CAT 2003 | Question: 2-65

Question

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 cm, 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 $n$?

• A. $h=\sqrt{2}n$
• B. $h=\sqrt{17}n$
• C. $h=n$
• D. $h=\sqrt{13}n$

Answer 1

for solving this question we will take a smaller example first.

suppose, we have a cylinder of $radius = r$ and $height = h$. a string of $length = l$, when wound on its cylindrical surface, starting at point $A$ and ending at point $B$, gives $n=3$ turns.

we will assume $2$ more points $p_1$ and $p_2$ co-linear to $A$ and $B$. now, we will take single full turns of the cylinder $( Ap_1, p_1 p_{2}$ and $p_2B )$ to $n=3$ similar cylinders of $radius = r$ and $height = h$.

now, when we open the paper through which cylinder was formed (it is a rectangle) we will get the length of one turn of the string like in the image for all of the 3 cylinders.

now, when we join these $n=3$ rectangles we will get the whole length of string $l$ = length of diagonal of resultant rectangle.

$l = \sqrt{(3.2\pi r)^{2} + h^{2}}$

 

so for $n$ turns we get the formula,

$l = \sqrt{(n.2\pi r)^{2} + h^{2}}$         ..............................(1)

 

now, similar story for the cube,

we get the formula,

$l = \sqrt{(4a)^{2} + a^{2}}$       ............................(2)

where, $a$ is the length of side of the cube.

here, $a = n$. so, from equation (2),

$l = \sqrt{(4n)^{2} + n^{2}}$          .............................(3)

from equation (1) and (3),

$\sqrt{(n.2\pi r)^{2} + h^{2}} = \sqrt{(4n)^{2} + n^{2}}$ 

both sides are +ve. so, squaring on both sides,

$(n.2\pi r)^{2} + h^{2} = (4n)^{2} + n^{2}$ 

$4n^{2}\pi^{2}r^{2} + h^{2} = 16n^{2} + n^{2}$ 

$4n^{2}\pi^{2}r^{2} + h^{2} = 17n^{2}$ 

putting, $r = 2/\pi$

$4n^{2}\pi^{2}.(4/\pi^{2}) + h^{2} = 17n^{2}$ 

$16n^{2} + h^{2} = 17n^{2}$ 

$h^{2} = 17n^{2} - 16n^{2}$

$h^{2} = n^{2}$

both sides are +ve. so, taking square root of both sides.

$h = \pm n$

$h$ and $n$ are both +ve. so, eliminating the -ve possibility.

hence, $h = n$.