Name: Conversion of Units Using Dimensional Analysis
Uploaded: 2019-07-19
Duration: 6 min 14 s
Description: How do you solve problems if the units of quantity are different? Is there any conversion method? Let us discuss the conversion using dimensional analysis in this video.

Question

If instead of mass, length and time as fundamental quantities we choose velocity, acceleration and force as fundamental quantities and express their dimensional symbols as

v, a

and

F

respectively. Show that the dimensions of Young's modulus can be expressed as

[{Fa}^{2} v^{- 4}]

.

Accepted Answer

According to question the Young's modulus $(Y)$ depends upon velocity $(v)$ , acceleration $(a)$ and force $(F)$ then

$Y \propto {(v)}^{x} {(a)}^{y} {(F)}^{z} \Rightarrow Y = k {(v)}^{x} {(a)}^{y} {(F)}^{z}$

After putting the dimensional formula of physical quantities used in equation

$Y = [M^{1} L^{- 1} T^{- 2}] = {[{LT}^{- 1}]}^{x} {[{LT}^{- 2}]}^{y} {[{MLT}^{- 2}]}^{z} \Rightarrow [M^{1} L^{- 1} T^{- 2}] = [M^{z} L^{x + y} T^{- x - y - 2 z}]$

Now by using the Principle of homogeneity,

$M^{1} = M^{z} \Rightarrow z = 1 L^{- 1} = L^{x + y + z} \Rightarrow x + y + z = - 1 \Rightarrow x + y = - 2 And T^{- 2} = T^{- x - y - 2 z} \Rightarrow - x - 2 y - 2 z = - 2 \Rightarrow - x - 2 y - 2 = - 2 \Rightarrow - x - 2 y = 0 \Rightarrow - y = - 2 \Rightarrow y = 2 Also, x = - 2 y = - 4$

After putting the values of $x, y and z$ and $k = 1$

$Y = [{Fa}^{2} v^{- 4}]$

If instead of mass, length and time as fundamental quantities we choose velocity, acceleration and force as fundamental quantities and express their dimensional symbols as $v, a$ and $F$ respectively. Show that the dimensions of Young's modulus can be expressed as $[{Fa}^{2} v^{- 4}]$ .

Important Questions on Physical World and Measurement

Learn and Prepare for any exam you want !

If instead of mass, length and time as fundamental quantities we choose velocity, acceleration and force as fundamental quantities and express their dimensional symbols as v, a and F respectively. Show that the dimensions of Young's modulus can be expressed as Fa2v-4.

Important Questions on Physical World and Measurement

Learn and Prepare for any exam you want !

If instead of mass, length and time as fundamental quantities we choose velocity, acceleration and force as fundamental quantities and express their dimensional symbols as $v, a$ and $F$ respectively. Show that the dimensions of Young's modulus can be expressed as $[{Fa}^{2} v^{- 4}]$ .