Question

10 drops each of radius r = 0.2 mm dropped on water surface of oil forms a circular film of radius 2 cm find thickness of film

Answer 1

Answer:

No of olive drops = 10

Radius of  olive drops, r = 0.2 mm = 0.02 cm

Radius of circular film on water, R = 14.6 cm

To Find:

The thickness (or height) of molecule of olive oil drops, i.e., H.

Calculations:

Volume of circular oil film = Volume of 10 olive oil drops

πR²H = 10 × 4/3 πr³

⇒ H = (40 r³)/(3 R²)

⇒ H = {40 × (0.02)³} / {3 × (14.6)²}

⇒ H = (40 × 8 × 10⁻⁶) / (3 × 213.16)

⇒ H = (32 × 10⁻⁴) / (639.48)

⇒ H = 0.05 × 10⁻⁴ cm

⇒ H = 5 × 10⁻⁵ mm

- So the thickness of the olive oil film on water will be 5 × 10⁻⁵ mm.

Answer 2

Answer:

thickness will be 66.7 nano meters

Explanation:

we know that volume of the 10 drops and of the film will be constant :

therefore 4/3 π(0.2/10)³×10=π(2)²×h

since film will be a thin cylinder

so 4/3×(0.008)/10²=4h

1/3×(0.008)/100=4h

so

h=(0.008)/1200=6.67×10^(-6)cm thick

or 6.67×10^(-8) m or 66.7nm