WBJEE · Physics · Mechanical Properties of Fluids
1000 droplets of water having \(2 \mathrm{mm}\) diameter each coalesce to form a single drop. Given the surface tension of water is \(0.072 \mathrm{Nm}^{-1}\). The energy loss in the process is
- A \(8.146 \times 10^{-4} \mathrm{J}\)
- B \(4.4 \times 10^{-4} \mathrm{J}\)
- C \(2108 \times 10^{-5} \mathrm{J}\)
- D \(4.7 \times 10^{-1} \mathrm{J}\)
Answer & Solution
Correct Answer
(A) \(8.146 \times 10^{-4} \mathrm{J}\)
Step-by-step Solution
Detailed explanation
Let the radius of single drop = \(r\) Radius of small drop \(=\frac{2 m m}{2}\) \[ \begin{array}{l} =1 \mathrm{mm} \\ =1 \times 10^{-3} \mathrm{m} \end{array} \] Surface tension of water \(=0.072 \mathrm{N} / \mathrm{m}\) The volume of large drop must be equal volume of all…
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