Eight spherical drops of the same mass and radius are falling down with a terminal velocity of 6cms if they coalesce to form a big drop the terminal speed of the bigger drop will be (Neglect the buoyancy of air)

VBig=n2/3Vsmall (8)2/3×6cm/s=4×6=24 cm/s

Eight spherical drops of the same mass and radius are falling down with a terminal velocity of 6cms if they coalesce to form a big drop the terminal speed of the bigger drop will be (Neglect the buoyancy of air)

VBig=n2/3Vsmall (8)2/3×6cm/s=4×6=24 cm/s

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Eight spherical drops of the same mass and radius are falling down with a terminal velocity of $6 \frac{c m}{s}$ if they coalesce to form a big drop the terminal speed of the bigger drop will be (Neglect the buoyancy of air)

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$6 \frac{c m}{s}$

$32 \frac{c m}{s}$

$1.5 \frac{c m}{s}$

$24 \frac{c m}{s}$

answer is C.

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$V_{B i g} = n^{2 / 3} V_{s m a l l} \begin{array}{l} {(8)}^{2 / 3} \times 6 c m / s \\ = 4 \times 6 \\ = 24 c m / s \end{array}$

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Eight spherical drops of the same mass and radius are falling down with a terminal velocity of 6cms if they coalesce to form a big drop the terminal speed of the bigger drop will be (Neglect the buoyancy of air)

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Eight spherical drops of the same mass and radius are falling down with a terminal velocity of $6 \frac{c m}{s}$ if they coalesce to form a big drop the terminal speed of the bigger drop will be (Neglect the buoyancy of air)