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