(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

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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, $p_0 = 1.5 \,N / cm^2 = 1.5 \times 10^4 \,N / m^2$. 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: $p = 1.5 \times 10^4 + \frac{2 \times 0.08}{10^{-4}}$ $p = 1.5 \times 10^4 + 0.16 \times 10^4$ $p = 1.66 \times 10^4 \,N / m^2$ Converting back to $N / cm^2$: $p = 1.66 \,N / cm^2$.

Class 11 Physics Fluid Mechanics and Surface Tension Excess Pressure and coalesce of Bubble and drop

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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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$0.32 \,N / cm^2$
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$1.18 \,N / cm^2$
C
$1.82 \,N / cm^2$
D
$1.66 \,N / cm^2$

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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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