A) \[20\text{ }Pa\]
B) \[40\text{ }Pa\]
C) \[2.5\text{ }Pa\]
D) \[5\text{ }Pa\]
Correct Answer: D
Solution :
Excess of pressure inside the soap bubble \[p=\frac{4S}{R}\] \[\therefore \] \[\frac{{{p}_{2}}}{{{p}_{1}}}=\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{R}{2R}=\frac{1}{2}\] \[{{p}_{2}}=\frac{1}{2}{{p}_{1}}=\frac{1}{2}\times 10Pa=5\,Pa\]
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