JEE Mains · Physics · STD 11 - 2. motion in straight line
Water droplets are coming from an open tap at particular rate. The spacing between a droplet observed at \(4^{{th}}\;second\) after its fall to the next droplet is \(34.3 \,{m}\). At what rate the droplets are coming from the tap ? (Take \(g=9.8\, {m} / {s}^{2}\))
- A \(1 \,drop/ second\)
- B \(2 \,drops / second\)
- C \(1\, drops / 7 seconds\)
- D \(3\, drops / 2 seconds\)
Answer & Solution
Correct Answer
(A) \(1 \,drop/ second\)
Step-by-step Solution
Detailed explanation
In 4 sec. \(1^{\text {st }}\) drop will travel \(\Rightarrow \frac{1}{2} \times(9.8) \times(4)^{2}=78.4 m\) \(\therefore 2^{\text {nd }}\) drop would have travelled \(\Rightarrow 78.4-34.3=44.1 m\). Time for \(2^{\text {nd }}\) drop \(\frac{1}{2}(9.8) t ^{2}=44.1\) \(t=3 sec\)…
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