Correct answer is (c) \( \frac{∂\overline{u}}{∂t}=\frac{k}{m_v γ_w} \frac{∂^2\overline{u}}{∂z^2}\)
The explanation: The decrease in volume is given by,
∆v=-mv V0 ∆σ’—————–(1)
also \(\frac{∂v}{∂z}=m_v \frac{∂\overline{u}}{∂t}\) ——————————-(2)
∴ from (1) and (2), we get,
\(\frac{∂\overline{u}}{∂t}=\frac{k}{m_v γ_w}\frac{∂^2 \overline{u}}{∂z^2}.\)