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NEET PHYSICSMedium

The amount of energy required to form a soap bubble of radius 2cm2 \text{cm} from a soap solution is nearly (surface tension of soap solution =0.03N m−1= 0.03 \text{N m}^{-1})

1

30.16×10−4J30.16 \times 10^{-4} \text{J}

2

5.06×10−4J5.06 \times 10^{-4} \text{J}

3

3.01×10−4J3.01 \times 10^{-4} \text{J}

4

50.1×10−4J50.1 \times 10^{-4} \text{J}

Step-by-Step Solution

Energy W=T×ΔAW = T \times \Delta A. For a soap bubble, there are two surfaces, so ΔA=2×(4πr2)=8πr2\Delta A = 2 \times (4\pi r^2) = 8\pi r^2. Given T=0.03N/mT = 0.03 \text{N/m}, r=0.02mr = 0.02 \text{m}. W=0.03×8×3.14×(0.02)2=0.03×8×3.14×0.0004=3.0144×10−4JW = 0.03 \times 8 \times 3.14 \times (0.02)^2 = 0.03 \times 8 \times 3.14 \times 0.0004 = 3.0144 \times 10^{-4} \text{J}.

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