A spherical liquid drop of radius 7cm is divided to 27equal droplets.if the surface tension is 2×10-3N/m,then the work done in the process is?
Answers
Answer:
Explanation:
=> A spherical liquid drop of radius 7cm is divided to 27 equal droplets.
Thus, Surface area at initial, Ai = 4 π R²
Final surface area Af = 27 *4 * π * r²
Change in surface area:
ΔA = Af - Ai
= 27 * 4π * r² - 4π * R²
= 4π (27r² - R²)
=> Here, Initial volume is equal to the final volume, vi = vf
4/3 * π * R³ = 27 * 4/3 * π * r³
R³ = 27r³
R = 3r
r = R/3
r = 7/3
=> Now, Total surface area,
ΔA = 4π (27r² - R²)
= 4π (27(7/3)² - (7)²)
= 4 * 3.14 * (27 *( 49 / 9) - 49)
= 4 * 3.14 * (3 * 49 - 49)
= 4 * 3.14 (147 - 49)
= 4 * 3.14 * 98
= 1230.88
=> The work done in the process, W:
W = surface tension * change in surface area
W = T * ΔA
= 2×10⁻³ * 1230.88
= 2461.76 * 10⁻³ J
Thus, the work done in the process is 2461.76 * 10⁻³ J
Answer:A spherical liquid drop of radius 7cm is divided to 27 equal droplets.
Thus, Surface area at initial, Ai = 4 π R²
Final surface area Af = 27 *4 * π * r²
Change in surface area:
ΔA = Af - Ai
= 27 * 4π * r² - 4π * R²
= 4π (27r² - R²)
=> Here, Initial volume is equal to the final volume, vi = vf
4/3 * π * R³ = 27 * 4/3 * π * r³
R³ = 27r³
R = 3r
r = R/3
r = 7/3
=> Now, Total surface area,
ΔA = 4π (27r² - R²)
= 4π (27(7/3)² - (7)²)
= 4 * 3.14 * (27 *( 49 / 9) - 49)
= 4 * 3.14 * (3 * 49 - 49)
= 4 * 3.14 (147 - 49)
= 4 * 3.14 * 98
= 1230.88
=> The work done in the process, W:
W = surface tension * change in surface area
W = T * ΔA
= 2×10⁻³ * 1230.88
= 2461.76 * 10⁻³ J
Thus, the work done in the process is 2461.76 * 10⁻³ J
Explanation: