Consider a water droplet of diameter 0.2 ,mm | vedclass

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(D) Given, diameter of water droplet, $d = 0.2 \,mm$. Radius, $r = 0.1 \,mm = 10^{-4} \,m$. Pressure outside the droplet, $p_0 = 1.5 \,N / cm^2 = 1.5

Vedclass · Accepted Answer

$1.66 \,N / cm^2$

Vedclass · Answer

(D) Given, diameter of water droplet, $d = 0.2 \,mm$. Radius, $r = 0.1 \,mm = 10^{-4} \,m$. Pressure outside the droplet, $p_0 = 1.5 \,N / cm^2 = 1.5 	imes 10^4 \,N / m^2$. Surface tension, $T = 0.08 \,N / m$. The excess pressure inside a spherical droplet is given by $\Delta p = \frac{2T}{r}$. Therefore, the total pressure inside the droplet is $p = p_0 + \frac{2T}{r}$. Substituting the values: $p = 1.5 	imes 10^4 + \frac{2 	imes 0.08}{10^{-4}}$ $p = 1.5 	imes 10^4 + 0.16 	imes 10^4$ $p = 1.66 	imes 10^4 \,N / m^2$ Converting back to $N / cm^2$: $p = 1.66 \,N / cm^2$.

Column-$I$	Column-$II$
$(a)$ Liquid drop in air	$(i)$ $\frac{4T}{R}$
$(b)$ Bubble of liquid in air	$(ii)$ $\frac{2T}{R}$
	$(iii)$ $\frac{2R}{T}$

Consider a water droplet of diameter $0.2 \,mm$ where the outside pressure is $1.5 \,N / cm^2$ at $25^{\circ} C$. The pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 \,N / m$ is

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