GATE 2024 IN GA Q.7: solution and answer

Official answer: BSource: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. Let each chalk-stick have radius $r$ and let the common length be $h$. The container has the same length $h$, so each volume is (cross-sectional area) $\times h$, and $h$ cancels in any ratio. A ratio of volumes is therefore the same as a ratio of cross-sectional areas.

2. Two equal circles of radius $r$ that touch have centres $2r$ apart. Each of the six outer circles touches the central one, so every outer centre is $2r$ from the centre of the container. (The six outer centres form a regular hexagon with side $2r$, so neighbouring outer circles touch each other too, which agrees with the tight fit in the figure.)

3. Each outer circle also touches the container wall, so the wall is $2r+r=3r$ from the centre. The container's radius is $R=3r$.

4. Container cross-section: $\pi R^2=\pi (3r{)}^2=9\pi r^2$.

5. Occupied cross-section (7 chalk-sticks): $7\times \pi r^2=7\pi r^2$.

6. Empty cross-section: $9\pi r^2-7\pi r^2=2\pi r^2$.

7. Ratio of occupied space to empty space: $$\frac{7\pi r^2}{2\pi r^2}=\frac{7}{2}$$ This is option (B).