Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

If L, M, N are physical quantities with different dimensions, then which of the following combinations can never be a meaningful quantity?
$\begin{align}
  & \text{A}\text{. }\dfrac{LM}{N} \\
 & \text{B}\text{. LN-M} \\
 & \text{C}\text{. }\dfrac{LN-{{M}^{2}}}{N} \\
 & \text{D}\text{. }\dfrac{L-M}{N} \\
\end{align}$

seo-qna
Last updated date: 25th Apr 2024
Total views: 396.6k
Views today: 3.96k
Answer
VerifiedVerified
396.6k+ views
Hint: We know that pure quantity doesn’t have units and therefore doesn’t have a dimension that’s why we can add or subtract it to any type of unit less quantity. In addition to this concept, we can only add or subtract only those quantities which have the same unit and therefore dimension. Find out the quantity in the given options which are violating the principle of homogeneity of dimensions.

Complete step-by-step solution:
We know that an equation contains several terms which can be separated from each other by using the symbols of equality like plus or minus and also the dimensions of all the terms in an equation must be identical to do mathematical operation. This is another way of saying that one can add or subtract similar physical quantities. Thus, addition of velocity to a force or subtraction of electric current from the thermodynamic temperature. This is nothing but the principle of homogeneity of dimensions in an equation and is an extremely useful method to check whether an equation may be correct or not. The equation must be wrong if the dimensions of all the terms are not the same. Let us check the equation \[~x=ut+\dfrac{1}{2}a{{t}^{2}}\], for the dimensional homogeneity. Here $x$ is the distance travelled by a particle in time $t$ which starts at a speed $u$ and has an acceleration $a$ along the direction of motion.
\[\begin{align}
  & \left[ x \right]=L \\
 & \left[ ut \right]=\text{velocity}\times \text{time=}\dfrac{\text{length}}{\text{time}}\text{ }\!\!\times\!\!\text{ time}=L \\
 & \left[ \dfrac{1}{2}a{{t}^{2}} \right]=\left[ a{{t}^{2}} \right]=\text{acceleration}\times {{\left( \text{time} \right)}^{2}} \\
 & \left[ \dfrac{1}{2}a{{t}^{2}} \right]=\dfrac{\text{velocity}}{\text{time}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{time} \right)}^{2}}=\dfrac{\text{length/time}}{\text{time}}\times {{\left( \text{time} \right)}^{2}}=L \\
\end{align}\]
Thus the equation is correct as far as the dimensions are concerned.
We have already stated in the question that L, M, and N are different physical Quantities and they have different dimensions, and also by the principle of homogeneity, we know that only like quantities can be added or subtracted i.e. can do mathematical operatic like plus and minus. So, $\dfrac{L-M}{N}$ this combination is not meaningful because two different physical quantities with different dimensions are subtracted (or added) which is not possible.
Therefore option (D) is the correct option.

Note: The dimension of $\dfrac{1}{2}a{{t}^{2}}$ is same as that of $a{{t}^{2}}$. Pure numbers are dimensionless. Dimension does not depend on the magnitude. Due to this reason the equation $x=ut+\dfrac{1}{2}a{{t}^{2}}$ is also dimensionally correct. Thus, a dimensionally correct equation need not be actually correct but a dimensionally wrong equation must be wrong.