SATHEE: Properties of Matter 4 Question 11

Properties of Matter 4 Question 11

13. 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 $Δ U = 10^{- 3} J$ . If $K = 10^{α}$ , then the value of $α$ is

(2017 Adv.)

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Answer:

Correct Answer: 13. 6

Solution:

From mass conservation,

$\begin{aligned} ρ \cdot \frac{4}{3} π R^{3} = & ρ \cdot K \cdot \frac{4}{3} π r^{3} \Rightarrow R = K^{1 / 3} r \\ Δ Δ U & = T Δ A = T (K \cdot 4 π r^{2} - 4 π R^{2}) \\ = T (K \cdot 4 π R^{2} K^{- 2 / 3} - 4 π R^{2}) \\ Δ U & = 4 π R^{2} T [K^{1 / 3} - 1] \end{aligned}$

Putting the values, we get

$10^{- 3} = \frac{10^{- 1}}{4 π} \times 4 π \times 10^{- 4} [K^{1 / 3} - 1]$

$\begin{aligned} 100 & = K^{1 / 3} - 1 \\ \Rightarrow & K^{1 / 3} & ≅ 100 = 10^{2} \\ Given that & K & = 10^{α} \\ ∴ & 10^{α / 3} & = 10^{2} \\ \Rightarrow & \frac{α}{3} & = 2 \Rightarrow α = 6 \end{aligned}$

Properties of Matter 4 Question 11

13. 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 $Δ U = 10^{- 3} J$ . If $K = 10^{α}$ , then the value of $α$ is

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Properties of Matter 4 Question 11

13. A drop of liquid of radius R=10−2 m having surface tension S=0.14πNm−1 divides itself into K identical drops. In this process the total change in the surface energy ΔU=10−3 J. If K=10α, then the value of α is

Answer:

Engineering Preparation

Medical Preparation

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Sample Mock Test

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NCERT Exercise Video Solution

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13. 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 $Δ U = 10^{- 3} J$ . If $K = 10^{α}$ , then the value of $α$ is