A drop of liquid of radius having surface tension divides itself into identical drops. In this process, the total change in the surface energy is If , then the value of is _______. [2017]

Question

A drop of liquid of radius $R = 10^{- 2} m$ having surface tension $S = \frac{0.1}{4 π} {Nm}^{- 1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $Δ U = 10^{- 3} J .$ If $K = 10^{α}$ , then the value of $α$ is _______. [2017]

Smart Achievers · Accepted Answer

(6)

$\frac{4}{3} π R^{3} = k \times \frac{4}{3} π r^{3}$ $∴$ $R = K^{1 / 3} r$

$Δ U = S [k \times 4 π r^{2} - 4 π R^{2}]$

$∴$ $Δ U = 4 π S [k \times \frac{R^{2}}{k^{2 / 3}} - R^{2}] = 4 π S R^{2} [k^{1 / 3} - 1]$

$∴$ $Δ U = 4 π S R^{2} [10^{α / 3} - 1]$ $[∵ K = 10^{α}]$

$∴$ $10^{- 3} = 4 π \times \frac{0.1}{4 π} \times {(10^{- 2})}^{2} [10^{α / 3} - 1]$

$∴$ $10^{2} = 10^{α / 3} - 1 Neglecting 1$

$10^{2} = 10^{α / 3} \Rightarrow \frac{α}{3} = 2$ $∴$ $α = 6$

Solution

Q.
A drop of liquid of radius $R = 10^{- 2} m$ having surface tension $S = \frac{0.1}{4 π} {Nm}^{- 1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $Δ U = 10^{- 3} J .$ If $K = 10^{α}$ , then the value of $α$ is _______. [2017]

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Solution

Q. A drop of liquid of radius R=10-2 m having surface tension S=0.14π Nm-1 divides itself into K identical drops. In this process, the total change in the surface energy is ΔU=10-3 J. If K=10α, then the value of α is _______. [2017]

Ans. (6) 43πR3=k×43πr3 ∴ R=K1/3r ΔU=S[k×4πr2-4πR2] ∴ ΔU=4πS[k×R2k2/3-R2]=4πSR2[k1/3-1] ∴ ΔU=4πSR2[10α/3-1] [∵K=10α] ∴ 10-3=4π×0.14π×(10-2)2[10α/3-1] ∴ 102=10α/3-1 Neglecting 1 102=10α/3⇒α3=2 ∴ α=6

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Q.
A drop of liquid of radius $R = 10^{- 2} m$ having surface tension $S = \frac{0.1}{4 π} {Nm}^{- 1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $Δ U = 10^{- 3} J .$ If $K = 10^{α}$ , then the value of $α$ is _______. [2017]