Small water droplets of radius 0.01 ~mm are formed in the upper atmosphere and falling with a terminal velocity of 10 ~cm / s . Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new

Correct answer: 0 m= mass of small drop M= mass of bigger drop aligned & V_t=(2)/(9) (R^2( -σ) g)/( ) & 8 m=M & 8 r^3=R^3 R=2 R aligned as V_ t × R^2 Radius double so V_ t becomes 4 time 4 × 10=40 ~cm / s

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Mechanical Properties of Fluids

JEE Main 2024 · 8 April · Evening Shift · Physics · Mechanical Properties of Fluids

Small water droplets of radius 0.01 ~mm are formed in the upper...

Small water droplets of radius

0.01 \mathrm{~mm}

are formed in the upper atmosphere and falling with a terminal velocity of

10 \mathrm{~cm} / \mathrm{s}

. Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new terminal velocity will be _____

\mathrm{cm} / \mathrm{s}

Correct answer

Step-by-step solution

\mathrm{m}=

mass of small drop

\mathrm{M}=

mass of bigger drop

\begin{aligned} & V_t=\frac{2}{9} \frac{R^2(\rho-\sigma) g}{\eta} \\ & 8 \propto m=M \\ & 8 r^3=R^3 \Rightarrow R=2 R\end{aligned}

\mathrm{V}_{\mathrm{t}} \times \mathrm{R}^2 \quad \because

Radius double so

\mathrm{V}_{\mathrm{t}}

becomes 4 time

\therefore 4 \times 10=40 \mathrm{~cm} / \mathrm{s}

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