Question

The ring of radius 1m is lying on the surface of liquid. It is lifted from the liquid
surface by a force of 4N in such a way that the liquid film in its remains intact. The
surface tension of liquid will be - (Ring weight negligible)

Answer 1

Explanation:

$\huge\fcolorbox{black}{orange}{Given}$

• RADIUS OF RING = 1m

• FORCE ACTED ON IT = 4N

$\huge\fcolorbox{black}{orange}{Formula used}$

Contect area = 2 × [Areaofring] because, water will pull it from both sides.

T= F / L

$\huge\fcolorbox{black}{orange}{Solution}$

$➣\: \: \: T = 4 / 2 \: * \: 2πr \\ \\ </p><p></p><p> ➣ \: \: \: 4 = 4πr \\ \\</p><p></p><p></p><p> ➣ \: \: \: 1 / π \: * \: n / m$

Answer 2

$\huge{ \underline{ \boxed{ \bold{ \star{ \blue{ \green{ \orange{ \pink{ \purple{ \red{ \mathfrak{ \pink{ \red{ \orange{ \green{a}{n} }{s} }{w} }{} }{} }{e} }{r} }{} }{} }{} }{} }{} }{} }{} }{} }{}$

Contect area =2×[Area of ring]

because, water will pull it from both sides.

T= L / F

T= 4/2×2πr

4 = 4πr

1/π × N/m

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