JEE Advance - Physics (2010 - Paper 2 Offline - No. 4)
After the drop detaches, its surface energy is
1.4 $$ \times $$ 10−6 J
2.7 $$ \times $$ 10−6 J
5.4 $$ \times $$ 10−6 J
8.1 $$ \times $$ 10−6 J
Explanation
We have
Surface energy = Tension $$\times$$ Area
That is, Esurface = T $$\times$$ 4$$\pi$$R2
where T = 0.11 N/m and R = 1.4 $$\times$$ 10$$-$$4 m. Therefore, the surface energy is
$${E_{surface}} = T \times 4\pi {R^2} = 0.11 \times 4 \times 3.14 \times {(1.4 \times {10^{ - 3}})^2} = 2.7 \times {10^{ - 6}}J$$
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