A mad engineer builds a cube, 2.5 m on a side, in which 6.2-cm-diameter rubber balls are constantly sent flying in random directions by vibrating walls. He will award a prize to anyone who can figure out how many balls are in the cube without entering it or taking out any of the balls. You decide to shoot 6.2-cm-diameter plastic balls into the cube, through a small hole, to see how far they get before colliding with a rubber ball. After many shots, you find they travel an average distance of 1.8 m. How many rubber balls do you think are in the cube?

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Hi, everyone in this practice problem, we're being asked to estimate the total number of silicon particles in a tank. We will have an eccentric scientist building a large tank with a dimension of three m times two m times 1.5 m where tiny spherical silicon particles are constantly moving around because of thermal agitation. She's challenging her colleagues to determine the total number of particles in the tank without removing any of them or using any specialized equipment. And to tackle this, you decide to release many small spherical plastic particles into the tank and measure their average distance of travel before colliding with one of the metallic particles. After the trials, we find that the plastic particles travel on average distance of 1. m and considering the diameter of plastic and silicon particle is 4.8 centimeter. We're being asked to estimate the total number of silicon particles in the tank. The options given are a 733 B 100 and C 2056 D 1000 and 300. So we have the equation for the mean free path to be U to be equals to Lambda, which is going to equals to one divided by four, multiplied by squared of two, multiplied by pi multiplied by open parentheses and divided by V closed parentheses multiplied by R squared. So this is the equation that we have for uh mean free path. And we are given in the problem statement that our lambda is going to be equal to 1.2 m, which is the um average distance that the particle is traveling to. And then uh R is going to be 2.4 centimeter which is given because the diameter of the plastic and silicon particle is 4.8 centimeters. So the radius is going to be half of that which is 2.4 centimeter. I'm gonna write down equals to D divided by two or equals to 4. centimeter divided by two. Next, you want notice that the R is in centimeters. So we want to convert that into meter by multiplying our 2.4 with 10 to the power of negative two to obtain the meter or the si uh radius. So that will give us 0.24 m. Lastly, the thing, the last thing that we know uh that we need it is the uh volume of the tank which is going to be given by three m times two m times 1.5 m. And I'm just gonna leave it as that uh when we are substituting into our equation here. So now we are actually ready to um substitute all of our values into our equation. But remember what we are looking for here is the total number of silicon particles inside of the tank or essentially N. So rearranging our equation here N can then be obtained by actually taking V and dividing off that with four N squared of two pi multiplied by Lambda multiplied by R squared. So and after some rearrangement will be equals to P divided by four, multiplied by squared of two, multiplied by pi multiplied by lambda and R squared. So let's substitute all of our values into this equation that we have this final equation and will then be equals to three m times two m times 1.5 m for the volume and divide that with four multiplied by squared of two multiplied by pi multiplied by a Lambda, which is given to be 1.2 m and then R squared which is given to be 0.24 m squared. So after two, uh some calculations, we can uh determine that the total number of particles for silicon in the tank is going to be 732.68 or rounding it up and will equals to 733 particles. So 733 will correspond to option A in our answer choices which will be the answer to this practice problem. So that'll, that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our other lesson videos on similar topics available on our website and that will be it for this one. Thank you.

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Key Concepts

Volume of the Cube

Volume of a Sphere

Collision Probability and Average Distance