# Video: CBSE Class X • Pack 5 • 2014 • Question 33

CBSE Class X • Pack 5 • 2014 • Question 33

04:31

### Video Transcript

Sushant has a vessel shaped as an inverted cone, where the top of the cone is open and has a radius of 2.5 centimeters. The inverted cone has a height of 11 centimeters and is full of water. If metallic spheres, each of diameter 0.5 centimeters, are placed in the vessel and cause two-fifths of the water within it to over flow, how many spheres are there?

Let’s start by finding the initial volume of water in the vessel. And since the vessel is in the shape of a cone, we simply need to find the volume of this cone. We have that the volume of the cone is equal to one-third 𝜋𝑟 squared ℎ, where 𝑟 is the radius of the cone and ℎ is the height of the cone.

We’re told in the question that the radius of the cone is 2.5 centimeters and the height of the cone is 11 centimeters. If we substitute these values in here, we’ll find the volume of water that was initially in the cone. And we find that this volume is equal to one-third 𝜋 times 2.5 squared times 11 centimeters cubed.

We’re told in the question that two-fifths of the water overflows when the metallic spheres are placed into the vessel. And so if we multiply our initial volume of water by two-fifths, we’ll find the volume of water which is displaced by the spheres. This gives us that the volume of water displaced is equal to two-fifth times one-third 𝜋 times 2.5 squared times 11 centimeters cubed.

And now, we can spot that we can do a small amount of cancelling here. Since we can write 2.5 as 0.5 times five, 2.5 squared can become 0.5 times five squared. And so one of the fives from this square can cancel with the five in the denominator of the fraction. And this gives us that the volume of water displaced is equal to two multiplied by one-third 𝜋 times five times 0.5 squared times 11 centimeters cubed. And we can write this all in one fraction as 110𝜋 times 0.5 squared all over three centimeters cubed.

Next, we need to consider the spheres. Let’s let the number of spheres which are added to the vessel be 𝑥. Then, if we use this along with the equation for the volume of a sphere which is four-thirds 𝜋𝑟 cubed and the diameter given in the question which is 0.5 centimeters, then we can find the total volume of the metallic spheres added to the vessel. And this volume must be equal to the volume of water displaced.

So our diameter of 0.5 centimeters means that the radius of the spheres is half of that, which is 0.25 centimeters. And so if we substituted this value of 𝑟 into the equation for the volume of a sphere, we’ll find the volume of one of our spheres. However, we have 𝑥 spheres. So the total volume of all of the spheres is 𝑥 multiplied by four-thirds 𝜋 times 0.25 cubed centimeters cubed. And so we obtain that 𝑥 multiplied by four-thirds 𝜋 times 0.25 cubed is equal to 110𝜋 times 0.5 squared over three since the volume of the metallic spheres is equal to the volume of the water displaced.

And now, we notice that on both sides of the equation, we have 𝜋. So this cancels. And we are also dividing by three on both sides of the equation. So these cancel also. And so we end up with four 𝑥 times 0.25 cubed is equal to 110 times 0.5 squared. And next, we’ll divide both sides of the equation by 0.25 squared. And what this leaves us with is four 𝑥 times 0.25 is equal to 110 multiplied by 0.5 squared over 0.25 squared.

And on the right-hand side, we notice that we have a square over a square. So we can take the square out of the fraction and then 0.5 over 0.25 is simply two since 0.5 is twice 0.25. We also notice that on the left-hand side of the equation 0.25 can be written as one-quarter. And so the equation can be simplified to 𝑥 is equal to 110 times two squared and then two squared is simply four.

And so this gives us that the number of spheres that were placed in the vessel was 440.