Question

2.] Six small raindrops each of radius 1.5mm, come down with a terminal velocity of 6cm^ -1. They coalesce to form velocity of the bigger drop?​

Answer 1

࿊ Solution ࿊

• $Radius \: of \: each \: small \: drop \: r \: = 1.5mm \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0.15cm.$
• $Terminal \: velocity \: for \: small \: drops \: Vt \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6cm \: s {}^{ - 1}$
• $Let \: the \: density \: of \: water \: = ρ \\ and \: the \: density \: of \: air \: = ρ'.$

$Then \: Vt \: = \frac{2r {}^{2} }{9η} (ρ - ρ')g$

$6cm \: s {}^{ - 1} \frac{2 \times (0.15) {}^{2} }{9η} (ρ - ρ')g \: ... \: (i)$

When the six drops combine, let the radius of the bigger drop be R.

Then volume of bigger drop

= 6 × (Volume of a small drop)

$➡ \: \frac{4}{3} \pi R {}^{3} = 6 \times \frac{4}{3} \pi r {}^{3}$

$➡ \: R {}^{3} = 6r {}^{3}$

$➡ \: R = (6) { \frac{}{} }^{ \frac{1}{3} }r$

$\: \: \: \: \: = (6) { \frac{}{} }^{ \frac{1}{3} } (0.15)cm$

Let the terminal velocity for this drop be VT.

$Then \: VT = \frac{2 \times R {}^{2} }{9η} (ρ - ρ')g$

$= > \: VT = \frac{2 \times 6 {}^{ \frac{2}{3}}(0.15) {}^{2} }{9η} (ρ - ρ')g \: ... \: (ii)$

Dividing equation (ii) by equation (i),

$we \: get \: \frac{VT}{6} = (6) { \frac{}{} }^{ \frac{2}{3} }$

$VT \: = 6 { \frac{}{} }^{ \frac{5}{3} } ms {}^{ - 1}$

$Answer :- \: VT \: = 19.81 \: ms {}^{ - 1}$

Explanation:

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Answer 2

Answer:

HOPE SO YOUR DOUBT IS CLEARED THANK YOU