If force (F), velocity (V) and time (T) are taken as fundamental units, the dimensions of mass are
A. $[FV{T}^{-1}]$
B. $[FV{T}^{-2}]$
C. $[F{V}^{-1}{T}^{-1}]$
D. $[F{V}^{-1}T]$
Answer
656.1k+ views
Hint: To solve this problem, use the formula for Newton’s second law of motion which gives relation between force, mass and acceleration. But acceleration can be written in terms of velocity and time. So, substitute this relation in the formula for force. Rearrange the equation and get the expression for mass of a body. Then write this obtained equation in dimensions form. This will be the dimensions of mass if force, velocity and time are taken as fundamental quantities.
Formula used:
$F= MA$
$A= \dfrac {V}{T}$
Complete step by step answer:
According to Newton’s second law of motion, force is given by,
$F= MA$ …(2)
Where, M is the mass of the body
A is the acceleration acting on the body
We know, acceleration is given by,
$A= \dfrac {V}{T}$ …(2)
Where, V is the velocity of the body
T is the times taken by the body
Substituting equation. (2) in equation. (1) we get,
$F= M\dfrac {V}{T}$
Rearranging the above equation we get,
$M= F \dfrac {T}{V}$
$\Rightarrow [M]= [FT{V}^{-1}]$
Hence, if force (F), velocity (V) and time (T) are taken as fundamental units, the dimensions of mass are $[F{V}^{-1}T]$.
So, the correct answer is “Option D”.
Note: There is an alternate method to solve this problem which takes much longer time to be solved as compared to the above solution.
According to the alternate method, take $[M] = [{F}^{x}{V}^{y}{T}^{z}]$
Now, substitute the dimensions of force, velocity and time in it.
$[{M}^{1}{L}^{0}{T}^{0}]= {[ML{T}^{-2}]}^{x} {[L{T}^{-1}]}^{y} {[T]}^{z}$
$[{M}^{1}{L}^{0}{T}^{0}]= [{M}^{x} {L}^{x+y}{T}^{-2x-y-+z}]$
Equating the powers on the both the sides we get,
$x=1$ …(1)
$x+y=0$ …(2)
$-2x-y+z=0$ …(3)
Substituting equation. (1) in equation. (2) we get,
$y=-1$ …(4}
Substituting equation. (1) and (4) in equation. (3) we get,
$z=1$
$\Rightarrow [M]= [F{V}^{-1}T]$
Hence, the dimensions of mass are $[F{V}^{-1}T]$.
Formula used:
$F= MA$
$A= \dfrac {V}{T}$
Complete step by step answer:
According to Newton’s second law of motion, force is given by,
$F= MA$ …(2)
Where, M is the mass of the body
A is the acceleration acting on the body
We know, acceleration is given by,
$A= \dfrac {V}{T}$ …(2)
Where, V is the velocity of the body
T is the times taken by the body
Substituting equation. (2) in equation. (1) we get,
$F= M\dfrac {V}{T}$
Rearranging the above equation we get,
$M= F \dfrac {T}{V}$
$\Rightarrow [M]= [FT{V}^{-1}]$
Hence, if force (F), velocity (V) and time (T) are taken as fundamental units, the dimensions of mass are $[F{V}^{-1}T]$.
So, the correct answer is “Option D”.
Note: There is an alternate method to solve this problem which takes much longer time to be solved as compared to the above solution.
According to the alternate method, take $[M] = [{F}^{x}{V}^{y}{T}^{z}]$
Now, substitute the dimensions of force, velocity and time in it.
$[{M}^{1}{L}^{0}{T}^{0}]= {[ML{T}^{-2}]}^{x} {[L{T}^{-1}]}^{y} {[T]}^{z}$
$[{M}^{1}{L}^{0}{T}^{0}]= [{M}^{x} {L}^{x+y}{T}^{-2x-y-+z}]$
Equating the powers on the both the sides we get,
$x=1$ …(1)
$x+y=0$ …(2)
$-2x-y+z=0$ …(3)
Substituting equation. (1) in equation. (2) we get,
$y=-1$ …(4}
Substituting equation. (1) and (4) in equation. (3) we get,
$z=1$
$\Rightarrow [M]= [F{V}^{-1}T]$
Hence, the dimensions of mass are $[F{V}^{-1}T]$.
Recently Updated Pages
Which will be the least stable resonating structure class 11 chemistry CBSE

How many 5 digit telephone numbers can be construc-class-11-maths-CBSE

How do you find the angle of the resultant vector class 11 physics CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

What is the need and importance of classification class 11 biology CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference between physical and chemical change class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

