A thin, rectangular film of liquid is extended from (2cm×3cm) to (4cm×6cm). If the work done in the process is 6×10−4J, the value of surface tension of the liquid is
A
0.133N/m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.33N/m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.166N/m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
0.12N/m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C0.166N/m The thin rectangular film of liquid has two surfaces (Upper and inner surfaces). Total Initial area Ai=2×2×3cm2=12cm2 Total final area Af=2×4×6cm2=48cm2 Increase in area of film, ΔA=(48−12)cm2=36cm2 ΔA=36×10−4m2 Hence, work done during the process, W=T×ΔA ⇒T=WΔA=6×10−436×10−4N/m ∴T=0.166N/m