Question

If the units of force, velocity and energy are given as 100 dyne, 10 cm/s and 500 erg, respectively, what are the units of mass, length and time?
A. 5 g, 5 cm, 5 s
B. 5 g, 5 cm, $ {\text{0}}{\text{.5 s}} $
C. $ 0.5{\text{ g}} $ ,5 cm, 5 s
D. 5 g, $ 0.5{\text{ cm}} $ , 5 s

Answer 1

Hint
When the unit of a physical quantity is written in a form that is different from the standard, the fundamental units of length, mass, and time changes. The new units for these base quantities can be found by performing dimensional analysis.

Complete step by step answer
Dimensional analysis is performed by equating the units of known physical quantities with the dimensions of others, and using comparison to find the new units.
In this question, we are provided with the following data:
Unit of force = 100 dyne
Unit of velocity = 10 cm/s
Unit of energy = 500 erg
We now write all these quantities in the form of fundamental units:
 $\Rightarrow F = [{M^1}{L^1}{T^{ - 2}}] $
 $\Rightarrow V = [{L^1}{T^{ - 1}}] $
 $\Rightarrow E = [{M^1}{L^2}{T^{ - 2}}] $
Here, $ F $ stands for force, $ V $ stands for velocity and $ E $ stands for energy. Since we need to find the values for $ M $ , $ L $ and $ T $ , we rearrange the given quantities in a way to get our desired results.
Let us start with $ E $ and $ F $ . Upon closer look, one can see that the quantities only vary slightly and can provide us with a single fundamental result if we perform the right operations on it.
So, dividing $ E $ by $ F $ we get,
 $ \dfrac{E}{F} = \dfrac{{[{M^1}{L^2}{T^{ - 2}}]}}{{[{M^1}{L^1}{T^{ - 2}}]}} $
This quantity is also equal to $ \dfrac{{500{\text{erg}}}}{{100{\text{dyne}}}} = 5{\text{cm}} $
We know that $ \dfrac{{{\text{erg}}}}{{{\text{dyne}}}} $ gives us cm which can also be checked from the unit of velocity in this question.
Comparing the two we get,
 $ {L^1} = 5{\text{ cm}} $
Now, dividing $ E $ by the square of $ V $ we get,
 $ \dfrac{E}{{{V^2}}} = \dfrac{{[{M^1}{L^2}{T^{ - 2}}]}}{{[{L^2}{T^{ - 2}}]}} $
This is also equal to $ \dfrac{{500}}{{{{10}^2}}} = \dfrac{{500}}{{100}}g $
Cancelling the appropriate fundamental units gives us:
 $ {M^1} = 5{\text{ g}} $
Lastly, we put our new values in the unit of velocity to find the new unit of time as:
 $ V = [{(5cm)^1}{T^{ - 1}}] = 10cm/s $
Cross-multiplying gives us:
 $\Rightarrow {T^{ - 1}} = \dfrac{{10}}{5}$
 $\Rightarrow T = \dfrac{5}{{10}}s $
Hence, the unit of time is $ 0.5s $ .
 $ \therefore $ The units of length, mass and time are 5cm, 5gm and $ 0.5s $ .
The answer is option (B).

Note
There are seven fundamental units. They are very useful because they provide an equal scale for measuring all quantities at a universal level. Imagine that every country had its own measurement system, then trade would become inconvenient because of the constant conversion required (as is with currencies). Thankfully, the SI units are widely accepted as units of measurements for all quantities.