Question

A mercury drop of radius 0.5 cm falls from a height onto a glass plate and breaks up into a million droplets, all of the same size. Find the height from which the drop must have fallen. (Density of mercury = 13,600 kg/m^3. Surface tension of mercury = 0.465 N/m) (Ans: 0.2072 m)

Name: a mercury drop of radius 05 cm falls from a height on a glass plate and breaks up into a million droplets all of the same size find the height from which the drop must have fallendensity of 16758
Uploaded: 2022-07-15T14:25:49-08:00
Duration: 3 min 51 s
Channel: Animesh Raj
Description: a mercury drop of radius 05 cm falls from a height on a glass plate and breaks up into a million droplets all of the same size find the height from which the drop must have fallendensity of 16758

          A mercury drop of radius 0.5 cm falls from a height onto a glass plate and breaks up into a million droplets, all of the same size. Find the height from which the drop must have fallen. (Density of mercury = 13,600 kg/m^3. Surface tension of mercury = 0.465 N/m) (Ans: 0.2072 m)

Added by -Ngeles M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Animesh Raj

07/15/2022

Step 1

The volume \( V \) of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] where \( r \) is the radius of the sphere. Given \( r = 0.5 \) cm = 0.005 m, \[ V = \frac{4}{3} \pi (0.005)^3 \] \[ V = \frac{4}{3} \pi \times 1.25 \times 10^{-7} \] Show more…

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A mercury drop of radius 0.5 cm falls from a height onto a glass plate and breaks up into a million droplets, all of the same size. Find the height from which the drop must have fallen. (Density of mercury = 13,600 kg/m^3. Surface tension of mercury = 0.465 N/m) (Ans: 0.2072 m)