The correct option is B 0.063 N/m
Given,
Diameter of soap bubble (d)=10 mm
∴ Radius (r)=d2=5×10−3 m
Excess pressure, ΔP=50 N/m2
From the definition, excess pressure of a soap bubble is given by
ΔP=Pi−P0=4Tr
From the data given in the question,
50=4T5×10−3
⇒T=50×5×10−34=0.0625 N/m≈0.063 N/m
=0.063 N/m
Thus, option (c) is the correct answer.