The energy stored in a soap bubble of diameter 6 cm and T = 0.04 N/m is nearly ______.

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The energy stored in a soap bubble of diameter 6 cm and T = 0.04 N/m is nearly 0.9 $\times$ 10–3J Explanation: Increase in surface area of the sphere is given by: A = 4pir^2 ......(i) Here, r is the radius of the sphere and A is the increase in the surface area. The bubble contains two surfaces. So multiply Equation (i) by 2 A = 2 xx 4pir^2 .....(ii) Diameter = 6 cm i.e. r = 3 cm Substitute the value of r = 3 cm to find the value of the area. A = `2 xx 4pi xx (3 " cm" xx (1" m")/(100 " cm"))^2` A = 0.023 m2 The formula for stored energy is: E = T $\times$ A Here, T is the surface tension. Substitute the value of T = 0.04 N/m and A = 0.023 m2 to find the stored energy. E = T $\times$ A E = 0.04 N/m $\times$ 0.023 m2 E = 0.9 $\times$ 10–3J

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