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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Physics
Chapter
Mechanical Properties of Fluids

BITSAT 2023 · Paper 1 · Physics · Mechanical Properties of Fluids

If 1000 droplets of water of surface tension 0.07 N / m . having...

If 1000 droplets of water of surface tension 0.07 N/m\mathrm{N} / \mathrm{m}. having same radius 1 mm1 \mathrm{~mm} each, combine to from a single drop. In the process the released surface energy is( Take π=227)\left.\pi=\frac{22}{7}\right)

Options

  1. A7.92×10−6J7.92 \times 10^{-6} \mathrm{ J}
  2. B7.92×10−4J7.92 \times 10^{-4} \mathrm{ J}
  3. C9.68×10−4J9.68 \times 10^{-4} \mathrm{ J}
  4. D8.8×10−5J8.8 \times 10^{-5} \mathrm{ J}

Correct answer

B. 7.92×10−4J7.92 \times 10^{-4} \mathrm{ J}

Step-by-step solution

We have Vf=ViV_f=V_i
\Rightarrow \frac{4}{3} \pi r_{\mathrm{f}^3=1000 \times \frac{4}{3} \pi \mathrm{r}_{\mathrm{i}^3 \Rightarrow \mathrm{r}_{\mathrm{f}^3=1000 \mathrm{r}_{\mathrm{i}^3
⇒rf=10ri\Rightarrow r_f=10 r_i
So, released energy = Initial surface energy - final surface energy
=1000 \times \mathrm{T} \times 4 \pi \mathrm{r}_{\mathrm{i}^2-\mathrm{T} \times 4 \pi \mathrm{r}_{\mathrm{f}^2
=4 \pi \mathrm{T} \left(1000 \mathrm{r}_{\mathrm{i}^2- \mathrm{r}_{\mathrm{f}^2\right)<br />=4 \pi \times0.07 \left(1000 \mathrm{r}_{\mathrm{i}^2-100 \mathrm{r}_{\mathrm{i}^2\right)
=4π×0.07×900ri2=4 \pi \times0.07 \times900 r_i^2
=4π×63×10−6=7.92×10−4J=4 \pi \times63 \times10^{-6}=7.92 \times10^{-4} \mathrm{ J}
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