Question

If C, R, L and I denote capacity, resistance, inductance and electric current respectively, the quantities having the same dimensions of time are
1. \[CR\]
2. \[L/R\]
3. \[\sqrt {LC} \]
4. \[L{I^2}\]

A. 1 and 2
B. 1 and 3
C. 1 and 4
D. 1, 2 and 3

Answer 1

Hint: In this question, we need to choose the correct option/s for the quantities having the same dimensions of time. For this, we need to use the dimensions of C, R, L, and I. After that, we need to find the dimensions of \[CR,L/R,\sqrt {LC} ,L{I^2}\] and have to check out of four how many are matching.

Formula used:
The dimensions of capacity (C), resistance (R), inductance (L) and electric current (I) are given below.
\[\left[ C \right] = \left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]\]
\[\Rightarrow \left[ R \right] = \left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]\]
\[\Rightarrow \left[ L \right] = \left[ {M{L^2}{T^{ - 2}}{A^{ - 2}}} \right]\]
\[\Rightarrow \left[ I \right] = \left[ {{M^0}{L^0}{T^0}A} \right]\]
Here, \[M\] is the mass, \[L\] is the length, \[T\] is the time and \[A\] is the electric current.
Also, we will use exponential properties to solve this question.
\[{a^m} \times {a^n} = {a^{m + n}}\] and \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]

Complete step by step solution:
We know the dimensions of capacity, resistance, inductance and electric current. Let us find the dimensions of \[CR,L/R,\sqrt {LC} ,L{I^2}\]. First one is \[CR\].
\[CR = \left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]\]
But we know that \[{a^m} \times {a^n} = {a^{m + n}}\]
By simplifying, we get
\[CR = \left[ {{M^{ - 1 + 1}}{L^{ - 2 + 2}}{T^{4 - 3}}{A^{2 - 2}}} \right]\]
\[\Rightarrow CR = \left[ {{M^0}{L^0}{T^1}{A^0}} \right]\] …… (1)
That is, the dimension of \[CR\] is \[\left[ {{M^0}{L^0}T{A^0}} \right]\]

The next one is \[L/R\].
\[\dfrac{L}{R} = \dfrac{{\left[ {M{L^2}{T^{ - 2}}{A^{ - 2}}} \right]}}{{\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]}}\]
But we know that \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
By simplifying and using exponential property, we get
\[\dfrac{L}{R} = \left[ {{M^1}{L^2}{T^{ - 2}}{A^{ - 2}}} \right]\left[ {{M^{ - 1}}{L^{ - 2}}{T^3}{A^2}} \right]\]
\[\Rightarrow \dfrac{L}{R} = \left[ {{M^{1 - 1}}{L^{2 - 2}}{T^{ - 2 + 3}}{A^{ - 2 + 2}}} \right]\]
By simplifying further, we get
\[\dfrac{L}{R} = \left[ {{M^0}{L^0}{T^1}{A^0}} \right]\] …… (2)
That is, the dimension of \[L/R\] is \[\left[ {{M^0}{L^0}{T^1}{A^0}} \right]\]

Now, third one is \[\sqrt {LC} \]
So, \[\sqrt {LC} = \sqrt {\left[ {M{L^2}{T^{ - 2}}{A^{ - 2}}} \right]\left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]} \]
Let us simplify this.
By applying the exponential properties, we get
\[\sqrt {LC} = \sqrt {\left[ {{M^{1 - 1}}{L^{2 - 2}}{T^{ - 2 + 4}}{A^{ - 2 + 2}}} \right]} \]
\[\Rightarrow \sqrt {LC} = \sqrt {\left[ {{M^0}{L^0}{T^2}{A^0}} \right]} \] …… (3)
That means, the dimension of \[\sqrt {LC} \] is \[\left[ {{M^0}{L^0}{T^2}{A^0}} \right]\]

And the last one is \[L{I^2}\].
\[L{I^2} = \left[ {M{L^2}{T^{ - 2}}{A^{ - 2}}} \right]{\left( {\left[ {{M^0}{L^0}{T^0}A} \right]} \right)^2}\]
\[\Rightarrow L{I^2} = \left[ {M{L^2}{T^{ - 2}}{A^{ - 2}}} \right]\left[ {{M^0}{L^0}{T^0}A} \right]\left[ {{M^0}{L^0}{T^0}A} \right]\]
By simplifying and using exponential properties, we get
\[L{I^2} = \left[ {{M^{0 + 0 + 0}}{L^{2 + 0 + 0}}{T^{ - 2 + 0 + 0}}{A^{ - 2 + 0 + 0}}} \right]\]
\[\Rightarrow L{I^2} = \left[ {{M^0}{L^2}{T^{ - 2}}{A^{ - 2}}} \right]\] ……. (4)
Thus, the dimension of \[L{I^2}\] is \[\left[ {{M^0}{L^2}{T^{ - 2}}{A^{ - 2}}} \right]\]
So, from (1), (2), (3) and (4), we get
So, we can say that Hence (1), (2), (3) have some dimension of time t.

Therefore, the correct option is (D).

Note:Many students generally make mistakes in finding dimensions of given quantities and also get confused with the exponential properties specifically with signs.