Answer
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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 $[F{V}^{-1}T]$.
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