Question

If force F, mass M and time T are chosen as fundamental quantities the dimensional formula for length is:
A. F M T
B. ${\text{F }}{{\text{M}}^{ - 1}}\,{{\text{T}}^2}$
C. ${\text{F }}{{\text{L}}^2}\,{{\text{T}}^{ - 2}}$
D. ${{\text{F}}^{ - 1}}{\text{ }}{{\text{L}}^{ - 2}}\,{{\text{T}}^{ - 2}}$

Answer 1

Recall the S.I units of length, force, mass and time. Express the dimensions of these physical quantities. As given in the question, equate the dimensions of length with the dimensions of product of F, M and T. Equate the powers on both sides of the equation.

Complete step by step answer:
We know that the S.I unit of length is m. Therefore, the dimensions of length are,
${\text{L}} = \left[ {{{\text{M}}^0}{{\text{L}}^1}{{\text{T}}^0}} \right]$
The S.I unit of force is ${\text{kg}}\,{\text{m/}}{{\text{s}}^2}$. Therefore, we express the dimensions of the force as,
${\text{F}} = \left[ {{{\text{M}}^1}{{\text{L}}^1}{{\text{T}}^{ - 2}}} \right]$
The S.I unit of mass is kg. Therefore, we can express the dimensions of mass as,
${\text{M}} = \left[ {{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^0}} \right]$
The S.I unit of time is s. Therefore, we can express the dimensions of time as,
${\text{T}} = \left[ {{{\text{M}}^0}{{\text{L}}^0}{{\text{T}}^1}} \right]$
Now, we can write,
${\text{L}} = {{\text{F}}^{\text{x}}}{{\text{M}}^{\text{y}}}{{\text{T}}^{\text{z}}}$

Now, using the dimensions of the fundamental quantities in the above equation that we have determined, we get,
$\left[ {{{\text{M}}^0}{{\text{L}}^1}{{\text{T}}^0}} \right] = {\left[ {{{\text{M}}^1}{{\text{L}}^1}{{\text{T}}^{ - 2}}} \right]^x}{\left[ {{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^0}} \right]^y}{\left[ {{{\text{M}}^0}{{\text{L}}^0}{{\text{T}}^1}} \right]^z}$
$\Rightarrow \left[ {{{\text{M}}^0}{{\text{L}}^1}{{\text{T}}^0}} \right] = \left[ {{{\text{M}}^x}{{\text{L}}^x}{{\text{T}}^{ - 2x}}} \right]\left[ {{{\text{M}}^y}{{\text{L}}^0}{{\text{T}}^0}} \right]\left[ {{{\text{M}}^0}{{\text{L}}^0}{{\text{T}}^z}} \right]$
$\Rightarrow \left[ {{{\text{M}}^0}{{\text{L}}^1}{{\text{T}}^0}} \right] = \left[ {{{\text{M}}^{x + y}}{{\text{L}}^x}{{\text{T}}^{ - 2x + z}}} \right]$

Equating the powers of M, L and T in the both sides, we get,
$x + y = 0$ …… (2)
$x = 1$ …… (3)
$- 2x + z = 0$ …… (4)
Substituting $x = 1$ in equation (2), we get,
$1 + y = 0$
$\Rightarrow y = - 1$

Also, substituting $x = 1$ in equation (4), we get,
$- 2\left( 1 \right) + z = 0$
$\Rightarrow z = 2$
Substituting $x = 1$, $y = - 1$ and $z = 2$ in equation (1), we get,
$\therefore{\text{L}} = {{\text{F}}^{\text{1}}}{{\text{M}}^{ - 1}}{{\text{T}}^2}$

Hence, the correct answer is option B.

Note: To solve such types of questions, the crucial step is to remember the S.I units of all the fundamental quantities. Students should be able to determine the dimensions from the unit of the fundamental quantity. Since we have determined the dimensional formula for length as ${\text{L}} = {{\text{F}}^{\text{1}}}{{\text{M}}^{ - 1}}{{\text{T}}^2}$, this can also be written as ${\text{L}} = \dfrac{{{\text{F }}{{\text{T}}^2}}}{{\text{M}}}$. If you substitute the units of force, time and mass in the above equation, you will get the unit of length as meter.