
If the unit of force were kilo newton, that of time millisecond and that of power kilowatt, what should be the units of mass in kg?
Answer
510.6k+ views
Hint: In the given question, we can calculate the unit of mass by using the formula of power and putting all the given values in it. After this, we will get the value of length and put it in Newton's 2nd law of motion for the unit of mass.
Complete Solution:
Given: Force, F = kN = ${10^3}N$
Time, t=Milliseconds =${10^{ - 3}}s$
Power, P=kW = ${10^3}W$
We need to find the units of mass in kg.
Now, we know
$Power = \dfrac{{Energy}}{{Time}} = \dfrac{{Force \times Displacement}}{{Time}}$
Putting all the values in above equation,
⇒${10^3}W = \dfrac{{{{10}^3}N \times L}}{{{{10}^{ - 3}}s}}$
⇒$L = {10^{ - 3}}m$
Now, to find the units of mass in kg we will use the relation,
$Force = Mass \times Acceleration$
⇒$F = M \times a$
⇒$F = M \times \dfrac{v}{t}$
⇒$F = M \times \dfrac{d}{{t \times t}}$
Here, we can take d=L
⇒$F = \dfrac{{ML}}{{{t^2}}}$
Put all the values in above equation, we get
⇒${10^3}N = \dfrac{{M \times ({{10}^{ - 3}}m)}}{{{{({{10}^{ - 3}}s)}^2}}}$
⇒$M = \dfrac{{{{10}^3}N \times {{({{10}^{ - 3}}s)}^2}}}{{{{10}^{ - 3}}m}}$
⇒$M = 1kg$
Hence, the units of mass is in kg.
Note: Alternative Method:
We can also solve this question by using dimensional analysis.
We know, dim P= $[M{L^2}{T^{ - 1}}]$ and ….eq i)
dim F=$[ML{T^{ - 2}}]$
${F^2} = {[ML{T^{ - 2}}]^2}$ ….eq ii)
Now divide eq i) and eq ii),
⇒\[\dfrac{P}{{{F^2}}} = \dfrac{{M{L^2}{T^{ - 1}}}}{{{M^2}{L^2}{T^{ - 4}}}}\]
⇒$\dfrac{P}{{{F^2}}} = \dfrac{{{T^3}}}{M}$
⇒$M = \dfrac{{{F^2}{T^3}}}{P}$
Put all the values in above equation, we get
⇒$M = \dfrac{{{{({{10}^3})}^2}{{({{10}^{ - 3}})}^3}}}{{{{10}^{ - 3}}}}$
⇒M= 1kg
Therefore, the unit of mass is in kg.
Complete Solution:
Given: Force, F = kN = ${10^3}N$
Time, t=Milliseconds =${10^{ - 3}}s$
Power, P=kW = ${10^3}W$
We need to find the units of mass in kg.
Now, we know
$Power = \dfrac{{Energy}}{{Time}} = \dfrac{{Force \times Displacement}}{{Time}}$
Putting all the values in above equation,
⇒${10^3}W = \dfrac{{{{10}^3}N \times L}}{{{{10}^{ - 3}}s}}$
⇒$L = {10^{ - 3}}m$
Now, to find the units of mass in kg we will use the relation,
$Force = Mass \times Acceleration$
⇒$F = M \times a$
⇒$F = M \times \dfrac{v}{t}$
⇒$F = M \times \dfrac{d}{{t \times t}}$
Here, we can take d=L
⇒$F = \dfrac{{ML}}{{{t^2}}}$
Put all the values in above equation, we get
⇒${10^3}N = \dfrac{{M \times ({{10}^{ - 3}}m)}}{{{{({{10}^{ - 3}}s)}^2}}}$
⇒$M = \dfrac{{{{10}^3}N \times {{({{10}^{ - 3}}s)}^2}}}{{{{10}^{ - 3}}m}}$
⇒$M = 1kg$
Hence, the units of mass is in kg.
Note: Alternative Method:
We can also solve this question by using dimensional analysis.
We know, dim P= $[M{L^2}{T^{ - 1}}]$ and ….eq i)
dim F=$[ML{T^{ - 2}}]$
${F^2} = {[ML{T^{ - 2}}]^2}$ ….eq ii)
Now divide eq i) and eq ii),
⇒\[\dfrac{P}{{{F^2}}} = \dfrac{{M{L^2}{T^{ - 1}}}}{{{M^2}{L^2}{T^{ - 4}}}}\]
⇒$\dfrac{P}{{{F^2}}} = \dfrac{{{T^3}}}{M}$
⇒$M = \dfrac{{{F^2}{T^3}}}{P}$
Put all the values in above equation, we get
⇒$M = \dfrac{{{{({{10}^3})}^2}{{({{10}^{ - 3}})}^3}}}{{{{10}^{ - 3}}}}$
⇒M= 1kg
Therefore, the unit of mass is in kg.
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