OFFICIAL EXPLANATION




\(\text { Let the radius of the circle above be ' } r \text { ' }\)

The area of the right triangle $\(A O B\)$ is $\(\frac{1}{2} \times\)$ Base $\(\times\)$ Height $\(=\frac{1}{2} \times A O \times O B=\frac{1}{2} \times r \times r=\frac{1}{2}\left(r^2\right)\)($ We have $\(\mathrm{AO}=\mathrm{OB}=\)$ radii $\(=\mathrm{r}\)$ )

Now, the area of the shaded portion $=$ area of sector $\(\operatorname{COD}=\frac{\theta}{360} \times \pi r^2=\frac{60}{360} \times \frac{22}{7} \times \mathrm{r}^2=\frac{11}{21}\left(\mathrm{r}^2\right)\)$

As $\(\frac{1}{2}\)$ is less than $\(\frac{11}{21}\)$, we get $\(\frac{1}{2}\left(\mathrm{r}^2\right)<\frac{11}{21}\left(\mathrm{r}^2\right)\)$

Hence column $\(B\)$ has higher quantity when compared with column $\(A\)$, so the answer is ( $\(B\)$ ).