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# Dimensions $[M{L^{ - 1}}{T^{ - 1}}]$ are related to(A) Torque(B) Work(C) Energy(D) Coefficient of viscosity

Last updated date: 21st Jun 2024
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Hint:Here, we will find the dimensional formula for all the given four multiple choices and its general formula. The dimensional formula is written in the form of $Q = {M^a}{L^b}{T^c}$ where Q is the unit of the derived quantity, M (Mass), L (Length) and T (Time) and the exponents a, b and c are called the dimensions.

(A) Torque: It is the rotational force which causes an object to rotate about an axis.
$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F$
Where, “r” is measured in meters and “F” in Newton
Dimensions of torque $\left[ \tau \right] = \left[ r \right] \times \left[ F \right]$
Dimensions of torque$\left[ \tau \right] = \left[ {{L^1}} \right] \times \left[ {{M^1}{L^1}{T^{ - 2}}} \right]$ (When base are same, powers are added)
Dimensions of torque$\left[ \tau \right] = \left[ {{M^1}{L^2}{T^{ - 2}}} \right]$ ...........$(1)$
(B)Work: Work is said to be done when the object displaces the certain distance when the force is applied.
$W = F \times d$
Force is measured in Newton and the displacement is measured in metre.
Dimensions of Work$\left[ W \right] = \left[ F \right] \times \left[ d \right]$ Dimensions of Work$\left[ W \right] = \left[ {{M^1}{L^1}{T^{ - 2}}} \right] \times \left[ {{L^1}} \right]$
Dimensions of torque$\left[ W \right] = \left[ {{M^1}{L^2}{T^{ - 2}}} \right]$ ...........$(2)$
(C)Energy: Energy is the capacity to do work. It can be mechanical energy, the potential energy and kinetic energy.
$E = \dfrac{1}{2}m{v^2}$ “m” mass measured in kilogram and “v” is the velocity measured in metres per second
Dimensions of the Energy $[E] = [m][{v^2}]$
Dimensions of Energy$\left[ E \right] = \left[ {{{({M^0}{L^1}{T^{ - 1}})}^2}} \right] \times \left[ {{M^1}} \right]$
Dimensions of Energy$\left[ E \right] = \left[ {{M^1}{L^2}{T^{ - 2}}} \right]$ ...........$(3)$
(D)The coefficient of the viscosity: It is the measure of the resistance to the flow of the fluid. $\eta = \dfrac{F}{{A \times velocity{\text{ gradient}}}}$
Dimensions of $\eta = \dfrac{{{M^1}{L^1}{T^2}}}{{\left[ {{L^2}} \right] \times \left[ {{T^{ - 1}}} \right]}}$
Dimensions of $\eta = {M^1}{L^{ - 1}}{T^{ - 1}}{\text{}}......{\text{(4)}}$
From equations $(1),{\text{ (2), (3) and (4)}}$
We can say that - Dimensions $[M{L^{ - 1}}{T^{ - 1}}]$ are related to the coefficient of the viscosity.

Hence, from the given multiple choices – the option D is the correct answer.

Note:Always remember the standard formula of all the physical quantities to solve these types of questions. Such as definition of force and its formula to get the dimensional formula for force. Also, know the basics of power and the exponents to solve complex formulas to get the dimensional equations.