A bird in air is diving vertically over a tank with speed 6 cm/s. Base of the tank is silvered. A fish in the tank is rising upward along the same line with speed 8 cm/s. (Take μ water = 4/3)
Column I | Column II |
(i) Speed of the image of fish as seen by the bird directly | [A] 16 |
(ii) Speed of the image of fish formed after reflection from the mirror as seen by the bird | [B] 0 |
(iii) Speed of image of bird relative to the fish looking upwards | [C] 12 |
(iv) Speed of image of bird relative to the fish looking downwards in the mirror | [D] 8 |
Text Solution
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Ans.
(i) [C]
(ii)[B]
(iii)[A]
(iv)[A]
Sol. (i) Velocity of fish in air = 8 × 
= 6
Velocity of fish w.r.t. bird = 6 + 6 = 12
(ii) Velocity of image of fish after reflection
from mirror in air = 
= 6 ↓
w.r.t. bird = – 6 + 6 = 0
(iii) Velocity of bird as seen from water = 6 × 
= 8 ↓ w.r.t. fish = 8 + 8 = 16 ↓
(iv) Velocity of bird in water after reflectionfrom mirror = 8 ↓ w.r.t. fish = 8 – 8 = 0
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