The surface tension of a liquid is 5 N/m. If a film is held on a ring of area 0.02 m2, its total surface energy is about:

Question

The surface tension of a liquid is 5 N/m. If a film is held on a ring of area 0.02 m2, its total surface energy is about: (A) 2 × 10-2 J (B) 2.5 × 10-2 J (C) 2 × 10-1 J (D) 3 × 10-1 J

Tardigrade Admin · Accepted Answer

Surface energy is related to the surface tension by the relation U=Td A Given, T=5 N/m dA=2A=2× 0.02=0.04 m2 ∴ U=5× 0.04 =0.20 J =2× 10-1 J

Q. The surface tension of a liquid is $5 N / m .$ If a film is held on a ring of area $0.02 m^{2},$ its total surface energy is about:

$2 \times 1 0^{- 2} J$

$2.5 \times 1 0^{- 2} J$

$2 \times 1 0^{- 1} J$

$3 \times 1 0^{- 1} J$

Surface energy is related to the surface tension by the relation
$U = T d A$
Given, $T = 5 N / m$
$d A = 2 A = 2 \times 0.02 = 0.04 m^{2}$
$∴$ $U = 5 \times 0.04$
$= 0.20 J$
$= 2 \times 10^{- 1} J$

Q. The surface tension of a liquid is 5N/m. If a film is held on a ring of area 0.02m2, its total surface energy is about:

2×10−2J

2.5×10−2J

2×10−1J

3×10−1J