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PhysicsFluidSurface TensionMedium2 minPYQ_2023

PhysicsMediumnumerical

The surface tension of soap solution is $3.5 \times 10^{- 2} N m^{- 1}$ . The amount of work done required to increase the radius of soap bubble from $10 cm$ to $20 cm$ is _____ $\times 10^{- 4} J$ . (take $π = \frac{22}{7}$ )

Answer:

264.00

Solution:

The amount of work done is equal to the change in the surface energy to increase the radius of the bubble.

The formula to calculate the change in surface energy is given by

$Δ U = 2 (T \times Δ A) . . . (1)$

Substitute the values of the known parameters into equation (1) to calculate the required work done.

$\begin{array}{rcl} ∆ U & = & 3.5 \times 10^{- 2} \times 4 π [(0.2)^{2} - (0.1)^{2}] \times 2 J \\ = & 3.5 \times 10^{- 2} \times 4 \times \frac{22}{7} \times 0.03 \times 2 \\ = & 2 \times 1.32 \times 10^{- 2} \\ = & 264 \times 10^{- 4} J \end{array}$

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: PYQ_2023

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