MCQ
In order to float a ring of area $ 0.04\, m^2$ in a liquid of surface tension $75\, N/m$, the required surface energy will be ....... $J$
- ✓$3$
- B$6.5$
- C$1.5$
- D$4$
✓
Answer
Correct option: A.
$3$
a
(a) $E = T \times \Delta A = 75 \times 0.04 = 3\,J$
(a) $E = T \times \Delta A = 75 \times 0.04 = 3\,J$
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