Question

A mercury drop of radius 0.5 cm falls from 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 fallen. (Density of mercury = 13600 kg/m³, . Surface tension of mercury = 0.465 N/m) (Ans : 0.2072 m)

Answer 1

a/c to question,
A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.

so, volume of a Mercury drop = one million × volume of each droplet

Let radius of each droplet is r

then, 4/3 πR³ = 10^6 × 4/3 πr³

R = 10²r = 100r

r = R/100 = 0.5 × 10^-2 cm = 5 × 10^-5 m

now, change in surface area = 10^6 × surface area of each droplet - surface area of big drop

= 10^6 × 4πr² - 4πR²

= 10^6 × 4πr² - 4π(100r)²

= 4πr² [ 1000000 - 10000 ]

= 990000 × 4πr²

now, surface energy = potential energy

990000 × 4πr² × T = Mgh

990000 × 4πr² × T = 4/3 π(100r)³dgh

9.9 × 10^5 × T = 1/3 × 10^6 × r × d × g × h

29.7 × T = 10 × r × d × g × h

h = 29.7 × 0.465/(10 × 5 × 10^-5 × 13600 × 9.8)

= 0.2072 m

Answer 2

Answer:

Answer:a/c to question,

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r then, 4/3 πR³ = 10^6 × 4/3 πr³

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r then, 4/3 πR³ = 10^6 × 4/3 πr³ R = 10²r = 100r

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r then, 4/3 πR³ = 10^6 × 4/3 πr³ R = 10²r = 100r r = R/100 = 0.5 × 10^-2 cm = 5 × 10^-5 m

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r then, 4/3 πR³ = 10^6 × 4/3 πr³ R = 10²r = 100r r = R/100 = 0.5 × 10^-2 cm = 5 × 10^-5 mnow, change

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r then, 4/3 πR³ = 10^6 × 4/3 πr³ R = 10²r = 100r r = R/100 = 0.5 × 10^-2 cm = 5 × 10^-5 mnow, change = 10^6 × 4πr² - 4πR²

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r then, 4/3 πR³ = 10^6 × 4/3 πr³ R = 10²r = 100r r = R/100 = 0.5 × 10^-2 cm = 5 × 10^-5 mnow, change = 10^6 × 4πr² - 4πR² = 10^6 × 4πr² - 4π(100r)²

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r then, 4/3 πR³ = 10^6 × 4/3 πr³ R = 10²r = 100r r = R/100 = 0.5 × 10^-2 cm = 5 × 10^-5 mnow, change = 10^6 × 4πr² - 4πR² = 10^6 × 4πr² - 4π(100r)² = 4πr² [ 1000000 - 10000 ]

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r then, 4/3 πR³ = 10^6 × 4/3 πr³ R = 10²r = 100r r = R/100 = 0.5 × 10^-2 cm = 5 × 10^-5 mnow, change = 10^6 × 4πr² - 4πR² = 10^6 × 4πr² - 4π(100r)² = 4πr² [ 1000000 - 10000 ] = 990000 × 4πr²

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r then, 4/3 πR³ = 10^6 × 4/3 πr³ R = 10²r = 100r r = R/100 = 0.5 × 10^-2 cm = 5 × 10^-5 mnow, change = 10^6 × 4πr² - 4πR² = 10^6 × 4πr² - 4π(100r)² = 4πr² [ 1000000 - 10000 ] = 990000 × 4πr² now, surface energy = potential energy

Answer:a/c to question, A mercury drop of radius 0.5 cm falls from height on a glass plate and breaks up into a million droplets, all of the same size.so, volume of a Mercury drop = one million × volume of each droplet Let radius of each droplet is r then, 4/3 πR³ = 10^6 × 4/3 πr³ R = 10²r = 100r r = R/100 = 0.5 × 10^-2 cm = 5 × 10^-5 mnow, change = 10^6 × 4πr² - 4πR² = 10^6 × 4πr² - 4π(100r)² = 4πr² [ 1000000 - 10000 ] = 990000 × 4πr² now, surface energy = potential energy = 0.2072 m