A small hole is made at the bottom of a hollow sphere the water enters into it when it is taken to depth of 40cm under water if the surface tension of water is 0.07 N/m diameter of hole is
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The diameter of the hole is 7.14 x 10^-3 cm
Explanation:
The pressure within the droplet due to surface tension = 2γ/r
Surface tension of water "γ" = 0.07 N/m
= 0.07 x 10^5 dyne /100 cm
= 70 dyne /cm
Radius of water = r
Hydro-static pressure at h depth of water = h x d x g
Where
"h" depth of water = 40 cm
"d" density of water = 1 g/cm^3
"g" acceleration due to gravity = 980 cms^-2
Now applying the condition.
Extra pressure = Hydro static pressure
Pst = Phy
2γ/r = h x d x g
r = 2γ / h x d x g
r = 2 x 70/ 40 x 1 x 980
r = 3.57 x 10^-3 cm
Diameter "d" = d x r = 2 x 3.57 x 10^-3
Diameter = 7.14 x 10^-3 cm
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