GATE Mechanical 2022 Set 2 | GA Question: 10

Question

Equal sized circular regions are shaded in a square sheet of paper of $1$ cm side length. Two cases, case $\text{M}$ and case $\text{N}$, are considered as shown in the figures below. In the case $\text{M}$, four circles are shaded in the square sheet and in the case $\text{N}$, nine circles are shaded in the square sheet as shown.

What is the ratio of the areas of unshaded regions of case $\text{M}$ to that of case $\text{N}$?

 

• A. $2 : 3$
• B. $1 : 1$
• C. $3 : 2$
• D. $2 : 1$

Answer 1

Let the radius of the big circle be $\text{R cm},$ and the radius of the small circle be $r \;\text{cm}.$

We can get from $\text{case M}$ and $\text{case N.}$

• $\text{R + R + R + R} = 1 \Rightarrow \text{4R} = 1 \Rightarrow {\color{Blue}{\boxed{\text{R} = \frac{1}{4}\;\text{cm}}}}$
• $r+r+r+r+r+r = 1 \Rightarrow 6r = 1 \Rightarrow {\color{Purple}{\boxed{r = \frac{1}{6}\;\text{cm}}}}$

Now, we can find the area of the circles in both the cases.

• $\text{Case M:}$
  • Area of the four circles $ = 4 \times \pi \text{R}^{2}$
• $\text{Case N:}$
  • Area of the nine circles $ = 9 \times \pi \text{r}^{2}$

The area of the square in both the cases $ = 1^{2} = 1$

Now, we can find the areas of unshaded regions.

• $\text{Case M:}$
  • The area of unshaded regions $ = 1-4\pi\text{R}^{2}  = 1 – 4\pi \left(\frac{1}{4}\right)^{2} = 1 – \frac{\pi}{4}$
• $\text{Case N:}$
  • The area of unshaded regions $ = 1-9\pi\text{r}^{2} = 1-9\pi \left(\frac{1}{6}\right)^{2} = 1 – \frac{\pi}{4}$

$\therefore$ The ratio of the areas of unshaded regions of $\text{case M}$ to that of $\text{case N}$ is $1:1.$

Correct Answer $:\text{B}$

${\color{Magenta}{\textbf{PS:}}}$

• ${\color{Green}{\text{The area of square}  = \text{(side)}^{2}}}$
• ${\color{Lime}{\text{The area of circle}  = \pi \times (\text{radius})^{2}}}$