First slide

Surface tension and surface energy

Question

A film of water is formed between two straight parallel wires of length 10cm each separated by 0.5 cm. If their separation is increased by 1 mm while still maintaining their parallelism, how much work will have to be done (Surface tension of water = $7 .2 \times 10^{- 2} N / m)$

Moderate

A

$7 .22 \times 10^{- 6} Joule$
B

$1 .44 \times 10^{- 5} Joule$
C

$2 .88 \times 10^{- 5} Joule$
D

$5 .76 \times 10^{- 5} Joule$

Solution

Increment in area of soap film = $A_{2} - A_{1}$
$= 2 \times [(10 \times 0 .6) - (10 \times 0 .5)] \times 10^{- 4} = 2 \times 10^{- 4} m^{2}$
Work done = $T \times ΔA$
$= 7 .2 \times 10^{- 2} \times 2 \times 10^{- 4} = 1 .44 \times 10^{- 5} J$

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