Question

The given dimensional formula $\left[ {{L}^{1}}{{M}^{1}}{{T}^{-2}} \right]$ is for
A) Velocity
B) Acceleration
C) Force
D) Work

Answer 1

To identify quantity for the unknown dimension given in the question first define the quantities given in the option write formula for each of the quantity and using formula, calculate dimension. As you can see, we have terms of the length, mass and time. Release these three quantities with formulas of quantities given in option and calculate answer.

Formula Used:
$E=P\times t$
E is the energy in kWh
P is the power in kW
t is the total time in hours.

Complete step by step answer:
In the question we have been provided with dimensional formula which is given as $\left[ {{L}^{1}}{{M}^{1}}{{T}^{-2}} \right]$. Now we need to calculate quantity whose dimension is $\left[ {{L}^{1}}{{M}^{1}}{{T}^{-2}} \right]$ according to newton’s second law of motion force is nothing but best product of magnitude of mass and acceleration.
Mathematically,
$F=m\times a......\left( 1 \right)$
Where, F = force, m = mass, a = acceleration
We know that acceleration is the rate of change of the velocity per unit time.
Mathematically,
$a=\dfrac{dv}{dt}$
Where, v = velocity, t = time.
Now velocity can be defined as the rate of change of displacement per unit of time.
Mathematically,
$v=\dfrac{ds}{dt}$
Where, s = displacement
So, equation (1) can be written as,
$F=m\dfrac{{{d}^{2}}s}{d{{t}^{2}}}$
(Since$a=\dfrac{dv}{dt}=\dfrac{d}{dt}\left( \dfrac{ds}{dt} \right)=\dfrac{{{d}^{2}}s}{d{{t}^{2}}}$)
We know that, standard format of dimension is
$\left[ Q \right]=\left[ {{M}^{a}}{{L}^{b}}{{T}^{c}} \right]$
So force can be written as
$\left[ Force \right]=\dfrac{\left[ mass \right]\left[ displacement \right]}{\left[ time \right]\left[ time \right]}.....(2)$
Unit of mass is kg and in dimensional form, it can be represented as (M). Unit of displacement is meter and in dimensional form it can be represented as (L).
Unit of time is second and in dimensional form it can be represented as (T)
So, equation (2) implies,
$\left[ force \right]=\dfrac{[M][L]}{[T][T]}=[[{{M}^{1}}][{{L}^{1}}][{{T}^{-2}}]] =[{{M}^{1}}{{L}^{1}}{{T}^{-2}}]$

Hence $\left[ {{L}^{1}}{{M}^{1}}{{T}^{-2}} \right]$ is the dimensional formula for force.

Therefore the correct option is (c).

Note:
Standard formula of dimension is $\left[ {{M}^{a}}{{L}^{b}}{{T}^{c}} \right]$. So in this case the value of a, b, c is 1, 1and -2 respectively.
Dimensional formula of velocity: velocity can be defined as rate of change of displacement with respect to time mathematically,
$\begin{aligned} & v=\dfrac{ds}{dt} \\\ & (velocity)=[L{{T}^{-1}}] \\\ \end{aligned}$
Dimensional formula of acceleration: acceleration can be defined as rate of change of velocity with respect to time. Mathematically,
$\begin{aligned} & v=\dfrac{dv}{dt}=\dfrac{{{d}^{2}}s}{d{{t}^{2}}} \\\ & \left[ acceleration \right]=(L{{T}^{-2}}) \\\ \end{aligned}$
Dimensional formula of work: work can be defined as product of force and displace mathematically,
$\begin{aligned} & \omega =Fds \\\ & (work)=[{{L}^{1}}{{M}^{1}}{{T}^{-2}}][{{L}^{1}}] \\\ & (work)=[{{L}^{2}}{{M}^{1}}{{T}^{-2}}] \\\ \end{aligned}$
Remember, force is the vector quantity i.e. it is having both magnitude and direction.