Animated Solution for Physics - Physics and Measurement: If $C$ and $V$ represent capacity and voltage respectively, then what are the dimensions of $\lambda$, where $\frac{C}{V}=\lambda$?

Dimensional analysis is a powerful tool in physics that allows us to understand the fundamental nature of physical quantities. In this problem, we are introduced to a new quantity, $\lambda$, which is defined as the ratio of capacitance $C$ to voltage $V$.

Our mission is to find the dimensional formula for $\lambda$. To do this, we must first determine the individual dimensional formulas for both capacitance and voltage.

If you don't have the dimensional formulas for $C$ and $V$ memorized, don't worry! You can easily derive them from basic principles.

Let's start with voltage, $V$. We know that voltage is defined as work done per unit charge:

The dimensions of work $W$ are $[M{L}^2{T}^{-2}]$, and the dimensions of charge $q$ are $[IT]$. Substituting these in, we get:

Next, let's find the dimensions of capacitance, $C$. Capacitance is defined as charge stored per unit voltage:

Using the dimensions of charge $[IT]$ and the dimensions of voltage we just found, we get:

Now that we have the dimensional formulas for both $C$ and $V$, we can substitute them into our expression for $\lambda$:

This is where we need to be careful with our exponent rules. When dividing terms with the same base, we subtract the exponent of the denominator from the exponent of the numerator.

Let's process each fundamental dimension one by one:

For Mass (M): $-1-1=-2⟹M^{-2}$

For Length (L): $-2-2=-4⟹L^{-4}$

For Time (T): $4-(-3)=4+3=7⟹T^7$

For Current (I): $2-(-1)=2+1=3⟹I^3$

Combining all these together, we arrive at our final dimensional formula:

This perfectly matches option (d). Dimensional analysis problems like this are excellent for building your confidence in algebraic manipulation and reinforcing your understanding of how different physical quantities are interconnected.