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JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF FLUIDS

1 Medium

1 of 1

Application +4 | -1

The amount of work done in increasing size of soap film

6 \mathrm{~cm} \times 4 \mathrm{~cm}

to

12 \mathrm{~cm} \times 8 \mathrm{~cm}

is: (surface tension

30 \mathrm{~dyne}/ \mathrm{cm}

)

View answer and explanation

Correct option: B

4320 erg

Option B (4320 erg) is correct because the work done is calculated by multiplying the change in area of the soap film by the surface tension.

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