Question

How watt is related to fundamental units?
A. $kg{m^2}{s^{ - 3}}$
B. $kg{m^2}{s^{ - 2}}$
C. $Nm{s^{ - 1}}$
D. $Nm$

Answer 1

Hint: As we know that SI unit of power is watt. Power is the amount of energy transferred per unit time. In this question, we need to tell how the SI unit of power is related to the fundamental units. As we know that Fundamental units are the units of fundamental quantities. They don't depend upon any other quantities and all other units are derived from them.

Complete step by step solution:
(a) As we know that Watt is a unit of Power. Power is the amount of energy transferred per unit time. It is a scalar quantity, as it depends only on the magnitude.
If we define one watt, it will be one joule of energy transferred in one second.
\[{\text{Power }} = {\text{ }}Force \times velocity\]………………………………………………………... (I)
(b) Hence the force applied is the external power that makes a body of mass ‘$m$’ move from rest.
\[Force{\text{ }} = {\text{ }}mass\; \times acceleration\]
Hence equation (I) becomes-
Power = mass $ \times $ acceleration$ \times $velocity……………………………………………. (II)
Where SI unit of the mass of object is in the Kilogram (Kg)
(c) The acceleration is change in velocity with respect to time.
Acceleration =\[\dfrac{{{V_2} - {V_1}}}{t} = \dfrac{{\Delta V}}{t}\]
Where $\Delta V = $ change in the velocity and t = time
SI unit of time is second (s).
Hence equation (II) becomes:
\[Power = {\text{ }}mass\]$ \times $\[\dfrac{{\Delta V}}{t}\]$ \times $\[velocity\] ………………………………………………. (III)
(d) Velocity is distance covered per unit time.
$Velocity{\text{ }} = \dfrac{{distance}}{{time}}$
Where SI unit of distance is meter m.
Hence Equation (III) becomes-
\[{\text{Power }} = {\text{ }}mass\]$ \times $\[\dfrac{{\Delta V}}{t}\]$ \times $$\dfrac{{distance}}{{time}}$……………………………………………… (IV)
Using SI units of mass, velocity, time, distance,
SI unit of Power will be-
\[
  {\text{Power }} = {\text{ Kg}}\; \times \dfrac{{m/s}}{s} \times \dfrac{m}{s} \\
 \Rightarrow {\text{Power }} = {\text{ Kg}}\; \times \dfrac{m}{{{s^2}}} \times \dfrac{m}{s} \\
 \Rightarrow Power = {\text{ Kg}}\; \times {m^2}{s^{ - 3}} \\
\]
Hence, option (A) is the correct.

Note: There are seven fundamental quantities:- Mass, Length, Time, temperature, amount of substance, electric current, luminous intensity. We need to find how this watt is related to them.