Question

10 drops of olive oil of radius 0.2mm is spread into a circular film of radius 14.6 cm on water Surface. Find the thickness of molecule of olive oil drops.

Answer 1

Given:

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.