Eight drops of a liquid of density ρ and each of radius a are falling through air with a constant velocity 3.75 cms-1 . When the eight drops coalesce to form a single drop the terminal velocity of the new drop will be

Question

Eight drops of a liquid of density ρ and each of radius a are falling through air with a constant velocity 3.75 cms-1 . When the eight drops coalesce to form a single drop the terminal velocity of the new drop will be (A)  1.5× 10-2ms-1  (B)  2.4× 10-2ms-1  (C)  0.75× 10-2ms-1  (D)  25× 10-2ms-1

Tardigrade Admin · Accepted Answer

Terminal velocity of a single drop, v=3.75 text cms-1=3.75× 10-2ms-1 Terminal velocity of the big drop V=n2/3v=(8)2/3× 3.75× 10-2 =4× 3.75× 10-2 =15× 10-2ms-1

Q. Eight drops of a liquid of density $ρ$ and each of radius a are falling through air with a constant velocity $3.75 c m s^{- 1}$ . When the eight drops coalesce to form a single drop the terminal velocity of the new drop will be

$1.5 \times 10^{- 2} m s^{- 1}$

$2.4 \times 10^{- 2} m s^{- 1}$

$0.75 \times 10^{- 2} m s^{- 1}$

$25 \times 10^{- 2} m s^{- 1}$

$15 \times 10^{- 2} m s^{- 1}$

Terminal velocity of a single drop,
$v = 3.75 c m s^{- 1} = 3.75 \times 10^{- 2} m s^{- 1}$
Terminal velocity of the big drop
$V = n^{2/3} v = (8)^{2/3} \times 3.75 \times 10^{- 2}$
$= 4 \times 3.75 \times 10^{- 2}$
$= 15 \times 10^{- 2} m s^{- 1}$

Q. Eight drops of a liquid of density ρ and each of radius a are falling through air with a constant velocity 3.75cms−1 . When the eight drops coalesce to form a single drop the terminal velocity of the new drop will be

1.5×10−2ms−1

2.4×10−2ms−1

0.75×10−2ms−1

25×10−2ms−1

15×10−2ms−1

Terminal velocity of a single drop, v=3.75cms−1=3.75×10−2ms−1 Terminal velocity of the big drop V=n2/3v=(8)2/3×3.75×10−2 =4×3.75×10−2 =15×10−2ms−1