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Question: R, L and C represent the physical quantities resistance, inductance and capacitance respectively. Wh...

R, L and C represent the physical quantities resistance, inductance and capacitance respectively. Which one of the following combinations has a dimension of frequency?
(A) 1RC\dfrac{1}{{RC}}
(B) RL\dfrac{R}{L}
(C) 1LC\dfrac{1}{{LC}}
(D) CL\dfrac{C}{L}

Explanation

Solution

According to the principle of homogeneity of equations, in an equation of physical quantities the dimensional formula of the expressions on both the sides of the expressions must be the same. Go through each option and check which one obeys the principle of homogeneity.

Complete step by step solution:
According to the principle of homogeneity, in the equations of physical quantities, the dimensional formula of the expression on both the sides of the expressions must be the same.
For example, speed=distanceTimespeed = \dfrac{{distance}}{{Time}}
According to question,
There are three physical quantities to represent: resistance, inductance and capacitance.
Resistance(R)- The unit of resistance is volt/Amp.
Capacitance(C)- the unit of capacitance is Amp.sec/Volt
Inductance(L)- The unit of inductance is Volt.Sec/Volt
Now when we calculate , LR\dfrac{L}{R}= sec
When we calculate, RC\dfrac{R}{C}=sec
Firstly we can write the dimensions of these three physical quantities.
Firstly, we calculate the dimension of resistance[R]
R=PotentialDiffCurrent=ML2T−3A−1A2=ML2T−3A−2R = \dfrac{{PotentialDiff}}{{Current}} = \dfrac{{M{L^2}{T^{ - 3}}{A^{ - 1}}}}{{{A^2}}} = M{L^2}{T^{ - 3}}{A^{ - 2}}
Now, we calculate the dimension of capacitance[C]
C=QPotentialDiff=ATML2T−3A−1=M−1L−2T4A2C = \dfrac{Q}{{PotentialDiff}} = \dfrac{{AT}}{{M{L^2}{T^{ - 3}}{A^{ - 1}}}} = {M^{ - 1}}{L^{ - 2}}{T^4}{A^2}
Now, we calculate the dimension of inductance[L]=
L=ϕCurrent=ML2T−2A−1A=ML2T−2A−2L = \dfrac{\phi }{{Current}} = \dfrac{{M{L^2}{T^{ - 2}}{A^{ - 1}}}}{A} = M{L^2}{T^{ - 2}}{A^{ - 2}}
Now let us check with options, whether they obey principle of homogeneity
OPTION(A) =1RC=1ML2T−3A−2×M−1L−2T4A2=1T1=s−1=f\dfrac{1}{{RC}} = \dfrac{1}{{M{L^2}{T^{ - 3}}{A^{ - 2}} \times {M^{ - 1}}{L^{ - 2}}{T^4}{A^2}}} = \dfrac{1}{{{T^1}}} = {s^{ - 1}} = f
Option (A) doesn’t satisfy the conditions.
OPTION(B) RL=ML2T−3A−2ML2T−2A−2=1T1=s−1=f\dfrac{R}{L} = \dfrac{{M{L^2}{T^{ - 3}}{A^{ - 2}}}}{{M{L^2}{T^{ - 2}}{A^{ - 2}}}} = \dfrac{1}{{{T^1}}} = {s^{ - 1}} = f
Option (B) satisfies the conditions for frequency.
OPTION(C) 1LC=1ML2T−2A−2×M−1L−2T4A2=1T2=s−2≠f\dfrac{1}{{LC}} = \dfrac{1}{{M{L^2}{T^{ - 2}}{A^{ - 2}} \times {M^{ - 1}}{L^{ - 2}}{T^4}{A^2}}} = \dfrac{1}{{{T^2}}} = {s^{ - 2}} \ne f
Option (C) doesn’t satisfy the conditions.
OPTION(D) CL=M−1L−2T4A2ML2T−2A−2=M−2L−4T6A−4≠f\dfrac{C}{L} = \dfrac{{{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}}}{{M{L^2}{T^{ - 2}}{A^{ - 2}}}} = {M^{^{ - 2}}}{L^{ - 4}}{T^6}{A^{ - 4}} \ne f
Option (D) doesn’t satisfy the conditions.

Hence, option (B) is correct because they have the dimension of frequency but option A,C,D is not correct because they do not have dimension of frequency.

Note: In these questions, we can use symbols instead of the names of the base quantities and they are represented in square brackets [M] denotes the mass, [L] denotes the length, [K] denotes the temperature, [T] denotes the time and [A] for current etc.