JEE MAIN - Physics (2023 - 13th April Morning Shift - No. 21)
A fish rising vertically upward with a uniform velocity of $$8 \mathrm{~ms}^{-1}$$, observes that a bird is diving vertically downward towards the fish with the velocity of $$12 \mathrm{~ms}^{-1}$$. If the refractive index of water is $$\frac{4}{3}$$, then the actual velocity of the diving bird to pick the fish, will be __________ $$\mathrm{ms}^{-1}$$.
Answer
3
Explanation
The bird's diving velocity is given relative to the fish. In order to find the actual velocity of the bird, we need to consider the refractive index of the water.
The fish sees the bird diving with a velocity of $$12 \ \text{m/s}$$. We can write the equation considering the velocities relative to the fish:
$$\frac{V_{\text{b/f}}}{\frac{4}{3}} = \frac{-8}{\frac{4}{3}} + \frac{-v}{1}$$
Here, $$V_{\text{b/f}}$$ is the bird's diving velocity relative to the fish, and $$v$$ is the actual velocity of the bird.
Now, let's solve for $$v$$:
$$\frac{-12}{\frac{4}{3}} = \frac{-8}{\frac{4}{3}} - v$$
$$v = 3 \ \text{m/s}$$
So, the actual velocity of the diving bird to pick the fish relative to the fish is $$3 \ \text{m/s}$$.
The fish sees the bird diving with a velocity of $$12 \ \text{m/s}$$. We can write the equation considering the velocities relative to the fish:
$$\frac{V_{\text{b/f}}}{\frac{4}{3}} = \frac{-8}{\frac{4}{3}} + \frac{-v}{1}$$
Here, $$V_{\text{b/f}}$$ is the bird's diving velocity relative to the fish, and $$v$$ is the actual velocity of the bird.
Now, let's solve for $$v$$:
$$\frac{-12}{\frac{4}{3}} = \frac{-8}{\frac{4}{3}} - v$$
$$v = 3 \ \text{m/s}$$
So, the actual velocity of the diving bird to pick the fish relative to the fish is $$3 \ \text{m/s}$$.
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