Q159 / 180 · COMEDK 2022 · Physics
Two guns $P$ and $Q$ can fire bullets at speeds $2 \mathrm{~km} / \mathrm{s}$ and $4 \mathrm{~km} / \mathrm{s}$, respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum or areas covered by the bullets fired by the two guns, on the ground is
- A$1: 2$
- B$1: 4$
- C$1: 8$
- Correct answerD$1: 16$
Solution
The maximum horizontal range $R$ of a projectile fired with speed $v$ is given by the formula $R = \dfrac{v^2}{g}$, where $g$ is the acceleration due to gravity.
The bullets are fired in all possible directions, so the area covered on the ground is a circle with radius $R$. The area $A$ is given by $A = \pi R^2 = \pi \left( \dfrac{v^2}{g} \right)^2 = \dfrac{\pi v^4}{g^2}$.
For gun $P$, the speed is $v_P = 2 \text{ km/s}$. The area covered is $A_P = \dfrac{\pi (2)^4}{g^2} = \dfrac{16\pi}{g^2}$.
For gun $Q$, the speed is $v_Q = 4 \text{ km/s}$. The area covered is $A_Q = \dfrac{\pi (4)^4}{g^2} = \dfrac{256\pi}{g^2}$.
The ratio of the areas covered is $\dfrac{A_P}{A_Q} = \dfrac{16\pi / g^2}{256\pi / g^2} = \dfrac{16}{256} = \dfrac{1}{16}$.
Answer: $1: 16$
The bullets are fired in all possible directions, so the area covered on the ground is a circle with radius $R$. The area $A$ is given by $A = \pi R^2 = \pi \left( \dfrac{v^2}{g} \right)^2 = \dfrac{\pi v^4}{g^2}$.
For gun $P$, the speed is $v_P = 2 \text{ km/s}$. The area covered is $A_P = \dfrac{\pi (2)^4}{g^2} = \dfrac{16\pi}{g^2}$.
For gun $Q$, the speed is $v_Q = 4 \text{ km/s}$. The area covered is $A_Q = \dfrac{\pi (4)^4}{g^2} = \dfrac{256\pi}{g^2}$.
The ratio of the areas covered is $\dfrac{A_P}{A_Q} = \dfrac{16\pi / g^2}{256\pi / g^2} = \dfrac{16}{256} = \dfrac{1}{16}$.
Answer: $1: 16$
Hence, the correct answer is option: 'D'.