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JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2021
  • Hard
  • MCQ

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 \(g=9.8\, {m} / {s}^{2}\))

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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

\(\frac{1}{2}(9.8) t ^{2}=44.1\)

\(t=3 sec\)

\(\therefore\) each drop have time gap of \(1 sec\)

\(\therefore 1\) drop per sec

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