Respuesta :
Answer:
The surface tension of the water is 6.278×10⁻² N/m
error = 13.65%
Explanation:
The surface tension of water is given by
[tex]$ \gamma = \frac{F}{L} $[/tex]
Where F is the force acting on water and L is the length over which is force is acted.
We are given the mass of 100 droplets of water
M = 3.78 g
n = 100
The mass of 1 droplet is given by
[tex]m = \frac{M}{n} \\\\m = \frac{3.78}{100}\\\\m = 0.0378 \: g \\\\m = 3.780\times10^{-5} \: kg[/tex]
The force acting on a single droplet of water is given by
[tex]F = m \cdot g[/tex]
Where m is the mass of water droplet and g is the acceleration due to gravity
[tex]F = 3.780\times10^{-5} \cdot 9.81[/tex]
[tex]F = 3.708\times10^{-4} \: N[/tex]
The circumferential length of the droplet is given by
[tex]L = \pi \cdot d[/tex]
Where d is the diameter
[tex]L = \pi \cdot 1.88\times10^{-3}\\\\L = 5.906 \times10^{-3} \: m[/tex]
Now we can find out the required surface tension of the water
[tex]\gamma = \frac{3.708\times10^{-4} }{5.906 \times10^{-3}} \\\\\gamma = 0.06278\: N/m\\\\\gamma = 6.278 \times10^{-2} \: N/m\\\\[/tex]
Therefore, the surface tension of the water is 6.278×10⁻² N/m
The tabulated value of the surface tension of water at 20 °C is given by
[tex]$ \gamma_t = 0.0727 \: N/m $[/tex]
The percentage error between tabulated and calculated surface tension is given by
[tex]$ error = \frac{\gamma_t - \gamma }{\gamma_t} $[/tex]
[tex]$ error = \frac{ 0.0727 - 0.06278}{0.0727} \times 100\% $[/tex]
[tex]$ error = 13.65 \%[/tex]
