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Question: Eight drops of water, each of radius 2 mm are falling through air at a terminal velocity of \(8 \mat...

Eight drops of water, each of radius 2 mm are falling through air at a terminal velocity of 8 cm s−18 \mathrm {~cm} \mathrm {~s} ^ { - 1 } If they coalesce to form a single drop, then the terminal velocity of combined drop will be

A

32 cm s−132 \mathrm {~cm} \mathrm {~s} ^ { - 1 }

B

30 cm s−130 \mathrm {~cm} \mathrm {~s} ^ { - 1 }

C

28 cm s−128 \mathrm {~cm} \mathrm {~s} ^ { - 1 }

D

24 cm s−124 \mathrm {~cm} \mathrm {~s} ^ { - 1 }

Answer

32 cm s−132 \mathrm {~cm} \mathrm {~s} ^ { - 1 }

Explanation

Solution

Let the radius of bigger drop is R and smaller drop is r then

Or R = 2r ….. (i)

Terminal velocity,

∴v′v=R2r2=(2rr)2=4\therefore \frac { \mathrm { v } ^ { \prime } } { \mathrm { v } } = \frac { \mathrm { R } ^ { 2 } } { \mathrm { r } ^ { 2 } } = \left( \frac { 2 \mathrm { r } } { \mathrm { r } } \right) ^ { 2 } = 4 (Using (i))

Or v′=4v=4×8=32 cm s−1\mathrm { v } ^ { \prime } = 4 \mathrm { v } = 4 \times 8 = 32 \mathrm {~cm} \mathrm {~s} ^ { - 1 }