Question

If C and L denote the capacitance and inductance, the dimensions of LC are
A. \[\left[ {{M^0}{L^0}{T^{ - 1}}} \right]\]
B. \[\left[ {{M^0}{L^{ - 1}}{T^0}} \right]\]
C. \[\left[ {{M^{ - 1}}{L^{ - 1}}{T^0}} \right]\]
D. \[\left[ {{M^0}{L^0}{T^2}} \right]\]

Answer 1

Hint: The term used for measuring the quantity or the portion of any object in certain units is known as the dimensions of that object. The fundamental way to write the dimension of any unit is \[\left[ {{M^a}{L^b}{T^c}} \right]\], where M is mass, L is length and T is time, and a, b and c are integers.

Formula Used:
\[f = \dfrac{1}{{2\pi \sqrt {LC} }}\]
where f is frequency, L is inductance and C is the capacitance.

Complete step-by-step solution:
As we have to find the dimensions of the product of inductance and capacitance LC, we will use the formula that contains that product or those terms. If a circuit contains an inductor and a capacitor, then we can find its inductance and capacitance, and using it we can also find its frequency. As we know the dimension of frequency, which is the inverse of time, which is \[\left[ {{M^0}{L^0}{T^{ - 1}}} \right]\], we can use it to find the dimension of the product LC.

Using the formula of frequency in a circuit, we get the dimension of LC as,
\[f = \dfrac{1}{{2\pi \sqrt {LC} }} \\
\Rightarrow \left[ {{M^0}{L^0}{T^{ - 1}}} \right] = \dfrac{1}{{\sqrt {LC} }} \\
\Rightarrow \dfrac{1}{{LC}} = \left[ {{M^0}{L^0}{T^{ - 2}}} \right] \\
\therefore LC = \left[ {{M^0}{L^0}{T^2}} \right] \]

So, option D, \[\left[ {{M^0}{L^0}{T^2}} \right]\] is the required solution.

Note: When we have to find the dimension of any value the constants are neglected as they do not have any unit, and only the terms containing the units are considered. Also, for finding the dimensions, the operations on the exponents are done like any other mathematical operation. Using the formula of frequency in the circuit, we can only find the dimensions of the product of LC and cannot find the dimensions of L and C separately. To find them separately, we will have to think and use some other formula.