The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = ) [2023]

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Q.
The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = $0.03 N$ $m^{- 1}$ ) [2023]

1 $3.01 \times 10^{- 4}$ $J$

2 $50.1 \times 10^{- 4}$ $J$

3 $30.16 \times 10^{- 4}$ $J$

4 $5.06 \times 10^{- 4}$ $J$

Ans.
(1)

We know that surface tension,

$S = \frac{Surface energy}{Surface area} \Rightarrow S = \frac{U}{A} \Rightarrow U = S A$

$= (0.03) \times 2 \times (4 π r^{2}) = (0.03) \times 2 \times (4 \times 3.14) \times {(2 \times 10^{- 2})}^{2}$

$= 3.014 \times 10^{- 4}$ $J .$

Solution

Q.
The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = $0.03 N$ $m^{- 1}$ ) [2023]

1 $3.01 \times 10^{- 4}$ $J$

2 $50.1 \times 10^{- 4}$ $J$

3 $30.16 \times 10^{- 4}$ $J$

4 $5.06 \times 10^{- 4}$ $J$

Ans.
(1)

We know that surface tension,

$S = \frac{Surface energy}{Surface area} \Rightarrow S = \frac{U}{A} \Rightarrow U = S A$

$= (0.03) \times 2 \times (4 π r^{2}) = (0.03) \times 2 \times (4 \times 3.14) \times {(2 \times 10^{- 2})}^{2}$

$= 3.014 \times 10^{- 4}$ $J .$

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Solution

Q. The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = 0.03Nm-1) [2023]

1 3.01×10-4J

2 50.1×10-4J

3 30.16×10-4J

4 5.06×10-4J

Ans. (1) We know that surface tension, S=Surface energySurface area⇒S=UA⇒U=SA =(0.03)×2×(4πr2)=(0.03)×2×(4×3.14)×(2×10-2)2 =3.014×10-4J.

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Q.
The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = $0.03 N$ $m^{- 1}$ ) [2023]

1 $3.01 \times 10^{- 4}$ $J$

2 $50.1 \times 10^{- 4}$ $J$

3 $30.16 \times 10^{- 4}$ $J$

4 $5.06 \times 10^{- 4}$ $J$

Ans.
(1)

We know that surface tension,

$S = \frac{Surface energy}{Surface area} \Rightarrow S = \frac{U}{A} \Rightarrow U = S A$

$= (0.03) \times 2 \times (4 π r^{2}) = (0.03) \times 2 \times (4 \times 3.14) \times {(2 \times 10^{- 2})}^{2}$

$= 3.014 \times 10^{- 4}$ $J .$