
The fundamental physical quantities that are having the similar dimension in the dimensional formula of the Torque and the Angular Momentum are:
A. mass, time
B. time, length
C. mass, length
D. time, mole
Answer
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Hint: The dimensional formula of torque can be found out using the basic equation. That is the torque is given as the product of force acting and the perpendicular distance. The angular momentum can be calculated by taking the product of mass, velocity and the radius of the path. Using this find out the dimensional formula and compare the powers of each of the fundamental quantities. Hope this will help you in solving this question.
Complete answer:
$\tau =F\times d$
Where $F$be the force acting and $d$be the perpendicular distance.
We know that the dimensional formula of the force can be given as,
$F=ML{{T}^{-2}}$
And the dimension of the distance will be written as,
$d=L$
Taking the product of these quantities will give the dimension of torque.
$\left[ \tau \right]=\left[ ML{{T}^{-2}} \right]\times \left[ L \right]$
That is,
$\left[ \tau \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$
Now let us calculate the dimensional formula of the angular momentum.
The angular momentum is given by the equation,
${{L}_{P}}=MVR$
Where $M$ be the mass of the particle, \[V\] be the velocity of the particle and \[R\] be the radius.
The dimensional formula of each of this can be written as,
\[\left[ M \right]=\left[ M \right]\]
\[\left[ V \right]=\left[ L{{T}^{-1}} \right]\]
\[\left[ R \right]=\left[ L \right]\]
Therefore let us substitute all these in the equation of angular momentum,
$\left[ {{L}_{P}} \right]=\left[ M \right]\times \left[ L{{T}^{-1}} \right]\times \left[ L \right]$
That is the dimension of angular momentum is,
$\left[ {{L}_{P}} \right]=\left[ M{{L}^{2}}{{T}^{-1}} \right]$
Now let us compare both the equations of torque and angular momentum. From this we can see that the mass in both the equations are having a power of one. And the length also in both the equations are having the same power of two. Mass and length will be the answer.
So, the correct answer is “Option C”.
Note:
Torque is defined as the measure of the force which is able to rotate the particle about a particular axis. This is defined as the rotational equivalent of the force. Angular momentum is defined as the rotational equivalent of linear momentum.
Complete answer:
$\tau =F\times d$
Where $F$be the force acting and $d$be the perpendicular distance.
We know that the dimensional formula of the force can be given as,
$F=ML{{T}^{-2}}$
And the dimension of the distance will be written as,
$d=L$
Taking the product of these quantities will give the dimension of torque.
$\left[ \tau \right]=\left[ ML{{T}^{-2}} \right]\times \left[ L \right]$
That is,
$\left[ \tau \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$
Now let us calculate the dimensional formula of the angular momentum.
The angular momentum is given by the equation,
${{L}_{P}}=MVR$
Where $M$ be the mass of the particle, \[V\] be the velocity of the particle and \[R\] be the radius.
The dimensional formula of each of this can be written as,
\[\left[ M \right]=\left[ M \right]\]
\[\left[ V \right]=\left[ L{{T}^{-1}} \right]\]
\[\left[ R \right]=\left[ L \right]\]
Therefore let us substitute all these in the equation of angular momentum,
$\left[ {{L}_{P}} \right]=\left[ M \right]\times \left[ L{{T}^{-1}} \right]\times \left[ L \right]$
That is the dimension of angular momentum is,
$\left[ {{L}_{P}} \right]=\left[ M{{L}^{2}}{{T}^{-1}} \right]$
Now let us compare both the equations of torque and angular momentum. From this we can see that the mass in both the equations are having a power of one. And the length also in both the equations are having the same power of two. Mass and length will be the answer.
So, the correct answer is “Option C”.
Note:
Torque is defined as the measure of the force which is able to rotate the particle about a particular axis. This is defined as the rotational equivalent of the force. Angular momentum is defined as the rotational equivalent of linear momentum.
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