Which of the following equations are dimensionally correct?
(a) ${v_f} = {v_i} + ax$
(b) $y = (2m)\cos (kx)$ , where $k = 2{m^{ - 1}}$
Answer
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Hint:In order to answer this question, to know which given equation is dimensionally correct, we will write the dimension of each of the given variables on the basis of their units. And then we will compare the dimensions of Left hand member quantities and the dimensions of Right hand member quantities.
Complete step by step answer:
(a) We will write the dimensions for each of the given quantity in the above equation-
${v_f} = {v_i} + ax$
The variables ${v_f}\,and\,{v_i}$ are expressed in units of $m.{s^{ - 1}}$ , therefore-
$[{v_f}] = [{v_i}] = L{T^{ - 1}}$
The variable $a$ is expressed in units of $m.{s^{ - 2}}$ ;
$[a] = L{T^{ - 2}}$
And, the variable $x$ is expressed in meters.
Therefore, $[ax] = {L^2}{T^{ - 2}}$
No, we will compare the dimensions of Left hand member quantity and the dimensions of Right hand member quantities:
$ \Rightarrow L{T^{ - 1}} = L{T^{ - 1}} + {L^2}{T^{ - 2}}$
Hence, the dimensions of both the sides are not the same.
Hence, the given equation ${v_f} = {v_i} + ax$ is not dimensionally correct.
(b) We will write the dimensions for each of the given quantity in the above equation-
$y = (2m)\cos (kx)$
For $y$ , $[y] = L$
For $2m$ , $[2m] = L$
And, for $kx$ , $[kx] = [(2{m^{ - 1}})x] = {L^{ - 1}}L$
Therefore, we can think of the quantity $kx$ as an angle in radians, and we can take its cosine. The cosine itself will be a pure number with no dimensions. For the left hand members and the right hand member of the equation, we have-
$[LHS] = [y] = L,\, \\
and \\
[RHS] = [2m][\cos (kx)] = L \\ $
Hence, the given equation $y = (2m)\cos (kx)$ is dimensionally correct.
Note:The units on both sides of a dimensionally accurate equation must agree.However, it is still possible that it is inaccurate. $E = 2m{c^2}$ is a basic example of an incorrect Einstein equation (the proper Einstein equation is $E = m{c^2}$ ). The source of the error in this example is a dimensionless factor that doubled one hand side.
Complete step by step answer:
(a) We will write the dimensions for each of the given quantity in the above equation-
${v_f} = {v_i} + ax$
The variables ${v_f}\,and\,{v_i}$ are expressed in units of $m.{s^{ - 1}}$ , therefore-
$[{v_f}] = [{v_i}] = L{T^{ - 1}}$
The variable $a$ is expressed in units of $m.{s^{ - 2}}$ ;
$[a] = L{T^{ - 2}}$
And, the variable $x$ is expressed in meters.
Therefore, $[ax] = {L^2}{T^{ - 2}}$
No, we will compare the dimensions of Left hand member quantity and the dimensions of Right hand member quantities:
$ \Rightarrow L{T^{ - 1}} = L{T^{ - 1}} + {L^2}{T^{ - 2}}$
Hence, the dimensions of both the sides are not the same.
Hence, the given equation ${v_f} = {v_i} + ax$ is not dimensionally correct.
(b) We will write the dimensions for each of the given quantity in the above equation-
$y = (2m)\cos (kx)$
For $y$ , $[y] = L$
For $2m$ , $[2m] = L$
And, for $kx$ , $[kx] = [(2{m^{ - 1}})x] = {L^{ - 1}}L$
Therefore, we can think of the quantity $kx$ as an angle in radians, and we can take its cosine. The cosine itself will be a pure number with no dimensions. For the left hand members and the right hand member of the equation, we have-
$[LHS] = [y] = L,\, \\
and \\
[RHS] = [2m][\cos (kx)] = L \\ $
Hence, the given equation $y = (2m)\cos (kx)$ is dimensionally correct.
Note:The units on both sides of a dimensionally accurate equation must agree.However, it is still possible that it is inaccurate. $E = 2m{c^2}$ is a basic example of an incorrect Einstein equation (the proper Einstein equation is $E = m{c^2}$ ). The source of the error in this example is a dimensionless factor that doubled one hand side.
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