0 0 votes 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). The length of the string is$\sqrt{2}n$ $\text{cm}$$\sqrt{17}n$ $\text{cm}$$\text{n cm}$$\sqrt{13}n$ $\text{cm}$ Quantitative Aptitude cat2003-2 quantitative-aptitude mensuration + – Lakshman Bhaiya 12.2k points 5.4k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
Best answer 1 1 vote 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, $\sqrt{17 n^2} = \sqrt{h^2 + 16n^2} \implies h=n$ Nikhil_dhama answered Apr 5, 2022 • selected Apr 11, 2022 by Arjun Nikhil_dhama 946 points comment Share Follow 0 reply Please log in or register to add a comment.