Analysis

If units of mass, length and gravitational constant are chosen to fundamental units, the dimensions of time would be

Choose the correct answer: