### Video Transcript

A pool containing a liquid with a density of 1000 kilograms per metre cubed at the surface of an unknown planet produces a pressure of 8400 pascals at a depth of 2.4 metres. What is the acceleration due to gravity at the surface of this planet?

Okay, so this is an interesting question because we’re studying the behaviour of a liquid on an unknown planet. Specifically, we’re told that the liquid has a density of 1000 kilograms per metre cubed. And we’re told that it produces a pressure of 8400 pascals at a depth of 2.4 metres. We need to use this information to find out the acceleration due to gravity.

To do this, we can recall the equation that gives us the pressure exerted by a liquid onto an object that’s immersed in the liquid. The equation in question tells us that the pressure exerted by the liquid onto the object is equal to the density of the liquid, not of the object, of the liquid, multiplied by the acceleration due to gravity of the planet, which by the way we’ve called 𝑔 sub planet.

Now normally, we just call this 𝑔. But 𝑔 usually represents the acceleration due to gravity on Earth. So we call this 𝑔 sub planet. And we also need to remember to multiply this expression by the distance below the surface that the object is in the liquid.

So if this is the surface of the liquid and the liquid has a density 𝜌 and we’ve got an object here at a depth which we’ll call ℎ and the acceleration due to gravity on this planet is given by 𝑔 sub planet, then the pressure exerted by the liquid onto the object is given by 𝜌 multiplied by 𝑔 planet multiplied by ℎ.

Now in this question, we’re trying to find out the value of 𝑔 sub planet, the acceleration due to gravity on the planet. Specifically, we’re trying to find it at the surface of the planet. But this is not a problem because we’ve been told that the pool containing the liquid is also at the surface of the planet. So we can ignore this piece of information.

So we need to rearrange this equation for 𝑔 sub planet. We do this by dividing both sides of the equation by 𝜌ℎ. This way, the 𝜌s on the right-hand side cancel and the ℎs on the right-hand side cancel. And we’re just left with 𝑔 sub planet on the right. Hence, our equation becomes 𝑃 divided by 𝜌ℎ is equal to 𝑔 sub planet.

At this point then, we can substitute in the values that we’ve been given. So we can say that 𝑔 sub planet is equal to, firstly, the pressure, which we know is 8400 pascals. And we divide this by the density of the liquid, which is 1000 kilograms per metre cubed, times the depth of that random object below the surface of the liquid, which in this case is 2.4 metres. And this will give us a value for the acceleration due to gravity on the surface of the planet.

Now what we’re trying to find is an acceleration. So it should have the units of metres per second squared because those are the standard units of acceleration. However, we’ll only have these standard units if we’ve used all the values in our equation in their own standard units as well. So let’s check that we’ve done this.

Firstly, the pressure: the standard unit of pressure is pascals. And we have given our value in pascals. So this is fine. Secondly, the density: the standard unit of density is kilograms per metre cubed. And that’s exactly what we’ve done here as well. So this value is also fine. Thirdly, the depth: now depth is a type of length. And length has a standard unit of metres, which is exactly what we’ve given our value in. So this is also in its standard unit. Hence, our final answer will be in metres per second squared.

So now that we know this, let’s evaluate the right-hand side of the equation. When we do this, we find our value of 𝑔 sub planet to be 3.5 metres per second squared. Hence, our final answer is that the acceleration due to gravity at the surface of this planet is 3.5 metres per second squared.