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)
Answers
Given :
R = 0.5cm = 0.5 × 10⁻²m
n = No. of drops formed = 10⁶
ρ = 13600 kg/m3 ,
T = 0.465 N/m
W = T∆A
P.E = mgh
Volume of single big drop
V = 4 / 3 π R ³
Volume of single small droplet = 4 /3 π r³
Volume of n droplets = n × 4/ 3 πr³
∴ 4/ 3 πR³ = n 4/ 3 π r³
∴ R³ = nr³
∴ r = R /∛n = 0.5x 10⁻² /∛ 10⁶
∴ r = 0.5 × 10⁻²m /10²
r=0.5 x 10⁻⁴m
The single drop is fallen from height h, hence its P.E. = mgh
But, P.E = Work done due to change in area ...(i)
Change in surface area ∆A = (n4πr²2 – 4πR²)
Also, W = T∆A =
T (n4πr ² – 4πR ² )
W = 4πT (nr² – R² )
∴ eq., (i) becomes,
mgh = 4πT (nr²– R² )
But, m = 4 /3 πR³ρ
∴ 4 /3 πR³ρgh = 4πT (nr² – R² )
R³ gh ρ/3 = T (nr² – R²)
∴ h = 3T (nr² – R² )/ R ³ ρ g
=3x0.465[10 ⁶ (0.5x10⁻⁴)² – (0.5x10⁻²)²]/ (0.5x10⁻²)3x13600x9.8
H=3x0.465[0.25x10⁻²)- (0.25x10⁻⁴)]/0.125x1.36x10⁴ x 9.8
On solving
=3x0.465x25x0.09/12.5x1.36x9.8
H=0.2072m
Answer:
Given :
R = 0.5cm = 0.5 × 10⁻²m
n = No. of drops formed = 10⁶
ρ = 13600 kg/m3 ,
T = 0.465 N/m
W = T∆A
P.E = mgh
Volume of single big drop
V = 4 / 3 π R ³
Volume of single small droplet = 4 /3 π r³
Volume of n droplets = n × 4/ 3 πr³
∴ 4/ 3 πR³ = n 4/ 3 π r³
∴ R³ = nr³
∴ r = R /∛n = 0.5x 10⁻² /∛ 10⁶
∴ r = 0.5 × 10⁻²m /10²
r=0.5 x 10⁻⁴m
The single drop is fallen from height h, hence its P.E. = mgh
But, P.E = Work done due to change in area ...(i)
Change in surface area ∆A = (n4πr²2 – 4πR²)
Also, W = T∆A =
T (n4πr ² – 4πR ² )
W = 4πT (nr² – R² )
∴ eq., (i) becomes,
mgh = 4πT (nr²– R² )
But, m = 4 /3 πR³ρ
∴ 4 /3 πR³ρgh = 4πT (nr² – R² )
R³ gh ρ/3 = T (nr² – R²)
∴ h = 3T (nr² – R² )/ R ³ ρ g
=3x0.465[10 ⁶ (0.5x10⁻⁴)² – (0.5x10⁻²)²]/ (0.5x10⁻²)3x13600x9.8
H=3x0.465[0.25x10⁻²)- (0.25x10⁻⁴)]/0.125x1.36x10⁴ x 9.8
On solving
=3x0.465x25x0.09/12.5x1.36x9.8