# Video: CBSE Class X • Pack 2 • 2017 • Question 20

CBSE Class X • Pack 2 • 2017 • Question 20

04:24

### Video Transcript

The dimensions of a solid iron cuboid are 4.4 metres by 2.6 metres by one metre. It is melted and recast into a hollow cylindrical pipe of inner radius 30 centimetres and thickness five centimetres. Find the length of the pipe.

In this scenario, we have two different shapes. First, we start out with a cuboid with given dimensions, which we’re told is made of metal and melted down into a totally different shape — a hollow cylinder. The hollow cylinder has an inner radius — we’ve called 𝑟 sub 𝑖 — which is given as 30 centimetres and a wall thickness of five centimetres, which we’ve called 𝑡.

Knowing all this, we want to solve for the length of the hollow cylinder 𝑙. If we label the volume of the cuboid 𝑉 sub 𝑐, we’re told that volume is 4.4 times 2.6 times one cubic metres. When it comes to the volume of the hollow cylinder, we can start calculating that by recalling that the area of a circle is equal to 𝜋 times the radius of this circle squared.

On the end of the cylinder, we want to solve for the shaded area, which is the difference between the area of two circles. If we call the volume of the hollow cylinder 𝑉 sub 𝑝 for the volume of the pipe, that volume is equal to the area of the larger circle on the end of the pipe 𝜋 times the inner radius plus the pipe thickness all squared minus the area of the smaller hollow circle on the end of the pipe 𝜋 times the inner radius squared all multiplied by 𝑙, the length.

We know that the volume of the cuboid is equal to the volume of the pipe. So we can say that the volume of the cuboid is equal to 𝜋 times the radius of the outer circle squared minus the radius of the inner circle squared all multiplied by the length of the pipe 𝑙.

Looking at the different units in this expression, we see that on the left-hand side we have length units of metres, while on the right we have length units of centimetres. Recalling that one metre is equal to 100 centimetres, we can rewrite our left-hand side as 440 times 260 times 100 cubic centimetres. And our right-hand side simplifies to 𝜋 times 35 squared square centimetres minus 30 squared square centimetres all multiplied by 𝑙. If we multiply 35 by itself, we find a result of 1225. And then, when we square 30, we find a result of 900.

When we then subtract the area of the inner circle from the area of the outer circle on the end of the pipe, we find a result of 325 square centimetres. Then, if we let the number 𝜋 equal exactly 3.14, we can multiply that value by the number of square centimetres on the right-hand side of our equation. When we do, we find a result of 1020.50.

And next, to convert this into an integer, we can multiply both sides of our equation by two. This leaves us with the equation 880 times 260 times 100 cubic centimetres is equal to 2041 square centimetres multiplied by 𝑙. We can rearrange the numbers on the left-hand side of our equation so that it reads 88 times 26 times 10000. Then, when we multiply 88 by 26, we find a result of 2288.

With that number on the left-hand side of our expression, we’re now ready to solve for the length 𝑙. To do that, we’ll divide both sides of our equation by 2041 square centimetres. This cancels this term out on the right-hand side. And considering the units on the left-hand side, we see that two factors of centimetres cancel out, leaving us with units of centimetres for our length.

We’re now ready to calculate the pipe length 𝑙. And we’ll do it by dividing 2041 into 22880000. To start off, 2041 goes into 2288 one time. Then, it divides into 2470 one time, it divides into 4290 two times, into 2080 one time, and finally into 390 zero times.

This means that 𝑙 is equal to 11210 centimetres or 112.10 metres. That’s the length of the pipe.