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Consider a cylinder of height $h$ cms and radius $r=\frac{2}{\pi}$ cms as shown in the figure (not drawn to scale). A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at point B, gives a maximum of $n$ turns (in other words, the string‟s length is the minimum length required to wind $n$ turns.)

The same string, when wound on the exterior four walls of a cube of side $n$ cms, starting at point C and ending at point D, can give exactly one turn (see figure, not drawn to scale).

How is $h$ related to $n?$

- $h=\sqrt{2}n$
- $h=\sqrt{17}n$
- $h=n$
- $h=\sqrt{13}n$

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Unfold the cylinder along the line AB,

It’ll form a rectangle with width $=2\pi r = 2 \pi \times \frac{2}{\pi} = 4$ cms, and length$=h$ cms.

length of $1$ roll of string $=\sqrt{(\frac{h}{n})^2 + 4^2} = \frac{1}{n} \sqrt{h^2 + 16n^2}$

length of string $=n \times \frac{1}{n}\sqrt{h^2 + 16n^2} = \sqrt{h^2 + 16n^2}$

Unfold the cube, then the length of the string is nothing but the hypotenuse of rectangle with height=$n$ and width=$4n$.

length of string $ = \sqrt{n^2 + (4n)^2} = \sqrt{17 n^2}$

So, $\sqrt{17 n^2} = \sqrt{h^2 + 16n^2} \implies h=n$

It’ll form a rectangle with width $=2\pi r = 2 \pi \times \frac{2}{\pi} = 4$ cms, and length$=h$ cms.

length of $1$ roll of string $=\sqrt{(\frac{h}{n})^2 + 4^2} = \frac{1}{n} \sqrt{h^2 + 16n^2}$

length of string $=n \times \frac{1}{n}\sqrt{h^2 + 16n^2} = \sqrt{h^2 + 16n^2}$

Unfold the cube, then the length of the string is nothing but the hypotenuse of rectangle with height=$n$ and width=$4n$.

length of string $ = \sqrt{n^2 + (4n)^2} = \sqrt{17 n^2}$

So, $\sqrt{17 n^2} = \sqrt{h^2 + 16n^2} \implies h=n$