00:01
The objective of this problem is to compare the weight of two beams.
00:08
It is known that the hollow circular section with external diameter capital d and internal diameter small d as 5 by 32 times d power 4 minus d power 4 divided by and and the solid circular section is equal to pi by 32 times d cube.
00:53
By the given condition, what we can say is the value of small d is equal to 0 .78 times capital d.
01:03
Consider z 1 to be equal to 5 divided by 32 times d cube.
01:10
Let's consider this as equation number 1.
01:13
Similarly, let z2 is equal to pi by 32 times d to the power 4 minus small d to the power 4 divided by capital d.
01:24
And this will be our second equation.
01:28
Now substitute this value of small d in equation number 2.
01:33
So equation number 2 becomes z 2 is equal to 5 by 32 times d power 4 minus 0.
01:45
Times capital d to the whole power 4 divided by and if we simplify this expression we will get pi by 32 times d cube times 0 .63.
02:04
Now as the two beans are same material and equal strength what we can say is the value of z1 is equal to z2.
02:14
So we can say pi times capital small d cube divided by 32 is equal to pi by 32 times d cube times 0 .63.
02:28
Here observe that we can cancel out these two pies and these two 32s...