Excess pressure in a drop of a liquid is 400 n/m2. Eight such identical drops coalesce to form a single drop. Then excess pressure in the drop is

43πR3=8×43πr3⇒R=2rNow ΔP∝1r⇒Δp'Δp=rR⇒ΔP'=400×r2rNm2=200 Nm2

Excess pressure in a drop of a liquid is 400 n/m2. Eight such identical drops coalesce to form a single drop. Then excess pressure in the drop is

43πR3=8×43πr3⇒R=2rNow ΔP∝1r⇒Δp'Δp=rR⇒ΔP'=400×r2rNm2=200 Nm2

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Excess pressure in a drop of a liquid is 400 n/m². Eight such identical drops coalesce to form a single drop. Then excess pressure in the drop is

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100 N/m²

400 N/m²

50 N/m²

200 N/m²

answer is B.

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$\frac{4}{3} {πR}^{3} = 8 \times \frac{4}{3} {πr}^{3} \Rightarrow R = 2 r$

Now $ΔP \propto \frac{1}{r} \Rightarrow \frac{{Δp}^{'}}{Δp} = \frac{r}{R} \Rightarrow {ΔP}^{'} = 400 \times (\frac{r}{2 r}) \frac{N}{m^{2}} = 200 \frac{N}{m^{2}}$

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Excess pressure in a drop of a liquid is 400 n/m2. Eight such identical drops coalesce to form a single drop. Then excess pressure in the drop is

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Excess pressure in a drop of a liquid is 400 n/m². Eight such identical drops coalesce to form a single drop. Then excess pressure in the drop is