Question Video: Solving for the Dimensions of an Unknown Quantity in a Compound Quantity Physics

Consider the four quantities 𝑔, 𝐻, π‘š, and π‘˜, where [𝑔] = 𝑀𝐿²𝑇, [𝐻] = 𝑀⁻¹𝐿⁻³, and [π‘š] = 𝑀²𝑇⁻¹. The compound quantity π‘˜π‘”π»/π‘š is dimensionless. What are the dimensions of π‘˜?

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Video Transcript

Consider the four quantities 𝑔, 𝐻, π‘š, and π‘˜, where the dimensions of 𝑔 are mass times length squared times time, the dimensions of 𝐻 are mass to the negative one times length to the negative three, and the dimensions of π‘š are mass squared times time to the negative one. The compound quantity π‘˜ times 𝑔 times 𝐻 over π‘š is dimensionless. What are the dimensions of π‘˜?

Of the four quantities named here, we’re given the dimensions of three of them: 𝑔, 𝐻, and π‘š. We’re also told that the quantity π‘˜ times 𝑔 times 𝐻 divided by π‘š is dimensionless. Symbolically, we can write that as the dimensions of π‘˜ times 𝑔 times 𝐻 divided by π‘š equals one. Knowing all this, we want to solve for the dimensions of π‘˜. Another way to write this fraction is as the dimensions of π‘˜ times the dimensions of 𝑔 times the dimensions of 𝐻 all divided by the dimensions of π‘š. If we substitute in all the known dimensions of these quantities, we get this result. What we’ll do is simplify this fraction as far as possible. And that will help us understand the dimensions of π‘˜.

Notice that in our numerator we have mass times mass to the negative one. When multiplied together, these values equal one. Similarly, we have a length squared multiplied by a length to the negative three. The overall result of this is length to the negative one. This gives us the dimensions of π‘˜ times 𝐿 to the negative one times 𝑇 all divided by 𝑀 squared times 𝑇 to the negative one. Note that if we multiply numerator and denominator by the time 𝑇, then in the denominator, 𝑇 cancels with one over 𝑇. Our result can be written this way: the dimensions of π‘˜ times 𝐿 to the negative one times 𝑇 squared over 𝑀 squared.

Let’s now recall that this product on the right is equal to one. Therefore, the dimensions of π‘˜, whatever they are, must effectively cancel out these dimensions. This means that the dimensions of π‘˜ are equal to the inverse of these dimensions. Therefore, π‘˜β€™s dimensions are 𝑀 squared divided by 𝐿 to the negative one times 𝑇 squared. This is equal to 𝑀 squared times 𝐿 all divided by 𝑇 squared. So, in our dimensionless expression, π‘˜ times 𝑔 times 𝐻 divided by π‘š, the dimensions of π‘˜ are 𝑀 squared times 𝐿 divided by 𝑇 squared.

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