Question

For which of the following values of n, the formula ${\text{W}} = \left( {{\text{F}} + {\text{2Ma}}} \right){{\text{v}}^{\text{n}}}$, where W is the work, F is the force, M is the mass, a is the acceleration and v is the velocity can be made dimensionally correct ?
${\text{A}}{\text{.}}$ n = 0
${\text{B}}{\text{.}}$ n = 1
${\text{C}}{\text{.}}$ n = -4
${\text{D}}{\text{.}}$ no value of n

Answer 1

Hint: Here, we will proceed by writing the dimensions corresponding to all the quantities which are there in the given formula in terms of fundamental dimensions and then, we will equate the dimensions on both the sides.

Formulas Used: F = $\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]$, M = $\left[ {\text{M}} \right]$, a = $\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right]$, v = $\left[ {{\text{L}}{{\text{T}}^{ - 1}}} \right]$ and W = $\left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}} \right]$.

Complete Step-by-Step solution:
Given, ${\text{W}} = \left( {{\text{F}} + {\text{2Ma}}} \right){{\text{v}}^{\text{n}}}$ where W is the work, F is the force, M is the mass, a is the acceleration and v is the velocity
As we know that
Dimension of force F = $\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]$
Dimension of mass M = $\left[ {\text{M}} \right]$
Dimension of acceleration a = $\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right]$
Dimension of velocity v = $\left[ {{\text{L}}{{\text{T}}^{ - 1}}} \right]$
Dimension of work W = $\left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}} \right]$
By taking the dimensions of any the quantities in the equation ${\text{W}} = \left( {{\text{F}} + {\text{2Ma}}} \right){{\text{v}}^{\text{n}}}$, we get
\[
   \Rightarrow \left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}} \right] = \left[ {\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right] + {\text{2}}\left[ {\text{M}} \right]\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right]} \right]{\left[ {{\text{L}}{{\text{T}}^{ - 1}}} \right]^{\text{n}}} \\
   \Rightarrow \left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}} \right] = \left[ {\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right] + {\text{2}}\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]} \right]\left[ {{{\text{L}}^{\text{n}}}{{\left( {{{\text{T}}^{ - 1}}} \right)}^{\text{n}}}} \right] \\
 \]
Since, in the above dimensional analysis 2 which is a constant can be neglected because it doesn’t affect the dimensions.
\[ \Rightarrow \left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}} \right] = \left[ {\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right] + \left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]} \right]\left[ {{{\text{L}}^{\text{n}}}{{\text{T}}^{ - {\text{n}}}}} \right]{\text{ }} \to {\text{(1)}}\]
As we know that in order for addition to occur between two quantities, there dimensions should be equal and the dimension of the final quantity obtained will be same as the dimensions of the two quantities that are being added.
So, \[\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right] + \left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right] = \left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]{\text{ }} \to {\text{(2)}}\]
By substituting equation (2) in equation (1), we get
\[
   \Rightarrow \left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}} \right] = \left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]\left[ {{{\text{L}}^{\text{n}}}{{\text{T}}^{ - {\text{n}}}}} \right] \\
   \Rightarrow \left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}} \right] = \left[ {{\text{ML}}{{\text{T}}^{ - 2}}{{\text{L}}^{\text{n}}}{{\text{T}}^{ - {\text{n}}}}} \right] \\
   \Rightarrow \left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}} \right] = \left[ {{\text{M}}{{\text{L}}^{1 + {\text{n}}}}{{\text{T}}^{ - 2 - {\text{n}}}}} \right]{\text{ }} \to {\text{(3)}} \\
 \]
For the given equation to be true, that equation should be dimensionally correct which means the dimensions on both the sides of the equation should be equal to each other.
By comparing the dimensions of L on both sides of equation (3), we get
$
  2 = 1 + {\text{n}} \\
   \Rightarrow {\text{n}} = 2 - 1 = 1 \\
 $
By comparing the dimensions of T on both sides of equation (3), we get
$
   - 2 = - 2 - {\text{n}} \\
   \Rightarrow {\text{n}} = - 2 + 2 = 0 \\
 $
Clearly, in order to make the dimensions of L same on both the sides, the value of n should be 1 but in order to make the dimensions of T same on both the sides, the value of n should be 0. Since, both of these are different. So, there can be no value of n possible which can make the given equation dimensionally correct.
Therefore, the required answer will be no value of n
Hence, option D is correct.

Note- The dimension of velocity = $\dfrac{{{\text{Displacement}}}}{{{\text{Time}}}}$ is given by v = $\dfrac{{\left[ {\text{L}} \right]}}{{\left[ {\text{T}} \right]}} = \left[ {{\text{L}}{{\text{T}}^{ - 1}}} \right]$, dimension of acceleration = $\dfrac{{{\text{Velocity}}}}{{{\text{Time}}}}$ is given by a = $\dfrac{{\left[ {{\text{L}}{{\text{T}}^{ - 1}}} \right]}}{{\left[ {\text{T}} \right]}} = \left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right]$, dimension of force = (mass)(acceleration) is given by F = $\left[ {\text{M}} \right]\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right] = \left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]$ and dimension of work = (Force)(Distance) is given by W = $\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]\left[ {\text{L}} \right] = \left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}} \right]$.