Mechanical Properties of Fluids

Question: A mercury drop of 1 cm radius is broken into ${{10}^{6}}$ small drops. The energy used will be (surface tension of mercury is $35\times {{10}^{-3}}N/cm)$

1) $4.4\times {{10}^{-3}}J$

2) $2.2\times {{10}^{-4}}J$

3) $8.8\times {{10}^{-4}}J$

4) ${{10}^{4}}J$

Solution:

Explanation: No Explanation

Surface tension surface energy

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