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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Physics
Chapter
Mechanical Properties of Fluids

BITSAT 2023 · Paper 2 · Physics · Mechanical Properties of Fluids

Eight equal drops of water are falling through air with a steady...

Eight equal drops of water are falling through air with a steady speed of 10cm/s10 \mathrm{ cm} / \mathrm{s}. If the drops coalesce, the new velocity is:

Options

  1. A10cm/s10 \mathrm{ cm} / \mathrm{s}
  2. B40cm/s40 \mathrm{ cm} / \mathrm{s}
  3. C16cm/s16 \mathrm{ cm} / \mathrm{s}
  4. D5cm/s5 \mathrm{ cm} / \mathrm{s}

Correct answer

B. 40cm/s40 \mathrm{ cm} / \mathrm{s}

Step-by-step solution

Let r\mathrm{r} be the radius of small drops of water.R=\mathrm{R}= radius of big drop formed as volume remain same.∴8.43πr3−43πR3⇒R=2rTerminalvelocity,\therefore 8. \frac{4}{3} \pi \mathrm{r}^3-\frac{4}{3} \pi \mathrm{R}^3 \Rightarrow \mathrm{R}=2 \mathrm{r} Terminal velocity,vT=29η(ρ−σ)r2g∴vT∝r2∴v1v2=(rR)2⇒10v2=(12)2(∵v1=10cm/s given  )⇒v2=40cm/s\begin{aligned} \\ & \mathrm{v}_{\mathrm{T}}=\frac{2}{9 \eta}(\rho-\sigma) \mathrm{r}^2 \mathrm{ g} \\ \\ & \therefore \mathrm{v}_{\mathrm{T}} \propto \mathrm{r}^2 \\ \\ & \therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\left(\frac{\mathrm{r}}{\mathrm{R}}\right)^2 \\ \\ & \Rightarrow \frac{10}{\mathrm{v}_2}=\left(\frac{1}{2}\right)^2\left(\because \mathrm{v}_1=10 \mathrm{ cm} / \mathrm{s} \text{\text{ given } }\right) \\ \\ & \quad \Rightarrow \mathrm{v}_2=40 \mathrm{ cm} / \mathrm{s} \\ \end{aligned}$$
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