Question

The pressure outside droplet of water of diameter 0.04 mm is 10.32. N\cm^2.calculate the presure within the droplet if surface Tension is given bar has 0.725 N\m of water​

Answer 1

Answer:

240Nm

step by step explanations

given:s = 0.07² N/m

diameter of water drop D=1.2mm =0.012 m

thus, radius of drop r= 0.012/ 2 = 0.006m

Excess pressure inside water drop ∆p =25/ r

• ∆p = 2x 0.072/0.006 =24 N/m²

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Answer 2

Answer:

The pressure outside the droplet of the water is $1.757\times 10^-^3 Nm^-^2$.

Explanation:

The excess pressure of  the droplet water is given as,

$\Delta P=\frac{2T}{r}$              (1)

Where,

ΔP=pressure difference inside and outside the droplet

T=tension in the droplet

r=radius of the droplet

From the question we have,

Diameter=0.04mm

Radius=0.02mm

P(outside)=10.32 N/cm²=10.32 × 10⁻⁴ N/m²

Tension=0.725N/m

Also,

$\Delta P=P_i-P_o$              (2)

Pi=inside pressure of the droplet

Po=outside pressure of the droplet

So, we can write equation (1) as,

$P_i-P_o=\frac{2T}{r}$           (3)

By substituting the required values in equation (3) we get;

$P_i-10.32\times 10^-^4=\frac{2\times 0.725}{0.02\times 10^-^3}$

$P_i-10.32\times 10^-^4=7.25\times 10^-^4$

$P_i=7.25\times 10^-^4+10.32\times 10^-^4=17.57\times 10^-^4=1.757\times 10^-^3 Nm^-^2$

Hence, the pressure outside the droplet of the water is $1.757\times 10^-^3 Nm^-^2$.