Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4?? second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap? (Take g = 9.8 m/s²)
Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4?? second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap? (Take g = 9.8 m/s²)
Option 1 -
1 drop / 7 seconds
Option 2 -
3 drops / 2 seconds
Option 3 -
1 drop / second
Option 4 -
2 drops / second
Asked by Shiksha User
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1 Answer
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Correct Option - 3
Detailed Solution:Displacement in 4? second
= S? - S?
= (1/2)g [2n - 1]
= (1/2)g [2×4 - 1]
= (1/2) × 9.8 × 7
= 34.3m
As this distance matches with data given in question for position of next drop. So drops are falling at the rate of 1 drop/second.
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