Question

Time(T), Velocity(V) and angular momentum (h) are chosen as the fundamental quantities instead of mass, length and time. In terms of these, dimension of mass would be:
(a)$\left[ M \right] = \left[ {{T^{ - 1}}{C^{ - 2}}{h^{ - 1}}} \right]$
(b) $\left[ M \right] = \left[ {{T^{ - 1}}{C^2}h} \right]$
(c) $\left[ M \right] = \left[ {{T^{ - 1}}{C^{ - 2}}h} \right]$
(d) $\left[ M \right] = \left[ {T{C^{ - 2}}h} \right]$

Answer 1

Hint: We can always express a physical quantity with some power of dimension of fundamental quantities. Then, we can equate both sides, using the principle of homogeneity of dimension on both sides of the equation.
Principle of homogeneity says that in any verified and valid physical equation the dimension of both sides of the equation should be the same.

Complete step by step answer:
We know, in terms of fundamental quantities: dimension of
$\left[ C \right] = \left[ {L{T^{ - 1}}} \right]$ …… (1)
$\left[ h \right] = \left[ {{L^2}{T^{ - 1}}M} \right]$ …… (2)
$\left[ T \right] = \left[ T \right]$ (same) …… (3)

Step 1:
We can write any physical quantity in terms of considered fundamental quantities dimensions.
Therefore, $\left[ M \right] = \left[ {{T^a}{C^b}{h^c}} \right]$ …… (4)
Where, a,b and c are arbitrary numbers.

Step 2:
Replace values from equation (1), (2) and (3) in equation (4) we get-
$\left[ M \right] = \left[ {{T^a}{{\left[ {L{T^{ - 1}}} \right]}^b}{{\left[ {{L^2}{T^{ - 1}}M} \right]}^c}} \right]$
Simplifying and rearranging variables we get-
$\left[ M \right] = \left[ {{T^{a - b - c}}{L^{b + 2c}}{M^c}} \right]$ …… (5)
Step 3:
Using principle of homogeneity, and equating powers of same base from both sides we get set of three linear equations-
c=1 …… (6)
a-b-c=0 …… (7)
b+2c=0 …… (8)
Step 4:
On solving (6), (7) and (8) we will get: a=1, b=-2 and c=1
On replacing values of a, b and c in equation (4) we will get dimension of M in terms of angular momentum, velocity and time taken as fundamental quantity:
After replacing we get $\left[ M \right] = \left[ {T{C^{ - 2}}h} \right]$
Correct Answer:
(d) $\left[ M \right] = \left[ {T{C^{ - 2}}h} \right]$

Note: Dimensional analysis can be used to verify the correctness of physical laws. Principle of dimensional homogeneity is used in various physical phenomena to predict the new physical quantities in classical physicals (even without knowing much about the underlying physics behind). It would be an empirical guess but very reliable.