Question

$ Erg - {m^{ - 1}} $ can be a measure of:
A. Force
B. Momentum
C. Power
D. Acceleration

Answer 1

Hint: To solve the question firstly use the mathematical formula of force, momentum, power, and acceleration to determine each of their dimensional formulas in terms of the basic units. Then, the answer can be solved in two ways. The first and simple method is to write $ Erg - {m^{ - 1}} $ in terms of the quantities these units represent and perform the derivation to get the answer. Another method is to represent $ Erg - {m^{ - 1}} $ in terms of the quantities these units represent and then perform dimensional analysis to get the answer.

Complete step by step answer:
In this question we have been asked whether $ Erg - {m^{ - 1}} $ is a measure of force, momentum, power or acceleration. Firstly, we will define all the terms and see their mathematical formulation in terms of basic unit. In the discussion below the used notations represent the physical quantities as follows:
 $ F $ represents force; $ m $ represents mass; $ v $ represents velocity; $ a $ presents acceleration ; $ d $ represents distance or displacement; $ l $ represents length; $ t $ represents time; $ p $ represents linear momentum; $ P $ represents power; $ W $ represents work done.
 $ M, L $ , and $ T $ are the dimensional representations of mass, length, and time respectively.
Force: Force can be defined as the cause that results in the change in state of a body of mass $ m $ . Mathematically, force is given by the product of mass $ m $ and acceleration $ a $ of the body, i.e., $ F = ma $ . In terms of fundamental units, we can write,
 $ F = ma = m \times \dfrac{v}{t} = m \times \dfrac{{\dfrac{d}{t}}}{t} = m \times \dfrac{l}{{{t^2}}} $
 $ \therefore \left[ F \right] = \left[ {{M^1}{L^1}{T^{ - 2}}} \right] $
Momentum: Momentum $ (p) $ is defined as the product of mass $ m $ of the body and its velocity $ v $ . Mathematically, we can formulate , $ p = mv $ . So, we can write:
 \[p = mv = m \times \dfrac{d}{t} = m \times \dfrac{l}{t}\]
 $ \therefore \left[ p \right] = \left[ {{M^1}{L^1}{T^{ - 1}}} \right] $
Power: Power $ P $ is the rate of work done $ W $ per unit time $ t $ . Mathematically, we write: $ P = \dfrac{W}{t} $ . So, we have:
 $ P = \dfrac{W}{t} = \dfrac{{F \times d}}{t} = \dfrac{{m \times a \times d}}{t} = \dfrac{{m \times \dfrac{v}{t} \times d}}{t} = \dfrac{{m \times \dfrac{d}{t} \times d}}{{{t^2}}} = \dfrac{{m \times {l^2}}}{{{t^3}}} $
 $ \therefore \left[ P \right] = \left[ {{M^1}{L^2}{T^{ - 3}}} \right] $
Acceleration: Acceleration is defined as the rate of change of velocity per time. Mathematically, we have: $ a = \dfrac{v}{t} $
So, we can write:
 \[a = \dfrac{v}{t} = \dfrac{{\dfrac{d}{t}}}{t} = \dfrac{l}{{{t^2}}}\]
 $ \therefore \left[ a \right] = \left[ {{M^0}{L^1}{T^{ - 2}}} \right] $
Now, we will do the dimensional analysis of $ Erg - {m^{ - 1}} $ . We know $ erg $ is the unit of work done and $ m $ is the unit of length. So, mathematically we get a ratio of work done to length. So,
 $ \dfrac{W}{l} = \dfrac{{F \times d}}{l} = \dfrac{{F \times l}}{l} = F $
So, $ Erg - {m^{ - 1}} $ is a measure of force. Thus, option $ (A) $ is the correct answer.

Note: Derived units are certain units that are calculated or derived in terms of the seven fundamental units,i.e., length, mass, time, luminous intensity, electric current, temperature, and amount of substance. All the above-given quantities like force, power, momentum, and acceleration have derived units. So, for solving the sum it was necessary to derive the units of these quantities in terms of fundamental units. It might be a question of why only these seven units are known as fundamental units. This is because they are independent of any other unit and hence cannot be related to other units.