Two water drops each of radius 'r' coalesce to form a bigger drop. If 'T' is the surface tension, the surface energy released in this process is: [2025]

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Q.
Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is: [2025]

1 $4 π r^{2} T [2 - 2^{\frac{2}{3}}]$

2 $4 π r^{2} T [2 - 2^{\frac{1}{3}}]$

3 $4 π r^{2} T [1 + \sqrt{2}]$

4 $4 π r^{2} T [\sqrt{2} - 1]$

Ans.
(1)

$2 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 2^{1 / 3} r$

$E_{i} = 2 \times 4 π r^{2} T$

$E_{f} = 4 π R^{2} T = 4 π r^{2} 2^{2 / 3} T$

Heat lost $△ Q = E_{i} - E_{f} = 4 π r^{2} T [2 - 2^{2 / 3}]$

Solution

Q.
Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is: [2025]

1 $4 π r^{2} T [2 - 2^{\frac{2}{3}}]$

2 $4 π r^{2} T [2 - 2^{\frac{1}{3}}]$

3 $4 π r^{2} T [1 + \sqrt{2}]$

4 $4 π r^{2} T [\sqrt{2} - 1]$

Ans.
(1)

$2 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 2^{1 / 3} r$

$E_{i} = 2 \times 4 π r^{2} T$

$E_{f} = 4 π R^{2} T = 4 π r^{2} 2^{2 / 3} T$

Heat lost $△ Q = E_{i} - E_{f} = 4 π r^{2} T [2 - 2^{2 / 3}]$

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Solution

Q. Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is: [2025]

1 4πr2T[2–223]

2 4πr2T[2–213]

3 4πr2T[1+2]

4 4πr2T[2–1]

Ans. (1) 2×43πr3=43πR3 ⇒ R=21/3r Ei=2×4πr2T Ef=4πR2T=4πr222/3T Heat lost △Q=Ei–Ef=4πr2T[2–22/3]

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Q.
Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is: [2025]

1 $4 π r^{2} T [2 - 2^{\frac{2}{3}}]$

2 $4 π r^{2} T [2 - 2^{\frac{1}{3}}]$

3 $4 π r^{2} T [1 + \sqrt{2}]$

4 $4 π r^{2} T [\sqrt{2} - 1]$

Ans.
(1)

$2 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 2^{1 / 3} r$

$E_{i} = 2 \times 4 π r^{2} T$

$E_{f} = 4 π R^{2} T = 4 π r^{2} 2^{2 / 3} T$

Heat lost $△ Q = E_{i} - E_{f} = 4 π r^{2} T [2 - 2^{2 / 3}]$