Question

Two guns A and B can fire bullets at speeds $1\mathrm{km}{\sec }^{-1}$ and $2\mathrm{km}{\sec }^{-1}$ respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is :

• A. $1:2$
• B. $1:16$
• C. $1:8$
• D. $1:4$

Answer 1

The correct option is B

$1:16$

Explanation for correct option:

Option (b) $1:16$

Step 1: Given data

Speed of bullet A: ${\mathrm{u}}_1=1\mathrm{km}{\sec }^{-1}$

Speed of bullet B: ${\mathrm{u}}_2=2\mathrm{km}{\sec }^{-1}$

Step 2: Formula used

We know that the formula for horizontal range is: $\mathrm{R}=\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}$
The horizontal range for gun ${\mathrm{A}}_{,}{\left({\mathrm{R}}_{\max }\right)}_{\mathrm{A}}=\frac{{\mathrm{u}}_{\mathrm{A}}^2}{\mathrm{g}}$
The horizontal range for gun $\mathrm{B},{\left({\mathrm{R}}_{\max }\right)}_{\mathrm{B}}=\frac{{\mathrm{u}}_{\mathrm{B}}^2}{\mathrm{g}}$

Now, we know that:
Area, $\mathrm{A}={\mathrm{\pi R}}^2$
$\mathrm{A}\propto {\mathrm{R}}^2$

$\frac{(\text{Area}{)}_A}{(\text{Area}{)}_{\mathrm{B}}}=\frac{{\left(R_{\max }\right)}_A^2}{{\left(R_{\max }\right)}_B^2}$$={\left(\frac{{\mathit{u}}_{\mathit{A}}^{\mathit{2}}}{{\mathit{u}}_{\mathit{B}}^{\mathit{2}}}\right)}^2$

To find the required answer, we will use the formula given below:

$\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{{\mathrm{u}}_1^4}{{\mathrm{u}}_2^4}$

Here, ${\mathrm{A}}_1$ is the maximum range for gun A

${\mathrm{A}}_2$ is the maximum range for gun B

${\mathrm{u}}_1$ is the speed of bullet A

${\mathrm{u}}_2$ is the speed of bullet B

Step 3: Find ratios of maximum areas covered by the bullets

By substituting the given values in the above-mentioned formula, we get:

$\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{1_{}^4}{2_{}^4}$

$\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{1}{16}$

Thus, the ratio of maximum areas covered by the bullets fired by the two guns, on the ground is $1:16$