Question

$9$ balls of diameter $2$ cm each are packed tightly in a box as shown. The height of the box is :

• A. $2\sqrt{3}+3$
• B. $2\sqrt{3}+2$
• C. $2+\sqrt{3}$
• D. $4$

Answer 1

The correct option is B $2\sqrt{3}+2$

Consider the three balls with centers $C,E$ and $D$ as shown in figure.

Line $PQ$ is a straight line parallel to side of the box, or, $PQ$ is the height of the box.

From fig,

$PQ=PC+CA+AD+DQ$

Given, diameter of balls is $2$.

$\therefore radius,r=\frac{diameter}{2}=1$

$PC=DQ=1$

$CA=AD$

$CE$ and $DE$ are the sum of radius of two balls. In other terms, the diameter.

$\therefore CE=ED=2$

Join $A$ and $E$.

$△AEC$ is right angled at $A$.

By Pythagoras Theorem,

$C{E}^2=C{A}^2+E{A}^2$

$4=C{A}^2+1$

$CA=\sqrt{3}$

$\therefore CA=AD=\sqrt{3}$

$PQ=PC+CA+AD+DQ$

$PQ=1+\sqrt{3}+\sqrt{3}+1$

$PQ=2+2\sqrt{3}$

$\therefore Heightofbox=2\sqrt{3}+2$