Question

a large ball 2m in radius is made up of rope of square cross section with edge length 4mm,Neglecting the air gaps in the ball what is the,total length of the rope of the nearest order of magnitude ​

Answer 1

Solution:

We know that:

$\boxed{\sf{Volume\:of\:ball = Volume\:of\:rope}}$

Let length of rope be 'L' m.

So,

$\boxed{\sf{Volume = Area \times Length\:of\:rope}}$

$\implies$ $\sf{(edge length)^{2} \times L\:m}$

$\implies$ $\sf{(40mm)^{2} \times L\:m}$

$\implies$ $\sf{(40 \times 10^{-3} m)^{2} \times L\:m}$

$\implies$ $\sf{(4 \times 10^{-2})^{2} \times L\:m^{3}}$

$\implies$ $\sf{16 \times 10^{-4} \times L\:m^{3}}$

We know that:

$\boxed{\sf{Volume \: of \: ball = \frac{4}{3} \pi {r}^{3}}}$

$\implies$ $\sf{4/3 \pi\:(2m)^{3}}$

$\implies$ $\sf{4/3 \pi \:8m^{3}}$

$\implies$ $\sf{32/3 \pi\:m^{3}}$

Now,

$\boxed{\sf{L=\frac{32 \pi}{(3\times16\times10^{-4})}}}$

$\implies$ $\sf{2 \times 3.14/3 \times 10^{4}}$

$\implies$ $\sf{6.28/3 \times 10^{4}}$

$\implies$ $\sf{2.093 \times 10^{4}}$

$\implies$ $\sf{2093\:m}$

Hence,

The total length of the rope of the nearest order of magnitude is 2093 m.