Animated Solution for Physics - Physics and Measurement: If $E,L,M$ and $G$ denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimension of $P$ in the formula $P=E{L}^2{M}^{-5}{G}^{-2}$ is

Imagine you are faced with a complex-looking formula that combines several fundamental physical quantities. The problem asks us to find the dimensional formula of a new quantity $P$, which is defined by the equation:

Here, $E$ stands for energy, $L$ for angular momentum, $M$ for mass, and $G$ for the universal gravitational constant. At first glance, this might look intimidating, but the beauty of dimensional analysis is that it breaks down even the most complex expressions into simple algebraic operations on the base dimensions of Mass ($M$), Length ($L$), and Time ($T$).

Gathering Our Tools

To solve this, we first need to recall or derive the dimensional formulas for each of these fundamental quantities. Let's list them out:

1. Energy ($E$): Energy is equivalent to work done, which is force times displacement. Therefore, its dimensional formula is $[M{L}^2{T}^{-2}]$. 2. Angular Momentum ($L$): Angular momentum is given by the product of mass, velocity, and radius ($mvr$). Its dimensional formula is $[M{L}^2{T}^{-1}]$. 3. Mass ($M$): This is a base quantity, so its dimension is simply $[M]$. 4. Gravitational Constant ($G$): From Newton's law of gravitation, $F=\frac{G{m}_1{m}_2}{r^2}$, we can rearrange to find $G=\frac{F{r}^2}{m_1{m}_2}$. Substituting the dimensions gives us $[M^{-1}{L}^3{T}^{-2}]$.

Now, let's carefully substitute these dimensional formulas back into our original equation for $P$. We must be very careful with the exponents:

Let's simplify the numerator first. Squaring the angular momentum gives us $[M^2{L}^4{T}^{-2}]$. Multiplying this with the energy dimension, we add the powers of $M$, $L$, and $T$:

Next, let's look at the denominator. We have $M$ to the power five. Squaring the gravitational constant gives $[M^{-2}{L}^6{T}^{-4}]$. Combining the $M$ terms ($5-2=3$), we get:

Look closely at what we have now. The numerator and the denominator are exactly the same!

When we divide them, all the terms cancel out perfectly. This leaves us with $M^0$, $L^0$, and $T^0$.

So, $P$ is a dimensionless quantity. This is a classic JEE problem where complex-looking expressions simplify beautifully. Always double-check your dimensional substitutions and exponent rules to avoid silly mistakes.